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Full text of "Practical mathematics"

PRACTICAL MATHEMATICS 



NEW EDITION 



REVISED UNDER THE SUPERVISION OF 



C. G. KNOTT, D.SC. (EDIN.), F.R.S.E., 
LECTURER ON APPLIED MATHEMATICS 
IN THE UNIVERSITY OF EDINBURGH ; 
FORMERLY PROFESSOR OF PHYSICS IN AND 
THE IMPERIAL UNIVERSITY OF JAPAN ; 
AUTHOR OF ' PHYSICS I AN ELEMEN- 
TARY TEXT -BOOK FOR UNIVERSITY 
CLASSES' 



J. S. MACKAY, M.A..LL.D, 
HEAD MATHEMATICAL MASTER 
IN THE EDINBURGH ACADEMY ; 
AUTHOR OF 'THE ELEMENTS 
OF EUCLID" AND 'ARITHMETIC, 
THEORETICAL AND PRACTICAL' 



LONDON : 38 Soho Square, W. 

W. & R. CHAMBERS, 

EDINBURGH : 339 High Street 

D. VAN NOSTRAND COMPANY 
NEW YORK 




ARITHMETIC, Theoretical and Practical. By J. 8. MACKAY, M.A., LL.D., 

Author of ' Mackay's Euclid.' 4/6. 

ALGEBRA FOR SCHOOLS. By WILLIAM THOMSON, M.A., B.Sc., Registrar, 
University of the Cape of Good Hope, formerly Assistant- Professor of Mathe- 
matics and Mathematical Examiner, University oi~ Edinburgh. 576 pages. 
Cloth, 4/6. 

CHAMBERS'S ELEMENTARY ALGEBRA. By WILLIAM THOMSON, M.A., 
B.Sc. Up to and including Quadratic Equations. 288 pages. Cloth, 2/. 
With Answers, 2/6. 

THE ELEMENTS OF EUCLID. Books I. to VI., and parts of Books XI. and 
XII. With Numerous Deductions, Appendices, and Historical Notes, by 
J. S. MACKAY, LL.D., Mathematical Master in the Edinburgh Academy. 
412 pages. 392 diagrams. 3/6. Separately, Book I., I/; II., 6d. ; III., 9d. : 
Books XI. XII., 6d. Key, 3/6. 

MATHEMATICAL TABLES. By JAMES PBYDE, F.E.I.S. These comprehend 
the most important Tables required in Trigonometry, Mensuration, Land- 
Surveying, Navigation, Nautical Astronomy, &c. The tables of Logarithms 
(1 to 108000), Logarithmic Sines, &c., are carried to seven decimal places. 
496 pages. 4/6. 

W. & R. CHAMBERS, LIMITED, LONDON AND EDINBURGH. 



EXTRACT FROM ORIGINAL PREFACE 



IN preparing this treatise on Practical Mathematics con- 
siderable pains have been taken to explain, in the clearest 
manner, the method of solving the numerous problems. 
The rules have been expressed as simply and concisely as 
possible in common language, as well as symbolically by 
algebraic formulae, which frequently possess, on account 
of their conciseness and precision, a great advantage over 
ordinary language; they have also in many instances been 
given logarithmically, because of the facility and expedi- 
tion of logarithmic calculation. To understand the algebraic 
formulae, nothing more is necessary than a knowledge of 
the simple notation of algebra ; the method of computa- 
tion by logarithms is explained in the Introduction to 
Chambers's Mathematical Tables. 



PREFACE TO THIRD EDITION 



IN general plan the present edition of Practical Mathematics 
does not differ from its predecessors, which were prepared 
and edited by Dr Piyde. The aim has been to illustrate 
the use of mathematics in constructing diagrams; in 
measuring areas, volumes, strengths of materials ; in calculat- 
ing latitudes and longitudes on the earth's surface ; and in 
solving similar problems. There is no attempt at a syste- 
matic development of any part of mathematics, except to 



PREFACE 

a certain extent in the sections on plane and spherical 
trigonometry. The plane trigonometry has been remodelled; 
but the spherical trigonometry, which is required for 
navigation and geodesy, has been left as it was. The 
greatest changes will, however, be found in the section 
upon strength of materials and associated problems in 
elasticity. New tables of constants have been added, 
and new types of problems have been worked out or 
indicated. Throughout this section graphical solutions are 
occasionally given; and a final section has been added 
to the book in which the elements of curve-tracing, a 
growingly important part of mathematics, both practical and 
theoretical, are discussed and illustrated by examples. 

One great branch of Practical Mathematics, that dealing 
with electricity and magnetism, has not been included in 
the present work. It would have been impossible to give 
this peculiarly modern subject adequate space without 
increasing the volume to an inconvenient size. 

When logarithms are required in solving any problem, 
the seven-place Logarithmic Tables are used; but since 
for many purposes a rough approximation is all that is 
needed, it has been thought advisable to make the book 
more complete in itself by the addition of six pages of 
four-place logarithms of the natural numbers and the 
trigonometrical ratios. 

It should be stated that in the section dealing with 
strength of materials, as well as in other parts of the work, 
valuable aid has been rendered by Mr Forrest Sutherland, of 
JUoemfontejn, 



CONTENTS 



DESCRIPTIVE GEOMETRY PAQ1S 

DEFINITIONS . . ' . . . .1 

PROBLEMS ........ 7 

CONSTRUCTION OF SCALES, AND PROBLEMS TO BE SOLVED 

BY THEM . . . . . . . 23 

COMPUTATION BY LOGARITHMS 
DEFINITIONS . . . . . . .34 

LOGARITHMIC SCALES 
CONSTRUCTION AND USE . . . . .35 

THE LINES OF THE SECTOR 
DEFINITIONS ....... 39 

LINE OF LINES . . . . . . .40 

i. CHORDS. . . . . . .41 

ii SINES . . . . . . .41 

TANGENTS . . . . . .42 

ii SECANTS . . . . . .42 

PLANE TRIGONOMETRY 
DEFINITIONS, FORMULAE, &c. . . . . .43 

RELATION BETWEEN THE SIDES AND ANGLES OF TRIANGLES 54 
SOLUTION OF TRIANGLES RIGHT-ANGLED TRIANGLES . 59 
OBLIQUE-ANGLED TRIANGLES . . . . .62 

PROMISCUOUS EXERCISES IN TRIGONOMETRY . . 71 

MENSURATION OF HEIGHTS AND DISTANCES 
DEFINITIONS . . . . . . .72 

COMPUTATION OF HEIGHTS . . ... .74 

ii ii DISTANCES . . . . .81 

ADDITIONAL EXERCISES . . . . . .90 

MENSURATION OF SURFACES 

DEFINITIONS AND EXPLANATIONS . . .92 

PROBLEMS AND EXERCISES . . . . .93 

TABLES OF AREAS, &c., OF REGULAR POLYGONS WHOSE 
SIDES ARE 1 . 114 



VI CONTENTS 

LAND-SURVEYING 

DEFINITIONS AND EXPLANATIONS 
SURVEYING WITH THE CHAIN AND CROSS . 
COMPUTATION OF ACREAGE .... 

OBSTACLES IN RANGING SURVEY LINES 

To SET OUT A RIGHT ANGLE WITH THE CHAIN 

USEFUL NUMBERS IN SURVEYING 

SURVEYING WITH THE CHAIN, CROSS, AND THEODOLITE 

tr ii PLANE-TABLE . . . 

DIVISION OF LAND . ... 

INCLINED LANDS . . . . . . 

CHAINING ON SLOPES ..... 

TABLE SHOWING VALUES OF K . . 

SURVEY OF A ROAD AND ADJOINING FIELDS 

n n SMALL FARM .... 

EXTENSIVE SURVEYS WITH THE THEODOLITE 

MENSURATION OF SOLIDS 

DEFINITIONS AND EXPLANATIONS . . . .160 

PROBLEMS AND EXERCISES . . . . .164 

MENSURATION OF CONIC SECTIONS 
DEFINITIONS . . . . . . .186 

PROBLEMS AND EXERCISES . . . . .188 

SOLIDS OF REVOLUTION OF THE CONIC SECTIONS 
DEFINITIONS ....... 203 

PROBLEMS AND EXERCISES ..... 204 

REGULAR SOLIDS 
DEFINITIONS . . . . . . .213 

PROBLEMS AND EXERCISES ..... 214 

TABLE OF SURFACES AND SOLIDITIES WHEN EDGE=! . 218 

CYLINDRIC RINGS 
DEFINITIONS ....... 219 

PROBLEMS AND EXERCISES ..... 219 

SPINDLES 

DEFINITIONS . . . . . . 221 

THE CIRCULAR SPINDLE PROBLEMS AND EXERCISES . 221 

THE PARABOLIC . 223 

THE ELLIPTIC n . 224 

THE HYPERBOLIC n n n n , 227 



CONTENTS vii 

UNGULAS 

DEFINITION . . . . . . . .228 

PYRAMIDAL AND PRISMOIDAL UNGULAS . . . 228 
CYLINDRIC UNGULAS ...... 230 

CONIC UNGULAS . . . . ... . 232 

IRREGULAR SOLIDS 

PROBLEM AND EXERCISES 237 



ADDITIONAL EXERCISES IN MENSURATION . . . 238 

THE COMMON SLIDING-RULE 
EXPLANATIONS . . . . . . .241 

PROBLEMS AND EXERCISES ..... 242 

MEASUREMENT OF TIMBER 

PROBLEMS AND EXERCISES ..... 243 

MEASURES OF TIMBER ...... 248 

RELATIONS OF WEIGHT AND VOLUME OF BODIES 

EXPLANATIONS ....... 249 

INSTRUMENTS USED FOR FINDING THE SPECIFIC GRAVITY. 249 

TABLE OF SPECIFIC GRAVITIES .... 253 

USEFUL MEMORANDA IN CONNECTION WITH WATER . 256 

PROBLEMS AND EXERCISES ..... 256 

ARCHED ROOFS 

EXPLANATIONS ....... 261 

VAULTS PROBLEMS AND EXERCISES .... 261 

DOMES H ii i, .... 262 

SALOONS .1 n . . 263 

GROINS .1 H .... 264 

GAUGING 

PRINCIPLES AND DEFINITIONS OF TERMS . . . 266 

TABLES OF MULTIPLIERS, DIVISORS, AND GAUGE-POINTS . 268 

PROBLEMS AND EXERCISES ..... 270 

CASK-GAUGING ....... 277 

MEAN DIAMETERS OF CASKS ..... 280 

TABLE OF MEAN DIAMETERS WHEN THE BUNG DIAMETER 

= 1 281 

CONTENTS OF CASKS WHOSE BUNG DIAMETERS AND 

LENGTHS ARE UNITY 282 



Vlll CONTENTS 

PAGE 

TABLE OF CONTENTS IN IMPERIAL GALLONS OF STANDARD 

CASKS C' 283 

GENERAL METHOD FOR A CASK OF ANY FORM . . 285 

ULLAGE OF CASKS ...... 285 

MALT-GAUGING . . . . . . . 287 

THE DIAGONAL ROD . . . . . . 289 

BAROMETRIC MEASUREMENT OF HEIGHTS 
THE THERMOMETER . . . . . .291 

COMPARISON OF DIFFERENT LINEAL MEASURES . . 293 

OLD AND NEW DIVISIONS OF THE CIRCLE . . . 293 

THE BAROMETER ....... 294 

RELATION OF VOLUME AND TEMPERATURE OF AIR . 296 

MEASUREMENT OF HEIGHTS ..... 298 

TABLES FOR COMPUTING HEIGHTS . . . .301 

MEASUREMENT OF DISTANCES BY THE VELOCITY 

OF SOUND 
PRINCIPLES ....... 308 

VELOCITY OF SOUND IN VARIOUS SUBSTANCES . . 308 

DISTANCES SOUND MAY BE HEARD . . . .308 

FORMULAE AND EXERCISES ..... 309 

MEASUREMENT OF HEIGHTS AND DISTANCES 
PROBLEMS AND EXERCISES ..... 310 

REFRACTION ....... 312 

CONCISE FORMUL/E FOR HEIGHTS .... 316 

CURVATURE AND REFRACTION. .... 317 

LEVELLING 

DEFINITIONS AND EXPLANATIONS . . . .318 

PROBLEMS AND EXERCISES ..... 319 

STRENGTH OF MATERIALS AND THEIR ESSENTIAL 

PROPERTIES 
ESSENTIAL PROPERTIES . . . . . .325 

CONTINGENT PROPERTIES . . . . .325 

MELTING-POINTS OF METALS . . ... . 327 

MOMENT OF INERTIA AND RADIUS OF GYRATION . . 327 

LOAD, STRESS, AND STRAIN ..... 328 

TENSILE STRESS AND STRAIN . . . . .328 

YOUNG'S MODULUS OF ELASTICITY .... 328 

TABLE OF YOUNG'S MODULUS OF ELASTICITY 331 



CONTENTS ix 

PAGE 

LIMITING STRESS, OR ULTIMATE STRENGTH . . .331 

TABLE OF ULTIMATE STRENGTH AND WORKING STRESS 

OF MATERIALS ...... 332 

EXAMPLES OF STRESS AND STRAIN .... 333 

COMPRESSIVE STRESS AND STRAIN .... 334 

RESILIENCE OF A BAR ...... 334 

DEFINITION OF SHEARING FORCE AND BENDING MOMKNT 335 

PROBLEMS AND EXERCISES ..... 338 

STRENGTH OF SHAFTING TO RESIST VARIOUS STRESSES 

PROBLEMS AND EXERCISES .... 339 

STIFFNESS OF SHAFTS: ANGLE OF TWIST . ^ . 344 

TORSIONAL MOMENT OF RESISTANCE FOR SHAFTS (TABLE) 348 

STRENGTH OF COLUMNS ...... 349 

TABLE OF CRUSHING WEIGHTS OF A FEW MATERIALS . 353 

MISCELLANEOUS FORMULAE AND TABLES . . . 353 

AVEIGHT AND STRENGTH OF ROPE AND CHAINS . . 353 
BREAKING WEIGHT OF BEAMS ON THE SLOPE . .355 

REACTIONS OF BEAMS . . . . . 356 

STIFFNESS OF BEAMS ...... 360 

TRANSVERSE STRESS OR BENDING MOMENT OF BEAMS . 361 

PROBLEMS ........ 362 

EXERCISES ON THE BENDING MOMENT AND SHEARING 

FORCE OF BEAMS ...... 369 

FORMULA . . . . . . .369 

TABLE OF STRENGTH AND WEIGHT OF MATERIALS . 372 

DEFLECTION OF BEAMS AND GIRDERS . . 376 

BREAKING WEIGHT OF CAST-IRON GIRDERS . . . 378 

DEFLECTION OF IRON AND STEEL GIRDERS . . . 379 
DETERMINATION OF MOMENT OF INERTIA . . .381 

To FIND THE STRENGTH OF THIN WROUGHT-!RON GIRDERS 389 
COLLISION OF BODIES . . . . . .390 

MOMENT OF INERTIA, MODULUS, &c., OF SOME SECTIONS . 392 
STRENGTH AND STIFFNESS OF BEAMS UNDER A LOAD OF 

WLBS. 396 

PROJECTILES AND GUNNERY 

THE PARABOLIC THEORY OF PROJECTILES . . . 400 

PROJECTILES ON HORIZONTAL PLANES . . . 403 

PRACTICAL GUNNERY . . . . . .408 

TABLE OF ACTUAL AND POTENTIAL RANGES IN TERMS OF F 415 
THE FLAT TRAJECTORY THEORY . . . .420 

A r ELOCITY AND MOMENTUM OF RECOIL . 422 



X CONTENTS 

PAQK 

GRAVIMETRIC DENSITY OF A CHARGE . . . 422 

TABLE OF WORK DONE BY EXPLODING POWDER . . 422 

PENETRATION OF ARMOUR . . . . . 423 

PENETRATION OF RIFLE-BULLETS .... 423 

BASHFORTH'S TABLE OF TIME AND VELOCITY . . 424 

.1 n DISTANCE AND VELOCITY . . 425 

EXAMPLES . . . . . . .426 

STRENGTH OF GUNS . . . . . . 429 

CONCLUDING REMARKS, AND MOMENTUM OF RECOIL . 430 

PROJECTIONS 
GENERAL DEFINITIONS . . . . . .433 

STEREOGRAPHIC PROJECTION OF THE SPHERE . . 434 

STEREOGRAPHIC PROJECTION OF THE CASES OF 
TRIGONOMETRY 

PROJECTION OF THE CASES OF RIGHT-ANGLED TRIGO- 
NOMETRY . . . . . . .441 

PROJECTION OF CASES OF OBLIQUE-ANGLED SPHERICAL 
TRIGONOMETRY ...... 442 

SPHERICAL TRIGONOMETRY 

DEFINITIONS ....... 444 

RIGHT-ANGLED SPHERICAL TRIANGLES . . . 448 

QUADRANTAL SPHERICAL TRIANGLES .... 459 

OBLIQUE-ANGLED SPHERICAL TRIGONOMETRY . . 460 

OTHER SOLUTIONS OF THE CASES OF OBLIQUE-ANGLED 

SPHERICAL TRIGONOMETRY .... 469 

ASTRONOMICAL PROBLEMS 

CIRCLES AND OTHER PARTS OF THE CELESTIAL SPHERE. . 475 
DEFINITIONS ....... 475 

PRELIMINARY PROBLEMS ..... 482 

PROPORTIONAL LOGARITHMS ..... 485 

AUGMENTATION OF THE MOON'S SEMI-DIAMETER . . 488 

CONTRACTION OF THE SAME . . . . .489 

THE SUN'S SEMI-DIAMETER .. . . . .489 

To FIND THE PARALLAX IN ALTITUDE OF A CELESTIAL 
BODY ........ 489 

REDUCTION OF THE EQUATORIAL PARALLAX . 490 



CONTENTS XI 

PAGE 

To FIND THE REFRACTION OF A CELESTIAL BODY. . 491 
n M DEPRESSION OF THE VISIBLE HORIZON . 492 

PROBLEMS REGARDING ALTITUDE .... 493 
ii M SIDEREAL TIME AND MEAN TIME . 496 

n n CULMINATIONS . . . 502 

PROBLEMS REGARDING ALTITUDES, DECLINATIONS, LATI- 
TUDES, &c., OF CELESTIAL BODIES 

METHODS OF DETERMINING TIME .... 

THE EQUATION OF EQUAL ALTITUDES 

METHODS OF FINDING THE LATITUDE 

LUNAR DISTANCES ...... 

THE LONGITUDE FOUND BY LUNAR DISTANCES 

NAVIGATION 
DEFINITIONS OF TERMS . . . . . .535 

INSTRUMENTS USED IN NAVIGATION .... 536 

PRELIMINARY PROBLEMS ..... 539 

PLANE SAILING . . . . . . .543 

TRAVERSE SAILING . . . . . . 546 

GLOBULAR M ...... 548 

PARALLEL ...... 549 

MIDDLE LATITUDE SAILING . . . . . 550 

MERCATOR'S SAILING . . . . . .551 

MISCELLANEOUS EXERCISES . . . . . 560 

NAUTICAL ASTRONOMY . . . . . .561 

To FIND THE VARIATION OF THE COMPASS . . .561 

REDUCTION OF ONE OF Two ALTITUDES TO THE PLACE 

AT WHICH THE OTHER WAS TAKEN . . .561 

CONSTRUCTION OF MAPS AND CHARTS 

PLANE CONSTRUCTION . . . . . .563 

CONICAL PROJECTION ...... 564 

STEREOGRAPHIC PROJECTION ..... 565 

GLOBULAR PROJECTION ...... 566 

MERCATOR'S CONSTRUCTION ..... 567 

GEODETIC SURVEYING 

PRELIMINARY PROBLEMS ..... 570 

REDUCTION OF A BASE TO THE LEVEL OF THE SEA . 575 

PROBLEMS REGARDING THE SPHERICAL EXCESS 576 



Xli CONTENTS 

PAGfc 

LEGENDRE'S METHOD BY MEANS OF AN EQUAL-SIDED 

PLANE TRIANGLE ...... 580 

EXAMPLE OF TRIANGULATION ..... 683 

To FIND THE LENGTH OF AN ARC OF THE MERIDIAN . 586 
TABLE OF LENGTHS OF A DEGREE IN DIFFERENT 

LATITUDES ... .... 588 

THE METHOD OF CHORDAL TRIANGLES . . . 588 

DISTRIBUTION OF ERRORS ACCORDING TO THE WEIGHTS . 590 
FIGURE OF THE EARTH ITS ELEMENTS . . .591 
To FIND THE ELLIPTICITY BY MEANS OF THE LENGTHS 

OF Two ARCS OF A MERIDIAN .... 593 

To FIND THE ELLIPTICITY BY OBSERVATIONS OF THE 

PENDULUM ....... 594 

PROBLEMS REGARDING THE LENGTH, RADIUS OF CURVA- 
TURE, AND NUMBER OF DEGREES IN AN ARC . . 595 
To FIND THE POLAR RADIUS OF THE EARTH . . 597 
n n RADIUS OF CURVATURE OF A MERIDIAN . 598 
To FIND THE BEARINGS AND DIFFERENCE OF LATITUDE 

AND LONGITUDE OF TWO PLACES GEODETICALLY . 599 

THE RELATIVE AND ABSOLUTE HEIGHTS OF STATIONS . 602 
To FIND THE REFRACTION AND THE CORRECT VERTICAL 

ANGLES ....... 602 

THE ORIGINAL BASE REDUCED TO THE SURFACE OF AN 

IMAGINARY SPHERE. ..... 604 

ERRORS CAUSED BY INCORRECT ESTIMATION OF REFRACTION 605 

CURVE-TRACING 
PRINCIPLES AND FORMULA . . . . .606 

TABLES 
THE METRIC SYSTEM . . . . . .616 

LENGTHS OF CIRCULAR ARCS ..... 617 

AREAS OF SEGMENTS OF A CIRCLE .... 618 

NUMBERS OF FREQUENT USE IN CALCULATION . . 620 

FOUR-PLACE LOGARITHMS OF NUMBERS AND 
CIRCULAR FUNCTIONS 

TABLES OF LOGARITHMS OF NUMBERS . . . 622 

ii it H SINES AND COSINES . . 624 

n i, TANGENTS AND COTANGENTS 626 



PRACTICAL MATHEMATICS 



DESCRIPTIVE GEOMETRY 

DESCRIPTIVE GEOMETRY explains the methods of performing 
certain geometrical operations, such as the construction of 
mathematical figures, the drawing of lines in certain posi- 
tions, and the application of geometrical principles to the 
accurate representation of plane surfaces and solids. Hence 
it is treated of under two heads PLANE DESCRIPTIVE 
GEOMETRY, and SOLID DESCRIPTIVE GEOMETRY. 

There are three kinds of geometrical magnitudes lines, 
surfaces, and solids. Lines have one dimension length; 
surfaces have two dimensions length and breadth; and 
solids have three dimensions length, breadth, and thickness. 

I. PLANES 

Plane Descriptive Geometry treats of the relations and 
dimensions of lines and figures formed by their combina- 
tions on planes or plane surfaces. 

DEFINITIONS 

1. A point has position only, but no magnitude. 

2. A line has length without breadth. 

Hence the extremities or ends of a line are points ; and if 
two lines intersect ov cross each other, the intersections 
ftre points, 




J DESCRIPTIVE GEOMETRY 

A line is named by two letters placed one at each of its ex- 

1 tremities. Thus, the line drawn here is named 

the line AB. 

7 c B 3 - A straight line is that which lies evenly 

between its extreme points. 

If a straight line, as AB, revolve like an axis, its two extremi- 
ties, A and B, remaining in the same position, any other 
point of it, as C, will also remain in the same position. 
4. A point of section is any point in a line, and the two parts 
into which it divides the line are called segments. Thus the 
point C in the preceding line AB is a point of section, and AC, 
BC, are segments. 

It is evident that two straight lines cannot enclose a space ; 
and that two straight lines cannot have a common segment, 
or cannot coincide in part without coinciding altogether. 

5. A crooked line is composed of two or more straight 
lines. 

6. A curve, or a curved line, is one of Avhich no 
part is straight. 

7. Parallel straight lines are such as are in the same plane, 
and are at all points equidistant ; hence, if they are 

produced indefinitely in either direction, they do not 
meet. 

8. Convergent lines are those in the same plane, but not 
parallel, while they are supposed to be produced in the direction in 

__ which they would meet. Such lines are said to be 

~-~~- divergent, when considered as receding from the same 

point. 

9. A surface has only length and breadth. 

The boundaries of a surface are lines ; and the intersection 
of one surface with another is also a line. 

10. A plane surface, or a plane, is a surface such that, if 
any two points are taken on it, the straight line joining them lies 
i > / wholly on that surface. 

ll. A plane rectilineal angle is the in- 
clination of two straight lines that meet, 
but are not in the same straight line. 
12. The angular point is the point at which an angle is formed, 
as E or B. 

When there is only one angle at a point, it may be named, 
by one letter, as angle E. 




DESCRIPTIVE GEOMETRY 6 

But when there are more angles than one at a point, they 
are named by three letters, the letter at the angular point 
being put in the middle. Thus the angle contained by the 
lines DB and BC is named the angle DEC or CBD. So the 
angle contained by the lines AB and DB is named the angle 
ABD or DBA. 

An angle may also be named by means of a small letter 
placed in it. Thus angle ABD may be named angle m ; 
angle DEC, n ; and ABC, m+n. 

The two lines containing an angle are called its sides. Thus 
DB, BC, are the sides of the angle DBC, or n. 

13. Supplementary angles are the two adjacent / 

angles formed by one straight line standing upon / 

another. 

Thus the angles ACD, DCB, are said to be supplementary to 
one another ; or the angle ACD is called the supplement 
of the angle DCB ; and DCB is called the supplement of 
ACD. 

14. A right angle is one of two supplementary 
angles which are equal; and the line which sepa- 
rates them is said to be a perpendicular to the 
line on which it stands. 

15. An obtuse angle is greater than a right angle, as O ; and an 
acute angle is less than a right 

angle, as A. 

16. A figure is that which is 
enclosed by one or more boun- 
daries. The space contained within 

the boundary of the figure is called its surface ; and the quantity 
of surface in reference to that of some other figure with which it 
is compared is called its area. 

17. A circle is a figure formed on a plane by causing a line to 
revolve round one of its extremities which remains 

fixed. 

18. The circumference of a circle is the line 
that bounds it. 

19. The centre of a circle is the fixed point of 
the revolving line which describes it, as C. 

An eccentric point in a circle is one which is not the centre of 
the circle, but spoken of in reference to it, 





DESCRIPTIVE GEOMETRY 




20. A radius is a straight line drawn from the centre to the 
circumference of a circle ; CB, CD, and CE are radii. 

It i evident that all radii of the same circle are equal in 
length. 

21. A diameter is a straight line passing through the centre of 
a circle, and terminated at each extremity by the circumference, 
as BE. The radius is sometimes called the semi -diameter. 

22. An arc of a circle is any part of the circumference. 

23. The chord of an arc is a straight line joining its extremities, 

as AB. 

24. A segment of a circle is a figure contained 
by an arc and its chord. 
25. A Semicircle is a segment having a diameter for its chord, 
and is evidently half of the whole circle. 

26. An angle in a segment of a circle is an angle 
contained by two straight lines, drawn from any point 
in the arc of the segment to the extremities of its 
chord, as m in the segment CED ; the angle m is also 
said to stand on the arc CD. 

27. A sector of a Circle is a figure contained by two radii and 
the intercepted arc, as AOB. 

28. A quadrant is a sector Avhose bounding radii 
are perpendicular to each other, and is evidently the 
fourth part of a circle. 

29. A quadrantal arc, or the arc of a quadrant, is 
the fourth part of the circumference, and is sometimes merely 
called a quadrant. 

30. Similar segments of circles are those 
that contain equal angles. 

31. Similar arcs of circles are those that subtend or are 
opposite to equal angles at the centre. 

32. Similar sectors are those that are bounded by similar 

arcs. 

33. Equal circles are those that have equal 
radii. 

34. Concentric circles are those that have 
the same centre, and eccentric circles are those 
which have different centres. 

36. A tangent is a straight line which meets a circle or 
curve in one point, and being produced, does not cut it, as AT. 




A' B 




DESCRIPTIVE GEOMETRY 





36. Tangent circles are those of which the circumferences meet, 
but do not cut one another. 

37. The point of contact is that point in which a tangent and a 
curve, or two tangent curves, meet; thus the points A, B, and C 
are points of contact. 

38. Rectilineal figures are those contained by straight lines. 

39. Trilateral figures,- or triangles, are contained by three 
straight lines. 

40. Quadrilateral figures are contained by four straight lines. 

41. Multilateral figures, or polygons, are contained by more 
than four straight lines. 

42. Of three-sided figures, an equilateral 

triangle has three equal sides, as E ; an / E \ / l\ / s 
isosceles triangle has two equal sides, as I ; t 
and a scalene triangle has three unequal sides, as S. 

43. A right-angled triangle has one 
right angle, as R ; an obtuse-angled tri- 
angle has one obtuse angle, as O ; and 
an acute-angled triangle has all its 
angles acute, as A. 

44. Of quadrilateral figures, a square has all 
its sides equal, and its angles right angles, as S. 

45. A rectangle has all its angles right angles, 
but its sides are not all equal, as R. 

46. A rhombus has all its sides equal, but its 
angles are not right angles, as B. 

47. A parallelogram has its opposite sides 
parallel, as P. 

48. A trapezium has only two sides parallel, 
asD. 

49. An angle of a rectilineal figure which is 
right angles is said to be re-entrant, as B. 

50. Any side of a rectilineal figure may be 
called its base. In a right-angled triangle, the 
side opposite to the right angle is called the 
hypotenuse ; either of the sides about the right 

angle, the base ; and the other, the perpendicular. In an 
isosceles triangle, the unique side is called the base ; the 
angular point opposite to the base of a triangle is called the 
vertex ; and the angle at the vertex, the vertical angle. 




reater than two 





6 DESCRIPTIVE GEOMETRY 

51. The altitude of a triangle is a perpendicular drawn from 
the vertex to the base. The altitude of a parallelogram is a 
perpendicular to the base from any point in the opposite side. 
The altitude of a trapezium is a perpendicular from any point 
in one of its parallel sides to the opposite side. 

52. A diagonal of a quadrilateral is a straight line joining two 
of the opposite angular points. 

A diagonal of any polygon is a straight line joining any two 
of its angular points which are not consecutive. 

53. A rectangle is said to be contained by two lines when its 
two adjacent sides are these lines, or lines equal to them. 

54. A line is said to be cut in medial section, or in extreme and 
mean ratio, when the rectangle contained by the whole line and one 
of its parts is equal to the square on the other part. 

55. A rectilineal figure is said to be inscribed 
in another rectilineal figure when all the angular 
points of the inscribed figure are upon the sides of the 
figure in which it is inscribed. 

56. A rectilineal figure is said to be circumscribed about 
another when its sides respectively pass through the angular 
points of the other figure about which it is circum- 
scribed. 

57. A rectilineal figure is said to be inscribed in 
a circle when all the angular points of the inscribed 
figure are upon the circumference of the circle. 

58. A rectilineal figure is said to be circum- 
scribed about a circle when each side of the recti- 
lineal figure touches the circumference of the circle. 

59. A circle is said to be inscribed in a recti- 
lineal figure when the circumference of the circle touches each of 
the sides of the rectilineal figure. 

60. A circle is said to be circumscribed about 
a rectilineal figure when the circumference of the 
circle passes through all the angular points of the 
figure. 

61. A regular polygon has all its sides and angles equal ; or 
it is both equilateral and equiangular. 

62. A polygon of five sides is called a pentagon ; of six, a 
hexagon; of seven, a heptagon; of eight, an octagon ; of nine, 
a nonagon ; of ten, a decagon ; of twelve, a dodecagon. 





DESCRIPTIVE GEOMETRY 7 

63. The centre of a regular polygon is a point equidistant 
from its sides, or from its angular points. 

64. The apothem of a regular polygon is a perpendicular from 
its centre upon any of its sides. 

65. The perimeter of any figure is its circumference or whole 
boundary ; it is also called the periphery. 

66. The ratio of any two quantities to one another is the 
number of times that the former contains the latter. 

Thus, if a line A contain a line B three times, the ratio of 

A to B is 3, or the ratio of B to A is 3. The ratio of A to 

B is denoted by A : B, or A -f B, or g. 

67. A proportion consists of two equal ratios. 




PROBLEMS 

68. Problem L To describe a circle with a given radius 
about a given point as a centre. 

Let AB be the given radius, and C the given 
point. 

Place one point of the compasses on A, and 
extend the other point to B ; then with that 
distance as a radius, and placing one point of 
the compasses on C, describe with the other 
point the circumference DEF ; and the required 
circle will be formed. 

69. Problem II. To bisect a given straight line ; that is, 
to divide it into two equal parts. 

METHOD L Let AB be the given straight 
line. 

From A and B as centres, with a radius 
greater than the half of AB, describe arcs EC, 
FC, intersecting in C (68) ; describe arcs simi- 
larly intersecting in D ; and join the points 
C, D, and CD will bisect AB in H. 

METHOD 2. As before, describe arcs inter- 
secting in C, and describe similarly two arcs intersecting in 
G ; and if GC be then drawn and produced, it will bisect AB 
inH. 

The first method can be proved by joining with straight lines 



DESCRIPTIVE GEOMETRY 




the points A and C, C and B, B and D, D and A. For then the 
two triangles thus formed namely, ADC and BDC would be 
equal in every respect (Eucl. I. 8) ; and hence the two angles 
thus formed at D would be equal. Then the two triangles ADH, 
BDH, would be equal (Eucl. I. 4), and hence AH = HB. 
The second method can be similarly proved. 

70. Problem III. To describe a semicircle on a given 

finite straight line as a diameter. 

Let AB be the given straight line. 

Bisect it in C (69), arid from C as a centre, 
with a radius equal to AC or CB, describe the 
semicircle ADB (68), and it will be the required 
semicircle. 

71. Problem IV. From a given point in a given straight 

line, to erect a perpendicular. 

TLet AB be the given straight line, and C the 
given point. 
CASE 1. When the point is near the middle 
A 5 c E of the line. 

On each side of C lay off equal distfinces 

CD, CE ; and from D and E as centres, with a radius greater 
than DC or CE, describe arcs intersecting in F ; draw CF, which 
is the required perpendicular. (Eucl. I. 11.) 

CASE 2. When the point is near one of the extremities of the 
line. 

METHOD 1. From C as a centre, with 
any radius, describe the arc DEF, and from 
D lay off the same radius to E, and from E 
to F ; then from E and F as centres, with 
the same or any other radius not less than 
half the former, describe arcs intersecting 
in G ; draw GC, and it will be perpendicular 
to AB. 

This is evident from Eucl. IV. 15, Cor. 
METHOD 2. From any point D as a centre, 
and the distance DC as a radius, describe 
an arc ECF, cutting AB in E and C ; draw 
ED, and produce it to cut the arc in F ; then draw FC, 
which is the required perpendicular. (Eucl. III. 31.) 



/Ts 




DESCRIPTIVE GEOMETRY 



9 



72. Problem V. From a given point without a given 
straight line, to draw a perpendicular 
to it. 

Let AB be the given line, and P the given 
point. 

CASE 1. When the point is nearly opposite 
to the middle of the line. 

From P as a centre, with any convenient 

radius, describe arcs cutting AB in C and D ; and from these 
two points as centres, witli a radius greater than the half of 
DC, describe arcs cutting in the point E ; draw PE, and PF 
will be the required perpendicular. 

This may be proved as Prob. II. 

CASE 2. When the given point is nearly 
opposite to one end of the line. 

METHOD 1. From any point C in AB as a 
centre, with the radius CP, describe an arc on 
the other side of AB ; and from any other 
point D in AB, with the radius DP, describe 
an arc cutting the former in E ; then draw PE, 
and PG is the perpendicular. 

This is proved as the preceding case. 

METHOD 2. Take any point C in AB, and 
join PC, and on PC describe a semicircle 
(III.) PDC intersecting AB in D ; draw PD, 
which is the perpendicular required. (Eucl. 
III. 31.) 



73. Problem VI. On a given straight line, 
to describe an equilateral triangle. 

Let AB be the given line. 

From A and B as centres, with a radius equal 
to AB, describe arcs intersecting in C, and draw 
AC, BC ; then ABC is the required triangle. 
(Eucl. I. 1.) 



74. Problem VII. To describe a triangle whose three 
sides shall be respectively equal to three given lines, of 
which the length of any two together is greater than the 
third. 

Let AB, CD, and EF be the three given lines. 





10 



DESCRIPTIVE GEOMETRY 




P N 
B 



Draw any line MN, and from it cut off a part MP equal 
to AB ; then from M as centre, with CD as 
radius, describe an arc at Q ; and from P 
as centre, with EF as radius, describe another 
arc cutting the former in Q ; and draw MQ 
and PQ ; then MPQ is the required triangle. 
(Eucl. I. 22.) 

75. Problem VIII. On a given straight 
line, to describe a square. 

Let MN be the given straight line. 

From M drawMP perpendicular to MN (71), 
and from MP cut off a part MQ equal to MN ; 
then from Q and N as centres, with a radius 
equal to MN, describe arcs intersecting in R ; 
draw QR and NR ; MR* is the required 
square. 

This is easily proved by Eucl. I. 8 and 32. 

76. Problem IX. To describe a rectangle whose length 
and breadth shall be respectively equal to two given 
straight lines. 

Let HI and KL be the given straight lines. 
Draw a line MN equal to HI ; and draw 
MP perpendicular to MN (71), and equal 
to KL ; from P as a centre, with a radius 
equal to MN, describe an arc at Q ; and 
from N as centre, with a radius equal to 
MP, describe an arc cutting the former in 
Q ; draw PQ, NQ ; and MQ is the required 
rectangle. 
This may be proved by Eucl. I. 8, 27 and 29. 

77. Problem X. To find the centre of a 
given circle. 

Let PQX be the given circle. 

Draw any chord PQ in the circle ; bisect 
the chord by the perpendicular XY, which 
is a diameter ; then bisect XY in the point 
W, and the point W is the centre of the 
circle. (Eucl. III. 1.) 

* Quadrilateral figures are thus concisely named by the letters at two opposite 
angular points. 





DESCRIPTIVE GEOMETRY 



11 




78. Problem XI. To describe a circle through three given 
points, not in the same straight line. 

Let P, Q, and R be the three points. 

Join PR and QR ; bisect PR by the per- 
pendicular ST, and QR by the perpendicular 
VT ; then from T as centre, with any of the 
distances TP, TR, TQ, describe a circle, and 
it will pass through the points P, Q, R, and be 
the required circle. (Eucl. IV. 5.) 

79. Problem XII. Given a segment of a circle, to com- 
plete the circle of which it is a segment. 

Let P, Q, and R (fig. Prob. XI.) be any three points in the 
arc of the segment. 

As in the preceding problem, find T the centre of the circle 
that passes through P, Q, and R, and it is the centre of the 
required circle, which can be described as in that problem. 

80. Problem XIII. To draw a tangent to a given circle 
from a given point in its circumference. 

Let PRS be the given circle, and P the 
given point. 

Find the centre of the circle, and from 
the point P draw the radius PQ ; then draw 
a line TV through P perpendicular to PQ, 
and TV is the required tangent. (Eucl. 
HI. 16.) 

81. Problem XIV. To draw a tangent 
to a given circle from a given point 
without it. 

Let P be the given point, and RVS tbe 
given circle. 

METHOD 1. Find the centre Q, and join 
P and Q ; on PQ describe a semicircle PRQ, 
cutting tbe given circle in R ; draw PR, and 
it is the required tangent. 

For if RQ is joined, then PRQ, being an 
angle in a semicircle, is a right angle. 
(Eucl. III. 31.) 

METHOD 2. Find Q the centre of the 
circle, and with the radius PQ describe the 
arc QUTj with the diameter of the circle 
VS as a radius, and Q as a centre, cut the arc QUT in T ; 




12 



DESCRIPTIVE GEOMETRY 




draw TQ, intersecting the given circle in R; draw PR, and 
it is the required tangent. 

For PR bisects QT, and is therefore perpendicular to it. 
(Eucl. III. 3.) 

82. Problem XV. To bisect a given angle. 

Let MDN be the given angle. 

Lay off on the sides of the angle any equal 
distances DP, DQ ; from P and Q as centres, 
describe arcs with equal radii intersecting in 
R ; draw DR, and it bisects the given angle, 
or divides it into the two equal angles MDR 
andNDR. (Eucl. I. 9.) 

83. Problem XVI. To bisect an arc of a circle. 
Let PSQ (fig. Art. 82) be the arc, of which D is the centre. 

Find the point R, as in the preceding problem ; and then the line 
DR divides the arc in S into the two equal arcs PS and SQ. 
(Eucl. III. 26.) 

84. Problem XVII. To trisect a right angle; that is, 
to divide it into three equal parts. 

Let MON be the right angle. 

From the point O, with any radius, describe 
an arc MPN, cutting the sides of the angle 
in M and N ; with the same radius and the 
centres, M and N, cut the arc in P and Q ; 
draw OP, OQ, which trisect the angle. This 
is evident from Eucl. IV. 15, Cor. 

85. COR. The quadrantal arc NPQM is 
evidently trisected in the points P and Q. 

86. Problem XVIII. At a given point 
in a given straight line, to make an angle 
equal to a given angle. 

Let O be the given angle, QP the given 
straight line, and Q the given point. 

From the centres O and Q, with the same 
radius, describe arcs MN and PS ; with a 
radius equal to the chord of the arc MN, 
and P as centre, cut the arc PS in R ; draw 
QR, and PQR is the required angle, being equal to angle MON. 






DESCRIPTIVE GEOMETRY 13 

For if MN are joined, and also PR, the two triangles MON, 
PQR, will be equal in every respect (Eucl. I. 8) ; and hence 
angle Q = O. 

87. Problem XIX. Through a given point to draw a 
straight line parallel to a given straight line. 

Let AB be the given line, and P the given 
point. 

METHOD 1. Take any point Q in AB, and 
draw PQ ; make the angle RPQ equal to the 
angle PQA (86), and PR is parallel to AB. 
(Eucl. I. 27.) _P .a 

METHOD 2. In AB take any two points 
M and N ; from P as centre, with the radius 

MN, describe an arc at Q ; from N as centre, r , . - 

with the radius MP, describe an arc cutting 
the former in Q ; draw PQ, and it is parallel to AB. 

For it is easily proved that if PM, NQ were joined, PN would 
be a parallelogram. 

88. Problem XX. To draw a straight line parallel to a 
given straight line, and at a given distance from it. 

Let KL be the given line, and D the given distance. 

From any two points M and N in KL as 

centres, with a radius equal to D, describe R ^- p --^ ^- a ---^ 5 

the arcs P and Q ; draw a line RS to touch 
these arcs ; that is, to be a common tangent 

to them ; and RS is the line required parallel * M N L 

toKL. 

89. Problem XXI. To divide a given straight line into 
any number of equal parts. 

Let AB be the given straight line, and let the number of equal 
parts into which it is to be divided be five. 

Draw a line AC through A at any incli- 
nation to AB, and through B draw another 
line BD parallel to AC ; take any distance 
AE, and lay it off four times on AC, forming 
the equal parts AE, EF, FG, GH ; lay off the D 
same distance four times on BD in the same manner, from the 
point B ; draw the lines HI, GK, FL, and EM, and they will divide 
AB into five equal parts. For AB, AH, and BM are cut pro- 
portionally. (Eucl. VI. 10.) 




14 



DESCRIPTIVE GEOMETRY 





90. Problem XXII. To find a third proportional to two 
given straight lines. 

Let A and B be the given lines. 
Draw a line CD equal to A, and through 
C draw a line CQ inclined at any angle to 
CD; make CE and CF each equal to B; 
join DF, and through E draw EG parallel 

A to DF ; and CG is the third proportional to 

A and B ; that is, 
A:B = B:CG. (Eucl. VI. 2.) 

91. Problem XXIIL To find a fourth proportional to 
three given straight lines. 

Let A, B, and C be the three given 
straight lines. 

Draw two lines DE, DF, forming any 
angle ; make DG equal to A ; DH equal 
to B ; DI equal to C ; join G and H, and 
through I draw IK parallel to GH, cutting 
DF in K ; then DK is the required fourth 
proportional ; that is, 
A:B = C:DK. (Eucl. VI. 2.) 

92. Problem XXIV. To find a mean proportional between 
two given straight lines. 

Let A and B be the given straight lines. 

Draw any line CP, and lay off on it CE 
equal to A, and ED to B ; on CD describe 
a semicircle CFD (70) ; from E draw EF 
perpendicular to CD, and EF is the mean 
proportional ; that is, 
A : EF = EF : B. (Eucl. II. 14, and VI. 17.) 

93. COR. If A and B be two adjacent sides of a rectangle, the 

line EF is the side of a square equal in area 
to it. 

94. Problem XXV. To find a square 
that shall be equal to the sum of two 
given squares. 

Let A and B (fig. in 95) be the sides of the 
two given squares. 
Draw any line CD and DE perpendicular to it; make DF = A, 





DESCRIPTIVE GEOMETRY 15 

and DG = B; join F, G, and FG is the side of the required 
square ; for 

FG 2 = FD 2 + DG 2 = A 2 + B 2 . (Eucl. I. 47. ) 

95. Problem XXVI. To find a square that shall be equal 
to the difference between two given squares. 

Let A and B be the sides of the two given * 

squares. 

Draw any line CP (fig. Prob. XXIV.); make CG and GD 
each = A, and GE = B ; from centre G, with radius GD, describe the 
circle CFD, and draw FE perpendicular to CE, and FE is the side 
of the required square ; for 

EF 2 = GF 2 - GE 2 = A 2 - B 2 . (Eucl. I. 47, Cor. ) 

96. Problem XXVII. To divide a given straight line 
similarly to a given divided straight line. 

Let AB be the given divided line, C and D being its points of 
section ; and MN the given line to be divided. 

Draw through M a line MP at any incli- 
nation to MN, and equal to AB, and make 
its segments equal respectively to those of 

AB namely, MQ to AC, QR to CD, and > - * g 

RP to DB. Join P and N, and draw through 

R and Q the lines RT, QS, each parallel to PN ; and MN is divided 

in S and T similarly to AB ; that is, 

MS : ST = AC : CD, and ST : TN = CD : DB. (Eucl. VI. 10.) 

97. Problem XXVIII. To cut a given straight line in 
medial section. 

Let PQ be the given line. 

Erect a perpendicular QR equal to the half 
of PQ ; join PR ; from R as a centre, with 
the radius RQ, describe an arc cutting PR in 
S ; from P as a centre, with the radius PS, 
describe an arc cutting PQ in T ; then PQ is cut medially in T ; 
that is, 

PQ:PT=PT:TQ. 

For PR 2 =PQ 2 + QR 2 , 

or PS 2 + SR 2 + 2RS -PS = PQ QT + PQ PT + QR 2 , 

and RS 2 = QR 2 , also 2RS PS = PQ PT ; 

hence PQ QT = PS 2 =PT 2 , 

or PQ:PT = PT:QT. 




16 



DESCRIPTIVE GEOMETRY 




98. Problem XXIX. To produce a line, so that the pro- 
duced line may be cut medially at the extremity of the 
given line. 

Let AB be the given line. 

Bisect AB in D ; draw BC perpendicular 
to AB, and equal to it ; from centre D, with 
radius DC, cut AB produced in E ; and AE 
is cut medially in B ; that is, 
AE:AB = AB:BE. 

For DB 2 + BC 2 =DC 2 =DE 2 =DB 2 + BE 2 + 2DB-BE, and taking 
away DB 2 from both, 

BC 2 or AB 2 =BE 2 + AB BE = AE BE ; 
or AE:AB = AB:BE. 

99. Problem XXX. To describe an isosceles triangle 
having each of the angles at the base double of the third 
angle. 

CASE 1. When one of the sides of the triangle is given. 
Let AB be the given side. 

Cut AB medially in C (97), so that AC may 

be the greater segment ; then construct an 
isosceles triangle on AC as a base, and having each of its sides 
equal to AB. (74.) 

CASE 2. When the base is given. 

Let AB be the given base. 

Produce AB to C, so that AC may be cut medially in B (98) ; 
then construct an isosceles triangle on AB as a base, and having 
each of its sides equal to AC. (Eucl. IV. 10.) 

100. Problem XXXI. On a given straight line, to describe 
a segment of a circle containing an angle equal to a given 
angle. 

Let AB be the given line, and C the given 
angle. 

Draw AD, making angle BAD equal to C ; 
draw AE perpendicular to AD, and GF bisect- 
ing AB perpendicularly ; from centre G, with 
radius GA, describe the circular segment 
AHB, and it is the segment required ; for 
any angle in it, as AEB, is equal to C. (Eucl. III. 33.) 




DESCRIPTIVE GEOMETRY 



17 





101. Problem XXXII. From a given circle, to cut off a 
segment that shall contain an angle equal to a given 
angle. 

Let ABC be the given circle, and D the 
given angle. 

At any point B in the circumference, draw 
a tangent EBF ; draw a chord BC, making 
angle CBF equal to D ; and BAG is the re- 
quired segment ; for any angle in it, as BAG, is equal to D. 
(Eucl. III. 34.) 

102. Problem XXXIII. In a given circle, to inscribe a 
triangle equiangular to a given triangle. 

Let ABC be the given circle, and DEF 
the given triangle. 

Draw a tangent GAH to the circle at the 
point A ; draw the chord AC, making the 
angle HAG equal to E, and the chord 
AB, making angle GAB equal F ; join BC, 
and ABC is the required triangle similar to DEF, having angle 
B equal to E, angle C to F, and BAC to D. (Eucl. IV. 2.) 

103. Problem XXXIV. To describe a 
circle about a given triangle. 

Let MON be the given triangle. 

Bisect the side MN by the perpendicular 
PR ; bisect NO similarly by the perpendicular 
QR ; from R, the point of intersection, as a 
centre, with any of the distances RM, RN, 
or RO, describe the circle MNO, which is the required circum- 
scribing circle. (Eucl. IV. 5.) 

Compare Problem XI. Y 

104. Problem XXXV. To inscribe a 
circle in a given triangle. 

Let WXY be the given triangle. 

Bisect the angle XWY by the line WZ, and 
the angle WXY by the line XZ (82) ; from 
the intersection Z draw ZV perpendicular 
to WX, with VZ as a radius and Z as a 

centre, describe the circle VTU, and it is the required inscribed 
circle. (Eucl. IV. 4.) 





18 



DESCRIPTIVE GEOMETRY 




105. Problem XXXVI. To inscribe an equilateral triangle 
in a given circle. 

Let WXY be the given circle, and V its 
centre. 

Draw a diameter ZY, and from Z as a centre, 
and the radius ZV, cut the circumference in W 
and X ; draw WX, WY, and XY, and WXY is 
the equilateral triangle. 

For the arcs WZ, ZX are each one-sixth of 
the circumference. (Eucl. IV. 15, Cor.) 

106. Problem XXXVII. In a given circle, to inscribe a 
regular hexagon. 

A Let WXY be the given circle (fig. Prob. XXXVI. ) 

With the radius of the given circle, and any 
point Z in the circumference as a centre, cut 
the circumference in W ; draw WZ, and it is a 
side of the regular hexagon, which may be laid 
off six times on the circumference of the circle, 
and every two successive points of section being 
joined, the resulting figure will be the regular 
hexagon. 

In this manner, the regular hexagon in the adjoining figure is 
described. (Eucl. IV. 15.) 

107. Problem XXXVIII. In a given circle, to inscribe a 
regular dodecagon. 

Let WXY be the given circle (fig. Art. 105). 

Let WZ be a side of the inscribed regular hexagon ; bisect the 
arc W T Z in U, and the distance WU being laid off twelve times on 
the circumference, and every two successive points of section 
being joined, the resulting figure is the regular 
dodecagon. 

108. Problem XXXIX. To inscribe a 
square in a given circle. 
Let PRQS be the given circle. 
Draw two diameters PQ, SR perpendicular 
to each other ; and join their extremities by 
PS, SQ, QR, and RP; and PSQR is the required square. 
(Eucl. IV. 6.) 





DESCRIPTIVE GEOMETRY 



19 




109. Problem XL. To inscribe a regular octagon in a 
given circle. 

Let the given circle be PSQ (fig. Art. 108). 

Find a side PR of the inscribed square, and bisect the arc PUR 
in U (83) ; join R and U, and RU may be laid off eight times on 
the circumference, and the adjacent points of section being joined, 
the regular octagon will be formed. 

110. Problem XLL To inscribe a regular pentagon in a 
given circle. 

Let SLR be the given circle. 

Draw two perpendicular diameters IK, 
LM ; bisect the radius OI in N ; from N as a 
centre, with NL as a radius, cut OK in P ; 
with radius LP, and centre L, cut the cir- 
cumference in Q ; join LQ, and other four 
chords equal to it being drawn in succession 
in the circle, the required polygon will be 
formed. 

This construction depends on this theorem : The square of a 
side of a regular pentagon inscribed in a circle is equal to the sum 
of the squares of the sides of the inscribed regular hexagon and 
decagon. 

in. Problem XLIL To inscribe a regular decagon in a 
given circle. 

Let SLR be the given circle (fig. Art. 110). 

Find a side LQ of the inscribed regular pentagon ; bisect the 
arc LQ in V, and the chord LV being drawn, it is a side of the 
regular decagon ; and ten chords equal to it being successively 
placed in the circle, will form the polygon. 

112. Problem XLIIL To describe a regular polygon about 
a given circle. 

Let WVY be the given circle. 

Find the angular points of the correspond- 
ing inscribed polygon of the same number 
of sides by the preceding problems ; let W, 
X, Y be three of these angular points ; 
through these points draw the tangents WTJ, 
UT, TY; and UT is a side of the required 
polygon ; in the same manner, the other 
sides are found, and the circumscribing polygon is thus described. 




20 




113. Problem XLIV. On a given straight line, to con- 
struct a regular pentagon. 

Let PQ be the given line. 
Produce PQ to S, so that PS may be cut 
medially in Q (98) ; then with a radius equal 
to PS, from P and Q as centres, describe 
arcs cutting in II ; from P and U as centres, 
with the radius PQ, describe arcs cutting in 
a s V ; from Q and U as centres, with the same 
radius, describe arcs cutting in W ; join in order the points Q, 
W, U, V, and P, and the required pentagon will be formed. 

Since PS is cut in medial section in Q, an isosceles triangle, 
of which PQ is the base, and PS the sides, has each of its 
angles at the base double of the vertical angle (Eucl. IV. 10) ; 
hence if UP and UQ were joined, UPQ would be such a tri- 
angle, and hence the figure PQWUV is the required pentagon. 
(Eucl. IV. 11.) 

114. Problem XLV. On a given straight line, to construct 

a regular hexagon. 

Let GH be the given line. 

From G and H as centres, with the radius 
GH, describe arcs intersecting in X, and X is 
the centre of the circumscribing circle ; hence 
from the centre X, with the radius XG, de- 
scribe a circle, and apply GH six times along 
the circumference, and GHKL is the required 
hexagon. This is evident. (Eucl. IV. 15, Cor.) 

115. Problem XL VI. On a given straight line, to con- 
struct a regular octagon. 

Let LM be the side of the octagon. 
METHOD 1. Draw from L and M two 
perpendiculars of indefinite length, LO and 
MN ; produce LM in both directions, and 
bisect the angles QLO, PMN by the lines 
LU, MR, which are to be made equal to 
LM ; from U and R draw UT, RS parallel 
to LO and MN, and equal to LM ; from 
T and S as centres, with the radius LM, 

describe arcs cutting LO and MN in O and N ; draw TO, SN, and 

ON, and the octagon is then constructed. 




V 



bESCRIfTIVE GEOMETRY 



21 




METHOD 2. After drawing the lines LU and MR, as in the 
first method, bisect the angles ULM, LMR, by LV and MV, 
and V is the centre of the circumscribing circle, and LV its 
radius. Hence if this circle be described, and the line LM be 
applied eight times along the circumference, the required octagon 
will be constructed. 

Since the angles at L and M in triangle VLM are together 
= 135, therefore V=45 = of 360, or four right angles. Hence 
LM is the side of an octagon inscribed in a circle, of which VL is 
the radius. 

116. Problem XL VII. To describe a circle about any 
given regular polygon. 

Let ABODE be the regular polygon. 

Bisect the angles BCD and CDE by the straight 
lines CF, DF intersecting in F ; from the centre F, 
with the radius FC, or FD, describe a circle, and 
it will pass through all the angular points of the 
polygon, and be described about it. In the same 
manner, a circle may be described about any other polygon. 

117. Problem XLVIIL To inscribe a circle in any given 
regular polygon. 

Let ABODE be any regular polygon. 

Bisect the angles BCD and CDE by the 
straight lines CF and DF, intersecting in F ; 
and from F draw FK perpendicular to CD. 
From the centre F, with radius FK, describe 
a circle, and it will touch all the sides of 
the polygon, and therefore be inscribed in the 
polygon (59). 

118. Problem XLIX. On a given straight line, to con- 
struct a triangle similar to a given triangle. 

Let DE be the given line, and 
ABC the given triangle. 

At the point D draw DF, making 
angle D equal to angle A (86) ; 
and at E draw EF, making angle E 
equal to B ; then DEF is the triangle required. 

For the third angle F is then = C. (Eucl. I. 32.) 





22 



DESCRIPTIVE GEOMETRY 




119. Problem L. On a given straight line, to construct a 
figure similar to a given rectilineal figure. 

Let FG be the given line, and 
ABCDE the given figure. 

Divide the given figure into 
triangles by drawing diagonals 
AC, AD; on FG describe a 
triangle FGH similar to ABC 
(118); on FH describe the triangle 
FIH similar to ADC ; on FI describe a triangle FKI similar to 
AED : then the whole figure FGHIK is similar to ABCDE. 
(Eucl. VI. 18.) 

120. Problem LI. To construct a rectangle equivalent 

to a given triangle. 

Let MNO be the given triangle. 

Through the vertex O draw OR parallel to 
the base MN ; through P, the middle point of 
the base, draw PQ perpendicular to it; through 
N draw NR parallel to PQ, and PQ11N is the 
required rectangle equal to the triangle MNO. (Eucl. I. 42.) 

121. Problem LIL To construct a triangle equivalent to 

a given quadrilateral. 

Let ABCD be the given 
quadrilateral. 

Join DB, and through C 
draw CE parallel to DB ; then 
join DE, and ADE is the 
required triangle equivalent to ABCD. 

122. Problem LIIL To rectify a crooked boundary; that 
is, to find a straight line that will cut off the same surface 
on each side of it that a given crooked boundary does. 

CASE 1. Let ABC be the crooked 
boundary, and DE a fixed line. 

Join A and C, and through B draw 
BF parallel to AC ; join AF, and AF is 
the required boundary, the triangle ABG, 
taken from the original space on one side 
of AF, being equivalent to FGC added on 
the other ; or the triangle ABC is = AFC. (Eucl. I. 37.) 






DESCRIPTIVE GEOMETRY ZO 

CASE 2. Let ABCDE be the crooked boundary, and MN the 
fixed line. 

Join EC, and through D draw DF 
parallel to CE ; then join C and F, and 
the line CF may now be taken instead 
of the lines CD, DE (Case 1). Again, 
join BF, and through C draw CG paral- A.. 
lei to BF ; join BG ; then BG may now 
be taken instead of BC, CF. Again, join 
AG, and through B draw BH parallel to 
AG ; join AH; and AH may now be 
taken instead of AB and BG. Hence AH is the required line. 

The method may easily be extended to a crooked boundary con- 
sisting of any number of lines. 

Instead of drawing the lines DF, CG, &c., only the points F, G, 
&c. may be jnarked. 

CONSTRUCTION OF SCALES, AND PROBLEMS TO BE 
SOLVED BY THEM 

Scales are lines with divisions of various kinds marked 
upon them, according as they are to be used for measuring 
lines or angles. In Geometry, they are employed for the 
construction and measurement of mathematical figures. 

The values of the magnitudes of lines or angles are 
numbers representing the number of times that some unit 
of the same kind is contained in them. 

The unit of measure for lines is some line of given 
length as a foot, a yard, a mile, and so on. 

The unit of measure of an angle is the 90th part of a right 
angle. Hence a quadrant of a circle which measures a right 
angle is supposed to be divided into 90 equal parts called 
degrees. The whole circumference of a circle, therefore, is 
supposed to be divided into 360 degrees ; each degree into 
60 equal parts called minutes ; and each minute into 60 
equal parts called seconds ; and so on. Degrees, minutes, 
and seconds are respectively denoted by the marks , ', " \ 
thus 23 27' 54" denotes 23 degrees, 27 minutes, and 54 
seconds. 



24 



DESCRIPTIVE GEOMETRY 



An angle is measured by the number of degrees, minutes, 
&c., in the circular arc intercepted by the sides of the angle, 
the angular point being the centre of the arc. The numerical 
measure of the angle in degrees, &c., will evidently be the 

same, whatever be the length of the 

radius of the arc. 

Thus, if E is the centre of the arcs AB, 
CD, and if AB contain 30, CD will also 
contain 30, for these arcs are the same 
parts of the circles to which they belong. 

123. Problem LIV. To construct a scale of equal parts. 
Lay off a number of equal divisions, AB, BC, CD, &c., and AE, 
and divide AE into 10 equal parts (89). When a large division, as 

4- s 2 i 

''ii 




AB, represents 10, each of the small divisions in AE will repre- 
sent 1. When each of the large divisions represents 100, each of 
the small divisions in AE represents 10. Hence, on the latter 
supposition, the distance from C to n is = 230; and on the former 
supposition, it is = 23. 

If the large divisions represent units, the small ones on AE 
represent tenths ; that is, each of them is T V or !. On this 
supposition, the distance Cn is = 2'3. 

124. Problem LV. To construct a plane diagonal scale. 

1. A diagonal scale for two figures. 

Draw five lines parallel to DE and equidistant, and lay off the 
equal divisions AE, AB, BC, CD, &c., and make EP, AQ, Bl, 













\ / 




B: 








h / 












"V 



C2, &c., perpendicular to DE. Find m the middle of AE, and 
draw the lines Qm, mP. 

The mode of using this scale is evident from the last. If the 
large divisions denote tens, then from n to o is evidently = 34. 



DESCRIPTIVE GEOMETRY 



25 



2. A diagonal scale for three figures. 

Draw ten lines parallel to DE, and equidistant. Lay off the 
equal parts AB, BC, CD, &c., and AE, and draw EP, AQ, Bl, 
C2,...&c., perpendicular to DE. Divide QP and AE into 10 equal 
parts. Join the 1st, 2nd, 3rd,... divisions on QP with the 2nd, 3rd, 
4th,... divisions on AE respectively. 



_Q2 4 6 8 




If the divisions on AD each represent 100, each of those on QP 
will represent 10. Thus from 3 on AD to 8 on QP is = 380; but 
by moving the points of the compasses down to the fourth line, 
and extending them from n to o, the number will be = 384. For 
the distance of 8 on QP from Q is = 80, and of r from A is = 90 ; and 
hence that of o from the line AQ is = 84. 

When the divisions on AD denote tens, those on QP denote 
units ; from n to o would then = 38 '4. 

NOTE. When the numbers representing the lengths of the sides 
of any figure would give lines of an inconvenient size taken from 
the scale, they may be all multiplied or all divided by such a 
number as will adapt the lengths of the lines to the required 
dimensions of the figure. 

125. Problem LVL To construct a vernier scale adapted 
to a scale of equal parts. 

Let AB be a part of the scale, and mu the vernier. 

1. When the divisions of the vernier lie in a direction opposite to 
those of the principal scale. 

The zero or of the vernier, in the accompanying diagram, lies 
between 42 and 43 on the scale, and the use of the vernier is to 



TTT 



determine what part rv is of a division of the scale. If it be required 
to estimate rv in tenth-parts of a division of the scale, then let 
10 divisions on the vernier mu be = 11 on the principal scale; 
then 1 division of the vernier will exceed 1 on the principal scale 



26 DESCRIPTIVE GEOMETRY 

by T V of a division of the latter. Therefore the 7 divisions of the 
vernier from s to v exceed the 7 on the principal scale from s to r 
by A 5 that is, rv is T V Hence this rule : 

RULE. -The number of the division on the vernier that coin- 
cides with a division on the principal scale shows the numerical 
value of the part to be measured ; that is, the part between the 
zero of the vernier and the preceding division on the principal 
scale. 

2. When the divisions on the vernier lie in the same direction as 
those on the principal scale. 

Let the vernier m'u' have its zero at v', and let it be required to 
estimate the part r'v', as in the former case, in tenths of a division 
of the principal scale. Make 10 divisions of the vernier equal to 
9 on the principal scale ; then each division on the latter will 
exceed each on the vernier by T V. Therefore the 4 divisions on 
the principal scale from s' to r' will exceed the 4 on the vernier 
from s' to v' by T 4 ff ; that is, r'v' is T \ ; and the rule thus obtained 
is the same as in the former case. 

The zero of the vernier stands, therefore, opposite to 42 - 7 in the 
former case, and to 45'4 in the latter. 

Let d, d', denote the divisions on the principal scale and vernier 
respectively; then in the first case, 10d' = lld, ord'=d + ^d; and 
in the second case, 9d=10d", or d=d' + %d' = d' + T V^ 

If the divisions 40, 50, &c. , are reckoned as 400, 500, &c., the 
small divisions on the principal scale will denote 10, and the part 
rv will then be estimated in units. The zeros of the verniers are 
thus opposite to 427 and 454 on the principal scale. 

In the same manner, a vernier for a circular arc is constructed. 
If the arc be divided into half-degrees, and the distance of 29 of 
these divisions on the vernier be divided into 30 equal parts, the 
angle can then be read to minutes, when the vernier reads in the 
same direction as the readings on the arc; and generally if r 
represent the value of one division on the arc, and n the divisions 
on the vernier for the length of (n-1) divisions on the arc, then 

- is the value of one division, as read by the vernier. 
n 

126. Problem LVIL To construct a scale of chords. 

Let AB be the radius to which the scale is to be adapted. 

With centre A and radius AB describe a quadrant BEG. 
Divide the quad ran tal arc BEC into 9 equal parts, BD, DE, &c., 
which is easily done by first dividing it into 3 equal parts, BF, 



DESCRIPTIVE GEOMETRY 



27 



FG, GC (XVII.), and then trisecting each of these parts by 

trial, as no direct method is known. Draw the chord of the 

quadrant BC ; from B as a centre, and 

the chord of BD as a radius, describe an arc 

cutting BC at 10 ; with the chord of BE as 

a radius, describe an arc cutting BC in 20 ; 

with the chord of BF, describe an arc cutting 

BC in 30 ; and in a similar manner, find 

the divisions 40, 50, 60, 70, 80. Then the 

arcs BD, BE, BF, &c., being arcs of 10, 20, 

30, &c., respectively, the distances from B 

to 10, 20, 30, &c., are the chords of arcs of 10", 20, 30, &c. ; 

so that BC is a scale of chords for every 10, from to 90. 

127. Problem L VIII. To construct scales of tangents, 
secants, and rhumbs. 

For definitions of sines, tangents, secants, see TRIGONOMETRY. 

Rhumbs are lines drawn to the points of the compass ; there 
are eight points in each quadrant, so that one point contains 





11 15'. The scale of rhumbs consists of the chords of one, 
two, three, &c., points, or of 11 15', 22 30', 33 45', 45, &c. 
(See NAVIGATION, page 535). 

The line of tangents is constructed by dividing the quadrant 
into 9 equal parts, each = 10, and drawing through the points 
of division, from the centre, straight lines ; then these lines 
produced will cut a tangent drawn at the extremity of the 
quadrant in the required points 0, 10, 20, &c. 



28 



DESCRIPTIVE GEOMETRY 



The distances from the centre to the divisions on the scale of 
tangents, being laid off on a straight line, give the divisions 0, 10, 
20, 30, &c., of the scale of secants. 

128. Problem LIX. To construct a scale of sines, and of 
semitangents, and a line of longitudes. 

Divide the quadrant BC into 9 equal parts ; through them draw 
parallels to OB, meeting the perpendicular BD, and it will be a 
scale of sines. 

The semitangent of an arc is the tangent of half that arc. Thus 
the semitangent of 48 is = tangent of 24. 

A line of longitudes is a scale of the lengths of a degree of 
longitude at different latitudes. For example, the length of a 
degree of longitude at latitude 60 is = 30 geographical miles, being 
exactly = the half of a degree of longitude at the equator. 




Join A with the divisions of the quadrant BC, and the con- 
necting lines will cut OC in the divisions required for a line of 
semitangents, which are just the tangents of half their respective 
arcs ; which the divisions on OC evidently are, for the angle at the 
circumference is half of the angle at the centre. (Eucl. III. 20.) 

To construct a line of longitudes first make OB a line of 
sines, beginning at O, and then it will also be a line of cosines ; 
thus from O to 80 is the cosine of 80 ; from O to 70, the cosine 
of 70 ; and so on. Now if OB be divided into 60 equal parts, 
represented by the numbers marked above it, they will show the 
number of miles in a degree of longitude corresponding to the 
latitude denoted by the numbers marked below OB. Thus 30 
above OB is opposite to 60 below it, denoting that in latitude 
60 the length of a degree of longitude is = 30 miles. If now, 
instead of the divisions on OB, there be taken on BC the chords 



DESCRIPTIVE GEOMETRY 29 

of the corresponding arcs determined by drawing perpendiculars 
to OB from these divisions, as from 70 to 70, then the higher 
divisions on BC being taken for latitudes, the divisions below 
BC show the length of a degree of longitude corresponding 
to the latitude on the upper side. Thus 40 below BC corre- 
sponds to about 48 above ; that is, in latitude 48, the length 
of a degree of longitude is = 40 miles ; and this appears also from 
the divisions on OB. . 

The lengths of a degree of longitude in two different latitudes 
are evidently proportional to the circumference of the parallels of 
latitude at these places, or proportional to their radii, which are 
just the cosines of the latitudes, the radius of the earth being 
radius, and the earth being supposed a sphere. 

Note. All the above scales are laid down on the common 
Gunter's scale. 

129. Problem LX. At a given point in a given straight 
line, to make an angle of a given number of degrees and 
minutes. 

Let AB be the given line, and A the given point, 
and the number of degrees = 38 30'. 

With a radius equal to the chord of 60, taken 
from the scale of chords, describe an arc CD from A 
A as centre ; with a radius equal to the chord of 
38 30', taken from the same scale, and from C as centre, cut the 
arc CD in E ; draw AE, and A is the required angle. 

When the angle exceeds a right angle ( = 90), lay off on the arc, 
first the chord of 60, and then the chord of the remaining number 
of degrees ; or lay off the chords of any two numbers of degrees 
whose sum is = the given number of degrees. 

130. Problem LXL To measure a given angle; that is, 
to find the number of degrees, &c., it contains. 

Let BAF (fig. Art. 129) be the given angle. 

With the chord of 60 as radius, and A as centre, describe an 
arc CE ; take the chord CE of the arc in the compasses, and 
apply it to the scale of chords ; one point of the compasses being 
placed at 0, the other point will extend to the number of degrees 
and minutes which the angle contains. 

When the chord of the intercepted arc exceeds the chord of 90, 
lay off the chord of 90 on the arc ; take the chord of the remaining 
arc, and find on the scale of chords the number of degrees belonging 




30 DESCRIPTIVE GEOMETRY 

to it, and this number added to 90, will give the whole number 
of degrees in the angle. 

131. Problem LXIL To find a third proportional to two 
given straight lines. 

Measure the two given lines by any scale of equal parts ; then 
find a number that is a third proportional to these two numbers ; 
this number, taken on the scale, will be the length of the third 
proportional. 

Generally, let a and b represent in numbers the measures of the 
two given lines on the scale, and let x be the unknown number 
which gives on the scale the length of the third proportional ; then 

a : b = b : x ; .'. x = . 
a 

Hence, divide the square of the number denoting the length of 
the second line by the number denoting the length of the first, 
and the quotient is the number that denotes the length of the 
required line. 

Let a = 128, and b = 160; then 

IP- 160 2 25600 OAA ,. .... 
x= = - = =200, the third proportional. 

'' I -^ I LS 

132. Problem LXIII. To find a fourth proportional to 
three given straight lines. 

Measure the three given lines on the scale ; then find a number 
that is a fourth proportional to these three ; and this number on 
the scale will give the line that is the fourth proportional. 

Let the measures of the three given lines be denoted by a, b, and 
c, and the required line by x, then 

i be 

a : b = c : x; .. x= . 
a 

Let = 225, 6 = 270, c = 235 ; then x = -= 27 * 235 = 282, 

a 225 

the fourth proportional. 

133. Problem LXIV. To find a mean proportional be- 
tween two given straight lines. 

Find on a scale of equal parts the numbers that represent the 
lines; find their product, and its square root will be the number 
that expresses the length of the required line. 

Generally, let a and b be the numbers representing the two given 
lines, and x the mean proportional ; then 

a : xx -. b ; .-. x 2 = ab, or x=\Jab. 



DESCRIPTIVE GEOMETRY 



31 



Hence, find the product of the numbers representing the given 
lines, ami the square root of this product will be the number 
denoting the length of the mean proportional. 

Let a = 240, and 6 = 364 ; then 

x = V& = V240 x 364 = V87360 = 295 "5. 

134. Problem LXV. To inscribe any regular polygon in 
a given circle. 

Let ABE be the given circle, and let the polygon to be 
inscribed be a regular heptagon. 

Divide 360 by 7, and the quotient 51 26' 
nearly is the angle at the centre of the circle 
subtended by the side of the polygon. 

Hence, make an angle APB at the centre 
= 51 26'; then the chord AB, laid off 7 times 
along the circumference, will form the heptagon. 

If the number of sides of the polygon be denoted by n, then the 

360 
angle at the centre is denoted by ; hence, 

To find the central angle of any regular polygon Divide 
360 by the number of its sides. 

135. Problem LXVI. On a given straight line, to describe 
any regular polygon. 

Let AB be the given straight line, and let the 
polygon to be described upon it be a regular 
heptagon. 

Multiply 90 by 5, and divide by 7 ; then 
the half of one of the interior angles BAH is 

90 x 5 450' 





7 7 

Make the angles BAG, ABC each = 64 17', and C is the centre 
of the circumscribing circle ; and AB being applied 7 times along 
the circumference, will form the heptagon ABDEFGH. 

For any regular polygon, let n = the number of its sides, and 

(n 2)90 
a = the half of one of its interior angles, then = ; hence, 

72- 

To find the half of one of the interior angles of any regular 
polygon Multiply 90 by the number of sides diminished by 2, 
and divide the product by the number of sides, and the quotient is 
the required angle in degrees. 



32 



DESCRIPTIVE GEOMETRY 





136. Problem LX VII. Given the hypotenuse and a side 
of a right-angled triangle, to construct it, and to measure 
the other parts of the triangle. 

Given the hypotenuse AC = 326, and the side 
AB = 200. 

Draw a line AB = 200; draw BC perpendicular to 
AB ; and from centre A, with radius = 326, cut BC in 
C ; draw AC, and ABC is the required triangle. 

By measurement, it will be found that BC = 257, 
angle A = 52 9', and C = 37 51'. 

137. Problem LX VIII. Given the two sides about the 
right angle of a right-angled triangle, to construct it, 

c and to measure the other parts. 

Given the side AB = 162, and BC = 216. 
Make AB = 162; draw BC perpendicular to it, and 
=216 ; and draw AC ; ABC is the required triangle. 

By measurement, it is found that AC = 270, angle 
A = 53 8', and C = 3652'. 

138. Problem LXIX. Given the hypotenuse and one of 
the acute angles of a right-angled triangle, to construct 
it, and measure the other parts. 

c Let the hypotenuse AC = 324, and angle A = 48 17'. 

Draw any line AB ; then draw AC, making angle 
A = 48 17' ; make AC = 324 ; from C draw CB 
perpendicular to AB, and ABC is the required 
triangle. 

By measurement, AB = 215, BC = 242, angle C = 
41 43'. 

139. Problem LXX. Given a side and an acute angle 

of a right-angled triangle, to construct it, 
and measure the other parts. 

Given the base AB = 125, and the adjacent angle 
A = 51 19'. 

Make AB = 125; draw AC, making angle A = 
51 19' ; from B draw BC perpendicular to AB, and 
ABC is the required triangle. 

By measurement, AC = 200, BC = 156, and angle C = 38 41'. 
Note. When a side, as AB, and the opposite acute angle C 





DESCRIPTIVE GEOMETRY 



33 




are given, the triangle may be constructed in the same manner. 
For the three angles of every triangle are equal to two right angles, 
or 180 (Eucl. I. 32) ; and as B is one right angle, A and C 
together are equal to one right angle, or = 90 ; hence when C is 
given, A is found by subtracting C from 90. Thus, let there be 
given AB = 125, and angle C = 38 41' ; then angle A is found thus, 
A=90-C = 90 -3841' = 51 19'. 

140. Problem LXXL Given a side and two angles of a 
triangle, to construct it. 

Given angle A =49 25', B = 66 47', and AB=275. 

As the three angles of every triangle are equal to 
180, therefore 

Angle C = 180 - ( A + B) = 180 - (49 25' + 66 47') = 
180 -116 12' -63 48'. 

Make AB = 275, angle A = 49 25', and angle B = 66 47'. 

By measurement, AC =282, BC=233, and it was found above 
that angle C=6348'. 

Note. When any other two angles of a triangle are given, the 
third angle may be found in the same manner ; that is, by sub- 
tracting the sum of the two given angles from 180 ; and the 
triangle can then be constructed as shown above. 

141. Problem LXXIL Given two sides of a triangle, and 
an angle opposite to one of them, to construct the triangle. 

Given AB = 345, BC = 232, and angle A = 
37 20'. 

Make AB=345, angle A = 37 20', and from 
B as a centre, with a radius BC = 232, describe 
an arc cutting AC in C, and C' ; then either of 
the two triangles ABC, ABC' is the required 
triangle. (See Art. 187.) 

By measurement, it is found that in triangle ABC, AC = 174, 
angle ABC =27 4', angle ACB = 115 36'; also in triangle ABC', 
AC' = 375, angle ABC' = 78 16', angle AC'B = 64 24'. 

142. Problem LXXIIL Given two sides of a triangle, 
and the contained angle, to construct the 

triangle. 

Let AB = 176, BC = 133, and angle B = 73. 

Make side AB=176, and then angle B = 73, 
and side BC = 133. 

By measurement, angle C = 64 9', angle A = 42 49', and AC = 187. 





34 DESCRIPTIVE GEOMETRY 

143. Problem LXXIV.^G-iven the three sides of a triangle, 
to construct it, and measure its angles. 

Let AB = 345, AC = 232, and BC = 174. 
Make AB = 345; from A and B as centres, 
with the respective radii 232, 174, describe arcs 
cutting in C ; then draw AC, BC, and ABC is 
the required triangle. 

By measuring the three angles of the triangle, it is found that 
A = 27 2', B = 37 20', and C = 115 38'. 



COMPUTATION BY LOGARITHMS 

144. The logarithm of a number is the exponent of the 
power to which another given number must be raised in 
order to produce the former number. 

Thus 1000= 10 x 10 x 10 ; that is, 10 3 = 1000, and the exponent 3 is 
the logarithm of 1000. 

So 100 = 10x10; that is, 10 2 =100, and 2 is the L 100, where L 
denotes logarithm. 

Or lO 1 ^ 10, 10 2 =100, 103 = 1000, 10 4 = 10,000, and 1 = L 10, 

2=L 100, 3 = L 1000, 4 = L 10,000, 

145. Since the logarithms of 100 and 1000 are respectively 2 and 
3, the logarithm of some intermediate number, as of 856, will be 
between 2 and 3, or=2 + a fraction. So the logarithm of a number 
between 1000 and 10,000 is between 3 and 4, or = 3 + a fraction. 

146. The integral part of a logarithm is called the charac- 
teristic, and is one less than the number of integral figures 
in the number ; thus, if the number contains 5 integral 
figures, the characteristic of its logarithm is 4 ; if it contains 
4 integral figures, the characteristic is 3 ; if 7, the charac- 
teristic is 6 ; and so on. 

147. Since 10- 2 =1/10 2 = 1/100= '01, the log. of "01 is -2; so 10" 3 

= 001, 10- 4 = -0001, 10- 5 = -00001, and -3 = L'001, -4 = L-0001, 

-5 = L -00001, 

Since the logarithms of -01 and -001 are -2 and -3, the logarithm 
of an intermediate number, as -00754, will be between - 2 and - 3, 
or = -3 4- a fraction. So the logarithm of a number between -001 



COMPUTATION BY LOGARITHMS 35 

and "0001, as "000754, is between -3 and -4, or=~4 + a fraction; 
hence the characteristic of the logarithm of a decimal fraction is 
a negative number, a unit greater than the number of prefixed 
ciphers. 

148. The decimal parts of the logarithms of numbers 
that consist of the same figures are the same wherever 
the decimal place is marked, for they differ only in their 
characteristics. 

Thus the logarithms of 43625, 4362 '5, 436 '25, 43 '625, 4 '3625, 

43625, -043625, '0043625, are the same in the decimal part, 

but the characteristics are respectively 4, 3, 2, 1, 0, T, 2, 3 ; the 
negative sign being written over the characteristic. 

Note. For additional information in reference to the nature and 
construction of logarithms, see the Introduction to Chambers's 
Mathematical Tables and Chambers's Algebra for Schools, by 
W. Thomson. 



LOGARITHMIC SCALES 

149. These scales are constructed by making the distances 
of the divisions from one extremity equal to the logarithms 
of the numbers marked on the divisions ; and by means of 
them, several processes of arithmetical and trigonometrical 
calculation can be easily performed approximately, and the 
results may be used as a check against errors in the ordinary 
methods of calculation. A scale of this kind is usually 
called Gunter's scale. The logarithmic lines of numbers, 
sines, and tangents are laid down on sectors, and are marked 
respectively N, 8, and T. 

150. Problem I. To construct a line of logarithmic 
numbers. 

The line of logarithmic numbers is constructed by making the 
distances from the extremity of the scale marked 1 equal to the 
logarithms of the series of natural numbers 1, 2, 3, 4, &c. ; that is, 
to 0, -301, -477, '602, &c. ; and if a scale of equal parts be con- 
structed of the same length as the line of logarithmic numbers, and 



36 LOGARITHMIC SCALES 

divided into 1000 equal parts, then the division marked 1 on the 
logarithmic line Avill be distant from its extremity ; that is, 1 will 
be at its extremity ; the division 2 will be at the distance 301 from 
the extremity, or from 1 ; 3 will be at the distance 477 ; 4 at the 
distance 602 ; and so on for the other divisions of this line, that 
marked 10 being at the distance 1000. 

Let a : b=c : d, and consequently j- = -j, 

o ct 

Then La - L6 = Lc - Lc? ; hence 

151. The difference between the logarithms of the first 
and second terms of a proportion is equal to the difference 
between the logarithms of the third and fourth. 

Since 1 :2 = 10:20, and 2 : 3 = 20 :30; .'. L 1~L2=L 10-L20, 
and L 2~L 3 = L 20- L 30. 

Hence the line may be easily extended beyond the division 10, 
for the extent from 10 to 20 is equal to that from 1 to 2 ; the 
extent from 20 to 30 is equal to that from 2 to 3 ; and so on. 

The divisions reckoned above, as 1, 2, 3, 4,... may also be con- 
sidered as 10, 20, 30, 40,... or as 100, 200, 300, 400;... or, in fact, 
any numbers proportional to these. 

152. Problem II. To perform proportion by the line of 
numbers. 

RULE. Extend the points of the compasses from the first to the 
second term, and this extent will reach in the same direction from 
the third term to the fourth. 

EXAMPLE. Find a fourth proportional to 124, 144, and 186. 

The distance from 124 to 144 on the line of numbers will extend 
from 186 to 216, the term required. 

153. Problem III. To construct the line of logarithmic 
sines, cosines, secants, and cosecants. 

On Gunter's scale the logarithm of 100, and the sine of 90, which 
is equal radius, are the same length, and therefore the sines are laid 
down on the scale to radius 100, of Avhich the logarithm is 2 ; but 
in the tables of logarithmic sines, the logarithm of radius is 10 ; 
hence, if 8 be subtracted from the logarithmic sines, the remainders 
will be the length of the logarithmic sines on the scale, taken from 
a scale of the same length divided into 200 equal parts. 

Or, the natural sines may be multiplied by 100, and the logarithms 
of the products taken from the scale will give the same results. 



LOGARITHMIC SCALES 37 

This scale may also be used as a scale of logarithmic cosines, for 
the cosine of an angle is the sine of its complement. It may also 
be used as a scale of logarithmic cosecants and secants, for sin A x 
cosec A = R 2 , therefore since on this scale 11 = 100, log. sin A + log. 
cosec A =4, hence log. cosec A =4 -log. sin A = the whole length 
of the scale + the excess of the whole scale above the logarithmic 
sine A. In the same manner, it can be shown that the logarithmic 
secant A=the whole scale + its excess above the logarithmic cos A. 
Generally, for any radius, since sin A x cosec A = R 2 , and cos 
Ax sec A = R 2 ; log. cosec A = 2 log. R-log. sin A, and log. sec 
A =2 log. R-log. cos A. Also tan Ax cot A = R 2 ; hence log. 
cot A = 2 log. R - log. tan A. 

154. Problem IV. Given two numbers and an angle, to 
find another angle such that the two numbers and the 
sines of the angle shall be proportional. 

The distance between the numbers on the line of numbers will 
extend in the same direction on the line of sines from the sine of 
the given angle to the sine of the required angle. 

T , , . , a sine c , T T , T . 

If a : b = sine c : sine a, - i = - / and La - Lo = L sine c - L sine d. 

b sine d 

EXAMPLES. 1. Given the numbers 121 and 100, and an angle of 
90, to find another angle a such that 

121 : 100= sine 90 : sine a. 

The extent from 121 to 100 on the line of numbers reaches from 
90 on the line of sines to 55 44'. 

2. Find an angle such that 121 is to 68 '5 as the sine of 90 to the 
cosine of the required angle. 

Let a be the required angle, then its complement 6 = 90-, and 
cos a = sine (90 - a) or sine b. Hence 

121 : 68 -5 = sine 90 : sine b, 

b is found, as in the preceding example, to be = 34 29' ; hence 
a=90-6 = 90-34 29' = 55 31'. 

3. Find an angle a such that 

135 : lll = sine 79 23' : sine a. 

The distance from 135 to 111 extends from 79 23' to 53" 55', which 
is therefore the value of a. 

155. Problem V. To construct a scale of logarithmic 
tangents and cotangents. 

The arcs (45 -A) and (45 + A) are evidently complements of 

Pra?. J) 



38 LOGARITHMIC SCALES 

each other, for their sum is 90; and (Art. 153) tan (45-A)x 
tan (45 + A) = R 2 , and therefore log. tan (45 + A) =2 log. R- 
log. tan (45 -A); hence, since tan 45 = R, and on the scale 
log. R = 2, and in the logarithmic tables log. R = 10 ; if 8 be sub- 
tracted from the tabular log. tangents, the remainders will be the 
lengths of the log. tangents on the scale. These being laid down 
on the scale, from a scale of the same length divided into 200 
equal parts, will give the scale of log. tangents up to 45. It is 
also evident from the above, that the log. tangents of angles 
greater than 45 may be found from the same scale by taking 
the tangent of 45 + the distance from the tangent of 45 to the 
tangent of the complement of the angle. 

This will also be a scale of logarithmic cotangents, for the 
cotangent of an angle is the tangent of its complement. 

156. Problem VI. Given two numbers and an angle, to 
find another angle such that the lines shall be propor- 
tional to the tangents of the angles. 

The distance between the numbers on the Line of Numbers will 
extend in the proper direction from the given number of degrees on 
the line of tangents to the required number of degrees. 

Note. When the distance extends beyond 45, take the excess 
in the compasses, and apply it back upon the scale from 45, and it 
will reach to the complement of the angle sought. 

EXAMPLES. 1. Given two numbers 420 and 650, and an angle 
45, to find another angle, such that the numbers shall be propor- 
tional to the tangents of the angles. 

420 : 650 = tan 45 : tan a, where a = the required angle. 

The extent from 420 to 650 on the line of numbers reaches from 
45 to 32 52' on the line of tangents ; but as the second term 
exceeds the first, the fourth will exceed the third ; therefore 32 52' 
is the complement of the angle sought; hence it is 90 -32 52' = 
57 8'. 

2. Given two numbers 142 and 42, and an angle of 71 34' to find 
another angle a, such that 

142 : 42= tan 71 34' : tan a. 

In this example, either 71 34' or its complement is taken 
namely, 18 26', and the distance from 142 to 42 on the line of 
numbers extends from 18 26' beyond 45 on the line of tangents, 
and the distance beyond it being taken with the compasses, will 
extend from 45 back to 41 35', the angle required. 



LOGARITHMIC SCALES 39 

THE LINES OF THE SECTOR 

157. Besides the logarithmic lines already explained, 
there are also on each of the legs of the sector a line of 
equal parts, as well as one of chords, of sines, of tangents, 
and of secants ; there is also a line of polygons. 

These lines proceed from the centre of the sector, which is 
the centre of the joint about which its legs are movable. 
The line of lines is a line of equal parts, the number of 
large divisions being 10, beginning at the centre, and marked 
10 at the extremity ; the line of chords is a scale of chords, 
as far as 60 ; the line of sines is a scale of sines, as far as 
90; and the line of tangents is a scale of tangents, as far 
as 45. The lengths of these four lines of lines, chords, 
sines, and tangents are the same; the chord of 60, the 
sine of 90, and the tangent of 45 being all equal to 10 
on the line of lines, which is the radius of the sector. 
Another scale for tangents above 45 begins at one-fourth 
of the radius 10 ; that is, at 2 '5, and is extended beyond 75. 
The line of secants also begins at the distance 2*5, which 
is the radius of the circle to which these secants and the 
tangents above 45 belong; the beginning of the line of 
secants being marked 0, for the secant of is = the radius 
= 2-5. The line of polygons is of the same length as that 
of lines or chords, being marked 4 at the extremity, and 5, 
6, 7, &c., towards the centre. 

The general principle on which the use of the sector is founded 
is this : 

Let ACB, DCE be two similar isosceles triangles, so that CA 
= CB, and CD = CE ; then CA : AB = CD : DE. 

Now, CA and CB being two lines of lines, 
of chords, of sines, or tangents, it is evident 
from the above proportion, the distance CA 
being the radius of the sector, and AB the 
radius of any other circle, the extremity of the sector being 
opened to this distance, that 




40 THE LINES OF THE SECTOR 

158. The radius of the sector is to the radius of the circle, as the 
length of any line (CD) belonging to the circle whose radius is that 
of the sector, to the length of a corresponding line (DE) of the other 
circle, whether that line is a chord, a sine, a tangent, or the side of 
an inscribed regular polygon. 

159. The two lines of lines, of chords, sines, and tangents, have 
the same inclination, so that when the sector is opened till the 
distance between 10 and 10 on the lines of lines is any given 
distance, the distances between 60 and 60 on the lines of chords, 
90 and 90 on the lines of sines, and 45 and 45 on the lines of 
tangents will all be the same. 

The distance from the centre of the sector on any of the lines 
proceeding from its centre is called the lateral distance. 

The distance from any point in one of the lines of the sector, to 
the corresponding point in the similar line on the other leg, is called 
the parallel distance. 

Any two lateral distances are evidently proportional to their 
corresponding parallel distances, as appears from the preceding 
proportion. 

THE LINE OF LINES 

The line of lines is one of equal parts. The two following are 
the most useful problems to be performed by this line. 

160. Problem I. To find a fourth proportional to three 
given numbers or lines. 

RULE. Make the parallel distance of the first term = the lateral 
distance of the second, then the parallel distance of the third term 
will be = the fourth term. 

EXAMPLE. Find a fourth proportional to 72, 48, and 60. 

Considering the large divisions as each = 10, take the lateral 
distance of 48, and make the parallel distance of 72 equal to it ; 
then the parallel distance of 60 applied to the line of lines will 
give 40, the fourth term required. 

If the lengths of the lines represented by the given numbers are 
too large, any parts of them may be taken, and then the fourth 
term will be the same part of the number sought. 

161. Problem II. To divide a line into any number of 
equal parts. 

RULE. Find some number on the line of lines which is a multiple 
of the number of parts into which the line is to be divided, and 
make the parallel distance of this number = the given line; then 



THE LINES OF THE SECTOR 41 

the parallel distance of the corresponding aliquot part of the latter 
number will be the required aliquot part of the given line. 

EXAMPLE. Let it be required to find the 8th part of the line AB. 

Since 80 is divisible by 8, make the parallel 

distance of 80 = the line AB ; then as 10 is the A c 

8th part of 80, the parallel distance of 10 will be = AC, the 8th part 

of the given line. 

Instead of 80, 48 may be taken, and its parallel distance being 
made = AB, the parallel distance of 6, the 8th part of 48, will 
be -AC, 

THE LINE OF CHORDS 

The chord of an arc is double the sine of half that arc ; 
hence the chord of 30 is twice the sine of 15. If, therefore, the 
natural sine of 15, to a radius = 10, which in a table of natural 
sines is 2'5882, be doubled, the product 5'17 will be the length of 
the chord of 30 to the same radius. Hence, take 5 '17 from the 
line of lines, and lay it off on the line of chords, and this will 
determine the division for 30. The divisions for the other degrees 
are found in the same manner. 

162. Problem III. To cut off an arc of any number of 
degrees from the circumference of a given circle. 

RULE. Open the sector till the parallel distance of 60 on the 
line of chords is = the radius of the given circle ; then the parallel 
distance of the required number of degrees will 
be the chord of the required arc, which can there- 
fore be cut off. 

EXAMPLE. Cut off from the circumference of 
the circle ABD an arc = 48. 

Make the parallel distance of 60 on the scale 
of chords = the radius AC, then the parallel distance of 48 D will 
be AB, the chord of the required arc. 

THE LINE OP SINES 

163. The line of sines is constructed by taking the numerical 
values of the sines from a table of natural sines, supposing 
the radius =10, and then taking in the compasses the lateral 
distances from the line of lines corresponding to these values, 
and laying them off on the line of sines. 

Thus, the natural sine of 40 to a radius = 10 is, by the table, = 
6 '43; hence, if 6 "43 be taken from the line of lines, and laid off on 




42 THE LINES OP THE SECTOR 

that of sines, it will determine the division for 40. In the same 
manner the other divisions are found. 

The sine of any arc of a circle, whose radius is given, is found 
exactly in the same manner as the chord of an arc was found in the 
preceding problem ; observing that the distance between 90 and 
90 is to be made = the radius. The same remark applies to the 
line of tangents, the distance between 45 and 45 being made = the 
radius of the given circle. 

THE LINE OP TANGENTS 

164. The line of tangents is constructed in the same 
manner as that of sines, taking the tangents of the degrees 
from a table of natural tangents ; or since tangent = R 
sin /cos , the tangents can be found by dividing the sine by 
the cosine, and multiplying the quotient by the given radius. 

Thus, tangent 30 to a radius R of 10 is = 5'77, and this distance 
taken on the line of lines, and laid on that of tangents, gives the 
division of 30. 

For tangents above 45, the radius is taken one-fourth 
of 10 or=2'5, and the numbers for the natural tangents 
to R = 10 are divided by 4. Thus, tangent 60 = 17 '32, the 
4th of which is 4'33 ; and this distance taken from the line 
of lines, and laid off on that of tangents above 45, gives the 
division for 60. 

The tangent of any arc of any circle above 45 is found from 
the lines of tangents in the same manner as those below 45 ; 
observing that it is the distance between 45 on the two lines that 
is made = the radius of the given circle. 

THE LINE OP SECANTS 

The line of secants is constructed exactly as that of tangents 
above 45 ; only in this case secan 1 = radius = one-fourth of 10 = 2'5. 
So for the division of 60 on this line, the secant of 60 by the table 
is=20 to radius 10, and one-fourth of this is 5; and the distance 
5 taken from the line of lines, and applied to that of secants, gives 
the division of 60 ; and in a similar manner the other divisions are 
found. 

This line is used for finding the secant of any arc of -a given 
circle, in the same manner as those for chords, sines, and tangents ; 
observing that the distance from to on the lines of secants 
is to be made = the radius of the given circle. 



PLANE TRIGONOMETRY 



43 



PLANE TRIGONOMETRY 

DEFINITIONS 

165. The object of Plane Trigonometry was originally 
the calculation of the sides and angles of plane triangles. It 
now investigates the general relations that subsist between 
any angles and their trigonometrical functions. 

166. In trigonometry, for the purposes of calculation, the 
circumference of a circle, and also all the angles round a 
point, are each divided into 

360 equal parts called de- 
grees ; each degree is sub- 
divided into 60 equal parts 
called minutes; and each 
minute into 60 equal parts 
called seconds : degrees, 
minutes, and seconds are 
thus indicated 5 17' 28"; 
which is read five degrees, 
seventeen minutes, and 
twenty-eight seconds. 

167. The complement of 
an angle is its difference 
from a right angle. 

168. The supplement of 

an angle is its difference from two right angles. 

169. Let XX' and YY', which are perpendicular to each other, 
be two diameters of the circle whose centre is ; the quadrants 
XOY, YOX', X'OY', Y'OX are called respectively the 1st, 2nd, 
3rd, 4th quadrants. 

170. If OP be supposed to start at OX and to revolve counter- 
clockwise round O, it will generate with OX angles of all sizes, 
the angles increasing the more OP revolves. Suppose OP to 
have generated the four angles XOP, one in each quadrant. 
From P draw PM perpendicular to XX' ; then the lines MP, 
OM, OP are called respectively the ordinate, the abscissa, the 
radius vector of the point P. 




44 PLANE TRIGONOMETRY 

171. The trigonometrical functions, or trigonometrical ratios, of 
the angle XOP are called sine, cosine, tangent, cotangent, 
secant, cosecant (abbreviated into sin, cos, tan, cot, sec, cosec), 
and are defined as follows : 

17-^Tn ordinate MP y abscissa OM x 

sin XOP = r =7S5 =2 cos XOP = r = 7^5 = - 

radms OP r radius OP r 

^/-vr. ordinate MP ?/ abscissa OM x 

tanXOP=-j : = T^ = - cot XOP = TT T- = ^TB=- 

abscissa OM x- ordinate MP y 

^/^T. radius OP r radius OP r 

secXOP = -j : - = T^T T = - cosec XOP = T . = ^r^=- 

abscissa OM x ordinate MP y 

172. From these definitions the following formuhe are derived. 
For shortness, let L XOP be denoted by A. 



Sill A 7- 

cosec A 


cusec A 


sin A 

1 


siu A cosec j 
cos A sec A 


L 1 


sec A 




cos A 

1 


1 



173. Other useful formulae are also obtained immediately from 
the definitions. 

, sin A 11 x y sec A r 'r y 

Thus r = - : -=^ = tan A, r =- : -=^ = tan A ; 

cos A r r x cosec A x y x 

, cos A cosec A 

and consequently 7- = cot A, r =cot A. 

sin A sec A 

174. Again, from any one of the four right-angled triangles OMP 
we obtain by Pythagoras's theorem (Eucl. I. 47), 

7/ 2 + a3 2 =r 2 . 

Divide both sides of this equality successively by j- 2 , a- 2 , y\ 
There are obtained : 

(1)^+^ = 1; that is, sin 2 A + cos 2 A = 1. 

(2) ^+ !=-; that is, tan 2 A + 1 = sec 2 A. 

*C iC 

(3) l+~ 9 = 9 ; that is, 1 + cot 2 A = cosec 2 A. 

2/ 2 Z/ 2 

175. It will be observed that sin 2 A, cos 2 A, &c., are written for 
(sin A) 2 , (cos A) 2 , &c. 

176. The relations just found may be put in other forms. For 
example, 



PLANE TRIGONOMETRY 



45 



os A 




corA 



SEC 



SECA 



sin A = Vl - cos 2 A, cos A = Vl - sin 2 A, tan A = Vsec 2 A - 1, 
cot A = Vcosec 2 A - 1 , sec A = Vl + tan 2 A, cosec A Vl + cot 2 A. 

177. All the formulae which 
occur in Articles 172, 173, 
174, and many other forms 
of them, may be obtained 
from the inspection of a 
figure which is not difficult 

to construct, and which may TANA 
be called Alison's Diagram. 

Take any three consecutive 
values, either diametrically 
or circumferentially ; then 
the middle one is equal to 
the product of the other 
two. Thus, of the three values diametrically, sin A, 1, cosec A, 

sin A cosec A = l. 

Of the three values circumferentially, sin A, tan A, sec A, 
sin A sec A = tan A. 

In each triangle the square of the value written at the vertex 
turned downwards is equal to the sum of the squares of the values 
written at the other two vertices. Thus, ! 2 =sin 2 A + cos 2 A. 

178. The following convention regarding the signs to be attri- 
buted to the four sets of three lines MP, OM, OP is observed by 
all mathematicians. 

Of the four MP lines, those situated above XX' are considered 
positive, those situated below XX' negative ; of the four OM lines, 
those situated to the right of YY' are considered positive, those 
to the left of YY' negative ; OP in any of its positions is always 
considered positive. 

179. To find the signs of the trigonometrical functions in the 
four quadrants. 

Take first sin XOP or sin A and find the sequence of signs for 
it, according as the angle is in the 1st, 2nd, 3rd, 4th quadrant. 
MP . 



Sin A = 



all cases. 



x MP positive positive negative negative 
wow, 7^=- r-: , - T-. , ^-r-. ^-rr: according as i. A 
OP positive positive positive positive 

is in quadrant 1234 
Hence the sequence of signs for sin A is, + + - - . 



46 PLANE TRIGONOMETRY 

OM. 
Again, cos A = -?yp in all cases. 

XT OM positive negative negative positive 
Now, 7^5 =* T-. , ^-r-. , - > - ^r- according as L A 
OP positive positive positive positive 

is in quadrant 1234 
Hence the sequence of signs for cos A is, + - - + . 

Since tan A = sin A/cos A, the sequence of signs for tan A may be 
obtained from the sequences of signs for sin A and cos A. 

It is ; that is, + - + - . 

The sequences of signs for cosec A, sec A, cot A are the same 
as those for sin A, cos A, tan A. 

180. To trace the variation in magnitude of the various 
functions. 

Sin A=MP/OP in all cases, and as OP does not vary in length, 
the variation of sin A will depend on MP. 

Now, when L A is very small, MP is very small ; therefore 
sin A is very small. Also, we can make sin A as small as 
we please (that is, less than any assignable magnitude) by 
diminishing / A. This statement is usually expressed by saying 
sin 0=0. 

If / A from being very small increases up to 90, MP increases 
till it equals OP ; hence sin A increases till sin 90 = 1. 

If L A increases beyond 90, MP begins to diminish, till when 
L. A is 180 MP has vanished. Hence sin 180 = 0. 

In the 3rd quadrant, as /A increases MP increases in 
magnitude, till when zA is 270 MP equals OP; hence 
sin 270= -1. 

In the 4th quadrant, as z A increases MP diminishes in 
magnitude, till when /A is 360 MP disappears ; hence 
sin 360 =0. 

A similar discussion may be made of the variations of cos A 
which are dependent on the variations of OM. 

Because tan A=MP/OM, the variations of tan A will not 
be so easy to trace, since they depend on the simultaneous 
variations of MP and OM. If z A is very small, MP is very 
small, and OM is nearly equal to OP ; therefore tan A is 
very small. Also, we can make tan A as small as we please 
by diminishing L A. This statement is usually expressed by 
saying tan = 0. 

If z A from being very small increases up to 90, MP increases 
till it equals OP, and OM shrinks down to 0. Hence, as the 



PLANE TRIGONOMETRY 



numerator of tan A increases and the denominator diminishes, 
tan A itself increases rapidly. Also, we can make tan A as great 
as we please (that is, greater than any assignable magnitude) by 
making L A as near as we please to 90. This statement is usually 
expressed by saying tan 90 = 00 (infinity). 

As z A increases from 90 to 180, MP diminishes and OM 
increases ; that is to say, tan A diminishes in magnitude. When 
i. A is 180 MP has vanished, and OM has become equal to OP ; 
hence tan 180 = 0. 

In the 3rd quadrant, as / A increases MP increases, and OM 
diminishes ; hence tan A increases and tan 270 = oo . 

In the 4th quadrant, as L A increases MP diminishes, and OM 
increases ; hence tan A diminishes and tan 360=0. 

The variations in magnitude of the functions cosec A, sec A, 
cob A may be investigated directly from the figure, or indirectly 
from the variations of their reciprocals sin A, cos A, tan A. 

181. The two following Tables give the variations of all the 
functions, first in sign and second in magnitude. 

SIGN 



A beiiig) 
between) 


and 90 


90 and 180 


180 and 270 


270 and 360 


Sin A, cosec A 


are 

+ 


are 
+ 


are 


are 


Cos A, sec A 


+ 


- 


- 


+ 


Tan A, cot A 


+ 





+ 






MAGNITUDE 



A being \ 
between ) 


and 90* 


90 and 180 


180 and 270 


270 and 360 




Varies from 


Varies from 


Varies from 


Varies from 


Sin A 


to 1 


1 to 


to-1 


-1 to 


Cos A 


1 ii 


-1 


-1 


1 


Tan A 


.1 oo 


oo M 


ii oo 


00 U 


Cot A 


oo H 


M oo 


oo ,i 


ii oo 


Sec A 


1 II 00 


00 II - 1 


- 1 it oo 


oo 1 


Cosec A 


00 ii 1 


1 ,, 00 


00 M -1 


-1 oo 



48 



PLANE TRIGONOMETRY 



182. To find the relations between the functions of an angle 
and the functions of its comple- 
ment. 

Let / XOP he denoted by A ; then 
if L YOQ be equal in magnitude 
to L XOP, the complement of A 
(namely, 90 - A) will be / XOQ. 

'Draw the ordinates PM, QN. 

Then the triangles ONQ, OMP 
are equal in all respects (Eucl. I. 
26); 
therefore NQ = OM, ON = MP. 




Hence 



. , ftno .. NQ OM . 

sin (90-A) = ^| = = cosA, 

ON MP 
cos (90 - A) = = = sin A, 



as far as magnitude is concerned. 

But since (in the figure) A and 90 -A are both in the 1st 
quadrant, therefore their sines and cosines are all positive, and 
consequently 

sin (90 - A) = cos A, cos (90 - A) = sin A. 

The values of the other functions of 90 -A may be deduced 
from those just found. For example, 

/nn A \ s i n (90 - A) COS A 

tan 90 -A )= - ^5 j-=- -r = cotA; 
cos (90 - A) sin A 

and so on. 

183. To find the relations between the functions of an 
angle and the functions of its 
supplement. 

Let zXOP be denoted by A; 
then if L X'OQ be equal in mag- 
v* nitude to L XOP, the supplement 
, v of A (namely, 180 - A) will be 
/XOQ. 

Draw the ordinates PM, QN. 
Then the triangles ONQ, OMP 
are equal in all respects (Eucl. I. 
26); 
therefore NQ = MP, ON = OM. 

H ' Q80-A)- - - ' A 




PLANE TRIGONOMETRY 49 

cos (180 - A) = ^ =^p = cos A, 

as far as magnitude is concerned. 

But since (in the figure) A and 180- A are respectively in 
the 1st and 2nd quadrants, sin A and sin (180 - A) will both be 
positive, while cos A will be positive and cos (180- A) negative. 

Consequently, sin (180-A)=sin A; cos (180 - A) = -cos A. 

The values of the other functions of 180 -A may be deduced 
from those just found. For example, 

sec(180-A) = - rs r-. = ' r-=-secA; 

cos (180 -A) -cos A 

and so on. 

184. When A is associated, either by addition or subtraction, 
with an odd number of right angles (90 -A, 90 + A, A -90, 
270- A, 270 + A, A -270, &c. ), the sin, tan, sec of such angle is 
equal to the cos, cot, cosec of A, and the cos, cot, cosec of such 
angle is equal to the sin, tan, sec of A. 

Thus, as far as magnitude is concerned, 

sin (90 - A) = cos A, sin (90 + A) = cos A, sin (A - 90) = cos A ; 
and so on. 

The proper signs to be affixed may be determined from the table 
of sequence for sines. Thus, if A is an angle in the 1st quadrant, 
say 20, 90 -A is in the 1st quadrant, 90 + A in the 2nd, A -90 
in the 4th ; consequently, the signs of the sines of these angles 
are + + - , and 
sin (90- A) = cos A, sin (90 + A) = cos A, sin (A -90)= -cos A. 

185. When A is associated with an even number of right angles 
(180- A, 180 + A, 360- A, 360 + A, &c.), the sin, tan, sec of such 
angle is equal to the sin, tan, sec of A ; and similarly for the cos, 
cot, cosec of such angle. 

Thus, as far as magnitude is concerned, 

cos (180- A)=cos A, cos (180 + A) = cos A, cos (360- A) = cos A. 
The proper signs to be affixed may be determined from the table 
of sequence for cosines. 

Thus, if A is an angle in the 1st quadrant, say 20, 180- A is 
in the 2nd quadrant, 180 + A in the 3rd, 360 -A in the 4th; 
consequentl y, the signs of the cosines of these angles are - - + , 
and 

cos (180- A)= -cos A, cos (180 + A)= -cos A, 
cos (360 - A) = cos A, 



50 



PLANE TRIGONOMETRY 



186. To find the sine of the sum of two angles, having 
given the sine and cosine of each of the angles. 

Let / BAC be denoted by A, and L CAD by B; 
then / BAD will be denoted by (A + B). 

From any point D in AD draw DE perpendicular 
to AB, and DC perpendicular to AC ; also through 
C draw CB and CF perpendicular to AB and DE ; 
then FB is a rectangle ; hence FE = CB, and FC = EB. 
And since the triangles AHE, DHC, are right-angled at E and 
C, and have the angles at H vertically opposite, the third angle 
EAH = CDH or CDF ; hence angle CDF = A. 
. T . /A , . DE CB + DF CB , DF 
Now, Bin (& + *) = =-- = + 




_CB AC DF DC 
"AC ' AD + DC 'AD 
= sin A cos B + cos A sin B. 

187. To find the cosine of the sum of two angles, having 
given the sine and cosine of each of the angles. 
From the above diagram, 

AE AB-FC AB FC 



_AB AC FC DC 
~AC 'AD DC ' AD 
= cos A cos B - sin A sin B. 

188. To find the sine of the difference of two angles, 
having given the sine and cosine of each of 
the angles. 

Let L BAC = A, and / CAD = B, 
then /BAD = (A-B). 

From D, any point in AD, draw DB and DC 
perpendicular to AB and AC, and from D draw 
DF perpendicular to CE. 
BF is a rectangle, therefore DB = FE, and FD = EB; and since 
the angles EAC and ACE are together equal to the right angle 
ACD, from each take the angle ACE, and there remains the angle 



DB CE 




CF CE CF 

~AD~ AD ~AD AD 
_CE AC_CF DC 
~AC ' AD~DC ' AD 
=sin A cos B - cos A sin B. 



PLANE TRIGONOMETRY 51 

189. To find the cosine of the difference of two angles, 
having given the sine and cosine of each of the angles. 

From the diagram to Art. 188, 

AB AE + FD AE FD 
C08(A - B) = AD = -^D- = AD + AD 
_AE AC FD CD 
~AC ' AD + CD' AD 
= cos A cos B + sin A sin B. 

190. Adding and subtracting the values in Arts. 186 and 188, 
and also those in Arts. 189 and 187, gives 

Sin (A + B) + sin(A-B) = 2 sin A.cos B []. 

Sin (A + B)-sin(A-B)=2cos A. sin B [&]. 

Cos(A-B) + cos(A + B) = 2cos A.cosB [c]. 

Cos(A-B)-cos(A + B)=2sin A.sinB [d]. 

191. If A + B = S, and A-B=D, then A = 4(S + D), 

and B = J(S-D); and substituting these values in the last four 
expressions, they become 

SinS +sinD=2 sin (S + D) cos (S-D) [a]. 

SinS -sinD=2cosi(S + D)sin i(S-D) [6]. 

CosD + cosS=2cos(S + D)cos(S-D) [c]. 

CosD-cos S=2sin^(S + D) sin |(S-D) [rf]. 

192. These four expressions prove thefourfollowing propositions : 
(a) The sum of two sines is equal to the product of twice the 

sine of half the sum of the angles, into the cosine of half 
their difference. 

(6) The difference of two sines is equal to the product of twice 
the cosine of half the sum of the angles, into the sine of 
half their difference. 

(c) The sum of two cosines is equal to the product of twice the 

cosine of half the sum of the angles, into the cosine of 
half their difference. 

(d) The difference of two cosines is equal to the product of twice 

the sine of half the sum of the angles, into the sine of 
half their difference. 

193. But S and D are any two arcs, and may therefore be repre- 
sented by any two letters ; hence, putting A for S, and B for D, 

Sin A + sin B=2sin |(A + B) cos(A-B) [a]. 

Sin A-sin B=2cos4(A + B) sin i(A-B) ..". [&]. 

CosB + cos A=2cos|(A + B)cosJ(A-B) [c]. 

CosB-cos A=2sini(A + B) sini(A-B) [efj. 



52 PLANE TRIGONOMETRY 

194. In Arts. 186-189 let B = A, then Arts. 186, 187, 188, and 
189 become 

Sin (A + A) = sin A cos A + cos A sin A; 

.*. sin 2 A 2 sin A cos A ... ... ... [a], 

Cos ( A + A) = co 2 A - sin 2 A ; 

. . cos 2 A = cos 2 A - sin 2 A ... ... ... [6]. 

Sin ( A - A) = sin A cos A - cos A sin A ; 

.-. sinO = 0. ......... [cj, 

Cos (A- A) = cos 2 A + sin 2 A = l (Art. 174, (1)) ; 

.-. cosO = l ......... Of). 

Also the expressions (Arts. 190, c, d) become 
CosO + cos2A = 2cos 2 A; .-. 1 + cos 2A=2 cos 2 A ...... \_e]. 

CosO-cos2A = 2sin 2 A; .'. 1 -cos 2 A =2 sin 2 A ...... [/]. 

From b, c, and /of this Art. transposed, we obtain 
Cos 2 A = cos 2 A - sin 2 A = 2 cos 2 A -1 = 1-2 sin 2 A ...... [#]. 

195. To find the sine and cosine of 3A ; in Arts. 186 and 
187, for B put 2A, and reduce by Art. 194 ( and g). 

Sin 3A = sin (A + 2A) = sin A cos 2A + cos A sin 2 A 

= sin A(cos 2 A - sin 2 A) + cos A x 2 sin A cos A 
= sin A cos 2 A - sin 3 A + 2 cos 2 A sin A 
= 3 sin A cos 2 A - sin 3 A = 3 sin A(l - sin 2 A) - sin 3 A ; 
. '. sin 3A = 3 sin A -4 sin 3 A ......... ... ... [a]. 

Cos 3A = cos (A + 2A) = cos A cos 2A-sin A sin 2A 
= cos A(cos 2 A - sin 2 A) - 2 sin 2 A cos A 
= cos 3 A - 3 cos A sin 2 A = cos 3 A - 3 cos A( 1 - cos 2 A) ; 
.'. cos 3A=4 cos 3 A-3 cos A ............... [b]. 

196. To find the tangent of the sum and difference of 
two angles. 

-, . . _,. sin (A + B) sin A cos B + cos A sin B , 
Tan A + B ) = - ^r ~ - - -, by 186 and 

cos (A + B) cos A cos B-sm A sin B J 

187. 
Dividing both numerator and denominator of the last value by 

cos A cos B, and remembering that = tan, gives 

COS 

. tan A + tan B 
tan 



7 - ~ ...... . 

1 - tan A tan B 

sin (A - B) sin A cos B - cos A sin B 

Tan (A - B) = -- ;-; - ={ = -- r - ^ - : - r : 5 ; and 
cos (A - B) cos A cos B + sin A sin B 

dividing as in the last, it becomes 

tan A- tan B 
feW(A-B)=- ^ ...... [61. 

1 + tan A tan B 



PLANE TRIGONOMETRY 53 

If in (a) B be made equal to A, we obtain 

. 2 tan A 

Tan2A=- - 5-7- ............ [c]. 

1 - tan 2 A 

197. From Art. 193 (a, b, c, d) tbe following six expressions may 
be easily derived : 

Sin A + sin B = 2 sin j(A + B) cos %(A-B) = tan j(A + B) 

Sin A-sin B~2 cos (A + B) sin (A-B)~tan (A-B) '" ' 



Sin A + sin B_2 sin j(A + B) cos j(A-B)_, 1/A . W 
Cos A + cosB~2 cos i(A + B) cos 4(A-B) 



Sin A + sin B _ 2 sin j( A + B) cos j( A - B) f i/4_-o 
Cos B- cos A 2sini(A + B)sin(A-B)~ ( 



Sin A-sin B^ 2 cos j(A + B) sin j(A-B) _,,*_,>. r 71 

CosA + cosB 2cosi(A + B)cos(A-B) 



Sin A-sin B_2 cos j(A + B) sin j(A-B) 
Cos B- cos A~2 sin 4(A + B) sin i(A-B)~' 



Cos B + cos A_2cos|(A + B)cos ^(A-B)_co 
~ ~ 



CosB-cosA~2sin^(A + B)sin i(A-B)~tan 4(A-B) 
^cot^(A-B) 



If in the last three expressions B = 0, they become 

sin -A. 

-r = tan % A, for sin 0=0, and cos 0=1 ......... [a]. 

1 + cos A 

sin A 

-r = cotiA ..................... [M 

1 - cos A 

1 + cos A cot \ A 

-. - T=- f-r- = cot 2 A ...... []. 

1 - cos A tan \ A 

Also by inverting, we have 

1-cosA tan ^A r7l 

- - r = -4-r= tan 4A ...... [*l 

1 + cos A cot \ A 

198. Again, in the expressions of Art. 193 (a, b, c), let A = 90, 
and we shall have 

1 + sin B = 2 sin (45 + B) cos (45 - B) 

= 2sin 2 (45 + JB) ...... [a]. 

1 - sin B = 2 cos (45 + B) sin (45 - B) 



cos B = 2 cos (45 + JB) cos (45 - |B) 

= 2cos 2 B-l ...... [c]. 



54 PLANE TRIGONOMETRY 



199. If in Art. 194 (a) we put (A + B) for A, we shall have 
Sin (A + B)=2 sin J(A + B) cos (A + B); and dividing this by 
each of the expressions in Art. 193 (a, b t c, d) successively, 
there results 

Sin(A + B) = cos ^( A + B) 
Sin A + sin B~cos |(A-B) 



Sin (A + B) s , 



Sin A-sin B~sm (A-B) 

Sin (A + B) _ sin \( A + B) r , 

Cos B + cos A ~ cos i( A - B) 

Sin (A + B) _cosJ(A + B) , 



Cos B- cos A sinJ(A-B) 

RELATION BETWEEN THE SIDES AND ANGLES 

OP TRIANGLES 

200. To investigate the relation that subsists between the sides 
and the trigonometrical functions of the angles of 
a plane triangle, 

Let ABC be any plane triangle, having the three 
angles A, B, and C ; calling the sides opposite to 
these angles respectively a, 6, and c. Draw BD =p 
A u * u perpendicular to AC. 

From the right-angled triangles ABD and CBD, we have 
sin A pic, and sin C=pla ; and dividing the former by the latter, 

Sin A p a a 

a-. ^ = - x - = Similarly, 

Sin C c p c 

Sin A a , 
= r, and 



Sin B 6' 

SinB b .. . 

= - ; that is, 



iSin C c 
the sides are proportional to the sines of the opposite anglee. 

a sin A 

201. Again, since T= 
b sin B 

+ &_sin A + sin B tan ^(A + B) 
-6~sin A-sin B~tan |(A-B) 

From the above diagram, we have also 

AD = c cos A, and CD= cos C ; 
therefore AD + CD = b - c cos A + a cos C. 



PLANE TRIGONOMETRY 55 

202. In the same manner, by drawing perpendiculars on each of 
the other sides, we obtain, 

\a=b cos C + c cos B,' 



{a= b cos C + c cos B, "I 
b=a cos C + c cos A, j- []. 

ca cosB + 6 cos A, J 

st of the above by , the s 

(d?=cib cos C + oc cos B,~j 
c 2 = ac cos B + be cos A, J 



Multiplying the first of the above by a, the second by b, and the 
third by c, gives 

a?=ab cos C + c cos B,~ 



. c 2 = ac cos B + be cos A, . 

Adding the second and third, and subtracting the first, gives 
b 2 + c 2 - a 2 = 2bc cos A ; hence, 

Cos A= ^r ; and similarly ... [c], 

Cos B=" n . [d], 

2ac 

,,2j 

Cos C=- 



203. Since 2bc cos A = 6 2 + c 2 - a 2 by transposition, 

a 2 =6 2 +c a -26ccos A. 
Similar] y , b- = a 2 + c 2 - 2ac cos B. 

and c 2 = a 2 + 6 2 - 2ab cos C. 

Whence 1 + cos A = 1 + 



2bc 2bc 



2bc 2bc 

But since cos 2A = 2 cos 2 A-l, 1 + cos 2A=2 cos 2 A, and hence 
1 + cos A = 2 cos 2 A. Therefore 



2bc 



rr 2 4 A 2\"'^ "/ r _-i 

6c 



204. Put ^(a + 6 + c)=, then %(b + c-a)=s-a, %(a + c-b)=s-b ; 
and ^(a + b- c)=s-c ; and inserting these values in the above, 
it becomes 

21 . s(s-a) . .. , , 1T) s(s-b) , s(s-c) 

cos 2 i.A = - 1 -T ; similarly, cos 2 iB = , cos 2 C = j 

oc oc ao 



56 PLANE TRIGONOMETRY 

Extracting the square root of each of the above, we have 

'-) _,TJ_ ks-b)\ 
be ' 



and 

205. Again, since cos A = 



cos4C =J^, 

V ab 
, subtracting both sides from 1, 






26c 



2bc 



_ ( + b - c)(a + c - b) 
2bc 

But 1 - cos A = 2 sin 2 ^A ; hence, 

(a + b-c)(a + c-b) 



therefore 



2 i A __%(a + b - c) x \(a + c - b) _ (s - b)(s - c) 
* ~ be be 



Sin ^A = 



and sin iC = 



; similarly, sin JB = 



(s-a)(s-c) 



(s-a)(s-i 
ab 



206. Now, since tan |A = -f-r, by dividing the value of 

COS 2^*- 

sin JA by that of cos ^A, we obtain 



tan JA = 
Similarly, 

tan 4B = 



(s-l>)(s-c)_ /I (s-a)(s-b)(s-c) 
s(s - a) 



s (s- a) 



taniB = A / ( ^ 



Extracting the root of the square factor in the denominator of 
each of the last values of the tangent given above, we obtain 



Tan JA = 
Tan 



PLANE TRIGONOMETRY 

1 



=-1 /V 

s-b^J s (s 



-(s - a)(s - b)(s - c), 



a)(s - b)(s - c), 



Tan JO = 






-(s - a)(s - b)(s - c), 



57 



... [&]. 



and by 



From Art. 194 (a) we have sin A=2 sin 4 A cos 
204 (a), cos 4A = / T ; and by 205, sin 
Substituting these values in Art. 194 (a), 

sin A=2 /*(*) x fr- 6 X*- c ) = 2 
AJ 6c AJ 6c fo 

And from the symmetry of the expression, the sines of the other 
angles may be written ; hence, 



Sin A = 
Sin B = 



! - a)(s - b)(s - c), 



2 / 

= A/ s(s - 



a)(s - b)(s - c), 



... [c]. 



207. To find the numerical values of sine, cosine, and 
tangent of 30. 

Let a;=sin 30; then cos 30 = Vl^, but cos 30=sin 60 
=2 sin 30. cos 30; 



^~a? = VI - 2 . Divide by 2 VI - 2 , 
x =4; .'. sin 30 = 4, 



and 
also 



cos 30 = VI - sin 2 30 = VI - i = - ; 

sin 30 2 1 

^= TS- 
v<> V3 



, 
tan 30 = 



cos 30 

COR. Since sin 30 = cos 60, cos 60 = 4; and similarly, we find 
sin 60 = -^, and tan 60 = \/3. 



58 



PLANE TRIGONOMETRY 



208. Otherwise thus : 

Let ABC be an equilateral triangle, and let BD be drawn per- 
pendicular to CA ; then BD bisects CA and 
also angle ABC. 

Hence angle ABD = 30. 
Denote AD by 1, then AB will be 2 and 
D (Eucl. I. 47) BD = V3. 



AT) 




209. To find the numerical values of the sine, cosine, and 
tangent of 45. 

By Art. 174 (1), sin 2 45 + cos 2 45= 1, but sin 45 = cos 45 ; 
..2 sin 2 45 = 1, and hence sin 45 ^ cos 45. 

inn. A ro sin 45 1 V2 , 

Whence, tan 45 = -7^ = ^ x -^ = 1. 

cos 45 V2 1 
Otherwise thus : 

Let ABC be a right-angled triangle having AC 
= BC; then /B=45. 
Denote AC and BC by 1 ; then (Eucl. I. 47) 



. ,_ AC 1 ._ BC 1 

Sin 45 = -j-s = = cos 45 = -ps = = 
AB 2 AB 2 




AC 



EXERCISES 



1. Prove that tan A + cot A 2 cosec 2A. 

2. Prove that sec A 1 + tan A . tan 
l+cot 2 A 



and cosec 2A 



~ 2 cot A ' 

3. Prove that cot 2 A . cos 2 A = cot 2 A - cos 2 A ; and cosec 2 A . sec 2 A 
= sec 2 A + cosec 2 A. 

,. , tan A + tan B , cos 2 A-sin 2 B 

4. Prove that - - -. - ^ = tan A . tan B ; and . . - ^-^5 

cot A + cot B smfA . sinrB 

= cot 2 A . cot 2 B-l. 

5. Prove that cos 2A + cos 2B = 2cos (A + B) . cos (A-B). 

6. Prove that cos (A + B) . cos (A-B) = cos 2 A -sin 2 B; and 
sin ( A + B) . sin ( A - B) = sin 2 A - sin 2 B. 



PLANE TRIGONOMETRY. 59 

7. If A + B 4-0 = 180, prove (1) that sin A + sin B + sin C = 
4 cos A . cos B . cos C ; (2) that cos A-t-cos B + cos C = 
4 sin JA . sin B .sin |C + 1 ; (3) that tan A + tan B + tan C 
=tan A . tan B . tan C ; (4) that cot A + cot B + cot C = 
cot A . cot B . cot C + cosec A . cosec B . cosec C. 

8. Determine the value of A in degrees from the following equa- 
tions : (1) Sin A=sin 2A; (2) tan 2A=3 tan A; (3) tan A + 
3 cot A=4 ; (4) 2 sin 2 3A + sin 2 6A=2. 

9. In any right-angled triangle ABC, in Avhich C is the right 
angle, c the hypotenuse, a the side opposite the angle A, and b 

a 2 -ft 2 
the side opposite the angle B, prove (1) that sin (A-B) = ^ ; 

<2ah ffl _ 

(2) cos (A-B) = ; (3) tan (A-B)= 



(5) cos 2 4A=j and (6) tan 2 \^=-~ 

10. If a line CD bisect the angle C of any triangle, and meet the 

base in D ; tan ADC = - r tan AC, and CD = -^-7 cos 4C. 
a - b a+b 

11. Prove (1) that tan 2 (45 + A) = } + S !" A ; (2) sec (45 + A) . 

j. sin A. 

sec (45 - A) = 2 sec 2A ; (3) tan (30 + A) . tan (30 a - A) = 

2 cog 2A 1 

2cos2A + l ; and {4) Sin ( 60 + A )- sin (60- A) = sin A. 

12. If the sides a, b, of a triangle include an angle of 120, 
show that c J = 2 + 6 + 6 2 ; and if they include an angle of 60, 



SOLUTION OF TRIANGLES 

210. First, let ABC be a right-angled triangle. 

When the two sides are given, the hypotenuse may be found 
by taking the sum of the squares of the sides and 
extracting the square root. 

When the hypotenuse and a side are given, the 
other side may be found by taking the difference of 
the squares of the hypotenuse and the given side 
and extracting the square root ; or, what comes to _ 
the same thing, find the product of the sum and dif- 
ference of the hypotenuse and the given side, and extract the 
square root. 




60 



PLANE TRIGONOMETRY 



211. CASE 1. Given the hypotenuse and a side of a right-angled 
triangle, to find the other parts. 

EXAMPLE. Given AC =415, and AB = 249, to find the other 
parts of triangle ABC. 

1. To find angle A 
AB 249 . 

COS A = ~r-^ = : 7T?= '6 j 

AC 415 

therefore, from a table of natural cosines, the value of A= 
53 7' 48-4". 
Or thus : 

. AC 415 
sec A=- r5 = ; 



therefore 



AB~249 
L sec A = L 415 - L 249 + 10 

= 2-6180481-2-3961993+10 

= 10-2218488 

= L sec 53 7' 48-4". 



BC 



2. To find BC 



therefore 



-= = tan A ; therefore BC = AB tan A ; 

\.X> 

L BC = L AB + L tan A- 10 

= L 249 + L tan 53 7' 48 -4" - 10 
=2-3961993 + 10-1249388-10 
=2-5211381 
= L 232. 



3. To find angle C 

C = 90 - A = 90 - 53 7' 48 -4" = 36 52' 1 1 -6". 
But BC may be found, independently of any of the angles, thus, 

BC 2 = AC 2 - AB 2 =415 2 - 249 2 = 172225 - 62001 = 110224, 
and BC = VI 10224 =332. 

Or BC 2 =(AC + AB)(AC-AB) = (415 + 249) 

(415 - 249) = 664 x 166 = 1 10224, 
and BC = VI 10224 =332. 

This latter method is well adapted to logarithmic calculation ; 
thus, 

L(AC + AB)664, . 2-8221681 

L(AC-AB) 166, . 2-2201081 

LBC 2 , . . . . , . . = 5-0422762 
hence (I. T.*), L BC 332, . . . = 2-5211381 

* I. T. refers to the Introduction to the Mathematical Tables. 




PLANE TRIGONOMETRY 61 

212. CASE 2. Given a side and one of the oblique angles. 

EXAMPLE. In the right-angled triangle ABC, given 
the hypotenuse AC 324 feet, and angle A 48 17', to 
find the other parts. 

1. To find angle C 

C = 90-A=90-4817' 
=41 43'. 

2. To find BC 

BC 

-rp = sin A ; therefore BC = AC sin A 

= 324 sin 48 17'; 

therefore L BC = L 324 + L sin 48 17' - 10 

=2-5105450 + 9-8729976 - 10 
=2-3835426 
= L 241 -848. 

3. To find AB 
-rp=cos A ; therefore AB = AC cos A 

= 324 cos 48 17'; 

therefore L AB = L 324 + L cos 48 17' - 10 

= 2-5105450 + 9-8231138-10 
=2-3336588 
= L 215 -605. 

These sides may also be found by natural sines, independently 
of logarithms ; thus, 

1. To find BC 

T5C 1 

-r-^=sinA; therefore BC = AC sin A 

A\J 

= 324 sin 48 17' 
= 324x -746446 
= 241-848. 

2. To find AB 
-rp=cos A; therefore AB = AC cos A 

= 324 cos 48 17' 
= 324x -6654475 
= 215-605. 



62 PLANE TRIGONOMETRY 

EXERCISES 

1. In a right-angled triangle, the hypotenuse is 1246, and one 
of the oblique angles 25 30' ; find the other angle and the two 
sides =64 30', 5364168, and 1124-621. 

2. The hypotenuse is 645, and an oblique angle 39 10' ; find the 
other sides =500-076, and 407 '368. 

3. In a triangle right-angled at B, given the side AB 125, and 
angle A 51 19', to find the other parts. 

C = 3841', BC = 156-1186, AC = 199'9949. 

4. In a right-angled triangle ABC, having a right angle at B, 
the side AB is 180, and angle A 62 40', find other parts. 

C = 27 20', AC = 392 -0147, BC = 348 -2464. 

5. In a right-angled triangle ABC, given the hypotenuse AC 645, 
and the base AB 500 ; required the other parts. 

BC = 407'459, angle A = 39 10' 38", and angle C = 50 49' 22". 

6. Given the base and hypotenuse 288 and 480, to find the other 
parts. 

The perpendicular = 384, and the oblique angles = 53 7' 48", 
and 36 52' 12". 

7. The two sides about the right angle of a right-angled triangle 
are 360 and 270 ; required the hypotenuse and the oblique angles. 

=450, 36 52' 12", and 53 7' 48". 

8. What are the hypotenuse and oblique angles in a right-angled 
triangle, of which the two sides are 389 and 467 ? 

= 60779, 30 47' 37", and 50 12' 23". 

9. Given the base = 530, the perpendicular = 670, to find the hypo- 
tenuse and the acute angles. =854-284, 51 39' 16", 38 20' 44". 

COMPUTATION OF THE SIDES AND ANGLES OF 
OBLIQUE-ANGLED TRIANGLES 

213. CASE 1. When two angles and a side opposite to one of 
them are given. 

RULE. The sides are proportional to the sines of the opposite 
angles. Hence, 

To find a side, begin with an angle namely, the angle oppo- 
site to the given side ; thus, the sine of the angle opposite to the 
given side is to the sine of the angle opposite to the required side 
as the given side to the required side. 

to 



PLANE TRIGONOMETRY 



63 



When two angles of a triangle are known, the third 
is found by subtracting their sum from two right 
angles. 

Let the three angles of any triangle be represented by the letters 
A, B, C, and the sides opposite to these angles respectively by the 
letters a, b, c ; and let A, B, and a be given, to find the other 
parts. 

1. To find angle C 
C = 180-(A + B). 



2. To find the side b or AC 

Sin A : sin B = a : b. 
a sin B 
sin A 
L6 = La + L sin B - L sin A. 



b = 



By natural sines, 

By logarithms, 

By the same means, the side c can be found. The two preceding 
formulae can be adopted for finding c, by merely changing b into c, 
and B into C ; thus, 

, a sin C 

sin A : sin C = a : c, and c= : r > 
sin A 

or Lc= La + L sin C - L sin A ; 

also Lc = L cosec A + L sin C + La - 20, 

which is the most convenient formula for calculating this case. 

EXAMPLE. In the triangle ABC, there are given angle 
A = 63 48', angle B = 49 25', and BC=275. 



1. To find angle C 
C = 180-(A + B) = 
180 -(63 48' + 49 25') = 
180 -113 13' = 66 47'. 




2. To find the side AC 
Sin A : sin B = BC : AC = a : b. 

L cosec A 63 48', . . . ' . = 10-0470825 
L sin B 49 25', . 9-8805052 

LBC275, = 2-4393327 

LAC 232 7665, . 



2-3669204 



64 PLANE TRIGONOMETRY 

3. To find AB 

Sin A : sin C = BC : AB = a : c. 

L cosec A 63 48', . 10 '0470825 

L sin C 66 47', = 9 '9633253 

LBC275, = 2-4393327 

LAB 281-67,. . . . _. = 2-4497405 

214. CASE 2. When two sides and an angle opposite to one of 
them are given. -^ 

RULE. The sides are proportional to the sines of the opposite 
angles. Hence, 

To find an angle, begin with a side namely, the side opposite 
to the given angle ; thus, the side opposite to the given angle is to 
the side opposite to the required angle as the sine of the given 
angle to the sine of the required angle. 

When two of the angles are known, the third is found 
by subtracting their sum from two right angles. 

Let a, b, and A be given, to find the other parts. 

To find angle B 
a : 6= sin A : sin B, 

, . . _ b sin A 

and by natural sines, sin B = 

ct 

By logarithms, L sin B = L sin A + L6 - La, 

or L sin B=ar. co. La + L6 + L sin A - 10, 

which is the most convenient formula for calculating this case. 

EXAMPLE. In the triangle ABC are given the sides AB and BC 
345 and 232 feet, and angle A 37 20'. 

In this case, when the side opposite to 
the given angle is greater than the other 
given side, only one triangle can be formed ; 
but when less, there can be two con- 
structed. 

1. In the triangle ABC. 

1. To find angle C 
BC : AB = sin A: sin C, 
or a : c = sin A : sin C. 




PLAN'S TRIGONOMETRY 



65 



Ar. co. L BC 232, 
L AB 345, 
L sin A 37 20', 



7-6345120 
2-5378191 
9-7827958 

9-9551269 



L sin C 64 24' 1", . , ; , 

2. To find angle B 
B=180-(A + C) = 180-10144' 1" = 78 15' 59". 

3. To find the side AC 
Sin A :sinB = BC : AC, 

sin A : sin B=a : b. 
L cosec A 37 20', . . . . 
L sin B 78 15' 59", .... 
L BC 232, . . _ . . 



= 10-2172042 
= 9-9908287 
= 2-3654880 

2-5735209 



L AC 374-559, .... 

2. In the triangle ABC'. 

The first proportion above gave angle C, but it gives also angle 
C' in triangle ABC', observing that, instead of the angle 64 24' 1", 
its supplement must be taken ; for angle AC'B is the supplement 
of BC'C, which is equal to C. Hence angle C' = 180 - 64 24' 1" 
= 115 35' 59". 

Then angle ABC' = 180-(A + C') = 180- 152 55' 59" =27 4' 1". 

The last proportion will then give AC', if for angle ABC 
78 15' 59" the angle ABC' 27 4' 1" is substituted. The student 
will find, by making this substitution, that AC' = 174'0738. 

EXERCISES. 

1. In a triangle ABC are given the angles A and C 59 and 52 15', 
and also the side AB 276'5, to find its other parts. 

AC = 325-9183, BC = 2997469, and angle B = 68 45'. 

2. In a triangle ABC, the angles A and B are respectively 54 20' 
and 62 36', and the side AB is 245 ; required the other parts of the 
triangle. AC = 243-978, BC = 223'26, and angle C = 63 4'. 

3. In a triangle ABC, the angles A and B are = 56 6' 13" and 
59 50' 27", and the side AB is = 130 ; required the remaining parts 
of the triangle. Angle C = 64 3' 20", AC = 125, and BC = 120. 

4. In a triangle ABC are given the side AB = 142'02, AC = 104, 
and angle B=44 12'. 

BC = 133-639 or 69-992, and angle C = 72 10' 55", or 107 49' 5". 

5. In a triangle ABC are given the side AB = 456, AC = 780, and 
angle B = 125 40' ; required the other parts. 

Angle C = 28 21' 23", A=25 58' 37", and BC = 420'529. 



66 PLANE TRIGONOMETRY 

6. In a triangle ABC are given AB = 520, BC = 394, and the 
angle C = 64 20' ; required the other parts. 

A =43 4' 23", B = 72 35' 37", and AC = 550'507. 

215. CASE 3. Given two sides and the contained angle. 
Let the given sides be a and b, and C the given contained 
angle. 

1. To find the sum of the angles opposite to the given 
sides, or A+B. 

RULE. The sum of the angles opposite to the given sides is 
found by subtracting the given angle from two right angles. 
Or, A + B = 180 - C, and J( A + B) = 90 - JC. 

2. To find the angles opposite to the given sides, or 
A and B. 

RULE. The sum of the two sides is to their difference as the 
tangent of half the sum of the angles at the base to the tangent of 
half their difference. 
Or a + b : -6 = tan i(A + B) : tan (A-B). 

_. (a-b) . tan (A + B) 
By natural sines, tan $( A - B) = - ^- ; 

and by logarithms, 

L tan (A-B) = L tan i(A + B) + L(a-6)-L( + 6), 
or L tan J( A - B) = ar. co. L(a + 6) + L(a - b) + L tan ( A + B) - 10. 

When A - B is thus found, then 

Half the difference of the two angles, added to half their 
sum, gives the greater ; and taken from half the sum, gives 
the less. 

Or 
and B 



216. When only the third side C is wanted, it can be found by 
the formula 



When C is obtuse, the upper sign + is to be used ; and when C is 
acute, the lower sign - is to be taken. 

The natural cosine of C is of course to be used. The third term 
of the value of c 2 may be found by logarithms ; thus, let it = M, ... 
then L M = La + L6 + L cos C- 10. 

EXAMPLE. Of a triangle ABC, given the sides AC, BC re- 
spectively =176 and 133, and the contained angle C = 73. 



PLANE TRIGONOMETRY 67 

1. To find the angles A and B 

The side AC being greater than BC, angle B opposite to the 
former exceeds angle A opposite to the latter ; 
also A + B = 180 - C = 180 - 73 = 107, and (A + B) 
=53 30'. 

And AC + BC : AC - BC = tan \( A + B) : tan i(B - A), 




Ar. co. L(AC + BC)309, . 7-5100415 

L(AC-BC)43, . 1-6334685 

L tan(A + B)5330', . . = 10-1307911 



Ltan^(B-A) 10 39' 3" . . = 9'2743011 

Hence, angle B = 64 9' 3" 

A = 42 50 57 

2. To find AB 
SinB :sinC = AC : AB. 

L cosec B 64 9' 3", . 10*0457840 

L sin C 73, . 9-9805963 

LAC 176, . . ... = 2-2455127 

L AB 187-022, . 2-2718930 

When the third side AB only is wanted, it may be found thus 

AB 2 =BC 2 + AC 2 -2BC . AC . cos C, 

or c 2 = a 2 + 6 2 - 26 .cos C = 133 2 + 176 2 - 2 x 133 x 176 x cos 73 = 
17689 + 30976-46816 x -2923717 = 48665- 13687-67=34977-33, and c= 
V34977-33 = 187-022. 
The sign - is used above because C is acute. 

EXERCISES 

1. In a triangle ABC, given AB and BC respectively = 180 and 
200, and angle B = 69, to find the other parts. 

Angle A = 59 52' 45", C=51 7' 15", and AC =215 -864. 

2. Two sides of a triangle are respectively = 240 and 180, and 
the contained angle is =25 40'; required the other angles and the 
third side. 

The angles are = 109 15' 30" and 45 4' 30", and the third 
side = 110-114. 

3. Two sides of a triangle are respectively = 3754 and 3226 "4, 
and the contained angle = 58 53' ; i-equired the other angles, and 
the third side. 

The angles = 68 11' 8" and 52 55' 52" ; third side =3461 '75. 



68 PLANE TRIGONOMETRY 

4. Two sides of a triangle are respectively = 375 '4 and 327763, 
and the contained angle = 57 53' ; required the other parts. 

The third side=342'818, and the angles=68 2' 35" and 54 4' 25". 

217. CASE 4. When the three sides of a triangle are given. 

This case may be solved by any of the following five rules : 

RULE I. Draw a perpendicular from one of the angles upon 
the opposite side, or this side produced ; then calling this side 
the base twice the base is to the sum of the two sides as the 
difference of these sides to the distance of the perpendicular 
from the middle of the base ; then the sum of half the base 
and this distance is = the greater segment, and their difference 
is = the less. 

The given triangle is thus divided by the perpendicular into two 
right-angled triangles, in each of which two sides are known ; and 
hence the angles at the base can be found, and consequently the 
third angle. 

RULE II. From half the sum of the three sides subtract each 
of the sides containing the required angle ; then add together the 
logarithms of the two remainders, and the arithmetical comple- 
ments of the logarithms of these two sides, and half the sum is 
the logarithmic sine of half the required angle. 

Let C be the required angle, and s=\(a + b + c) ; 

(s (t)(s b) 
then, by natural sines, sin 2 C= -- ^V - , 

and by logarithms, 

2LsinJjC = L(* - a) + L(* - 6) + ( 10 - La) + ( 10 - L6). 

RULE III. From half the sum of the three sides subtract the 
side opposite to the required angle ; then add together the loga- 
rithms of the half sum and of this difference, and the arithmetical 
complements of the logarithms of the other two sides, and half the 
sum is the logarithmic cosine of half the required angle. 

By natural sines, cos 2 C = , ; and by logarithms, 



RULE IV. From half the sum of the three sides subtract each 
side separately ; then subtract the logarithm of the half sum 
from 20, and under the result write the logarithms of the three 
remainders ; half the sum of these will be a constant, from 
which, if the logarithms of the three remainders be successively 
subtracted, the new remainders will be the logarithmic tangents 
of half the angles of the triangle. 



PLANE TRIGONOMETRY 69 

r> 1 i i i rt ( s ~ a )( s ~ b)(S - C) 

By natural tangents, tan 2 C = ^ 

RULE V. The angle may also be found by the formula, 

cos C = ^-r . 

2ab 

This method is simple when the sides are small numbers ; when 
c 2 is less than a? + b*, angle C is acute ; but when greater, this angle 
is obtuse. When c 2 = a 2 + 6 2 , angle C is a right angle. 

The proof of these rules is given in the articles 201-205. 

The second method ought not to be used when the angle is a 
large obtuse angle, for then | C will be nearly a quadrant, and the 
sines of angles near 90 vary sloAvly, and the seconds will not be 
accurately obtained. For a similar reason, the third method ought 
not to be used Avhen C is very small. When all the angles are 
required, the fourth method is much more 
expeditious than any of the others. 

EXAMPLE. In the triangle ABC there 
are given the three sides AB, BC, and AC 
respectively = 150, 130, and 140, to find the 
angles. A Q 

BY RULE I 

1. To find the difference of the segments AD, DB 

2 AB : AC + CB = AC - CB : DE. 

300 : 270 = 10 :9 = DE; and 




2. To find angle A 
CosA-AD- 84 - 

~AC~140' 
L cos A = 10 + L 84 - L 140 

= 11 -9242793 -2-1461280 

= 9-7781513 

=L cos 53 7' 48". 

3. To find angle B 
n _ BD 66 
CosB = BC = 130 ; 

therefore L cos B = 10 + L 66 - L 1 30 

= 11-8195439-2-1139434 

= 9-7056005 

= L cos 59 29' 23". 

Pnc. p 



70 



PLANE TRIGONOMETRY 



4. To find angle ACB 
= 180-(A + B) = 180-112 37' 11" = 67 22' 49". 



BY RULE II 

This rule may be used exactly as the following, taking (s - a) and 
(s - b) instead of * and (s - c), and sin C for cos J C. 



BY RULE III 
To find angle C 



*-c=210- 150 = 60. 
L*210, 

L(-c)60, . . 
10- La 130, . 
10 - L b 140, 



= x 420 = 210, 

= 2-3222193 

= 1-7781513 

7-8860566 

= 7-8538720 



L cos 



C 33 41' 24" -2 
2 



2)19-8402992 
= 9-9201496 



C = 67 22' 48" -4 

By Rule IV. all the angles may be found, and being added 
together, when the work is correct, their sum will be = 180. 

BY RULE IV 



a=130 






6 = 140 






c=150 






2s =420 






*=210, 


and 20 - Ls = 


17-6777807 


*-= 80 


L(s-a) = 


1-9030900 


s-b= 70 


L(s-6) = 


1-8450980 


s-c= 60 


L(s- c) = 


1-7781513 



Tan 
Tan 
Tan 



Hence 



2)23-2041200 

Constant = 11-6020600 
A = 26 33' 54 -2" = 9 -6989700 {cont. - L(s - a)}. 
B =29 44 41 -6 =9 '7569620 {cont. - L(s - 6)}. 
C=33 41 24-2 = 9 -8239087 {con t.-L(s-c)}. 

A= 53 7' 48-4" 
B= 59 29' 23 -2" 
C = 67 22' 48-4" ; and adding 

= 180 



PLANE TRIGONOMETRY 71 

2 1302+1402-150 2 



By Rule V.,... Nat. cos C -^- 2 x 130x140 

36500-22500 14000 5 _,.._. ^ A0 ,, _ 

= 2^T30TT40 = 2x 130x 140 = I3 = 3846154 = nat C * 67 ffl 48 5 " 

When the sides are small numbers, as in the present example, 
this method is very expeditious. The angle C being found, A and 
B may be similarly calculated by these formulae, 



_ _ 

cos A= m- , and cos B = s - . 
26c 2ac 

EXERCISES 

1. The three sides of a triangle AB, BC, AC are = 100, 80, and 
60 ; find the angles. A=53 7' 48", B = 36 52 12", and C = 90. 

2. The three sides of a triangle AB, AC, BC are = 457, 368, and 
325 ; find the angles. 

A=44 48' 15", B=52 55' 56", and C = 82 15' 49". 

3. The three sides of a triangle are AB = 562, BC = 320, and 
AC = 800 ; required the angles. 

A =18 21' 24", B = 128 3' 49", C = 33 34' 47". 

PROMISCUOUS EXERCISES IN TRIGONOMETRY 

1. Given the hypotenuse of a right-angled triangle = 774, and one 
of the oblique angles = 57 8' ; to find the other parts. 

The other acute angle is = 32 52', and the other two sides 
are=420'039 and 650'11. 

2. Given one of the sides about the right angle of a triangle 
= 2456, and the opposite angle = 44 26' ; to find the other parts. 

The other acute angle is = 45 34', and the other sides are 
=2505-068 and 3508-176. 

3. Given the hypotenuse and another side = 3604'5 and 2935'2; 
to find the other parts. 

The angles are = 35 28' 48" -8 and 54 31' 11 "2", and the other 
side is = 2092 -13. 

4. Given the two sides about the right angle = 1260 and 1950 ; to 
find the other parts. 

The angles are =57 7' 53" and 32 52' 7", and the hypotenuse 
is =2321 -66. 

5. Given two angles, A and C, of a triangle = 32 42' and 28 58', 
and the side AC = 6364 ; to find the other parts. 

The angle B is = 118 20', the side AB= 3501 -57, and BC 
is =3906 -02. 



72 PLANE TRIGONOMETRY 

6. The sides AB, BC of a triangle are = 1000 and 1200, and angle 
A is = 36 50' ; required the other parts. 

Angle B is = 113 11' 41", angle C = 29 58' 19"; and the side 
AC is =1839 -909. 

7. The two sides AC, BC of a triangle are = 281 '67 and 275, and 
angle C is = 49 25' ; required the other parts. 

Angles A and B are = 63 48' and 66 47', and the side BC 
is -232 -7665. 

8. The three sides of a triangle are = 133, 176, and 187 '022; 
required the angles. 

The angles are = 73, 64 9' 3", and 42 50' 57". 



MENSURATION OF HEIGHTS AND DISTANCES 

218. For the measurement of lines, some line of a de- 
terminate length is assumed as an inch, a foot, a yard, &c. 
The assumed line is called the lineal unit. The number 
of lineal units contained in a line is its measure or numerical 
value. 

The heights and distances of objects are represented by 
lines, and are therefore expressed in terms of some lineal 
unit. 

The measure of any height or distance might be ascertained 
by applying the lineal unit to its length, were it possible to 
reach it; but many heights and distances are of such a 
nature that their measures can be obtained only by the 
application of the principles of trigonometry. 

219. Heights and distances are said to be accessible or 
inaccessible according as it is possible or not to reach the 
base of the perpendiculars that measure the heights, or accord- 
ing as the distance between two objects can be directly measured 
or not. 

220. A vertical line is the direction of the plumb-line. 

221. A vertical plane is a plane passing through a vertical 
line. 

222. A horizontal plane is perpendicular to a vertical line. 



MENSURATION OF HEIGHTS AND DISTANCES 



73 



223. A horizontal line is one in a horizontal plane. 

224. An oblique plane is one that is neither vertical nor 
horizontal. 

225. A vertical angle is an angle in a vertical plane. 

226. A horizontal angle is an angle in a horizontal plane. 

227. An inclined angle is an angle in an oblique plane. 

228. An angle of elevation of one point above another is 
the vertical angle formed by a line joining the two points and a 
horizontal line passing through the latter point. 

An angle of elevation is also called an angle of altitude. 

229. An angle of depression of one point below another 
is the vertical angle contained by a line joining the two points 
and a horizontal line passing through the latter point. 

230. The angular distance between two objects at any point is 
the angle formed at that point by two lines drawn from it to the 
objects ; this angle is therefore the angle of a triangle opposite to 
the line joining the objects. 

Horizontal and vertical angles can be measured most conveni- 
ently by means of the Theodolite ; for an account and engraving 
of which, see LAND-SURVEYING. 

When much accuracy is not required, vertical angles can be 
measured by means of a quadrant of simple construction, repre- 
sented in the adjoining figure. The arc AB is a quadrant, 
graduated into degrees from B to A ; C, the point from which the 
plummet P is suspended, 
being the centre of the 
quadrant. 

When the sights A, C 
are directed towards any 
object, S, the degrees in 
the arc BP are the measure 
of the angle of elevation 
SAD of the object. For 
AD being a horizontal line, 
and SD (supposing S and 
D joined) a vertical line, 
and therefore CP parallel 

to SD, the angle ACP = ASD; now BCP is the complement of 
ACP, and SAD of ASD ; therefore angle SAD = BCP, which is 
measured by the arc BP. 




74 



MENSURATION OF HEIGHTS AND DISTANCES 



231. Problem I. To compute the height of an accessible 
object. 

Let the object whose height is required be a tower BC. 
Measure a horizontal line AB from the base of the object to any 
convenient distance A, and then measure 
the angle of elevation of the top of the 
object at A. 

Then if AD denote the height of the 
eye, the angle CDE is the given angle, 
DE being parallel to AB. Hence, in the 
triangle DEC, the side DE = AB, and 
angle D are given ; therefore CE can be found by 180. 

EXAMPLE. Required the height of the tower BC, having given 
the horizontal line DE = 120 feet, the angle of elevation CDE 
=39 49', and the height of the eye =5 feet 2 inches. 

To find CE in triangle CDE 
CE 




therefore CE = DE tan D. 

Tan D 39 49', 
DE 120, 



CE 100-039, 

Height of eye = 5-166 

it tower =105 -205 



9-9209898 

= 2-0791812 

12-0001710 



= 2-0001710 



EXERCISES 

1. The breadth of a ditch in front of a tower is=48 feet; and 
from the outer edge of the ditch the angle of elevation of the top 
of the tower is = 53 13' ; what is the height of the tower? 

= 64 -20184 feet. 

2. Required the height of an accessible building, the angle of 
elevation of its top being =41 4' 34" at a point =101 -76 feet distant 
from it, the height of the eye being = 5 feet . . =93 '696 feet. 

3. At the top of a ship's mast = 120 feet high, the angle of de- 
pression of another ship's hull was = 15 45' ; what is the distance 
between the ships ? =425 -49 feet. 

4. Required the height of a tower, a horizontal base of 245 feet 
being measured, and the angle of elevation being =35 24'. 

= 174 -112 feet. 




MENSURATION OF HEIGHTS AND DISTANCES 75 

232. Problem II. To compute an inaccessible height, 
when a "horizontal line in the same level with its base, 
and in the same vertical plane with its top, can be 
found. 

Let the object whose height is wanted be a hill, CDE, such that 
the foot D of the vertical line CD that measures its altitude is 
inaccessible. 

Measure a horizontal base ^^' \ m 

AB, and the angles of elevation ^^""^X/C I 

of the top C at A and B namely, ~__ 

n and m. A B 

In the triangle ABC angle C=m-n (Eucl. I. 32), and is hence 
known. Then to find BC in the same triangle, sin C : sin n 

AB : BC ; thus BC is found to be A ^ S1 , *. Again, to find CD 

sin C 

in the triangle BCD : 

CD 

^=sin m; therefore CD = BC sin nt. 
r>U 

Hence, using the value of BC, 

~~ AB sin n sin m . . 

CD = : ^ AD sin n sin m cosec C, 

sin C 

,_ AB sin n . sin m ~ . 

CD = ^ : ~ , LD = AB sin n . sin m . cosec C ; 

R sin C 

. *. L . CD = L . AB + L sin n + L . sin m + L cosec (m - n) - 30. 

Since C = (m-n) ; and 30 has to be subtracted, because each of 
the logarithmic trigonometrical functions is 10 greater than the 
logarithm of the natural function. 

EXAMPLE. From the base of a hill a horizontal line of 384 feet 
was measured in a direction from the hill, and such that the line 
and the top of the hill were in one vertical plane, the angles of 
elevation of the top of the hill, taken at two stations at the 
extremities of this base-line, were =40 12' and 50 42'; required 
the height of the lull. 

Here C=m-n=5Q 42'-40 12' = 10 30 7 . 

LAB 384, = 2-5843312 

L sin n 40 12*, . . . . = 9 '8098678 
L sin m 50 42', = 9'8886513 

L cosec (m-n) 10 30 7 , = 10'7393670 

L . CD, = 3-0222173 

.-. CD =1052-488. 



76 MENSURATION OP HEIGHTS AND DISTANCES 

The two proportions may also be wrought separately. 
In the following exercises the base-line is measured as in the 
above example : 

EXERCISES 

1. In order to find the height of a hill, a base-line was measured 
= 130 feet, and the angles of elevation of the top of the hill, 
measured at the extremities of the base, were = 31 and 46; 
required its height. =186 '089 feet. 

2. Required the height of an inaccessible tower on the opposite 
side of a river, the length of the base being = 170 feet, and the 
angles of elevation at its extremities = 32 and 58; the height of 
the eye being = 5 feet =179-276 feet. 

3. Required the height of a hill from these measurements : AB 
= 1356, angle m=36 50', and n=25 36'. . . =1803 '06 feet. 

233. Problem III. To measure the height of an object 
situated on an inaccessible height, when a horizontal base 

can be measured in the same 
vertical plane with the top 
of the object. 

Let EC be the object situated 
on the lull ED, AB the horizon- 
tal base; measure the angles of 
elevation CBD, EBD of the top and bottom of the tower at 
B ; then measure at A, the angle of elevation of the top of the 
tower C. 

Find BC in the triangle ABC thus : Angle C = CBD-A, and 
sin C : sin A=AB : BC. Then find EC in triangle BCE thus : 
Angle BEC = D + EBD = 90 + EBD, and angle B = CBD - EBD, 
then sin EEC : sin CBE = BC : CE. 

EXAMPLE. Required the height of a fort CE, situated on the 
top of a hill, the angles of elevation of the top of the hill and the 
top of the fort at B being =48 20' and 61 25'; at A, the elevation 
of the top of the fort, being = 38 19' ; and the base AB = 360 feet. 

1. To find BC in triangle ABC 

C = CBD -A =61 25' -38 19' = 23 6'. 

L cosec C 23 6', . . . . = 10-4063406 

L sin A 38 19', . . ... = 9 '7923968 

L AB 360, = 2-5563025 

.'. LBC, . . . . . . = 2-7550399 




MENSURATION OP HEIGHTS AND DISTANCES 



77 



2. To find CE in triangle BOB . 

E = 90 + EBD = 90 + 48 20' = 138 20'. 

B = CBD-DBE = 6125'-48 20' = 13 5'. 

L cosec E 138 20' (41 40'), . . = 10-1773117 

L sin B 13 5', . . . . = Q'3548150 

LBC, ...... = 2-7550399 

LCE, ...... = 2-2871666 

.-. CE = 193-7165. 

These two calculations may also be easily combined into one by 
compounding the two proportions from which they arise. 

EXERCISES 

1. Find the height of a tower on the top of a hill from these 
measurements : The angles of elevation of the top of the hill, and 
the top of the tower at the nearer station, are = 40 and 51 ; at the 
farther station, the angle of elevation of the top of the tower is 
=33 45', and the horizontal base = 240 feet. . . =111-9978. 

2. In order to determine the height of a lighthouse, situated 
on the top of an inaccessible eminence, the following data were 
obtained : A base-line = 368 feet, the angles of elevation at the 
nearer station of the top and bottom of the lighthouse = 36 24', 
and 24 36', and the angle of elevation of the top of the light- 
house at the farther station = 16 40' ; what was its height? 

= 70 -304 feet. 

234. Problem IV. To calculate the height of an object 
standing on an inclined plane. 

Let CD be the object, and AC 
the inclined plane. 

Measure a base AB on the plane, 
and the angles of elevation DBF, 
DAE of the top of the steeple, 
taken at the extremities of the base, 
and also the angle of inclination i 
of the plane with the horizon. 

To find the angles m, n, r, and v 
?=DBF-t, = DAE-', and r = m-n; 




also 

To find BD in triangle ABD, sin r:sin = AB:BD; and to 
find CD in triangle BCD, sin v : sin m = BD : CD. 



78 



MENSURATION OF HEIGHTS AND DISTANCES 



EXAMPLE. Required the height of the steeple CD, situated on 
the inclined plane AC, from these measurements: AB = 112, the 
angles of elevation at A and B=44 25' and 63 40', and the 
inclination of the plane = 15 20'. 

To find angles m, n, r, and v 

TO = DBF - i = 63 40' - 15 20'= 48 20'. 
n = DAE - i = 44 25' - 15 20'= 29 5'. 
r = m - n = 48 20' - 29 5' = 19 15'. 
v = 90 +i = 90 + 15 20' = 105 20'. 



To find BD 

L cosec r 19 15', = 10-4818934 
L sin n 29 5', . = 9-6867088 
L AB 112, . = 2-0492180 

L BD . 165-12, . = 2-2178202 



To find CD 

L cosec v 105 20', = 10-0157411 
L sin m 48 20', = 9 -8733352 
L BD 165-12, . = 2-2178202 



LCD, 



= 2-1068965 
CD = 127 -9077. 



EXERCISE 



Required the height of an object standing on an inclined plane 
from these data : AB = 124 feet, angle DBF = 58 20', DAE =40 30', 
and the inclination of the plane = 14 10'. . . =129 -068 feet. 

235. Problem V. To find the height of an inaccessible 
object, when only one station can be taken on the same 
horizontal plane with its base, and a base-line on an 
inclined plane, and in the same vertical plane with 
its top; and also to find the distances of the stations 
from the object. 

Let DE be the inaccessible object ; A, the station in a horizontal 

plane with D ; AB, the 
acclivity. 

Measure a base AB in 
the same vertical plane 
with DE ; and BF being 
a horizontal line, measure 
the angles of depression 
FBA, FBD, and FEE. 
Then in triangle ABD, angle D = FBD (Eucl. I. 29), and angle 
B = FBA-FBD, and A=180-ABF; for ABF = BAC. Again, in 
triangle BDE, B = FED -FEE, and E = F + FBE = 90 + FEE. 




MENSURATION OF HEIGHTS AND DISTANCES 79 

Hence, in triangle ABD, find BD from sin D:sin A=AB:BD; 
then in triangle BDE, find DE from sin E : sin B = BD : DE. 

If the distance of A from D were required, it could also be found 
from sin D : sin B= AB : AD. 

EXAMPLE. In order to find the height of a tower on the other 
side of a river, a base of 204 feet was measured up an acclivity from 
a station on the same horizontal plane with the bottom of the 
object, and at the upper station the angles of depression of the 
first station, and of the bottom and top of the object, were = 47 42', 
17 52', and 11 40'; required the height of the tower, and the 
distance of the two stations from its bottom. 

In triangle ADB, angle D = FBD = 17 52', and B=FBA-FBD= 
47 42*- 17 52 / =29 50', and A=180- ABF = 180-47 42' = 132 
18'. And in triangle BDE, angle B= FED -FEE = 17 52' -11 40' 
=6 12', and E = F + FBE = 90 + 11 40' = 101 40'. 



1. To find BD in triangle ABD 

L cosec D 17 52', . = 10-5131405 

L sin A 132 18' (47 42'), . . = 9'8690152 
LAB 204, ..... = 2-3096302 

BD 491 -797, . ' . '. . . = 2-6917859 

2. To find DE in triangle DBE 

L cosec E 101 40' (78 20'), . . = 10-0090662 
L sin B 6 12', = 9-0334212 

LBD491-797, . 2-6917859 

DE 54-2342, .-..-; . , .. = 1-7342733 

3. To find AD in triangle ADB 

L cosec D 17 52', . 10-5131405 

L sin B 29 50', = 9 '6967745 

LAB 204, ..... = 2-3096302 

AD 330-7846, . 2*5195452 

EXERCISE 

Required the height of an inaccessible object from measurements 
similar to those in the above example, and also the distance of the 
lower station from the object, the angles of depression of the first 
station, and of the bottom and top of the object, taken from the 
upper station, being 42, 27, and 19, and the distance between 
the stations = 165 yards. 

The height = 35 -796, and distance = 94 -066 yards. 



80 



MENSURATION OP HEIGHTS AND DISTANCES 




236. Problem VI. To measure an inaccessible height 
when a horizontal base can be obtained, but not in the 
same vertical plane with the top of the object. 

Let BC, the altitude of a hill, be the inaccessible height, and 
AD the horizontal base. 

c Measure- the base AD and the 

angle of elevation of the top C of 
the hill at the station A ; let AB, 
DB be horizontal to the vertical 
line CB ; and measure the horizon- 
tal angles ADB and DAB with the 
theodolite, as well as the vertical 
angle CAB. 

In the triangle ADB the angles at" A and D are known, and the 
side AD ; hence angle B = 180-(A + D), and AB is found by the 
proportion sin B : sin D = AD:AB. Then in the triangle ABC, 

"RO 
having the right angle B, BC is found from -r>. = tan A. 

Using the value found for AB, we obtain 
, AD sin D tan A 

tft_/ = : -fj ; 

sin B 
hence L . BC = L . AB + L sin D + L tan A + L cosec B-30. 

EXAMPLE. Required the height of a hill from these measure- 
ments : AD =1284, angle ADB = 74 15', DAB = 85 40', and 
angleBAC = 2556'. 

In triangle ADB 

) = 180-159 55'=205'. 



3-1085650 

10-4642168 

9-9833805 

9-6868981 



L AB 1284, . 
L cosec B 20 5', 
L sin D 74 15', 
L tan A 25 56', 

LBC, 



To find BC 



3-2430604 



BC = 1750-09. 



EXERCISES 

1. In order to find the height of a mountain, a base of 1648 feet 
was measured in a valley, and at the station at one of its ex- 
tremities, the angle of elevation of the top of the hill was found to 
be=3225'; and the horizontal angle at it, formed by the base, 
and a horizontal line drawn from this station to the vertical line 



MENSURATION OP HEIGHTS AND DISTANCES 81 

from the top of the hill, was = 78 16'; also the horizontal angle, 
similarly formed at the other station, was = 80 12'; what is the 
height of the hill ? .... The altitude is = 2809 '63 feet. 

2. Find the height of a mountain BC from these measure- 
ments : In the horizontal triangle ABD, the base AD = 1245 
feet, angle A=74 12', D = 84 20', and the angle of elevation 
CAB = 25 45' Height = 1632 -92 feet. 

If the base AD (last figure) were not horizontal, and if AE is a 
horizontal line, and DE a vertical one, the angle of acclivity DAE 
could be measured, and the base AD ; and then the horizontal base 
AE could be found thus: AE = AD cos DAE; then the base AE 
is known, and the horizontal angle measured at D, as formerly 
described, is = to the horizontal angle contained by AE, and a 
horizontal line from E to a point in CB. Considering, therefore, 
AE as the base, the height of the hill could be found exactly as 
before. 

237. Problem VII. To measure a distance inaccessible at 
one extremity. 

Let AD be the inaccessible distance 
between two objects at A and D, on 
opposite sides of a river. 

Measure a base AB, and the angles 
A and B at its extremities. 

Angle D = 180-(A + B); and AD is 
found by the proportion 

Sin D: sin B=AB:AD. 

EXAMPLE. Required the distance between a tree and a wind- 
mill on the other side of a river, a base of 1140 feet being measured 
from the tree to another station ; the angular distance of the tree 
and windmill, measured at the latter station, being =43; and the 
angular distance of the windmill and second station, measured at 
the tree, being = 60. 

Angle D = 180 - (A + B) = 180 - 103=77. 

To find AD 

L cosec D 77, = 10-0112761 

L sin B 43, = 9-8337833 

L AB 1140, = 3-0569049 

AD 797 -929, ..... = 2-9019643 




MENSURATION OF HEIGHTS AND DISTANCES 



EXERCISES 

1. Having taken two stations on the side of a river, and 
measured a base between them of 440 yards, and also the angles 
at the stations formed by the base and lines drawn from the 
stations to a house on the other side of a river, which were 
= 73 15' and 68 2'; what are the distances of the stations from 
the house? =673-624 and 652-4 yards. 

2. A line was measured on the side of a lake of 500 yards, and 
the angles at its extremities contained by it and lines drawn to 
a castle on the other side of the lake were = 79 23' and 54 22'; 
what is the distance of the castle from the extremities of the 
base? =680-323 and 562-57 yards. 

238. Problem VIII. To find the distance between two 
objects that are either invisible from each other or inac- 
cessible in a straight line. 

Let A and C be the two objects, 
c inaccessible in a straight line from each 
other on account of a marsh. 

Measure two lines AB, BC to the 
objects and the contained angle B. 
In the triangle ABC, two sides AB, BC, and the contained angle 
B, are known ; hence AC may be found. 

EXAMPLE. Given the two lines AB, BC, 562 and 320, and the 
contained angle B 128 4', to find AC. 




1. To find the angles at A and C 

A + C = 180 -128 4' = 51 56'. 
Ar. co. L(AB + BC)882, . 

L(AB-BC)242, . 

2558', . 



Tan(A~C) 7 36' 40", 

.*. Angles C= 33 34' 40" 
n A= 18 21 20 



7 '0545314 
2-3838154 
9-6875402 



= 91258870 



2. To find AC 



L cosec C 33 34' 40" 
L sin B 128 4', . 
L AB 562, . 

L AC 800-008, 



= 10-2572213 
9-8961369 
2-7497363 



2-9030945 




MENSURATION OP HEIGHTS AND DISTANCES 

EXERCISES 

1. Find the distance between two objects that are invisible from 
each other, having given their distances from a station at which 
they are visible = 882 and 1008 yards, and the angle at this station, 
subtended by the distance of the objects = 55 40'. =889*45 yards. 

2. The distance of a given station from two objects situated at 
opposite sides of a hill are =564 and 468 fathoms, and the angle 
at the station, subtended by their distance, is =64 28'; what is 
their distance ? Distance = 556'394 fathoms. 

239. Problem IX. To find the distance between two 
inaccessible objects. 

Let M and K be the two objects 
on the opposite side of a river from 
the observer. 

Measure a base AB ; and at 
each station measure the angular 
distances between the other station 
and the two objects namely, the 
angles BAK, BAM, ABM, and 
ABK. 

Then in the triangle ABK, angle K = 180-(A + B), and the side 
AB is known ; hence find AK thus : sin K : sin B = AB : AK. 

Again, in triangle AMB, angle M = 180-(A + B), and AM is 
found by the proportion sin M : sin B = AB : AM. 

Hence, in triangle AMK, the two sides AM, AK are known, and 
the contained angle A = BAM - BAK ; therefore AK + AM : AK ~ AM 
= tan (M + K) : tan (M~K) ; and M~K being thus found, each 
of the angles M and K can then be found. Hence MK will now be 
found in triangle AMK by the proportion sin K : sin A= AM : MK. 

EXAMPLE. Kequired the distance between the two objects M 
and K in the preceding figure from the following data : 

BAK = 64 25', ABM =56 15', BAM = 104 25', ABK =106 23', and 
the base AB=520 yards. 

In triangle ABK, angle K = 180-(A+B) = 180- 170 48' = 912'. 

In triangle ABM, angle M = 180 - (A + B) = 180 - 160 40' = 19 20'. 

And in triangle AMK, angle A = BAM -BAK =40. 

1. To find AK in triangle ABK 
L cosec K 9 12, . . . = 10-7962026 

L sin B 106 23' (73 37'), . . = 9 "9819979 

LAB 520, = 2-7160033 

LAK 3120-35, = 3-4942038 



84 MENSURATION OF HEIGHTS AND DISTANCES 

2. To find AM in triangle AMB 
L cosec M 19 C 20', ... = 10-4800888 
L an B 56 15', . ' . . . = 9-9198464 
LAB 520, . . . . . = 2-7160033 

LAM 130538, . . . . . = 3-1159385 

3. To find the angles M and K in triangle AMK 

M + K=180 -A=180-40=140. 

AT. oa L(AK + AM) 4426133, . -,., . = 6-3539563 
HAK- AM) 1814-37, . ' '. fc . = 3-2587259 

Ltan J(M + K)70, . 104389341 



Tan KM -K) 48 23' 48", . . = 10-0516163 
Hence K= 21 36' 12" 

4. To find MK in triangle AMK 
L eosec K 21 36' 12", ... = 10-4339415 
Lain A 40, . . . . = 9-8080675 

L AM 1305-98, = 3-1159385 



LMK 2280-06, . 33579475 

It is evident that if the sides MB, BK had been found instead 
of MA, AK, the distance MK could have been found in a similar 
manner in the triangle MBK. 

EXERCISES 

L Find the distance between two objects situated as in last 
example, from these measurements : 

BAR = 58 30', ABM = 53 3<r, 

BAM = 95 20', ABK = 98 45', and 

the base AB = 375 yards ; .*. MK = 599 "742. 

9L Find the distance between the two objects M and K from 
these data : 



In triangle MAT^ In triangle KAB, 

A =199 f, A =W12', 

B = 45 30', B =98 1ST, 

AB = 1248 feet; .-. MK =3581*2 feet. 

240. Problem X. To find the distance between two objects, 
having given the angles formed at each of them by lines 



MENSURATION OF HEIGHTS AND DISTANCES 



85 



drawn to the other, and to two given stations, the distance 
between which is also given. 

Let C and D be the two objects, and A and B the two stations 
which may be invisible from each other. 

Measure the angles at C and D, formed by lines joining them 
with the two stations, and with each 
other. 

The angles at C and D are known, 
and were the distance CD known, AB 
could be found by calculation by the 
last problem. CD, however, is un- 
known ; assume it equal to some 
number, as 1000, and compute by the 
preceding problem the value of AB 

on this supposition ; then the length of CD will be found by 
this proportion the computed value of AB is to its real value 
as the assumed value of CD to its real value. 

The distance between A and B, supposing CD = 1000, can be 
first found by means of Prob. IX. 

EXERCISES 

1. Required the distance between the two objects C and D 
from these measurements : 




Angle ACB=36 15' 5", 
BCD = 33 7' 40", 
AB =1410-4; 



ADB = 45 1' 3", 
ADC = 302'0", 
.-.CD =2080-88. 



2. Find the distance between the objects C and D from the 
following measurements : 



In triangle ACD, 
C =110 50', 
D = 38 45', 
AB=1540; 



In triangle BCD, 
C = 43 30*, 
D =115 21', 

. CD = 661 -78. 



241. Problem XI. To find the distance between two 
inaccessible objects, so situated that a base cannot be 
obtained from the extremities of which both objects are 
visible, but which are both visible from one point. 

Let A and B be the two objects, and C the point from which 
both are visible. 

tnc. Q 



86 



Measure two bases DC, CE, and the angles at their ex- 
tremities. 

In triangle ADC, angle A = 180-(C + D) ; hence find AC 
thus : sin A : sin D = CD : CA. In the triangle BCE, find 
BC in a similar manner. Then in the triangle ABC are given 

AC, CB, and angle C ; hence 
find the angles at A and B by 
AC + CB : AC~CB = tan J(A + B) : 
tan |(A~B); then find A and B; 
and AB is found thus: sin A: 
sinC = BC :AB. 

After finding AC and CB in 
triangle ABC, AB may also be 
found independently of the angles 
A and B, for AB 2 = AC 2 + CB 2 2AB . BC . cos C. 

EXAMPLE. Find the distance between the two objects A and 
B, and their distances from C, from these measurements : 




In triangle ADC, 
CD =456 links, 
C =44 20 7 , 
D =87 56', 
ACB=8850'. 



In triangle BCE, 
CE = 524, 
C =50 24', 
E =89 40', 



1. In triangle ADC, 
A=180 - (C + D) = 180 - 132 16' = 47 44'. 



To find AC 



L cosec A 47 44', 
L sin D 87 56', 
L CD 456, . 

L AC 615-797, 



= 10-1307551 
9-9997174 
2-6589648 

2-7894373 



2. In triangle BCE, 
B=180 - (C + E) = 180 - 140 4' = 39 56'. 



To find BC 



L cosec B 39 56', 
L sin E 89 40', 
L CE 524, . 

L BC 816-318, 



= 10-1925354 
9-9999927 
2-7193313 



2-9118594 



MENSURATION OP HEIGHTS AND DISTANCES 87 

3. To find AB in triangle ABC 
By Art. 189, AB 2 = AC 2 + CB 2 - 2AC . BC . cos C. 

LAC 2 =2LAC, . . = 5 -5788746 = L 379206 
LBC 2 =2LBC, . . = 5 -8237 188 = L 666375 

Hence AC 2 + BC 2 , . . = 1045581 

L2, . 0-3010300 

L AC . BC = L AC + L BC, = 5 -7012967 
L cos C 88 50', . . = 8-3087941 

- 10- 



4-31 1 1208 = L 20470 -14 

Hence AB 2 , = 1025110'86 

And AB = \/1025110-86=1012-48. 

EXERCISES 

1. Find the distance of the two inaccessible objects A and B, 
and their distances from the station C, from these data : 

In triangle ADC, In triangle BCE, 

CD =424, CE = 640, 

C =40 10', C =56 10', 

D =85 25', E =84 30', 

ACB = 8920'; 
AB =1126-1, AC=519-685, and BC = 1005'08. 

2. Required the distance of the two inaccessible objects, A and 
B, from these data : 

In triangle ADC, In triangle BCE, 

C = 40 20', C = 36 25', 

D =11240', E =118 15', 

CD =1256, CE = 1480, 

and angle ACB = 108 24' ; . . AB = 4550 -92. 

242. Problem XII. Given the distances between three 
objects, and the angles subtended by them at a station, 
to find the relative position of the station, and its dis- 
tance from the objects. 

CASE 1. When the station is out of the triangle formed by 
lines joining the given objects, and the middle object is beyond 
the line joining the other two. 

Let A, B, C be the three objects, E the station, and m, n the 
given angles. 




88 MENSURATION OP HEIGHTS AND DISTANCES 

Describe the triangle ABC with the given distances, and make 
the angles m', n' respectively equal to the given angles m, n. Then, 
through ABD, describe a circle ABE ; draw CD, and produce it 
to E, and this point will be the station. Draw AE and BE. 

In triangle ACB the there sides are given ; 
hence angle A can be found. 

In triangle ADB the angles and AB are 
given ; hence AD can be found. 

In triangle ADC, AC and AD are given, 
and angle A= CAB - DAB ; hence angle v can 
be found. 

In triangle ACE the angles and AC are 
known ; hence AE and CE can be found ; 
then in triangle ABE the sides AB, AE and 
the angles are known ; and hence BE can be found. 

EXAMPLE. Let the distances AB, BC, and CA be respectively 
= 1727, 1793, and 1540, and the angles subtended at the station E 
by BC and AB respectively = 25 40' and 53 24'; what is the 
distance between the station and each of the objects ? 

Angle n = 25 40', and m=53 24' - 25 40' =27 44'. 

1. To find angle A in triangle ABC 
a=1793, & = 1540, c=1727, s=2530. 

Ls2530, . .'.-.. = 3-4031205 

L(s-)737, . . . 2-8674675 

10- L b 1540, . 6-8124793 

10-Lcl727, . 6-7627077 

2)19-8457750 

LcosJA33 9 8' 33", . . . = 9-9228875 
.-.A=6617' 6" 



2. To find AD in triangle ADB 
Angle D = 180 -(m' + ') = 180 -53 24' = 126 36'. 

L cosec D 126 36', . 10-0953832 

L sin B 27 44', . 9-6677863 

L AB 1727, = 3-2372923 

L AD 1001-06, = 3-0004618 



MENSURATION OF HEIGHTS AND DISTANCES 89 

3. To find angle C in triangle ADC 
r=A-n' = 66 17' 6" - 25 40' = 40 37' 6". 

C + D = 180 - r = 139 22' 54". 

AT. co. L( AC + AD) 2541 -06, . 6-5949850 

L( AC -AD) 538 94, . 2-7315404 

Ltan(C + D)6941'27", . . = 10'4316890 

Ltan JHC~D)2948'58", . . = 97582144 
.-. C = 39 52' 29" 

4. To find AB and CE in triangle ACE 

L cosec in 27 44', = 10-3322137 

L sin v 39 52' 29", . 9 '8069333 

L AC 1540, . 3-1875207 

LAE 2121-62, . 3-3266677 

Angle A = 180 - (m + v) = 180 - 67 36' 29" = 1 12 23' 31". 

L cosec m 27 44', . 10-3322137 

Lsin A 112 23' 31", . 9'9659537 

L AC 1540, . 3-1875207 

L CE 3059-76 = 3-4856881 

5. To find BE in triangle ABE 

Angle A = CAE-CAB = 112 23' 31" -66 17' 6"=46 6' 25". 
L cosec E 53 24', = 10'0953832 

L sin A 46 6' 25", . 9-8577155 

L AB 1727, = 3-2372923 

L BE 1550-21, = 3-1903910 

The distances, therefore, are AE = 2121-62, CE = 3059'76, and 
BE = 1550 -21. 

EXERCISE 

A, B, and C are three conspicuous objects in three towns. The 
distance of A from B = 125'6 furlongs, B from C = 130'4, and C from 
A = 112 furlongs; and at a station E, the distances AB and AC 
subtend angles = 48 58' and 25 52' ; required the distances of the 
station from the three objects. 

AE = 165-357, BE = 123'25, and CE = 234'462 furlongs. 

CASE 2. When the station is outside the triangle, and the 
middle object is on the same side of the line joining the other two. 



90 MENSURATION OP HEIGHTS AND DISTANCES 

Let the middle object C be between the station E and the line 
AB ; then the points E and D will be both outside the triangle 
ABC, and on opposite sides of it, and the solution will be analogous 
to that of the first case.* 

EXERCISE 

Let the distances of the objects be AB = 106, AC =65 '5, and 
BC = 53-25, angle BEC= = 13 30', and AEC = m 29 50'; what 
are the distances of E from A, B, and C ? 

= 131-06, 151-428, and 107 '42. 

CASE 3. When the station is inside the triangle. 

Let D be the station, then the angles ADC, BDC being given, 
their supplements ADE, BDE are also given. 

Make angles ABE, BAE respectively = ADE and BDE; then 
describe a circle about ABE ; draw CE, and it will cut the circle 
in the station D. 

If the station D is now marked E, and E is changed to D, the 
method described in the preceding case is exactly applicable to 
this; excepting that now angle CAD = CAB + DAB, and angle 
= 180 3 -(ACE + AEC), and angle BAE = CAB-CAE. 



EXERCISE 

The distances between three objects, taken in order, are BC = 
5340, AC = 6920, and AB = 4180 feet; and the angles, subtended by 
these distances at a point inside the triangle formed by them, are 
respectively EEC = 128 40', AEC = 140, and AEB=91 20' ; what 
are the distances of the objects from the station ? 

AE = 3577-1, CE = 3786-2, and BE = 2080. 

ADDITIONAL EXERCISES IN MENSURATION OF HEIGHTS AND 
DISTANCES 

1. From the bottom of a tower a horizontal line was measured 
= 230 links, and at its extremity the angle of elevation of the top 
of the tower was = 43 30' ; required its height? . =218 '262 links. 

2. At a horizontal distance of 170 feet from the bottom of a 
steeple, the angle of elevation of its top was =52 30'; what was 
the height of the steeple ? ..... =221-548 feet. 

3. Find the height of a precipice, its angle of elevation at two 
stations in a horizontal line with its base, and in the same vertical 
plane with its top, being=39 30' and 34 15', and the distance 
between the stations = 145 feet ..... = 567 '293 feet. 

* The student may draw a diagram to enable liiin to understand this case. 



MENSURATION OF HEIGHTS AND DISTANCES 91 

4. In order to find the height of a steeple, measurements were 
taken as in the preceding example ; the base was = 90 feet, and the 
angles of elevation were =28 34' and 50 & ; required its height, 
and the distance of the nearer station from it. 

Height=89-818 feet, and distance = 74 '9666. 

5. From the top of a tower 136-5 feet high, the angle of 
depression of the root of a tree at a distance on the same plane 
was = 22 40' ; what was the distance of the tree from the bottom 
of the tower? =326 '848 feet. 

6. From the summit of a hill, 360 feet high above a plain, the 
angles of depression of the top and bottom of a tower standing 
on the same plain were = 41 and 54; required the height of the 
tower =132 -63 feet. 

7. From the summit of a lighthouse 85 feet high, standing on a 
rock, the angle of depression of a ship was =3 38', and at the 
bottom of the lighthouse the angle of depression was =2 43' ; find 
the horizontal distance of the vessel and the height of the rock. 

=5296-47 and 251-319 feet. 

8. In order to find the distance between two objects, A and B, 
and their distances from a station C, the following measurements 
were taken, as in Prob. XL namely, CD = 200 yards, CE = 200 
yards, angle ACD = 89, ADC =53 30', BCE=5430', EEC = 88 30', 
and ACB = 72 30' ; what are the distances ? 

AB = 356-86, AC = 264'096, and BC = 332'214 yards, 

9. The distances between three objects, A, B, C, are known 
namely, AB = 12 miles, BC = 7'2 miles, and AC = 8 miles; and at a 
station between A and B, in the line joining them, from which 
the three objects were visible, the distance AC subtended an 
angle of 107 56' ; required the distances of this station from the 
three objects. . BD = 6'9984, DA = 5-001 6, and DC = 4-8908 miles. 

10. Three conspicuous objects, A, B, C, whose distances are 
AB = 9, BC = 6, and AC = 12 miles, were observed from a station 
D, from which B appeared to be the middle object, and lay beyond 
the line joining A and C ; at this station the distances AB, BC 
subtended respectively angles of 33 45' and 22 30' ; what is the 
distance of the station from the objects ? 

AD = 10-663, BD = 15-641, and CD = 14-0107 miles. 

11. From the top of a mountain I observe two milestones on the 
level ground in a straight line from one another, and I find their 
angles of depression to be 5 and 15 respectively. Determine the 
height of the mountain. =228 '64 yards. 

12. The cone of dispersion of a shrapnel-shell bursting 100 yards 



92 MENSURATION OF HEIGHTS AND DISTANCES 

short of an object is found to be 30. What is the front covered ? 
Neglect height of burst above horizontal plane. . = 53'72 yards. 

13. A castle is situated on the top of a hill whose angle of in- 
clination to the horizon is 30 ; the angle subtended by the castle 
to the foot of the hill is found to be 15, and on ascending 485 feet 
up the hill the castle is found to subtend an angle of 30. Find the 
height of the castle, and the distance of its base from the foot of 
the hill. 

Height of castle = 280'02 feet. Distance of its base from foot 
of hill = 765-01 feet. 

14. A and B are two stations on a hillside ; the inclination of the 
hill to the horizon is 30; the distance between A and B is 500 
yards. C is the summit of another hill in the same horizontal 
plane as A and B, and on a level with A ; but at B its elevation 
above the horizon is 15. Find the distance between A and C. 

= 1366 -025 yards. 

15. A man standing at a certain station on a straight sea-wall 
observes that the straight lines drawn from that station to two 
boats lying at anchor are each inclined at 45 to the direction of 
the wall, and when he walks 400 yards along the wall to another 
station he finds that the former angles of inclination are changed 
to 15 and 75 respectively. Find the distance between the boats, 
and the perpendicular distance of each from the sea-wall. 

= 156 '4 yards; 556 '4 yards. 

16. If the ratio of two sides of a triangle is 2 + V3, and the 
included angle is 60, find the other angles. . =105 and 15. 



MENSURATION OF SURFACES 

243. The length and breadth of a surface are called its 
dimensions. 

The dimensions of a surface are straight lines, and are 
therefore measured by some lineal unit as an inch, a foot, a 
yard (Art. 218). 

244. The unit of measure of surfaces, called the unit of 
superficial measure, or the superficial unit, is the square 
of the lineal unit. 

Thus, if an inch is the lineal unit, a square inch that is 
(Art. 44), a square whose sides are each one inch is the 



MENSURATION OF SURFACES 93 

superficial unit; when the lineal unit is a foot, the super- 
ficial unit is a square foot ; so if the former unit is a yard, 
the latter unit is a square yard. 

245. The quantity of surface of a figure is called its area 
or superficial content, and is the number of superficial units 
it contains. 

If a surface is = 20 square feet that is, 20 times a square 
whose side is one foot its content is = 20 square feet. The 
superficial content of any plane figure is not immediately 
found, however, by applying to it the superficial unit ; as, for 
instance, a square foot, in the way that a line is measured by 
directly applying to it the lineal unit ; for this method would 
be very tedious, and incapable of much accuracy ; but the 
content can be computed by certain rules, given in the follow- 
ing problems, with the greatest precision, when the dimensions 
of the figure are accurately known. 

246. It is necessary to make a distinction between some 
expressions relating to superficial measure that on first con- 
sideration appear to be equivalent. Thus, 2 square inches and 
2 inches square are very different ; for the former expression 
can mean only one square inch taken twice, whereas the 
latter means a square described on a line 2 inches long, so 
that its sides are each 2 inches, and its content, as will be 
found by Problem IV., is 4 square inches. So 10 square 
inches are very different from 10 inches square, which, 
according to the same problem, contains 100 square inches. 

247. Problem I. To find the area of a rectangle when 
its length and breadth are given. 

RULE. Multiply the base by the perpendicular height, and the 
product is the area. 

Let JR = the area, b = the base, and h = the height ; 
then M = bh, or LJR = Lb + U. 

Hence b=-^--, and h= -=-. 

ti b 

If CE is a rectangle, and M the lineal unit as, for example, 
a foot and if the base CD contains M 4 times, and the side DE 




94 MENSURATION OF SURFACES 

contains it 3 times, the number of squares described on M that are 
contained in CE is just=4 x 3 = 12 square feet. 

For, by laying off parts on CD, DE equal to 
M, and drawing through the points of division 
lines parallel to the sides of the figure, it will 
evidently be divided into 3 rows of squares, each 
containing 4 squares; that is, 3x4=12 squares 
or square feet. 

If the side CD contained 4^ inches, and DE 
3 inches, it would similarly be found that the number of square 
inches in the figure would be = 4x 3 = f x 3 = 13J square inches; 
or 4'5x3 = 13'5 square inches; and whatever is the length of the 
sides, the area is found always in the same manner. 

EXAMPLES. 1. How many square inches are in a leaf of paper 
which is = 10 inches long and 6 broad ? 

^H = bh = 10 x 6J = 65 square inches. 

2. How many square feet are there in a table which is = 10 feet 
5 inches long and 3 feet 8 inches broad ? 

b = 10 ft. 5 in. = 10 T \ ft., and h = 3 ft. 8 in. = 3 ft. 
l=M= 10A x 3 = Vz (l * -V- =J i" = 38-194 sq. feet 

= 38 sq. feet 28 sq. inches. 
For -194x144=28. 

Or 6 = 125 inches, and A=44 inches. 

JR = bk = 125x44 =5500 sq. in. =4,^ = 38 sq. ft. 28 sq. in. 

3. Find the number of square yards in the ceiling of a room 
which is = 24 feet 9 inches long and 15 feet 6 inches broad. 

6 = 24 ft. 9 in. =24 -75, and h = 15 ft. 6 in. = 15 -5. 
JR = bh = 21-75 x li> '5 = 383-625 sq. ft. =383 sq. ft. 90 sq. in. 

=42 sq. yd. 5 sq. ft. 90 sq. in. 
Or L^l = L6 + LA = L24-75 + L 15 -5 = 1-3935752 + 1-190331 7 

= 2-5839069 ; . -. ^R = 383'625 sq. ft. =383 sq. ft. 90 sq. in. 

4. Required the area of a rectangular field whose length is =24 -5 
chains and breadth = 8 '5 chains. 

-l=&/i=24-5x8-5=208-25 sq. chains =20 '825 acres 

= 20 acres 3 roods 12 sq. poles. 

Or M = 2450x850 sq. links = 2082500 sq. links = 20'825 ac. ; for 10 
square chains or 100,000 square links make one acre. 

EXERCISES 

1. Required the number of square inches in a sheet of paper 
which is =20 inches long and 15 inches broad. =300 square inches. 



MENSURATION OF SURFACES 95 

2. How many square feet are in a rectangular table, the length 
of which is =10 feet 6 inches and breadth =4 feet 3 inches? 

= 44 square feet 90 square inches. 

3. Required the number of square feet in a rectangular board 
whose length is = 12 feet 6 inches and breadth = 9 inches. 

= 9 '375 square feet. 

4. What is the number of square feet in a piece of carpeting 
= 14 feet 6 inches long and 4 feet 9 inches broad ? 

= 68 square feet 126 square inches. 

5. How many square yards of painting are there in the ceiling 
of a room whose length is=24 feet and breadth = 15 feet 6 inches? 

=41 square yards 3 square feet. 

6. Required the number of square yards in a floor = 16^ yards 
long and 10* yards broad =168 -3 square yards. 

7. Find the area of a rectangle = 27 feet 3 inches long and 1 foot 
6 inches broad. . . . =40 square feet 126 square inches. 

8. Find the number of square yards in a rectangular piece of 
ground = 162 feet 3 inches long and 32 feet 5 inches broad. 

=584 square yards 3 square feet 87 square inches. 

9. What is the number of square yards of painting in the side 
and end walls of a room, the circumference of the room being=103 
feet 2 inches and height = 10 feet ? 

= 114 square yards 5 square feet 96 square inches. 

10. Find the area of a rectangular field whose length is = 33 '4 
chains and breadth = 7 '5 chains. . =25 acres 8 square poles. 

11. How many acres are in a rectangular field, the length of 
which is = 2750 links and breadth = 190 links? 

= 5 acres 36 square poles. 

248. Problem II. To find the area of a rectangle when 
the base and diagonal are given. 

RULE. Find the other side (Art. 182), and then find the area by 
last problem. 

Let the diagonal AD = d ; 
then h= \J(d? ft 2 ), or h = \J(d+b}(d-b). 
Then JR = bh = b\/(d+b)(d-b) ; 

EXAMPLE. Find the area of a rectangle whose base and 
diagonal are respectively = 100 and 125 feet. 

h= V(P - & 2 ) = V(125 2 - 100 2 ) = V(15625 - 10000) 




and 2Si = bh = 100 x 75 = 7500 sq. feet = 833 sq. yards 3 sq. feet. 



96 MENSURATION OP SURFACES 

Or 



= L 100 + i{L 225 + L 25} = 2 + (2-3521825 + 1-3979400) 

= 2 + x 3-7501225 = 3-8750612 = L 7500. 
Hence .51 = 7500 sq. feet = 833 sq. yards 3 sq. feet. 

EXERCISES 

1. Find the area of a rectangle whose base and diagonal are 
respectively = 21 and 35 feet. . ... . =588 square feet. 

2. How many acres are contained in a rectangle whose diagonal 
is = 320 yards and base = 240 yards? 

= 10 acres 1 rood 39 '28 square poles. 

3. Find the area of a rectangular field whose base and diagonal 
are = 480 and 720 links. . =2 acres 2 roods 12-152 square poles. 

249. Problem III. To find the area of a rectangle when 
a side, or the diagonal, and the inclination of the diagonal 
and a side are given. 

RULE. Find the other side or the other two sides by Art. 180, 
and then find the area by Prob. I. 

Let angle DAB = v (fig. to Prob. II.), and let the other symbols 
remain as before. 

1. To find h when b and v are given. 

Taking natural tangents (Chambers's Math. Tables, Art. 29), 

1 : tan v = b : h, and h = b tan v ; 
then JR = bh = b 2 tan v ......... [1], 

or L^l=2L6 + L tan v-10 ......... [2]. 

2. To find b and h when d and v are given. 
By natural sines, b = d cos v, and h = d sin v ; 

then 1R = bh = d 2 sin v . cos v = \(J? sin 2v (Art. 204, a) ... [3], 
or L2^l = 2LflJ + Lsm2w-10 ...... [4]. 

Use the formula [1] or [2] when 6 and v are given, and [3] or 
[4] when d and v are given. 

EXAMPLES. 1. Find the area of a rectangle, the base of which 
is 36 feet, and the inclination of the base and diagonal = 32 25'. 
By [1], M = 6 2 tan v = 36 2 tan 32 25' = 36 2 x -6350274 

= 1296x -6350274 = 822-9955 sq. feet. 
Or by [2], L^l = 2L6 + L tan v - 10 = 2L 36 + L tan 32 25' - 10 

= 3-1126050 + 9-8027925-10 = 2-9153975; 
hence ;R = 822 -995 sq. feet. 

2. Find the area of a rectangular field, the diagonal of which is 
= 475 links, and its inclination to the longer side = 36 45'. 



MENSURATION OP SURFACES 97 

Here rf=475, and v=3Q 45', or 2v = 73 30'. 
By [3], ^B-X sin 2v=% x 475 2 x -9588197 

= i x 225625 x -9588197 = 108166 '85 sq. links. 
Or L 2.3l = 2Lrf+L sin 2y- 10 =5 -3533872 + 9 '98 17370- 10 

= 5-3351242 = L 216334 ; 
And .51 = 108167 sq. links = 1 ac. 13 sq. poles. 

EXERCISES 

1. What is the area of a rectangle, the base of which is = 14 '4 
yards, and the inclination of the base and diagonal = 35 40' ? 

= 148 '82 square yards. 

2. Find the number of acres in a rectangular field, its diagonal 
being = 470 links, and its inclination to the longer side =42 45'. 

= 1 acre 16'17 square poles. 

250. Problem IV. To find the area of a square when the 
side of it is given. 

RULE. The square of the side is the area, or twice the logarithm 
of the side is the logarithm of the area. 

Let s= a side of a square, 

then ^R=s 2 , or L^=2Ls. 

EXAMPLES. 1. Required the area of a square whose side is 
=25 feet 

JR = 52 = 25 2 = 625 sq. feet. 

2. What is the area of a square whose side is = 425 links ? 

^R=2=425 2 = 180625 sq. links = l ac. 3 ro. 9 sq. poles. 
Or L^R = 2Ls = 5 -2567778 = L 180625. 



EXERCISES 

1. What is the area of a square, the side of which is = 11 feet 
6 inches? ....... Area =132^ square feet. 

2. What is the number of square yards in a square whose side 
is = 16 feet 8 inches ? 

= 30 square yards 7 square feet 112 square inches. 

3. How many square yards are contained in a square, the side of 
which is = 31 feet? . . . = 106 square yards 7 square feet. 

4. What is the area of a square whose side is = 12 feet 6 inches? 

= 156 square feet 36 square inches. 

5. Find the number of square yards in a square court, the side of 
which is = 160 feet 6 inches. . . . = 2862^ square yards. 

6. How many acres are in a square field whose side is = 705 
links? .... =4 acres 3 roods 35 -24 square poles. 



98 MENSURATION OP SURFACES 

7. Find the number of acres in a farm of a square form, the side 
of winch is = 1 mile ........ =640 acres. 

251. Problem V. To find the area of a square when its 
diagonal is given. 

RULE. The area is equal to half the square of its diagonal ; or 
the logarithm of twice the area is equal to twice the logarithm of 
the diagonal. 

For (Eucl. I. 47), d?=2s 2 =2M, 
or JR = |cP, and L 2^1 = 2Ld. 

EXAMPLES. 1. Find the area of a square whose diagonal is 
= 45 feet. 

. feet. 



2. Find the area of a square field, its diagonal being = 524 links. 

JR = \&= J x 524 2 =i x 274576 = 137288 sq. links. 
Or L 2^51 = 2 Ld= 5 "4386626 ; 

hence 2^1 = 274576, and .31 = 137288 sq. lk. = l ac. 1 ro. 19'66 sq. po. 

EXERCISES 

1. What is the area of a square whose diagonal is = 25 yards? 

= 312 J square yards. 

2. How many square yards in a courtyard, the diagonal of which 
is = 124 feet? .... = 854 square yards 2 square feet. 

3. How many acres are in a square field, the diagonal of which 
is = 786 links? .... =3 acres 14'2368 square poles. 

252. Problem VI. To find the area of a parallelogram 
when its base and altitude are given. 

RULE. Multiply the base by the height, and the product is 
=the area. 

This is evident from Eucl. I. 35. 

JR=l)h, or L^B = L6 + LA. 

EXAMPLES. 1. Find the area of a parallelogram whose base and 
perpendicular breadth are respectively =24 and 18 feet. 
JR = bh-24: x 18 = 432 sq. feet =48 sq. yards. 
2. Find the area of a field in the form of a parallelogram whose 
base and height are respectively = 428 and 369 links. 
M = bh = 428 x 369 = 1 57932 sq. links. 

Or L^l = L& + L/t=2-6314438 + 2-5670264 = 5-1984702; 

hence .51=157932 sq. links = l acre 2 roods 12-6912 sq. poles. 



MENSURATION OP SURFACES 99 

EXERCISES 

1. What is the area of a parallelogram whose length is =25 feet 
3 inches, and height =13 feet? . . . = 328 J square feet. 

2. How many square yards in a parallelogram whose length is 
=45 feet, and breadth = 24 feet? . . = 120 square yards. 

3. What is the area of a parallelogram whose base and height 
are = 625 and 240 links ? =1 acre 2 roods. 

4. How many acres are contained in a farm of the form of a 
parallelogram, the length and breadth of which are = 48 and 28 
chains? .... =134 acres 1 rood 24 square poles. 

5. What is the area of a field in the form of a parallelogram 
whose length and perpendicular breadth are = 2102 and 1284 links? 

=26 acres 3 roods 38-3488 square poles. 

The examples under Prob. I. are performed by the same rule, 
for parallelograms and rectangles whose lengths and perpendicular 
breadths are equal have equal areas. (Eucl. I. Prop. 35, 36.) 

253. Problem VII. To find the area of a parallelogram, 
when there are given two of its adjacent sides and the 
contained angle. 

RULE. The area is equal to the continued product of the two 
sides and the natural sine of the contained angle ; or, 

The logarithm of the area is equal to the sum of the logarithms 
of the two sides and of the sine of the contained angle diminished 
by 10. 

Let AB = 6, AC=s ; if these and the contained c p 

angle i are given, then 

M = bs sin t, 
or L^3l = L6 + Ls + L sin f - 10. 

For if CE = A, then h=s sin t. 

But (Art. 252), M=bh=bs sin i [1], 

from which L^l = L6 + Ls + L sin e-10 [2]. 

EXAMPLE. What is the number of square feet in a parallelo- 
gram whose length is = 42 feet 6 inches, the adjacent side =21 feet 
3 inches, and the contained angle = 53 30' ? 
Here b = 42 feet 6 inches = 42 -5, 

s=21 feet 3 inches = 21'25. 

Hence & = bs sin t = 42 -5 x 21 25 x -8038569 = 725 -983 sq. feet. 
Or L^l = L6 + L + Lsin A -10 = 1-6283889 + 1-3273589 

+ 9 -90;") 1 787 - 1 = 2 "8609265 ; 
hence .31 = 725 -983 sq. feet. 



100 MENSURATION OP SURFACES 

EXERCISES 

1. Find the area of a rhomboid, two of whose adjacent sides are 
= 18 feet and 25 feet 6 inches, and the contained angle = 58. 

= 389 -254 square feet. 

2. What is the area of a field of the form of a parallelogram, 
two of whose adjacent sides are =1200 and 640 links, and the 
contained angle = 30? . . =3 acres 3 roods 14 '4 square poles. 

3. What is the area of a rhombus whose sides are = 42 feet 
6 inches, and the acute angles = 53 20' ? =1448 '835 square feet. 

4. Find the area of a field in the form of a rhombus, the sides of 
which are = 420 links, and the acute angles = 54 30'. 

= 1 acre 1 rood 29 '776 square poles. 

5. Find the area of a field in the form of a parallelogram, two of 
whose adjacent sides are = 750 and 375 links, and the included 
angle = 60 ..... =2 acres 1 rood 29 '7 square poles. 

6. What is the area of a rhomboidal field whose sides are = 5070 
and 1040 links, and the acute angles = 30 ? 

= 26 acres 1 rood 18 '24 square poles. 

7. Find the area of a field in the form of a parallelogram, two of 
whose adjacent sides are = 1245 and 864 links, and the contained 
angle = 65 40'. . . . = 9 acres 3 roods 8'192 square poles. 

254. Problem VIII. Given the diagonals of a parallelo- 
gram and their inclination, to find its area. 

RULE. Half the continued product of the diagonals and the 
natural sine of the contained angle is equal to the area ; or, 

Add together the logarithms of the two diagonals and the 

logarithmic sine of the contained angle, and from their sum 

subtract 10 ; the remainder will be the logarithm of twice the area. 

c D Let the diagonals AD, CB be denoted by d 

/"x ..-'"/ an d ^' an d their inclination or angle DIB by i ; 
...-' x: ... / then A. = \dd' sin i ... [1], 

A - B or L2A = Ld+Ld' + ~Lsmi-10 ... [2]. 



EXAMPLES. 1. Find the area of a square whose diagonal is 
= 20 feet 3 inches. 

Here d=d' =20-25 feet, and i=90 ; 

hence M = $dd' sin 90 = ^ x 1= x 20'25 2 =205'03125 sq. feet. 
Or L 21R = 2Ld + L sin *- 10'=2'6128500 + 10- 10 

=2-6128500 = L 410'0625, and ^ = 205 "03125 sq. feet. 

2. What is the area of a parallelogram whose diagonals are 
= 1245 and 1040 links, and the contained angle =28 45'? 



MENSURATION OF SURFACES 101 

&=\dd' sin i=i x 1245 x 1040 x -4809888 

=31 1392 sq. links, 
L 2&=Ld+Ld' + L sin i- 10 = 3-0951694 + 3-0170333 

+ 9-6821349- 10 =5 '7943376 = L 622784, 

and yR = 311392- sq. links = 3 acres 18-23 sq. poles. 

The diagonals AD, BC bisect each other; and hence AI = ID, 

and therefore (Eucl. I. 38) the triangles AIC, IDC are equal, and 

also AIB and IDE ; but (Eucl. I. 34) the diagonal AD bisects 

the parallelogram, and therefore these four triangles are equal. 

But (Prob. X.) the area of the triangle DIB is = BI.ID sin i 

=^.5. sin i; hence the area of the four triangles is four times 

' 



this quantity, or 2& = \dd' sin i; 

hence L 21R=Ld + Ld' + L sin i- 10. 



EXERCISES 

1. Find the number of square yards of pavement in a square 
court, the diagonal of which is = 36 feet 8 inches. 

= 74*69 square yards. 

2. How many square yards are contained in a rectangular field 
whose diagonals are each = 96 feet, and contain an angle =30? 

=256 square yards. 

3. Find the area of a rhombus whose diagonals are = 75 and 
60 feet ..... . ., . . =250 square yards. 

4. What is the number of square feet in a rhomboidal piece of 
ground, the diagonals of which are = 90 aud 50 feet, and their 
inclination = 60 ? . -.* . . * ' .. =1948-557 square feet. 

5. How many acres are contained in a rhomboidal field whose 
diagonals are inclined at an angle of 36 40', and their lengths 
= 875 and 480 links ? . =1 acre 1 rood 0'645 square pole. 

255. Problem IX. To find the area of a triangle when 
its base and altitude are given. 

RULE. The area is equal to half the product of the base and 
height ; or, 

The logarithm of twice the area is equal to the sum of the 
logarithms of the base and height. 

Let 6 = the base, and A = the altitude ; 
then JR=\bh, and L 2JR = Lb + Lh. 

The truth of the rule is evident from the rule in Prob. VI. , and 
the fact that a triangle is half of a parallelogram of the same base 
and altitude. (Eucl. L 41.) 



102 MENSURATION OF SURFACES 

EXAMPLES. 1. Required the number of square feet in a triangle 
whose base is = 25 feet and altitude = 36 feet. 

JR = %bh = $x 25 x 36 = 25 x 18=450 sq. feet. 

2. Find the number of square yards in a triangular field, one of 
whose sides is = 240 feet, and the perpendicular upon it from the 
opposite angle = 125 feet. 

2&=lbh = ix 240 x 125 = 15000 sq. feet = 1666 sq. yards. 

3. How many acres are contained in a triangular field, one of its 
sides being = 1248 links, and the perpendicular upon it from the 
opposite corner = 945 links? 

& = $h = b x 1248 x 945=589680 sq. links 
= 5 acres 3 roods 23 '488 sq. poles. 

Or L 21R=U x LA = 3-0962146 + 2 -9754318 

= 6-0716464 = L 1179360, 

and .1=589680 = 5 acres 3 roods 23 '488 sq. poles. 

EXERCISES 

1. What is the area of a triangle whose base is = 128 feet, and 
height =40 feet ? =2560 square feet. 

2. What is the area of a triangle whose base is = 21 feet 6 inches, 
and height=14 feet 6 inches? =155 square feet 126 square inches. 

3. Required the number of square yards in a triangle, the base of 
which is = 49 feet 6 inches, and the perpendicular =42 feet 9 inches. 

= 117'5625 square yards. 

4. Find the area of a triangle whose base is = 60 feet, and 
perpendicular =10-25 feet. .... =307 '5 square feet. 

5. The length of one side of a triangular field is = 160 yards, and 
the perpendicular on it from the opposite angle is = 140 yards; 
required its area =11200 square yards. 

6. Required the area of a right-angled triangle whose base is 
= 225 feet, and perpendicular = 160 feet. =2000 square yards. 

7. How many acres are contained in a triangular field whose 
base and height are = 1225 and 425 links ? 

= 2 acres 2 roods 16'5 square poles. 

8. Required the area of a triangular field whose base is = 10 
chains, and height = 726 -184 links. 

= 3 acres 2 roods 20 '9472 square poles. 

9. One side of a triangular court is = 97 feet, and the perpendicular 
on it from the opposite angle is = 61 feet ; required the expense of 
paving it, at 2s. 3d. the square yard. . , . =36, 19s. 7d. 



MENSURATION OF SURFACES 103 

256. Problem X. To find the area of a triangle when 
two of its sides and the contained angle are given. 

RULE. The area is equal to half the product of the two sides 
and the natural sine of the contained angle ; or, 

The logarithm of twice the area is equal to the sum of the 
logarithms of the two sides, and of the sine of the contained angle, 
diminished by 10. 

Or if b base, s=a side, and z= included angle, then 

JR^bs. sin i. 
Or L2M = Lb + Ls + Lsini-lO. 

The rule is evident from Prob. VII., for the area of the triangle 
ABC is half that of the parallelogram AD. 

EXAMPLES. 1. Find the area of a triangle which has two sides 
= 125 and 80 feet, and the contained angle = 28 35'. 

JR=$s sin 1 = 4 x 125 x 80 x -4784364 =2392 '182 feet. 

2. How many acres are contained in a triangular field, two of 
whose sides are = 625 and 640 links, and the contained angle = 
40 25' ? 

1R = $bs sin i=\ x 625 x 640 x -6483414= 129668 sq. links 

= 1 acre 1 rood 7 '469 sq. poles. 
Or L 2^R=L6 + Ls + L sin i- 10=2-7958800 + 2-8061800 + 9-8118038 

- 10=5-4138638 = L 259336; 
hence .1=129668 -5 = 1 acre 1 rood 7 '469 sq. poles. 

EXERCISES 

1. Required the area in square yards of a triangle, two of whose 
sides are = 50 feet and 42 feet 6 inches, and the contained angle 
=45 =83-47788 square yards. 

2. How many square yards are contained in an isosceles 
triangle, the equal sides being =50 -49 feet, and the contained 
angle =45? .... =100 square yards 1 '29 square feet. 

3. Find the area of a triangle, two of whose sides are = 80 and 90 
feet, and the contained angle =28 57' 18". =1742-84 square feet. 

4. How many square yards are contained in a triangle, two of 
whose sides are=42J and 75 yards, and the included angle = 50? 

= 1220-88 square yards. 

5. How many square yards are contained in a triangular field, 
two of whose sides are = 204J and 146 yards, and the contained 
angle = 30? =7498J square yards. 

6. How many acres are contained in a triangular field, two of 



104 MENSURATION OP SURFACES 

whose sides are = 1500 and 6400 links, and the contained angle 
= 39 36' ? . . . . =30 acres 2 roods 15-416 square poles. 

257. Problem XI. To find the area of a triangle when 
the three sides are given. 

RULE. Find half the sum of the three sides, and also the differ- 
ence between the half-sum and each of the sides ; then find the 
continued product of the half-sum and the three differences, and 
the square root of the product will be the area ; or, 

Add together the logarithms of the half-sum and of the three 
differences, and half the sum will be the logarithm of the area. 

Let the three sides be denoted by a, b, and c, and half their sum 
bys;thatis, s = ^(a + b + c); 

then M = \/{s(s-a)(s-b)(s-c)}. 

Or L, JR = %{Ls + L(s - a) + L(s-b) + L(s- c)}. 

EXAMPLE. Find the area of a triangular field whose three sides 
are =4236, 2544, and 3650 links. 

Let a = 4236 then s = 5215 

b = 2544 s-a = 979 

c = 3650 s-b = 2671 

2s = 10430 S ~ C = 



And &, = \Js(s -a)(s- b)(s -c) = \J5215 x 979 x 2671 x 1565 
= V21341514430775 = 4619687. 

Or Ls5215, = 3-7172543 

L(s-a)979, = 2-9907827 

L(s-6)2671 = 3-4266739 

L(s-c)1565, = 3-1945143 

2)13-3292252 

LM 4619687, . 6'6646126 

Hence area = 46 acres 31 -5 sq. poles. 

For the demonstration, see Eucl. IT. 13, Note. 
Or if the sides and angles of the triangle ABC 
be denoted, as in Art. 210, and CT) h, it is 
proved in Trigonometry (Art. 215, c) that 

2 
sin A = j- \/{s(s - a)s( - b)(s - c)}. 

OC 

But if CD = h, then h = b sin A, 
and ^l=^AB. CD = icA = |6c.sin A; 

hence, substituting for sin A the above value, 
JR = \/{s(s-a)(s-b)(s-c)}. 




MENSURATION OP SURFACES 105 

EXERCISES 

1. What is the number of square yards in a triangle whose sides 
are = 90, 120, and 150 feet? . . . . = 600 square yards. 

2. Find the area of a triangle whose sides are = 200, 150, and 
250 feet ......... = 15000 square feet. 

3. How many square yards are in a triangular field whose sides 
are = 126, 247, and 296 yards ? =15328 square yards. 

4. Find the number of square yards in a triangular court whose 
sides are = 45, 42, and 39 yards. . . . =756 square yards. 

5. What is the area of a triangular field whose sides are = 1000, 
1500, and 2000 links ? . =7 acres 1 rood T89 square poles. 

6. Find the area of a triangular field whose sides are = 1200, 
1800, and 2400 links. . =10 acres 1 rood 33'128 square poles. 

7. What is the area of a triangular field whose sides are = 2569, 
5025, and 4900 links? . . =61 acres 1 rood 39 '68 square poles. 

258. When the triangle is equilateral, if 6=its base, then 



hence JR= b = z =b^Z= '433^, nearly. 



And if M = the area, b = V^V27 = 1 '52 \JJR, nearly. 

Find the area of an equilateral triangle whose side is = 16. 
M= -4336 2 = -433 x 16"= '433 x 256 = 110-848. 

Find the area of a field in the form of an equilateral triangle, the 
side of which is = 12 '5 yards. . . . =67 '658 square yards. 



259. Problem XII. To find the area of a trapezium. 

RULE. Multiply the sum of the parallel sides by the perpendicu- 
lar distance between them, and half the product is the area ; or, 

Add together the logarithms of the sum of the parallel sides and 
of the perpendicular distance between them, and the sum is the 
logarithm of twice the area. 

Let ABDC be a trapezium, and let the 
parallel sides AB, CD be denoted by b and *, 
and their perpendicular distance CH by h ; 
then 



r~ti 

or L 2JR = L 



EXAMPLES. l^Wtiat is the area of a trapezium whose parallel 
sides are = 34 andife.feet, and their perpendicular distance = 25 feet? 
s)h = i(34 -f- 26)25 = i x 60 x 25 = 750 sq. feet. 



106 MENSURATION OF SURFACES 

2. What is the area of a trapezium of which the parallel sides 
are = 1025 and 836, and their perpendicular distance = 650 links? 

M = l(b + s)h = % x 1861 x 650 = 1861 x 325 = 604825 sq. links. 
Or L2^l = L 1861 + L 650 = 3-2697464 + 2-8129134 

= 6 -0826598 = L ] 209650 ; 
and hence JR= 604825 sq. links = 6 acres 7 '72 sq. poles. 

If DB be bisected in E, and FG be drawn parallel to AC, then 
GB will be equal to DF, and triangle DEF to BEG (Euclid I. 15, 
29, and 26) ; and hence AG is half the sum of AB and CD, and the 
area of the parallelogram AF is equal to that of the trapezium. 
But the area of AF is = AG. CH ; hence the rule is evident. 

EXERCISES 

1. Find the area of a trapezium whose parallel sides are = 30 and 
40 feet, and perpendicular breadth = 15 feet. . =525 square feet. 

2. How many square yards of paving are contained in a court 
of the form of a trapezium, the parallel sides being = 45 and 63, and 
their perpendicular distance = 25? . . . = 150 square yards. 

3. How many square feet are there in a trapezium whose parallel 
sides are = 643-02, 428-48, and perpendicular distance 342-32? 

= 183397-95 square feet. 

4. Find the area of a trapezium whose parallel sides are = 41 and 
24'5 feet, and perpendicular distance = 43. =1408-25 square feet. 

5. How many square feet are contained in a trapezium whose 
parallel sides = 24 feet and 36 feet 8 inches, and the perpendicular 
distance between them = 21 feet? . . . =637 square feet. 

6. How many square yards are contained in a trapezium whose 
parallel sides are = 54 and 60 feet, and their perpendicular distance 
= 30 feet? -'''- . " =190 square yards. 

7. The parallel sides of a trapezium are = 45 and 50 feet, and their 
perpendicular distance =25 feet; how many square yards does it 
contain? =131 square yards 8'5 square feet. 

8. The parallel sides of a trapezoidal field are =2482 and 1644 
links, and its perpendicular breadth is = 1030 links; what is its 
area? =21 acres 39 '824 square poles. 

9. Find the area of a trapezoidal field whose parallel sides are 
= 1500 and 2450 links, and breadth 770 links. 

= 15 acres 33-2 square poles. 

10. What is the area of a trapezoidal field whose parallel sides 
are = 750 and 975 links, and perpendicular breadth = 700 links? 

=6 acres 6 square poles. 



MENSURATION OF SURFACES 107 

260. Problem XIII. To find the area of any quadrilateral 
when its diagonals and their inclination are given. 

RULE. Multiply half the product of the diagonals by the natural 
sine of their inclination ; or, 

Add together the logarithms of the two diagonals and of the 
sine of the contained angle ; from the sum subtract 10, and the 
remainder will be the logarithm of twice the area. 

Let d and d' be the diagonals, and * the included angle ; 
then &'=\ dd' sin i , 

or L2^l=Lc?+Lc?' + Lsinz-10. 

EXAMPLE. Find the number of square yards in a quadrilateral 
whose diagonals are =420 and 325 feet, and the contained angle 
=40 25'. 

M=^dd' sin t = i x 420 x 325 x -6483414 = 44249-3 sq. feet. 
Or L 2M = Ld + LeP + L sin t - 10 = 2 -6232493 + 2 -51 18834 

+ 9-8118038- 10 =4 -9469365 = L 88498 "6 ; 
and .51 = 44249 -3 sq. feet = 4916 sq. yards 5-3 sq. feet. 

Let ABDC be the given quadrilateral, and AD, BC its diagonals. 

Through its angular points draw lines parallel to its diagonals, 
and they will form the parallelogram EG, whicli is evidently 
double of the quadrilateral ; for the parallelogram IG is double 
of the triangle AIB (Eucl. I. 34), and so of the other four 
parallelograms that compose EG ; also angle F = DIB = t. 

Now, the area of EG (Article 253) is = EF . FG . sin F ; 
hence ABDC = \ AD . CB . sin i, 

for AD = EF, and BC = FG (Eucl. I. 34). 

Hence & = \dd' sin i ; 

or L 2M = Ld+Ld' + L sin i-10. 




EXERCISES 

1. What is the area of a quadrilateral whose c 
diagonals are = 50 and 40 feet, and the included 

angle = 60? . . . . . . =866-0254 square feet. 

2. How many square yards are contained in a court, the diagonals 
of which are = 180 and 210 feet, and the contained angle 30? 

= 1050 square yards. 

3. Find the number of acres contained in a quadrilateral field 
whose diagonals are = 1500 and 2000 links, and their inclination 
= 48 =11 acres 23-55 square poles. 

4. How many acres are contained in a quadrilateral field whose 
diagonals are = 30 and 40 chains, and the contained angle = 60? 

= 51 acres 3 roods 33-84 square poles. 



108 MENSURATION OF SURFACES 

261. Problem XIV. To find the area of a quadrilateral 
that can be inscribed in a circle ; that is, one whose oppo- 
site angles are supplementary. 

RULE. From half the sum of the four sides subtract each side 
separately ; find the continued product of the four remainders, and 
the square root of this product is the area ; or, 

Add together the logarithms of the four remainders, and half 
their sum is the logarithm of the area. 

Let a, b, c, d denote the sides, and s half their sum ; 
then s=%(a + b+c + d), 

and JR = \/{(s-a)(s-b)(s-c)(s-d)}; 

or L, JR = 4{L(s - ) + L(s - b) + L(s -c) + L(s - d)}. 

EXAMPLE. What is the area of a quadrilateral inscribed in a 
circle whose four sides are =24, 26, 28, and 30 chains? 
a = 24 s - a = 30 

b = 26 s - b = 28 

c = 28 s - c = 26 

d = 30 s - d = 24 

2)108 



and area = V30 x 28 x 26 x 24 = V524160 = 723 -989 sq. chains 
=72 acres 1 rood 23 - 824 sq. poles. 

EXERCISES 

1. The four sides of a quadrilateral inscribed in a circle are = 40, 
75, 55, and 60 feet ; required its area. . =3146 '4265 square feet. 

2. How many acres are contained in a quadrilateral field whos? 
opposite angles are supplementary, its sides being = 600, 650, 700, 
and 750 links? . . . =4 acres 2 roods 3 '988 square poles. 

262. Problem XV. To find the area of a quadrilateral 
when one of its diagonals and the perpendiculars on it 
from the opposite angles are given. 

RULE. Multiply the diagonal by the sum of the perpendiculars, 
and half the product is the area ; or, 

Add the logarithms of the diagonal and of the sum of the per- 
pendiculars ; the sum will be the logarithm of twice the area. 

Let d be the diagonal, and p, p' the two perpendiculars on it; 
then M = \d( p +p') ; 

or L 2 M = Ld + L(p +p'). 




MENSURATION OP SURFACES 109 

EXAMPLE. How many acres are contained in a quadrilateral 
field, a diagonal of which is = 1245 links, and the perpendiculars 
on it from the opposite angles = 675 and 450 links? 
& = bd(P+P r ) = $x 1245(675 + 450) = | x 1245 x 1125 = 700312-5 sq. Ik. 

= 3-0951694 + 3'0511525 = 6-1463219 = L 1400625, 
and -51 = 700312-5 sq. links = 7 acres 0-5 sq. pole. 

Let ABCD be the quadrilateral, DB its diagonal, and CF, 
AE the two perpendiculars on it ; then (Art. 
255), t>_ 

triangle ADB = DB x |AE, 
and triangle DCB = DB x CF ; 
hence ABCD = i x DB(AE + CF). 

A 8 

EXERCISES 

1. How many square yards are contained in a quadrilateral, one 
of its diagonals being = 60 yards, and the perpendiculars upon it 
= 12'6 and 11-4 yards? =720 square yards. 

2. Find the area of a quadrilateral, one of its diagonals and the 
perpendiculars on it being respectively = 168, 42, and 56 feet. 

= 8232 square feet. 

3. Find the number of square yards in a quadrilateral which has 
a diagonal = 70 feet, and the perpendiculars upon it = 28 and 35 feet. 

=245 square yards. 

4. How many square yards in a quadrilateral, one of its diagonals 
being = 40 feet, and the perpendiculars on it=21'6 and 13 feet? 

= 76 square yards 8 square feet. 

5. Find the number of acres in a quadrilateral field, one of whose 
diagonals is = 4025, and the perpendiculars on it = 1225 and 1505 
links =54 acres 3 roods 30 '6 square poles. 

263. Problem XVI. To find the area of a quadrilateral 
when the four sides and the inclination of the diagonals 
are given. 

RULE. Add the squares of each pair of opposite sides together ; 
subtract the less sum from the greater ; then multiply the difference 
by the tangent of the angle formed by the diagonals, and one-fourth 
of this product is the area ; or, 

Add the logarithm of the remainder to the logarithmic tangent 
of the inclination of the diagonals, and the sum diminished by 10 
will be the logarithm of four times the area. 

Let the sides be denoted by a, b, c, d, and the inclination of the 



110 MENSURATION OP SURFACES 

diagonals by i; then if a and d are the opposite sides whose squares 
exceed those of the other two, 

M = {( 2 + #<) _ (#2 + C 2)j _ tan i . 
or L 41R = L{( 2 + d 2 ) - (6 2 + c 2 )} + L tan - 10. 

EXAMPLE. Find the area of a quadrilateral figure, two of whose 
opposite sides are = 10 and 12 chains, the other two sides = 9 and 18, 
and the inclination of the diagonals = 84 25'. 

a 2 + (^ = 81 +324=405, 6 2 + c 2 = 100 + 144 = 244; 
hence 2 + d 2 -(6 2 + c 2 ) = 405-244 = 161, 

and JR = x 161 tan 84 25' = i x 161 x 10-229428 = J x 1646 -9379 

= 411-7345. 
Or L 4^l = L 161 + L tan 84 25' - 10=2-2068259 + 11 "0098513 - 10 

= 3-2166772 = L 1646-94; 
and -1 = 411-735 sq. chains -41 acres 27 '76 sq. poles. 

EXERCISES 

1. Find the area of a quadrilateral, two of whose opposite sides 
are =500 and 400 links, the other two =450 and 350 links, and the 
inclination of the diagonals = 80. . =1 acre 32 -8 square poles. 

2. What is the area of a quadrilateral field, two of whose opposite 
sides are =450 and 900 links; the other two = 600 and 500 links; 
and the inclination of its diagonals = 78 40'? 

=5 acres 3*29 square poles. 

264. Problem XVII. To find the area of any quadri- 
lateral. 

RULE. Divide the quadrilateral into triangles, or triangles and 
trapeziums, calculate the areas of these component figures by 
former rules, and the sum of these partial areas will be the area 
of the whole figure. 

EXERCISES 

1. Find the area of the quadrilateral ABDC, the lines AE, EF, 

FB being =40, 64, and 28 feet, and the perpen- 
diculars CE, DF = 50 and 40 feet. 

Calculate the area of AEC by Prob. IX., that 
of CEFD by Prob. XII., and of DFB also by 
Prob. IX. ; and the sum is =4440 square feet, the 
area of ABDC. 

2. What is the number of acres in a quadrilateral field ABDC, 
the distances AE, AF, AB being = 420, 1160, and 1380 links, and 
the perpendiculars CE, DF = 840 and 680 links? 

= 8 acres 2176 square poles. 





MENSURATION OP SURFACES 111 

3. Given the four sides AB, BC, CD, DA of a quadrilateral 
field =650, 425, 470, and 580 ; the angle A = 85 40' 

and C = 112 15' ; to find its area. 

Find the area of the triangle ADB by Prob. X., 
and also that of DCB, and the sura of their 
areas is the area required. A 1 

= 2 acres 3 roods 8 '6 square poles. 

4. Find the area of the quadrilateral ABCD, its sides AB, BC, 
CD, and AD being = 720, 540, 520, and 600 links, and the angles A 
and C = 72 40' and 102 20'. =3 acres 1 rood 29 '36 square poles. 

5. Required the area of the quadrilateral figure ABCD, the sides 
AB, BC, CD, and AD being =1600, 1150, 1500, and 1650 links, and 
the diagonal AC = 1800 links. 

Find the areas of the two triangles ABC and ACD separately by 
Prob. XL, and their sum will be the area of the quadrilateral. 

=20 acres 2 roods 24^2 square poles. 

6. Find the area of the quadrilateral ABCD from these data : 

AB = 548 links, CD =751 links, 

BC=715 .. AD = 821 .. 

and the diagonal AC = 967 links. 

=4 acres 3 roods 27 "67 square poles. 

7. Find the area of the quadrilateral field ABCD, having given 

AB = 205 links, CD = 1000 links, 

BC = 700 AD= 600 , 

and the diagonal AC = 800 links. . =3 acres 10'37 square poles. 

8. How many acres are contained in a quadrilateral field, from 
these measurements : 

AB = 1 5 chains, CD = 14 chains, 

BC = 13 .. AD=12 

and the diagonal AC = 16 chains? 

= 17 acres 1 rood 0*396 square pole. 

9. Find the area of the quadrilateral field ABCD, having given 
AB=2000, AD = 1500, and AC the diagonal =2390 links; and each 
of the angles BAC, DAC = 30. =20 acres 3 roods 26 square poles. 

In this example, find the areas of the triangles BAC, CAD 
separately by Prob. X. 

10. Find the area of the quadrilateral ABCD from these measure- 
ments : 

AB=468 links, Angle ABC = 73, 

BC=395 .. , .. BCD = 87 30'; 

CD = 410 , 

Find the side DB and angle B in triangle DCB by Trigo- 



112 MENSURATION OF SURFACES 

nometry; then angle ABD = ABC-DBC is known. Hence the 
areas of the two triangles ABD, DBC can now be found by 
Prob. X., as in the preceding ninth exercise. 

= 1 acre 1 rood 19 '6 square poles. 

11. Find the area of the quadrilateral field ABCD, the four sides 
AB, BC, CD, DA being respectively = 750, 700, 650, and 600 links, 
and the angle A = 83 30'. 

In the triangle ADB find the angle at D or B, and then the side 
DB (Art. 187) ; next find the area of triangle ADB by Prob. X., and 
that of DBC by Prob. XI. . = 4 acres 1 rood 39 '4 square poles. 

265. Problem XVIII. To find the interior and central 
angle of any regular polygon. 

RULE. From double the number of the sides of the polygon 
subtract 4 ; multiply the remainder by 90 ; divide the product 
by the number of sides, and the quotient is the number of degrees 
in the interior angle. 

Divide four right angles, or 360, by the 
number of sides, and the quotient is the 
central angle. 

Let t = one of the interior angles DAB, 
e=one of the central angles at C, and n=the 
number of sides of the polygon. 

. 90 180. , 360 

Then ^= (2n-4) = (n-2), and c= 

n ^ n v n 

EXAMPLE. Find the interior and central angles of a regular 
pentagon. 

i=^(n-2) = "f (5 -y = W*.,c = 3 ^ = 3f f = 7Z>. 
n 5 v n 5 

EXERCISES 

1. Find the interior and central angles of a regular hexagon. 

Interior = 120, and central =60. 

2. What is the number of degrees contained in the interior and 
central angles of a regular heptagon ? 

Interior = 128 34' 17}", and central=51 25' 42f". 

3. Find the number of degrees in the interior and central angles 
of a dodecagon Interior =150, and central = 30. 

266. Problem XIX. To find the apothem of a regular 
polygon, its side being given. 

RULE. Multiply half the side of the polygon by the tangent of 
half its interior angle, and the product is the apothem ; or, 




MENSURATION OF SURFACES 113 

Add the logarithm of half the side to the logarithmic tangent 
of half the interior angle, and the sum, diminished by 10, is the 
logarithm of the apothem. 
Let jo = the apothem CF, 

s = one of the sides AB, 
i=the interior angle DAB; 

then p = \s . tan \i, and Lp = L %s + L tan it - 10. 

In the right-angled triangle AFC (fig. to Prob. XVIII.) 

1 : tan CAF=AF : FC = s : p ; 
therefore p = \s. tan Ji, or L/? = L s + L tan \i- 10. 

EXAMPLE. Find the apothem of a regular hexagon whose side 
is =120. 

p = \s. tan ^'=60 tan 60 = 60 x 1- 7320508 = 103 "923048. 
Or Lp = L 60 + L tan 60 -10 

= 1-7781513 + 10-2385606 -10=2-0167119; 
hence j9 = 103'923. 

EXERCISES 

1. Find the apothem of a regular pentagon whose side is = 10. 

= 6-8819. 

2. What is the length of the apothem of a regular heptagon 
whose side is 80? ..... . . . =83-0608 

267. Problem XX. Given a side of a regular polygon 
and its apothem, to find its area. 

RULE. Find the continued product of the side, the number of 
sides, and the apothem and half this product is the area ; or, 

Add together the logarithms of the side, the number of sides, and 
the apothem, and the sum is the logarithm of twice the area. 

Let s, n, and p denote the same quantities as in the two preced- 
ing problems ; 
then .51 = \nps, or L 2.51 = Ls + L + L/>. 

The area of the triangle ABC (fig. to Prob. XVIII.) is=JAB. FC 
= \sp. And there are as many triangles equal to ABC as the 
polygon has sides ; hence its area is = n . ^sp = ^nsp. 

EXAMPLE. The side of a regular hexagon is = 10, and its 
apothem is = 8 "66 ; what is its area? 

= % x 6 x 10 x 8'66 = 259'8. 



EXERCISES 

1. The side of a regular pentagon is = 5, and its apothem is = 3 "44 ; 
what is the area ? . =43. 



114 



MENSURATION OF SURFACES 



2. Find the area of a park in the form of a regular octagon whose 
side is = 12 chains, and apothem = 14-485 chains. 

= 69 acres 2 roods 4 "48 square poles. 

268. Problem XXL To find the area of a regular polygon 
when only a side is given. 

RULE. Find the interior angle, and then the apothem by 
Prob. XVIII. and XIX. ; then find the area by last problem. 

Or, by substituting the value of p in the expression for the area, 

we have /\ 2 i i /A 2 ,. i 

A\, = n\-\ tan $i = n( - I cot \c, 

Whence LJR - Lra + 2L %s + L tan \i - 10. 

EXAMPLE. Find the area of a regular hexagon whose side is= 10. 
L^=Ln + 2Lis + L tan |z-10 = L6 + 2L5 + L tan 60 -10= -7781513 

+ 1 '3979400 + 10-2385606 - 10 = 2 '414651 9 ; 
hence ^1 = 259-808. 

EXERCISES 

1. Find the area of a regular pentagon whose side is = 30 feet. 

= 1548-4275 square feet. 

2. What is the number of square yards in a regular heptagon 
whose side is = 20 yards? ... =1453-564 square yards. 

3. How many acres are contained in a field of the form of a 
regular octagon whose side is = 5 chains ? 

= 12 acres 11 '37 square poles. 

By means of the preceding problems regarding regular polygons, 
the following Table may be constructed : 



Name of 
Polygon 


No. of 
Sides 


Apothem 
when Side = l 


Area 
when Side = l 


Interior 
Angle 


Central 
Angle 


Triangle 


3 


0-2886751 


0-4330127 


60 0' 


120 0' 


Square 


4 


0-5 


1- 


90 


90 


Pentagon 


5 


0-6881910 


1-7204774 


108 


72 


Hexagon 


6 


0-8660254 


2-5980762 


120 


60 


Heptagon 


7 


1 -0382607 


3-6339124 


128 34? 


51 25f 


Octagon 


8 


1-2071068 


4-8284271 


135 


45 


Nonagon 


9 


1-3737387 


6-1818242 


140 


40 


Decagon 


10 


1-5388418 


7-6942088 


144 


36 


Undecagon 


11 


1-7028436 


9-3656399 


147 16 T * T 


32 43 T \ 


Dodecagon 


12 


1-8660254 


11-1961524 


150 


30 



MENSURATION OF SURFACES 115 

269. Problem XXII. To find the area of a regular poly- 
gon of not more than twelve sides, by the preceding Table. 

RULE. Multiply the tabular area for the corresponding polygon, 
whose side is = l, by the square of the side of the given polygon, 
and the product will be the required area ; or, 

To the logarithm of the tabular area add twice the logarithm of 
the given side, and the sum is the logarithm of the required area. 

Let ^B' = the tabular area; 

then 
and 

The areas of similar polygons are to one another as the 
squares of their sides (Eucl. VI. 20) ; hence JR': 1R = I 2 : s t ; there- 
fore ^H=sW. 

EXAMPLE. Find the area of a regular heptagon whose side is 
= 15 feet. 

J&=s>JK= 15 2 x 3-6339=225 x 3 -6339 = 817 '6275 sq. feet. 
Or L.5l = L 3-6339 + 2 L 15=0-5603730 + 2-3521826 = 2-9125556, and 
-51 =817 '628 square feet. 

EXERCISES 

1. How many square yards are contained in a regular hexagon 
whose side is = 50 feet ? . . . =721-688 square yards. 

2. Required the area of a regular pentagon whose side is = 50 
feet =4301 -1935 square feet. 

3. What is the area of a regular pentagon whose side is = 45 
feet? =3483-9667 square feet. 

4. What is the area of a regular hexagon whose side is=40 
yards? =4156 '92192 square yards. 

5. Find the area of a pentagon whose side is = 60 feet. 

= 6193-71864 square feet. 

6. Find the area of a regular octagon whose side is = 80 yards. 

= 30901-93 square yards. 

7. How many square yards are contained in a regular decagon 
whose side is = 12 feet? . . . =123-10734 square yards. 

8. How many acres are contained in a farm of the form of a 
regular decagon whose side is = 2050 links? 

= 323 acres 1 rood 15 "86 square poles. 

270. Problem XXIII. Given the diameter of a circle, to 
find the circumference. 

RULE. Multiply the diameter by 3-1416, and the product is the 
circumference ; or, 



116 MENSURATION OF SURFACES 

Add the. constant logarithm 0'4971509 to that of the diameter, 
and the sum is the logarithm of the circumference. 

Let d, r, and c denote the diameter, radius, and circumference 
of a circle, and <r = 3'1416 ; 

then c=3-1416e?=W, or c=2x3-1416r = 2*r. 

Or Lc=0 -4971509 + Ld. 

When greater accuracy is required, the number 3*14159 may be 
used instead of 3'1416 ; or, for still greater accuracy, the number 
3*1415926536. This number is nearly the length of the circum- 
ference of a circle whose diameter is 1. When less accuracy is 
required, the ratio of 1 to 3|, or 7 to 22, or of 113 to 355, may be 
taken for the ratio of the diameter to the circumference of a circle. 

EXAMPLE. Required the circumference of a circle whose dia- 
meter is = 25 feet. 

c=*=3*1416x 25 = 78-54 feet. 

EXERCISES 

1. Find the circumference of a circle whose diameter is = 28 feet. 

= 87-9648 feet. 

2. What is the circumference of a circle whose diameter is = 24 
feet 3 inches? =76 feet 2 '2 inches. 

3. Find the circumference of a circle whose diameter is = 120 
feet =376 -992 feet. 

4. If the mean diameter of the earth be = 7912 miles, what is 
its mean circumference ? =24856 miles. 

271. Problem XXIV. Given the circumference of a circle, 
to find the diameter. 

RULE. Divide the circumference by 3*1416, or multiply it by 
3183, and the result is the diameter ; or, 

From the logarithm of the circumference subtract the constant 
logarithm 0*4971509, and the remainder is the logarithm of the 
diameter. 

For d= - = , ,** , or d= *3183c. 

te 3*1416 

EXAMPLE. Find the diameter of a circle whose circumference 
is =45 feet. 

c=*3183c = -3183 x 45 = 14-3235 = 14 feet 3'882 inches. 

EXERCISES 

1. Find the diameter of a circle whose circumference is = 177 
feet =56-3391 feet. 



MENSURATION OF SURFACES 117 

2. What is the diameter of a circle whose circumference is = 32 
feet? ......... =10-1856 feet. 

3. What is the diameter of a wheel whose rim is = ll feet? 

= 3-5013 feet. 

4. What is the diameter of a circular pond whose circumference 
is=200feet? ........ =63-66 feet. 

5. What is the diameter of a circular plantation whose circum- 
ference is =1250 yards? ..... =397 "875 yards. 

272. Problem XXV. To find the area of a circle when 
the diameter and circumference are given. 

RULE. Multiply the diameter by the circumference, and one- 
fourth of the product will be the area ; or, 

Add the logarithm of the diameter to that of the circumference, 
and the sum is the logarithm of four times the area. 

Or 1R = \cd, and LJR = LeZ + Lc - '6020600. 

EXAMPLE. Find the area of a circle whose diameter is = 12'732 
feet, and circumference = 40 feet. 

.1 = |ce?=ix 40 x 12-732 = 127-32 sq. feet. 

EXERCISES 

1. Find the area of a circle whose diameter is =21, and circum- 
ference = 65 -973 ......... =346-358. 

2. What is the area of a circle whose diameter is = 20, and 
circumference = 62-8318? ...... =314-159. 

3. Find the area of a circle whose diameter is = 226 links, and 
circumference = 710 ...... =40115 square links. 

4. Find the area of a circular plantation whose diameter is = 640 
links, and circumference 2010 - 6. . =3 acres 34-71 square poles. 

273. Problem XXVI. To find the area of a circle when 
the diameter is given. 

RULE. Multiply the square of the diameter by -7854, or the 
square of the radius by 3*1416, and the product is the area. 



EXAMPLE. What is the area of a circle whose diameter is = 120 
feet? 

1R = 7854d 2 = -7854 x 1 20 2 = '7854 x 14400 = 1 1 309 '76 sq. feet. 
Or l=*T 2 =3-1416x 60 2 = 3'1416x 3600 = 11309-76 sq. feet. 

Frac. I 



118 MENSURATION OF SURFACES 

For by last problem, M = cd; and by Prob. XXIII., c = 3-141Gd; 
therefore JR= x 3'1416e^= -7854^. 
And since d=2r, or cP-^r" ; therefore .1 = 3-1416^= srr 2 . 

EXERCISES 

1. What is the area of a circle whose diameter is = 60 feet? 

= 2827-44 square feet. 

2. Find the area of a circle whose diameter is = 35 feet. 

= 962-115 square feet. 

3. Find the area of a circle whose diameter is = 397 "885 feet. 

= 124338-4 square feet. 

4. What is the area of a circle whose diameter is = 50 yards ? 

= 1963'5 square yards. 

5. Find the area of a circle whose diameter is = 450 links. 

= 1 acre 2 roods 14-5 square poles. 

6. How many square yards are contained in a circle whose 
diameter is = 350 feet? . . . =10690-16 square yards. 

274. Problem XXVII. To find the area of a circle when 
the circumference is given. 

RULE. Multiply the square of the circumference by '0795775, 
and the product is the area. 

Or ^l=-0795775c 2 =^. 

4<r 

If no great accuracy be required, ^H= -0796c 2 . 

EXAMPLE. What is the area of a circle whose circumference 
is = 20 feet 3 inches? 

M= -0795775c 2 = -0795775 x 410^ = 32-63 square feet. 

By the former problems, JR=vd 2 , and c = vd, hence d- ; 

5T 

and therefore JB, = & x -- ---^ -0795775C 2 . 



EXERCISES 

1. Find the area of a circle whose circumference is = 25 feet. 

= 49-73594 square feet. 

2. What is the area of a circle whose circumference is = 15 -708? 

= 19-635. 

3. Find the area of a circular field whose circumference is = 50 
chains. .... =19 acres 3 roods 23 '1 square poles. 

4. Find the area of a circle whose circumference is = 200 yards. 

= 3183-1 square yards. 



MENSURATION OF SURFACES 119 

5. What is the number of square yards in a circle whose circum- 
ference is 25-1328 yards? .... =50-2656 square yards. 

275. Problem XXVIII. To find the area of a circular 
annulus or ring. 

RULE. Multiply the sum of the diameters by their difference, 
and this product by '7854, and the result will be the area ; or, 

Multiply the sum of the circumferences by their difference, and 
this product by '0795775, and the result will be the area ; or, 

Multiply the sum of the circumferences by the difference of the 
diameters, and one-fourth of the product will be 
the area. 

Let d and d' be the diameters AB, A'B' of the 
greater and less circle, and c, c' their circum- 
ferences ; then 



.5l=-0796(c + c')(c-c') 




EXAMPLES. 1. Find the area of a circular annulus contained 
between two concentric circles whose diameters are = 10 and 12. 
& = -7854(10 + 12)(12 - 10)= -7854 x 22 x 2 = 34-5576. 

2. Find, the area of a circular annulus, the circumferences of the 
containing circles being = 30 and 40. 

M = -0796(c + c')(c - c') = -0796 x 70 x 10 = 55 72. 

3. Find the area of a circular annulus, the diameters of the con- 
taining circles being = 50 and 60, and their circumferences = 157 '08 
and 188-496. 

l& = (c + c')(d-d')=x 345-576x10=863-94. 

If JR', JR" be the areas of the greater and less circles, and M 
that of the annulus, then is JR=JR'-&"= '7854^- '7854d' 2 = 
7854(d 2 -d' 2 )= 7854:(d+d')(d- d') ; since d?-d' z = (d+d')(d - d'). 

Again : JR' - .51" = "0796c 2 - -0796c' 2 = -0796(02 - c' 2 ) = -0796(e+c') 



EXERCISES 

1. What is the area of a circular annulus, the diameters of the 
containing circles being = 30 and 40 feet? , =549 '78 square feet. 



120 MENSURATION OP SURFACES 

2. The circumferences of two concentric circles are = 62 -832 and 
37 "6992 ; required the area of the annulus contained by them. 

=201-063. 

3. The diameters of two concentric circles are =20 and 32, and 
their circumferences are = 62 -832 and 100 '531 ; what is the area of 
the annulus contained between them ?. . . . =490'089. 

4. The diameters of two concentric circles are = 19 and 43 "5 feet ; 
what is the area of the included annulus ? =1202-64 square feet. 

5. The circumferences of two circles are = 62'832 and 94-248 feet ; 
what is the area of the contained annulus ? =392 -7 square feet. 

276. Problem XXIX. Of the chord, height, and apothem 
of an arc of a circle, any two being given, to find the 
radius of the circle. 

Let MN = c, PR=A, and the apothem RQ=^. 

1. When PR and RQ, or h and p, are given, then PQ = QR + RP, 
or rp + h, 

2. In the triangle MQR, when MR and RQ are given, MQ 

can be found by Trigonometry. 

Thus, MQ 2 =MR 2 + RQ 2 , or r 2 =|c 2 +.p 2 . 
. 3. When MR and RP, or c and h, are 
given, then (Eucl. III. 35) RS.PR = MR 2 , or 



hence 2r-h = -r; ; hence r=^r, and d= 
s 4A 8^ 



277. If the chord of MP, half the arc, is given, and the height 
PR, then PS. PR=MP 2 ; or if chord MP=c', then since PS=2r, 
2rA=c' 2 ; 

c' 2 
therefore r=ni' 

2n, 

EXAMPLES. 1. Given the apothem and height of an arc =3 -5 
and 8'7, to find the radius of the circle. 

r=p + h=87 + 3-5=12-2. 

2. Given the chord =20 and apothem = 12 of an arc, to find the 
radius of the circle. 

r 2 = Jc 2 +p* = J x 20 2 + 12 2 = 100 + 144 = 244, 
and r = V244 = 15-6204994. 

3. Given the height and chord of an arc = 4 and 30 respectively, 
to find the radius of the circle. 



,__ __ 

~~ ~~~"~ ' 



MENSURATION OP SURFACES 121 

4. The height of an arc is = 4, and the chord of its half is = 20; 
find the radius. 

_c' 2 _20 2 _400_ 
'~2h~ 8 -"8~- 

EXERCISES 

1. What is the radius of a circle, the height of an arc of which 
is=5'6, and apothem = 8-4? =14. 

2. What is the radius of a circle, the chord of an arc of which 
is = 12, and the apothem = 10? =11-6619. 

3. The chord of an arc is = 36, and its apothem is = 25; find the 
radius of the circle =30 '8058. 

4. The height and chord of an arc are = 10 and 24 respectively ; 
find the radius of the circle =12 '2. 

5. Find the radius of a circle, the chord and height of an arc of it 
being = 24 and 4 =20. 

6. The height of an arc is = 2, and its chord is = 15; find the 
diameter of the circle . = 30'125. 

7. What is the diameter when the height is = l and the chord 
= 12? =37. 

8. Find the radius when the chord is = 40 and the height 
= 5 =42-5. 

9. What is the radius of an arc whose height is = 6, and the 
chord of its half = 15 ? = 1875. 



278. Problem XXX. Of the chord, height, and apothem 
of an arc, and the radius of the circle, any two being 
given, to find the number of degrees in the arc. 

1. In the triangle MRQ, when any two of its sides r, p, 
and |c are given, the angle MQR, or \n, can be found by 
Trigonometry. 

2. When QR and RP that is, p and h are given, then, since 
p + h = r, in this case r and p, or MQ and QR, are given, and this 
case is reduced to the former. 

3. When MQ and PR that is, r and A are given, then, since 
QR = QP-PR, orp=r-h ; therefore r andja are again known, and 
this case is reduced to the first. 

4. When any two sides of the triangle MPR are given that is, 
any two of the quantities \c, c', and h angle M, which is = i, can 
be found by Trigonometry. 



122 



EXAMPLES. 1. The chord of an arc is =40, and the radius of 
the circle = 60 ; how many degrees does it contain ? 

MQ:MR=l:sinQ. 
Or r : |c=l :sin $n ; 

c 40 . A 




= sin 19 28' 16". 
L sin n = L c + 10-Lr 
= 1-3010300 + 10-1-7781513 = 9-5228787; 
and n = 19 28' 16", and % = 38 56' 32". 

2. Find the number of degrees in a circular arc whose apothem 
and height are = 24 and 6. 
Here p = 24, h = 6 ; therefore r =p + h = 30 ; 



hence 
and 



cos \n = = = ~ = -8 = cos 36 
n = 7344' 24". 



12", 



3. The radius of a circle is = 25, and the height of an arc of it 
is = 5 ; required the number of degrees in it. 



and 
and 



l= r = fjl = ^=-& = caB 36 52' 12", 
w=73 44' 24". 



4. The chord of an arc is = 36, and its height is =4; how many 
degrees are contained in it ? 



and 



tan 4 = = = == -2 = tan 12 31' 43"'7, 
T^C c ou y 

n = 50 6' 54" -8. 



EXERCISES 

1. The chord of an arc is = 36, and the radius of the circle is 
=54 ; what is the number of degrees in the arc ? . =38 56' 32". 

2. The apothem and height of an arc are = 50 and 12; required 
the number of degrees in it =72 29' 55" -2. 

3. What is the number of degrees in an arc whose height is = 12, 
the radius of the circle being = 56? .... =76 25' '6. 

4. How many degrees are contained in an arc whose chord is = 40, 
and height =5? =56 8' *7. 

5. The chord of half an arc is = 20, and the height of the arc is = 2; 
how many degrees are contained in it ? . . . =22 57' '2. 



MENSURATION OP SURFACES 123 

279. Problem XXXI. To find the length of a circular 
arc, the number of degrees in it and the diameter being 
given. 

RULE. Multiply the number of degrees in it by the diameter, 
and the product by -008727 ; the result will be the length of the arc. 

Let n = the number of degrees in the arc, 

I = the length of the arc ; 
then 1= -008727nd. 

Or 1= 0174533nr, where r radius. 

For c = 3-1416rf, and 360 : =3'1416rf : I ; 

, 3-14l6nd 

hence I = ^ = -008727d. 

280. When, instead of the diameter being given, the chord or 
apothem is given, the radius can be found. For 

in the triangle MQR a side is then given, and 
angle Q=|, to find the radius MQ. Also, when 
the chord and apothem are given, two sides of 
the triangle MQR are given ; and hence the 
radius and number of degrees can be found by 
Trigonometry. 

EXAMPLES. 1. Find the length of a circular arc containing 30, 
the diameter being 50. 

1= -008727 nd= -008727 x 30 x 50 = 13-0905. 

2. The number of degrees in the arc of a circle, whose radius 
is =25, is 25 30' ; what is the length of the arc? 

/= -0174533nr= '0174533 x 25'5 x 25 = 11-126. 

EXERCISES 

1. What is the length of a circular arc of 45, the diameter being 
= 12? =4-71258. 

2. What is the length of a circular arc of 32, the radius of the 
circle being = 20? =11'17. 

3. Find the length of a circular arc containing 120 40', the radius 
of the circle being =50. . . . =105*3. 

4. Required the length of an arc of a circle whose diameter is 
= 125, and the number of degrees in the arc = 54 '6, or 54 36'. 

= 59-559. 

5. Find the length of an arc whose chord is = 12, and radius = 18. 

= 12-234. 




124 MENSURATION OP SURFACES 

6. What is the length of an arc whose chord is = 25'4, and the 
chord of its half = 15 -3? =32-461. 

7. Find the length of an arc whose chord is=4'8, and that of its 
half=2-443. . ... -. . -. /. . = 4'9159. 

281. The lengths of arcs may also be easily computed by means 
of a Table containing the lengths of arcs of any number of degrees 
belonging to a circle whose radius is = l. Such a Table can be 
calculated by this problem. The rule by this method is : 

Multiply the tabular length of the arc of the same number of 
degrees by the radius of the given arc, and the product will be 
its length. 

Let V = length of arc in Table ; then lrl'. 

Thus, for the first example given above, where n = 3Q, and d=5Q, 
it is found that l'= '5235988 ; 
hence I = rl' = 25 x -5235988 = 13-08997. 

And, for the second example, where = 25, /'= -4363325, and for 
30', /'= -0087266; hence, for 25 30', I' is the sum of these two, or 
= -4450591 ; 
hence I =rl'=25 x -4450591 = 11-12648. 

282. Problem XXXII. To find the area of a circular 
sector. 

RULE. Multiply the length of the arc of the sector by the radius, 
and half the product will be the area. 

Or M=tfr. 

For the area of the whole circle is equal to the product of its 
circumference into the radius divided by two ; and hence the area 
of the sector is also the product of its arc into the radius divided 
by two. 

EXAMPLES. 1. The length of a circular arc is =24, and the 
diameter of the circle is = 30 ; find its area. 

M = tfr= | x 24 x 30=180. 

2. The number of degrees in a circular arc is = 30, and the radius 
is =25 ; what is its area ? 

1= -0174533nr= -0174533 x 30 x 25 = 13-08997, 
and JR = tfr=lx 13 '08997 x 25 = 163 -62468. 

283. When the number of degrees in the arc is given, as in the 
last example, the formula may be a little improved. 

Thus, if in M = $lr, the value of I found in Art. 279 be substi- 
tuted, it becomes JR = % x > 0174533;m-= 008727nr z . 



MENSURATION OF SURFACES 125 

The last example, calculated by this formula, gives 

^1= 008727r 2 = -008727 x 30 x 252= 163 '625. 

When the radius and the length of the arc, or the number of 
degrees in it, are not given, they must be found by preceding 
problems. 

EXERCISES 

1. The length of a circular arc is =50, and its radius is = 30; 
what is the area of the sector ? = 750. 

2. The length of a circular arc is = 10 75, and its radius = 12 -5; 
what is the area of the sector ? =67 '1875. 

3. The number of degrees in a circular arc is = 40, and the 
diameter is = 60 ; find the area of the sector. . . =314-172. 

4. What is the area of a sector, the arc of which contains 50 42', 
the radius of the circle being =28? .... =346'8877. 

5. What is the area of a sector, the length of its arc being = 78 -14, 
and the diameter of the circle = 70? .... =1367*45. 

6. What is the area of a sector whose radius is = 18, and its chord 
= 12? =110-106. 

7. Find the area of a sector whose arc contains 27, its radius 
being = 6 feet. = 8 -4826 square feet. 

8. Find the area of a sector whose arc contains 36, its radius 
being = 50 =785 '4. 

9. Find the area of a circular sector, the chord of the arc being 
= 8, and that of half the arc =5 =22-344. 

10. What is the area of a sector whose chord is = 30, and height 
= 4? =473-015. 

11. The height of the arc of a sector is = 2'5, and the chord of its 
halfis = 5; what is its area ? =26 '18. 

284. Problem XXXIII. To find the area of a circular 
segment. 

RULE I. Find the area of the sector that has the same arc as 
the segment ; find also the area of the triangle whose vertex is 
the centre and whose base is the chord of the segment ; then the 
area of the segment is the difference or sum of these two areas, 
according as the segment is less or greater than a semicircle. 

EXAMPLE. The chord and height of a segment are = 24 and 6; 

find its area. 

cyi, 10 
By Art. 278, tan J=^=g= -5 =tan 26 33' 54" ; 

and hence n = 106 15' 36" = 106 -26. 



126 MENSURATION OP SURFACES 

r 2 144 
Also (Art. 276), 2r-& = ^r = ^=24, and r = 15. 

Then (Art. 283), sector = -008727m' 2 - -008727 x 106 '26 x 225 = 208 "64 ; 

also, triangle \cp = ^ x 24 x 9 = 108 ; 

hence segment - sector - triangle = 208 '64 - 108 = 100 '64. 

285. When either the chord or apothem is unknown, and the 
radius is either given or found, and also the number of degrees 
in the arc, the area of the triangle is to be found by Prob. X. 

EXERCISES 

1. Find the area of a circular segment, its chord being=40, and 
height =4 =107 '56. 

2. The chord of a segment is = 20, and its height = 5; what is 
its area? =69'896. 

3. Find the area of a segment whose chord is = 24 feet, and 
height =9. =159'1 square feet. 

4. What is the area of a segment whose chord is = 30, and 
diameter = 50? =102-188. 

5. Required the area of a segment whose chord is = 16, and 
diameter = 20 =44-7293. 

6. The chord of a segment is = 24, and the radius =20; what is 
its area? =65-401. 

7. What is the area of a segment, the arc of which is a quadrant, 
and the diameter = 12 feet? .... =10'274 square feet. 

8. Find the area of a segment whose arc contains 280, the 
diameter being = 10 feet. .... =73-3966 square feet. 

9. The height of a segment is = 18, and the diameter of the 
circle=50; what is the area of the segment ?. . . =636'376. 

10. The diameter of a circle is = 100 feet, and the height of a 
segment of it is 6-5 ; what is its area? . =216-597 square feet. 

286. The area may also be found by means of a Table containing 
the areas of segments of a circle, whose diameter is = l, and whose 
heights are all the numbers between and -5 carried to any 
number of decimal places, as to two or four, or any other number, 
according to the degree of accuracy required. Such a Table can 
be calculated by means of the preceding rule. The rule by this 
method is : 

RULE II. Divide the height by the diameter, the quotient is 
the height of the similar segment when the diameter is = l ; take 
the tabular area corresponding to this height, and multiply it 



MENSURATION OP SURFACES 127 

by the square of the diameter, and the product is the area of the 
given segment. 

EXAMPLE. For the above example, h' = ~ f ^--2. 

The tabular area is then &' = -111824 ; 

and ^fl = ^^l'=30 2 x -111824 = 100-6416. 

The exercises given above may be performed in the same manner 
to exemplify this rule. 

Before this method can be used, h must be known. 

When r and c are known, then p 2 r z - %c 2 , from which p is found, 
then h=r-p. 

When the chord of half the arc is given and the diameter, then 
(Art. 277) 2rh = c' 2 , where c' is the chord of half the arc ; and from this, 

r 1 ^ 

h = C -. 
2r 

When the chord of the arc and that of half the arc are given, 
or c and c'; then in triangle MPR (fig. to Prob. XXX.), PR 2 = 
MP 2 - MR 2 , or A 2 =c' 2 - c 2 . 

287. Problem XXXIV. To find the area of a circular 
zone that is, the figure contained by two parallel chords 
and the intercepted arcs. 

Find the area of the trapezium ACGF, and 
of the segment AIC, and double their sum will 
be the area of the zone ABDC ; or, 

Find the areas of the two segments AHB, 
CHD (Art. 285), and their difference will be the 
area of the zone ABDC. 

Let the chord AB = c, CD = c', and AC = c"; 
and the distance GF = 6. 

When b, c, and c' are given, the diameter d will be found thus 

Let CL=m, then m = b + (c + '} ( ~ ^ . [1], 

40 

and d 2 =?n 2 + c' 2 ............... [2]. 

For CK . KL = AK . KB (Eucl. III. 35) ; hence 
^ T _AK . KB 
CK ; 
that is, KL = ^^ 




o 46 

But LCD being a right angle, LD is the diameter, and 
DL 2 =CL 2 + CD 2 . 

Also C "2 = 2 + (C _ C ')2 

For AC 2 = CK 2 + AK 2 , and AK = J(e - c'). 



128 MENSURATION OP SURFACES 

The diameter d and c" being known, the area of the segment 
AIC can be found by Prob. XXXIII. ; and if = the area of the 
trapezium ACGF, it is 

= J(AF + CG)CK, or*=J(c + c')4 ... [4]. 

The diameter and the chords c and c' being known, the areas of 
the segments AHB, CHD can be found by Prob. XXXIII. 

Hence, if a, a', and a" denote the areas of the segments, whose 
chords are c, c', and c", and ^R that of the zone, then 

JR = 2(t + a"), or M-a-a' ...... [5]. 

288. Instead of finding the areas of the segments by the first rule 
of Prob. XXXIII., they may be found by the second that is, by 
means of a Table. The heights, however, of the segments must 
be known before the rule can be applied. For the methods of 
finding h, see end of Art. 286. 

When the zone contains the centre of the circle, the areas of the 
two segments on its opposite sides may be found, and their sum 
being taken from the area of the whole circle, will give that of the 
zone. 

EXAMPLE. Find the area of a circular zone, the parallel chords 
of which are = 90 and 50, and the distance between them = 20. 

The areas of the segments may be calculated by either of the 
two rules of the last problem. They are calculated here by the 
second rule. 

Here c = 90, c' = 50, and b = 20. 



by [2], d 2 = TO 2 + c' 2 = 90 2 + 50 2 = 10600, and d = 102 -956. 

Let p, p' and h, h' be the apothems arid heights of these two 
segments, then (Prob. XXXIII.) 

p'2 = ,.2 _ (^ C ')2 = 2650 - 25 2 = 2025, and p' = 45 ; 
hence h'=r-p' = 51-438 -45 = 6 '478. 

Also h =b + h' = 2Q+ 6 -478 = 26 '478. 



The tabular height for h' is=-= = -06292. 

d 10^ 'Doo 

h 26-478 



Tabular area for -0629 is = -020642 
H for -2572 is =-159811 

Difference, = -139169 

Hence JR =-139169^= -139169 x 10600 = 1475-19. 



MENSURATION OF SURFACES 129 

EXERCISES 

1. The chords of a circular zone are = 30 and 48, and the distance 
between them is = 13; required its area. . . . =534-19. 

2. The chords of a zone that contains within it the centre of the 
circle are = 30 and 40, and their distance is = 35 ; what is its area? 

= 1581-7475. 

3. The diameter of a circle is =25, and two parallel chords in it, 
on the same side of the centre, are = 20 and 15 ; find the area of the 
zone contained by them. =44-343. 

289. Problem XXXV. To find the area of a lune that 
is, the space contained between the arcs of two circles 
that have a common chord. 

RULE. Find the areas of the two segments that stand on the 
same side of the chord, and their difference is the area of the luue. 

EXERCISES 

1. The length of the common chord AB is =40, the heights CE 
and CD = 10 and 4 ; what is the area of the lune 

AEBD? =172-05. 

2. The chord is = 30, and the heights = 3 and 
15 ; find the area of the luue . . =292-954. 

3. The chord is =48, and the heights are = 7 ' 

and 18; what is the area? =408 '609. 

290. Problem XXXVI. To find the area of any irregular 
polygon. 

RULE. Divide the polygon, by means of diagonals, into tri- 
angles, or into triangles and trapeziums, and find the areas of 
these component figures by former problems, and the sum of their 
areas will be the area required. 

1. Find the area of a hexagonal figure from these measure- 
ments : 

AC =525 links 

BG =160 

DF =490 ., 

FH . . . . . =210 i. 

El =100 

CK =300 

= 1 a/:re 3 roods 32 "2 poles, 





130 MENSURATION OF SURFACES 

2. Kequired the area of the irregular hexagon ABCDEF from 
these data : 

The side AB = 690 links the side FA = 630 links 
I. ii BC = 870 the diagonal AE = 1210 ., 

ii CD = 770 AD = 1634 

n DE = 510 BD = 1486 

ii ii EF = 670 M 

= 11 acres 18-46 square poles. 

In this example the polygon is divided into triangles of 
which the three sides are known ; and their areas are found by 
Prob. XI. 

3. Find the area of the figure ABCDEF from these measure- 
ments : 

The side AB = 2000 links the angle BAG = 40 
ii n AF = 1800 CAD=43 

the diagonal AC =2500 . ., ., DAE =40 30' 

AD = 2750 ., EAF=4820' 

AE=3450 

= 93 acres 2 roods 2 -67 square poles. 

The areas of the triangles in the preceding question are to be 
found by Prob. X. 

4. Find the area of the field ABCDE from these data : 

D The sideAB . . . =450 links 

ii ii BC . . . =365 n 

n CD : =324 i. 

n DE ... =428 n 

the angle ABC . . . =110 14' 

ii BCD . = 84 30' 

n n CDE . . . =140 24' 

= 2 acres 21-27 square poles. 

Divide the polygon into triangles by means of the diagonals AC 
and CE. In triangle ABC, calculate the angle C and side AC 
(Art. 187) ; and similarly in triangle EDC, calculate angle C 
and side EC ; then, if the sum of these two angles be subtracted 
from the whole angle BCD, the remainder is angle ACE. Two 
sides and a contained angle are then known in each of the three 
triangles ; and hence their areas can be found (Art. 256). 

291. When all the sides but one of any polygon are known, and 




MENSURATION OF SURFACES 131 

also all the angles except the two at the extremities of that side, 
the area may be calculated in a manner similar to the method used 
in the solution of the preceding example. 

292. Problem XXXVII. To find the area of any curvi- 
lineal space by means of equidistant ordinates. 

Let ACDB be the given space. 

Draw the perpendiculars or ordinates GC, HD, &c. ; then, if 
the curves AC, CD, DE, &c. are suffi- 
ciently short, they may be considered 
as straight lines without any material 
error, and then the figure will be divided A G H ' K B 
into triangles and trapeziums, whose areas can be found as 
formerly. 

I. When the curve meets the base at both extremities, and the 
base is divided into a number of equal parts, and ordinates are 
drawn from the points of division, multiply the sum of the ordi- 
nates by the common distance between them, and the product is 
the area. 

Or, if the common distance =1, and the sum of the perpendiculars 
=s, then &, = ls. 

For let the perpendiculars be a, b, c, d, taken in order; then 
the areas of the triangle AGC, of the trapeziums, and of triangle 
FKB are 



that is, I is multiplied twice by \a, twice by %b, &c., or by the sum 
of a, b, c, and d. 

When the figure is bounded by two perpendiculars, as by CG 
and KF, let them be denoted by a and z, and the sum of all the 
perpendiculars by s', as above ; then if 



EXAMPLE. Let the perpendiculars of the figure ABD be = 10, 
12, 13, and 11, and the equal divisions of AB = 9, what is its area ? 

5=10 + 12+13 + 11 = 46; 
hence ^H=?s=9x46=414. 

EXERCISES 

1. The perpendiculars are = 12, 20, 26, 30, and 24, and the 
common distance is = 14 ; find the area. . =1568. 



132 MENSURATION OP SURFACES 

2. What is the area of the figure CGKF, terminated by the 
perpendiculars CG and FK, the four perpendiculars being = 14, 15, 
16, and 18, and the common distance = 12? . . . =564. 

II. When the surface is terminated at its two extremities by 
ordinates, divide the base into an even number of equal parts ; 
find the sum of the first and last ordinates ; also the sum of the 
even ordinates that is, the second, fourth, &c. and also the sum 
of the remaining ordinates ; then add together the first sum, four 
times the second, and twice the third ; and the resulting sum, 
multiplied by one-third of the common distance of the ordinates, 
will give the area. 

Let A=the sum of the first and last ordinates, 

B= it it even ordinates, the second, fourth, &c., 
C = it M remaining ordinates, 
and D = the common distance between the ordinates ; 
then the area = J(A + 4B + 2C)D. 

For twice the area by last case = (A + 2B + 2C)D ; and supposing 
the second ordinate to be equal to half the first and third, the 
fourth equal to half the third and fifth, and so on, the area will 
equal 2BD ; and adding these two, gives 

3JR= (A + 4B + 2C)D; 
hence the M = i( A + 4B + 2C)D. 

EXAMPLE. Find the area of a surface, the ordinates being in 
order=10, 11, 14, 16, and 16, and the common distance between 
them = 5. 

Here A = 26, B = 27, and C = 14, 

and area . = i(26 + 108 + 28) x 5 = $ x 162 = 270. 

EXERCISES 

1. What is the area of a surface, the common distance between 
the ordinates being = 10, and the ordinates in order = 20, 22, 28, 32, 
and 32? =1080. 

2. Find the area of a field, one side of it being = 198 links, and 
seven ordinates to it measured at equal distances to the opposite 
curvilineal boundary being in order=60, 75, 80, 82, 76, 63, and 50. 

= 14322 square links. 

3. One side of a field is = 60, and five equidistant ordinates are 
measured perpendicular to it, extending to the curvilineal boun- 
dary, which are = 30, 33, 42, 48, and 48; what is the area of the 
field? =2430. 



LAND-SURVEYING 133 

4. Find the area of a field, one side of it being = 990 links, and 
seven equidistant ordinates from it to the opposite curvilineal 
boundary being = 300, 375, 400, 410, 380, 315, and 250. 

= 3 acres 2 roods 12 -88 square poles. 



LAND-SUEVEYING 

293. Land-surveying is the method of measuring and 
computing the area of any small portion of the earth's 
surface as a field, a farm, an estate, or district of moderate 
extent. 

294. The quantity of surface to be ascertained in any case 
by this species of surveying is comparatively so limited 
that the spherical form of the earth is seldom taken into 
consideration. 

295. The surfaces to be measured are divided into triangles 
and trapeziums, as in Articles 290 and 292 in 'Mensuration 
of Surfaces.' Various instruments are used for obtaining the 
measurements necessary for the computation of the areas, 
and for the construction of plans of the surfaces. The most 
common instruments are the chain, the surveying-cross, a 
theodolite, and a plane table. 

296. The chain, called also Gunter's chain, is 22 yards or 
66 feet long, and is composed of 100 equal links, the length 
of each being 7 '92 inches. At every tenth link is a mark 
made of brass, to assist the eye in reckoning the number of 
links measured off. An acre consists of 10 square chains, or 
100,000 square links. There are 80 chains in a mile, and 
640 acres in a square mile. 

297. Ten iron pins, called arrows, with pieces of red 
cloth attached to them, are used for sticking in the ground 

Prac. J 



134 LAND-SURVEYING 

at the end of each chain-length when measuring in the 
field. 

298. Offset-staffs are wooden rods ten links long divided 
into links for measuring offsets (Art. 305). 

299. Other staffs, about six feet long, called picket-staffs 
or station-staffs, with small red flags attached, are used for 
marks to be placed at the corners of fields and other places 
called stations (Art. 303). 

300. The surveying-cross, or cross-staff, consists of two 
bars of brass placed at right angles, with sights at their 
extremities, perpendicular to the plane of the bars. There are 

narrow slits at A and C, to which the eye is 
applied, and wider openings at B and D, 
with a fine wire fixed vertically in the 
middle of them. The cross is supported on 
a staff E, about 4J feet high, which at the 
lower end is pointed and shod with brass, so 
that it can be easily stuck in the ground. 

The sights are placed on the top of the staff, and fixed in any 

position by a screw F. 

301. A simple cross-staff may be made by cutting two 
grooves with a saw along the diagonals of a square board, to 
be fixed on the top of the staff. 

302. It can easily be ascertained if the sights are at right 
angles, by directing one pair of them, as AB, to one object, 
and observing to what object the other pair, CD, are then 
directed ; then by turning the sights till the second object is 
seen through the first pair of sights AB, if the first object is 
then visible through the second pair of sights and is exactly 
in apparent coincidence with the wire, the sights are at right 
angles ; if not, they must be adjusted. 

303. The angular points of the large triangles or polygons 
into which a field is to be divided for the purpose of taking 




LAND-SURVEYING 135 

its dimensions are called stations, and are denoted by the 
mark ; thus, 1 is the first station, 2 the second, and 
so on. 

304. The stations are joined by lines, which are measured 
by the chain ; hence called chain-lines or station-lines. 

305. Lines measured perpendicularly to chain-lines, to the 
angular points, and other points of the boundary of a field 
are called offsets. 

306. The cross-staff is used for finding the position of 
offsets. The point in the chain-line from which an offset is 
to be measured to any point in the boundary is found by 
fixing the staff in the chain-line so that one pair of sights 
may coincide with it; then, if the point in the boundary 
coincides with the other sight, the cross is at the proper 
point for an offset. Thus, the cross being placed at g (fig. to 
Art. 310), and one pair of sights coinciding with AB, the 
other will coincide with gC. 

307. The theodolite is one of the most common and useful 
angular instruments. It consists of two graduated circles 
perpendicular to each other, one of which is fixed in a 
horizontal and the other in a vertical plane, and is used for 
measuring horizontal and vertical 

angles. 

In the figure, WPB represents a 
side view of the horizontal circle, 
and FTP a direct view of the ver- 
tical one, which extends to little 
more than a semicircle. The vertical 
circle is movable about an axis, 
coinciding with the centre of the 
circular arc FTP. 

On the vertical circle is fixed a 
telescope, \V, furnished with a spirit-level, ~N the tele- 
scope moves vertically about a horizontal axis which 




136 LAND-SURVEYING 

passes through the centre of the vertical arc ; and it 
moves horizontally by turning the upper horizontal plate 
on which it is supported, the lower plate B remaining 
fixed. 

Both the horizontal and vertical circles are graduated 
to half-degrees, and hy means of verniers, which are applied 
to them, angles can be read to minutes. Two levels are 
placed on the top of the horizontal plate, and when the 
instrument is to be used it is placed on a tripod stand, the 
horizontal circle being brought to a horizontal position by 
means of adjusting screws, H, and two spirit-levels, n, fixed 
on the circular plate. 

308. To measure a horizontal angle subtended at the 
instrument by the horizontal distances of two objects : direct 
the telescope to one of the objects, and observe the number of 
degrees at the vernier on the horizontal circle ; then turn the 
vertical circle, which is supported on the upper horizontal 
plate, till the other object is visible through the telescope, and 
in apparent coincidence with the intersection of the cross 
wires, and note the number of degrees on the horizontal 
circle ; then the difference between this and the former 
number is the required horizontal angle. 

309. To measure a vertical angle : direct the telescope to 
the object whose angle of elevation is required ; then the arc 
intercepted between the zero of the arc and that of the 
vernier is the required angle. An angle of depression is 
similarly measured. 

310. Problem I. To survey with the chain and cross- 
staff. 

RULE. Divide the field into triangles, or into triangles and 
quadrilaterals, the principal triangles or trapeziums occupying 
the great body of the field, and the rest of it containing 
secondary triangles and trapeziums formed by offsets from 
the chain-lines. Measure the base and height, or else the three 
sides of each of the principal triangles, then calculate their 



LAND-SURVEYING 



137 



areas by the rules in 'Mensuration of Surfaces,' and also the 
offset spaces, and the sum of all the areas will be that of the 
entire field. 

EXAMPLE. Find the contents of the adjoining field from 
these measurements, A being the first and B the second 
station : 



On chain-line 
Ag =150 
Ah =323 
Ai = 597 
A&=624 
AB = 769 



Offsets 

#0=141 to left 
/iE = 180 to right 
iD= 167 to left 
F = 172 to right 




The doubles of the areas of the component triangles and trape- 
ziums are found, in order that there may be only one division by 2 
namely, that of their sum. 



gi = Ai- Ag = 447, 
B/,;=AB-A = 145. 



= AB-Ai = 172, and hk = Ale- Ah = 301, 



Twice 
the 



of the 



'triangle AgC = Ag . #0 = 150x141 , . . 
trapezium CgiD =gi(Cg + Di) =447 x (141 + 167), 



triangle DiE = Eix iD = 172x167, 

triangle AAE = AAx AE = 323x1 80, 

trapezium 

.triangle BF = Bx &F = 145x172, 

Twice area 



= 21150 
= 137676 
=28724 
=58140 
=105952 
=24940 



=376582 



therefore area = 188291 = 1 acre 3 roods 21 - 2656 square poles. 

311. Instead of writing the measurements as above, they are 
usually registered in a tabular form, called a field-book, as 
follows. The beginning of the field-book is at the lower end 
of the table, as this arrangement suggests more readily the 
direction of the measurements. The middle column of the field- 
book contains the lengths measured on the chain-lines, and tire 
columns to the right and left of it contain respectively the right 
and left offsets. 

The station from which the measurements are begun is called 
the first station ; that next arrived at, the second ; and so 
on. 

The field-book of the measurements of a field similar to that of 
the last example is given below in the following exercise, in which 
A is Oi and B is O-j. 



138 



LAND-SURVEYING 



EXERCISES 

1. Find the area of a field, the dimensions of which are given in 
the following field-book : 



Left Offsets 


Chain-line 


Right Offsets 




1538 to O 2 






1248 


344 


334 


1194 






646 


360 to road. 


To fence, 282 


300 






From Oi 





7 acres 2 roods 5'06 square poles. 

2. Find the area of the subjoined field from the following 
measurements : 




Chain-lines 


Offsets 


AO = 291 links 


Bre = 155 


An =430 it 


DO = 160 


AC =450 ii 
Dp = 65 ,. 
Dq =210 ., 
DA = 325 


ps = 30 
qr - 25 
gh = 50 
ik = 55 


Ag = 180 
Ai =410 




AB = 460 ii 





Twice 
the 



of the 



quadrilateral ABCD 

=AC(En + DO) =450(155 + 160), =141750 
triangle Agh = Ag . gh = 180 x 50, . 
trapezium gikh=gi(ik+gh) = 230(55 + 50), . 
triangle ~Bik=Bi . iJc= 50 x 55, 
triangle Dps =Dp . ps=65 x 30, . 
trapezium pqrs=pq(ps + qr) = 145(30 + 25), . 
triangle Aqr =Aq . qr= 115 x 25, . 

Twice area =190450 

therefore area = 95225 square links = 3 roods 32 '36 square poles. 

3. Find the area of a field similar to the preceding from the 
measurements given in the subjoined field-book. 

= 1 acre 1 '3184 square poles. 



= 9000 
= 24150 
= 2750 
= 1950 
= 7975 
= 2875 



LAND-SURVEYv^G 



139 



Left Offsets 


Chain-line 


Right Offsets 


To O 3 276 


368 to O 4 
328 
144 
From G! on R of O 2 


44\\ 
40 j tonver 

20 [f\r to gate 
24 

192^ toO 4 


360 to Oj 
168 
52 
From O 3 on L of O 2 


360 to O 2 
344 
248 
From 0, 



312. The initial letters R and L are used for right and left, to 
denote the direction in which a line is to be measured. Sometimes 
the marks f and ] are used to denote respectively a turning to the 
right and left. The expression in the above field-book ' From O 3 
on L of Go' means that a chain-line is to be measured from the 
third station, and that it is situated to the left of the second 
station, in reference to the direction in which the first chain-line, 
AC, is measured ; so ' From Oj on R of O 2 ' means that the next 
chain-line extends from O a to a point on the right of O 2 namely, 
toO 4 . 

When the field to be surveyed is 
not very extensive, or the measure- 
ments not complex, they may be 
marked on a rough sketch of the 
field instead of in a field-book, as 
in the figure to Example 2. 

313. On the left of the numbers 
denoting the left offsets, and to the 
right of those denoting the light 
offsets, lines are sometimes made, to 
represent in a general way the form 
of the boundary to which the offsets 
are drawn. 

4. Find the area of the adjoining field ABCIHDFGE from 




140 LAND-SURVEYING 

the measurements in the following field-book, A, B, C, D, E 
being respectively the 1st, 2nd, 3rd, 4th, and 5th stations. 



Offsets on Left 


Chain-lines 


Right Offsets 




From O 3 802 to O 5 
From G! 760 to O 3 
444 to Oj 
From 6 


112] 
120 J 


585 to O 5 
426 
136 
From 4 


[110 
1 80 


474 to O 4 
310 
120 
From O 3 






623 to O 3 
From O 2 


547 to O 2 
From Oj 



= 4 acres 3 roods 23 '7 square poles. 

Find the areas of the principal triangles ABC, ACE, and CDE, 
in the above exercise, by Article 257 in ' Mensuration of Surfaces ; ' 
then find the areas of the triangles and trapeziums composing 
the offset spaces EGFD and DHIC, the former of which is to 
be added to the areas of the principal triangles, and the latter 
to be deducted, in order to give the area of the given field 
ABCIHDFGE. 

The crooked boundary, DHI, may be reduced to a straight 
line DL, meeting CI produced in L (see Art. 122, ' Descriptive 
Geometry '), and then a triangle CLD is formed equal to the 
irregular space CIHD, the area of which is = iLS . CD. The 
length of LS can be found by means of the scale used in construct- 
ing the figure. The offset space EGFD, with the curvilineal 
boundary, can also be reduced to a triangle EKD of equal area, 
which can be calculated like that of triangle CLD. The straight 
lines EK, KD can be determined with sufficient accuracy by the 
eye, so as to cut off as much space from the inside of the curved 



LAND-SURVEYING 



141 



boundary EGFD as is added on the outside. A ruler made of 
transparent horn is used for this purpose, or a fine wire stretched 
on a whalebone bow. 

314. The practice of constructing a plan of any surface, the 
dimensions of which are taken, and reducing the crooked and 
curved boundaries in the manner stated above, is very common 
with the best surveyors, on account of its expedition and sufficient 
accuracy. It is also usual to measure on the plan the altitudes of 
the principal triangles, and to calculate their areas by the simple 
rule in Article 255 of ' Mensuration of Surfaces.' 

Thus, by drawing the perpendicular BV on AC, and measuring 
it, the area of triangle ABC is = AC . BV ; and in a similar manner 
the areas of the other principal triangles are found. 

COMPUTATION OF ACREAGE 

Divide the area into convenient triangles, and multiply the base 
of each triangle in links by half the perpendicular in links ; cut 
off 5 figures to the right, and the remaining figures will be acres. 
Multiply the 5 figures so cut off by 4, and again cut off 5 figures, 
and the remainder is in roods. Multiply the 5 figures by 40, and 
again cut off for square poles. 

OBSTACLES IN RANGING SURVEY LINES 

If it be possible to see over the obstacle, but not to chain over 
it, lay off AC and BD (fig. 1) equal to each other, and at right 




Fig. 1. 

angles to the line ; then CD = AB. If it be not possible either to 
chain or see over the obstacle, lay off the lines EF, AC equal to each 
other, and at right angles to the line (fig. 2) range the points DH 
in line with EC, and set off the lines DB, HG equal to AC and EF, 



142 



LAND-SURVEYING 



and at right angles to the line EH ; then B and G are points 
for ranging the continuation of the line FA, and AB = CD. 




Kg. 2. 



TO SET OUT A RIGHT ANGLE WITH THE CHAIN 

Take 40 links on the chain for the base, 30 links for the perpen- 
dicular, and 50 for the hypotenuse. 

USEFUL NUMBERS IN SURVEYING 



For Converting 


Multiplier 


Converse 


Feet into links, 


1-515 


66 


Yards into links, . 
Square feet into acres, . -,- 
Square yards into acres, 
Feet into miles, 


4-545 
0000229 
0002066 
00019 


22 
43560 
4840 
5280 


Yards into miles, . 
Chains into miles, . 


00057 
0125 


1760 

80 



TO SURVEY WITH THE CHAIN, CROSS, AND 
THEODOLITE 

315. Although it frequently happens that the most expeditious 
mode of surveying is by the chain and cross, yet in the case of 
large surveys the theodolite is very advantageously combined with 
them for measuring angles. When some of the angles of a triangle 
are known, its area can be found without knowing all its sides ; 
and the tedious process of measuring them all by the chain is 
thus dispensed with, unless the measuring of offsets or some other 
cause requires all the sides to be measured. It is often useful to 
measure more lines and angles than are necessary for determining 




LAND-SURVEYING 143 

the area, for the purpose of serving as a check to ensure accuracy 
in the results. 

316. Problem II. To survey a field by taking a single 
station within it, and measuring the distances to its 
different corners, and the angles at the station contained 
by these distances. 

The field is thus divided into triangles, in each of which two 
sides and the contained angle are known ; and their areas may 
therefore be found by Article 256 in ' Mensuration of Surfaces ; ' 
their sum will be the area of the field. 

EXERCISES 

1. From a station O within a pentagonal field, the distances to 
the different corners A, B, C, D, E were 

measured and found to be respectively 1469, 
1196, 1299, 1203, and 1410; and the angles 
AOB, BOG, &c. contained by them were in 
order 71 30', 55 45', 49 15', and 81 30'; re- 
quired the area of the field. 

= 39 acres 30 "2 square poles. 

2. From a station near the middle of a field of six sides, 
ABCDEF, the distances and angles, measured as in the preceding 
exercise, were as below : 

AO = 4315 links Angle AOB = 60 30' 

OB = 2982 I. i. BOC = 47 4tf 

OC = 3561 H COD = 49 50' 

OD = 5010 DOE = 57 10' 

OE = 4618 u EOF = 64 15' 

OF = 3606 ., FOA = 80 35' 

What is the area? . . =412 acres 1 rood 17 '3 square poles. 

317. Problem III. To survey a polygonal field by measur- 
ing all its sides but one, and all its angles except the 
two at the extremities of that side. 

From the data it will always be possible, by applying trigono- 
metrical calculation, to find two sides and the contained angle of 
each of the component triangles, the areas of which can be calcu- 
lated as in last problem. 

Let ABODE be the polygonal field ; and let the sides AB, BC, 
CD, DE be given, and also the angles B, C, and D. Join CE 
and CA; then, in triangle ABC, find AC and angle C; and in 



144 



LAND-SURVEYING 



triangle CDE, find CE and angle C; then angle ACE = BCD 
-(ACB + DCE). There are therefore now known two sides and 
a contained triangle in each triangle ; and hence their areas can be 
found, the sum of which is that of the given field. 

EXERCISE 

Find the area of the subjoined field ABCDE from these 
measurements : 




Side AB = 388 
it BC = 311 
CD = 425 
DE = 548 

=2 acres 2 roods 24 '68 square poles. 



Angle B = 110 30' 
,, C = 117 45' 
ii D = 91 20' 



318. Problem IV. To survey a field from two stations 
in it by measuring the distance between them, and all 
the angles at each station contained by this distance, 
and lines drawn from the stations to the corners of the 
field. 

From the data all the lines drawn from one of the stations to 
the corners of the field can be calculated by trigonometry ; and 
then the areas of the triangles contained by these lines, and the 
sides of the figure, can be calculated 'as in the last problem. 

Let ABCD be the given figure, and OQ the 
stations ; measure all the angles at O and Q ; 
then in triangle DOQ the angles are known, 
and the side OQ ; hence find OD ; similarly 
in triangle OQC find OC ; then find OB in 
triangle OBQ ; and, lastly, OA in triangle 
OAQ. Then the areas of the four triangles 
AOB, BOG, COD, DOA can be found as in 
the last problem. 

EXERCISES 

1. Find the area of the field ABCD from these measure- 
ments : 

Angle ra = 120 40' 




n n = 85 30' 

,, x = 25 50' 

v = 20 40' 

and hence u = 107 20' 

and OQ = 1440 links. . 



Angle w = 36 10' 
y = 86 45' 
ti e = 115 16' 
M r = 94 30' 
and hence z = 27 19' 
= 61 acres 1 rood 6 '448 square poles. 



LAND-SURVEYING 



145 



2. Find the area of the field ABCDEF from the subjoined 
measurements : 



Angles at O 



AOB=w=49' 
BOG =x =57' 
COD=w=29< 
DOE=v=64< 
EOF =w=79' 



20' 
10' 
12' 
40' 
25' 
16' 




FQE=e= 



and the distance OQ = 500 links. 

= 12 acres 3 roods 1-18 square poles. 



319. It is evident that, by the preceding method, a field may be 
surveyed from two stations situated outside the field, its area 
computed, and a plan of it made. But in this case the area of 
some of the triangles will have to be subtracted from the sum 
of the areas of the others. 



SURVEYING WITH THE PLANE-TABLE 

320. By means of the plane-table, a plan of a field or estate is 
expeditiously made during the survey, from which the contents 
may be computed by the method described in Article 314. 

321. This instrument consists of a plain and smooth rect- 
angular board fitted in a movable frame 

of wood, which fixes the paper on the 
table, PT, in the adjoining figure. The 
centre of the table below is fixed to a 
tripod-stand, having at the top a ball- 
and-socket joint, so that the table may 
be fixed in any required position. 

The table is fixed in a horizontal posi- 
tion by means of two spirit-levels lying 

in different directions, or by placing a ball on the table, and 
observing the position of it in which the ball remains at rest. 

The edges of one side of the frame are divided into equal parts, 
for the purpose of drawing on the paper lines parallel or perpen- 
dicular to the edges of the frame ; and the edges of the other side 
are divided into degrees corresponding to a central point on the 
board for the purpose of measuring angles. 

A magnetic compass-box, C, is fixed to one side of the table for 
determining the bearings of stations and other objects, and for 




146 LAND-SURVEYING 

the purpose of fixing the table in the same relative position in 
different stations. 

There is also an index-rule of brass, IR, fitted with a telescope 
or sights, one edge of which, called the fiducial edge, is in the 
same plane with the sights, and by which lines are drawn on the 
paper to represent the direction of any object observed through the 
sights. This rule is graduated to serve as a scale of equal parts. 

322. Problem V. To survey with the plane-table from a 
station inside the field. 

Place the table at the station O (fig. to Art. 316) ; adjust it so 
that the magnetic needle shall point to north on the compass card, 
or else observe the bearing of the needle, and fix on some point in 
the paper on the table for this station ; bring the nearer end of the 
fiducial edge of the index-rule to this point, and direct the sights 
to the corner A, and draw an obscure line with the pencil or a 
point along this edge to represent the direction OA ; measure OA, 
and from the scale lay down its length on the obscure line, and 
then the point A is determined. Draw on the table the lines OB, 
OC, OD, and OE exactly in the same way. The points A, B, C, 
D, E being now joined, the plan of the field is finished, and its 
contents may be computed as explained in Article 314, by measure- 
ments taken on the plan. 

The angles at O, subtended by the sides of the field, can also be 
measured at the same time by placing the frame with that side 
uppermost which contains the angular divisions, and then the 
contents of the field can be calculated independently of the plan. 

323. Problem VI. To survey with the plane-table by 
taking stations at all the corners of the field but one, 
and measuring all its sides. 

Let ABODE (fig. to Art. 317) be the field ; place the table at 
some corner, as A, and mark a point in the paper where most 
convenient to represent that station ; adjust the instrument, as 
to the direction of the magnetic-needle, as in last problem. Apply 
the nearer end of the index-rule to this station point, and direct 
the sights to the station E, and draw an obscure line as before to 
denote the direction AE ; then, in a similar manner, through the 
station point, draw a line for the direction AB ; measure AE and 
AB, and with the scale lay off these measures on the obscure lines 
denoting AE and AB. Remove the instrument now to the second 
station B, and place it so that the needle shall rest at the same 



LAND-SDRVEYINO 147 

point of the compass-card as before. If the index-rule is now 
laid along the direction of AB, the first station A will coincide 
with the sights, if the table is properly placed. With the index- 
rule draw an obscure line through the second station point to 
represent BC ; measure BC, and lay the distance oft' on the line 
BC on the paper ; remove the table to the third station C, adjust 
its position as before, and draw a line to represent the direction 
CD ; measure CD, and, by the scale, lay this length off on CD on 
the paper ; and, lastly, place it at D as before, and draw a line to 
represent DE ; this line will meet AE in E, if the work has been 
correctly performed. The plan of the field is now completed. 

324. Problem VII. To survey a field from two stations. 
Let and Q (fig. to Art. 318) be the two stations. Fix the table 

at O, and adjust it as formerly ; assume a convenient point on 
the paper for the first station O ; draw an obscure line to represent 
OQ ; as before, measure OQ, and, with the scale, lay this length off 
on OQ on the paper, and the point for the station Q is determined. 
Then draw obscure lines to represent the lines OC, OB, &c., drawn 
from O to the angles of the field, without measuring these lines, as 
in Article 322 ; and having placed the table at Q, and adjusted it, 
draw obscure lines from Q to represent the lines drawn from Q to 
the corners of the field ; and the intersections of these lines, with 
the former lines from 0, will determine the corners C, B, A, &c., 
and the plan will be completed. 

325. Problem VIII. To survey more than one field with 
the plane-table. 

Having surveyed one of the fields according to any of the 
methods in the three preceding problems, fix on a station in this 
field, whose position is known on the paper, and take some station 
in the adjoining field at a sufficient distance ; then, from the 
former station, draw an obscure line in the direction of the latter, 
measure the distance between them, and lay it off from a scale on 
the paper ; and thus the new station in the adjoining field is 
determined on the plan. Place the table in this station, adjust it, 
and if it is correctly placed, and the index-rule placed on the line 
joining the two last stations, the sights will coincide with the 
station in the first field. Proceed to the planning of this second 
field ; then, in a similar manner, plan the next ; and so on till the 
whole survey is finished, and then measure it as before by means 
of the plan (Art. 314). 



148 LAND-StJRVEtING 

When a new sheet of paper is required in consequence of that 
on the table being filled, some line must be drawn on the latter at 
the most advanced part of the work, and the edge of the former 
being applied to it, the station lines must be produced on this 
sheet. Before drawing the line, the latter sheet must be held in 
such a position as is most convenient for continuing the next part 
of the work upon it. The first sheet being removed from the table, 
and this one, previously moistened, fixed in by means of the frame, 
the work may be continued after the paper has got dry. When 
this sheet is filled, another is similarly fixed on the table ; and 
when the survey is completed, the sheets can all be accurately 
joined by means of the connecting lines. 

At the beginning of the work, the position of some conspicuous 
object or mark may be laid down on the paper, and at any stage 
of the subsequent operation its position may be ascertained ; and 
if it coincide with the first position, it is a proof that the work 
is correct. If not, some error must have been committed, which 
must be rectified before proceeding further ; with this check, the 
greatest accuracy may be secured in the survey. 

EXERCISE 

From a station within a hexagonal field the distances of each of 
its corners were measured, and also their bearings ; required its 
plan and area, the measurements being as below. 

= 12 acres 3 roods 6 '448 square poles. 

Distances Bearings 

To first corner = 708 NE. 

it second u = 957 N E. 

it third . 783 NW by W. 

M fourth ,, . 825 SW by S. 

M fifth 406 SSE7E. 

ii sixth u 589 E by S 3i E. 

This exercise is to be solved like that under Problem II. 



DIVISION OF LAND 

326. It frequently becomes a problem in land-surveying to cut 
off a certain portion from afield. When the field is of a regular 
form, this process may be frequently effected by a direct method ; 
but in the case of irregular fields, it can be accomplished only by 
indirect or tentative methods. 




LAND-SURVEYING 149 

327. Problem IX. To cut off a portion from a rectangular 
field by a line parallel to its ends. 

Find the area of the field ; then as its area is to that of the 
part to be cut off, so is the leiigth of the field to the length of 
the part. 

Let AD be the field, AF the part to be cut c r _ D 
off, then 

Divide the area of the required part by the 
breadth of the field, and the quotient will be 
the length. 

Let = area of AF, and 6 = the breadth AC, and J=the length 
AE; then l j- 

Or, if A area of the field AD, and L = its length AB, then 
A : = L : I, and =-r- = AE. 

EXAMPLE. The area of a rectangular field is = 10 acres 3 roods 
20 square poles, its length is = 1500 links, and breadth = 725 links ; 
it is required to cut off a part from it of the contents of 2 acres 
28 square poles by a line parallel to its side. 

A = 10 acres 3 roods 20 sq. poles = 1087500 links, 
o= 2 M M 28 ii = 217500 



hence, J= = ~x 1500=300 links = AE. 
Or, 1=1 =^|^ = 300 links=AE. 

When any aliquot part is to be cut off from the field, find the 
same part of the base, and it will be the length of the required 
part. 

EXERCISE 

A rectangular field is = 1250 links long and 320 broad; it is 
required to cut off a part of it, to contain 1 acre 2 roods 16 square 
poles, by a line parallel to one of its ends ; what is the length 
of this part? ........ =500 links. 

328. Problem X. To cut off any portion from a triangle 
by a line drawn from its vertex. 

Let ABC be the triangle, and APC the part to be cut off, 
then 

Pne. K 




150 LAND-SURVEYING 

Divide the area of the required part by the altitude of the 
triangle, and the quotient will be half the length of its 
base. 

If a = area of the required part APC, 
l=AP, and h = altitude of the triangle, then 
a ,_2a 

B Or, if A = area of the given triangle ABC, 
L=its base AB, then A : = L : I, and J=-r- = AP. 

EXAMPLE. The length of one side of a triangular field 
is=2500 links, and the perpendicular upon it from the opposite 
corner is = 1240 links; it is required to cut off a triangular 
portion from it, by a line drawn from the same angle to this 
side, so that its contents shall be = 5 acres 16 poles. 

= 5 acres roods 16 sq. poles = 510000 sq. links ; 

. 2a 1020000 , 

hence, I = -r = 10 . A = 822 '6 links = AP. 
n 1240 

EXERCISE 

Cut off from a triangular field, as in the preceding exercise, a 
part containing 2 acres 1 rood 24 square poles, the length of one 
side of the triangle being = 1280 links, and the perdendicular on it, 
from the opposite corner, = 1500. . . Length of base = 320 links. 

329. Problem XI. To cut off any portion from a 
triangular field by a line drawn from a point in 
one of its sides. 

Let ABC be the given triangle, and D 
the given point. 

Cut off a part ACE, by last problem, of 
the required content. Join DE, and through 
C draw CF parallel to DE ; draw DF, and 
it is the line required. 

For triangle DFE=DEC (Eucl. I. 37); and hence triangle 
ADF = ACE = the required area. . 

330. Problem XII. To cut off a part from a triangle 
by a line parallel to one of its sides. 

Let ACB be the given triangle, and AB the side to which 
the required line is to be parallel. 




LAND-SURVEYINa 151 

Let A = area of the given triangle ACB, a = area of the 
required triangle CDE, S = the side CB, s=the 

side CE. Then A : a=S 2 : s 2 , and s i =^ , or 




Hence find s, and make CE equal to it, and through E draw DE 
parallel to AB, and it will be the required triangle. 

EXAMPLE. The area of a triangle ACB is = 5 acres 2 roods 
15 square poles, the side CB is = 1525 links; required the length 
of CE, so that the triangle CDE shall contain 2 acres 1 rood 
10 poles. 

A=5 acres 2 roods 15 sq. poles = 559375 sq. links, 
a=2 1 rood 10 =231250 u 



hence, s=SVT= 152 w = 980-4 Hnks = CE. 



EXERCISE 

The area of a triangle is = 12 -96 acres, its side CB is = 1200 links ; 
required the length of CE, so that the triangle CDE shall contain 
3 -24 acres ..... ."V" . : ' " . . =600 links. 

331. Problem XIII. To cut off from a quadrilateral any 
portion of surface by a line drawn from one of its 
angles, or from a point in one of its 
sides. 

Let ABCD be the quadrilateral. 

1. Let A be the angle from which the line is 
to be drawn. 

Draw the diagonal DB, and cut off a part, 
DE, from it that has the same proportion to 
DB as the required part has to the quadri- 
lateral ; draw AE, EC ; then AECD is equal 
to the required area. Rectify the crooked boundary AEC by 
drawing AF (Prac. Geom.), and it is the required line which cuts 
off the part AFD= AECD = the given area. 

2. When it is required to draw the dividing line from a point, G, 
in one side. 

Draw AF, by the preceding case, then join GF, and through A 
draw a line parallel to GF, cutting CD in H, and a line joining G 
and H will be the required line. 




152 



LAND-SURVEYING 




332. Problem XIV. To cut off any part of the area of 
a given polygon by a line drawn from any of its angles, 

or from a point in one of its sides. 
Let ABCDEF be the given polygon. 
1. Let A be the point from which the line 
is to be drawn. 

Draw the diagonals FB, FC, FD. Cut BF 
in G, so that A : a = BF : BG, A and a being 
the areas of the polygon and of the part re- 
quired. Join AG and GC. Cut DF in H, so 

that A : a=DF : DH, and join CH and HE. Then the crooked 
boundary AGCHE evidently cuts off an area equal to that re- 
quired ; for the triangle AGB is the same part of ABF that a is 
of A, and BGC the same part of FBC, and so on. Hence, rectify 
the crooked boundary AGCHE by drawing from A the straight 
line AI, and AICB is the required part. 

2. When the line is to be drawn from a point P in one of the 
sides. 

Draw AI, as in the first case, then from P draw another line to 
be determined, as GH in the preceding problem. 

333. Problem XV. To cut off any proposed portion from 
a field with curvilineal boundaries by a line from a point 
in its boundary, or by a line parallel to a given line. 

Let ABCE be the given field. 
1. When the line is to be drawn from a point 
in the boundary A. 

Draw a trial line AC, and measure the area 
of the part cut off, AEDC. If it is too great, 
divide the excess in square links by the length 
of AC in links, and make the GF perpendicular 
to AC equal to twice the quotient ; draw GD 
M N parallel to AC ; join AD, and AD is the required 

line. 

For the area of the triangle ADC, considering CD as a straight 
line, is=AC . GF, and therefore equal to the excess. 

2. When the dividing line is to be parallel to a given line MN. 

Draw a trial line PQ parallel to MN, to cut off a portion PBQ 
equal to the required part, and measure it. If it is too small, find 
the defect in square links, and divide it by the length of PQ 
in links, and make the VW perpendicular to PQ equal to the 




LAND-SURVEYING 153 

quotient ; and through W draw RS parallel to PQ, and it will be 
the required line when PR and QS are either parallel or equally 
inclined to PQ. When they are not so, a small correction may 
require to be made, by drawing RS a little nearer to or a little 
farther from PQ. 

INCLINED LANDS 

334. When the surface of a field is inclined, it is not that 
surface, but the surface of its projection on a horizontal plane, 
that is laid down on the plan as its area. This projection is 
just the quantity of surface on a horizontal plane, determined 
by drawing perpendiculars upon it from every point in the 
boundary of the field, or, in other words, by projecting its 
boundaries on a horizontal plane ; and a plan of this projection 
only is made : it is impossible to construct a plan of a curved 
surface on one plane. 

The area of the horizontal projection can easily be computed by 
measuring the angle of acclivity of the field at different places. 
Thus, if ABCD is a vertical section of 
the field, then if AB is measured, and 
the angle of elevation A, the horizontal 
projection AE of AB is AE = AB cos A 
when rad. = l. Thus, if AB = 1200 links, 

and angle A = 15 40', AE = 1 200 x -9628490 = 1155 "41 88, or 1155 
links, is the length of AE on the plan, which must also be 
taken for its length in computing the area. So if BC and angle 
CBF be measured, BF, or its projection EG, can be found ; then 
AG = AE + EG, is the projection of AB and BC. In the same way 
the other dimensions of the projection can be found ; and if a 
theodolite is used for measuring any of the angles contained by 
lines measured on the field, these being horizontal angles on the 
instrument, are just the angles of the projection, and are to be 
used unaltered for constructing the plan. 



CHAINING ON SLOPES 

A = Angle of slope with horizon. 

L = Length of line chained on the slope. 

?=length of line reduced to the horizontal. 

J=LK. 
K=cos A, 




154 



LAND-SURVEYING 

TABLE SHOWING VALUES OF K 



A 


K 


A 


K 


A 


K 


5 


996 


19 


945 


33 


839 


6 


994 


20 


94 


34 


829 


7 


992 


21 


933 


35 


819 


8 


99 


22 


927 


36 


809 


9 


988 


23 


92 


37 


799 


10 


985 


24 


913 


38 


788 


11 


982 


25 


906 


39 


111 


12 


978 


26 


899 


40 


766 


13 


974 


27 


891 


41 


755 


14 


97 


28 


883 


42 


743 


15 


966 


29 


875 


43 


731 


16 


961 


30 


866 


44 


719 


17 


956 


31 


857 


45 


707 


18 


951 


32 


848 







SURVEY OF A ROAD AND ADJOINING FIELDS 
335. Construct a plan of a road and adjoining fields from the 
subjoined field-book and following sketch : 
Plan from following Field-book 




Scale of Chains 



155 



Left Offsets 


Chain-lines 


Bight Offsets 


Along the road 

Cross / 
92 


at O 7 

1404 
840 
725 
300 

From O 6 to R 


48 to corner of field. 

/ 
fence to outside. 

g 


1 

To river 

l>- 




120 
40 

140 
42 

inn 


925 to O 6 
340 


96 40' 


From O 5 to R 


1256 to O 5 

780 
640 

300 
180 


Cross \ 

Cross / 
120 

To road ... 160 
Cross the 

5 
25 

54 
4 

20 


fence to inside. 
> 
fence to outside. 

road to inside of field. 

56 
35 

6 
58 to O 7 
42 


138 15' 


From O 4 to R 


328 to O 4 
78 


122 40' 


From O 3 to R 


605 to O :{ 
320 


156 15' 
From O 2 to L 


625 to O 2 
240 



12 10' NE. 


From Oj 



156 LAND-SURVEYING 

The chain-lines in this field-book are the sides of a polygon, and 
the lengths of all these sides except one, and all its interior angles 
except the two at the extremities of the unknown side, are given ; 
and these are sufficient for constructing it, or for calculating its 
contents. 

The first angle 12 10' gives the bearing of the first chain-line 
that is, its inclination to the meridian ; and as the inclination of 
each of the successive chain-lines to the preceding is given, the 
bearings of all the rest are given. The direction of the meridian 
can therefore be drawn through the first station, as it will lie to 
the left of the first chain-line, making with it an angle of 12 10', 
as AN' in the plan. Then any line, NS, on any convenient part 
of the plan parallel to AN', will be the direction of the meridian. 

The second chain-line lies to the left of the first ; and hence the 
angle of the polygon is here a re-entrant angle, and = 360 - 156 15' 
= 203 45'. 

When any boundary-line crosses a chain-line, as KL in the 
plan, an oblique line is drawn on the right and left opposite to the 
number which denotes the distance of the point of intersection 
from the station at the beginning of the line. Thus, opposite to 
300 and 640, between O 4 and O 3 , oblique lines are drawn for this 
purpose. When any internal boundary, fence, or other important 
line is passed, as at E in DF, a straight line is drawn on both 
sides of the corresponding distance in the chain-line ; and when the 
line is straight to the. end, these lines are marked S, as in the 
field-book opposite to 780, between O 4 and O 3 ; and opposite to 725, 
between O 6 and 7 ; and these two points determine the line. 
Thus, RS is determined by the points E and H. When an offset 
is not at right angles, but in another direction, as along a fence, 

the mark ~^-> is placed over it, as 100 at 780 between O 4 and O 5 . 

SURVEY OP A SMALL FARM 

336. Construct the plan and compute the area of a small farm 
from the following field-book. Area = 66 ac. 3 ro. 13'46 sq. po. 

337. In the following field-book, the expression 'in line with 
fence,' with dots ... on the left or right of the chain-lines, means that 
the point arrived at in the chain-line is in the same line with some 
straight fence on the left or right, which does not extend so far as 
the chain-line. In the subjoined plan the continuous lines represent 
the fences, and the dotted lines represent the lines that are measured, 
and also the extension of the lines of some of the fences. 



Left Offsets 


Chain-lines 


Bight Offsets 




3208 to 3 

OAfid. 





In line with 


2145 
909ft 


...fence. 




In line with... 
To corner where 240 
To end of fence 88 

<> Diagonal 

S? y 


1904 
1324 
1080 
620 
From G! to O 3 


fence, 
three fences meet. 

91 to end of fence. 


s/ 120 


1776 toO! 

1 14-ft 




H O ^ 2 \tjt\ own! rf c 1 ^fi 


KOK 




44 



From O 4 




To road 1 1 1 352 


3040 to O 4 

">>< K 




To NW. corner of stables 92 
To NE. corner of barn 94 
365 


1868 
1704 
1560 

i j.7x 




Cross fence \ 

Cross fence / 
104 


925 

584 

252 

From O 3 


to inside. 
I&t> 
to outside. 


S"S 102 

* tn r>nd nf ? PQ7 


1896 to O 3 
i Qftn 




o c to end of s 401 
150 


534 

From O 2 




i 112 
100 


2340 to O 2 
1760 

1 KQ1 


. . .in line with fence. 




MO 




100 
64 


600 







E by S 6 24' S 
From O x 





158 



LAND-SURVEYING 
Plan from preceding Field-book 




EXTENSIVE SURVEYS WITH THE THEODOLITE 

338. In large surveys with the theodolite, as that of an estate or 
the mapping of a district, extensive chain-lines are run through 
the country, joining a series of successive stations conveniently 
chosen for observing the important or conspicuous objects within 
the limits of the survey ; the bearing of each station-line is also 
observed that is, its inclination to the meridian (see Art. 335) 
and the bearings of each of the distant and important points or 
objects of the survey are observed at least at two different stations. 
Offsets are also measured, in the usual way, for determining the 
positions of objects not far distant from the chain-line. 

Let ABCD (fig. to Art. 335) represent a portion of a chain-line, 
and AN' the direction of the meridian passing through the first 
station A, this direction being determined either by means of the 
magnetic compass, or more accurately by astronomical methods, 
as by the position of the pole-star when on the meridian, or by 
means of the computed culminations of any other star (Astrono- 
mical Prob. XVIII.). The theodolite is first placed in the first 
station, A, and the horizontal circle is brought to a level position 
by means of the adjusting screws ; the index or zero of the vernier 
is now brought to 360 on the limb of the horizontal circle, which 
is now to be fastened to the other part of the head by means of the 
clamping-screw, and the whole head is now turned till 360 is in 
the direction of the meridian line, which is determined by directing 
the telescope till the centre of the QVOSS wires coincides wilh*the 



LAND-SURVEYING 159 

picket placed at N'. The head is now fixed in this position by 
means of the locking-screw below the head of the tripod. The 
upper part of the head to which the vernier is attached is now 
set free by unclamping the screw, the theodolite is directed to 
the picket at the second station, and the upper part is again 
clamped ; and the degree opposite to the vernier, which is under 
the eye-glass of the telescope, is noted, as it measures the bearing 
of the second station from the first. The first station-line, AB, is 
now measured, and the instrument removed to the second station 
at B ; then, after being levelled, and the locking-screw unscrewed, 
the whole head is turned till the telescope is directed back on the 
first station A, when the whole head is again locked. In this 
position of the head, the division 360 is evidently again situated 
in the meridian line passing through the second station, B, on the 
south of this station, for the angle contained by this portion of the 
meridian through B, and the line AB, is equal to the alternate 
angle A. The clamping-screw is now unscrewed, and the tele- 
scope is directed to the third station, and the angle noted as 
before, which in this instance measures the bearing of the second 
station-line, BC, from the meridian drawn through B towards the 
south. The upper part of the head is now clamped, and the second 
station-line measured, the instrument being placed in the third 
station, and the telescope directed back to the second as before. 
This process is continued throughout the survey. 

The measurements of lines and angles thus obtained are suffi- 
cient for determining, on a plan or map, the position of the chain- 
lines and stations. The usual mode of plotting these lines is this : 
a straight line is drawn in the direction in which the meridian line 
is intended to lie on the plan, and the central point of a protract- 
ing scale, or of a protractor (the former of which is divided into 
180, and the latter is a complete circle divided into 360), is 
placed on some convenient point of this line with the degree 
coinciding with the line ; fine marks are then made on the paper 
at the various divisions of the protractor, corresponding to the 
bearings observed ; then the protractor being removed, lines are 
drawn from the assumed central point through these marks, and 
are produced in both directions. These lines will evidently be 
parallel to the directions of the various station-lines joining the 
successive stations. In order to plot these lines on the plan, a 
convenient point is chosen for the first station, and through it 
a line is drawn parallel to the first line of bearing, and in this 
instance between the N and E, because AB (fig. to Art. 335) is in 



160 LAND-SURVEYING 

this direction ; the length of this first station-line is then laid off, 
and the second station is thus determined. Proceed in the same 
manner with the second station -line, and the third station will be 
determined. In this manner the chain-lines and all the stations 
are plotted. 

In order to lay down the positions on the plan of any important 
points or objects in the country, the bearings of each from at least 
two stations are to be observed ; these bearings are also to be 
drawn through the formerly assumed central point by means of 
the protractor ; and then lines being drawn from eacli of the two 
stations parallel to these bearing-lines (or rhumb-lines) respectively, 
their intersections determine the positions of the corresponding 
points. In this manner the positions of the tops of hills, of con- 
spicuous buildings, of a succession of points on the banks of rivers, 
and of other objects are determined. 

The bearings will sometimes exceed 180, but this is no incon- 
venience when a complete or circular protractor is \ised ; but if the 
horizontal circle of the theodolite is graduated into 180 twice 
instead of 360, or if both an ocular and objective vernier are 
attached to the instrument that is, verniers under the eye and 
object-end of the telescope the bearings can all be got in angles 
not exceeding 180. 



MENSURATION OF SOLIDS 

339. In Solid Geometry the magnitudes have three dimen- 
sions namely, length, breadth, and thickness. They do not, 
therefore, exist in one plane, but they can be represented by 
diagrams drawn on a plane. 

DEFINITIONS 

340. When a straight line is at right angles to every line it 
meets in a plane, it is said to be perpendicular to the plane ; 
and if it be at right angles to two straight 
lines in the plane, it can be proved to be at 
right angles to every straight line that meets it 
in that plane. 

Let PL be a plane, CD and EF any two 
straight lines in it, and AB a line perpendicu- 
lar to both these lines ; then AB is perpen- 
dicular to the plane. 





MENSURATION OP SOLIDS 161 

341. The inclination of a straight line and a plane is the 
acute angle contained by that line and a line drawn from the 
point in which the former meets the plane to the foot of the 
perpendicular to the plane, from any point in the 

first line. 

Thus, if AC is a line, and PL a plane, and AE 
a perpendicular on the plane, the angle ACE is 
the inclination of the line AC to the plane PL. 

342. The inclination of one plane to another o 
is the acute angle formed by two lines, one in 

each plane, drawn from any point in their line of common 
section, and perpendicular to this line. This angle is called a 
dihedral angle. 

Let PL and BD be two planes, and CS their line of common 
section, and CL, CA lines in these planes perpendicular to CS ; 
then ACE is the inclination of the planes. 

343. One plane is perpendicular to another when its angle of 
inclination to it is a right angle. 

344. Parallel planes are such as do not meet though produced. 

345. A straight line and plane are said to be parallel if they 
do not meet though produced. 

346. A solid is a figure that has length, breadth, and thickness. 

347. A solid angle is formed by more than two plane angles in 
different planes meeting at a point. 

348. The boundaries of solids are surfaces. A surface no part 
of which is plane is called a curve surface. 

349. Any solid contained by planes is called a polyhedron. 

350. When the solid is contained by four planes it is called a 
tetrahedron ; by six, a hexahedron ; by eight, an octahedron ; 
by twelve, a dodecahedron ; and by twenty, an icosahedron. 

351. The planes containing a polyhedron are called its sides or 
faces, and the lines bounding its sides, its edges. 

352. Two polyhedrons are said to be similar when they are 
contained by the same number of similar sides, similarly situated, 
and containing the same dihedral angles. 

353. A polyhedron is said to be regular when its sides are equal 
and regular figures of the same kind, and its solid angles equal. 

There are only five regular polyhedrons, of 4, 6, 8, 12, and 20 
sides, which are named, as in the definition in Article 350. The 
first is contained by equilateral triangles, the second by squares, 



162 MENSURATION OP SOLIDS 

the third by equilateral triangles, the fourth by pentagons, and the 
fifth by equilateral triangles. 

354. A prism is a solid contained by plane figures, of which two 
are opposite, equal, similar, and having their sides parallel ; and 
the others are parallelograms. 

The two parallel similar sides are called the ends, or terminat- 
ing planes, either of which is called the base ; the other sides are 
called the lateral sides, and constitute the lateral or convex 
surface. The edges of the lateral surface are called lateral edges, 
and those of the terminating planes are called terminating 
edges. The altitude of a prism is the perpendicular distance 
of its terminating planes. The prism is said to be triangular, 
rectangular, square, or polygonal according as the ends are 
triangles, rectangles, squares, or polygons. When the lateral edges 
are perpendicular to the base, it is said to be a right prism ; 
in other cases it is said to be oblique. 

355. A right prism, having regular polygons for its terminating 
planes, is said to be regular. 

The line joining the centres of the ends of a regular prism is 
called its axis. 

356. A parallelepiped is a solid contained by six quadrilateral 
figures, every opposite two of which are parallel. 

It can be proved that these sides are parallelograms. A paral- 
lelepiped is a prism having parallelograms for its terminating 
planes. The other terms applied to a prism, respecting the sides, 
edges, and altitude, are applicable to the parallelepiped. 

357. A cube is a solid contained by six equal squares. 

358. A pyramid is a solid having any rectilineal figure for its 
base, and for its other sides triangles, having a common vertex 
outside the base, and for their bases the sides of the base of the 
solid. The altitude of a pyramid is a perpendicular from its vertex 
on the plane of the base, and the apothem is a perpendicular from 
the vertex on a side of the base. 

The pyramid is said to be triangular, quadrilateral, poly- 
gonal, &c., according as its base is a triangle, a quadrilateral, a 
polygon, &c. 

359. When the base is regular, a line joining its centre and the 
vertex is called the axis of the pyramid. 

360. When the axis of a pyramid having a regular base is 
perpendicular to the base, it is called a regular pyramid. 

361. A cone is a solid contained by a circle as its base, and a 



MENSURATION OF SOLIDS 163 

curve surface, such that any straight line drawn from a certain 
point in it, called its vertex, to any point in the circumference of 
the base, lies wholly in that surface. 

362. The line joining the vertex and centre of the base of a cone 
is called its axis ; and when the axis of a cone is perpendicular 
to its base, it is called a right cone. Other cones are said to be 
oblique. 

The axis of a right cone is also its altitude. A line from the 
vertex of a right cone to any point in the circumference of its base 
is called its slant Side. A right cone may be described by the 
revolution of a right-angled triangle about one of the sides of the 
right angle. 

363. A cylinder is contained by two equal and parallel circles 
and a convex surface, such that any straight line that joins two 
points in the circumferences of these circles, and is parallel to the 
axis, lies wholly in the curve surface. 

The circles are called the bases, ends, or terminating planes 
of the cylinder ; the line joining their centres, its axis. 

364. When the axis of a cylinder is perpendicular to the plane of 
one of its bases, it is called a right cylinder. 

365. A wedge is a solid having a rectangular base, and two 
opposite sides terminating in an edge. 

366. A prismoid is a solid whose ends are any dissimilar parallel 
plane figures, having the same number of sides. 

When the ends of a prismoid are rectangles, it is said to be 
rectangular. 

367. A sphere, or globe, is a solid such that every point in its 
surface is equidistant from a certain point within it, and may be 
generated by the revolution of a semicircle about its diameter. 

The point within the sphere is called its centre ; any line drawn 
from the centre to the circumference, a radius ; and any line 
through the centre, terminated at both extremities by the surface, 
a diameter. 

A cylinder circumscribing a sphere is a cylinder of the same 
diameter as the sphere, whose ends touch the sphere, and whose 
axis passes through its centre. 

368. Circles of the sphere, whose planes pass through the centre, 
are called great circles ; other circles of the sphere are called small 
circles. 

369. A segment of a sphere is a portion of it cut off by a 
plane ; and a segment of a cone, pyramid, or solid with a plane 



164 MENSURATION OF SOLIDS 

base is a portion of it cut off from the top by a plane parallel to 
the base. 

370. A frustum of a solid is a portion contained between the 
base and a plane parallel to it when its base is plane ; or between 
two parallel planes when the solid has no plane base. The frustum 
of a sphere is also called a zone ; and when the ends of a spherical 
zone are equidistant from the centre, it is called a middle zone. 

371. A sector of a sphere is composed of a segment and a cone 
having the same base and its vertex in the centre of the sphere ; 
or it is the difference between these two solids, according as the 
segment is greater or less than a hemisphere. 

372. The unit of measure for solids is a cube, the length of whose 
edge is the lineal unit. Thus, if the lineal unit is 1 inch, a cube 
Avhose edges are each 1 inch is the unit of measure for solids, or, in 
other words, a cubic inch is the cubic unit. So, if the lineal unit 
is a foot, the cubic unit is a cubic foot ; and so on. 

373. The number of cubic units contained in a solid, or in a 
vessel, is called its volume. 

The volume of a solid is also called its solidity, or solid con- 
tents, or cubic contents ; and that of a vessel is called its cubic 
contents, or capacity. 

374. Problem I. To find the solidity of a rectangular 
parallelepiped. 

RULE. Find the continued product of the length, breadth, and 
height, and the result is the solidity. 

Let /, b, and h be the length, breadth, and the height, and V 
the volume or solid contents, then 

V V V 

V = Ibh ; . . I rr, b = - T r, and h = rr- 
bh Ih Ib 

Let AF be a right rectangular parallelepiped. Let its length 
AB be 4 lineal units, as 4 inches, its breadth BC, 2 inches, and its 
height AD, 3 inches. The solid can evidently be divided into 
three equal portions by planes through G and H 
parallel to the base AC ; and into four equal 
portions by means of planes through K, L, M, 
parallel to the side BF ; and into two equal 
portions by a plane through I parallel to BD. 
Each of the small cubes into which the solid is 
now divided is a cubic inch ; the number of cubic 
inches in the lowest portion HC is 4 x 2, or 8, and in the second and 




MENSURATION OF SOLIDS 165 

uppermost portion there are as many ; and in them all, therefore, 
there are 4x2x3, or 24 ; that is, to find the cubic contents of the 
solid, find the continued product of the length, breadth, and height. 

EXAMPLES. 1. Find the number of cubic feet in a parallele- 
piped whose length is = 15 feet, breadth 12 feet, and height 
= 5 feet 6 inches. 

V = Ibh = 15 x 12 x V- = 990 cubic feet. 

2. How many solid feet are contained in a square parallelepiped, 
each side of its base being = 1 foot 4 inches, and its height =5 feet 
6 inches ? 

V = lbh = 1J x 1 x 5 = f x | x Y-=-=9 cubic feet 
=9 cubic feet 1344 cubic inches. 

EXERCISES 

1. Find the solid contents of a block of granite =25 feet long, 
4 broad, and 3 thick =300 cubic feet. 

2. The length of a square parallelepiped is = 15 feet, and each 
side of its base = 1 foot 9 inches ; what are its contents ? 

=45-9375 cubic feet. 

3. Find the number of cubic yards in a rectangular block of 
sandstone, the length of which is=16 feet, its breadth = 9 feet, and 
height = 6 feet 9 inches = 36 cubic yards. 

4. What is the number of cubic feet in a log of wood = 10 feet 
long, 1 foot 6 inches broad, and 1 foot 4 inches thick? 

= 20 cubic feet. 

5. Find the contents of a parallelepiped whose length, breadth, 
and thickness are respectively = 30 '5 feet, 9 '5 feet, and 2 feet. 

=579 '5 cubic feet. 

6. Find the solidity of a block of marble whose length, breadth, 
and thickness are respectively = 10 feet, 5| feet, and 3^ feet. 

=201-25 cubic feet. 

375. Problem II. To find the solidity of a cube. 

RULE. Find the cube of one of its edges, and the result is the 
solidity. 

Let e = an edge of a cube, 

then V^e 3 . 

The reason of the rule is evident, since a cube is 
just a parallelepiped whose length, breadth, and 
height are equal. 

EXAMPLE. How many cubic feet are contained in a block 

Prac. L 




166 MENSURATION OF SOLIDS 

of granite of the form of a cube, one of its edges being =2 feet 
6 inches ? 

V = e?=(2%) 3 = (%)*= ^ = 15-625 cubic feet. 

EXERCISES 

1. Find the solidity of a cube whose edge is = 4 feet. 

= 64 cubic feet. 

2. How many cubic feet are contained in a cube whose edge is 
= 7 feet 6 inches? =421'875 cubic feet. 

3. The edge of a cube is = 12 feet 9 inches ; required its volume. 

=2072-671875 cubic feet. 

4. Find the contents of a cube whose edge is = 6*5 yards. 

=274-625 cubic yards. 

376. Problem III. To find the solidity of a prism, or of 
any parallelepiped. 

RULE. Multiply the area of the base by the 
height, and the product will be the solidity. 

Let b denote the base, and h the height, 
then V = bh. 

EXAMPLE. Find the solidity of a regular tri- 
angular prism, a side of its base being = 5 feet, and 
its length = 20 feet. 

By Art. 268, area of base = '433 x 5 2 = 10'825 ; 
hence V = &A = 10'825x20 = 216'5 cubic feet. 

EXERCISES 

1. What is the solidity of a triangular prism whose length is 
= 10 feet 6 inches, one side of its base being = 14 inches, and the 
perpendicular on it from the opposite angle = 15 inches? 

= 7-65625 cubic feet. 

2. Find the solidity of a regular triangular prism whose length 
is = 9 feet, and one side of its base = l foot 6 inches. 

= 8 -76825 cubic feet. 

3. Find the contents of a square prism whose length is = 20 "5 feet, 
and one side of its base = 2'5 feet. . . . =128-125 cubic feet. 

4. What is the solidity of a regular pentagonal prism whose 
length is = 25 feet, and a side of its base = 10 feet ? 

= 4301 -1935 cubic feet. 

5. Find the contents of a regular hexagonal prism whose length 
is = 18 feet, and a side of its base = 16 inches, =83-138 cubic feet. 




MENSURATION OF SOLIDS 167 

6. What is the solidity of a regular octagonal prism = 20 feet 
long, and a side of its base = 10 feet? . . = 9656 - 854 cubic feet. 

377. Problem IV. To find the surface of a cube, paral- 
lelepiped, or prism. 

RULE I. When the prism or parallelepiped is right, multiply 
the perimeter of the base by the height of the solid, and the pro- 
duct will be the lateral surface, to which add double the area of 
the base, and the sum is the whole surface of the solid. 

RULE II. When the prism or parallelepiped is oblique, 
its lateral surface is found by multiplying the perimeter of a 
section perpendicular to one of the lateral edges by that edge. 

The surface of a cube can be found by the first rule ; but it is 
more readily found by taking six times the square of one of its 
edges. 

Let e =one of the lateral edges of a prism or parallelepiped, 
p = the perimeter of the base when the solid is right, 
p'= it it of a section perpendicular to one of the 

edges, UVW (fig. to Prob. III.), 
b =area of the base, 
s = whole surface ; 

then s =pe + 26, when the solid is right, 
and s =p'e + 2b, is oblique ; 

then s =6e 2 , when the figure is a cube. 

The reason of the rule is evident from those for the Mensuration 
of Surfaces. 

EXAMPLES. 1. Find the surface of a cube, one of its edges 
being =18 inches. 

s = 6e 2 = 6x(l-5) 2 = 6x 2-25 = 13'5 square feet. 

2. "What is the surface of an oblique prism =20 feet long, the 
perimeter of a section perpendicular to one of its lateral edges being 
=25 feet, and its base a rectangle = 6 feet long and 4 broad? 
=25x20 + 2x4x 6=500 + 48=548 square feet. 



EXERCISES 

1. Find the surface of a cube whose edges are each = 10 feet. 

= 600 square feet. 

2. What is the surface of a cube whose edge is = 2 feet 4 inches? 

= 32 square feet. 

3. Find the number of square yards in the surface of a cube 
whose edge is = 1 1 feet. . = 80 square yards 6 square feet. 



168 MENSURATION OF SOLIDS 

4. What is the surface of a right rectangular parallelepiped 
whose length is = 36 feet, breadth = 10 feet, and thickness = 8 feet? 

= 1456 square feet. 

5. The length of a rectangular cistern within is = 3 feet 2 inches, 
the breadth = 2 feet 8 inches, and height = 2 feet 6 inches ; required 
the internal surface, and also the expense of lining it with lead at 
2d. per lb., the lead being 7 Ib. weight per square foot. 

= 37H square feet, and 2, 3s. 10|d. 

6. Find the surface of a right triangular prism, its length being 
= 20 feet, and the sides of its base respectively = 6, 8, and 10 feet. 

= 528 square feet. 

7. What is the surface of a regular pentagonal prism whose 
length is = 32 feet, and a side of its base = 6 feet? 

= 1150-037 square feet. 

8. What is the surface of an oblique prism, having a regular 
hexagonal base whose side is = 10 inches, the lateral edges of the 
prism being =20 feet, and the perimeter of a section perpendicular 
to them = 4 feet? = 93'6084 square feet. 



378. Problem V. To find the solidity of a cylinder. 

RULE. Multiply the area of the base by the altitude 
of the cylinder, and the product will be the solidity. 
Or, y=bh, where b= 7854^, or irt a by Art. 273. 

EXAMPLE. What is the solidity of a cylinder 
whose length is = 21 feet, the diameter of its base 
being = 15 inches ? 
Here b = -7854^ = -7854 x ( I -25) 2 = 1 -227 ; 

and V = bh = 1 -227 x 21 = 25 '767 cubic feet. 

EXERCISES 

1. Find the solidity of a cylinder the height of Avhich is = 25 
inches, and the diameter of its base = 15 inches. =2-5566 cubic feet. 

2. What is the volume of a cylinder whose altitude is = 28 feet, 
and diameter =2 feet? = 137 "445 cubic feet. 

3. The circumference of the base of an oblique cylinder is =20 
feet, and its perpendicular height = 19'318 ; what is its volume? 

= 614-91 cubic feet. 

4. The circumference of the base of an oblique cylinder is = 40 
feet, its axis = 22 feet, and the axis is inclined to the base at an 
angle of 75 ; what is its volume ? . . =2705-6818 cubic feet. 



MENSURATION OF SOLIDS 169 

379. Problem VI. To find the surface of a right cylinder. 
RULE. Multiply the circumference of its base into its height, 

and the product is the convex surface ; and double the area of 
the base being added, gives the whole surface of the cylinder. 
Let d, r, and c diameter, radius, and circumference of base, 

h = height of cylinder, 

6= its base, 

z= con vex surface ; 
then 
and 

EXAMPLE. The radius of the base of a right cylinder is =5 feet, 
and its height =20 ; what is its surface? 

z=ch=2wrh = 2 x 3'1416 x 5 x 20 = 628'32 ; 

2& = 2a-- 2 =2x3-1416x5 2 =157'08; 
hence s = z + 2b = 785 '4 square feet ; 

or s=2irr(h + r) = 2 x 3 '1416 x 5 x 25 = 785 "4 square feet. 

The curve surface of a right cylinder is evidently equal to the 
area of a rectangle whose height is that of the cylinder, and length 
equal to its circumference. 

EXERCISES 

1. Find the surface of a right cylinder whose length is =20 feet, 
and circumference = 6. .... = 125 72958 square feet. 

2. What is the convex surface of a right cylinder whose diameter 
is = 10 inches, and length = 14J feet? . . =37 '961 square feet. 

3. Find the convex surface of a cylinder whose length is = 40 
feet, and the diameter of its base =4 feet. =502 '656 square feet. 

4. What is the superficies of a right cylinder whose length is 
=40 feet 8 inches, and the diameter of its base = 10 feet 6 inches? 

= 1514-6439 square feet. 

380. Problem VII. To find the solidity of a 
pyramid. 

RULE. Multiply the area of the base of the pyra- 
mid by its perpendicular height, and one-third of the 
product is the solidity. 

V-iW, 

EXAMPLE. Find the solidity of a rectangular 
pyramid, the length and breadth of its base being = 6 and 4 feet 
respectively, and its altitude = 20 feet. 

V = JWt = x 6 x 4 x 20 = 160 cubic feet. 




170 MENSURATION OF SOLIDS 

EXERCISES 

1. What is the solidity of a square pyramid, each side of its base 
being = 3 feet, and its altitude = 10 feet? . . =30 cubic feet. 

2. Find the volume of a regular triangular pyramid, a side of its 
base being = 6 feet, and its altitude =60 feet. =311 '769 cubic feet. 

3. Find the solidity of a square pyramid, a side of its base being 
= 30 feet, and its apothem = 25 feet. . . =6000 cubic feet. 

4. What is the solidity of a pentagonal pyramid, with a regular 
base, each side of which is = 4 feet, and the altitude of the pyramid 
=30 feet? . . . . . . =275 -276 cubic feet. 



381. Problem VIII. To find the surface of a pyramid. 

RULE. When the pyramid is regular, multiply the perimeter of 
the base by the apothem of the pyramid, and half the product is 
the convex surface, to which add the area of the base, and the sum 
is the whole surface. 

When the pyramid is irregular, find separately the areas of 
the lateral triangles, and to their sum add the area of the 
base. 

Let c = the perimeter of the base of a regular pyramid, 

p = the apothem of the pyramid = VQ (fig. to last problem), 

and z and b, as in Prob. VI. ; 
then, for a regular pyramid, 

z = %pc, and s=z + b. 

For the area of one of the lateral triangles is evidently equal to 
half the product of the apothem by the base of the triangle ; hence 
the truth of the rule is evident. 

EXAMPLE. Find the surface of a square pyramid, its apothem 
being=40 feet, and each side of its base = 6 feet. 

z=\cp=\ x 24 x 40=480, 
and *=2 + 6 = 480 + 6 2 =516 square feet. 

EXERCISES 

1. What is the surface of a square pyramid, a side of its base 
being = 5 feet, and the apothem of the pyramid = 12 feet? 

= 145 square feet. 

2. Find the convex surface of a pyramid whose apothem is = 10 
feet, and its base an equilateral triangle whose side = 18 inches. 

22 - 5 square feet. 




MENSURATION OP SOLIDS 171 

3. What is the surface of a regular pentagonal pyramid whose 
apothem is = 10 feet, and each side of its base = l foot 8 inches? 

=46 - 4457 square feet. 

4. The apothem of a regular hexagonal pyramid is = 8 feet, and a 
side of its base=2 feet ; what is its surface ? =76-24 square feet. 

382. Problem IX. To find the solidity of a cone. 

RULE. Multiply the area of the base by the altitude of the 
cone, and one-third of the product is the solidity. 

Or, V^bh, where b is found by Art. 270. 

EXAMPLE. Find the solidity of a right cone, the slant side 
of which is = 5 feet, and the diameter of its base y 

= 6 feet. 

Here h = VD, and VD 2 =AV 2 -AD 2 , or if AV=p, 

and b = -7854^ = '7854 x 6 2 = 28 -2744 square feet ; 
hence V=6A = x28 -2744 x 4 = 37 '6992 cubic feet. 

EXERCISES 

1. The altitude of a right cone is = 30 feet, and the diameter of 
its base = 6 feet ; what is its volume ? . . =282-744 cubic feet. 

2. The diameter of the base of a cone is = 10, and its altitude 
= 12 ; what is its solidity ? =314-16. 

3. Find the volume of a cone whose altitude is = 10 feet, and 
the diameter of its base =2 feet 8 inches. . =18 '61 7 cubic feet. 

4. Find the solidity of a cone, the diameter of whose base is 
= 3 feet, and its altitude = 30 feet. . . =70*686 cubic feet. 

5. The diameter of the base of a cone is = 3 feet 4 inches, and 
its slant side = 16 feet ; what is its solidity ? =46-2856 cubic feet. 

6. The circumference of the base of a cone is =20 feet, and its 
height =25; required its volume. . . =265 -258 cubic feet. 

383. Problem X. To find the surface of a right cone. 
RULE. Multiply the circumference of the base of the cone by 

the slant side, and half the product will be the curve surface, to 
which add the area of the base, and the sum will be the whole 
surface. 

Or, z\cp, and 6='7854d 2 , orirr 2 , and s=z + b. 

It is evident that if the cone (last fig.) AVB be rolled on a plane, 
the curve surface will be equivalent to a circular sector whose 



172 MENSURATION OF SOLtt>8 

radius is BV, the slant side of the cone, and its arc the circum- 
ference of the base of the cone, from which the rule is evident. 

EXAMPLE. Find the surface of a cone whose base has a 
diameter of 12 feet, and whose height is = 8 feet. 

Here (last fig. ) AD = r = 6, and D V = h = S ; 
hence AV 2 =$* = W + r 2 = 64 + 36 = 100, and p = 10 ; 

therefore, z=\cp = \ x 3'1416 x 12 x 10= 188-496, 

and s = z + b = 188-496 + '7854 x 12 2 = 301 -5936 square feet. 

EXERCISES 

1. What is the surface of a cone, the diameter of its base being 
= 5 feet, and its slant height = 18 feet? . =161'007 square feet. 

2. The slant height of a cone is =40 feet, and the diameter of 
its base = 9 feet ; what is its surface ? . =629-1054 square feet. 

3. The diameter of the base of a cone is = 6 feet, and its slant 
height = 30 feet ; required its convex surface. =282 '744 square feet. 

4. The slant height of a cone is = 18J feet, and the circumference 
of its base = 10f feet ; find its convex surface. 

= 98-09375 square feet. 

384. Problem XI. To find the solidity of a frustum of a 
pyramid. 

RULE I. Add together the areas of the two ends and their mean 
proportional, multiply this sum by the altitude of the frustum, 
and one-third of the product will be the solidity. 

RULE II. Multiply the area of the greater end by one of its 
sides, and that of the smaller end by its corresponding side ; divide 
the difference of these products by the difference of the sides, and 
multiply the quotient by the height of the frustum, and one-third 
of this product will be the solidity. 

RULE III. When the ends are regular polygons, to the sum of 
the squares of the ends add the product of the ends ; 
multiply the sum by the tabular area correspond- 
ing to the polygons, and by a third of the height, 
and the result will be the solidity. 

Let MNPUTS be a frustum of a pyramid, the 
complete pyramid being VMNP. 

Let the heights of the whole pyramid and the 
smaller one VSTU be h', h", and that of the 
frustum h ; and let V, V", V denote the solidities 
of these three solids respectively ; B, b the greater and smaller 




MENSURATION OF SOLIDS 173 

ends of the frustum ; and E, e two of their corresponding sides, as 
MN, ST ; also, 

When the ends are regular polygons, let A' = the corresponding 
tabular area (Art. 268), then the three rules above can be expressed 
thus : 



EXAMPLE. Find the solidity of a frustum of a square pyramid, 
a side of the ends being = 6 and 4 feet, and the altitude = 10 feet. 
By the first rule 

VB6 = V(36 x 16) = V576 =24 ; 

hence V = ( B + b + VBfe) = ^(36 + 16 + 24) = 253$ cubic feet. 

o o 

By the second rule 

10/36x6-16x4\ 10 __ OK01 ,. . t 
^( - 6^4 - )=y*76=253i cubic feet. 

By the third rule 



= J(36 -f 24 + 16) x 10 = x 76 x 10 = 253J cubic feet. 

If V' = JBA', and V"= J6A"; 
then V = V'-V"=J(BA'-6A") ...... [1]. 

But the two ends are proportional to the squares of two of their 
corresponding sides, as the}' are similar, or of the edges MV, SV, 
or of the altitudes h', h"; 



hence, B : b=h^ : h"*; hence A"=/t'v ...... [2J 

Also, VB : \Jb = h' : h", 

and VB-V& = VB=A'-A" or h : h' ...... [3]. 

Therefore, h' = .^ TL 5 an( * hence h" = -^5 - 77. 
VB - V& V B - V o 

Substituting these values of h', h" in [1], it becomes 



This result is the first rule. In order to prove the second, 
substitute in the proportions [2], [3] above the quantities E and e, 
instead of \fR and \Jb, since they are proportional to them, and 
the expressions for h' and h" will then become 

' ; which, being substituted in [1], gives 



_ l EhE-bhe_fi/'BE-be\ 
~ * E-e ~3\ E-e )' 



174 MENSURATION OF SOLIDS 

When the ends are regular polygons, then A' being the tabular 
area, as in Art. 268, B = A'E 2 , and 6 = A'e 2 ; hence, substituting 
these values for B and 6, the last expression for s gives 



EXERCISES 

1. Find the solidity of a frustum of a square pyramid, the sides 
of its two ends being = 3 feet and 2 feet, and its height = 5 feet. 

= 37}| cul) ic feet. 

2. Find the solidity of a frustum of a square pyramid, the sides 
of its ends being = 10 and 16 inches, and its length = 18 feet. 

='J1 cubic feet. 

3. What is the solidity of a frustum of a regular hexagonal 
pyramid, the sides of its ends being=4 and 6 feet, and its length 
= 24 feet? ....... =1579-6303 cubic feet. 

4. Find the solidity of a frustum of a regular octagonal pyramid, 
the sides of its bases being = 3 and 5 feet, and its height = 10 feet. 

= 788-643 cubic feet. 

5. Find the number of solid feet in a piece of timber of the form 
of a frustum of a square pyramid, the sides of its ends being = 1 foot 
and 2 feet, and the perpendicular length of one of the sides = 48 
feet ..... .;_.', . =155-981 cubic feet. 

385. Problem XII. To find the surface of a frustum of a 
pyramid. 

RULE I. When the pyramid is regular, add together the peri- 
meters of the two ends ; multiply their sum by the lateral length, 
and half the product will be the lateral surface ; to which add the 
areas of the two ends, and the sum will be the whole surface. 

RULE II. When the pyramid is irregular, the lateral planes are 
trapeziums, and their areas being separately found by Art. 259, and 
those of the two ends added, the sum will be the whole surface. 

Let P and p be the perimeters of the two ends, and I the lateral 
length, or apothem, and B and b the areas of the two ends ; then 
the first rule is 



EXAMPLE. What is the surface of a frustum of a regular 
triangular pyramid, a side of its ends being = 3 and 2 feet, and 
the lateral length = 10 feet ? 



MENSURATION OP SOLIDS 175 

Here P=3x3=9,^?=2x3 = 6 ; 

hence a = J(P +/>)+ B + 6 = 4(9 + 6) x 10 + '433(3 2 + 2*) 
= 75+ -433 x 13=80-629 square feet. 

EXERCISES 

1. Find the surface of a frustum of a regular square pyramid, 
the sides of its ends being = 14 and 24 inches, and the lateral length 
=2 feet 3 inches ..... ' . =19 '61 square feet. 

2. Find the surface of a frustum of a regular square pyramid 
whose lateral length is = 5 feet, the sides of its ends being = 13 and 
20 inches ....... =31 '451 38 square feet. 

3. What is the surface of a frustum of a regular pentagonal 
pyramid, its lateral length being = 5 feet 10 inches, and the sides 
of its ends = 10 and 15 inches ? . . =34'26496 square feet. 

386. Problem XIII. To find the solidity of a frustum of 
a cone. 

RULE I. To the squares of the diameters of the two ends add 
the product of the diameters ; multiply the sum by the height of 
the frustum, and this product by '2618 ; the result will be the 
solidity ; or, 

RULE II. To the squares of the circumferences of the two ends 
add the product of the circumferences ; multiply the sum by the 
height of the frustum, and this product by -026526 ; the result will 
be the solidity. 

Let D and d be the diameters of the two ends, C and c their 
circumferences, and h the height of the frustum (fig. to Prob. IX.) ; 
then V=-2618A(D 2 + eP + DeO, 

and V = -026526A(C 2 + c 2 + Cc). 

EXAMPLE. What is the solidity of a frustum of a cone whose 
height is = 5 feet, the diameters of its two ends being respectively 
=2 and 3 feet. 



= 1-309x19=24-871. 

Let the notation used in Prob. XI. for the frustums of pyramids 
be similarly applied to the conic frustums in the figure, and by the 
same reasoning it is proved that 



But B= -7854D 2 , and b= 7854<2 2 ; and hence 



and therefore V = '7854(D 2 + cP + Dd) = -2618A(D 2 + d? + Dd) ; which 
is the first rule. 



176 MENSURATION OF SOLIDS 

And since '7854D 2 = -0795775C 2 by Articles 273 and 274, and 
7854^ = -0795775c 2 , and 7854Drf= -0795775Cc ; by substitution, 



V - -0795775(C 2 + c 2 + Cc) = -026526/t(C 2 + c 2 + Cc). 
o ^ 

EXERCISES 

1. What is the solidity of a frustum of a cone whose height 
is = 10 feet, and the diameters of its ends = 2 and 4 feet ? 

= 73-304 cubic feet. 

2. Find the solidity of a conic frustum of which the height is 
= 9 feet, and the diameters of its ends = l and 2 feet. 

=28-8634 cubic feet. 

3. What is the solidity of a conic frustum whose height is 
=4 feet, and the diameters of its two ends = 2 and 4 feet ? 

= 32-9868 cubic feet. 

4. What is the solidity of a conic frustum whose length is 
=25 feet, and the diameters of its two ends = 10 and 20 feet? 

=4581'5 cubic feet. 

5. Find the solidity of a conic frustum, its length being = 38 
inches, and the diameters of its ends = 18 and 32 inches. 

= 19140-72 cubic inches. 

6. How many cubic feet are contained in a ship's mast whose 
length is = 72 feet, and the diameters of its ends = l foot and 1? 

= 89 -5356 cubic feet. 

7. How many cubic feet are contained in a cask which is 
composed of two equal and similar conic frustums, united at 
their greater ends, its bung diameter being = 14 inches, its head 
diameter = 10 inches, and its length = 20 inches? 

= 1-32112 cubic feet. 

387. Problem XIV. To find the surface of a frustum of 
a right cone. 

RULE. Multiply the sum of the circumferences of the two ends 
by the slant side, and half the product will be the convex surface ; 
to which add the areas of the two ends for the whole surface. 

Let C, c be the circumferences of the ends, D and d their 
diameters, B and b their areas, and p the slant side AE (fig. to 
Prob. IX.); 

then s= ^p(C + c) + E + b; 

where C + c = 3'1416(D + d), and B + 6='7854(D 2 + rf 2 ). 

EXAMPLE. Find the whole surface of a frustum of a right 
cone, the slant side being=20 feet, and the diameters of the 
ends = 2 and 4 feet. 



MENSURATION OP SOLIDS 177 



6=18-8496, 
B + b = -7854(D 2 + d 2 ) = -7854 x 20 = 15 "708, 
and s = ip(C + c) + B + 6 = x20x 18-8496 + 15-708 

= 204-204 square feet. 

The rule is easily derived from that in Prob. X. For let C, c be 
the circumferences of the two ends, p the slant side AE (fig. to 
Art. 382) of the frustum, and p', p" the slant sides of the two 
cones VAB, VEG, and *', s" their convex surfaces, and s that 
of the frustum ; then (Art. 383) 

s'=^p'C, s" = %p"c ; therefore *=' -s" = ^(p'C -p"c), 
or S =${pC+p"C-p"c} = l{pC+p"(C-c)} ...... [1]. 

But C : c =p' : p" ; and hence C - c : c =p' -p", or p : p" ; therefore 

p" = n , and substituting this in [1] for p", it becomes 




) = the curve surface. 

EXERCISES 

1. What is the convex surface of a frustum of a right cone whose 
slant side is = 10 feet, and the circumferences of its two ends = 5 
and 15 feet ? ....... = 100 square feet. 

2. Find the convex surface of a frustum of a right cone whose 
slant side is = 39 feet, and the circumferences of its two ends 
= 15 feet 9 inches and 22 feet 6 inches. . =745-875 square feet. 

3. What is the surface of a frustum of a right cone, its length 
being =31 feet, and the diameters of its two ends = 12 and 20 feet? 

= 1985-4912 square feet. 

4. If a segment whose slant side is = 6 feet is cut off from the 
upper part of a cone whose slant side is = 30 feet, and the circum- 
ference of its base = 10 feet, what is the convex surface of the 
frustum ? ........ = 144 square feet. 

388. Problem XV. To find the solidity of a wedge. 

RULE. To twice the length of the base add the length of the edge, 
and find the continued product of this sum, the breadth of the base 
and the height of the wedge, and take one-sixth of this product. 

Let I, e, b, and h denote respectively the length of the base AB, 
the edge EG, the breadth of the base BD, and the height ; 
then v = $(e + 2l)bh. 

EXAMPLE. Find the solidity of a wedge, the length of which 
is =5 feet 4 inches, its base = 9 inches broad, the edge of the 
wedge = 3 feet 6 inches, and its height =2 feet 4 inches. 

v=$(e + 2l)bh = i(3 + 10)| x 2J = x 14 x f x = 4-132 cubic feet. 



178 



MENSURATfON OP SOLIDS 



Let ADE be a wedge. Through E let a plane EFH pass 
parallel to the end GDB. Then EDH is a 
prism, and is equal to three times the pyramid, 
whose base is the triangle DBH, and vertex G 
(Solid Geom. II. 17), or it is 




Also, the pyramid ECAF=AF. ^ = 
hence V={& + $(l-e)}b 

Were the edge longer than the base, the formula would be 
the same; for the expression \e + ^(l-e) would then become 

), as before. 



EXERCISES 

1. Find the contents of a wedge whose base is = 16 inches long 
and 2J broad, its height being = 7 inches, and its edge = 10J inches. 

= 111-5625 cubic inches. 

2. The length and breadth of the base of a wedge are = 5 feet 
10 inches and 2^ feet, the length of the edge is =9 feet 2 inches, 
and the height = 34 '29016 inches ; what is its solidity ? 

=24-8048 cubic feet. 




389. Problem XVI. To find the solidity of a prismoid. 
RULE. To the sum of the areas of the two ends add four times 
the area of the middle section parallel to the ends ; multiply this 
sum by the height, and take one-sixth of the product. 
Let L =the length of the base AB, 
B = M breadth of the base BD, 
I = M length of the top GH, 
b = breadth of the top GH, 
M= n length of middle section, 
m = M breadth of middle section, 
h = ii height of the prismoid ; 
then M = J(L + ), and wi=^(B + 6), 

and V = i(BL + bl + 4Mm)A. 

EXAMPLE. Find the solidity of a prismoid, the length and 
bread tli of its base being = 10 and 8, those of the top = 6 and 5, 
and the height = 40 feet. 

Here M = (L + /) = ^(10 + 6) = x 16 = 8, 

i = ^(B + 6) = ^( 8-f 5) = x 13 = 6*5, 

=$(10 x8 + 6x5 + 4x8x 6'5)40 = 2120 cubic feet. 



MENSURATION OF SOLIDS 179 

The prismoid ADG is evidently equal to two wedges ADG and 
GFC ; the base and edge of the former being AD and GH, and 
those of the latter FG and CD ; and their height that of the 
prismoid. 

Let V and V" be the volumes of these wedges ; 
then V'=$(l + 2L)Eh, V"=$(L + 2l)bh, and V=V' + V" ; 
also, 4M?M=(B + &)(L + J) = BL + W + BJ + &L; 

hence V = J(2BL + EI + 2bl + bL)h = J(BL + bl + Mm)h. 

EXERCISES 

1. What is the solidity of a log of wood of the form of a rect- 
angular prismoid, the length and breadth of one end being =2 feet 
4 inches and 2 feet, and those of the other end = 1 foot and 8 inches, 
and the height or perpendicular length = 61 feet? 

= 144 -592 cubic feet. 

2. Find the capacity of a trough of the form of a prismoid, its 
bottom being = 48 inches long, 40 inches broad, and its top = 5 feet 
long and 4 feet broad, and its depth = 3 feet. . =49| cubic feet. 

3. What is the volume of a prismoid, the length and breadth of 
its greater end being = 24 and 16 inches, those of its top = 16 and 12 
inches, and its length = 120 inches ? . . =19 '629 cubic feet. 

390. Problem XVII. To find the solidity of a sphere. 

RULE I. Find the solidity of the circumscribing cylinder 
that is, a cylinder whose diameter and height are equal to the 
diameter of the sphere and two-thirds of it will be the volume 
of the sphere. 

RULE II. Multiply the cube of the diameter of the sphere by 
5236, or more accurately by -5235988, and the product will be its 
solidity. 

Let rf=the diameter of the sphere ; 
then= V=-5236d 3 . 

EXAMPLE. Find the solidity of a sphere 
whose diameter is 2 feet 8 inches. 

V= -5236^= -5236 x (2|) 3 = '5236 x (f ) s 

= -5236 x 4jV-=9-929 cu bi c feet. 
The first rule is derived from a theorem 

discovered by Archimetfes. The second rule is easily derived 
from the first. For (Art. 378) the volume of a cylinder is 
r = JA='7854cT 2 A ; and when h = d, which is the case for the cylinder 




180 MENSURATION OP SOLIDS 

ABV circumscribing the sphere, then v=-7854e 2 e?=*7854eP, and 
two-thirds of this is the volume of the sphere, or V = '5236c? 3 . 

Note. A sphere may also be considered as composed of an 
indefinite number of minute pyramids, Avhose bases are in the 
surface, and vertices in the centre of the sphere, and the 
sum of their solidities would be equal to the surface of the 
sphere, multiplied by one- third of its radius. But (Art. 391) 
the surface = 4 x '7854G? 2 ; hence the volume = 4 x '7854rf 2 x \d 
= -5236a 3 . 

The volumes of spheres are proportional to the cubes of the radii 
of the spheres (Eucl. XII. 18). 

EXERCISES 

1. How many cubic inches are contained in a sphere =25 inches 
in diameter? ...... =8181 '25 cubic inches. 

2. Find the solidity of a sphere, the diameter of which is = 8 
inches ........ =321-5558 cubic inches. 

3. What is the solidity of a sphere whose diameter is=5 inches? 

= 65 '45 cubic inches. 

4. How many cubic feet of gas can a balloon of a spherical form 
contain, its diameter being = 50 feet? . . = 65450 cubic feet. 

5. Find the solidity of a sphere whose diameter is = 6 feet 
2 inches ........ =122 '7866 cubic feet. 

6. The diameter of a globe is = 4 feet 2 inches; what is its 
volume? ........ =37 '876 cubic feet. 

391. Problem XVIII. To find the surface of a sphere. 

I. The surface of a sphere is equal to four times the area of a 
great circle of the sphere ; or, 

II. The surface of a sphere is equal to the product of the square 
of its diameter by 3'1416 ; or, 

III. The surface of a sphere is equal to the product of its circum- 
ference by its diameter ; or, 

IV. The surface of a sphere is equal to the convex surface of the 
circumscribing cylinder. 

The surface and the volume of a sphere are two-thirds of the 
surface and the volume of the circumscribed cylinder. * 

s = cd, or s = 



Archimedes' Sphere and Cylinder, I. 34, Cor. (Heiberg's edition)i 



MENSURATION OF SOLIDS 181 

EXAMPLE. How many square feet of sheet-copper are contained 
in a hollow copper globe =25 inches in diameter? 

s = 3-1416d 2 =3'1416 x 25 2 =3-1416 x 625 
= 1963 "5 square inches = 13*6347 square feet. 

Let mnm'ri and rsr's' be two corresponding zones of the sphere 
GEVF and its circumscribing cylinder ABV (last fig. ). The area 
of the cylindric zone is equal to the circumference of the cylinder 
xrs=2ir . Kr. rs. Also, the surface of the spherical zone is, when 
its breadth is exceedingly small, equal very nearly to the surface 
of a frustum of a cone, and equal to the middle circumference of 
the zone x mn (Art. 387), or nearly =2ir . Km . mn. 

But, from similar triangles, mn : mo = Cm : Km ; hence Km . mn 
= Cm . mo=Kr . rs ; therefore, 2?r . Km . mn=2ir . Kr . rs. That is, 
the surfaces of the spherical and cylindric zones are equal. The 
same can be similarly proved of the surfaces of all the other corre- 
sponding small zones of these two solids. Hence the whole surface 
of the sphere is equal to the convex surface of the circumscribing 
cylinder 



This proposition can only be proved with rigorous accuracy and 
conciseness by means of the calculus. 

EXERCISES 

1. How many square inches of gold-leaf will gild a globe = 1 foot 
in diameter? ...... =452 '39 square inches. 

2. What is the surface of a sphere whose diameter is = 2 feet 
9 inches? ....... =23*75835 square feet. 

3. Find the surface of a globe whose diameter is = 51 inches. 

= 5674515 square feet. 

4. Find the surface of a ball whose diameter is = 5 inches. 

= 78 '54 square inches. 

5. What is the surface of a sphere whose diameter is = 2 feet 
8 inches? ...... . =22 -34 square feet. 

6. What is the surface of a globe whose diameter is = 9 inches ? 

= 1 '767 square feet. 

392. Problem XIX. To find the surface of any spherical 
segment or zone. 

RULE I. Multiply the circumference of the sphere by the 
height of the segment or zone, and the product will be the area ; 
or, 

RULE II. Multiply the diameter of the sphere by 3'1416, and 

Pw, M 



182 MENSURATION OF SOLIDS 

the product by the height of the segment or zone ; the product will 
be the area. 
f. Or, s=ch, or s= 



EXAMPLE. What is the surface of a spherical zone whose 
height is = 4 feet, the diameter of the sphere being =5 feet? 
s=3-1416c#t = 3-1416x 5x4 = 62'832 square feet. 

It was proved in Prob. XVIII. that the surfaces of any two corre- 
sponding zones of a sphere and its circumscribing cylinder are equal. 
Now, the surface of any zone of the cylinder is evidently equal to 
the circumference of the cylinder, or of the sphere, multiplied by 
the height of the zone ; and hence the surface of the spherical zone 
is found in the same manner. 

EXERCISES 

1. Find the convex surface of a spherical zone whose height is 
4: inches, the diameter of the sphere being =1 foot. 

= 150 '7968 square inches. 

2. Find the convex surface of a spherical zone, the height of 
which is =5 inches, and the diameter of the sphere =25 inches. 

= 392'7 square inches. 

3. What is the convex surface of a spherical segment whose 
height is = 3 feet 6 inches, the diameter of the sphere being 
= 10 feet? ....... =109 -956 square feet. 

4. Find the number of square inches in the convex surface of a 
spherical segment whose height is =2 inches, the diameter of the 
sphere being =6 inches. . . . =37 '6992 square inches. 

5. What is the convex surface of a spherical segment whose 
height is = 9 inches, the diameter of the sphere being = 3 feet 
Cinches? ...... =1187 '5248 square inches. 

393. Problem XX. To find the solidity of a spherical 
segment. 

RULE I. To three times the square of the radius of the base of 
the segment add the square of its height ; multiply the sum by 
the height, and the product by '5236, and the result will be the 
solidity. 

RULE II. From three times the diameter of the sphere subtract 
twice the height of the segment ; multiply the difference by the 
square of the height, and the product by '5236 ; the result will be 
the solidity. 




MENSURATION OF SOLIDS 183 

Let ACBV be a spherical segment, 

h = VC its height, 

r = AC the radius of its base, 
and d = the diameter of the sphere ; 

then V = -5236/i(3r 2 + A 2 ), 

or V= -5236^(3^ -2h). 

EXAMPLES. 1. The height of a spherical 
segment is = 8 inches, and the radius of its base = 14 inches; what 
is its solidity ? 

V = -5236A(3r 2 + A 2 ) = "5236 x 8(3 x 14 2 + 8 2 ) 
= 4-1888 x 652 = 2731-0976 cubic inches. 

2. The diameter of a sphere is =5 feet, and the height of a 
segment of it = 2 feet ; what is the solidity of the segment? 
V = -5236A 2 (3d - 2&) = '5236 x 2 2 (3 x 5 - 2 x 2) 

= 2-0944 x 11 =23-0384 cubic feet. 

The spherical segment ACBV is equal to the difference between 
the spherical sector OAVB and the cone OACB. But the spherical 
sector is to the sphere as the surface of the segment to the surface 
of the sphere. Hence, if S, s be the surfaces of the sphere and 
segment, and V, V the volumes of the sphere and sector, 

V 
S:*=V:V; hence V' = ~- 

o 

But S = 3'1416d 2 , = 3-1416rfA, 

w -5236^x3-1416^/1 
hence V = 



For the cone, volume = V"=r-A5236r 2 (rf-2A) ; hence 

volume of segment, 

V = V - V"= -52361^ -r 2 (d-2A)} ...... [1]. 

But AC 2 = VC . CF, or r 2 = h(d - h) ; 

r* + h? 

and hence d= ? 

h 

Substituting this value of r 2 in [1], it becomes 

V = 5236{rf 2 A -h(d-h)(d-2h)}= -5236A 2 (3d - 2h). 
Substituting in this last expression the above value for a, 

V= 



EXERCISES 

1. Find the solidity of a spherical segment whose height is 
=4 inches, and the radius of its base = 8 inches. 

=435-6352 cubic inches. 



184 MENSURATION OP SOLIDS 

2. Find the volume of a spherical segment, the diameter of the 
hase of which is = 20, and its height =9 . . . =1795 '4244. 

3. What is the solidity of a spherical segment, the radius of 
whose base is = 25 inches, and its height = 6 '75? 

= 6787'844 cubic inches. 

4. Find the solidity of a spherical segment, the height of which 
is =2 feet, the diameter of the sphere being = 10 feet. 

= 54 '4544 cubic feet. 

394. Problem XXI. To find the solidity of a spherical 
zone. 

RULE I. Add together the squares of the radii of the two ends 
and one-third the square of the height ; multiply the sum by the 
height, and this product by 1/5708, and the result will be the 
solidity. 

RULE II. For the middle zone, add together the square of the 
diameter of either end, and two-thirds of the square of the height, 
or find the difference between the square of the diameter of the 
sphere, and one-third of the square of the height of the zone ; then 
multiply the sum or the difference by the height, and the product 
by 7854, and the result will be the solidity. 

Let R and r be the radii of the ends ; then the first rule gives 
V = 1 -57087t(R 2 + r 2 + JA). 

And the rules for the middle zone give 

V= -7854A(D 2 + A 2 ), or V= 7854/i(d 2 - A 2 ), 

where d is the diameter of the sphere, and D that of either end of 
the zone. 

EXAMPLES. 1. Find the solidity of a spherical zone, the 
diameters of its ends being = 4 and 3 inches, and its height 
= 2 inches. 

V=l'5708A(R 2 + r 2 + /i 2 ) = 1-5708x2(4 + | + f) 

=3-1416 x f = 23-8238 cubic inches. 

2. Find the solidity of the middle zone of a sphere, the diameters 
of its ends being = 4 feet, and its height =6 feet. 



= 4-7124 x 40= 188-496 cubic feet. 

Or, since d? = JD 2 + J# or d = D 2 + A 2 = 4 2 + 6 2 = 16 + 36 = 52, 
V = -7854A(^ _ JA2) _ . 785 4 x 6(52 - i x 6 2 ) 

= 4-7124 x 40 = 188-496 cubic feet. 
The spherical zone ABED (fig. to Prob. XX.) is evidently equal 



MENSURATION OF SOLIDS 185 

to the difference between the two segments VDE and VAB. 

Let r' = the radius of the sphere, and h' the height of the less 

segment; A' + A, the height of the greater, then A=the height of 

the zone ; 

also 3R 2 = 6r'(h' + h)- 3(h' + A) 2 , 

and Sr 2 = 6r'A' - 3A' 2 ; hence, 

V = 6r'(A' + A) - 2(A' + A) 2 }(A' + h) - A'(6r'A' - 2A /a ) 



= ^{6r'A' 2 + 12/A'A + 6/A 2 - 2A* - 6h^h - 6A'A 2 - 2h s - 
= ^{6r'(A' + A) - 3(A' 4- A) 2 + 6/A' - 3A' 2 + A 2 } 
= ^{3R 2 + 3r + A 2 } = |A(R 2 + r 2 + JA 2 ) 



But for the middle zone R=r, and if D be the diameter of 
the end, then R^r^iD^D^D 2 ; hence, for the middle 
zone, 



V = (D 2 + A 2 ) = (D 2 + 1A 3 ) = -7854A(D 2 + A 2 ) 
= -7854A(P- JA 2 ), if rf=the diameter of the sphere. 

EXERCISES 

1. What is the volume of a spherical zone, the diameters of its 
ends being = 10 and 12 inches, and its height=2 inches? 

= 195-8264 cubic inches. 

2. Find the solidity of the middle zone of a sphere whose 
diameter is =40 inches, the diameter of its base being =24, and 
its height = 32 inches ..... =31633-8176 cubic inches. 

3. What is the volume of a spherical zone whose height is 
= 15 inches, and the diameters of its ends=20 and 30 inches? 

= 9424 "8 cubic inches. 

4. The diameters of the ends of a spherical zone are = 8 and 
12 inches, and its height = 10 inches ; what is its solidity? 

= 1340 "4 16 cubic inches. 

5. What is the volume of a middle zone of a sphere, its height 
being =8 feet, and the diameters of its ends = 6 feet? 

= 494-2784 cubic feet. 

6. Find the volume of a spherical zone whose height is =4 feet, 
and the end diameters = 6 feet. = 146-608 cubic feet. 



186 




MENSURATION OF CONIC SECTIONS 

395. The conic sections are the three curves the parabola, 
the ellipse, and the hyperbola. 

DEFINITIONS 

396. A parabola is a curve such that any point in it is equi- 
distant from a given point and a given straight line. 

Thus, if the curve DVE is such that any 
point in it, as D, is equidistant from a given 
point F and a given line AB that is, such that 
DF = DA the curve is a parabola. 

397. The given point is called the focus of 
the parabola, and the given line its directrix. 

Thus, F is called the focus of the parabola, 
and AB is its directrix. 

398. That part of a perpendicular to the directrix passing through 
the focus, which is contained within the curve, is called the axis, 
or principal diameter ; and the extremity of the axis is called 
the vertex of the parabola. 

Thus, VG produced indefinitely is the axis, and V the vertex of 
the curve. 

399. An ordinate is a perpendicular from any point in the curve on 
the axis, and when produced to meet the curve on the other side of 
the axis, it is a double ordinate ; and the portion of the axis inter- 
cepted between the ordinate and the curve is called the abscissa. 

Thus, DG is an ordinate to the axis, and GV is its abscissa ; also 
DE is a 'double ordinate. 

400. The principal parameter is four times the distance of the 
vertex from the directrix. 

Thus, four times CV is the parameter. It is also equal to 4 VF, 
or to the double ordinate through the focus. 

401. An ellipse is a curve such that the 
sum of the distances of any point in it from 
JB two given points is equal to a given line. 

Thus, if any point, as P, in the curve ACBD 
has the sum of its distances from two given 
points, E and F namely, PE + PF equal to a 
given line, the curve is an ellipse. 

402. The given points are called the foci ; and the middle of the 
line joining them, the centre. 




MENSURATION OP CONIC SECTIONS 



187 



Thus, E and F are the foci, and G the centre. 

403. The distance of the centre from either focus is called the 
eccentricity. 

EG or GF is the eccentricity. 

404. The major axis is a line passing through the foci, and 
terminated by the curve ; and a line similarly terminated, passing 
through the centre, and perpendicular to the major axis, is named 
the minor axis. The former axis is also called the transverse 
diameter ; and the latter axis, the conjugate diameter. 

Thus, AB is the major, and CD the minor axis. 

405. An ordinate to either axis is a line perpendicular to it from 
any point in the curve ; and this line produced to meet the curve 
on the other side of the axis is called a double ordinate ; also each 
of the segments into which the ordinate divides the axis is called 
an abscissa. 

Thus, PM is an ordinate to the axis AB ; and AM and MB 
abscissae. 

406. The parameter of either axis is a third proportional to it 
and the other axis. 

Thus, the parameter of AB is a third proportional to AB and 
CD, and is the same with the double ordinate through the focus, 
called the focal ordinate. 

407. A hyperbola is a curve such that the difference between 
the distances of any point in it from two given points is equal to a 
given line. 

Thus, if any point, as P in the curve PEN, has the difference of 
its distances from the two given points E and F namely, PE, PF 
equal to a given line AB, the curve is an hyperbola. 

If another curve, P'AN', similar to PBN, pass through A, these 
two branches are called opposite hyperbolas. 

408. The two given points are 
called the foci ; and the middle 
of the line joining them, the 
centre. 

Thus, E and F are the foci, and 
G the centre. 

409. The distance of the centre 
from either focus is called the 
eccentricity. 

Thus, GE or GF is the eccentricity. 

410. The major axis is that portion of the line joining the foci, 




188 



MENSURATION OP CONIC SECTIONS 



which is terminated by the opposite hyperbolas ; it is also called 
the transverse diameter. 
Thus, AB is the major axis. 

411. A line passing through the centre perpendicular to the 
major axis, and having the distance of its extremities from those 
of this axis equal to the eccentricity, is called the minor axis, or 
conjugate diameter. 

Thus, if the line CD is perpendicular to AB, and if the distances 
of C and D from A or B are equal to EG, CD is the minor axis. 

412. An ordinate to the major axis is a line perpendicular to it 
from any point in the curve, and this line produced to meet the 
curve on the other side of the axis is called a double ordinate ; 
and the segment of the axis between the ordinate and curve is 
called an abscissa. 

Thus, PM is an ordinate, and PN a double ordinate to the axis 
AB ; and BM an abscissa. 

413. A third proportional to the major and minor axis is called 
the parameter of the former axis. 

Thus, a third proportional to AB and CD is the parameter of 
AB, and is equal to the double ordinate through the focus. 

414. Problem I. Given the parameter of a parabola, to 
construct it. 

Let the parameter of a parabola be equal to the line N, it is 
required to construct it. 

Draw GH for the directrix, and DC perpendicular to it for the 
axis. Make DV and VF each = one-fourth 
of N, then V will be the vertex, and F in the 
axis the focus of the parabola. Draw any line 
LM parallel to GH, and with the distance 
DS for a radius, and F as a centre, cut LM in 
L and M, and these are two points in the para- 
bola. Draw any other parallel, as AB, and 
\B find the points A, B in a similar manner ; 
and so on. Then a curve APVQB, passing 
through all these points, will be a parabola. 
For the distance of L from GH namely, SD is equal to the 
distance of L from F ; and the same holds for the other points. 

When the length of the directrix is given in numbers, a line N 
must be taken from some convenient scale of equal parts of the 
required length, and the figure may then be constructed. 




MENSURATION OP CONIC SECTIONS 189 

The curve may also be described by means of a bar GW, moved 
parallel to the axis with its extremity G on the directrix, and 
having a thread FEW with one end F fixed in the focus, and a 
pencil at E, held so as to keep the thread tight, will describe the 
curve. 

EXERCISES 

1. Construct a parabola having a parameter equal to the given 
line A. A 

2. Construct a parabola whose parameter is 200 on a scale of 
half an inch to the hundred. 

415. Problem II. Given an ordinate of a parabola and 
its abscissa, to find the parameter. 

RULE. Divide the square of the ordinate by the abscissa, and 
the quotient will be the parameter. 
Let d= the ordinate BC (fig. to Prob. I.), 

a= ii abscissa CV, and 

p= ii parameter; 

d? , d? 

then a^pa ; hence =, and a 

a p 

EXAMPLE. Given an ordinate of a parabola =6 and its abscissa 
= 15, to find the parameter. 

cP 6 2 36 72 . 

*> O *A 

* a~15~15~30~' 

EXERCISES 

1. An ordinate of a parabola is = 20, and its abscissa = 36; find 
the parameter. = 11^. 

2. Find the parameter of a parabola, an ordinate and abscissa 
being respectively = 12 and 25 - : i' : . = 5'76. 

3. What is the parameter of a parabola one of whose ordinates 
is = 16, and the corresponding abscissa =18? . . . = 14f. 

4. Find the parameter of a parabola, one of its ordinates being 
= 25, and the corresponding abscissa =20. ." -. '_' . . =31 '25. 

416. Problem III. To construct a parabola, any ordinate 
and its abscissa being given. 

Find by Prob. II. the parameter, and then by Prob. I. construct 
the curve. 

EXERCISE 

Construct a parabola one of whose ordinates is = 120, and the 
corresponding abscissa =225. 



190 MENSURATION OF CONIC SECTIONS 

417. Problem IV. Of two abscissae and their ordinates, 
any three being given, to find the fourth. 

The abscissae are directly proportional to the squares of their 
ordinates by Prob. II. 
Hence, if A, a are the abscisses, and D, d their ordinates, 

A :a =D 2 :^; .-.cP = ~, and D 2 = . 
A a 

A.cP aD z 

Also D 2 : cP= A : a ; . . a = -j^-, and A =^~- 

EXAMPLE. Given the abscissa VS = 10 (fig. to Prob. I.), and the 
abscissa VC = 12, also the ordinate SL = 9; required the ordinate 
AC. 

VS :VC = SL 2 :AC 2 ; 
10 : 12 = 9 2 : AC 2 , 

. 12 x9 2 6x81 486 Q7 
and AC 2 : j^ = 5 =-^-=97 '2; 

hence AC = 



/7T1 2 1 9 v Q 2 

or ^=^=^^-=97-2, and d=9'859. 

EXERCISES 

1. Two abscissae of a parabola are = 18 and 32, and the ordinate 
of the former is = 12 ; find the ordinate of the latter. . =16. 

2. Two abscissae are = 3 and 6, the ordinate of the former is = 5; 
find that of the latter ........ =7 '07. 

3. Two abscissae are = 9 and 16, and the ordinate of the former 
= 6 ; find that of the latter. . . ..... . . =8. 

4. Two ordinates are = 6 and 8, and the abscissa of the former 
=9 ; find that of the latter. . ..... =16. 

5. Two ordinates are = 18 and 24, and the abscissa of the former 
= 18 ; find that of the latter. . . . .-*';.';*. =32. 

418. Problem V. To find the length of a parabolic curve 
cut off by a double ordinate. 

RULE. To the square of the ordinate add four-thirds of the 
square of the abscissa, and the square root of the sum, multiplied 
by two, will be the length of the curve nearly. 

Let d=ihe ordinate, and a=the abscissa, and 
1= it length of the curve ; 

then / 



MENSURATION OP CONIC SECTIONS 191 

EXAMPLE. The abscissa of a parabola is = 3, and its ordinate =9 ; 
what is the length of the arc ? 

2 =4x93, 



and I = 2 V93 = 2 x 9'6436 = 1 9*2872. 

EXERCISES 

1. The abscissa of a parabolic arc is =4, and the ordinate is = 8 ; 
what is its length ? ....... - . =18-47. 

2. The abscissa and ordinate of a parabola are = 10 and 8; what 
is the length of the curve ? . . . . . =28-09. 

419. Problem VI. To find the area of a parabola. 

RULE. Multiply the base by the height, and two-thirds of the 
product is the area. 

Let 6 = base or double ordinate, a = height or abscissa j 
then M = $ba. 

EXAMPLE. What is the area of a parabola whose base is =25, 
and height = 18? 

JR=%ba = $x 25x18 = 300. 

It was proved, by the method of exhaustions, by Archimedes, 
and can be easily proved by the integral calculus, that the area of 
a parabola is two-thirds of the circumscribing rectangle, which has 
the same base and height as the curve, its upper side being a 
tangent through the vertex. 

EXERCISES 

1. Find the area of a parabola whose base or double ordinate is 
= 36, and height or abscissa = 45 ...... =1080. 

2. What is the area of a parabola whose base and height are = 18 
and 28 respectively ? . . . ...... . . =336. 

3. Find the area of a parabola whose base is = 30, and height 
=44. ......... ....... =880. 

4. Find the area of a parabola whose base is = 3 - 6 feet, and height 
-5-6 feet. ... ....... . . =13-44 feet. 

420. Problem VII. To find the area of a zone of a 
parabola. 

RULE. Divide the cubes of the two parallel sides by the differ- 
ence of their squares ; multiply the quotient by the height of the 
zone, and two-thirds of the product will be the area ; or, 



192 MENSURATION OP CONIC SECTIONS 

Divide the sum of the squares of the parallel ends, increased 
by their product, by the sum of the parallel ends ; multiply this 
quotient by the altitude, and two-thirds of the 
product is the area. 

Let D, d denote the two ordinates AB, CD, and 
h = the height EF; 




(T)8_ 
D 






EXAMPLE. Find the area of a parabolic zone, the two terminat- 
ing ordinates being = 18 and 30, and the altitude = 9. 

303-18 3 . 900 + 540 + 324 

. - ___ _ fi v 



30 + 18 
1764 1764 



By means of the preceding problem, the above rule may be easily 
proved. Let A', A", and h', h", denote respectively the areas and 
heights of the two parabolas VAB, VCD; then (last problem) 

A'=DA', and A" = %dh"; 
hence M = A' - A" = (DA' - dh") = {DA + A"(D - d)}. 

But by a property of the parabola, 

ffii. 
D 2 : tP=h' : h"; hence D 2 - cP : d*=h : h", and *>"=&_# 

Substitute this value of h" in the above value of JR, 



then 



EXERCISES 



1. Find the area of a zone of a parabola whose parallel sides are 
=5 and 3, and its height =4 ...... . = 16. 

2. Find the area of a parabolic zone whose parallel ends are = 6 
and 10, and the height = 6 ..... ... =49. 

3. Required the area of a zone of a parabola whose height is = 11, 
and its two ends = 10 and 12 ....... =121. 

4. The ends of a parabolic zone are = 5 and 10, and its height = 6 ; 
what is its area ?. ........ =46. 

5. The ends of a zone of a parabola are = 6 and 9, and their 
distance = 8 ; what is its area? ...... =60 '8. 

6. The parallel sides of a parabolic zone are = 10 and 15, and their 
distance = 15; required its area. ..... =190. 




MENSURATION OP CONIC SECTIONS 193 

421. Problem VIII. To describe an ellipse, having given 
its major and minor axes. 

Let AB be the major axis. Draw a line CD, bisecting it per- 
pendicularly, and make GC, GD each equal to half of the minor 
axis, then CD is this axis. From C as a 
centre, with half the major axis AG as a 
radius, cut AB in E and F, and these 
points are the foci. 

Produce AB to Q till EQ = AB; then 
from E as a centre describe an arc PQ, 
and this arc is a species of directrix to the 
ellipse. With any radius El, from E as a centre, describe an arc 
HK ; and with the distance IQ as a radius, from F as a centre, cut 
HK in H and K, and these are points in the curve. Describe from 
E as a centre any other arc LM, and find as before the points L 
and M. Proceed in the same manner till a sufficient number of 
points are found, and the curve passing through them namely, 
ADBC is an ellipse. 

The construction of the ellipse by this method is exactly similar 
to that of the parabola, PR being considered the directrix, and the 
concentric arcs HK, LM, &c. as parallels to the arc PR. 

COR. 1. When the major axis and the eccentricity or the foci 
are given, the ellipse can be constructed nearly in the same manner 
as in the problem. 

COR. 2. When the minor axis and the eccentricity are given, 
the ellipse may be constructed thus : 

Draw AB, bisecting CD perpendicularly, and lay off GE, GF 
each equal to the eccentricity, then EC is equal to half the major 
axis. Hence, make GA, GB each = EC, and AB is the major axis ; 
and the ellipse can now be constructed as before. 

The ellipse may also be constructed by means of elliptic com- 
passes, which consist of two brass bare AB, CD, with grooves and 
a third bar OH = AG, half the major axis, a part of it, NH, being 
= CG, half the minor axis, with two pins at O and N; and OH 
being moved, so that the pins at N and O move respectively in the 
grooves of AB and CD, the extremity H will move in the curve of 
the ellipse. 

Or, if a thread EPF (fig. to Art. 401), equal in length to AB, 
have its extremities fixed in the two foci, and be drawn tight by 
means of a pencil moving in the angle P, the pencil will describe 
an ellipse. 




194 MENSURATION OF CONIC SECTIONS 

422. Problem IX. When the two axes and an abscissa 
are given, to find the ordinate. 

RULE. As the square of the major axis is to that of the minor, 
so is the rectangle under the two abscissas to the square of the 
ordinate. 

The halves of the two axes may be taken instead of the axes 
themselves in this rule. 

Let a = the major axis. =AB, 
b the minor , =CD, 
h = one abscissa = AM ; 

B then a -h= the other abscissa = MB, 
d= the ordinate =PM, 

and a 2 : b' 2 =(a-h)h :rf 2 , 

D //" 

or d?-i;(ct h)h. 

a* 

EXAMPLE. The axes are = 30 and 10, and one abscissa is =24; 
find the ordinate. 

a 30, 6 = 10, h = 24 ; hence a-^ = 30-24 = 6; 

10 2 144 

hence ^2=^x6x24=-g- =16; .'. rf=Vl6 = 4. 

EXERCISES 

1. The major and minor axes of an ellipse are = 60 and 20, and 
one abscissa is = 12; find the ordinate =8. 

2. The axes are =45 and 15, and one abscissa is = 9 ; what is the 
ordinate? =6. 

3. The axes are = 52 - 5 and 17'5, and the abscissa=42; find the 
ordinate. . =7. 

4. The axes are 17'5 and 12'5, and an abscissa =14; find the 
ordinate. ..'.' : '. . =5. 

423. Problem X. When the axes and an ordinate are 
given, to find the abscissae. 

RULE. As the square of the minor axis is to the square of 
the major axis, so is the product of the sum and difference of the 
semi-axis minor and the ordinate to the distance of the ordinate 
from the centre. 

This distance being added to the semi-axis major, and also 
subtracted from it, will give the greater and less abscissas. 

Let c = the distance MG (last fig.) from the centre, and a, b the 
semi-axes, and h AM, and dPM ; 



MENSURATION OF CONIC SECTIONS 195 

then b*:a?=(b + d)(b-d) re 2 , or c 2 =p(& + rf)(&-rf), 

and h=a + c, 2a-h=a-c. 

EXAMPLE. The axes are = 30 and 10, and the ordinate=4; 
what are the absciss* ? 



and c 

hence the greater abscissa AM = a + c = 15 + 9=24, 

and the less abscissa MB=a-c=15-9 = 6. 

The rule depends on the same principle as the last; for 
CG 2 : AG 2 = ON . ND : PN 2 , or MG 2 . 

EXERCISES 

1. The axes are=45 and 15, and the ordinate 6; what are the 
abscissae? =36 and 9. 

2. The axes are = 70 and 50, and an ordinate =20; find the 
abscissae .'"'.' : I .. =14 and 56. 

424. Problem XI. When the minor axis, an ordinate, 
and an abscissa are given, to find the major axis. 

RULE. Find the square root of the difference of the squares of 
the semi-axis minor and the ordinate, and, according as the less 
or greater abscissa is given, add this root to or subtract it from the 
semi-axis minor ; then, 

As the square of the ordinate is to the product of the abscissa and 
minor axis, so is the sum or difference found above to the major 
axis. 

Or, if a, b are the semi-axes, and h the abscissa, cP : 2bh 

= b\f(b*-d 2 ):2a, and < 2a = ^~{b> 



EXAMPLE. The minor axis is = 10, the smaller abscissa 6, and 
the ordinate =4 ; find the major axis. 

&2_d2 = 5 2 _42=:25-16 = 9, and \/(b*-d?) = 3; 
and cP:2bh = b>\/(b z -d 2 ):2a. 

Or, 4 2 :10x6=5 + 3:2a, or 16 :60 = 8 : 2a, and 2a=30. 

Or by the formula, 2a = f {& + V(& 2 - &)} 



The rule is derived from the same proposition as the last two- 



196 MENSURATION OF CONIC SECTIONS 

Thus, AG 2 : CG 2 = AM . MB : PM 2 , 

or a?:b z =(2a-h)h:cP; hence atd 2 = b 2 h(2a -h). 

From this quadratic equation, the value of a, the unknown 

quantity, is easily found, and the result is the above value. 

EXERCISES 

1. The minor axis is = 15, an ordinate = 6, and the less abscissa 
= 9; what is the major axis? =45. 

2. The minor axis is =50, an ordinate=20, and the less abscissa 
= 14; find the major axis =70. 

3. The minor axis is = 5, the greater abscissa = 12, and the ordinate 
=2; what is the major axis ? =15. 

425. Problem XII. When the major axis, an ordinate, 
and one of the abscissae are given, to find the minor axis. 

RULE. Find the other abscissa, then the product of the two 
abscissae is to the square of the ordinate as the square of the 
major axis to that of the minor axis. 

oV 2 

Or, h(a - h) : d?=a? : 6 2 , and 6 2 = ., ... 

i n(a n) 

When a and b are semi-axes, 6 2 =, /c . ... 

h(2a - h) 

EXAMPLE. The major axis is = 15, an ordinate = 2, and an 
abscissa =3 ; what is the minor axis ? 

a?cP 15 2 x2 2 75x4 

6 = ^31j = 30533J = -l2- = 25 ' &nd 6 = V25 = 5. 
The rule is derived from the same theorem as that in Prob. IX. 

426. If the abscissae were segments of the minor axis, and the 
ordinate a perpendicular to it, then the major axis could be found 

by the analogous formula, 2 = rrr rr- 

fl\0 ft/f 

EXERCISES 

1. The major axis is = 70, an ordinate =20, and one of the 
abscissae = 14 ; what is the minor axis ? . . . =50. 

2. The major axis is = 210, an ordinate = 28, and one of the 
abscissae = 1 68 ; what is the minor axis ? . . . . =70. 

427. Problem XIII. To find the length of the circumfer- 
ence of an ellipse when the axes are given. 

RUL.E. Multiply the square roqt of half the sum of the squares 



MENSURATION OF CONIC SECTIONS 197 

of the two diameters by 3-1416, and the product will he the 
circumference nearly. 

If J=the length of the curve, then Z=3'1416\/{4( 2 + & 2 )}. 

EXAMPLE. What is the length of the circumference of an 
ellipse whose axes are = 10 and 30 ? 



= 3-1416 x 22-3607 = 70-2484. 

The rule is derived by means of the calculus ; but it is only an 
approximation, though sufficiently accurate for practical purposes. 

EXERCISES 

1. The axes are = 10 and 12 ; what is the length of the curve of 
the ellipse? ...... . . . =34-7001. 

2. The axes are = 6 and 8 ; what is the length of the curve ? 

= 22-214. 

3. The axes are =4 and 6 ; what is the length of the curve? 

= 16-019. 

428. Problem XIV. To find the area of an ellipse. 
RULE. Multiply the product of the two axes by '7854, and the 
result will be the area. 

Or, ^l=-7854a6. 

EXAMPLE. What is the area of an ellipse whose axes are = 15 
and 20 feet ? 

JR= -7854a6= '7854 x 20 x 15=235-62 square feet. 

The rule can only be demonstrated rigorously by means of the 
integral calculus. The truth of it, however, will appear evident 
from the consideration that if a circle is described on the major 
axis, and an ordinate to this axis 'be produced to meet the circle, 
then if rf'=the ordinate of the circle, d'*=7i(a-h) by Eucl. III. 35. 
But (Prob. IX.) a? : V z =h(a -h):<P; and hence 

a?:b 2 = d'*:d 2 , ora:b = a':d-, 

that is, each ordinate of the circle is to the corresponding one of 
the ellipse as a : b ; hence, if A'=area of circle, 

A' :M=a : b, or ^l= = - x -7854a 2 = '7854a&. 
a a 

EXERCISES 

1. Find the area of an ellipse whose axes are = 5 and 10. =39*27. 

2. Find the area of an ellipse whose axes are = 5 and 7. =27 "489. 

Prac. N 



198 MENSURATION OP CONIC SECTIONS 

3. What is the area of an ellipse whose axes are = 12 and 16 ? 

= 150-7968. 

4. What is the area of an ellipse whose axes are = 6 and 7 ? 

=32-9868. 

429. Problem XV. To find the area of an elliptic 
segment. 

RULE. Find the area of the corresponding segment of the circle 
described upon that axis of the ellipse which is perpendicular to 
the base of the segment ; then this axis is to the other axis as the 
circular segment to the elliptic segment ; or, 

Multiply the tabular area belonging to the corresponding cir- 
cular segment by the product of the two axes of the ellipse, and 
the result will be the area. 

Let 2R=the area of the elliptic segment, A=its height, and 
A' = the area of a segment of the same height of a circle described 
on the axis, of which the height is a part ; then, when h is a part 
of the major axis , 

7, A' 

a : b = A':M and M = - = segment PBQ (fig. to Prob. IX.). 
ct 

When h is a part of the minor axis b, 

aA.' 
b : a = A' : M, and J3, = r- = segment RCP. 

EXAMPLE. What is the area of an elliptic segment whose base 
is parallel to the minor axis, the height of it being = 10 feet, and 
the axes of the ellipse = 35 and 25? 

Height of tabular circular segment = |f =f = -2857 ; 
area of tabular circular segment t -185154 ; 
then ^l = a^ = 35 x 25 x -185154=162-00975. 

The rule depends on the principle that an elliptic segment bears 
the same proportion to the corresponding circular segment that 
the whole ellipse does to the whole circle described on the axis of 
which the height is a part. 

EXERCISES 

1. Find the area of an elliptic segment whose base is perpen- 
dicular to the major axis, its height being =6, and the axes = 30 
and 10 =33-5472. 

2. Find the area of an elliptic segment whose base is parallel to 
the major axis, its height being=2, and the diameters=14 and 10. 

= 15-65536. 



MENSURATION OF CONIC SECTIONS 



199 




430. Problem XVI. To describe an hyperbola, its two 
axes being given. 

Make AB equal to the major axis ; bisect it perpendicularly by 
CD, and make CG and GD each equal to half the minor axis. The 
distance CA or DB being laid off from 
G to E and F, these two points will 
be the foci of the hyperbola. 

From E as a centre, with a radius 
=AB, describe an arc RST, and it 
will be a species of directrix. From 
E as a centre, describe any arc, as 
UMX ; and with the distance MS of 
U from the directrix, and with F as 
a centre, cut UMX in U and X, and 
these are points in the curve. Find other two points in the same 
manner, and so on till a sufficient number are found ; then the 
curve PEN passing through them all is an hyperbola. Another 
hyperbola similarly described, and passing through the point A, 
would be the opposite hyperbola. 

If a tangent IK to the curve at its vertex B be drawn, such that 
BI and BK are each = half the minor axis CG, and straight lines 
GH, GL be drawn from the centre through its extremities I and 
K, they are called asymptotes, and possess the singular property 
of continually approaching to the curve without ever meeting it. 

COR. 1. When the major axis AB and the eccentricity EG or 
GF are given, the minor axis CD can be found thus : Bisect AB 
perpendicularly by CD, and then from A as a centre, with EG as 
a radius, cut CD in the points C and D, and they will be the 
extremities of this axis. The curve can then be described as in 
the above problem. 

! COR. 2. When the minor axis CD and the eccentricity EG are 
given, the major axis can be found thus : Bisect CD perpendicu- 
larly by EF, with EG as a radius and C as a centre, cut EF in 
A and B, then AB is the major axis. The curve can then be 
described as above. 

431. Problem XVII. The axes of an hyperbola and an 
abscissa being given, to find the ordinate. 

RULE. As the square of the major axis is to that of the 
minor, so is the product of the two abscissae to the square of 
the ordinate. 



200 MENSURATION OF CONIC SECTIONS 

Let the axes AB, CD be denoted by a and 6, the abscissa 
p BM by h, and the qvdinate PM by d; 
then AM = + 7t, and 




EXAMPLE. The major axis of an 
hyperbola is = 15, the minor axis 9, and 
the less abscissa = 5 ; what is the ordinate? 



hence d*=- 

(f 

and 

EXERCISES 

1. The major and minor axes are = 48 and 42, and the less abscissa 
= 16; what is the ordinate? ....... =28. 

2. The major axis is = 25, the minor = 15, and the less abscissa 
= 8; what is the ordinate? . " . . . . .." .''." . , ' , =10. 

3. The major and minor axes ave = 15 and 7, and the less abscissa 
=5; what is the ordinate ? . -. . : / . ; '. :.:".'. =5. 



432. Problem XVIII. The two axes and an ordinate 
being given, to find the abscissae. 

RULE. As the square of the minor axis is to that of the major 
axis, so is the sum of the squares of the semi-axis minor and the 
ordinate to the square of the distance between the ordinate and 
the centre. 

The sum of this distance and the semi-axis major will give the 
greater abscissa, and their difference the less. 

Let c = this distance = GM (fig. to Prob. XVII.), and a, 6 half the 
axes ; 

then 4fe 2 : '4a?=b*+d? :c 2 , or c 2 =p(6 2 + rf 2 ), 

and AM=2a + h = a + c, BM = A = c-. 

EXAMPLE. The major and minor axes are =30 and 18, and the 
ordinate = 12 ; what are the abscissse? 



Hence +c = 15 + 25 = 40, e- = 25- 15 = 10; and the two abscissae 
are = 10 and 40. 



MENSURATION OP CONIC SECTIONS 201 

EXERCISES 

1. The major and minor axes are = 24 and 21, and the ordinate 
= 14; what are the abscissae ? . . . . . =32 and 8. 

2. The major and minor axes are = 55 and 33, and an ordinate is 
= 22; required the abscissae. . . . . =73J and 18 J. 

3. The major and minor axes are = 60 and 45, and an ordinate 
= 30; what are the abscissae? . ... =80 and 20. 

433. Problem XIX. The major axis, an ordinate, and 
the two abscissae being given, to find the minor axis. 

RULE. The product of the abscissae is to the square of the 
ordinate as the square of the major axis is to that of the minor 
axis. 

(a + h)h : d?=a? : 6 2 , or b 2 =- TTD or 6= ; 
(a + li)li v 

where a and b are the axes. 

EXAMPLE. The major axis is=30, the ordinate = 12, and the 
two abscissae = 40 and 10; what is the minor axis? 

ad 30x12 360 _ 1 _ 

V i, .1,7 - i.^r. OOU X ^: 1O. 



The rule depends on the same principle as that of Prob. XVII. 

EXERCISES 

1. The major axis is = 15, an ordinate = 6, and the two abscissae 
=20 and 5 ; what is the minor axis? =9. 

2. The major axis is = 36, and ordinate = 21, and the abscissae 
= 12 and 48; find the minor axis =31 '5. 

434. Problem XX. The minor axis, ordinate, and the two 
abscissae being given, to find the major axis. 

RULE. Find the square root of the sum of the squares of the 
semi-axis minor and the ordinate ; and, according as the less or 
greater abscissa is given, find the sum or difference of this root and 
the semi-axis minor ; then, 

As the square of the ordinate is to the product of the abscissa and 
minor axis, so is the sum or difference found above to the major 
axis. 

Let , b be the semi-axes ; 

then cP : 2bh = b i V(& 2 + &) 2> 

apd 2a= 

a 



202 MENSURATION OP CONIC SECTIONS 

EXAMPLE. The minor axis is = 18, the ordinate = 12, and the 
less abscissa = 10 ; what is the major axis? 



=x24=30. 

The rule is derived from the same theorem as that in last prob- 
lem. When , 6 are the semi-axes, the proportion in the last 
problem becomes 2 : b 2 =h(2a + h) : d 2 ; hence a?d?=b' 2 h(2a+h). 
From this quadratic equation, the value of a, the unknown 
quantity, is easily found, and the result is the value given above. 

EXERCISES 

1. The minor axis is = 45, the less abscissa = 30, and the ordinate 
= 30 ; required the major axis ....... =90. 

2. The minor axis is = 15, an ordinate = 10, and the less abscissa 
= 8; what is the major axis ? ...... =25. 

435. Problem XXI. To find the length of an arc of an 
hyperbola, reckoning from the vertex of the curve. 

RULE. To 15 times the major axis add 21 times the less abscissa, 
and multiply the sum by the square of the minor axis ; add this 
product to 19 times the product of the square of the major axis by 
the abscissa, and add it also to 9 times the same product ; divide 
the former sum by the latter, multiply the quotient by the ordinate, 
and the product will be the length of the arc. 

Let l t\ie length of the arc, 



_ . _ / 

'~ ~ V 



(15a + 21/i)6 2 + 

EXAMPLE. The major and minor axes of an hyperbola are = 15 
and 9, an ordinate at a point in it is = 6, and the abscissa = 5 ; what 
is the length of the arc to this point from the vertex ? 

(15x15 + 21 x5)9 2 + 19xI5 2 x5 
(15 x 15 + 21 x 5)9 2 + 9xl5 2 x5 X 
Expunge 15, which is a common factor to the terms of this 
fraction, 

(15 + 7)81 + 1425 3207 



The rale can be demonstrated by means of the integral calculus. 

EXERCISES 
1. The major and minor axes of an hyperbola are = 30 and 18, the 



MENSURATION OF CONIC SECTIONS 203 

ordinate = 12, and the smaller abscissa = 10 ; what is the length of 

the arc? = 15'663. 

2. The major and minor axes are = 105 and 63, a double ordinate 
= 84, and the less abscissa =35; what is the length of the whole 
arc? . . . . =109-641. 



436. Problem XXII. -To find the area of an hyperbola, 
the axes and abscissa being given. 

RULE. To 7 times the major axis add 5 times the abscissa; 
multiply the sum by 7 times the abscissa, and multiply the square 
root of this product by 3. 

To this last product add 4 times the square root of the product 
of the major axis and abscissa. 

Multiply this sum by 16 times the product of the minor axis and 
abscissa ; divide this product by 300 times the major axis, and the 
quotient will be nearly the required area. 
Or, JR = 166A{3 V7A(7 + 5h) + 4\/ah} -j- 300a. 

EXAMPLE. The major and minor axes of an hyperbola are = 10 
and 6, and the abscissa = 5 ; what is its area ? 

M = 166A{3 V7A(7a + 5h) + \/ah} -f- 300a 

= 16 x 6 x 5{3\/7 x 5(70 + 25) + 4\/10 x 5} -=- 300 x 10 

= 16 x 3(3 V3325 + 4 V50) -r 300 = 16 x 201 '273 -r 100 = 32 -20368. 

EXEECISES 

1. In an hyperbola the major and minor axes are = 15 and 9, and 
the abscissa =5 ; what is the area? .... =37 '92. 

2. The major and minor axes are = 20 and 12, and the abscissa 
6| ; find the area =67 '414. 



THE SOLIDS OF REVOLUTION OF THE CONIC 
SECTIONS 

437. The solids of revolution generated by the conic 
sections are the paraboloid, the spheroid or ellipsoid, and 
the hyperboloid. 

438. A paraboloid is a solid generated by the revolution 
of a parabola about its axis, which remains fixed. 



204 THE SOLIDS OF REVOLUTION OF THE CONIC SECTIONS 

The paraboloid is also called the parabolic conoid, as it is 
like a cone. 

439. A frustum of a paraboloid is a portion of it con- 
tained by two parallel planes perpendicular to its axis. 

440. A spheroid is a solid generated by the revolution of 
an ellipse about one of its axes, which remains fixed. 

The spheroid is said to be oblate or prolate according 
as the minor or major axis is fixed. The fixed axis is 
called the polar axis, and the revolving one the equatorial 
axis. 

441. A segment of a spheroid is a portion cut off by a 
plane perpendicular to one of its axes. 

When the plane is perpendicular to the fixed axis, the 
segment may be said to be circular, as its base is a circle ; 
and when the plane is parallel to the fixed axis, the 
segment may be said to be elliptical, as its base is an 
ellipse. 

442. The middle zone or frustum of a spheroid is a 
portion of it contained by two parallel planes at equal 
distances from the centre, and perpendicular to one of 
the axes. 

The frustum may be said to be circular or elliptic 
according as its ends are perpendicular or parallel to the 
fixed axis. 

443. An hyperboloid is a solid generated by the revolution 
of one of the opposite hyperbolas about its axis remaining 
fixed. 

This hyperboloid is also called a hyperbolic conoid. 

444. A frustum of a hyperboloid is a portion of it 
contained between two parallel planes perpendicular to 
the axis. 

445. Problem I. To find the solidity of a paraboloid. 
RULE. Multiply the area of the base by the height, and half 
the product will be the solidity ; or, 
Multiply the square of the diameter of the base by the 




THE SOLIDS OF REVOLUTION OF THE CONIC SECTIONS 205 

height, and this product by '7854, and half the result will be 
the solidity. v 

Let ABV be the paraboloid, 

b = the area of the base, 
A = the height ; 
then V=\bh. 

Or, if c?=the diameter of the base, 

b = -7854^, 
and V = x -7854^A = 3927c? 2 A. 

EXAMPLE. What is the solidity of a paraboloid the diameter 
of whose base is = 10, and its height = 15? 

V= -3927rf 2 A = "3927 x 10 2 x 15 = 589'05. 

Take any ordinate HG at a distance VH = A'from the vertex, 
and another at the same distance from the base at D, then the 
abscissa of the latter is h - h' ; and if d' denote the ordinate HG, 
and d" the other ordinate, then (Art. 415) if j9 = the parameter, 
ph' = d">, &n&p(h-h') = d" z ; hence d' 2 + d"*=ph = %d?, and therefore 
Trdt + ird" 2 =ird?; that is (Art. 273), the circular sections of the 
paraboloid perpendicular to the axis, of which d', d" are the radii, 
are equal to the base ACB ; and the same can be proved of every 
two sections of the paraboloid that are equidistant from the vertex 
and base. Therefore, if a cylinder were described on the base 
ACB, having a height = the half of VD, any horizontal section of 
it would be = the corresponding section of the paraboloid, at the 
same distance from the base, together with the section equidis- 
tant from the vertex. Hence the whole paraboloid is equal to 
a cylinder on the same base, and having half the altitude, which 
proves the rule. 

EXERCISES 

1. What is the volume of a paraboloid the height of which is 
= 10, and the diameter of its base =20? . . . . =1570'8. 

2. Find the solidity of a paraboloid whose altitude is =21, and 
the diameter of its base = 12 =1187 '5248. 

3. What is the solidity of a paraboloid whose height is = 15, and 
the diameter of its base =20? =2356 '2. 

446. Problem II. To find the solidity of a frustum of a 
paraboloid. 

RULE. Multiply the sum of the areas of the two ends by the 
height, and half the product will be the solidity ; or, 
Multiply the sum of the squares of the diameters of the two ends 



206 THE SOLIDS OP REVOLUTION OF THE CONIC SECTIONS 

by '7854, and this product by tlie height, and half the last product 
will be the solidity. 

Let D and d be the diameters of the ends, and h the height of 
the frustum ; 
then V = J x -7854(D 2 + d 2 )& = 3927(D 2 fd 2 )A. 

EXAMPLE. Find the solidity of a frustum of a paraboloid, the 
diameters of its ends being = 15 and 12, and its height = 9. 
V=-3927(D 2 + d 2 )A=-3927(15 2 +12 2 )9 

= 3-5343x369 = 1304-1567. 

Let h', h", and V, V", be the heights and solidities of the two 
paraboloids ABV, EGV (fig. to Prob. I.); 



then V'=DW, V"=PA", and V = V - V" = (DW - d?k"). 
08 8 

But D 2 : d?=h' : h" ; hence D 2 -d 2 : d?=(h'-h"), or h : h" ; 
and therefore h" = ~5 Also, h'=h + h"=^ -jy 

Substituting these values of h' and h" in the above value of V, it 
becomes 

(D 2 + d z )h = -3927(D 2 + eP)h. 

o 

EXERCISES 

1. What is the solidity of a frustum of a paraboloid, the 
diameters of its ends being = 30 and 24, and its height = 9? 

=5216-6268. 

2. Find the solidity of a frustum of a paraboloid, the diameters 
of its ends being = 29 and 15, and its height = 18. . =7535-1276. 

447. Problem III. To find the solidity of a spheroid. 
RULE. Multiply the square of the equatorial axis by the polar 
axis, and this product by -5236, and the result will be the solidity. 
Let PELQ be an oblate spheroid, the 
minor axis PL being the fixed axis, or 
that of the spheroid, and EQ the major 
) a axis being the revolving axis. 

Let the major axis = a, and the minor = 6 ; 
then V= -5236a 2 6 for an oblate spheroid, 
and V= -5236a6 2 for a prolate spheroid. 
EXAMPLE. What is the solidity of the oblate spheroid whose 
polar axis is = 30, and equatorial axis = 50 ? 

V= -5236a 2 6= -5236 x 50 2 x 30 = 39270. 




THE SOLIDS OF REVOLUTION OF THE CONIC SECTIONS 207 

If a sphere PMLN be described on the axis PL, and if a section 
AB of the spheroid be taken perpendicular to its axis by a plane 
passing through D, and a section of the sphere by the same plane 
be taken, the section of the spheroid namely, the circle AFB 
would be to that of the sphere, whose diameter is MN, as the 
squares of their radii ; that is, as the square of the ordinate AD 
to the square of the corresponding ordinate MD of the circle 
PMLN. But the squares of these ordinates are as a 2 : 6 2 ; hence 
any section of the spheroid perpendicular to the axis is to the 
corresponding section of the sphere as a? : ft 2 . And, therefore, if 
V' = the solidity of the sphere, 

n z\r' n i 

V : V'=a 2 : ft 2 , and V=~=^x -52366 3 = -5236 2 &. 
cr o* 

The rule for the prolate spheroid may be similarly proved. 

EXERCISES 

1. Find the solidity of an oblate spheroid whose polar axis is 
= 15, and equatorial axis = 25 =490875. 

2. The axes of an oblate spheroid are = 12 and 20; what is its 
solidity? = 2513'28. 

3. Find the solidity of the prolate spheroid whose polar axis is 
= 7, and equatorial axis = 5. . . ." . . . =91 '63. 

4. What is the solidity of the prolate spheroid whose axes are 
= 18 and 14? =1847'2608. 

448. Problem IV. To find the solidity of a segment of a 
spheroid whose base is perpendicular to one of the axes. 

1. When the segment is circular. 

RULE. Find the difference between three times the polar axis 
and twice the height of the segment, and multiply it by the square 
of the height, and the product by '5236 ; then 

The square of the polar axis is to that of the equatorial axis as 
the last product to the solidity of the segment. 

When the segment is a portion of an oblate spheroid, 
6 2 : o 2 = -5236(36 - 2A)A 2 : V ; 

hence V = '5236(36 - 2A)^- 

When the segment is a portion of a prolate spheroid, it is similarly 

6 2 A 2 
shown that V= -5236{3 - 2A) 3 

2. When the segment is elliptical. 

RULE. Find the difference between three times the equatorial 



208 THE SOLIDS OF REVOLUTION OF THE CONIC SECTIONS 



axis and twice the height of the segment, and multiply it hy the 
square of the height, and this product hy '5236 ; then 

The equatorial axis is to the polar axis as the last product to the 
solidity of the segment. 

When the segment is a portion of an ohlate spheroid, 
a:6=-5236(3-2A)A 2 :V; 



V = -5236(3a - pE. 



hence 

When the segment is a portion of a prolate spheroid, it is similarly 

shown that V 



ah 2 
'5236(36 - 2A)~ 



EXAMPLES. 1. The axes of an oblate spheroid are = 50 and 30, and 
the height of a circular segment of it is = 6 ; what is its volume? 

,-,2^2 Kf>2 x C2 

V= -5236(36 - 2h)~ = -5236(90 - 12F^- 

= -5236 x 78 x 100 = 4084-08. 

2. What is the solidity of an elliptical segment of a prolate 
spheroid, its height being=12, and the axes = 100 and 60? 

nlfl 100 v T>2 

V = -5236(36 - 2h)~ = -5236( 180 - 24) g Q 

= -5236 x 156 x 240 = 19603 "584. 

The first rule is easily derived thus : Let APB be a circular 
segment of an oblate spheroid (fig. to Prob, III.). Then it was 
shown in last problem that the corresponding sections of the 
spheroid and sphere, such as those whose diameters are AB and 
MN, were to one another as a 2 : 6 2 . Hence, if V' = the volume of 
the spherical segment MPN, V'= -5236(36 - 2A)A 2 (by Art. 393), and 



6 2 



l 

= V : V, and V = = -5236(36 - 



When the spheroid is prolate, 2 : 6 2 = V : V. 

Let the segment be elliptical, as AQB, 
the spheroid PELQ being oblate. Then, 
if HEGQ be a sphere described on the 
axis EQ, and ACBD, MCND be two 
corresponding sections of the spheroid and 
sphere, it is evident that CD is a diameter 
of each of these sections, and equal to 
MN. Also the diameter MN : AB = HG : PL 
= EQ:PL = a:6. 
Now, MN and AB are any corresponding 

chords of the sphere and spheroid parallel to PL ; hence any 




209 

other two corresponding chords in the plane of the section MCND, 
parallel to AB, have the same proportion. Hence ACBD is an 
ellipse. Therefore the area of the elliptic section ACBD is to 
that of the sphere MCND as b to a. Hence, V being the volume 
of the spherical segment, 

bV bh 2 

a:b = V'-.V, and V = = '5236(3rt - 2A) 
a 

When the spheroid is prolate, the rule for the elliptic segment 
may be proved in the same manner, by describing a sphere on the 
equatorial axis. 

EXERCISES 

1. Find the solidity of a circular segment of a prolate spheroid, 
the axes being =40 and 24, and the height = 4. . . =337 '7848. 

2. The axes of an oblate spheroid are = 25 and 15, and the height 
of a circular segment of it is = 3 ; what is the solidity of the 
segment? . . . . . . . . =510'51. 

3. What is the solidity of an elliptic segment of an oblate 
spheroid whose height is = 10, the axes of the spheroid being =100 
and 60? . . . ...... =8796-48. 

4. Find the solidity of an elliptic segment of a prolate spheroid 
whose axes are =25 and 15, the height of the segment being =3. 

=306-306. 

449. Problem V. To find the solidity of the middle 
frustum of a spheroid. 

1. When the frustum is circular. 

RULE. To twice the square of the middle diameter add the 
square of the end diameter ; multiply this sum by the length of 
the frustum, and then this product by '2618. 

When the frustum is a portion of an oblate spheroid, 



When the frustum is a portion of a prolate spheroid, 



2. When the frustum is elliptical. 

RULE. To double "the product of the axes of the middle section 
add the product of the axes of one end, multiply this sum by the 
length of the frustum, and by -2618. 

Let d and e be the greater and less axes of one end, then, whether 
the frustum is a portion of an oblate or prolate spheroid, 



EXAMPLES. 1. What is the solidity of a circular middle frustum 



210 THE SOLIDS OF REVOLUTION OP THE CONIC SECTIONS 

of an oblate spheroid, the middle diameter being = 25, the end 
diameters = 20, and the length = 9? 

V = -261 8(2a 2 + &}l = -2618(2 x 25 2 + 20 2 )9 = 2 "3562 x 1650 = 3887 73. 

2. Find the solidity of an elliptic middle frustum of a spheroid, 
the axes of the middle section being = 50 and 30, those of the ends 
=30 and 18, and the length =40. 

V = -2618(2a& + e?e)/ = "2618(2 x 50 x 30 + 30 x 18)40 
= 10-472 x 3540 = 37070-88. 

The rules are easily proved by means of those in Problems 
III. and IV. 

Let ABA'B' (fig. to Art. 447) be a circular middle frustum of 
the oblate spheroid PELQ ; then the volume of the middle zone 
of the sphere described on PL is V'= < 7854(& 2 -Z 2 K by Art. 394, 
where I = DD'. But PL 2 : EQ 2 = LD . DP : AD 2 , or & 2 : a 2 

7,2^72 

(&-*):iP, or 6 2 : 2 = 6 2 -Z 2 : c^ 2 ; hence, W-1 2 = ~ 



Hence b 2 -^l 2 = : 

and V = -2618(2a 2 + d 2 ) "-2. 

or 

But any section, as MN, of the sphere perpendicular to the axis 
PL, is to the corresponding section AB of the spheroid as 6 2 : a 2 , 
or 6 2 : a 2 =V : V 

therefore, V = -- = - 2618(2a 2 + d z )l. 

Again, let ABA'B' (fig. to Art. 448) be an elliptic middle frustum 
of an oblate spheroid PELQ ; then the volume of the middle frustum 
MNM'N' of the sphere described on EQ is 

V'=-7854(a 2 -^% where J = FF'. 
But EQ 2 : PL 2 = EF . FQ : AF 2 , 

or a 2 :6 2 =a 2 -/ 2 :e 2 ; 

therefore, a 2 - 1 2 = ^-^- 

Hence a 2 - i 

But EQ : PL = CD : AB, or a : b = d : e ; hence d 2 = ^J-, and 



V'= '7854 x J(2 2 + rf 2 )/. But any section of the spherical frustum, 
as MN, is to the corresponding one of the spheroidal frustum AB 
as a : b, by Prob. IV. ; 



THE SOLIDS OP REVOLUTION OP THE CONIC SECTIONS 211 

hence a : b = V : V, and V = = -2618(2 2 + rf 2 )- 

a a 

(*tf> 

= -2G\8(2ab + de)l, for d=^- 

EXERCISES 

1. Find the solidity of a circular middle frustum of a spheroid, 
the middle diameter being = 100, those of the ends = 80, and the 
length = 36 =248814 '72. 

2. What is the solidity of a circular middle frustum of a spheroid, 
its middle diameter being = 30, its end diameters = 18, and its 
length = 40? =22242-528. 

3. Find the solidity of an elliptic middle frustum of an oblate 
spheroid, the axes of the middle section being = 25 and 15, those of 
each end = 15 and 9, and the height = 20. . . =4633-86. 

4. What is the solidity of an elliptic middle frustum of an oblate 
spheroid, the axes of the middle section being = 100 and 60, those of 
each end = 60 and 36, and the length = 80? . . =296567 '04. 



450. Problem VI. To find the solidity of an hyperboloid. 

RULE. To the square of the radius of the base add the square 
of the middle diameter between the base and 
the vertex ; multiply this sum by the altitude, 
and then by -5236. 

Letr=the radius of the base = AE, c?=the 
middle diameter, and A = the height EV ; 
then V = -5236(7^ + d?)h. 

When the diameter of the base, the height, 
and the axes of the generating hyperbola are given, but not the 
middle diameter, it may be found by Art. 431 ; thus, if h'=\h, 

cP ; hence, rf 2 = 




EXAMPLE. Find the solidity of an hyperboloid, the altitude 
of which is =25, the radius of the base = 26, and the middle 
diameter =34. 

V = 5236(r 2 + d*)h = -5236(26 2 + 34 2 )25 

= 13-09 x 1832=23980-88. 

Let the figure in Art. 430 be supposed to revolve about its axis 
GM, and the cone, generated by the asymptotes GH, GL, is called 
the asymptotic cone. Let V denote the volume of the conoid 
liPN ; V that of the frustum of the cone of the same height with 



212 

the conoid, the diameter of its upper end being IK ; and V" that 
of the cylinder of the same height with these two solids, and whose 
radius is equal to half the minor axis CG or IB ; then it can be 
shown in the following manner that V = V - V". 

Let r and ' denote the radii of the bases of the conoid and conic 
frustum, and h their height, d the double ordinate at the middle 
point between the base and vertex, and a, b the semi-axes ; then 
(r' + r)(r'-*) = & 2 , or - 2 = r' 2 -6 2 ; hence the section of the conoid is 
equal to the difference between the corresponding section of the 
cone and of the cylinder whose volume is V" ; and as this can be 
shown to be the case for every section, therefore V = V - V". 

Now, V = ^(r' 2 + W + br' )h, V" = irtfh, 

therefore, V = ^(r' 2 + br' - 2& 2 )A. 

But since GB = , and BI = 6, from similar triangles, 

a : b=a + h : r', and r' = -(a + h) ; 
ct 

hence , V = ^(6a& 2 A + 2& 2 A 2 )A. 

Ct 

Now, a 2 : 6 2 = (la + %}% : \d\ and d? = -Ma + h)h. 

\ i)' Cl" 

Also, a 2 : b 2 =C2a + h)h : r 2 , and r*=^(2a + h)h, and if cP and r 2 be 



substituted in the preceding expression for V, by eliminating a 
and b, it becomes -5236(r* + d?)h. 



EXERCISES 

1. Find the solidity of an hyperboloid whose altitude is = 50, the 
radius of its base = 52, and the middle diameter = 68. =191847 '04. 

2. What is the solidity of an hyperboloid whose altitude is=20, 
the radius of its base = 24, and the middle diameter = 31 749 ? 

= 16587-6375. 

451. Problem VII. To find the solidity of a frustum of 
an hyperboloid. 

RULE. Add together the squares of the radii of the two ends, 
and of the middle diameter between them, multiply the sum by the 
altitude, and this product by '5236. 

Let R and r be the radii of the two ends AE, CF (fig. to Prob. VI.), 
d the middle diameter through G, and h the height EF ; 
then V = -5236( R 2 + ?- 2 + d 8 ) A. 

EXAMPLE. Find the solidity of a frustum of an hyperboloid, 



the diameters of its ends being = 6 and 10, the middle diameter = 8, 
and the height = 12. 

V= 5236(R 2 + r 2 + rf 2 )/t= -5236{5 2 + 3 2 + (8i) 2 }12 

= -5236 x 425 x 3 = '5236 x 1275 = 667 '59. 
Let the axes of the generating hyperbola be a, b, and h', the 

height VF, then a 2 : V 2 =(a + h')h' : r z ; and hence i s =-^(a + h')h'. 

And similar values can be found for R 2 and (^c?) 2 , in terms of , b, 
h', and h. From the three equations thus formed, if the values of 
the unknown quantities a, b, and h' be found in terms of R, r, d, 
and h, the solidities of the two hyperboloids VAB, VCD can then 
be found in terms of the same given quantities, and the difference 
of these solidities would give that of the frustum, and hence the 
formula for finding it. 

EXERCISES 

1. What is the solidity of a frustum of an hyperboloid, the 
diameters at its two ends and at its middle being 12, 20, and 17, 
and its height = 18? =4005'54. 

2. Find the solidity of a frustum of an hyperboloid, the diameters 
at its ends and middle being =3, 5, and 4 '25, and its height =8. 

= 111-265. 



REGULAR SOLIDS 

There are only five Regular Solids, or, as they are some- 
times called, Platonic Bodies, and it can be proved that no 
more can exist. 

DEFINITIONS 
The regular solids are the five following : 

452. The tetrahedron is a regular triangular pyramid whose 
sides are equilateral triangles. 

453. The hexahedron is a cube. 

454. The octahedron is contained by eight equilateral triangles. 

455. The dodecahedron is contained by twelve regular pen- 
tagons. 

456. The icosahedron is contained by twenty equilateral tri- 
angles. 

Each side of a regular solid, except the tetrahedron, has an 
opposite face parallel to it, and the edges of these faces are also 
respectively parallel. 

fnc. Q 



214 REGULAR SOLIDS 

457. Problem I. To find the solidity of a regular tetra- 
hedron. 

RULE. Multiply the cube of one of its edges by the square root 
of 2, or 1 '414214, and take one-twelfth of the product. 

Let e = one of the edges ; 

then V = T VV2=iV>< l'414214=-117851e 3 ; 

also, the surface s = e 2 \/3 (by Art. 257), or s = l'732e 2 . 

EXAMPLE. What is the solidity and surface of a tetrahedron 
whose edge is = 15 ? 

v= T VV2, or v=-117851e 3 = -117851 x 15 3 = -117851 x 3375 = 397 '747. 
The surface may be found by the rules formerly given for the 
areas of regular polygons. 

Thus, the surface of the four sides of this pyramid, as they are 
equilateral triangles, is (A' being the tabular area), 

= 4e 2 A' = 4 x 15 2 x -433 = 900 x -433 = 389 '7 ; 
or s = e 2 \/3 = 15 2 \/3= 225 x 1-732 = 389-7. 

Let VABC be a regular tetrahedron. Draw VG perpendicular 
to the base ; join AG, and draw DG perpendicular to AB. 

It can be easily shown that AG bisects the angle 
CAB, and that DG bisects AB. Hence angle 
GAD = 30, and therefore AG = 2DG, as may be 
easily proved ; and hence AG 2 = 4DG 2 . ' 
Now, AG 2 = AD 2 + DG 2 , 
or 4AG 2 =4AD 2 + 4DG 2 ; 

that is, 3AG 2 = AB 2 = AV 2 =AG 2 + GV 2 ; 
hence 2AG 2 = GV 2 , and GV = AG\/2. 

Now, the base 6 = ABC = 6ADG = 3AD . DG ; 
and hence V or $bh = AD . DG . GV = AB . JAG . AGV2 ; 
or 




EXERCISES 

1. Find the solidity of a tetrahedron whose edge is = 8. =60'3397. 

2. Find the solidity of a tetrahedron whose side is = 3. =3'1819. 

3. What is the volume and surface of a tetrahedron whose edge 
is=6? . . . . ..... =25-4558 and 62-352. 

The rules for finding the volume and surface of a cube were 
given in Articles 375 and 377. 

458. Problem II. To find the solidity of an octahedron. 
RULE. Multiply the cube of one of the edges by the square root 
of 2, and take one-third of the product. 



REGULAR SOLIDS 215 




2, or v=-471405e 3 , 
and s=8e 2 A', or s=2e 2 \/3 

by Articles 269 and 257. 

EXAMPLE. Find the volume and surface of an octahedron 
whose side is = 6. 

V = JeV2= '471405e 3 = -471405 x 6 3 = 101 -823, 
and s=Se' 2 A' =8x6 2 x '433 = 124-704; 

or s= 2e 2 V3 =2 x6 2 x 1-732= 124-704. 

Let AVCV be a regular octahedron. Draw VG perpendicular to 
the plane ADC ; join AC. v 

It is easily proved that AG = GC, 
and 4AG 2 = AC 2 = 2AB 2 ; 

hence AG 2 = JAB 2 , or AG = JAB V2. 

Also, VG 2 =AV 2 -AG 2 = AB 2 -AG 2 =AG 2 ; 
or VG=AG. 

Now, area of square AC = b = AB 2 =e 2 =2AG 2 . 
Volume of VABCD = J6A= J x 2AG 2 . AG=|AB 2 . 
hence the whole solid V = %e?\/2. 

Also the surface is s=2e 2 \/3 (by Art. 257), or it is = 8e 2 A' (by 
Art. 269). 

EXERCISES 

1. Find the solidity of an octahedron whose edge is = 16. 

= 1930-87. 

2. What are the volume and surface of an octahedron whose 
edgeis = 3? =127279 and 31-1769. 

459. Problem III. To find the solidity of a dodeca- 
hedron. 

RULE. To 47 add 21 times the square root of 5 ; divide this sum 
by 40 ; find the square root of the quotient, and multiply it by 
five times the cube of the edge ; or multiply the cube of the edge 
by 7-6631. 

Also the surface s=15e 2 V * ^ = 15e 2 x 1-376382. 
5 

EXAMPLE. What are the solidity and surface of a dodecahedron 
whose edge is =2? 

V=e3x 7-6631 = 8 x 7'663 1=6 1-3048, 
and s= 15e 2 x 1 -376382 = 15x4x1 -376382 = 82-58292. 




216 KBGULAK SOLIDS 

Let ABKL be a regular dodecahedron. 

Draw BD, DE, EB on three contiguous sides, and AC perpen- 
dicular to the plane DBE, and draw DC, CE. 
Then, in the isosceles triangle ADE, angle A 
= 108; hence angle E = 36; and hence by 
trigonometry DE can be found, the side AD 
being given. Hence the sides of the equilateral 
triangle DBE are known ; and C is evidently 
its centre, also angle CDE = CED = |BED 
=x60 = 30; hence C = 120; therefore, DE 
being known, CD can be found. Hence, in the right-angled 
triangle ACD, AC 2 = AD 2 -CD 2 , and AC can thus be found. 

Now AC, if produced, Avould evidently pass through the centre 
of the polyhedron, or of its circumscribing sphere ; AC is the 
versed sine of an arc of it passing through AD ; hence (as in 
Art. 276), if D = the diameter of the sphere, D. AC = AD 2 , there- 

AD 2 e 2 
fore D = -rp- = , if p = AC. 

Again, ADH being a regular pentagon, if G be its centre, and 
GP be perpendicular to FH, then angle GFP = 54, and FP=|e, is 
known ; hence FG can be found. 

Now, the lines joining the centre of the polyhedron, and the 
points G and F, are evidently the radii of the inscribed and 
circumscribing spheres, and with FG form a right-angled triangle. 
Hence, if R and r are their radii, and r' = FG, R 2 =r 2 + r' 2 ; and 
hence r^R 2 -?-' 2 ; and R = JD, and r' being known, r can thus 
be found. 

But every regular polyhedron is composed of regular pyramids, 
whose altitudes are the radius of the inscribed sphere, and base 
one of the sides of the solid, and their number is the number of its 
sides. Hence, if n,=the number of sides, A = area of one side, then 



By actually calculating the values of the preceding quantities, 
the result would be the rule 7'6631e 3 . 

The first expression in the rule given above would be found by 
using, instead of the natural sines of the various angles, the follow- 
ing values namely, sin 36 = iV(10-2\/5), sin 108 = i\/(10+2\/5), 
sin 30 =\, sin 120 = JV3, sin 72 D =i\/(10 + 2\/5) ) and sin 54= 



EXERCISES 
1. Find the solidity of a dodecahedron the side of which is = 6. 

= 1655-2296. 



REGULAR SOLIDS 217 

2. What are the solidity and surface of a dodecahedron whose 
side is =4? ...... =490 -4384 and 330-33168. 

460. Problem IV. To find the solidity of an icosahedron. 

RULE. To 7 add three times the square root of 5, divide this 
sum by 2, find the square root of the quotient, and multiply it by 
the cube of the edge, then take five-sixths of this product. 

Or, V = f6 3 V 7 + 2 X/ . or V = fe 3 x2-61803=2-18169e 3 , 

and s = 20e 2 A', or s = 5e VS. 

EXAMPLE. What are the solidity and surface of an icosahedron 
whose edge is =2? 

V=* eV^^=f x 2 3 V 7 + 3x2-23606 

=^V6 -85409 =-Y- x 2-61803 = 17-4535, 
and s= 5e 2 V3 = 5 x 2 2 x 1-73205 = 34-641 ; 

or s=20e 2 A' =20x22 x -433 = 34-64. 

Let fall from A a perpendicular AC upon the plane of the regular 
pentagon DFHEK ; then C will be the centre of the pentagon, 
and CD may be found as FG in the preceding E 

figure. Then AC 2 = AD 2 - CD 2 ; and AC is thus 
found. 

Let G be the centre of one of the triangular 
sides AFH, and find FG as CD was found in 
the preceding figure. 

Then D, R, r, p, and r' denoting the same 
quantities as in the preceding problem, D . AC 




= AD 2 , and D = = = -. Also, R = D, and r* = R 2 - r. 
AC p 

As r can thus be found, then V = nrA', where n=20. If the 
value of r be calculated, and substituted in this expression, the 
result will be the preceding formula. 

EXERCISES 

1. Find the solidity of an icosahedron whose edge is=6. 

= 471-24504. 

2. What are the volume and surface of an icosahedron whose 
edgeis=5? ...... =272-711 and 216-506. 

461. The five regular solids may be easily made by cutting a 
piece of pasteboard into the following figures. The pasteboard 
should be cut nearly half -through along all the lines of the 



218 



REGULAR SOLIDS 



figure, and it will then be easily folded up into the form of the 
solid. 




Tetrahedron 





Cube 



Dodecahedron 




Octahedron 



Icosahedron 



The solidities and surfaces of regular solids may also be found 
by means of a Table containing the surfaces and solidities of regular 
solids, whose edges are = 1. The Table may be calculated by means 
of the preceding rules. 



No. of Sides 


Names 


Surfaces 


Solidities 


4 


Tetrahedron 


1-7320508 


0-1178513 


6 


Hexahedron 


6- 


1- 


8 


Octahedron 


3-4641016 


0-4714045 


12 


Dodecahedron 


20-6457288 


7-6631189 


20 


Icosahedron 


8-6602540 


2-1816950 



The rules for finding the solidities and surfaces by means of this 
Table are : 

For the solidity of a regular solid, multiply the tabular solidity 
of the corresponding solid by the cube of the edge. 
Or, V = e 3 V, if V' = the tabular solidity for edge = l. 

For the surface of a regular solid, multiply the tabular surface 
of the corresponding solid by the square of the edge. 
Or, s=eV, if s' = the tabular surface for edge = l. 

EXAMPLE. What are the solidity and surface of an icosahedron 
whose edge is =2? 



CYLINDRIC RINGS 219 

V = e?V = 2 s x 2 -181695 = 17 '45356, 
and s=eV =22 x 8-660254 = 34-641016. 

The student may perform the preceding exercises in regular 
solids by means of this rule. 



CYLINDRIC RINGS 

462. A cylindric ring is a solid formed by the revolution 
of a circle about an axis in its own plane, the centre of 
revolution being without the circle. 

463. The circle described by the centre of the generating 
circle is the axis of the ring. 

The centre of the axis is the centre of the ring. 

464. A cross section of a cylindric ring is one perpen- 
dicular to the axis. This section is equal to the generating 
circle. 

465. The interior diameter of the ring is a line passing 
through its centre in the plane of its axis, and limited by its 
interior surface ; and an external diameter is one terminated 
by its exterior surface. 

466. Problem I. To find the solidity of a cylindric ring. 

RULE. Multiply the area of a cross section by the axis of the 
ring ; or, 

Multiply the square of the thickness by the diameter of the 
axis, and this product by 2 -4674. 

The diameter of a cross section of the ring is equal to half the 
difference of the interior and exterior diameters ; and the diameter 
of the axis is half the sum of these diameters. 

Let d' and d" be the exterior and interior diameters AB and 
DE of the ring, and d that of the axis GHK, 
and t the thickness of the ring EB ; 
then d=\(d' + d"), and t=\(a' -d"). 

Also, if .(Jl^ the area of a cross section, 

and c=the length of the axis ; 

then J ^,= -7854 2 J and c = 3'1416W, 

and V=,5te 




220 CYLINDRIC RINGS 

EXAMPLE. Find the solidity of a cylindric ring whose inner 
diameter is = 12, and its exterior diameter =16. 

#=|(16 - 12) =2, and e?=i(16 + 12) = 14; 
hence V=2-4674eft 2 =2'4674x 14 x2 2 = 138'1744. 

By what is sometimes called the theorem of Guldinus (it is in 
reality due to Pappus), it appears that the solidity of the ring is 
equal to the area of the generating circle, multiplied by the line 
described by its centre of gravity or centre that is, by the axis of 
the ring or = '7854 2 x 3-1416tf=2-4674eft 2 , as above. 

EXERCISES 

1. Find the solidity of a cylindric ring whose interior and 
exterior diameters are = 16 and 24 = 789'568. 

2. Find the solidity of a cylindric ring, its diameters being = 8 
and 14 inches =244-2726. 

3. The interior diameter of a cylindric ring is = 26 inches, and 
its thickness = 8 inches; what is its solidity? . . =5369 - 0624. 

467. Problem II. To find the surface of a cylindric ring. 

RULE. Multiply the circumference of a cross section of the ring 
by the axis. 

Or, s=3'1416x3-1416d=9-8696cft, 

where d and t are found, as in the last problem. 

EXAMPLE. What is the surface of a cylindric ring whose thick- 
ness is = l inch, and inner diameter =9? 

* = 9 -8696<ft = 9 -8696 x 10 x 1 = 98 '696. 

The rule for the area of the surface can also be proved by the 
theorem of Pappus ; for by it the surface is equal to the product 
of the circumference of the generating circle by the line described 
by its centre of gravity or centre that is, by the axis of the ring. 

EXERCISES 

1. Find the surface of a cylindric ring whose diameters are 
= 36 and 52. =3474-0992. 

2. What is the surface of a cylindiic ring whose thickness is 
= 6 inches, and inner diameter = 24 inches ? . . =1776 '528. 



SPINDLES 221 

SPINDLES 

468. A spindle is a solid generated by the revolution of 
an arc of a curve, cut off by a double ordinate, about that 
ordinate as an axis. 

469. The spindle is said to be circular, parabolic elliptic, 
or hyperbolic according as the generating arc is a portion of 
a circle, a parabola, an ellipse, or hyperbola. 

I. THE CIRCULAR SPINDLE 

The central distance of a circular spindle is the distance 
between the centre of the circle and the centre of the spindle. 

470. Problem I. To find the solidity of a circular spindle. 
RULE. From one-twelfth of the cube of the length of the 

spindle subtract the product of the central distance and the area 
of the generating segment, and multiply this remainder by 6'2832. 

The length of the spindle and half its middle diameter are the 
chord and height of the generating circular segment ; hence the 
radius of the circle can be calculated by Art. 276, and the area of 
the segment by Art. 285. 

From the radius of the circle subtract half of the middle 
diameter, and the remainder is the central distance. 

Let CKDL be a circular spindle, and 
Z=KL, the length of the spindle; rf=CD, 
the middle diameter; e = SM, the central 
distance; ^5l = KCL, the area of the gene- 
rating segment; A' = the tabular segment 
(Art. 286), and h' its height. 
Then V = 6'2832(^-^Elc), 

and h' = $d-r2r; hence A' is known, and ^?l = 

EXAMPLE. Find the solidity of a circular spindle Avhose length 
is = 40 inches, and middle diameter = 30 inches. 
By Art. 276, if r = radius of circle = SC, then 




2r ==4 ii, and r=ao| . 

hence e=r- rf=20- 15 = 5f, 

and A' = (irf)-r2r=15^41 = 15x T f T = 1! 9 5 = -36 = height of tabular 
segment, to which corresponds the tabular area 
A' ='254551. 



222 SPINDLES 

Therefore, &= -254551 x (41) 2 =441 "9288, 

and V = 6-2832( T y 3 - Me) = 6'2832( x 16000 - 441 '9288 x *f) 
= 6-2832 x 2755-415 = 17312-8235. 

EXERCISES 

1. The length of a circular spindle is = 8, and its middle 
diameter = 6 ; what is its solidity ? .... =138-503. 

2. The length of a circular spindle is = 24, and its middle 
diameter = 18; find its solidity. .... =3739-5696. 



471. Problem II. To find the solidity of the middle 
frustum of a circular spindle. 

RULE. From three times the square of the length of the spindle 
subtract the square of the length of the frustum, multiply the 
difference by the length of the frustum, and take one-twenty- 
fourth of this product ; from the last result subtract the product 
of the area of the generating surface by the central distance, and 
multiply this remainder by 6 '2832. 

Let EGHF be a middle frustum of the spindle CKDL (last fig.) ; 
and let CD = D, EG = d, AE = l, KL = L, h = CI, A = area of segment 
CEF, of which CEI is the half, <7 = area of generating surface 
CEABF, and c, r, and h', as in last problem ; 

P h 

then A = (D-rf), 2r = -j- + h, h' = ^-; hence A' = tabular area is 



known, and ^5l = 

Also, 2EM=c#; and hence g= 

Then c = r - D, and L 2 = r 2 - c 2 , 

and V = 6 2832{ T 1 j(3L 2 - P)l - eg}. 

EXAMPLE. Find the solidity of a middle frustum of a circular 
spindle, the middle and one of the end diameters being = 16 and 
12, arid the length of the frustum = 20. 



2r=5 + A = nr + 2=S2, andr=26. 

4/fc o 

'=^=J2= '038462 ; and hence A' = '009940. 

' =2704 x -009940 = 26-87776. 
= 26 -87776 + 6x20 = 146 '87776. 



= r 2 - c 2 = 26 2 - 18 2 = 676 - 324 = 352, 



SPINDLES 223 



and V = {^t(3L 2 - P)l - c#}6'2832 

= {^(4224 - 400)20 - 2643 '8} x 6 '2832 
= (3186-6 -2643 -8)6-2832 = 542 -866 x 6 '2832 =3410-93984. 

EXERCISES 

1. Find the solidity of a middle frustum of a circular spindle, the 
middle diameter being = 18, an end diameter =8, and the length of 
the frustum = 20 =3657 '142. 

2. What is the solidity of a middle frustum of a circular spindle, 
its middle diameter being = 32, an end diameter = 24, and the length 
of the frustum =40? . =27287 '54. 

II. THE PARABOLIC SPINDLE 

472. Problem III. To find the solidity of a parabolic 
spindle. 

RULE. Multiply the square of the middle diameter by the 
length of the spindle, and this product by '41888 ; or, 
Take eight-fifteenths of the circumscribing cylinder. 
Let CADB be a parabolic spindle, CD d 

then V= -41888^, 

or V= T \x -7854^. 

EXAMPLE. Find the solidity of a parabolic spindle whose length 
is=40, and middle diameter=16. 

V= -41888^ = '41888 x 16 2 x 40 = 4289'33. 

EXERCISES 

1. The length of a parabolic spindle is = 30, and its middle 
diameter = 17; what is its solidity? .... =3631-6896. 

2. Find the solidity of a parabolic spindle whose length is = 18, 
and middle diameter =6 feet =271 '434. 

3. W T hat is the solidity of a parabolic spindle whose length is 
= 50, and middle diameter =10? .... =2094 -4. 

473. Problem IV. To find the solidity of the middle 
frustum of a parabolic spindle. 

RULE. Add together 8 times the square of the middle diameter, 
3 times the sqiiare of an end diameter, and 4 times the product of 
these diameters ; multiply this sum by the length of the frustum, 
and then this product by -05236. 




224 SPINDLES 

Let D = CD, the middle diameter (last fig.), 

e?=EF, an end diameter, 
= NP, the length of the frustum ; 
then V = -05236(8D 2 + 3d 2 + 4Dd)l. 

EXAMPLE. Find the solidity of a middle frustum of a parabolic 
spindle, its middle and an end diameter being = 20 and 16, and its 
length = 20 feet. 

V = -05236(8 x 20 2 + 3 x 16 2 + 4 x 20 x 16)20 = 1 '0472(3200 + 768 + 1280) 
= 1 -0472 x 5248 = 5495 "7056. 

EXERCISES 

1. What is the solidity of a middle frustum of a parabolic 
spindle, its middle and an end diameter being = 16 and 12, and 
its length = 30? =5101'9584. 

2. Find the solidity of a middle frustum of a parabolic spindle, 
an end and its middle diameters being = 10 and 18, and its length 
= 40. =7564-9728. 



III. THE ELLIPTIC SPINDLE 

474. The central distance of an elliptic spindle is the distance 
from the centre of the generating ellipse to the centre of the 
spindle. 

475. A diameter at one-fourth the length from the end of a 
spindle or a frustum is called the quarter diameter. 

476. Problem V. To find the solidity of an elliptic 
spindle. 

RULE. Divide 3 times the area of the generating segment 
by the length of the spindle, and from the quotient subtract the 
middle diameter ; multiply this remainder by 4 times the central 
distance, and subtract this product from the square of the middle 
diameter ; multiply this remainder by the length of the spindle, 
and the product by -5236 for the solidity. 

The central distance is found thus : From 3 times the square 
of the middle diameter take 4 times the square of the quarter 
diameter, and from 4 times the latter diameter take 3 times the 
former ; divide the former difference by the latter, and one-fourth 
of the quotient will be the central distance. 

To the central distance add half the middle diameter, and the 
sum will be the semi-axis minor ; and the major axis can then 



SPINDLES 225 

be found by Art. 426, and the area of the generating segment 
by Art. 429. 

Let =AB, the length of the spindle; 
D = CD, the middle diameter; rf=EF, the A< 
quarter diameter ; c = IH, the central dis- 
tance ; 6 = 1C, the semi-axis minor. ' 

Then, if =semi-axis major, h = ^D, s=area of segment ACB, 
s' that of the corresponding circular segment, also h' and A' the 
height and area of the tabular segment, 

3D 2 -4^ bl 

. _> o +*u, " 




h' = -rr, s' = 4& 2 A' s = -r-, or s= 
46 b 

and V = -5236{D 2 -4^--D \c}l. 

EXAMPLE. Find the solidity of an elliptic spindle whose length 
is = 20, its middle diameter = 6, and its quarter diameter =4 7477. 



3D 2 -4^ 108-90. 1624 _ 
' - = *' 18.9908-18 ~ 



and 




*'=4-6 = 30 = - 2 ' andA ' = ' 111824 ' 
s=46A'=4x 12-5x7'5x -111824=41-934. 
'3s 



Then V = -5236{D 2 - 4^ y - ~D\c}l 



= -5236{36 - 



= -5236 x (36 - 18 x -2901)20= -5236 x 30 "7782 x 20 =322 '3093. 

Let h = CH = ^D, h' = CK = J(D - d) ; then EK = #, and by Art. 423, 
ft 2 : a?=(2b-h)h : (&)*, 6 2 : a?=C2b-h')h' : (J/) 2 . Hence (2b-h)h 

4^,^ _ ^2 
= <l(2b-h')h', from which 26 = -jT7 j~- Substituting the above 

3D 2 4^ 
values of 7i' and h, it appears that b - JD = c = ^ ^y ; and hence 

is known. Then (Art. 426) a = = 



The value of s is found by Art. 429 to be = 46A'. The aid of the 
calculus is required to prove the expression for the solidity. 



226 SPINDLES 

EXERCISE 

Find the solidity of an elliptic spindle whose length is =40, the 
middle diameter = 12, and the quarter diameter =9'49547. 

= 2578-4748. 

477. Problem VI. To find the solidity of the middle 
frustum of an elliptic spindle. 

RULE. Find the area of the elliptic segment whose chord is the 
length of the frustum ; divide 3 times this area by the length 
of the frustum, from the quotient subtract the difference of 
the middle and an end diameter, and multiply this remainder by 
8 times the central distance. 

Find the sum of the square of an end diameter and twice the 
square of the middle diameter ; from this sum subtract the product 
last found ; multiply this difference by the length, and this product 
by -2618 for the solidity. 

The central distance is found thus : From the sum of 3 times 
the square of the middle diameter and the square of an end 
diameter take 4 times the square of the quarter diameter ; and 
from 4 times the last diameter take the sum of an end diameter 
and three times the middle diameter ; divide the former difference 
by the latter, and one-fourth of the quotient will be the central 
distance. 

To the central distance add half the middle diameter, and the 
sum will be the semi-axis minor ; the major axis can then be 
found by Art. 426, and the area of the above elliptic segment by 
Art. 429. 

Let Z=AB, the length of the frustum; D = CD, the middle 
diameter ; d = EG, an end diameter ; 
q = MN, the quarter diameter ; c = IP, 
the central distance; h = CO, the height 
of segment EOF ; and s - the area of 
segment EOF. 




, , 

rrjfi > T ,, h =^r, s=4aoA, 
V(2o - h)h 2b 



and V 

EXAMPLE. Find the solidity of a middle frustum of an elliptic 
spindle whose length is = 14, its middle diameter = 12, an end 
diameter = 10 -8, and a quarter diameter = 1 1 '7045. 



' 



SPINDLES 227 

432 + 116 -64 -547 "9813 "6587 



and 



and 



and 



46-818-36-10-8 
= 15-l, and /t = (D 
x14 211-4 





~ 



-072 
-6; 
_ 
~' 



= 25'08; 



' = 4 x 25-08 x 15'1 x -003712=5-623, 



= -2618{288 + 1 16-64 - 



- 1 -2 



Vl}14 



= -2618(404-64 - -357)14= -2618 x 404-283 x 14 = 1481 "778. 

EXERCISE 

The length of a middle frustum of an elliptic spindle is = 20, its 
middle and an end diameter = 16 and 12, and a quarter diameter 
= 15-07878; what is its solidity? . . . . =3427 '4856. 

IV. THE HYPERBOLIC SPINDLE 

The formulae for the solidities of an hyperbolic spindle, and for 
the middle frustum of this spindle, are the same as for the elliptic 
spindle and its middle frustum, with a slight change in the signs. 
1. For the hyperbolic spindle. 
The notation remaining as in Prob. V., 

1.T =C = J ~T~5 o-r^ > 



a=c-JD = 



Then 
and then 



2. For the middle frustum of an hyperbolic spindle. 
The notation remaining as in Prob. VI., 



s = KCL is found by Art. 436, 
V= -5236{D 2 + 4 - D 




26 = 



; then * = AECFB = ECF + EB, 



and then V = -2618{2D 2 + o? + g^ - D + d}c}l. 



228 UNGtJLAS 



UNGULAS 

Ungulas are portions cut off from pyramids, prismoids, 
cylinders, and cones, by plane sections not parallel to the 
base. 

I. PYRAMIDAL AND PRISMOIDAL UNGULAS 

478. Problem I. To find the solidity of the two ungulas 
into which a frustum of a rectangular pyramid or a 
prismoid is cut by a plane inclined to its base. 

CASE 1. When the section passes through two opposite edges 
of its ends. 

Let ABDGF be the frustum or prismoid, and 
ACFH the section. The solid is thus divided 
into two wedges BCH, EHC ; hence, 

Find the solidities of the two wedges into 
which the frustum is divided, by Art. 388. 

For both wedges, V = 




EXAMPLE. Find the solidities of the two wedges, BCH, EHC, 
into which the frustum of a rectangular pyramid AF is divided, 
the length and breadth of its base = 30 and 20 inches, and of its 
top = 24 and 16 inches, and its height = 36 inches. 
For the wedge BCH, 

V = (e + 2/)6A = &( 16 + 2x20)30x36 =10080. 
For the wedge EHC, 

= 4(20 + 2 x 16)24 x 36 = 7488. 



EXERCISE 

Find the solidities of the two wedges into which a frustum of 
a rectangular pyramid is divided by a plane passing through two 
of the shorter opposite edges of its ends, the length and breadth 
of its base being = 45 and 30, those of its top = 36 and 24, and its 
height =40 ........ =25200 and 18720. 

CASE 2. When the section passes through an edge of one end 
and cuts off a part of the other end. 

Let the section be ACKI (last fig.). The frustum is thus divided 
into a wedge EIC, and a rectangular prismoid ADHK, the volumes 
of which can be found by Articles 388 and 3S9. 



UNGULA8 229 

For the wedge V = %(e + 2l)bh, 
the prismoid V = J(BL + bl + 4Mm)h, 

where M = (L + 1), and m = (B + 6). 

EXAMPLE. Find the solidities of the wedge EIC and the 
prismoid ADI, the dimensions of the frustum being the same as 
in the former example, and the distance of I from G = 10 inches. 
EIC = V = |(e + 2l)bh = J(20 + 32)10 x 36 = 3120, 

ADI = V = (BL + bl + 4Mm)h 
= J(30 x 20 + 14 x 16+4 x 22 x 18)36 = 2408 x 6 = 14448. 

EXERCISE 

Find the solidities of the wedge and prismoid into which a 
frustum of a rectangular pyramid is cut by a plane passing 
through one end of its base, and cutting off a portion of the top 
= 15 inches distant from its corresponding end ; the dimensions 
of the frustum being the same as in the exercise of the last case. 

= 7800 and 36120. 

CASE 3. When the section passes through an edge of one of the 
ends and cuts off a part from the opposite side. 

Let Bl be the section. The frustum is thus cut into a wedge 
ADK and an irregular polyhedron DKG, the 
volume of which is found by deducting that of 
the wedge from that of the frustum. 

Let H = height of frustum, and 
h= it wedge; then 
for the wedge v=&(e + 2l)bh, and when AG 
is a pyramidal frustum, its volume V is found 
by Art. 384, or when it is a prismoid, V is found by Art. 389, and 
then the polyhedron = V= V - v. 

Draw FL parallel to EC, then AL = AC-FE, and IK = IM + MK 
= EF + MK. 

Or, if AL = D, MK=e/, and A' = H-A, then H :h' = D :d, and d= 

EXAMPLE. Let the section DK cut the side AE in a line IK 
at a perpendicular distance of 27 inches from the base, to find the 
volumes of the wedge and polyhedron, the dimensions of the frustum 
being the same as in the example of the first case. 

Here P = 20- 16=4, rf=-^=-^- = l, 

Prac. P 




230 UNGULAS 



and e 

Then v=J(e + 2J)6A=J( 17 + 40)30x27 = 7695, 

, ,,, .(RE-be\ u , 600x20-384x16 _ 

and y= l(-^^-r^ ~ 20-16 
Hence V = V'-v = 9873. 

EXERCISE 

Find the solidities of the wedge and polyhedron into which the 
frustum AG is divided, the height of the wedge being = 30, and 
the dimensions of the frustum the same as in the exercise of the 
first case =19237 '5 and 24682-5. 

II. CYLINDRIC UNGULAS 

479. Problem II. To find the solidity of an ungula of 
a cylinder cut off by a section perpendicular to the base. 
Multiply the area of the circular segment, which is the base 
of the ungula, by the height of the cylinder. 
Let FGBDE be the ungula, and let 

d= AB, the diameter of the cylinder, 
h= BH, n height of the base, 
c = FG, n chord of the base, 
1= BD, n length of the ungula, 
c ^il = FBG, area of the base. 

Then A' = -5= height of tabular segment ; let its area = A' ; 
then , JR=d?A', and V = LH. 

EXAMPLE. The length of a cylindric ungula is = 10 feet, the 
diameter of the cylinder =18 inches, and the section = 6 inches dis- 
tant from the axis ; what is the solidity of the two ungulas? 

For the ungula EBFG, h' = -. = ^ = 1= -l6, and A'= -086042. 
a 18 o 

A = cPA' = 1 -5 2 x -086042 = -19359, 
and V = IK = 10 x -1935945 = 1 '935945. 

For the ungula EAFG, 

cylinder AD= -7854cPV = -7854 x l'5 2 x 10 = 17 '6715, 
ungula = 1 7 '67 1 5 - 1 '9359 = 1 5 '7356 - 1 "935945 = 15 '735555. 

EXERCISE 

What is the solidity of the two cylindric ungulas cut off by a 
plane parallel to the axis of the cylinder, at a distance of 2 feet 




UNGULAS 231 

from it, the diameter and length of the cylinder being = 6 and 20 
feet respectively ? =61*95 and 503-53776. 

480. Problem III. To find the solidity of a cylindric 
ungula cut off by a section inclined to the base. 

Let AD be the cylinder, and EFHB the ungula. 
/ = BE, the length of the ungula, 
^, = GB, ii height of the base, 
c = FH, ii chord of the base, 
2rore=AB, n diameter of the cylinder, 
.l = area of segment FBH. 

Then ^' = ^ an d ^R=c? 2 A' ) where A' = tabular area ; 

Uf 

EXAMPLE. Find the solidity of a cylindric ungula, the diameter 
of the cylinder being = 25, the length of the ungula = 60, and the 
height of its base = 5. 

By Art. 256, Jc 2 =AG. GR = (d- &)A=20x5 = 100 ; hence c=20, 

A' = - = = -2; hence A' =-11 1824, 
and JR = d 2 A' = 25 2 x. -111824 = 69-89, 

= i{Jx20 3 - 69-89(25 -10)}Y- = (1333-3- 1048-35)12= 1709-9. 

EXERCISES 

1. A cylindric vessel ACDB, 10 inches diameter, containing 
some fluid, is inclined till the horizontal surface of the fluid EFH 
meets the bottom in FH, leaving AG 8 inches of the diameter 
dry, and meets the side at E 24 inches from the bottom ; how 
many cubic inches of fluid is contained in it? . . =109-4334. 

2. Suppose that the vessel stated in last example is inclined till 
the surface of the fluid bisects the base, and that the surface rises 
to the same height on the side as before ; find the quantity of 
fluid , =400 cubic inches. 

In this example, 2h = d, and c = d, and the above formula becomes 



3. Suppose that the fluid in the same vessel leaves only 2 inches 
of the bottom diameter dry, and that it rises to the same height 
as before ; what is the quantity of fluid ? =734-218 cubic inches. 



232 UNGULAS 

III. CONIC UNGULAS 

481. Conic ungulas are elliptic, parabolic, or hyperbolic. 

An elliptic conic ungula is a portion of a cone cut off by a plane 
which, produced if necessary, would cut the opposite slant sides of 
the cone, and would form an elliptic section. 

A parabolic conic ungula is a portion of a cone cut off by a 
plane parallel to the slant side of the cone, and which forms a 
parabolic section. 

A hyperbolic conic ungula is a portion of a cone cut off by a 
plane which neither cuts the opposite slant sides nor is parallel to 
the slant side, and which forms an hyperbolic section. 

482. Problem IV. To find the solidity of the elliptic 
ungulas of a conic frustum made by a section passing 
diagonally through opposite edges of the ends. 

Let ACDB be a conic frustum, and ADB, ACD two ungulas 
into which it is divided by the section AD. 

Let D = AB, the diameter of the greater end, 
e?=CD, the diameter of the smaller end, 
l = DE, the perpendicular length, 
V = solidity of the greater ungula, 
v=solidity of the less ungula ; 

then V= '2618( y=r -j )D, and v = 




EXAMPLE. Find the solidity of the greater ungula ADB of a 
conic frustum ACB, the diameters of the ends being = 15 and 
9 "6 inches, and the height of the frustum being =20 inches. 

!=-2618x ! 



15-9'6 
2618 x 109-8 x 300 



= 1596-98. 



5-4 

EXERCISES 

1. A vessel of the form of a conic frustum is inclined till the 
surface of a quantity of fluid contained in it just covers the bottom 
and reaches the edge of its mouth ; how many cubic inches of fluid 
does it contain, the diameters of the mouth and bottom being = 38'4 
and 60, and the depth of the vessel = 40 inches? . . =51103-36. 

2. A vessel of the same dimensions as that of the preceding 
example, the bottom of which is the narrower end, contains a 



UNGULAS 



233 




quantity of fluid similarly disposed ; how many cubic inches of 
fluid are there ? ......... =26164 '92. 

483. Problem V. To find the solidity of the elliptic 
ungulas of a conic frustum made by a section cutting off 
a part of the base. 

Let D = AB, the diameter of the greater end of 

frustum, 
d CD, the diameter of the smaller end of 

frustum, 

1= perpendicular height of frustum, 
A=BG, height of base, 

A'=tabular area of segment for which h 1 = 

A'j= tabular area of segment for which the height is 
,,,_h-D + d 
~~d ' 

V= volume of ungula DEFB, 
v= H ii complemental ungula EAFCD, 
V'= u ,i the frustum AFBDC. 

Then V 



Or if * 

and v = V - V, where V = -2618(D + P + Vd)l. 

EXAMPLE. Find the volume of the ungula DEFB of a conic 
frustum ABCD, the diameters of its ends being=15 and 9-6, the 
perpendicular length = 20, and the height of the base of the ungula 
BG= 10 inches. 



Here 

Also, 
and 



and 



h' = = = , and A' = '556226. 



= = -9lft, 



A' 1= -371872. 
10 10 



V = 



x -556226 - 9-e 3 x -371872 x 2-1739V27139) x 



90 



90 

=J(1877 '2628 -1054 -537)^ = 822-7258 

O *4 



-= 1015-701. 



234 UNGULAS 

EXERCISES 

1. A vessel in the form of a conic frustum, whose bottom and 
top diameters are = 30 and 19 - 2 inches, contains a quantity of fluid, 
which, when the vessel is inclined, just reaches the lip, and leaves 
10 inches of the bottom diameter dry ; how many cubic inches of 
fluid are there, supposing the depth of the vessel to be =20 inches ? 

= 4062-787. 

2. If a vessel of the form of a conic frustum, equal in dimensions 
to that of the last example, but close at both ends, be so inclined 
that a quantity of fluid in it just covers the smaller end and 
10 inches of the diameter of the greater, what is the quantity of 
fluid contained in it ? .... =5595 '748 cubic inches. 

484. Problem VI. -To find the solidity of the parabolic 
ungulas of a conic frustum. 

Let D, d, V, v, V, I, and h have the same meaning as in the last 
problem, and A = the area of the base EBF of the ungula (last fig.) ; 
then A= D-d (Art. 481), 

and v 



also v = V - V, where V = -2618(D 2 + d? + Dd)l. 

EXAMPLE. Find the solidity of the parabolic ungula DEFB 
(last fig.) of a conic frustum, the diameters of the ends being 
= 15 and 9 '6 inches, and its height = 20 inches, and the upper 
edge of the ungula terminating in the edge of the upper end of 
the frustum. 

Here ^ = D-d=15-9-6 = 5'4, 

and h' = ~ . = ~ = -36, and A' = '254551 ; 

U 15 

hence A = cPA' = 15 2 x -254551 = 57 '273975, 

and V 



= (159-0943 - 92-16)20=446-229. 

EXERCISES 

1. Let a vessel in the form of a conic frustum, the diameters of 
its bottom and top being =30 and 19-2 inches, be inclined so that 
its upper slant side shall be parallel to the horizon ; to find the 
quantity of fluid it is capable of containing in this position, the 
depth of the vessel being =20 inches. . =1784-9166 cubic inches. 



DNGULAS 235 

2. Find the ungula which is complementary to that in the pre- 
ceding exercise =7873 '61 84 cubic inches. 

The preceding rales for finding the volumes of conic ungulas 
may be proved thus : 

Let VAB be a cone, and EFBD an ungula of the conic 
frustum ABDC. 

Produce GD to meet VA produced in L, draw LM parallel 
to AB, VK perpendicular to GD produced, and VH, DI per- 
pendicular to AB, and BN to GD, and join VE, VF. 

The ungula EFDB = conic solid EFBV- conic solid EFDV. 

Let L'=VH, D=AB, h =GB; \ 

then D h = AG ; \ 




andlet/=DI, d = CD, A = segment EFB, _ !V , 

/' = VP, d'=LM, Aj = segment EDF, 

r=VK, h'^GV, =LD, 
then a-h' GL. 

Also, let 6 = minor axis of the ellipse of 
which EFD is a segment, and V, V, v', 
the volumes of the ungula EFBD, and the 
solids VEFB, VEFD. /<" 

By the similar triangles, ABV, CDV, L ..... ~" 

AB:CD = VH:VP, orD:d=L':Z' ...... [1]. 

Hence D-d :D = l :L' ; therefore L' = jyr^ ...... PI; 

and by [1], *' = = jd ......... [3} 

Also, from the similar triangles GBN, GDI, and BDN, VDK, 
GB :BN = GD : DI, and BN :VK = BD :DV = DI : VP. 
Hence GB : GD = VK : VP, or h:h' = I" : I'. 

rri I in hi' Ml r ., 

Therefore, r = _ = __ ......... [4]. 



Now, V = AL', * = J A!*", and V = V - 

Or, V 



This is an expression for the volume of an ungula of any cone 
or pyramid. 

1. When angle VAB exceeds DGB, the section EDF is a seg- 
ment of an ellipse, of which DL is the major axis = , and the 
minor axis b = VCD . LM = *Jdd'. For if G were the middle of LD, 
then EF would be the minor axis, and CD = 2AG, LM=2GB, and 
EF 2 =4AG . GB = CD . LM or V-=dd'. 

From similar triangles LAG, LCD, LD : LG = CD : AG. 



236 tMGULAS 

Or, a : a-h' = d:D -h; hence a : h' = d:h- (D-d) ; 

hence a= 



Also, h' :a = h:d' ; 

, ,. ah dh -L j , h 

hence d' = - T7 = j- ^ -T. ; hence b-d\/-, ^ -r.. 
h h-(D-d) h-(D-d) 

Let A'^a circular segment, heigh t= A', and diameter = , 

1 A/ 6 A, 

then : 6 = A i : A, ; hence A, = -A , = 



Let A" x = a segment similar to A'^ but of a circle AEB, so that 
its height is = - =-j(h-D + d) ; 

Ct Ctr 

" 
then A"! : A\ = D 2 : a 2 ; hence, A^ = 

, 



Therefore, 



h 



And if A;> and A 3 denote the areas of segments of a circle whose 
diameter = 1, similar to A and A'^, then A = A2D 2 , and A" 1 = A 3 D 2 ; 

hence V = * . 



2. When the plane GD passes through A, then EFD or A be- 
comes a whole ellipse, and EFB becomes the circle AFB. Also, 
A=-7854D 2 , A! = -7854a&, h=T>, h' = a, b*=dd' = dD. 

Hence by [5], V = -261 8 f^(D 2 - d\/Dd) ; 

J ~~ Ct 

and this being subtracted from the volume of the frustum 

(D 3 -d 3 \ 
'26181 -=: T 1^, gives for the complementary nngnla 



3. When DG is parallel to AC, the section EFD is a parabola ; 
and hence its area A 1 = EF . GD. But in this case AG = CD = d, 
h = T>-d, and EF 2 = 4AG . GB = 4dh = 4d(D - d) ; hence Aj = |A' 
x 2V(D - d)d ; and substituting this value of Aj in [5], 

V= l {A ' D " * rf(D " d) V(D " 



UNGULAS 237 

4. When the angle DGB is greater than VAB, the section EDF 
becomes an hyperbola, and GD produced would then meet AV pro- 
duced in some point, as Q above V. And its major axis would 

be DQ = a =- -= r , and the minor axis would be b = 



=- -= r , fr j i 

D-d-h D-d-h 

for in this case h - (D - d) becomes (D - d) - h. The expression for 
the area of the now hyperbolic segment EDF being found, and 
substituted in [5] for A 1; the resulting expression would be the 
volume of the hyperbolic nngula. 



IRREGULAR SOLIDS 

485. Problem. To find the solidity of an irregular solid. 

RULE I. When the solid is of an oblong form, find the areas 
of several equidistant sections perpendicular to some line that 
measures the length of the solid, and proceed with these areas 
exactly as with equidistant ordinates (Art. 292), and the result 
will be the cubic contents. 

Or, V=!(A + 4B + 2C)D. 

RULE II. Divide the solid, by parallel sections, into portions 
nearly equal to frustums of conic solids, find the area of a middle 
section of each portion, and multiply it by the length of that por- 
tion, and the product will be nearly its solidity, and the sum of the 
solidities of all the portions will be nearly the solidity of the whole. 

RULE III. When the solid is not great, and is very irregular 
and insoluble in water, immerse it in water in some vessel of a 
regular form containing a sufficient quantity of water to cover the 
solid, then take out the body, and measure the capacity of that 
portion of the vessel which is contained between the two positions 
of the surface of the water before and after the body was removed. 

EXAMPLE. Find the solidity of an oblong solid whose length 
is = 100 feet, and the areas of five equidistant sections = 50, 55, 
70, 80, and 80 square feet. 

Here A = 50 + 80 = 130, 4B = 4 x 135 = 540, and 2C = 140 ; 
hence V = (130 + 540 + 140) x 25 = 6750 cubic feet. 

EXERCISES 

1. Find the quantity of excavation of a portion of a canal, the 
areas of five equidistant vertical sections being 200, 240, 360, 300, 



238 IRREGULAR SOLIDS 

and 280 square feet, and the common distance of the sections 
=25 feet =28000 cubic feet. 

2. What is the solidity of an oak-tree of irregular form, the 
lengths of four portions of it being respectively = 8, 5, 6, and 7 feet, 
and the areas of their middle sections = 10, 8, 7, and 5 square feet ? 

= 197 cubic feet. 

3. The surface of a portion of excavated earth is nearly of a 
rectangular form, its length is = 60 feet, its mean breadth = 40 feet, 
and the mean depth of the excavation =8 feet ; required the number 
of cubic yards of excavation =711*1. 



ADDITIONAL EXERCISES IN MENSURATION 

1. What is the difference between the superficial contents of a 
floor =28 feet long and 20 broad, and that of two others of only 
half its dimensions ? =280 feet. 

2. It is required to cut off a piece of a yard and a half from a 
plank =26 inches broad ; Avhat must be the length of the piece ? 

=6-23 feet. 

3. The area of an equilateral triangle is = 720 ; required its side. 

= 40-784. 

4. What must be the length of the radius of a circle which 
contains an acre ? =117*752 feet. 

5. A circular fish-pond is to be dug in a garden ; what must be the 
length of the cord with which its circumference is to be described, 
so that it shall just occupy half an acre? . . . =83 '263 feet. 

6. What length of a plank = 10 inches broad will make 4 square 
feet? =5-4 feet. 

7. A log of wood is = 15 inches broad and 11 thick ; what length 
of it will make 10 cubic feet ? . . . = 8 feet 8^ inches. 

8. A round cistern is = 26'3 inches in diameter ; what must be the 
diameter of another to contain twice as much, the depth being the 
same? =37'19 inches. 

9. What will be the expense of painting a conical church-spire, 
at 8d. per yard, the circumference of the base being = 64 feet, and 
its slant height = 118 feet? =13, 19s. 8d. 

10. What would be the expense of gilding a spherical ball of 
6 feet diameter, at 3d. the square inch? . . =237, 10s. l'19d. 

11. How many 3-inch cubes can be cut out of a cubic foot? =64. 

12. The numbers expressing the surface and solidity of a sphere 
are the same ; what is its diameter ? =6. 



ADDITIONAL EXERCISES IN MENSURATION 239 

13. To what height above the earth's surface must a person ascend 
to see one- third of its surface ? = A height equal to its diameter. 

14. A cylindric vessel = 3 feet deep is wanted that will contain 
twice as much as another = 28 inches deep and = 46 inches diameter ; 
what must be the diameter of the former ? . = 57'372 inches. 

15. A cubic foot of brass is to be drawn into a cylindric wire 
=tV of an inch in diameter ; what will be the length of the wire? 

= 97784-6 yards. 

16. A rectangular bowling-green, 300 feet long and 200 broad, 
is to be raised one foot higher by means of earth dug from a ditch 
to be made around it ; what must be the depth of the ditch, its 
breadth being = 8 feet? =7||feet. 

17. A frustum of a square pyramid is = 18 feet long, and the sides 
of its ends are = l and 3 feet, and it is to be divided into three equal 
portions ; what must be the length of each ? 

= 3-269, 4-559, and 10-172. 

18. A cone =40 inches high is to be cut into three equal parts by 
planes parallel to its base ; what must be their lengths ? 

= 5-057, 7-209, and 27 '734 nearly. 

19. The same number expresses the solidity and convex surface 
of a cylinder ; what is its diameter? =4. 

20. The base and head diameters of a tub are = 20 and 10 inches 
respectively ; what ought to be its depth in order that it may 
contain 9163 cubic inches ?. . . . '; . =50 inches. 

21. A circle = 60 inches in diameter is to be divided into three 
equal portions by means of two concentric circles ; what must be 
their diameters ? =34 -641 and 48-9898. 

22. A square inscribed within a circle contains 16 square yards ; 
what is the area of the circumscribed square ? =32 square yards. 

23. The side of the cubic altar of Apollo at Delphi was = l cubit ; 
what must be the side of the new cubic altar, which was to be 
twice the size of the former ? .... =1-259921 cubits. 

24. A pot of the form of a conic frustum is 5'7 inches deep, and its 
top and bottom diameters are = 3 '7 and 4-23 inches ; supposing it at 
first to be filled with liquid, and that a quantity of it is poured out 
till the remaining liquid just covers the bottom, what is the excess 
of the remaining quantity above that poured out ? =7 "0534 inches. 

25. A conical glass, whose depth is = 6 inches, and the diameter 
of its mouth = 5 inches, is filled with water, and a sphere 4 inches 
in diameter, of greater specific gravity than water, is put into it ; 
how much water will run over ? . . =26*2722 cubic inches. 

26. If a sphere and cone are the same as in the last exercise, and 



240 ADDITIONAL EXERCISES IN MENSURATION 

the cone only one-fifth full of water, what portion of the vertical 
diameter of the sphere is immersed ? . = '546 inch very nearly. 

27. A cone equal to that in the former exercise being one-fifth 
full of water, what is the diameter of a sphere which, when placed 
in it, would just be covered Avith the water? ;. ,, = 2'446 inches. 

28. A coppersmith proposes to make a flat-bottomed kettle, of 
the form of a conic frustum, to contain 13 '8827 gallons ; the depth 
of the kettle to be = 12 inches, and the diameters of the top and 
bottom to be in the ratio of 5 to 3 ; what are the diameters ? 

= 25 and 15 inches. 

29. A piece of marble, of the form of a frustum of a cone, has its 
end diameters = 1| and 4 feet, and its slant side is = 8 feet; what 
will it cost at 12s. the cubic foot ? . . . =30, Is. llfd. 

30. The price of a ball, at Id. the cubic inch, is as great as the gild- 
ing of it at 3d. the square inch ; what is its diameter ? =18 inches. 

31. A garden = 500 feet long and 400 broad is surrounded by a 
terrace- walk, the surface of which is one-eighth of that of the 
garden ; what is the breadth of the walk ? . . =13 '4848 feet. 

32. The paving of a square court at 6d. a yard cost as much as 
the enclosing of it at 5s. a yard ; what was its side? =40 yards. 

33. A reservoir is supplied from a pipe 6 inches in diameter ; how 
many pipes of 3 inches diameter would discharge the same quantity, 
supposing the velocity the same ? =4 pipes. 

34. A pipe of 4 inches diameter is sufficient to supply a town 
with water ; what must be the diameter of a pipe which, with the 
same velocity, will supply it when its population is increased by 
a half? =4-899 inches. 

35. The ditch of a fortification is = 1000 feet long, 9 deep, 20 
broad at bottom, and 22 at top ; how many cubic yards of excava- 
tion are there in it ? = 7000. 

36. When the pressure of the atmosphere is = 15 Ib. on the 
square inch, what would be the pressure on the surface of a man's 
body, supposing it to be = 16 square feet? . . = 34560 Ib. 

37. A silver cup, of the form of a conic frustum, whose top and 
bottom diameters are = 3 and 4 inches, and depth = 6 inches, being 
filled with liquor, a person drank out of it till he could just see 
the middle of the bottom ; how much did lie drink? 

=42-8567 cubic inches. 

38. The sanctuary of Butis, in Egypt, was formed of one stone, 
in the form of a cube of 60 feet, open at top, and hollowed so that 
it was every where = 6 feet thick ; required its weight, at the rate of 

. avoirdupois the cubic foot. ,_>' =6439 tons. 



THE COMMON SLIDING-RULE 241 



THE COMMON SLIDING-RULE 

486. The common or carpenter's sliding-nile consists of 
two pieces, each a foot long, connected by a folding-joint. It 
is used for computing the quantity of timber and the work of 
artificers. 

When the rule is opened out, one side or face of it is 
divided into inches and eighths of an inch, with other scales 
of parts of an inch ; and one-half of the other side contains 
several tables of practical use. But the part of it used for 
performing arithmetical operations is one face of one of the 
pieces, in the middle of which is a narrow slip of brass, 
which slides in a groove. 

487. On each of the two parts into which the stock is 
divided by the slider is a scale, and there are also two scales 
on the slider. The scales on the stock are named A and D, 
and those on the slider B and C, the scales A and B being 
contiguous, as are also C and D. The scales A, B, C are 
exactly equal, and are just a scale of logarithmic numbers 
like that in Article 150. The numbers on the scale D 
are the square roots of those opposite to them on the 
scale C. 

488. Suppose the slider in its place, with 1 on its ex- 
tremity, coinciding with 1 on the contiguous scales, and let 
the number d on D be opposite to the number c on C, then 
d 2 = c ; and were these numbers on scales of the same stan- 
dard, then Avould 2Ld = Lc ; but Ld on the scale D is = Lc on 
the scale C ; and hence 

489. The logarithms of the numbers on the scale D are 
double the logarithms of the same numbers on the scale C. 

In finding any number on any of the scales which is the 
result of some operation, as of multiplication, division, &c., it 
is necessary to know previously how many places of figures 
the number will contain ; but this is generally easily known. 



242 THE COMMON SLIDING-RULE 

490. Problem I. To find the product of two numbers. 
RULE. Set 1 on B to one of the numbers on A, then opposite to 

the other number on B is the product on A. 

EXAMPLE. Multiply 24 by 25. 

Set 1 on B opposite to 24 on A, then opposite to 25 on B is 600 
on A. 

For if a, b, and p are the two numbers and their product, 

b f) 
then pab, andy = -; .'. L6-Ll = Lp-La; 

that is, the extent from 1 to b on the line B is = that from a to p 
on the line A ; and, therefore, if 1 on 6 is opposite to a on A, then 
6 on B will be opposite to p on A. 

491. Problem II. To divide one number by another. 
RULE. Set 1 on B opposite to the divisor on A, then opposite to 

the dividend on A is the quotient on B. 

EXAMPLE. Divide 800 by 32. 

Set 1 on B opposite to 32 on A, then opposite to 800 on A is 25 
on B, which is the quotient. 

Let a, b, and q be the divisor, dividend, and quotient, 

then <7 = -> or- = =r; ' L6-L = Lo-Ll; 

a a 1 

that is, the distance from a to b on A is = that from 1 to q on B. 

492. Problem III. To perform proportion. 

RULE. Set the first term on B to the second on A, then opposite 
to the third on B is the fourth term on A. 

EXAMPLE. Find a fourth proportional to 20, 28, and 25. 

Set 20 on B opposite to 28 on A, then opposite to 25 on B is 35 
on A, which is the fourth term required. 

Let a, b, c, and d be four terms of a proportion, 
then a : b = c : d, or a : c = b : d ; .'. La - Lc = U - ~Ld ; 
that is, the distance between the numbers a and c on one loga- 
rithmic line is = the distance between b and d on the same or on 
an equal line, as in Art. 152. 

493. Problem IV. To find the square of a number. 
RULE. Set 1 on D to 1 on C, then opposite to the given number 

on D is its square on C. 
EXAMPLE. Find the square of 15. 
Set 1 on D to 1 on C, then opposite to 15 on D is 225 on C. 



THE COMMON SLIDING-RULE 243 

The reason of the rule is evident from Art. 487. 

The square of a number can also be found by Prob. I. Art. 490. 
For it is just the product of the number by itself. Thus, 
15 2 =15 x 15 ; and, by rule in Art. 490, this product is 225. 

494. Problem V. To find the square root of a given 
number. 

RULE. Set 1 on C to 1 on D, then opposite to the given number 
on C is its square root on D. 

EXAMPLE. Find the square root of 256. 

Set 1 on C to 1 on D, then opposite to 256 on C is 16 on D. 

Since the numbers on C are the squares of those opposite to them 
on D, therefore, according as 1 on D is reckoned 1, 10, 100,... the 1 
on C must be reckoned 1, 100, 10,000... 

495. Problem VI. To find a mean proportional between 
two numbers. 

RULE. Set one of the numbers on C to the same on D, then 
opposite to the other number on C is the mean proportional on D. 
EXAMPLE. Find a mean proportional between 9 and 16. 
Set 16 on C to 16 on D, and opposite to 9 on C will be 12 on D. 
The rule is proved thus : Since a : x=x : b ; therefore, 

a a x a a a 2 T T , nr nr 
-=-.-=-.-=; .-. La-U=2La-2Lx. 
o x b x x a? 

Therefore the distance between the numbers a and b on the line 
C will be equal to the distance between a and x on the line D 
(Art. 489). 



MEASUREMENT OF TIMBER 

496. The measurement of timber is merely a particular 
application of the principles of the mensuration of surfaces 
and solids ; but as approximate rules are sometimes adopted 
on account of their practical utility in measuring timber, it is 
necessary to treat this subject separately. 

497. Problem I. To find the superficial content of a 
board or plank. 

RULE. Multiply the length by the breadth, and the product is 
the area. 



244 MEASUREMENT OF TIMBER 

When the board tapers gradually, take half the sum of the two 
extreme breadths, or the breadth at the middle, for the mean 
breadth, and multiply it by the length. 

Let 6 = the breadth in inches, 

J= ii length in feet, 
and .51= superficial content in feet ; then JR =^ bl. 

By the Sliding-rule. Set the breadth in inches on B to 12 on 
A, and opposite to the length in feet on A will be the content on 
B in feet and decimal parts of a foot. 

EXAMPLE. How many square feet are contained in the surface 
of a plank = 10 feet 6 inches long and 8 inches broad ? 
M=bl=& x 104 = | x 3r=7 square feet. 
Or, set 8 on B to 12 on A, and opposite to 10-5 on A is 7 on B. 

The first rule depends on Art. 247. 

The reason of the method by the sliding-rule is this : The area 
or surface &, = bl, b and I being the breadth and length in feet. 

When b is given in inches, then JR=-r~-l, or bl = l2^R, which is 

\.i 

convertible into the proportion 12 : b = l : JR, and 12, b and I being 
given, M is found by Prob. III. 

EXERCISES 

1. Find the area of a board = 18 inches broad and 16 feet 3 inches 
long =24 square feet 54 square inches. 

2. What is the price of a plank, the length of which is = 12 feet 
6 inches, and breadth = 1 foot 10 inches, at l^d. per square foot ? 

=2s. 10|d. 

3. Find the price of a plank, the length of which is = 17 feet, and 
breadth = 1 foot 3 inches, at 2^d. a square foot. . =4s. 5&d. 

4. What is the superficial content of a board = 29 feet long and 
22 inches broad ? . . . . . = 53 J square feet. 

5. The length of each of five oaken planks is=17 feet, two of 
them have the mean breadth of 13 inches, one is = 14$ inches at 
the middle, and the remaining two are each = 18 inches at the 
broader end, and = ll inches at the other end ; what is their price 
at 3d. per square foot ? =1, 5s. 9d. 

498. Problem II. To find the cubic content of squared 
timber of uniform breadth and thickness. 

RULE. Find the continued product of the length, breadth, and 
thickness, and the result is the content. (See Art. 374.) 



MEASUREMENT OP TIMBER 245 

Let b, t, I, and V be the breadth, thickness, length, and volume 
or solidity ; then V = btl. 

By the Sliding-mle. Find the mean proportional between the 
breadth and thickness in inches (Art. 495), then set the length 
on C to 12 on D, and opposite to the mean proportional on D is the 
content on C in feet. 

When the timber is square, the mean proportional is = the side of 
the square. When the mean proportional is in feet, 1 on D is to be 
used instead of 12. 

EXAMPLE. Find the solidity of a squared log of timber, of the 
invariable breadth and thickness of 32 and 20 inches, its length 
being =40 feet 6 inches. 

V = btl = ft x f| x 40J = 180 cubic feet. 

Or, find (Art. 495) the mean proportional between 32 and 20, 
which is 25 '3 ; then set 40 '5 on C to 12 on D, and opposite to 25 '3 
on D is 180 on C, the content. 

The method by the sliding- rule is derived thus : Let m = the 
mean proportional between b and t, then itfbt, and as m is in 
inches, 

^ r , m m ,/7n\ a , V 



and LV-L6=2Lm-2L12; 

that is, the distance between 12 and m on D is equal to that 

between b and V on C. 

EXERCISES 

1. Find the solidity of a log of wood = 30 inches broad, 18 thick, 
and 16 feet long ........ = 60 cubic feet. 

2. What is the content of a log the end of which is = 30 inches 
by 20, and length =20 feet? ..... = 83 J cubic feet. 

3. What is the content of a square log of wood, the side being 
= 14 inches, and the length = 12 feet ? . . . = 16 cubic feet. 

4. The side of a square block of sandstone is = 3 feet, and its 
length = 6 feet; what is its content? . . . =54 cubic feet. 

5. Find the cubic content of a log of wood = 20 feet 3 inches long, 
its ends being = 32 by 20 inches. . . . . = 90 cubic feet. 

6. The side of a square log of wood is = 2 feet, and its length 
= 24 feet 1 inch ; what is its content? . . . =96 cubic feet. 

499. Problem III. To find the content of squared taper- 
ing timber. 

RULE. Find the mean breadth and thickness, and multiply 
their product by the length. 

Pra* Q 



246 MEASUREMENT OF TIMBER 

As in last problem, V = btl. 

By the Sliding-rule. The method is the same as that of last 
problem, using the mean breadth and thickness. 

EXAMPLE. The breadth of a tapering plank of wood at the 
two ends is = 18 and 12 inches, and its thickness at the ends=14 
and 10 inches, and its length = 19 feet 10 inches; Avhat is its 
solidity ? 

Here 6 = ^(18 + 12) = 15, and *=(14 + 10) = 12, 

and V = btl = f f . $% x 19f = 24^1 cubic feet. 

The above rule, though generally used, is correct 6nly in 
one case namely, when two of the sides are parallel and the 
other two converge ; for the solid is then a prism, having one 
of the parallel sides for its base. In other cases this rule gives 
the content a little less than the real solidity, and the error 
is greater the more the difference between the breadth and thick- 
ness. But the true solidity can always be found by consider- 
ing the log a prismoid, and calculating its content by the rule 
in Art. 389. 

The preceding example, calculated thus, gives for the content 
25-067 cubic feet instead of 24f. 

500. When the breadth is irregular, it may be measured at 
several places, and the sum of these breadths, divided by their 
number, may be taken for the mean breadth. In the same way 
the mean thickness may be found. 

EXERCISES 

1. Find the content of a squared tapering log of wood, the 
breadth and thickness at one end being =34 and 20 inches, and 
those at the other end = 26 and 16 inches, and the length = 32 feet. 

= 120 cubic feet. 

2. Find the cubic content of a log, the breadth and thickness at 
one end being =33 and 22 inches, and those at the other end =27 
and 18 inches, and the length = 40 feet. . . = 166 cubic feet. 

3. The breadth and thickness of one end of a piece of timber 
are =21 and 15 inches, and those at the other end are = 18 and 
12 inches, and the length is = 41 feet ; what is its solidity ? 

= 74-95 cubic feet. 

4. The breadth and thickness at the greater end of a piece of 
timber are = 1 "78 and 1 -23 feet, and at the smaller end = 1 '04 and 
0'91 feet ; what is its content, its length being =27 "36 feet? 

=41-278 cubic feet, 



MEASUREMENT OF TIMBER 247 

501. Problem IV. To find the content of round or un- 
squared timber. 

RULE I. Find the quarter girt that is, one-fourth of 
the mean circumference and multiply its square by the 
length. 

By the Sliding-rule. Set the length on C to 12 on D, 
and opposite to the quarter girt in inches on D is the content 
onC. 

RULE II. Find one-fifth of the girt, and multiply its square by 
twice the length. 

By the Sliding-rule. Set twice the length on C to 12 
on D, and opposite to one-fifth of the girt on D is the content 
onC. 

Let I and c denote the length and mean circumference of a piece 
of round timber, and V its volume ; then, 



by Rule I., V= l = 

by Rule II., V = 2^= -08c 2 J. 



EXAMPLE. The mean circumference of a piece of unsquared 
timber is = 6 feet 8 inches, and its length = 16 feet 4 inches; what 
is its content ? 

By Rule L, V= (|) *=(i !) 2 16J = (* ) 2 V =45'37 cubic feet. 
By Rule II., V=2^|) l = (i . f) 2 . 32 = ($) 2 . ^=58-074 cubic feet. 

Note. When the piece of timber is of a cylindric form, its 
volume, by Art. 378, is V=6A=-07958c 2 J, if l=h. Therefore the 
first rule in this case gives the content too small by more than 
one-fifth part of the true solidity ; and the second gives it too 
much by about the 191st part. 

When the tree tapers uniformly, it is then a frustum of 
a cone, and its true volume can be found by the rule in 
Art. 386. In this case the first rule gives a result still further 
from the truth, for a conic frustum exceeds a cylinder of the 
same length, whose circumference is the mean girt of the 
frustum. 

The first rule is generally followed in practice, and the deficiency 
in the content given by it is intended to be a compensation to the 
purchaser for the loss of timber caused by squaring it. 



248 



MEASUREMENT OP TIMBER 



The following formula commends itself : 

G = J girt of tree at middle in feet, 
g= ii M one end .n 
h = M M other end n 
L = length of log in feet, 
c = cubic contents of log in feet, 



Allowance is to be made for bark by deducting from each J girt. 
The allowance varies from half an inch in trees with thin bark to 
2 inches for trees with thick bark. 

MEASURES OF TIMBER 
100 superficial feet of planking = 1 square. 
120 deals . . . . =1 hundred. 

50 cubic feet of squared timber I load. 
40 feet of unhewn timber . = 1 load. 
600 superficial feet of inch planking = 1 load. 
Boards 7 inches wide . . = battens. 
it 9 ii . . =deals. 

ii 12 M - ;.- '_' " -. '' = planks. 

To cut the best beam from a log. 

Divide the diameter, ab, into 3 equal parts, of, fe, and eb, and 





Fig. 1. 



Fig. 2. 



from e and / draw the lines ed,fc at right angles to ab ; join ac, 
ad, be, and bd, then acbd is the cross section of the strongest beam 
(fig. 1). 

To cut the stiffest beam, divide the diameter into 4 instead of 
3 parts (fig. 2), 



MEASUREMENT OP TIMBER 249 

In the following exercises the first answer is the result by the 
first rule, and the other is that by the second rule : 

EXERCISES 

1. The mean girt of a tree is = 8 feet, and its length = 24 feet; 
required its content. . .. .' =96 cubic feet, or 122 '88 cubic feet. 

2. What is the content of a piece of round timber, the girt at the 
thicker end being = 16 feet, and at the smaller = 12 feet, and its 
length = 19 feet? . . =232| cubic feet, or 297 '92 cubic feet. 

3. Find the content of a tree whose mean girt is = 3 '15 feet, and 
length = 14 feet 6 inches. =8 '992 cubic feet, or 1P51 cubic feet. 

4. The girts of a piece of round timber at five different places 
are = 9-43, 7 '92, 6'15, 4 "74, and 3'16 feet, and its length is = 17 feet 
3 inches ; what is its content ? 

= 42-5195 cubic feet, or 54 '425 cubic feet. 



RELATIONS OP WEIGHT AND VOLUME OF BODIES 

502. The relations of the weights and volumes of bodies 

are determined by means of their specific gravities. 

503. The specific gravity, or specific density, of any solid 
or liquid is the ratio which its density bears to that of 
distilled water at its maximum density point (4 C., or 
39 R). 

Tables of specific gravities are formed for reference, of 
which a specimen is appended. Thus, the specific gravity of 
mercury is 13'6, by which we mean that any given bulk 
of mercury will weigh 13'6 times as much as an equal bulk 
of water at the same temperature. 

When a body is immersed in a liquid, it loses as much 
weight as the weight of the liquid displaced. This is the 
Principle of Archimedes, and is the foundation of the 
method adopted for finding the specific gravity of solids. 
See Prob. I., below. 

The following instruments are commonly used for finding 
specific gravities, namely : 



250 



RELATIONS OP WEIGHT AND VOLUME OF BODIES 




(1) Nicholson's hydrometer; (2) Tweddel's hydrometer; 
and (3) the specific gravity bottle, or Pyknometer. 

The second (Tweddel's) hydrometer is used for finding 
the specific gravity of liquids heavier than water, such as 
sulphuric acid. It acts by ' variable im- 
mersion' that is, measures specific gravities 
by the depth to which it sinks. It is made 
of glass, with two globes, one for flotation, 
the other for balancing it in an upright 
position ; and the stem is so graduated that 
the reading of the number of degrees mul- 
tiplied by 5 and added to 1000 gives the 
specific gravity of the liquid as compared 
with water, whose specific gravity is for 
convenience taken to be 1000. Thus, 15 
Tweddel represents the specific gravity of 
1075 ; or, calling the specific gravity of water 1, it represents 
a specific gravity of 1'075. 

There are other hydrometers constructed on 
the principle of variable immersion, such as 
those for determining the density of alcohol, 
which is lighter than water, and those employed 
for determining the density of salt water in a 
boiler, where the graduations are so arranged as 
to indicate in a ready manner either the strength 
of the alcohol or the quantity of salt held in 
solution in the water. 

The specific gravity bottle, in one form, is a 
v bottle (&) with a fine stem (c), ending in a wide 
tube (a) having a glass stopper. The bottle is 
first weighed when empty, then when filled with 
water up to the mark (c), and finally, when filled 
with a given liquid up to the same mark. 
We thus ascertain the weight (1) of a given volume 
of water, (2) of the same volume of a given liquid, and 



RELATIONS OP WEIGHT AND VOLUME OP BODIES 



251 




ratio of the second to the first gives the specific gravity of 
the liquid. 

Nicholson's Hydrometer and its Manipulation. Nichol- 
son's hydrometer consists of a hollow cylinder (B) which 
ensures flotation, having at its base a loaded 
pan (C) to keep it upright, and at the top a 
stem supporting a dish (A) ; upon the stem 
a standard point (m) is marked. 

This instrument may be used for finding 
the specific gravity of a solid or a liquid. 
For example, let the solid be a piece of 
sulphur. Put the hydrometer in water, 
when it will require a given weight placed 
in (A) in order to sink the hydrometer to 
(ra). Let this weight be 125 grs. Now 
place the sulphur in (A), and add, say, 55 grs. 
in order to sink the instrument again to (ra). It follows 
that the weight of the sulphur is 70 grs. 

Next place the sulphur in (C) as marked (o), and let 
34'5 grs. be placed in the dish (A), in addition to the 
55 grs., in order to sink the instrument to (o). 

Then weight of sulphur = 70 grs., 

weight of water displaced by sulphur = 34*5 grs. 

70 

.'. Specific gravity of sulphur = sj^ = 2 '03. 

In order to find the specific gravity of a liquid (B), let x be 
the weight which sinks the instrument to the point (o) in 
water, (?/) the weight which sinks it to the same point in the 
liquid (B), and let W be the weight of the instrument. 

Then weight of water displaced by instrument == W + x, 
weight of liquid displaced by instrument = W + y. 



. ' . Specific gravity of liquid (B) = == - 



EXAMPLE. The standard weight of a Nicholson's hydrometer 
is 1250 grs. ; a small substance is placed in the upper pan, and it is 



252 

found that 530 grs. are needed to sink the instrument to the stan- 
dard point ; but when the substance is put in the lower pan, 
620 grs. are required. What is the specific gravity of the substance ? 

Standard weight . = 1250 grs. 

Weight required to sink the instrument to 

standard point .....= 530 n 

.'. Weight of body = 720 

Extra weight required when body is placed in the lower pan 
= 620 -530 = 90 grs. 

720 
. . Specific gravity of the body = -^r = 8. 

yo 

Definition of a Perfect Fluid. A ' perfect ' fluid is 
defined to be a substance which offers no resistance to a 
continuous change of shape. There are two kinds of fluids 
those which are practically incompressible, termed liquids; 
and those which are easily compressed, called gases and 
vapours. We know of no substance which completely fulfils 
the above definition ; but water, many other liquids, and all 
gases so nearly comply with it that for many purposes we 
may in practice consider them as perfect fluids. 

Viscosity. All known fluids, however, do offer some resist- 
ance to a change of shape, although they have no elasticity 
of form or power of recovery when the stress that has 
produced the change is removed ; and the property in virtue 
of which they do so is called the viscosity of the fluid. The 
viscosity of a fluid is measured by the shearing stress required 
to deform it at the uniform rate of unit shear per unit time. 

In many investigations it is necessary, for simplicity, to 
assume that we are dealing with a perfect fluid ; it is there- 
fore of the first importance that these definitions be clearly 
understood. 

504. The following table contains the specific gravities of 
the most common solids and fluids, those for solids and 
liquids being referred to water as standard, those for gases 
to air as standard. Referred to water, the specific gravity 
of air at C. and 30 inches barometric pressure is 0'001293. 



RELATIONS OF WEIGHT AND VOLUME OP BODIES 



253 



TABLE OF SPECIFIC GRAVITIES 



METALS 


ffrom 


0-690 




Ueecli j 




Aluminium, sheet, . . 2-670 


\ to 


0-696 


ii cast, . . 2-560 


Birch / fl ' m 


0711 


Antimony, . . 6720 


\ to 


0-730 


Bismuth, n . . 9 "822 


Box, .... 


1 -280 


Copper bolts, . ' .-. - - . 8-850 


Cedar, West Indian, 


0748 


H wire, . . . 8-900 


n American, . 


0-554 


Gold, . . . .18-417 


n Lebanon, 


0-486 


Iron, cast, average, . 7 '248 


Chestnut, ;. .- . 


0-606 


M wrought, average, . 7780 


Cork, 


0-240 


Lead, cast, . . . 11-360 


Deal, Christiania, . ;>.' 


0-689 


sheet, . . . 11-400 


Ebony, . . ^ . ,/ . 


1-187 


Mercury, .' : . 13-596 


Elm, English, . 


0-553 


Platinum, . . . 21-531 


n n . . . 


0-579 


n sheet, . . 23-000 


n Canadian, 


0725 


Silver, .... 10-474 


Fir, spruce, 


0-512 


Steel, .... 8-000 


n male, 


0-550 


Tin, cast, . . . 7 '291 


M female, 


0-498 


Zinc, .. . . .7-000 


Hornbeam, . . 


0-760 




Iron wood, 


1-150 


ALLOYS 


Greenheart, 


1-143 


Aluminium bronze, 90 to 


Larch, . .., . . 


0-543 


95 per cent, copper, . 7 "680 


n .... 


0-556 


Bell-metal (small bells), . 8-050 


Lignum-vitae, . . '-. ;..- 


1-333 


Brass, cast, . . . 8-400 


Lime, . . }\ 


0-564 


ii sheet, . ..;.-, 8-440 


Mahogany, Nassau, 


0-668 


t . Avire, . ..--"" .; 8 '540 


M Honduras, . 


0-560 


Gold (standard), .. .17724 


n Spanish, 


0-852 


Gun -metal (10 copper, 


Maple, . . v . 


0-675 


1 tin) 8-464 


Oak, African, ' . 


0-988 


Silver (standard), . . 10-312 


M American, red, 


0-850 


Speculum (metal), . . 7'447 


M ii white, . 


0779 


White-metal (Babbett), . 7'310 


ii English, 


0777 




n ii . 


0-934 


TIMBER 


Pine red -! 


0-576 


ffrom 0710 
Acacia, . . { ^ ^ 


\ to 
... ffrom 


0-657 
0-432 


. , ffrom 0-690 


n white, . 
I to 


0-553 


SI1 ' * ' I to 0-760 


n yellow, . .... 


0-508 



254 



RELATIONS OP WEIGHT AND VOLUME OF BODIES 



Pine, Dantzic, 


0-649 


MISCELLANEOUS SUBSTANCES 


TIT i (from 
n Memel, . 


0-550 


Asphalt, . 


2-500 


I. to 


0-601 


. , ffrom 


1-600 


. /from 


0-466 


Brick, common, -c 


2-000 


" Ri s a ' i to 


0-654 


n London stock, 


1-840 


Satinwood, 


0-960 


n Red, 


2-160 


/from 


0-740 


n Welsh fire, . 


2-400 


Teak, . . | to 


0-860 


M Stourbridge fire, . 


2-200 






Cement, Portland, . from 


3-100 


STONES, &c. 




ii n in powder, 


3-155 


Agate, .... 


2-590 


M Roman, 


1-600 


Amethyst, common, 


2-750 


Clay, .... 


1-900 


Basalt, Scotch, 


2-950 


Coal, Anthracite, . 


1-530 


ii Greenstone, 


2-900 


i, Cannel, 


1-272 


Welsh, 


2-750 


M Glasgow, 


1-290 


. .. ffrom 


2-330 


n Newcastle, 


1-269 


Chalk, . . | to 


2-620 


Coke, . . . ,'rtjj.-i 


0-744 


Firestone, 


1-800 


Concrete, ordinary, 


1-900 


Granite, Aberdeen gray, 


2-620 


u in cement, 


2-200 


n M red, 


2-620 


tf 1.1 ffrom 
Earth, -1 


1-520 


n Cornish, . 


2-660 


\ to 


2-000 


n Mount Sorrel, . 


2-670 


Glass, flint, 


3-078 


Limestone, compact, 


2-580 


M crown, . 


2-520 


M Purbeck, 


2-600 


M common green, 


2-528 


n Blue Lias, 


2-467 


M plate, . 


2-760 


M Lithographic, 


2-600 


Gutta-percha, . 


0-966 


Marble, Statuary, . 


2-718 


Gypsum, .... 


2-286 


M Italian, 


2-726 


Ice, if from water purged 




n Brabant block, . 


2-697 


from air, 


0-954 


Oolite, Portland Stone, . 


2-423 


India-rubber, . 


0-930 


u Bath n 


1-978 


Ivory, .... 


1-820 


Sandstone (Arbroath 




Lime, quick, . . 


0-843 


pavement), 


2-477 


,, . ffrom 
Mortar. 


1-380 


n Bramley Fall, 


2-500 


\ to 


1-900 


ti Caithness, 


2-638 


M average, 


1-700 


ii Craigleith, 


2-450 


Pitch, . . . --V 


1-150 


n Derby grit, 


2-150 


Plumbago, 


2-267 


M Red (Cheshire), 


2'510 


Snow, . . . 


0-083 


Slate, Anglesea, 


2-870 


Sand, quartz, . 


2-750 


ii Cornwall, 


2-510 


n river, 


1-880 


n Welsh, . 


2-880 


n pit (coarse), . 


1-610 


Trap, 


2-720 


n (fine), 


1-520 



RELATIONS OF WEIGHT AND VOLUME OP BODIES 



255 



Sand, pit, Thames, . 


1-640 


Shingle, . 


1-420 


Tallow, . 


0-940 


Tar, . 


1-016 


Tile, common, 


1-810 


il ii '"'i^ ^ 1< 


1-850 


LIQUIDS, &c. 




Water, distilled, 


1-000 


ii sea, . . . 


1-027 


Acetic acid, 


1-060 


Alcohol, absolute, . 


0-792 


it of commerce, . 


0-800 


n proof, . -i'i 


0-916 


Chloroform, 


1-490 


Citric acid, . --., 


1-034 


Ether, . . . . 


0-716 


Fluoric acid, . 


1-060 


Hydrochloric acid, . 


1-200 


Milk, . . , l 


1-032 


Nitric acid, 


1-420 


Oil, Linseed, . k .. 


0-940 


it Olive, 


0-915 


Whale, . . . . 


0-923 


Petroleum, crude, . 


0-885 


n refined, -' 


0-910 


Sulphuric acid, , : . 


1-843 


Oil, Amber, . \ . 


0-868 


n Cinnamon, 


1-043 



1-640 


Oil, Lavender, 


0-894 


1-420 


ii Turpentine, 


0-864 


0-940 


it Sweet almonds, 


0-932 


1-016 


n Codfish, . 


0-923 


1-810 


ii Hempseed, . i 


0-926 


1-850 


Porter, brown stout, 


1-011 




Proof spirit, 


0-922 




Strong ale, . . ' i 


1-050 


1-000 


Wine, Port, . " : * 


0-997 


1-027 


it Champagne, j 


0-997 


1-060 


Brandy, French, . . 


0-941 


0-792 


Rectified spirit, 


0-838 


0-800 






0-916 


GASES 




1-490 


Atmospheric air, . . 


1-000 


1-034 


Ammoniacal gas, 


0-590 


0-716 


Carbonic acid gas, . * 


1-527 


1-060 


it oxide gas, 


0-972 


1-200 


Carburetted hydrogen 




1-032 


gas, .... 


0-972 


1-420 


Chlorine gas, . . i 


2-500 


0-940 


Cyanogen gas, 


1-805 


0-915 


Hydriodic Acid gas, 


4-340 


0-923 


Hydrogen gas, , 


0-069 


0-885 


Iodine, vapour of, . 


8-716 


0-910 


Nitrous oxide gas, . j 


1-527 


1-843 


Oxygen gas, . 


1-111 


0-868 


Prussic acid gas, . . 


0-937 


1-043 


Steam of water at 212, . 


0-623 



505. The weight of a cubic foot of water is very nearly 
1000 ounces, or 62 J Ib. avoirdupois, and therefore, if the 
decimal point in the numbers in the preceding table for 
solids and fluids is removed three figures to the right, the 
numbers will denote very nearly the weight in ounces of a 
cubic foot of the different substances. 

The weight of a cubic foot of water at the maximum 
density, and in a vacuum, is 999-278 ounces avoirdupois, 
and that of a cubic inch is 253 grains, or "527 ounce troy, 
or -5783 ounce avoirdupois. 



256 RELATIONS OP WEIGHT AND VOLUME OF BODIES 

The weight of a cubic foot of air at C. and 30 inches 
barometric pressure is '08071 Ib. 

USEFUL MEMORANDA IN CONNECTION WITH WATER* 

1 cubic foot of water =62 -425 Ib. = '557 cwt. = '028 ton. 

1 gallon ii =10 Ib. = '16 cubic foot. 

1 cubic inch , - '03612 Ib. 

1 H foot M = 6-24 gallons = say 6J gallons. 

1 cwt. M =1'8 cubic feet = 11*2 gallons. 

1 ton =35'9 =224 gallons. 

1 cubic foot of sea water = 64'11 Ib. 

Weight of sea water = 1 '027 weight of fresh water. ., 

1 cubic inch of ice at 32 F. = '0334 Ib. 

1 foot =57'8 Ib. 

lib. H H = 29 '94 cubic inches. 

Snow, 1 cubic inch = -003 Ib. 

n 1 foot = 5 -2 Ib. 

1 Ib. =332-3 cubic inches = -1923 cubic foot. 

Snowfall '433 Ib. per inch depth per superficial foot. 
Inches of rainfall x 2323200 = cubic feet per square mile, 
n n x 14J = millions of gallons n n 

506. Problem I. To find the specific gravity of a body. 

CASE 1. When the body is heavier than water. 

RULE. Find the weight of the body in air and also in water; 
then the difference of these weights is to the former weight as the 
specific gravity of water to that of the body. 

When water is the standard, its specific gravity is 1, and the 
specific gravity of the body is the quotient, obtained by dividing 
the whole weight by the difference of the weights. 

Let w, w' be the weights of the body in air and water, and s, s' 
the specific gravities of the body and of water ; 
then w - w' : w = s' : s, or w - w' : w = 1 : s, when s' = 1, 
nr ws/ u i i w 

s= - or when s =1, s= , 



EXAMPLE. A piece of silver weighs 33 '6 ounces in air, and 
29-56 ounces in water ; what is its specific gravity? 

* For ordinary calculations the weight Of a cubic foot of fresh water is assumed 
=to 62-5 Ib. or 1000 ounces. 



RELATIONS OP WEIGHT AND VOLUME OF BODIES 257 

ws' 33 -6s' 33 '6s' . 

* = ;^7 = 33-6-29-56 = -^04 = 11 ' 5 *- 
Or, when *' = !, *=11'05. 

507. The rule is founded on the hydrostatic principle that a 
body immersed in a fluid lighter than itself loses as much of its 
weight as that of an equal volume of the fluid. Hence, if w" 
=the weight of a portion of water equal in volume to that of 
the immersed body, then w" = w-w'', and hence w" : w =s':s, 
or w-w' : w=s' : s. 

The weight of a portion of air, equal in volume to that of the 
body, is here disregarded. 

EXERCISES 

1. A piece of limestone weighs in air 20 lb., and in water 
134 lb. ; what is its specific gravity ? .... =3'077. 

2. A piece of steel was found to weigh 78'5 lb. in air, and 
68 -5 lb. in watef ; what was its specific gravity? . . =7 '85. 

3. A bar of lead weighed 30 cwt. in air, and only 27 cwt. 1 quarter 
11 lb. 5 ounces in water ; required its specific gravity. . =11 '325. 

CASE 2. When the body is lighter than water. 
RULE. Find the weight of the body in air, and the weight in 
water of another body, which, when attached to the former, will 
make it sink ; find also the weight in water of the compound mass ; 
from the sum of the two former weights subtract the latter; then 

The remainder is to the weight of the given body as the specific 
gravity of water to that of the given body. 
Let w denote the weight of the given body, 
w' the weight in water of the attached body, 
W' ti M ii compound mass ; 

then, s and s' remaining as formerly, 

w + w' W' : w = s' : s, or = 1 : s, if s' = 1 , 

and i 



- --Vf 1 " A - / ITT 

-W w + vf - W 



EXAMPLE. A piece of ash weighs 60 lb. in air, and to it is affixed 
a piece of copper which weighs in water 40 lb., and the compound 
weighs also in water 25 lb. ; what is the specific gravity of the ash ? 

wsf 60s' 60s' , ., , . 

.=-^ = '8, if *' = !. 



w + w'-W 60 + 40-25 75 
Since the attached body is weighed in water both times, its 
weight remains the same ; hence the weight lost in water by the 
lighter body = (w + w' - W) ; therefore, by the last case, 
W + w'- W : w=s' : s=l : s, if ^ l, 



258 RELATIONS OF WEIGHT AND VOLUME OF BODIES 

EXERCISES 

1. If a piece of elm weighs 30 Ib. in air, and a piece of copper, 
which weighs 32 Ib. in water, be affixed to it, and the com- 
pound weigh 12 Ib. in water, what is the specific gravity of 
the elm? . \. ='6. 

2. If a piece of cork weighs 25 Ib. in air, and a piece of lead, 
weighing 91'17 Ib. in water, be attached to it, and the compound 
mass weigh 12 Ib. in water, what is the specific gravity of the 
cork? =-24. 

3. If a piece of beech weigh 42'6 Ib. in air, and a piece of iron, 
weighing 40*7 Ib. in water, be attached to it, and the compound 
mass weigh 33'3 Ib. in water, what is the specific gravity of the 
beech? '-.;. : . . . . =-852. 

508. Problem II. To find the weight of a body when its 
cubic content and specific gravity are given. 

RULE. Multiply the number of cubic feet in the volume by the 
specific gravity of the body, and this product by 1000, and the 
result is the weight in ounces ; or, 

w = lOOOsw ounces 62%sv Ib. 

For, by Art. 505, 1000 times the specific gravity is the weight 
of a cubic foot of the body in avoirdupois ounces ; hence the rule is 
obvious. 

EXAMPLE. Find the weight of a bar of cast iron, its breadth 
and thickness being =4 and 2 inches, and length = 8 feet. 

V=& = x-j B T x 8 = f cubic foot ; 
hence w=62'5 x 7'248 x =453 x f = 251 Ib. 

EXERCISES 

1. Find the weight of a block of marble of the specific gravity 
of 2'7, its length, breadth, and thickness being respectively = 6 feet, 

5 feet, and 18 inches. . =3 tons 7 cwt. 3 qr. 5 Ib. 12 ounces. 

2. One of the stones in the walls of Baalbec was a block of 
marble 63 feet long, its breadth and thickness being each 12 
feet ; what is its weight, the specific gravity being 2'7? 

= 683 tons 8 cwt. 3 qr. 

3. Find the weight of a log of oak 24 feet long, 3 broad, and 1 
foot thick, its specific gravity being -925. =37 cwt. 18 Ib. 8 ounces. 

4. How many male fir-planks 16 feet long, 9 inches broad, and 

6 inches thick will a ship 400 tons burden carry ? . . =4344/ 7 , 



RELATIONS OF WEIGHT AND VOLUME OF BODIES 259 

509. Problem III. To find the cubic content of a body 
when its weight is given. 

RULE. Divide the weight of the body in ounces by 1000 times 
its specific gravity, and the quotient is the content in feet ; or, 
Divide twice the weight in pounds by 125s. 
By last problem, w = lOOOsv ounces =&2\sv Ib. ; 

iv w 2w 
* = 1000* = 62in257 

The first value must be used when w is given in ounces, and the 
last when w is given in pounds. 

EXAMPLE. Find the content of an irregular block of sandstone 
weighing 1 cwt., its specific gravity being 2 - 52. 

w 112 x 16 , . , . , . . , 

v=- - = '7 cubic feet = 1228-8 cubic inches. 

lUl/US liO U 

EXERCISES 

1. How many cubic feet are in a ton-weight of male fir ? 

= 65-1636. 

2. How many cubic feet are contained in a block of sandstone 
weighing 8 tons, its specific gravity being 2'52 ? . . = 113 feet. 

3. Find the number of cubic feet contained in a ton of dry oak of 
the specific gravity -925. . . ,,.'.,'', = 38'746. 

510. Problem IV. To find the quantity of either of the 
ingredients in a compound consisting of two, when the 
specific gravities of the compound and of the ingredients 
are given. 

RULE. Multiply the weight of the mass by the specific gravity 
of the body whose quantity is to be found, and by the difference 
between the specific gravity of the mass and the other body ; 
divide this product by the difference of the specific gravities of the 
bodies, multiplied into the specific gravity of the compound mass ; 
and the quotient will be the quantity of that body. 

Let W, w, w' denote the weights of the compound and of 
the ingredients ; and S, s, s', their specific gravities respectively, 
s being that of the denser ingredient ; 

(S-s')s,, r , , (s-S)s' 

then w=] jTrrW, andw/ = ; ~ W. 

(s-s)S (s-s)S 

EXAMPLE. A composition weighing 56 Ib., having a specific 
gravity 8-784, consists of tin and copper of the specific gravities 



260 RELATIONS OF WEIGHT AND VOLUME OF BODIES 

7 '32 and 9 respectively; what are the quantities of the 
ingredients ? 

(S-*X, 7 1-464x9x56 13'176 _ _. 



4 76712 
and hence w'=W-w=5Q-50 Q. 

Or there are 56 Ib. of copper and 6 of tin. 
By Art. 505, the volume of the body whose weight is w in ounces 

; and the same applies to the bodies whose weights are 



w' and W, and hence multiplying by 1000, 

w >' W , , ,,,. 

-\ r=-cr> a l so W+W=W. 
s s ;! 

From these two equations are easily found the two formulae given 
above ; and hence the origin of the rule. 

Note. This rule in many cases of alloys gives only approximate 
results ; for experiment shows that in these cases the density is in 
some instances greater, and in other instances less, than what would 
result from a simple mixture of the ingredients. This indicates 
something of the nature of chemical action. 

EXERCISES 

1. An alloy of the specific gravity 7 '8 weighs 10 Ib., and is 
composed of copper and zinc of the specific gravities 9 and 7 '2; 
what is the weight of the ingredients ? 

= 3 '846 Ib. of copper, and 6-154 Ib. of zinc. 

2. An alloy of the specific gravity 7 '7, consisting of copper and 
tin of the specific gravities 9 and 7 '3, weighs 25 ounces ; what is 
the weight of each of the ingredients ? 

=6 '875 ounces of copper, and 18 '125 of tin. 

3. A circular piece of gold and a common cork have equal 
weights and diameters, and the cork is If inches long. How thick 
is the piece of gold, the specific gravity of the gold being 19 '25, 
and that of the cork -25 ? ...... = -fa inch. 

4. Given that the specific gravity of petroleum is 0'88, and that 
a quart of water weighs 40 ounces ; find how many gallons of 
petroleum will weigh 38 Ib ...... =4f gallons. 

5. Find the weight of a piece of oak 7 feet high, 3 feet wide, and 
1J inches thick, taking the specific gravity of oak as '93. 

= 152 -578 Ib. 

6. If the specific gravity of brass be taken as 8 '4, find the weight 
of a bar of the same material 10 inches long and 4 square inches in 
Section, , ...... 



ARCHED ROOFS 261 

7. The specific gravity of mercury is 13'6 ; find the length of a 
column of water 1 inch in diameter which shall be equal to a 
column of mercury of the same diameter which is 30 inches in 
length =34 feet. 



AECHED ROOFS 

511. Arched roofs are either vaults, domes, saloons, or 
groins. 

Vaulted roofs consist of two similar arches springing from 
two opposite walls, and meeting in a line at the top, or else 
forming a continuous arch. 

Domes are formed by arches springing from a circular or 
polygonal base, and meeting in a point above. 

Saloons are formed by arches connecting the side-walls 
with a flat roof or ceiling in the middle. 

Groins are formed by the intersection of vaults with each 
other. 

512. Arched roofs are either circular, elliptical, or Gothic. 
In the first kind the arch is a portion of the circumference 
of a circle ; in the second it is a portion of the circumference 
of an ellipse ; and in the third kind there are two arches 
which are portions of circles having different centres, and 
which meet at an angle in a line directly over the middle of 
the breadth, or span, of the arch. 

513. By the cubic content of arched roofs is to be under- 
stood the content of the vacant space contained by its arches, 
and a horizontal plane passing through the base of the arch. 

VAULTS 

514. Problem I. To find the cubic content of a vaulted 
roof. 

RULE. Multiply the area of one end, or of a vertical section, by 
the length. 

Let Jl = the area of the end, =the length ; then V=JRl. 

The areas of the ends are to be found by means of the rules in 
the ' Mensuration of Surfaces. ' 

Pmc. R 



2j5# ARCHED ROOFS 

,, EXAMPLE. Find the volume of a semicircular vault, the span 
of which is = 20, and its length =60 feet. 

l = -7854 x 20 2 x = 157-08, 
and V=JRl = 157 '08.x 60 = 9424-8 cubic feet. 

EXERCISES 

1. Find the cubic content of an elliptic vault whose span is = 30, 

height = 12, and length = 60 feet. . . . =16964 '64 cubic feet. 

, 2. What is the cubic content of a Gothic vault, its span being 

= 24, the chord of each arch = 24, and the distance of each arch from 

the middle of its chord = 9, and the length of the vault = 30 feet ? 

= 17028-1218 cubic feet. 

515. Problem II. To find the surface of a vaulted roof. 
RULE. Multiply the length of the arch by the length of the 

vault. 

Let = the length of the arch, = that of the vault, and s the 
surface ; then s=al. 

EXAMPLE. What is the surface of a semicircular vault, the 
span of which is = 20, and length = 60? 

=irr = 3-1416x 10=31 '416, 
and s = al=3l -416x60 = 1884 -96. 

EXERCISES 

1. What is the surface of a circular vaulted roof, the span of 
which is = 60 feet, and its length = 120 feet ? = 1 1309 '76 square feet. 

2. Find the surface of a vaulted roof, its length and that of its 
arch being = 106 and 42 '4 feet. .. . . =499 '38 square yards. 

DOMES 

516. A dome with a polygonal base and circular arches, whose 
radii are equal to the apothem of the base, is called a polygonal 
spherical dome. 

517. Problem III. To find the cubic contents of a dome. 
RULE. Multiply the area of the base by two- thirds of the 

height. 

Let 6 = the base, 7t = the height ; then V = 6A. 

EXAMPLE. What is the solidity of a hexagonal spherical dome, 
a side of its base being = 20 feet? 

Here 6=x6*A = 3x20x h = Wh (Art. 267)> 

and V = $Z>A 



ARCHED ROOFS 

and A 2 = $s?; for (fig. to Art. 265) ACB is in this case an equi- 
lateral triangle, and AC = s, AF = s, and CF = h, also CP 2 
= AC 2 -AF 2 =f* 2 ; 
hence V = 40^ = 30s 2 = 30 x 20 2 = 12000 cubic feet. 

EXERCISES 

1. Find the content of a spherical dome whose circular base has 
a diameter =30 feet =7068 -6 cubic feet. 

2. What is the content of an octagonal dome, eacli side of its 
base being=40 feet, and its height =42 feet ? =216313-53 cubic feet. 

518. Problem IV. To find the surface of a dome. 

RULE. When the dome is hemispherical, its surface is twice the 
area of the base ; or, s=2 x 7854<2 2 . 

When the dome is elliptical on a circular base, multiply twice 
the area of the base by the height, and divide the product by the 
radius of the base ; the quotient will be the surface. 

In other cases, multiply double the area of the base by the height 
of the dome, and divide the product by the radius of the base for 

an approximation to the surface ; or s=-(2bh). 

T 

EXAMPLE. Find the surface of a hexagonal spherical dome, 
each side of its base being =30 feet. 
Here h=r, and s=26=2x30 2 x2'598 = 4676-4 square feet. 

EXERCISES 

1. How many square yards of painting are contained in a hemi- 
spherical dome =50 feet diameter? . . =436 '3 square yards. 

2. Find the surface of a dome with a circular base = 100 feet 
circumference, its height being =20 feet. . =2000 square feet. 

SALOONS 

519. The vacuity of a saloon is the space contained by a hori- 
zontal plane through the base of the arches, the flat ceiling, and 
the arches. 

520. Problem V. To find the vacuity of a saloon. 
RULE. Find the continued product of the height of the arc, 

its breadth or horizontal projection, the perimeter of the ceiling, 
and -7854. 

From a side of the room, or its diameter when circular, take a 
like side or diameter of the ceiling, multiply the square of the 



264 ARCHED ROOFS 

remainder by the corresponding tabular area for regular polygons, 
or by 1 when the room is rectangular, or by '7854 when circular, 
and multiply this product by of the height. 

Multiply the area of the flat ceiling by the height of the arch, 
and the sum of this product, and the two preceding, will be the 
content. 

Let h, b, and p be the height and breadth of the arc and 
perimeter of ceiling ; S, s two corresponding sides of the room 
and ceiling ; a, a' the areas of the ceiling and of corresponding 
tabular polygon (Art. 268) ; and A, B, C the three products ; 
then A = -7854bhp, B = (S - s^a'h, C = ah, and V = A + B + C. 
For a square or rectangular room take 1 for a', and for a circular 
room take '7854. 

EXAMPLE. Find the cubic content of a saloon formed by a 
circular quadrantal arc of 2 feet radius, connecting a ceiling with 
a rectangular room = 20 feet long and 16 wide. 

A = 7854%? = '7854 x 2 x 2 x 56 = 1 75 '93 
B = (S-*)VA=(20-16) a xlx 2= 21-33 
C = ah. . = 16x12x2 . =384- 

Hence V = A + B + C . . =581-26 cubic feet. 

EXERCISE 

A circular building = 40 feet diameter, and =25 feet high to the 
ceiling, is covered with a saloon, the circular quadrantal arc of 
which is = 5 feet radius ; required the cubic contents of the room. 

= 30779-46 cubic feet, 

521. Problem VI. To find the curve surface of a saloon. 

RULE. Multiply the length of the arch by the mean perimeter. 

Let / = the length of the arc, and p = t\\e mean perimeter measured 
along the middle of the arch ; then s=pl. 

EXAMPLE. The breadth of the curve surface of a saloon is = 10 
feet, and the mean perimeter = 150 feet ; what is its curve surface? 
s=pl = 150 x 10 = 1500 square feet. 

EXERCISE 

Find the curve surface of a saloon, whose breadth is = 8 feet, 
and mean perimeter = 164 feet. . . . =1394 square feet. 

GROINS 

522. Problem VII. To find the cubic contents of the 
vacuity of a groin. 



ARCHED ROOFS 265 

RULE. Multiply the area of the base by the height, and this 
product by '904. 



EXAMPLE. Find the vacuity of a'square circular groin, the side 
of its base being = 24 feet, and its height = 12 feet. 

V= -9046A= -904 x 24 2 x 12=6248-4 cubic feet. 

EXERCISE 

Find the content of the vacuity of an elliptical groin with a 
square base, whose side is =20 feet, and the height of the groin 
= 6 feet ......... =2169-6 cubic feet. 

523. Problem VIII. To find the surface of a groin. 
RULE. Multiply the area of the base by 1-1416. 
This rule will give very nearly the surface for circular and 
elliptical groins of small eccentricity. 

s = 1-14166. 

EXAMPLE. Find the surface of a circular groin with a square 
base, whose side is =12 feet 

5 = 1-14166 = 1-1416 x 12 2 =164 -39 square feet 

EXERCISE 

What is the surface of a circular groin having a square base, 
whose side is = 9 feet? ..... =92 -4696 square feet. 



GAUGING 

524. Gauging is the art of measuring the dimensions and 
computing the capacity of any vessel or any portion of it. 

The vessels usually gauged are casks, tuns, stills, and ships. 
The dimensions of the three former kinds are generally taken 
in inches, as the object is to determine the numher of gallons 
of liquid contained in them. 

When the capacity of a vessel is known in cubic inches, 
the number of gallons contained in it could then be easily 
found by dividing the capacity by 277 '274, the number 
of cubic inches in an imperial gallon. The capacities of 
vessels can be found by means of the rules in the 'Men- 
suration of Solids,' but they can be found more readily by 



266 GAUGING 

means of certain numbers called divisors, multipliers, and 
gauge-points. 

PRINCIPLES AND DEFINITIONS OP TERMS 

525. For Eectilineal Figures. The number of cubic inches in 
the measure of capacity or quantity of any vessel or solid is called 
the divisor for that body. 

The number of cubic inches in the capacity being divided by the 
divisor, will give the capacity or quantity in the required denomina- 
tion. Thus, the number of cubic inches in the capacity of a vessel 
being divided by 277'274, gives the number of imperial gallons ; by 
2218 '192, gives the number of imperial bushels. So the number of 
cubic inches contained in a quantity of dry starch being divided 
by 40'3, will give the number of pounds, for 40'3 is the number of 
cubic inches in a pound of starch. 

526. The reciprocals of the divisors are the multipliers. 

It is evident that if, instead of dividing by the preceding 
divisors, we multiply by their reciprocals, the results will be the 
same. These multipliers will therefore be found by dividing 1 
by the preceding divisors. 

527. The square roots of the divisors are called gauge-points. 

The gauge-points are just the sides of squares, of which the con- 
tent at one inch deep is tl*e measure of capacity or of quantity 
that is, 1 gallon, 1 bushel, or 1 pound of starch, soap, tallow, or glass. 

528. By the content of any given surface at one inch deep 
is meant the content in cubic inches of a right prism or vessel 
whose height or depth is 1 inch, and base the given surface. 

Thus the content of a circular area is the content of a cylinder 1 inch 
high, whose base is the circle ; the content of a square is the content 
of a parallelepiped 1 inch high, whose base is the given square. 
Let V = the volume of a vessel or solid in cubic inches, 
c= ii capacity of it in the required denomination, 
m= it number of cubic inches in the measure of capacity, 

as in 1 gallon, 1 pound, &c., 
n= ti multiplier, 
g= ii gauge-point ; 

then c = = nV, forn = , 

m m 

also g*x 1 =i., and g = ^m; 

V V 

hence also c = = ,. 

m g* 



GAUGING 267 

529. For Circular Areas. If the number of cubic inches in the 
measure of capacity or quantity is divided by the number '785398 
or '7854, the quotients are the circular divisors. 

Let m 1 = this divisor, and dthe diameter of the area ; 

V d z m 

then V= '785398eP ; and hence, c = = , for ii= 

' 



, i - _ 0ft0 . 

m OT! '785398 

530. If the number '785398 is divided by the number of cubic 
inches in the measure of capacity or quantity, the quotients are 
the circular multipliers. 

It is evident that the multiplier % is the reciprocal of w^ ; hence 
c=nid 2 . 

531. The square roots of the circular divisors are the circular 
gauge-points. 

The gauge-points are the diameters of circles, of which the 
content at 1 inch deep is the number of cubic inches in the measure 
of capacity or quantity. 

Since c= , when e=l, = 1, therefore d? =m lt 
mj ' h 

or d=\/m 1 =g 1 . 

532. Polygonal Areas. If the number of cubic inches in the 
measure of capacity is divided by the tabular areas of polygons 
(Art. 268), the quotients are the polygonal divisors. 

Thus, if a =the area of any regular polygon, and s its side, 

!= M it a similar regular polygon whose side is 1, 
m%= ii polygonal divisor, 

m a fPa-, s 2 

then m= , a=s 2 a 1 , c= - L = . 





m 

533. The reciprocals of the divisors are the multipliers. 

If n 2 =the multiplier, then n^= , and hence c=n 2 s 2 . 

534. The square roots of the divisors are the gauge-points. 

The gauge-points are the sides of regular polygons whose areas 
are equal to the number of cubic inches in the measure of capacity. 

For if <7 2 =the gauge-point, then g 2 = \Jm. 2 , or g.? = m= 

j 

Hence m=g ! ?a 1 , or g. 2 is the side of the polygon, whose content 
is in. 

535. Spherical Areas. If the circular divisors are increased in 
the ratio of 2 to 3, the results are the spherical divisors ; the 
spherical multipliers are the reciprocals of the divisors ; and the 
spherical gauge-points are the square roots of Jhe divisors, ____ | 



268 



GAUGING 



c 
By Art. 531, c= h, if 7t = the height of the cylinder. Now, if 

rf=the diameter of a sphere, and m 3 the divisor, 

5236^ 2 -7854^ 2d 3 3 
c= = 5 . - v = w 3 -r^M 1 = a 3 -rn 3 . 
m 3 m Swij 2 x 

And if 3 is the reciprocal of m 3 , c = n 3 d?- 

Also the gauge-point g 3 = \Jm 3 is the diameter of a sphere whose 
volume is=mg 3 . 

o o 

For g s *=m t =m 1 = 7, or m= -5236<7 3 2 . 



536. For Conical Vessels. The conical divisors are three 
times those for cylinders, the multipliers are their reciprocals, 
and the gauge-points are the square roots of the divisors. 

The reason why the divisors are three times as great as those for 
cylinders is, that the volume of a cylinder is three times that of a 
cone of the same base and height. It can also be proved, as is 
similarly done in the preceding articles, that the gauge-point is 
the diameter of a cone which at one inch of height is equal to the 
measure of capacity. 

537. For Prismoidal Vessels. If the divisors for rectilineal 
and cylindric figures are multiplied by 6, the products will be 
prismoidal divisors; their reciprocals, the prismoidal multi- 
pliers ; and the square roots of the prismoidal divisors, the 
prismoidal gauge-points. 

TABLES OF MULTIPLIERS, DIVISORS, AND 
GAUGE-POINTS 

I. FOR PRISMATIC VESSELS WITH SQUARE BASES. 



Measures 


Divisors 


Multipliers 


Gauge-points 


Inches in the area of unity, 


1 


1 


1 


Superficial foot, 


144 


006944 


12 


A solid foot, 


1728 


000578 


41-57 


Imperial gallon, . - .' 


277-274 


003607 


16-65 


ii bushel, 


2218-192 


000451 


47-1 


A pound of hard soap, 


27-14 


036845 


5-21 


ti ii dry starch, . 


40-30 


024813 


6-35 


n ii green glass, . 


12-18 


082102 


3-48 



GAUGING 



269 



II. FOR CYLINDRIC VESSELS 



Measures 


Divisors 


Multipliers 


Gauge-points 


Inches in the area of unity, 


1-27324 


785398 


1-128 


A superficial foot, . 


183-34 


005454 


13-54 


A solid foot, 


2200-16 


000454 


46-91 


Imperial gallon, 


353-04 


002833 


18-79 


ii bushel, 


2824-29 


000356 


53-14 


A pound of hard soap, . 


35-65 


02805 


5-97 


ii it dry starch, . 


51-3 


019491 


7-16 


H M green glass, . 


15-5 


064516 


3-94 



III. FOR REGULAR POLYGONAL PRISMATIC VESSELS 



Measures 


Divisors 


Multipliers 


Gauge-points 


PENTAGONAL BASE 








Imperial gallons, 
ii bushels, . 


161-161 
1289-288 


006205 
000776 


12-69 
35-91 


HEXAGONAL BASE 








Imperial gallons, 
it bushels, . . 


106-723 

853-782 


00937 
001171 


10-33 
29-22 


HEPTAGONAL BASE 








Imperial gallons, 
ii bushels, . 


76-302 
610-414 


016106 
001638 


8-73 
24-71 


OCTAGONAL BASE 








Imperial gallons, 
it bushels, 


57-425 
459-403 


017414 
002177 


7-58 
21-43 



IV. FOR CONICAL VESSELS. 



Measures 


Divisors 


Multipliers 


Gauge- points 


Imperial gallons, 
ii bushels, . . 


1059-109 

8472-87 


000944 
000118 


32-54 
92-049 



270 



GAUGING 
V. FOE SPHERICAL VESSELS 



Measures 


Divisors 


Multipliers 


Gauge- points 


Imperial gallons, 
it bushels, 


529-554 
4236-434 


001888 
000236 


23-01 
65-09 



VI. PRISMOIDAL VESSELS, FRUSTUMS, OR CYLINDROIDS 



Measures 


Divisors 


Multipliers 


Gauge-points 


WITH SQUARE ENDS 








Imperial gallons, . 
it bushels, . . 


1663-644 
13309-15 


000601 
000075 


40-79 
115-36 


WITH CIRCULAR ENDS 








Imperial gallons, 
ii bushels, 


2118-217 
1694574 


000472 
000059 


46-02 
130-17 



538. Problem I. To gauge regular rectilineal and circular 
areas one inch deep. 

RULE. Find the square of the side or the diameter in inches, 
and multiply or divide it by the proper multiplier or divisor for 
the regular figure. 

a s* & 

c na, c = =ns*. or c= =** 
m m m t 

EXAMPLE. Find the content of a square cistern whose side is 
= 108 inches in imperial gallons. 



~ 



Or, 



__ 
' ~m~ 277-274 
c=nsP= -003607 x 108 2 =42'072. 



EXERCISES 

1. If the side of a square is =49 inches, what is its content in 
imperial gallons ? =8 -66. 

2. What is the content of a regular octagon whose side is = 150 
inches in imperial gallons ? =391*8. 

3. Find the content of a circular tun in imperial gallons, its 
diameter being = 72 inches. . ... . , . =14'684, 



GAUGING 271 

539. Problem II. To gauge areas one inch deep. 

RULE. Find the superficial content, and divide or multiply it 
by the proper divisor or multiplier, for the required denomination. 
EXAMPLE. Find the area of a rectangular cistern in imperial 
bushels, its length and breadth being = 72 and 42 inches. 
_V 72x42 _ 
-m-2218-192- 

EXERCISES 

1. Find the content in pounds of hard soap of an oblong vessel, 
its length being = 201, and its breadth = 60 inches. . . =444 -36. 

2. Find the area of a triangular vessel in imperial gallons, its base 
being = 100 inches, and the perpendicular on it =80 inches. =14-426. 

3. Required the content of a parallelogram in pounds of hard 
soap, the length being = 84 inches, and the perpendicular breadth 
= 32 inches. . = 99-04. 

4. What is the area in imperial bushels of a trapezoid, the parallel 
sides being=60 and 145, and the perpendicular breadth = 80 inches? 

= 3-698. 

5. Find the area of a quadrilateral in pounds of dry starch, one 
of its diagonals being = 80, and the perpendiculars on it from the 
opposite angles being =24 '6 and 14'4. . . . . =38 '7. 

6. What is the area in imperial gallons of an oval figure whose 
transverse diameter is = 85 inches, and six equidistant ordinates, 
whose common distance is = 15 inches, being in order 40*6, 44 '3, 
50'4, 50'1, 42'7, and 38'2, and two segments at each end, whose 
bases are the extreme ordinates, and heights = 5 inches, and nearly 
of a parabolic form ? . ..... . . . . =13 '22. 

540. Problem III. To find the area of an ellipse when 
its two axes are given. 

RULE. Find its area by the rule in Art. 428, and divide or 
multiply it by the proper divisor or factor, and the result will be 
the required area ; or, 

Find the product of the axes, and multiply or divide it by the 
circular factor or divisor, and the result is the area. 

EXERCISES 

1. Find the area, of an ellipse in imperial gallons, its axes being 
=99 and 75 =21 '03. 

2. Find the content of an ellipse iu imperial gallons, its axes 
being =108 and 75. . . . '..-.. .... =22-947. 



272 GAUGING 

541. Problem IV. To gauge solids whose bases are 
regular figures. 

RULE. Find the cubic content ; then multiply or divide it by 
the proper multiplier or divisor corresponding to the required 
measure or weight ; or, 

Multiply the square of the given side or diameter by the depth, 
and divide or multiply the product by the proper tabular divisor or 
multiplier for the given figure of the base (Art. 537). 

V 

If V=the volume, c = , or c=nV. 
m 

EXAMPLE. Find the content of an octagonal prism in imperial 
bushels, its side being = 60 inches, and depth = 75. 
By Art. 369, V = 4 -8284 x 60 2 x 75 = 1303668, 

V 1303668 . 

and c= = HST ^ = 587-7 bushels; 

m 2218'19 

or c = n.^h = '0021 77 x 60 2 x 75 = 587 '79 bushels. 

EXERCISES 

1. Find the content in imperial gallons and bushels of a vessel 
with a square bottom, each side being = 30 inches, and its depth 
= 40. =129 '83 and 16-236. 

2. What is the content in imperial bushels of a cylindric vessel 
whose diameter is =48 inches, and depth = 64 inches? 

=52-2 bushels. 

3. Find the content in imperial bushels of a regular pentagonal 
prismatic vessel, a side of its base being =54 inches, and its depth 
= 80 inches = 180'94. 

4. Find the content in imperial gallons of a conical vessel, the 
diameter of its base being = 27 inches, and its height =60 inches. 

=41-3. 

5. What is the content in imperial bushels of a pyramidal vessel 
whose base is a regular hexagon, the length of its side being =40 
inches, and the height of the vessel = 72 inches? . . =44-96. 

6. Find the content of a conical vessel in imperial gallons, the 
diameter of its base being = 60 inches, and its height = 60 inches. 

= 203-9. 

7. What is the content in imperial bushels of a pyramidal vessel, 
whose base is a regular octagon, its side being = 105 inches, and its 
depth = 120 inches? =960-5. 

542. Problem V. To find the content of a spherical vessel. 
RULE. Find the volume of the sphere, and multiply or divide 



GAUGING 273 

it by the proper multiplier or divisor for the required measure or 
weight ; or, 

Divide or multiply the cube of the diameter by the corresponding 
tabular divisor or multiplier (Art. 537). 

EXERCISES 

1. Find the content of a spherical vessel whose diameter is = 34 
inches in imperial gallons =74 '2. 

2. Find the content in imperial bushels of a spherical vessel 
whose diameter is = 68 inches. . ". . . . . =74 '2. 

543. Problem VI. To find the content of a spheroid. 
RULE. Find its volume, and multiply or divide it by the proper 

multiplier or divisor for the required measure or weight ; or, 
Multiply the square of the equatorial diameter by the polar 

diameter, and divide or multiply the product by the divisor or 

multiplier for spherical vessels. (See Art. 447.) 
For, if b is the equatorial diameter and a the polar diameter of 

a spheroid, and the diameter of a sphere ; v the volume of the 

spheroid, and v' that of the sphere ; 

then (Art. 447), v = -52366 2 , and v' = -5236a 3 ; 

and hence v : v' = 6 2 : a 2 , 

from which the rules are evident. 

EXERCISES 

1. Find the content in imperial gallons of a prolate spheroid, its 
polar diameter being = 72 inches, and its equatorial =50. =339'9. 

2. What is the content in imperial bushels of a prolate spheroid 
whose diameters are = 70 and 90? . ', . . . =104-1. 

544. Problem VII. To find the content of a frustum of a 
cone or pyramid, or of a prismoid or cylindroid. 

RULE. To the areas of the ends add four times the area of the 
middle section ; multiply the sum by one-sixth of the height, and 
the product is the volume. Divide or multiply the volume by the 
proper divisor or multiplier for the given denomination, and the 
result is the content. (See Art. 389.) 

V=Vi(B + 6 + 4M), and c = -, orc = nV. 
m 

For regular figures, to the squares of a side of each end add four 
times the square of the side of the middle section, multiply the sum 
by one-sixth of the height, and this product by the multiplier for 
the corresponding prismoid al vessels ; 



274 ^ GAUGING 

in which E = a side of the greater end, e=a side of the less end, 
e' = {E + e}, and n = the prismatic multiplier for the form of the 
base, or the cylindric multiplier if the frustum be that of a cone. 

EXAMPLES. 1. Find the content in imperial gallons of a vessel, 
which is a frustum of a square pyramid, the sides of its ends being 
= 78 and 42 inches, and its depth = 60 inches. 
V= 10(78 2 + 42 12 + 120 2 ) = 222480, 

V 222480 OAO . . . , 
and c= = = 802-4 imperial gallons. 

/ ' ' Zi t f L / 4 

2. What is the content in imperial gallons of a frustum of a 
regular hexagonal pyramid, the sides of its ends being = 72 and 
48 inches, and its deptli = 72 inches ? 
It is found that V=682400'28 cubic inches, 

y 

and c = = 2461'! imperial gallons. 

m 



Here n = '00937, and V = JA(E 2 + e 2 + 4e')A', 

or c =1 x 72(722 + 48 2 + 4 x 60 2 ) x -00937 

= 12 x 21888 x -00937 = 2461-09 imperial gallons. 

EXERCISES 

1. What is the content in imperial gallons of a frustum of a 
square pyramid, the sides of its ends being=52 and 28, and its 
deptli = 36 inches? ....... . =213'996. 

2. What is the content in imperial gallons of a frustum of a 
regular hexagonal pyramid, the sides of its ends being = 54 and 
36, and its depth = 48 inches? ..... =922-9. 

3. Find the content in imperial gallons of a frustum of a 
rectangular pyramid, the sides of its greater end being = 36 and 
16, those of its smaller end = 27 and 12, and its depth = 80 inches. 

= 128-12. 

4. What is the content of a conic frustum in imperial gallons, 
the diameters of its ends being=44 and 16, and its depth = 40 
inches? .......... =109-39. 

5. W r hat is the content in imperial gallons of a vessel of the 
form of an elliptic cone, the diameters of one end being =48 and 42, 
and those of the other =40 and 34 inches, the corresponding axes of 
the ends being parallel, and the depth = 30 inches ? . = 142-55. 

The contents of other solids can be found by determining their 
volumes by the usual rules, and then dividing by the proper number 
for the required measure of quantity. The contents of many solids 
of rather irregular figures can be calculated by means of the first 



GAUGING 



275 



and second rules of Art. 485, which may also be used for regular 
figures, as portions of conoids and spheres. 

545. Problem VIII. To gauge mash-tuns, stills, and other 
brewing and distilling vessels. 

Divide the vessel into small portions by means of planes parallel 
to its base j fmd the areas of the middle sections of these portions, 
and multiply these areas by the corresponding depths of the por- 
tions to which they belong : the products are the volumes of the 
portions, and the sum of these volumes is the whole volume ; 
divide the whole volume by the number corresponding to the re- 
quired measure or weight, and the result is the required content. 

The vessel is divided into portions of 6 or of 10 inches deep, 
according as the sides are more or less inclined ; and so that the 
difference of the corresponding diameters of two successive middle 
sections may not differ by more than 1 inch. 

When the vessel is nearly circular, cross that is, perpendicular 
diameters are taken at the middle of any portion, and the mean 
of them is considered to be the diameter of a cylinder of the same 
depth as that portion, whose volume is nearly equal to it. In 
this case the volumes of the different portions are calculated as 
cylinders. 

EXAMPLE. Find the content, in imperial gallons, of an under- 
back, the form of which is nearly the frustum of a cone, from 
the following dimensions, the cross diameters being measured at 
the middle of the several portions into which the vessel is divided, 
their depths being those in the first column : 



Depth 
of Portions 


Depth of 
Middle Sections 


Cross Diameters 


Mean Diameters 


8 


4 


70 


68-8 


69-4 


10 


13 


72 


72-2 


72-1 


10 


23 


73-6 


73-5 


73-5 


10 


33 


74 


73-8 


73-9 


L 











The whole depth is 38 inches. The area to 1 inch deep of the 
middle section of the first portion is found thus : 
a = <P -r m = 69'4 2 -f 353 "04 = 1 3'64. 

In the same manner, the areas to 1 inch deep for the other middle 
sections is found to be, in order, 14 '73, 15 - 3, 15'47 ; and each of 



276 GAUGING 

these being multiplied by the depths, and the sum of the results 

taken, it will be the content as under : 

For the 1st portion, content = 13'64x 8 = 109*12 
n 2nd ,. =14-73x10=147-3 

ii 3rd =15-3 x 10 =153-0 

4th n =15-47x10=154-7 

Content of vessel in imperial gallons = 564 -12 

It is usual to construct a table containing the contents of a 
fixed vessel, as of a mash-tun or still, for every inch in depth. 
The contents for the first inch of depth from the bottom of the 
preceding vessel is 15-47; for two inches, it is the double of this, 
or 30'94 ; for three inches, it is three times this, or 46 '41 ; and so 
on. For the area of each of the first ten inches, it is 15-47 ; for 
each of the next ten inches, it is 15 '3 ; and a table is thus easily 
constructed. 

EXERCISES 

1. Find the content in imperial gallons of a flat-bottomed copper, 
the mean diameters at the middle of four portions into which it is 
divided by horizontal sections being as under : 

Depth of Portions Mean Diameters of their 

in Inches Middle Sections 

12 54-4 

10 51-9 

10 49-6 

10 47-3 

Whole depth = 42 

Content in imperial gallons = 310. 

The bottoms of coppers are seldom flat; they are generally 
rising or falling that is, convex or concave internally. The 
content of a vessel with a rising or falling crown, as the bottom 
is in this case called, is found by calculating, as in the preceding 
example, the content above the centre of the crown when it is 
rising, and then adding the content of the space contained between 
the bottom and a horizontal plane touching its crown, from which 
the depth of the vessel is taken. The content of this portion is 
most easily found by measuring the quantity of water required to 
fill it, till the bottom is covered. A similar method is adopted for 
a falling crown. 

2. Find the content of a still, from the dimensions below, the 
uppermost portion being considered a frustum of a sphere, the 



GAUGING 27? 

Cross diameters at its two ends being given, and also those at the 
middle of other four portions, the quantity of water required to 
cover its rising crown being 35 gallons : 

Depth of Portions Cross Diameters Content in 

in Inches " N Imperial Gallons 

f27 27 \ 

\55-2 54-8/ 

9 59-8 60-2 91-77 

9 63-8 64-4 10474 

9 64 64-6 105-4 

10-5 62 62-4 115-07 

45 -5 = whole depth. = 495 -52. 

546. When the sides of the vessel are sloping and straight, 
though the vessel be circular or oval, if two corresponding 
diameters at the top and bottom are measured, those at any 
intermediate depth are easily found. Thus, if e denotes the 
excess of the top diameter above that at the bottom, and if h 
is the depth of the vessel, and h' the depth of any other place, 
reckoning from the bottom, and e' the excess of the diameter 
there above the bottom diameter ; then 

h : h' = e : e', and e' -rh' ; 

fl 

e' being thus found, if it is added to the bottom diameter, the 
result is the diameter at the given depth. If h' = 10 inches, then 
e' is the difference of diameters for every 10 inches ; and the 
diameters for every 10 inches of depth are therefore easily found. 
The cross diameters are computed in the same manner ; and then 
the content at every inch of depth can be found and registered in 
a table. 

CASK GAUGING 

547. Casks are usually divided into four varieties : The first 
variety is the middle frustum of a spheroid ; the second, the 
middle frustum of a parabolic spindle ; the third, two equal 
frustums of a paraboloid united at their bases ; and the fourth, 
two equal conic frustums united at their bases. 

The rules for calculating the contents of the middle frustums of 
circular, elliptic, and hyperbolic spindles are^too difficult for the 
purposes of practical gauging, and they are therefore omitted in 
treatises on this subject. 

When the cask is much curved, it is considered to belong to the 
first variety ; when less curved, to the second ; when still less, to 

Prac 8 



278 GAUGING 

the third ; and when it is straight from the bung to the head,* 
to the fourth variety. 

First Variety 

548. Problem IX. To find the content of a cask of the 
first or spheroidal variety. 

RULE. To twice the square of the bung diameter add the 
square of the head diameter, multiply the sum by the length of 
the cask, and divide the product by 1059'108, and the quotient 
is the content in imperial gallons, the dimensions being all taken 
in inches ; or, 

C = (2B 2 + H 2 )L -r 1059-108, 

where H, B are the head and bung diameters, and L the length of 
the cask. 

Note. This is just the rule given in Art. 449 in the first 
case ; only, instead of multiplying by -2618 or J of -7854, 
and then dividing by 277'274 for imperial gallons, the divisor 
1059 '108 is taken, which is 3 times the divisor 353'036 for circular 
areas. 

EXAMPLE. What is the content of a cask whose bung and head 
diameters are = 32 and 24, and length = 40 inches ? 

C = (2B 2 + H 2 )L-r 1059 -108 = (2 x 32*+ 24 2 ) x 40 -r 1059*108 -:, >,\ ' 
= (2048 + 576) x 40j- 1059'108 = 99-1 imperial gallons. 

EXERCISES 

1. Find the content in imperial gallons of a cask whose bung 
and head diameters are = 30 and 18, and length=40 inches. = 80'2. 

2. What is the content of a cask whose bung and hea<l dia- 
meters are = 24 and 20, and length = 30 inches? . . =43 '97. 



Second Variety 

549. Problem X. To find the content of a cask of the 
second variety. 

RULE. To twice the square of the bung diameter add the 
square of the head diameter, and from the sum subtract of the 
square of the difference of these diameters ; multiply the remainder 
by the length, and the product, divided by 1059-108, will give the 
content in imperial gallons. 

C = {2B 2 + H 2 - |(B - H) 2 }L -=-1059-108. 



GAUGING 279 

The rule in this case is the same as that in Art. 473, which is 
easily reduced to this form. 

EXAMPLE. Let the dimensions of a cask of the second variety 
be the same as those given in the example for the first variety, to 
find its content. 

C = {2B 2 + H 2 - |(B - H) 2 }L 4 1059-108 

= (2x32 2 + 24 2 -f x8 2 )x 40 -f 1059-108 

= (2624-25-6)40-r 1059-108 = 98-1 imperial gallons. 

EXERCISES 

1. Find the content of a cask whose Lung and end diameters 
are =48 and 36, and length = 60 inches. . . . . =331-24. 

2. What is the content of a cask whose bung and head dia- 
meters are = 36 and 20, and its length =40 inches? . . =109-14. 

Third Variety 

550. Problem XI. To find the content of a cask of the 
third variety. 

RULE. Add the square of the bung diameter to that of the 
head diameter, multiply the sum by the length, and divide the 
product by 706-0724 for its content. 

C = (B 2 + H 2 )L -r 706 -0724. 

The formula is the same as that in Art. 446 ; only, instead of 
multiplying by ^x '7854, and dividing by 277 '274, the equivalent 
divisor 706-0724 is used. 

EXAMPLE. What is the content in imperial gallons of a cask 
of the third variety, of the same dimensions as that in the example 
for the first variety ? 

C = (B 2 + H 2 )L -~ 706 -0724 = (32 2 + 24 2 ) x 40 -f 706 '0724 
= (1024 + 576) x 40 -f- 706 -0724 = 90 '64. 

EXERCISES 

1. Find the content of a cask whose bung and head diameters 
are = 30 and 24, and its length = 36 =75 -26. 

2. What is the content of a cask, whose bung and head dia- 
meters are=29 and 15, and its length = 24 inches? . . =36*23. 

Fourth Variety 

551. Problem XII. To find the content of a cask of the 
fourth variety. 

RULE. Add together the product of the bung and head 



280 GAUGING 

diameters, and their squares ; multiply the sum by the length, 
and divide the product by 1059-1086 for the content. 

C = (B 2 + BH + H 2 )L -f- 1059-1086. 

For this is the formula of Art. 386, except that, instead of the 
factor -2618 or Jx-7854, and the divisor 277 '274, the equivalent 
divisor 1059 '1086 is taken. 

EXAMPLE. Find the content of a cask of the fourth variety, 
whose bung and head diameters are = 32 and 24, and length 
=40 inches. 

C = (B 2 + BH + H 2 )L -f 1059 -1086 
= (1024 + 768 + 576) x 40 -f 1059 '1 1 = 89 "43. 

EXERCISES 

1. What is the content of a cask whose bung diameter is = 32 
inches, end diameter =18, and length = 38 inches? . . =69 '04. 

2. Find the content of a cask whose diameters are =40 and 20, 
and length =50 inches ........ =132'2. 

MEAN DIAMETERS OP CASKS 

552. The mean diameter of a cask is the diameter of a 
cylinder of the same length, whose capacity is equal to that 
of the cask. 

The mean diameter may be found by means of the following 
table, the construction of which is this : If the bung diameter be 
denoted by 1, and the head diameter, divided by the bung diameter, 
be denoted by H, the contents of the four varieties of casks will 
be expressed by 



multiplied by -7854L ; but if D = the mean diameters, the contents 
are also expressed by < 7854D 2 L ; hence, as each of the above 
expressions x -7854L is equal to the last, therefore these expres- 
sions themselves are = D 2 for the four varieties, or D for these 
varieties is equal to the square root of each of them. For example, 
when H= -5, then, for the first variety, D = Vi(2 + H 2 ) = W3= '866 ; 
which is the number under the first variety in the following Table 
opposite to -5 or H. In the same manner, all the other numbers 
in the Table are found, for H = -51, -52, -53,... up to 1. The 
numbers marked H are just the ratio of the bung to the head 
diameter, and the numbers under the different varieties are the 
mean diameters when the bung diameter is = l, and the ratio H is 
its head diameter. 



GAUGING 



281 



TABLE OF MEAN DIAMETEKS WHEN THE BUNG 
DIAMETER IS = 1 



H 


First 
Variety 


Second 
Variety 


Third 
Variety 


Fourth 
Variety 


H 


First 
Variety 


Second 
Variety 


Third 
Variety 


Fourth 
Variety 


50 


8660 


8465 


7905 


7637 


76 


9270 


9227 


8881 


8827 


51 


8680 


8493 


7937 


7681 


77 


9296 


9258 


8944 


8874 


52 


8700 


8520 


7970 


7725 


78 


9324 


9290 


8967 


8922 


53 


8720 


8548 


8003 


7769 


79 


9352 


9320 


9011 


8970 


54 


8740 


8576 


8036 


7813 


80 


9380 


9352 


9055 


9018 


55 


8760 


8605 


8070 


7858 


81 


9409 


9383 


9100 


9066 


56 


8781 


8633 


8104 


7902 


82 


9438 


9415 


9144 


9114 


57 


8802 


8662 


8140 


7947 


83 


9467 


9446 


9189 


9163 


58 


8824 


8690 


8174 


7992 


84 


9496 


9478 


9234 


9211 


59 


8846 


8720 


8210 


8037 


85 


9526 


9510 


9280 


9260 


60 


8869 


8748 


8246 


8082 


86 


9556 


9542 


9326 


9308 


61 


8892 


8777 


8282 


8128 


87 


9586 


9574 


9372 


9357 


62 


8915 


8806 


8320 


8173 


88 


9616 


9606 


9419 


9406 


63 


8938 


8835 


8357 


8220 


89 


9647 


9638 


9466 


9455 


64 


8962 


8865 


8395 


8265 


90 


9678 


9671 


9513 


9504 


65 


8986 


8894 


8433 


8311 


91 


9710 


9703 


9560 


9553 


66 


9010 


8924 


8472 


8357 


92 


9740 


9736 


9608 


9602 


67 


9034 


8954 


8511 


8404 


93 


9772 


9768 


9656 


9652 


68 


9060 


8983 


8551 


8450 


94 


9804 


9801 


9704 


9701 


69 


9084 


9013 


8590 


8497 


95 


9836 


9834 


9753 


9751 


70 


9110 


9044 


8631 


8544 


96 


9868 


9867 


9802 


9800 


71 


9136 


9074 


8672 


8590 


97 


9901 


9900 


9851 


9850 


72 


9162 


9104 


8713 


8637 


98 


9933 


9933 


9900 


9900 


73 


9188 


9135 


8754 


8685 


99 


9966 


9966 


9950 


9950 


74 


9215 


9166 


8796 


8732 


TOO 


1-0000 


1-0000 


1-0000 


1-0000 


75 


9242 


9196 


8838 


8780 













553. Problem XIII. To find the capacity of a cask 
of any of the four varieties by means of their mean 
diameters, found by the Table. 

RULE. Divide the head by the bung diameter, and find the 
quotient in the column marked H in the Table, and opposite to it 
and under the proper variety is the mean diameter of a similar 
cask, whose bung diameter is 1. 

Multiply this tabular mean diameter by the given bung diameter, 
and the product is the required mean diameter, the square of 



282 GAUGING 

which, multiplied by the length, and the product, divided by 
353-036, or multiplied by -0028325, is the content in imperial 
gallons; or, 

C = D 2 L-f 353-036, or C = -0028325D 2 L. 
For, if D=the mean diameter, the content is 



Instead of this divisor, the corresponding multiplier may be used, 
and then 

C = '0028325D 2 L. 

EXAMPLE. Find the content of a cask of the first variety, whose 
diameters are = 30 and 24, and length = 36 inches. 

H = $= -8, and opposite to -8 is -938 -D' ; 
then D =BD' = 30 x -938 = 28-14, 

and C= -0028325D 2 L= -0028325 x 28-14 2 x 36 = 807. 

EXERCISE 

What are the contents of each of four casks of the four varieties, 
their diameters being =32 and 24, and length = 40 inches? 

For the first, 99'1 ; for the second, 98'11 ; for the third, 90'62 ; 
and for the fourth, 89-44. 

CONTENTS OP CASKS WHOSE BUNG DIAMETERS AND 
LENGTHS ARE UNITY 

554. The contents of casks may be more readily computed by 
means of a Table of the contents of casks whose bung diameters 
and lengths are = l. 

Let D' have the same meaning as in the preceding problem 
that is, let it denote the numbers in the preceding Table under the 
different varieties, which are just the mean diameters of casks 
whose bung diameters are 1 and head diameters H ; also, let C' be 
the content of a cask whose mean diameter is D' and length 1, and 
which may be called the standard cask ; then 



=0^353-036, for L' = l. 

Hence, if the numbers in the preceding Table are squared, and the 
square divided by 353 '036, or multiplied by '0028325, the results 
will be the contents C' required. The following Table can thus be 
constructed from the preceding one. Thus, for example, when 
H= -75, then, by the preceding Table, D' for the second variety is 
9196, and 

C' = D' 2 -r 353-036 = '9196 2 ~ 353 "036 = '0023954, 
which is just the capacity opposite to -75 in the following Table : 



GAUGING 



283 



TABLE OF CONTENTS IN IMPERIAL GALLONS OF 
STANDARD CASKS C' 



H' 


First Variety 


Second Variety 


Third Variety 


Fourth Variety 


50 


0021244 


0020300 


0017704 


0016523 


51 


0021340 


0020433 


0017847 


0016713 


52 


0021437 


0020567 


0017993 


0016905 


53 


0021536 


0020702 


0018141 


0017098 


54 


0021637 


0020838 


0018293 


0017294 


55 


0021740 


0020975 


0018447 


0017491 


56 


0021845 


0021114 


0018604 


0017690 


57 


0021951 


0021253 


0018764 


0017891 


58 


0022060 


0021394 


0018927 


0018094 


59 


0022170 


0021536 


0019093 


0018299 


60 


0022283 


0021679 


0019261 


0018506 


61 


0022397 


0021823 


0019433 


0018715 


62 


0022513 


0021968 


0019607 


0018925 


63 


0022631 


0022114 


0019784 


0019138 


64 


0022751 


0022262 


0019964 


0019352 


65 


0022873 


0022410 


0020147 


0019568 


66 


0022997 


0022560 


0020332 


0019786 


67 


0023122 


0022711 


0020521 


0020006 


68 


0023250 


0022863 


0020712 


0020228 j 


69 


0023379 


0023016 


0020906 


0020452 


70 


0023510 


0023170 


0021103 


0020678 


71 


0023643 


0023326 


0021302 


0020905 ! 


72 


0023778 


0023482 


0021505 


. -0021135 


73 


0023915 


0023640 


0021710 


0021366 . 


74 


0024054 


0023799 


0021918 


0021599 


75 


0024195 


0023954 


0022129 


0021834 


76 


0024337 


0024120 


0022343 


0022071 


77 


0024482 


0024282 


0022560 


0022310 


78 


0024628 


0024445 


0022780 


0022551 


79 


0024777 


0024610 


0023002 


0022794 


80 


0024927 


0024776 


0023227 


0023038 


81 


0025079 


0024942 


0023455 


0023285 


82 


0025233 


0025110 


0023686 


0023533 


83 


0025388 


0025279 


0023920 


0023783 


84 


0025546 


0025449 


0024156 


0024035 



284 



GAUGING 



H' 


First Variety 


Second Variety 


Third Variety 


Fourth Variety 


85 


0025706 


0025621 


0024396 


0024289 


86 


0025867 


0025793 


0024638 


0024545 


87 


0026030 


0025967 


0024883 


0024803 


88 


0026196 


0026141 


0025131 


0025063 


89 


0026363 


0026317 


0025381 


0025324 


90 


0026532 


0026494 


0025635 


0025588 


91 


0026703 


0026672 


0025891 


0025853 


92 


0026875 


0026851 


0026150 


0026120 


93 


0027050 


0027032 


0026412 


0026389 


94 


0027227 


0027213 


0026677 


0026660 


95 


0027405 


0027396 


0026945 


0026933 


96 


0027585 


0027579 


0027215 


0027208 


97 


0027768 


0027764 


0027489 


0027484 


98 


0027952 


0027950 


0027765 


0027763 


99 


0028138 


0028137 


0028044 


0028043 


1-00 


0028326 


0028326 


0028326 


0028326 



555. Problem XIV. To find the content of a cask by 
means of the Table of Contents of Standard Casks. 

Divide the head by the bung diameter, and find the quotient in 
the column H, and opposite to it and under the proper variety is 
the content C' of the standard cask ; multiply this tabular content 
by the square of the bung diameter of the given cask, and this 
product by the length, both in inches, and the result will be the 
required content in imperial gallons. 

For, by Art. 553, C = D 2 L -f 353 -036 ; 
and D 2 = B 2 D' 2 , also (Art. 554) C' = D' 2 -=- 353 '036 ; 

hence C = C'B 2 L. 

EXAMPLE. Find the content of a cask of the first variety, 
whose diameters are = 30 and 24, and length = 36 inches. 

H = f$ = |= -8 ; and hence C' = -0024927, 
and C = C'B 2 L = -0024927 x 30 2 x 36 = 80 '7. 

The same answer as that to the example in Art. 553. 

EXERCISES 

1. What are the contents of each of four casks of the four 
varieties, their diameters being=32 and 24, and length = 40 inches? 
The contents will be the same as for the four casks in the 
exercise to the preceding problem. 



OA.UGING 285 

2. Find the contents of each of four casks of the four varieties, 
their diameters being = 31 "5 and 24 '5, and the length = 42 inches. 

Content for the first=102'6, the second = 101 '87, the third 
= 94-9, and the fourth = 93 '98. 

3. What is the content of a pipe of wine, whose length is = 50 
inches, head diameter=22'7, and bung diameter = 31*7, the cask 
being of the first variety ? =119'19. 

GENERAL METHOD FOR A CASK OP ANY FORM 

556. Problem XV. To find the content of a cask of any 
form, by one method, independently of tables. 

RULE. Add together 39 times the square of the bung diameter, 
25 times the square of the head diameter, and 26 times the product 
of the diameters ; multiply the sum by the length, and divide the 
product by 31773'25 for the content in imperial gallons. 

C = (39B 2 + 25H 2 + 26BH)L4-31773-25. 

EXAMPLE. Find the content of a cask whose diameters are 
= 32 and 24, and length = 40 inches. 

C = (39B 2 + 25H 2 + 26BH)L-f 31773-25 

= (39 x 32 s + 25 x 24 2 + 26 x 32 x 24) x 40 -r 31773 -25 

= 93-5 imperial gallons. 

EXERCISE 

Find the content of a cask whose diameters are = 36 and 48, and 
length = 60 inches. . . ...,-.,, . =315'7. 

Or the capacity in imperial gallons of any cask may be found as 
follows : 

Let D, d= inside diameters at the heads, B = inside diameter at 
the bung, and L the length, all in inches ; 
then the capacity in imperial gallons 

= -0014162L(D-d+B 2 ). 

The buoyancy in pounds equals ten times the capacity in gallons 
minus the weight of the cask itself. 

ULLAGE OF CASKS 

The ullage of a cask is the content of the part occupied by 
liquor in it when not full, or of the empty part. Only two cases 
are usually considered namely, when the cask is lying, or when 
it is standing. When the ullage of the part filled is found, that 
of the empty part can be obtained by subtracting the ullage found 
from the content of the whole cask. 



286 GAUGING 

557. Problem XVI. To find the ullage of the filled part 
of a lying cask in imperial gallons. 

RULE. Divide the number of wet inches by the bung diameter, 
and if the quotient is under '5, deduct from it J of what it wants 
of '5 ; but when the quotient exceeds '5, add of that excess to it ; 
then if the remainder in the former case, or the sum in the latter, 
be multiplied by the content of 'the whole cask, the product will be 
the ullage of the part filled. 
Let W=FK the wet inches, 

R = W-fB, 

C' = the content of the cask, 
U= ullage of EBDG, 




then U = (R+D)C, 

using - when R < '5, and + when R > *5. 

EXAMPLE. The content of a lying cask is = 98 gallons, the bung 
diameter =32, and wet inches = 10; required the ullage of the part 
filled. 

R=W-fB = ^='3125, D = -5 --3125= -1875, D=-0469; 
hence U = (R - JD)C = ( '3125 - -0469) x 98 = "2656 x 98 =26-03. 

Let L, L' = the length of the given and experimental cask used 
in constructing the lines S.S. and S.L. on the 
ganger's rule. 

C, C' = their capacities ; and hence C' = 100. 
U, U' = the capacity of a portion of the given cask when 
lying to be ullaged, and of a similar portion of 
experimental cask ; 

W, W=the wet inches for these portions. 
Then L:W = L':W, 

and log. L - log. W = log. L' - log. W. 

Hence, since the slider for the line S.L. is a logarithmic line, the 
distance from L to W on it is equal to that from L' to W; and 
when L on the slider is opposite to C' or 100 on S.L. , W on the 
slider will be opposite to the same number on S.L. that W would 
be opposite to when L' is opposite to 100 on S.L. ; that is, W 
would be opposite to U', the ullage of a similar portion of 
the experimental cask, which is therefore obtained by the above 
rule. 

Again, C' or 100 : C = U': U ; and since C', C, and U' are known, 
therefore U, their fourth proportional, can be found by means of 
the lines A, B, according to the rule in Art. 492. 



GAUGING 287 

EXERCISE 

The content of a lying cask is = 90, its bung diameter = 36, and 
the wet inches =27 ; find the ullage of the part filled. =73 -125. 

558. Problem XVII. To find the ullage of the filled part 
of a standing cask in imperial gallons. 

RULE. Divide the number of wet inches by the length of the 
cask, then if the quotient is less than '5, subtract from it ^V part 
of what it wants of '5 ; but if it is greater than '5, add to it T ^ of 
its excess above '5 ; then multiply the remainder in the former 
case, or the sum in the latter, by the content of the cask, and the 
product will be the ullage. 

Let W = GH the wet inches, 




and let C, U, and D have the same meaning as in 

last problem ; 

then U = (R+ T VD)C, 

using - when R < '5, and + when R > -5. 

This rule is proved in exactly the same manner as that of the 
preceding problem. 

EXAMPLE. The content of a standing cask is = 120 gallons, its 
length = 48, and the wet inches = 40 ; required the ullage of the part 
filled. 

R=W-rL = !=f=-83; hence D= -3, and T y)=-03. 
Hence U = (R + ^D)C = (-83 + -03) x 120= -86 x 120=104. 



EXERCISE 

The content of a cask is = 105 gallons, its length=45 inches, and 
the wet inches =25 ; what is the ullage of the part filled ? =58'9. 

MALT-GAUGING 

559. Barley to be malted is steeped in water in a cistern for not 
less than 40 hours. When sufficiently steeped, it is then removed 
to a frame called a couch-frame, where it remains without altera- 
tion for about 26 hours ; it is then reckoned a floor of malt, till it 
is ready for the kiln. 

During the steeping, the barley swells about \ of its original 
bulk, or \ of its bulk then ; after being less than 72 hours out of 
the cistern, it is considered to have increased \ of its bulk at that 
time ; and after being out a longer time, it is considered to have 
increased by \ of its bulk then ; and hence the rule in the following 
problem : 



288 GAUGING 

560. Problem XVIII. Having given the cistern, couch, 
or floor gauge of a quantity of malt, to find the net 
bushels. 

RULE. Multiply the cistern or couch bushels by - 8, and the floor 
bushels by , when it has been out of the cistern for less than 72 
hours, or by J when it has been out a longer time. 

EXERCISES 

1. The number of couch bushels of malt is =420; what are the 
net bushels ? = 336. 

2. If the number of floor bushels, which has been 30 hours out 
of the cistern, is = 524, what are the net bushels? . . = 349J. 

3. What is the number of net bushels corresponding to 636 
bushels that have been out of the floor for more than 72 hours ? 

= 318. 

During the process of malting, the malt is repeatedly gauged, 
and the duty is charged on the greatest gauge, after the legal 
deductions are made, whether that arises from the measurements 
taken in the couch, frame, or floor. The greatest gauge can be 
determined by the following problem : 

561. Problem XIX. Having given the greatest cistern 
or couch gauge, or the greatest floor-gauge, to determine 
which is the greatest or duty gauge. 

RULE. Multiply the greatest cistern or couch gauge by 1-2 if 
the floor-gauge has been taken before the malt was 72 hours out 
of the cistern, or by 1 -6 if taken after that time ; then, if this 
gauge is greater than the floor-gauge, it is to be taken for the 
duty-gauge, otherwise the floor-gauge is to be taken. 

Since the cistern and couch gauges are each to be multiplied 
by the same number 1-2 to obtain the floor-gauge, therefore, in 
this problem, the couch or cistern gauge is to be taken in prefer- 
ence, according as it is the greater. 

EXAMPLE. If the cistern and couch gauges, after being more 
than 72 hours out of the cistern, are respectively = 131 '2 and 132, 
and the floor-gauge = 205*2 bushels, which will afford the greatest 
or duty gauge ? 

The couch-gauge exceeds the cistern-gauge ; 
hence 132 x 1-6 = 211-2. 

So that the couch bushels produce 6 bushels more than the floor- 
gauge. 



GAUGING 289 

EXERCISES 

J. Whether will 120 cistern bushels or 146 floor bushels, which 
have been less than 72 hours out of the cistern, produce the 
greatest gauge ? The floor-gauge by 2 bushels. 

2. Whether will 145 couch bushels or 230 floor bushels, after the 
malt has been more than 72 hours out of the cistern, produce the 
greatest gauge ? . . . The couch -gauge by 2 bushels. 

562. Problem XX. To find the content of a cistern, 
couch, or floor of malt. 

RULE. Make the malt of as nearly a uniform depth as 
possible, then measure the length and breadth, and take a 
number of equidistant depths, the sum of which, divided by 
their number, will give the mean depth ; multiply the length, 
breadth, and depth together, and their product by '000451 for 
the content in bushels. 

If the base is not a rectangle, find its area, and multiply it by 
the depth and by the proper multiplier ; or calculate as in the 
previous rules. 

EXAMPLE. Find the quantity of malt in a rectangular floor, its 
length being = 48 inches, its breadth = 32, and depth, at six different 
places = 6'l, 5'8, 6'3, 5-9, 6'4, and 5'5. 



and c = 48x32x6x -000451 =4-156 imperial bushels. 

EXERCISES 

1. What is the content in imperial bushels of a cistern of malt 
whose length and breadth are = 160 and 108 inches, and mean depth 
=4-68 inches? ......... =36-47. 

2. What is the content of a floor of malt the length of which is 
= 280 inches, the breadth = 144 inches, and the depths at 5 places 
are = 21-6, 22-3, 22-9, 23-4, and 23-55? .... =413'69. 

3. Find the content of a regular hexagonal cistern of malt, the 
length of its side being = 269 inches, and its mean depth = 5 inches. 

=423-6. 
THE DIAGONAL ROD 

563. The diagonal gauging-rod is 4 feet long and -4 of an inch 
square. The four sides of it contain different lines ; the principal 
one of which is a line for imperial gallons for gauging casks. 

The use of the principal line is to determine the content of a cask 



290 GAUGING 

of the most common form by merely measuring with the rod the 
diagonal extending from the bung-hole to the opposite side of the 
head (BH, fig. to Art. 557) that is, to the part where the start' 
opposite to the bung-hole meets the head ; then the number on the 
rod at the bung-hole on the principal or first side is the number of 
gallons in the content of the cask. 

The principal line is constructed thus : It is found that for a cask 
of the common form, whose diagonal e?=40 inches, the content c is 
144 imperial gallons nearly, and therefore at 40 inches from the 
end of the rod is placed 144. Hence, if D, C are the diagonal 
and content of any other similar cask, then, since the contents of 
similar solids are proportional to the cubes of any two of their 
corresponding dimensions, 

d s : D 3 =c : C ; 
c 14.4. Q 

hence C^p-^WoO^^D*. 

From this formula the numbers showing the contents can easily 
be calculated. 
Thus, to find the content C for a diagonal D of 30 inches, 

C = * A*D 3 = T A? x 30 s = 60 -75. 

So that at 30 inches from the end of the rod is placed 60 '75 gallons, 
for the content of a cask whose diagonal is 30 inches. In a similar 
manner, the other numbers showing the content are calculated and 
marked on the rod. 

Another line on a different side of the rod is marked Seg. St. 
for ullaging a standing cask. Another side contains tables for 
ullaging lying casks. The remaining side contains lines for 
ullaging casks of known capacity as firkins, barrels, &c., either 
lying or standing. 

The diagonal dimension can easily be found by calculation when 
the usual dimensions are known namely, the head and bung 
diameters, and the length. For it is easily perceived, by the 
figure to Art. 557, when a line is drawn through H, parallel to 
EF, to meet BA produced in some point P, that PH = L half the 
length, and BP = |(B + H), or half the sum of the bung and head 
diameters HK, AB, and (Eucl. I. 47) 



or D 2 = JL 2 + JM 2 , if M = B + H, and D = the diagonal. 

The content obtained by using the diagonal rod will not be very 
correct unless the cask be of the most common form that is, 
intermediate between the second and third varieties. 



BAROMETRIC MEASUREMENT OF HEIGHTS 291 



BAROMETRIC MEASUREMENT OF HEIGHTS 

564. The difference of the heights of mountains, or of 
other situations, can be determined by means of the atmo- 
spheric pressure at these places ; and the absolute height of 
one of them above the level of the sea being known by the 
same or any other method, the height of the others above 
the same level is also known. 

The principle on which the method is founded is, that 
when various corrections are made for difference of tempera- 
ture and other variable elements, the differences of heights 
are proportional to the differences of the logarithms of the 
atmospheric pressures. 

THE THERMOMETER 

565. A thermometer is an instrument for measuring the tem- 
perature of bodies that is, their state with respect to sensible 
heat. 

Bodies are found to change their volume with a change of 
temperature, and the former change is adopted as a measure of 
the latter. The volumes of most bodies, for any increase or de- 
crease of temperature, undergo a corresponding expansion or con- 
traction. As the change of volume of fluids for a given change of 
temperature is greater than for solids, they are preferred 
for the construction of thermometers. But even a fluid 
expands so little for a moderate change of temperature 
that particular contrivances are resorted to to render more 
apparent the real expansion or contraction. The usual 
method is to enclose the fluid in a glass vessel, AB, 
consisting of a narrow-bored tube and a hollow bulb, B, 
formed on one of its extremities. Since the capacity 
of the bulb is many times greater than that of the 
tube, the rise or fall of the fluid in the tube, due to any 
change of volume, will be many times greater than if 
the tube had not a bulb. The fluid employed is coloured 
spirit of wine, or, more generally, mercury ; and a graduated 
scale, ED, is attached to the stem to show the expansion. 
Thus, if the upper part, C, of the mercury is opposite to 57 on 



292 BAROMETRIC MEASUREMENT OP HEIGHTS 

the scale, the temperature is said to be 57 degrees, or 57. Before 
the scale can be constructed, at least two points corresponding 
to two known temperatures must first be found. Two such 
points, called fixed points, can be determined as corresponding 
to the temperature of any fluid when freezing and boiling under 
given conditions. The freezing and boiling points of water are 
generally used. 

There are three different methods in use for graduating the scale 
of a thermometer. When the freezing-point is marked 32, and 
the boiling-point 212, the scale is called Fahrenheit's ; when 
the freezing-point is marked 0, and the boiling-point 100, it is 
called centigrade ; and when the freezing point is marked 0, and 
the boiling-point 80, it is called Reaumur's. 

Reaumur's thermometer is not in use in any English-speaking 
country. 

566. Problem I. To reduce degrees of temperature of the 
centigrade thermometer to degrees of Fahrenheit's scale; 
and conversely. 

RULE. Multiply the centigrade degrees by 9, and divide the 
product by 5 ; then add 32 to the quotient, and the sum is 
the temperature on Fahrenheit's scale. 

From the number of degrees on Fahrenheit's scale subtract 32, 
multiply the remainder by 5 ; and the product being divided by 9, 
will give the temperature in centigrade degrees. 

Let t = the temperature on Fahrenheit's scale, 

t' '= it M the centigrade scale ; 

then t =32 + f*', and *' = S(*-32). 

EXAMPLE. Find the number of degrees on Fahrenheit's scale 
corresponding to 20 on the centigrade scale. 

* = 32 + *' = 32 + |x 20 = 32 + 36=68. 

Between the freezing and boiling points, there are on the 
centigrade scale 100, and on Fahrenheit's 180; these numbers 
are proportional to 5 and 9 ; hence for corresponding tempera- 
tures t, t' there will be the proportion 

(<-32) : f = 9:5, 
from which the formulae are easily obtained. 

EXERCISES 

1. Find the number of degrees on Fahrenheit's scale correspond- 
ing to 25 on the centigrade scale =77. 



BAROMETRIC MEASUREMENT OF HEIGHTS 293 

2. Find the temperature on Fahrenheit's scale corresponding to 
14 '4 on the centigrade scale =57 - 92. 

3. Find the temperature on the centigrade scale corresponding 
to 80 on Fahrenheit's =26 '6. 

COMPARISON OF DIFFERENT LINEAL MEASURES 

567. Problem EL To reduce metres to imperial feet ; 
and conversely. 

RULE. Multiply metres by 3-2808, and the product will be the 
equivalent number of imperial feet. 

Multiply imperial feet by -3048, and the product will be the 
equivalent number of metres. 

Let F = the number of imperial feet, 

and M= n equivalent number of metres ; 

then F = 3 -2808M, and M = -3048F. 

EXAMPLE. Find the number of imperial feet in 3462 metres. 
F = 3462 x 3-2809 = 1 1358 "47. 

EXERCISES 

1. How many imperial feet and fathoms are contained in 6254 '6 
metres? =205207 feet, or 3420-1 fathoms. 

2. In 7645 metres how many imperial feet ? . = 25082-48 feet. 

OLD AND NEW DIVISIONS OF THE CIRCLE 

568. There are 100 centesimal degrees, called also grades, in a 
quadrant, 100 minutes in one of these degrees, and 100 seconds 
in a minute ; this division was used by some French authors ; 
the nonagesimal is the usual division of a quadrant into 90 
degrees. 

569. Problem III. To reduce the centesimal degrees of 
an arc to nonagesimal degrees ; and conversely. 

RULE. From the centesimal degrees subtract ^ of them, 
and the remainder is the equivalent number of nonagesimal 
degrees. 

To the nonagesimal degrees add of them, and the sum will be 
the equivalent number of centesimal degrees. 
Let d =the number of noiiagesimal degrees, 

d'= equivalent number of centesimal degrees ; 
then d = d' - fad' = &&', and d' = d + %d = ^d. 



294 



BAROMETRIC MEASUREMENT OF HEIGHTS 



The given minutes and seconds, if there are any, are to be 
reduced to the decimal of a degree before applying the rule. 

EXAMPLES. 1. Express 60 45' 24" of the centesimal division in 
degrees of the nonagesimal division. 

d=&d'= r \x 60-4524 = 54 -40716 =54 24' 25776". 
2. Convert 54 24' 25776" of the nonagesimal division into 
grades. 

25' 24". 



EXERCISES 

1. Convert 25 14' 25'4" of the centesimal division to degrees of 
the nonagesimal division ...... =22 37' 41 '83". 

2. Express 28 40' 28 % 64" of the nonagesimal division in terms of 
the centesimal division. ..... =31 86' 6 - 9". 

[In consequence of the 60 seconds and 60 minutes, the word 
sexagesimal is frequently used instead of nonagesimal.] 



THE BAROMETER 

570. The barometer is an instrument for measuring the weight 
or pressure of the atmosphere. Air is an elastic fluid, whose 
density is very sensitive to changes of pressure or of temperature, 
and is also sensibly affected by the quantity of water vapour 
present, though within the range of natural temperature this 
quantity is very small. Atmospheric air being a gravitating body, 
the pressure caused by it on any surface as, for instance, a square 
inch measures the weight of a column of air whose base is this sur- 
face, and whose height extends to the top of the atmo- 
sphere ; and it is found, by means of the barometer, 
that this pressure is, at its mean state, nearly equal 
to the weight of a column of mercury standing on 
the same base, and having a height of 30 inches. 

If HLS' represent a bent tube with parallel branches 
standing in a vertical position, and open at both ends 
at S' and T, then if mercury be poured into it till it 
stand at H in one branch, it will rise to S' in the other 
to a level with H. But let the branch TL be closed 
at the top, and let all the air be removed from the 
region HT above the mercury surface, then it will 
be found that the column of mercury MH does not 
require for its support the column in SS'. If the tubes are 
long enough, it will be found possible to retain in the closed 




BAROMETRIC MEASUREMENT OF HEIGHTS 295 

tube a column of mercury MH about 30 inches long, although the 
corresponding part SS' in the open branch is empty of mercury. 
In short, the column of mercury MH is supported by the pressure 
of the air on the surface S ; and since the weight of a column of 
mercury 1 square inch in section and 30 inches high is 14 '7 lb., it 
follows that this is the measure of the atmospheric pressure. A 
round bulb at S, with a small opening at e, is generally made on 
the end of the barometric tube in order that the surface of the 
mercury in it at S may be much greater than the surface at H. 
The surface S will consequently alter its position very little, while 
the surface H moves up or down over the range of variation 
corresponding to that of the atmospheric pressure, which is only 
between three and four inches. The atmospheric pressure changes 
continually from various causes, and therefore the length of the 
barometric column varies accordingly, its mean height being from 
29'5 to 30 inches at the sea-level according to locality. 

Since mercury is subject to a sensible variation of volume from 
change of temperature (Art. 565), the length of the barometric column 
must always be reduced to what it would be at some standard 
temperature, in order to express exactly the atmospheric pressure. 

571. Problem IV. To reduce the height of the barometer 
for a given temperature of the mercury to its height for 
any other proposed temperature. 

RULE. Multiply the height of the barometer by 10000, in- 
creased by the excess of the proposed temperature above 32 ; 
and divide the product by 10000, increased by the excess of the 
given temperature above 32, and the quotient will be the required 
height. 

Let h =the required height of barometer, 

h'= M given height of barometer, 
t = H temperature for height h, 
t'= a n ii ii h' ; 

, 10000 + (* -32),, 
* = 10000 + (f-32)*' 

EXAMPLE. If the height of the barometer is = 30 inches when 
the temperature of the mercury is = 52, what would its height 
be for the same atmospheric pressure if the temperature of the 
mercury were = 87 ? 

10000 + - 32) 10055 30 _ 30 . 1(M7 
h ~ 10000 + (*'-32f ~ 10020 x30 - 30 



296 BAROMETRIC MEASUREMENT OF HEIGHTS 

The volume of mercury varies -g^n f its volume at zero for 
every change of one centigrade degree of its temperature, or 
^*V!7 x = TTm5T; nearly for 1 Fahrenheit, the change of volume 
between the freezing and boiling points being assumed to be 
uniform for a mercurial thermometer.* 

Hence, if A 1 = the height of barometer at 32, its increase for 

(f -32) degrees is = 



,, , , (t'-32), 10000 + ('- 32). 
and hence h =h 1+ 



, 10000 + (t -32), 
Similarly, h= iww - h l ; 

h 10000 + (t - 32) 
and hence j-, = 10000 + {f _ 32) 5 

and from this expression the rule is obtained. 

When t, t' are within the limits of natural temperature, the 
more simple formula, 

' 



may be used, where k is the variation of h' for (t-f) degrees. 
The maximum error caused by using this formula will be simply 
0006 of an inch, if neither t nor t' should exceed 122, or the 
error is less than yA^ P ar ^ f an inch, or less than the errors of 
observation in noting the height of the barometer. The preceding 
example, calculated by this formula, gives h = 30 '105. 

EXEUCISES 

1. Find the height of the barometer by both formulae for the 
temperature of 85, when its height at 60 is = 30'2 inches. 

= 30 -2753 and 30 '2755. 

2. If, at the temperature of 87, the height of the barometer 
was observed to be = 29'75 inches, what would its height be at 
the temperature of 69 by both formulae? =29'6967 and 29'6964. 

RELATION OP VOLUME AND TEMPERATURE OF AIR 

572. Problem V. Given the volume of a quantity of air 
at the temperature of 32, to find its volume at any other 
temperature, the pressure being the same. 

Multiply the given volume by 9 times the excess of the given 

* Biot, Traite de Physique, vol. i. 



BAROMETRIC MEASUREMENT OP HEIGHTS 297 

temperature above 32, and divide the product by 4000, and the 
quotient will be the increase of volume. 
Let v 1 =the volume at 32, 

v = ,i it the given temperature, 

t = ti given temperature ; 
then v = w 



It could also be proved, as in the preceding problem, that, if vf 
be the volume at the temperature t' of the air whose volume is v^ 
at 32, 



-32)~3712+W 

EXAMPLE. The volume of a quantity of gas at the tempera- 
ture of 32 was = 1000 cubic inches; what was its volume when its 
temperature was raised to 52 ? 

v=v l + i-f fs (t-32)v 1 = 1000 + ^n^ 20 = 1000 + 45 
= 1045 cubic inches. 

Air, when heated from 32 to any higher temperature, ex- 
pands very nearly uniformly that is, for equal increments 
of temperature, there are equal increments of volume. For 
moderate heights in the atmosphere, the decrease of temperature 
may, for practical purposes, be assumed to be proportional to 
the increase of height. Then if /tj is the height of a column 
extending to a moderate height, when 32 is the mean tem- 
peraturethat is, the temperature at the middle point or half 
the sum of the extreme temperatures at the lower and upper 
extremities of the column its height h, when the mean tempera- 
ture has any other value t\, can be found in the same manner 
as v is found from v l ; or, 



where ^ = the mean temperature = \(t + t'), if t and t' denote the 
temperatures at the lower and upper ends of the column. 

EXERCISES 

1. If the volume of a quantity of air at the temperature of 
freezing is =2500 cubic feet, what would its volume be at the 
temperature of 87 ? ....... =2809 '375. 

2. If the height of a column of atmospheric air whose mean 
temperature is 32 is = 5000 feet, what would be its height were 
the mean temperature 57 ? ..... =5281-25. 



298 BAROMETRIC MEASUREMENT OP HEIGHTS 



MEASUREMENT OF HEIGHTS 

573. In the measurement of heights by the barometer, the 
thermometer by which the temperature of the air is measured is 
called the detached thermometer ; and that by which the tempera- 
ture of the mercury in the barometer is measured is called the 
attached thermometer. At the lower and upper stations, whose 
difference of level is to be determined, the pressure and temperature 
of the air in the shade, and the temperature of the mercury in the 
barometer, are observed ; and from these observations the differ- 
ence of level can be computed. The observations ought to be made 
during settled weather ; and the best time of the day for doing so 
is between eleven and twelve o'clock the morning and evening 
being unfavourable times for this purpose. 

574. Problem VI. Given the pressure and temperature of 
the air, and of the mercury in the barometer, at two 
stations, to find their difference of level. 

METHOD 1. RULE. Reduce the barometric column at the upper 
station to its length for the temperature of the mercury at the 
lower station by Art. 571. 

Find the difference of the common logarithms of this reduced 
column and that at the lower station ; and this difference, 
multiplied by 10000, will give the first approximate height in 
fathoms. 

Reduce this height, considered as the length of a column of air 
at the temperature of freezing, to its length for the mean tem- 
perature of the detached thermometers at the two stations by 
Art. 572; and this reduced length, multiplied by 6, will be the 
second approximate height in feet. 

To this last height add the ^$ part of the second approxi- 
mate, height in fathoms, and the sum is the required height 
in feet. 

Let p, p' = the barometric heights at the lower and upper station, 

S and S' suppose, 

t, ' = the temperatures of the air at S and S', 
T, T'= i. ,i i, mercury at S and S', 

=( + ') = the mean value of t and t', 
d=T - T' the difference of temperatures of mercury, 
p 1 = tlie reduced value of p' to temperature T, 
h",h l ,h',h= ii first approximate height in fathoms, the second 



BAROMETRIC MEASUREMENT OF HEIGHTS 299 

approximate height in fathoms and feet, and the required height 

in feet ; 

then Pi =p' + i-Q&rijdp', or 1 L(p l -p') = Ld + Lp' - 4 ; 

h" = l(mO(Lp-Lp 1 ), 

and h = 6{h" + ^^s - 32)h"} ; 

and h^ 



Instead of reducing p' to its value at the temperature T, p and 
p' may both be reduced to their values at any common tempera- 
tures, as their ratio will then be always the same ; and instead of 
multiplying the difference of the logarithms of p and p 1 by 10000, 
the decimal point may merely be removed 4 places to the right. 
In the preceding formula, p and p' may be expressed in any 
denomination, provided it be the same ; but the temperatures 
are according to Fahrenheit's scale. 

EXAMPLE. Find the difference of level between two places 
at which the barometric pressures were observed to be = 31 '725 
and 27 '84 inches, the temperatures of the air = 65 -75 and 54'25, 
and the temperatures of the mercury = 60 '05 and 50 '75. 

p =31-725, t =65-75, T =60 '05, 

p' = 27-84, t' = 54 -25, T' = 50 '75 ; 
hence 5 = 120, d= 9 '3, 

and jj =p' + -WOldp' = 27 '84 + -0001 x 9 -3 x 27 '84 = 27 '866. 

Lp . . =1-5014016 
L/J! . . =1-4450746 

0563270, or 563 -27= h" 

563 = 35-47 



598-74 = ^ 

Then 6^ . = 3592 '44= h' 

. = 5-99 



Required height h . . = 3598 '43 feet. 

575. The preceding rale has been derived from the formula 
h = 



in Avhich the coefficient is expressed in metres, and the tempera- 
tures on the centigrade scale, and p, p lt are the barometric heights 
reduced to a common temperature, as in the preceding article. 
When reduced to imperial fathoms the coefficient is = 10025 very 
nearly, and %(t + t') becomes on Fahrenheit's scale |s-32, where 
s = the sum of the temperatures of the air; also -r^ (Art. 572) 



300 BAROMETRIC MEASUREMENT OF 

must be used for ^ or y^ ; so that ^ x 
becomes ^Au{i s ~ 32}, and the formula is 

h = 10025(1 +-dW(4 - 32)}(Lp - Lp,). 

The term (Lp-Lpj), multiplied by 10000, will give the approxi- 
mate height in fathoms to within T \ of the whole at its mean value, 
and the result is therefore h". To h" is then to be added the term 
T$nr(i* ~ 32)A", and the sum is 7i,, which, multiplied by 6, will give 
h' the second approximate height in feet. Since 25=-j$ T of 10000, 
there ought now, for the omission of 25 in the coefficient, to be 
added to h' T $ T of h' to give h. But when the formula is complete 
.there is a term in the denominator = 1 - '0027 cos 2/, dependent on 
the variation of gravity with the latitude I, and for the mean 
latitude of Britain, or 54 30', this term is nearly = 1 + T^TT 5 hence 
the coefficient ought, for this latitude, to be reduced about TFQH 
part and increased ? ^ ; and TFIT ~ r^Vu = TS^TT > that is, the whole 
increase of h' ought to be -$%$ part of h', or merely y^y of h^ is to be 
added to h' for the required height in feet. 

The preceding method is applicable, with sufficient accuracy, for 
the height of any mountain in Great Britain. In other cases one 
of the two following methods must be used : 

576. METHOD 2. The second method is by means of Tables 
containing the logarithms of all the possible values of the terms of 
the formula for every integral value of s, d, and I, the latitude. 
This method is the most simple, concise, and expeditious. 

RULE. Using the same notation as in the last method, opposite 
to the value of d in Table II. find the value of B ; then let 

R = Lp-Lj'-B. 

Then, opposite to the value of s in Table I. find the value of A ; 
and opposite to the latitude I in Table III. find the corresponding 
value of C ; and let 



When the height does not exceed two or three thousand feet, the 
value of h' thus found will be sufficiently correct ; but when the 
height is considerably greater, find in Table IV. the number of 
thousands in h', denoted by n in the first horizontal line, and under 
it is a correction c ; also, when the value of s is different from 64, 
find its value in Table V., in the first horizontal line, and under 
it is a number k, which, multiplied by n, gives a second correction 
c'=nk, and then the required height is h = h' + c + c'. 

577. When the lower station is some thousands of feet above the 
level of the sea, a third correction c" will be found by multiplying 
the value of k^ in Table VI., which corresponds to n', the number 



BAROMETRIC MEASUREMENT OP HEIGHTS 



301 



of thousands in this height, by n, the number of thousands in 
the computed height h, and dividing the product by 10 ; that is, 
c" = ^nk l . This correction is always additive, and 



TABLE I 



s 


A 


s 


A 


* 


A 


s 


A 


44 


4-76943 


75 


4-78465 


105 


4-79890 


135 


4-81268 


45 


993 


76 


513 


106 


936 


136 


314 


46 


4-77042 


77 


562 


107 


983 


137 


359 


47 


092 


78 


610 


108 


4-80030 


138 


404 


48 


142 


79 


658 


109 


076 


139 


449 


49 


192 


80 


706 


110 


122 


140 


494 


50 


242 


81 


754 


111 


168 


141 


539 


51 


291 


82 


801 


112 


215 


142 


584 


52 


341 


83 


849 


113 


262 


143 


629 


53 


390 


84 


897 


114 


308 


144 


674 


54 


440 


85 


945 


115 


354 


145 


719 


55 


489 


86 


993 


116 


400 


146 


764 


56 


538 


87 


4-79040 


117 


447 


147 


808 


57 


588 


88 


088 


118 


494 


148 


853 


58 


637 


89 


136 


119 


539 


149 


898 


59 


686 


90 


183 


120 


584 


150 


942 


60 


735 


91 


231 


121 


630 


151 


987 


61 


784 


92 


278 


122 


676 


152 


4-82031 


62 


833 


93 


326 


123 


722 


153 


076 


63 


882 


94 


373 


124 


768 


154 


120 


64 


931 


95 


420 


125 


814 


155 


164 


65 


980 


96 


467 


126 


859 


156 


209 


66 


4-78029 


97 


515 


127 


905 


157 


253 


67 


077 


98 


562 


128 


951 


158 


298 


68 


126 


99 


609 


129 


997 


159 


342 


69 


175 


100 


656 


130 


4-81042 


160 


386 


70 


223 


101 


702 


131 


088 


161 


430 


71 


272 


102 


749 


132 


133 


162 


474 


72 


320 


103 


796 


133 


178 


163 


518 


73 


369 


104 


843 


134 


223 


164 


561 


74 


417 















302 



BAROMETRIC MEASUREMENT OP HEIGHTS 



TABLE II 



d 


B 


d 


B 


d 


B 


d 


B 





ooooo 


13 


00056 


26 


00113 


39 


00170 


1 


4 


14 


61 


27 


117 


40 


174 


2 


9 


15 


65 


28 


122 


41 


178 


3 


13 


16 


69 


29 


126 


42 


183 


4 


17 


17 


74 


30 


00130 


43 


187 


5 


22 


18 


78 


31 


134 


44 


192 


6 


26 


19 


83 


32 


139 


45 


196 


7 


30 


20 


87 


33 


144 


46 


200 


8 


35 


21 


91 


34 


148 


47 


205 


9 


39 


22 


96 


35 


152 


48 


209 


10 


43 


23 


100 


36 


156 


49 


213 


11 


48 


24 


104 


37 


161 


50 


217 


12 


52 


25 


109 


38 


165 







TABLE III 





I 


C 


I 


C 


1 


C 


I 


C 





00117 


33 


00048 


46 


9-99996 


59 


9-99944 


3 


116 


34 


44 


47 


92 


60 


41 


6 


114 


35 


40 


48 


88 


63 


31 


9 


111 


36 


36 


49 


84 


ee 


22 


12 


107 


37 


32 


50 


80 


t>9 


13 


15 


101 


38 


28 


51 


76 


72 


05 


18 


095 


39 


24 


52 


72 


75 


9-99899 


21 


087 


40 


20 


53 


68 


78 


893 


24 


00078 


41 


16 


54 


64 


81 


889 


27 


69 


42 


12 


55 


60 


84 


886 


30 


59 


43 


08 


56 


56 


87 


884 


31 


55 


44 


04 


57 


52 


90 


883 


32 


52 


45 


00 


58 


48 







BAROMETRIC MEASUREMENT OF HEIGHTS 



303 



TABLE IV 



n = 


1 


2 


4 


6 


8 


10 


12 


14 


16 


18 


20 


c= 


2-5 


5-2 


10-7 


16-7 


23 


29-8 


36-9 


44-4 


52-2 


60-5 


69-2 



TABLE V 



s= 


44 


84 


104 


124 


144 


164 


k= 


-06 


06 


11 


17 


22 


26 



TABLE VI 



n' = 


1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


*i = 


1 


1-9 


2-9 


3'8 


4-8 


5-8 


6-7 


7-7 


8'6 


9-6 



EXAMPLE. Find the altitude for the observations in the example 
of the first method, supposing the latitude to be = 55. 
p =31-725, t =65-75, T =60-05, 
2? &1 "o4 * 54 *o , A i)U * / o y 
Hence . . s =120, d = 9 '3, and Z=55. 

L.^> . . =1-50140 And . L.R= 275074 

L.y . . =1-44467 A =4-80584 

05673 = 9-99960 

B= -00040 Therefore, L. h' =3 '55618 

Hence, . . . R= -05633 And . . /t' = 3599 



By Table IV., the value of c 

V., : c 

Therefore, .... 



6 



h = 3609 -2 

Had the lower station been 6000 feet above the level of the sea, 
it would have been necessary to add to this value of h the correc- 
tion from Table VI. ; namely, A-J = 5 '8, under w' = 6, multiplied by 
T*U x 3 '6, and then the value of h would have been 

3609 -2 + 2 -1 = 361 1-3. 

The value of /i' = 3599 is jnst ^ of a foot greater than that found 
by the first method. But as the formula in the first method is 
adapted to the latitude 54 30', had the latitude in the example 
been assumed considerably different, the results would also have 
differed considerably. When either the latitude differs consider- 



304 BAROMETRIC MEASUREMENT of HEIGHTS 

ably from 54 30', or the height exceeds three or four thousand feet, 
the second or third method must be used. 

578. METHOD 3. The altitude may also be computed indepen- 
dently of the preceding Tables, and with equal accuracy, though 
with more calculation, by means of the complete formula, 

, 60160 9 

~ 1 - -0027 cos 21' ( + 4000 
in which h is expressed in imperial feet, ^ is = s-32, p^ is the 
reduced value of p' (Art. 571), and r is the radius of the earth in 
feet at the lower station. The quantity h in the last two factors 
may, without sensible error, be taken equal to the approximate 
value of h, as found from the two preceding factors. 

The above formula is that given by Poisson, with merely an adap- 
tation to imperial measures and to Fahrenheit's scale ; butp and^ 
may be in any denomination provided it is the same in both cases. 

579. Principles of the Method. It is proved in the principles 
of pneumatics that if the altitudes, reckoning vertically from the 
surface of the earth, are taken in arithmetical progression, the 
pressures of the air are in a diminishing geometrical progression, 
supposing the temperature uniform, and the tension of the vapour 
in it proportional to the pressure. Now, at the height h, let the 
pressure be p, 

and let h, h + k, h + 2k . . h + nk . . [1], 
and p, rp, r*p t M p . . . [2], 

be an arithmetical and a geometrical series, such that any term 
in the latter (as r n p) denotes the pressure at the height, denoted 
by the corresponding term of the former (as h + nk); then the 
difference between any two terms of the former series will be pro- 
portional to the difference of the logarithms of the corresponding 
terms of the latter. 

For let a be such a number that a*=p, and let m be such a 
quantity that h=mk, then the term h-mk of series [1] has corre- 
sponding to it the term v~ m p in [2], and the unit of measure of 
pressure being assumed equal to that corresponding to a height 
h o that is, to h-mk it follows that r~ m p = l, and therefore 
p=t m , and also p n / m =r n . But since h=mk, therefore nk = nhlm t 
and consequently a h+nk = a h+nh / m = p . p n / m = r^p ; and hence if 
h=~L'.p for the system whose base is a, then is h + nk = L'. r n p 
for the same system. 

If h + n'k and r n 'p are another two corresponding terms of these 
two series, it is similarly proved that 

h + n'k L'. r"'p ; 



BAROMETRIC MEASUREMENT OF HEIGHTS 305 

and therefore if the heights h + nk and h + n'k are denoted by /tj 
and /( 2 , and the pressures r"p, r"'p by p 1 and p 2 , then is 

h z -h 1 = 'L'.p^-L'.p 1 . 

And, since the logarithms of the same numbers in different 
systems are always proportional, if L denote common logarithms, 
then there is some number M', such that if N is any number, 
L'.N = M'L.N; hence 

A 2 - hi = M'(L . pi - L . ^ 2 )- 

As the pressures diminish while the heights increase, the ex- 
ponents of a that is, the logarithms will be negative ; but this 
circumstance does not affect the preceding reasoning. 

By means of this formula, then, the difference of level of the 
two stations could be computed, were the air always in the same 
state, and the mercury in the barometer at the freezing-point of 
water. But as this is not the case, the value of the pressure 
must be corrected, as in Art. 571, which introduces the term 
/^(l+TTroiru^). where rf=T-T". Again, the mean temperature 
of the air being different from 32, the column of air must be 
corrected for this temperature t t = ^ (t + f) -32 (by Art. 572); and 
hence the factor (1 + iA^i) i 8 formed. Again, the force of 
gravity at the mean latitude 45 l>eing considered = 1, at 
any other latitude I it is represented by 1 - '0027 cos 21, 
and this expression in the investigation becomes a factor of 
the denominator. The same investigation introduces also the 
last factor of the formula (1+A/r) and the term preceding it, 
or2L(l+A/>-). 

When the higher station is situated on a nearly level surface, 
such as tableland, and its height above the level of the sea is 
required, instead of the fraction h/r, only 5h/8r is to be taken ; 
and for an insulated mountain, $h/r would probably be more 
correct than h/r. The fifth of the following exercises affords an 
application of the former remark. 

When the term h/r of the formula, and that depending on the 
latitude, are omitted, and the coefficient increased from 18336 
metres to 18393 metres that is, from 10025 fathoms to 10060 
the results obtained will be sufficiently correct, if they do not 
exceed seven or eight thousand feet. This coefficient was empiri- 
cally determined by Ramond, from numerous barometric and 
trigonometric measurements of mountains in the Pyrenees. The 
formula is then 

h= 10060(1 + T *Wi) 



306 BAROMETRIC MEASUREMENT OP HEIGHTS 

This coefficient, however, is rather large for the mountains and 
latitude of Britain. 

580. The number M', which is termed the barometric modulus, 
is found by combined observation and theory to be 18336 metres, 
or 10025 fathoms nearly, supposing the temperature that of freez- 
ing, and the air in a medium hygrometric state. As the mean 
temperature of the two stations, however, is generally considerably 
above freezing, and the hygrometric state indicates more than a 
medium humidity, the air will be of less specific gravity than if it 
contained only the mean quantity of water vapour at freezing ; and 
if the consequently greater expansion of the aerial column is not 
taken into account, the computed altitudes will be rather less than 
the real, and at an average by about %%-$ part. The expansion 
from this cause can be computed by means of the principle ex- 
plained in Art. 572. But the diminution of the temperature in 
ascending is greater nearer the earth, and consequently the mean 
temperature %s = %(t + t') will exceed the temperature at the mean 
height. Were ^ only the half of a degree too great, the excess of 
the computed above the true height would be about ^J-j- part of the 
latter. It appears, then, that these two errors, being of opposite 
kinds, tend to compensate each other. 

In consequence of these two sources of error, great heights would 
be more accurately determined by taking observations at inter- 
mediate stations, and computing separately the difference of level 
of every two succeeding stations. 

In the following exercises, the heights in Britain are computed 
by the first two methods, and the corrections from Table IV. are 
added to the heights found by the first method. 

EXERCISES 

1. Find the height of Arthur Seat from these observations : 
At Leith Pier the height of the barometer was = 29 '567 inches, 
attached thermometer = 55 '25, detached thermometer = 54; on 
the summit of Arthur Seat the barometer was = 28 '704, attached 
thermometer = 5175, and detached thermometer = 50 '5 ; and 
;=56 =801 -3 feet. 

2. Find the height of the Erzegeberg, near Ilfeld, from these 
observations : 

p =326-5 lines, T =7 '6, t =7 '8, 
p' = 317-8 T' = 6-4, i!' = 6-2, and 1=51 34'. 
The height of the barometer is expressed in French lines, and the 
temperature in Reaumur's scale =724 -6 feet. 



BAROMETRIC MEASUREMENT OF HEIGHTS 307 

3. Required the height of Ben Lomond from these data : 

p =30-295 inches, T =75 '5, t =75 '5, 

/>' = 27-064 T' = 60-l, f = 60'2, and J=56. 

The height of the summit above the upper barometer that is, 
above the surface of its cistern being = 2 feet, the height of the 
lower barometer above the lake =2 feet, and the height of the 
lake above the sea = 32 feet . . . =3181 '3. or 3182 '6 feet. 

4. Find the height of the Pic de Bigorre from these observations, 
the temperatures being on the centigrade scale : 

p =73-558 centimetres, T =18-625, t = 19'125, 

y = 53'72 T'= 9-75, t' = 4, and Z=43. 

= 8579-9 feet. 

The altitude of the Pic de Bigorre was found trigonometrically 
by Ramond to be = 2613'137 metres, or 8573'4 feet, or 6'5 feet less 
than the preceding result. 

5. Required the height of Guanaxuato from these observations 
made by Humboldt : 

Centimetres Centigrade Degrees 

p =76-315, T =t =25-3, 

2/ = 60'095, T = t' =21-3, and 1 = 21. 

The height is 6846 '2 feet ; but if only | of the correction c + c' is 
taken (Art. 579), the height is = 6838-4 feet, which is = 8 feet 
more than the height found by Poisson (Mfcaniqtte, ii. 631). 

6. Required the height of Mont Blanc from these observations 
of Saussure : 

French Inches On Reaumur's Scale 

p =27-267, T =t = 22-6, 

p'= 16-042, T = t'= - 2-3, and /=45 45'; 

the summit of the mountain being = 3 '3 feet above the upper 
station, the lower station = 116'6 feet above the Lake of Geneva, 
and the lake = 1228 - 8 feet above the level of the sea. 

The height is = 15818'l feet. The same found geometrically 
by Corabeuf is = 15783, or 35 - l feet less. 

7. Find the height of Chimborazo from these observations of 
Humboldt : 

Centimetres Centigrade Degrees 

p =76-2, T =25-3, t = 25 '3, 

j' = 37-727, T' = 10, t'= - 1-6, and; =1 45'; 

the height of the summit above the upper station being = 2000 feet. 

= 21293 feet. 



308 MEASUREMENT OF DISTANCES BY THE VELOCITY OF SOUND 



581. Since the velocity with which sound passes through 
the atmosphere has been determined with considerable pre- 
cision, at least to within about a two-hundredth part, the 
formula expressing that velocity can therefore be employed 
to determine the distance of the source of any sound, such as 
the report of a gun or a peal of thunder. All we need to 
know is the time elapsed between the flash of the powder 
or of the lightning and the perception of the sound. But 
this time can easily be found ; for the velocity with which 
the light of the flash is conveyed (namely, 186,000 miles 
per second) is so great compared with the rapidity of the 
propagation of sound, that the time required for the former 
conveyance is practically insensible ; and therefore the time 
elapsed between the perception of the flash and the per- 
ception of the sound is just to be reckoned the time 
required for the propagation of the sound alone. 



= 1090 feet per second. 
= 4900 

= 825 



SOUND 

Velocity of sound in air at 32 F. 
water 
wet sand . 

contorted rock . = 1090 

discontinuous granite = 1306 

solid granite . . = 1664 

iron .... =17500 

copper . . . =10378 
wood . =11000 to 16700 

Distant sounds may be heard on a still day : 

Human voice, 160 yards. 

Rifle, 5350 ., 

Military band, .... 5200 
Artillery field-guns, . . . 35000 



MEASUREMENT OF DISTANCES BY THE VELOCITY OF SOUND 309 

582. Problem. Given the time required for the convey- 
ance of sound from one place to another, to determine 
their distance. 

RULE I. To 1090 add the product of 1*14 multiplied by the 
excess of the temperature above 32 of Fahrenheit's scale, or sub- 
tract it if below 32, and the sum or difference multiplied by the 
seconds in the given time will be the distance in feet. 

Let ti = t - 32, t being the temperature of the air, 

and v = the required velocity ; 

then 17=1090 + 1-14^. 

Also, let 2 = the number of seconds observed, 
and d= \\ required distance in feet ; 

then d=vt. 

583. RULE II. If much accuracy is not required, the velocity 
of sound may be considered as constant and = 1125, the velocity 
obtained from the preceding formula, for = 62f. 

For let ^ = 62| - 32 = 3075, 

then v= 1090+ 1-14 x 3075 = 1125 nearly. 

EXAMPLE. Find the distance of a ship, having observed that 
the report of a gun fired on board of it was heard 10 seconds after 
the flash was seen ; the temperature of the air being 52. 

Here ^ = 52 - 32 = 20, and t = 10 s. ; 

hence v = 1090+1 -14x20 = 1090 + 22-8 = 1112-8, 

and d=vt = lU2'8 x 10=11128. 



584. When there is wind, it will affect the velocity of the convey- 
ance of sound. If the direction of the wind is perpendicular to the 
direction of conveyance of the sound, it will not materially affect 
this velocity ; although it is found by experience that, from some 
peculiar influence, it is sensibly altered. If i = the inclination of the 
direction in which the wind is blowing to the direction in which the 
sound is moving, v' = the wind's velocity, and v" = the alteration pro- 
duced on the velocity of the sound by the wind, then it is easily proved 
that v': v"= 1 : cos t, and v" = v' cos i ; 

and hence v becomes v + v" = v + v' cos t. 

When z>90 its cosine is negative, and v becomes v v". These 
conclusions assume that the wind alters the velocity of sound by 
the quantity of the constituent of its motion, reckoned in the 
direction of the sound ; but it is found by experiment that this is 
not strictly the case. 

The above rules have been deduced from a great variety of ex- 
periments made by different philosophers ; see Herschel's Treatise 

Frac. V 



310 MEASUREMENT OF DISTANCES BY THE VELOCITY OF SOUND 

on Sound in the Encyclopaedia Metropolitana. The velocity of 
sound is also slightly affected by the hygrometric state of the 
atmosphere, but the results can only be taken into account in 
delicate philosophical experiments. 

EXERCISES 

1. Find the distance of a thunder-cloud when the time elapsed 
between the flash of lightning and the thunder is = 6 seconds (by 
Art. 583) =1 mile 490 yards. 

2. An echo of a sound was reflected from a rock in 4 seconds 
after the sound, the temperature of the air being =60; required 
the distance of the rock and the velocity of the sound by the first 
method.. . . The velocity = 1121 -92, and distance = 2243'84. 

3. Find the velocity of sound for a temperature = 69, and the 
distance of a gun when the sound is heard 12 seconds after seeing 
the flash. . Velocity =1132 -18, and distance = 13586-16 feet. 

4. When the temperature of the atmosphere was =25 centigrade, 
a peal of thunder was heard 13 seconds after seeing the flash ; find 
the velocity of the sound and the distance of the thunder in miles. 

Velocity = 1141 - 3, and distance 2 '81 miles. 



MEASUREMENT OF HEIGHTS AND DISTANCES 

585. Problem I. To find the diameter of the earth when 
the height of a mountain and the depression of the horizon 
from its top are given, supposing its form to be spherical. 
Let ADH be the earth, AB the height of the mountain, and BH 
a line drawn from it to the horizon at H. 

Measure AB, the height of the mountain, 
and the angle of depression of H, or its com- 
plement ABH. 

Draw AT and BR perpendicular to AB ; 
draw HC from H to the centre C of the earth, 
and produce the vertical BA through C to D. 

The angle HER of depression of H at B 
being known, its complement ABH = 90-HBR is known. 

In the triangle ABT, right-angled at A, the side AB and angle 
B are known ; hence find AT and BT thus : 
AT BT 




MEASUREMENT OF HEIGHTS AND DISTANCES 



311 



But AT=TH; 

hence BH = BT + TH = BT + AT. 

Then in the triangle BCH, BH is known, and also angle B ; 

r*TT 

hence find CH thus : =7^ = tan B; and HC being now found, the 
UH 

diameter AD = 2HC. 

EXAMPLE. If from a point 2 miles above the surface of a globe, 
the angle of depression of the horizon was found to be 2 2', 
required the diameter of the globe. 

Angle ABH = 90 - RBH = 90 - 2 2' = 87 58'. 

1. To find AT and BT in triangle ABT 



L, tan B 87 58', 
L, AB 2, . 

L, AT 56-33284, . 


= 11-4497317 
= 0-3010300 


L, sec B 87 58', . 
L, AB 2, . 

L, BT 56-36834, 


= 11-4500052 
= 0-3010300 


11-7507617 
10 


11-7510352 
10 


= 1-7507617 


= T7510352 



and BH = AT + BT = 1 12-7012. 



2. To find CH in triangle BCH 



L, tan B 87 58', 
L, BH 112-7012, 



= 11-4497317 
= 2-0519285 

13-5016602 

10 
3-5016602 



L, CH 3174-39, .... 

And diameter AD = 2CH = 6348'78. 

EXERCISE 

If the height of the Peak of Teneriffe is = 12350 feet, and the 
depression of the horizon from its summit = 1 58' 10", required the 
diameter of the earth. =7914-826 miles. 



586. Problem II. The converse problem may now be easily 
solved namely, To find the height of a mountain when 
the depression of the horizon from its summit and the 
diameter of the earth is given. 

For in the triangle BCH, right-angled at H, the side CH, the 
earth's radius is known, and also angle CBH the complement of 
the depression, and hence CB can be computed. 
Then AB = BC - AC = required height. 




312 MEASUREMENT OP HEIGHTS AND DISTANCES 

EEFRACTION 

587. The elevation of objects at a considerable distance is sensibly 
increased by atmospheric refraction ; for instead of a ray of light 
from any object moving in a straight line through the atmosphere, 
its path deviates a little from a rectilineal direction, and in ordi- 
nary states of the air it is a curve line, the concavity of which is 
turned towards the earth. 

Thus, let BS be the altitude of a mountain, then its summit S 
is seen from a point A, by means of a ray of light moving in a 
curvilineal direction SCA ; and if AS' is a 
tangent to this curve at A, the summit S 
will appear to be at S'; so that its apparent 
altitude exceeds its real altitude, or its angle 
of elevation is increased by the angle 
SAS', supposing AS joined by a straight 
line. 

In a similar manner the point A, when seen from S, appears to 
lie in the direction SA', and the real angle of depression HSA 
exceeds the apparent depression HSA' by the angle ASA'. 

The distance of the horizon, seen from any point above the earth's 
surface, is greater in consequence of refraction than it would be 
were there no refraction. The former may be called the actual, 
and the latter the tangential distance of the horizon. 

When the actual distance of the horizon is known, the tangential 
distance is found by subtracting ^ of the former from it ; and when 
the tangential distance is known, the actual distance is found by 
adding ^ of itself to it. 

Hence, also, a height just visible at a given distance when 
there is refraction would be lower than one just visible at the 
same distance when there is no refraction. Therefore, if a 
height be calculated by the preceding method (Art. 586) when 
the actual distance is given, the height thus computed must be 
diminished by about part of itself in order to obtain the true 
height. 

588. Problem III. Given the diameter of the earth and 
the height of a mountain, to find the distance of the visible 
horizon from its summit when the effect of refraction is 
considered. 

Let S be the summit of the mountain, E the earth's centre, 
AB its surface, and A the horizon seen from S, supposing no 
refraction. 



MEASUREMENT OP HEIGHTS AND DISTANCES 313 

In the triangle ASE the side ES is known, for ES = EB + BS, 
and EB and BS are known ; also the side AE is known ; and 
angle A is a right angle ; hence find angle E, 
and then AS ; then if <fa of AS be added to it, 
the sum is the visible distance of the horizon. 

EXAMPLE. Given the height BS of a moun- 
tain =8456 feet, and the diameter of the earth 
= 7912 miles, to find the distance of the horizon 
from its summit. 

BS = 8456 feet = ff miles = 1 '6015; 
hence ES = EB + BS=x 7912 + 1'6015 = 3957 '6015. 

1. To find angle B 2. To find AS 

L, ES 3957-6015, . = 3"5974321 I 10" 




L, EA 3956, . . = 3'5972563 
10- 



L, cos E 1 37' 48", = 9'9998242 



L, sin E 1 37' 48", = 8'4540028 
L, ES 3957-6, . = 3-5974321 
L, AS 112-57, . = 2-0514349 



The distance may also be found thus (Eucl. III. 36) : 
Let D the earth's diameter, 

then AS 2 = BS(D + BS) = 1 '6015(7912 + 1 -6015) = 1 '6015 x 7913'6015 ; 
hence AS = V12673'6328 = 112-57. 

The tangential distance of the horizon AS = 112-57 

S= 10-23 



The actual distance .... =122-8 

EXERCISE 

At what distance from the summit of Mount Etna is the apparent 
horizon, the height of the mountain being = 10963 feet? 

= 139 -846 miles. 

589. Problem IV. Given the distance at which a moun- 
tain is visible at sea, to find its height, the diameter of 
the earth being =7912 miles. 

From the given distance deduct ^ of it, and the remainder will 
be AS (fig. to Art. 588) ; then in the triangle EAS, EA, the radius 
of the earth, is known, and AS is given ; hence angle E and ES 
can be found. Then BS = ES - BE is known. 

EXAMPLE. If the distance at which a mountain is visible at 
sea be = 180 miles, required its height. 

The tangential distance of S from the horizon is T V less, or 
= 180-15 = 165; hence, in the triangle AES, AS must be con- 
sidered to be only 165 miles. 



MEASUREMENT OP HEIGHTS AND DISTANCES 



1. To find angle E 



2. To find ES 



L, AE 3956, . 
L, AS 165, . 

L, tan E 2 23' 18-12", 


= 3-5972563 
= 2-2174839 
10- 


L, sin E, 

L, AS 165, . 
L, ES 3959-437 
BE = 3956- 
BS= 3-437 


. = 8-6198504 
10- 
. = 2-2174839 


= 8-6202276 


= 3-5976335 

miles = 18147 ft. 



EXERCISES 

1. The distance at which a mountain is visible at sea is = 142 
miles; required its height ...... = 2-143 miles. 

2. The distance at which a mountain is visible at sea is = 120 
miles ; required its height ...... =1-852 miles. 

690. It can easily be proved that, if e, e' are the two angles of 
depression of two distant objects, taken at each other, and a the 
angle at the earth's centre, the refraction, supposing it the same 
for both, and denoting it by r, is found from the formula 



Thus, if a = angle at the earth's centre, found by reckoning 1' 
for every geographical mile, or 6076 feet, and if it = 40' 20", and 
if e, e' be respectively 20' 12" and 14' 2", then 2r=40' 20" 
-(20' 12" + 14' 2") = 40' 20" -34' 14" = 6' 6", and r=3' 3", or rather 
more than ^ of a. 

If one of the angles, instead of being one of depression, be one 
of elevation, its sign must be changed. Thus, if e' is an angle of 
elevation, then 

2r=a-(e-e') = a + e' -e. 

Any distance on the earth's surface may be converted into 
angular measure by allowing 1 for a geographical mile, or for 
69-05 English miles, or 1' for every 6076 feet. 

The effect of refraction varies very much with the state of the 
atmosphere. In extreme cases the variation is from to T V of the 
distance ; but in ordinary states of the atmosphere it varies from 
t*& to T \j, of which the mean is ^. By French mathematicians it 
is reckoned at about '079 of this distance. 

When great accuracy is required, small angles of elevation must 
be diminished or small angles of depression increased by ^ of the 
distance, or 5" for 6076 feet; that is, 1" for 1215'2 feet, or 1" for 
every 405 yards nearly. 

591. Another correction is also necessary, when great accuracy 
is required, on account of the earth's curvature. Thus, if CD 




MEASUREMENT OP HEIGHTS AND DISTANCES 315 

be the vertical height of an object, and AB a horizontal line 
from A, the angle of elevation CAB ought to be increased 
by BAD, which is half the angle at the 
earth's centre, subtended by the arc AD. 
Hence half-a-minute must be added for every 
6076 feet of distance, or 1" for every 202'5 feet, 
or 67^ yards. The angle ADC also exceeds a right 
angle by the same quantity, or it is =90 + half 
the angle subtended by the distance AD. 
Thus, if AD =5280 feet, or 1 English mile, and the angle of eleva- 

5280 
tion CAB 12 4', it must be increased by =- x 1"=26", and it 

.iOti'O 

becomes 12 4' 26" = angle CAD; also angle ADC = 90 0' 26". 
There are then two angles of the triangle ADC namely, A and 
D known, and the side AD ; hence CD can be calculated. But 
when CAB is. a small angle it must be diminished, on account of 
refraction, by 1" for every 405 yards, or Vs 6 / x 1" = 4'3". 

592. The following example is here given as an illustration of 
the application of these corrections. 

EXAMPLE. Given the angle of elevation = 15' 0", and the distance 
AD =20-17 nautical miles, to find the elevation of C above A. 

The angle CAB of true elevation is found by deducting the 
refraction from the observed elevation of C above A ; and as there 
are 60 nautical miles in 1, therefore 

AD=20-17 miles, =20' 10" 

Refraction=TVof AD, . = 1 41 

Observed elevation, .... = 15 
True elevation CAB, . 13 19 

Also BAD = ^(20' 10"), = IQ 5^ 

Hence CAD, = 23 24 

and ADC =90 + 10' 5" = 90 10' 5". 

To find CD in triangle ACD 
C = 180-(A + D)=180-90 33' 29" = 89 26' 31". 
L, cosec C 89 26' 31", . 10 -0000206 

L, sin A 23' 24", = 7 '8329386 

L, AD 20-17, = 1-3047059 

L, CD in nautical miles, . . = T-1376651 
L, 6076 feet (in a mile), . . = 3-7836178 

L, CD in feet 2-9212829 

Hence, CD = 834*2. 



316 MEASUREMENT OP HEIGHTS AND DISTANCES 

The relative height of the point C above A is therefore 834'2 
feet ; but if the latter point were, say, 240 feet above the level of 
the sea, then the absolute height of C above the sea would be 
= 834-2 + 240 = 1074-2. 

CONCISE FORMULA FOR HEIGHTS 

593. The distance at which the summit of an object may be seen 
at sea, when its height is known, and the height of an object when 
the distance at which its summit can be seen at sea is known, may 
be found more simply thus : 

Let D = the diameter of the earth, 
h = n height of the object, 
d = n distance at sea at which the summit of the object is 

visible ; 
then d? = (T) + h)h = Dh + h?. 

Now, h 2 will be very small compared with DA, for h is so com- 
pared with D. If h were 3'956 miles, it would just be ^^ part 
of D, and the error produced on the value of d 2 by rejecting 
the term A 2 would just be Tinnr P ai 't of A in defect. The formula 
then becomes 

d 2 d? 

d?=Dh, and h = ^, also D = -r- 

L) fl 

When d is 100 miles, it would give an error on the value of A of 
about gTjVfr part in excess. When A or d is less, the errors are also 
less. 

The formula may be simplified by taking d in miles and A in feet ; 
then since d 2 = mzVDA = BifA, and | = f nearly, 
therefore d?=%h, and h = %d 2 , 

where the denomination of d is miles, and that of h is feet. 

Since fH = l'4985, and f = l'5, therefore f is roVW^tfrnr too 
great. The value of d 2 , therefore, in miles will be too great by the 
TniVfr part of A in feet ; and the value of h in feet w 7 ill be about the 
WW part of d 2 in miles too small. 

The former simplification makes errors on the values of d and h 
respectively in defect and excess, and the latter in excess and 
defect ; and they thus to some extent compensate each other. 

594. When the summit of one object is just visible from that of 
another, the line joining them being a tangent to the surface of the 
sea, the distance at which each of the objects separately is visible 
must be calculated, and the sum of these is the whole distance at 
Which they are mutually visible. 

Thus, if AB is a lighthouse, just visible from the mast of a 



MEASUREMENT OF HEIGHTS AND DISTANCES 



317 



ship, the whole distance EB, or DCA, is just the sum of the 
distances DC and CA. Now, the height DE being given, the 
distance CD can be found ; and AB being 
known, CA can be calculated ; and hence the 
whole distance DCA can be found. 

The formula d?=%h gives the value of d, were there no refrac- 
tion ; and if to this value of d is added T J T of itself, the result will 
be the value of d, increased by the effect of mean refraction. So 
the formula h = %d? gives h too great when d is the apparent dis- 
tance ; and if ^ of d is subtracted from d, the formula, with 
this reduced value of d, will give the corrected value of h. Or the 
formula becomes, with this reduction, h = %(d- s d) 2 =%(^) 2 d 2 =%d? 
nearly, and eP=A; and h ^d 2 . Also, since $-|=> and %d*=? 
x|/i = |A, therefore the formula h^d 2 gives a value of h about 
\ of itself less than the other, or h \d?. 

EXAMPLES. 1. At what distance can an object = 24 feet high be 
seen at sea ? 

P=p=$x24 = 36, and d=6 miles. 

This is the distance were there no refraction ; but the distance is 
increased -^ by refraction ; hence the corrected distance = 6 '55 miles. 

Or, cP = %h = f x 24 = 43-2, and rf= 6 '57. 

2. From what height will the horizon be = 12 miles distant? 
A = |d 2 =f x!2 2 =96feet. 

But refraction makes it visible ^ farther ; hence h must be 
less, 

or h=l(d-^ = l x II 2 =80 -6 feet 

Or, 



= fP=x 122=80 feet. 



CURVATURE AND REFRACTION 



D 


C 


C-R 


D 


C 


C-R 


D 


C 


C-R 


1 


66 


57 


6 


24- 


20-57 


12 


96 


82 


2 


2-67 


2-29 


7 


32-67 


28-00 


14 


130 


112 


3 


6- 


5-14 


8 


42-67 


36-57 


16 


170 


146 


4 


10-67 


9-14 


9 


54- 


46-30 


18 


216 


185 


5 


16-67 


14-29 


10 


66-67 


57-14 


20 


266-7 


228-6 



D= distance in statute miles, C = curvature in feet = |D 2 approxi- 
mately, C-R= curvature less ref raction = f D 2 approximately. 



SI 8 MEASUREMENT OF HEIGHTS AND DISTANCES 

In the following exercises the effect of refraction is taken into 
account. 

EXERCISES 

1. At what distance can an object = 54 feet high be seen at sea? 

=9-8 miles. 

2. At what distance can the top of a lighthouse = 21 6 feet high 
be seen at sea? =19'7 miles. 

3. Required the distance of the visible horizon from the top of 
Arthur Seat, which is = 820 feet high? . . . =38 -26 miles. 

4. From what height will the horizon be = 36 miles distant? 

= 720 feet. 

5. From a ship's mast at the height of 120 feet, the top of a 
lighthouse = 240 feet above the level of the sea was just visible; 
required the distance of the ship and lighthouse. . =35*5 miles. 

6. If from the summit of a mountain = 11, 310 feet high, the 
distance of the visible horizon is = 142 miles, required the earth's 
diameter =7910 miles. 



LEVELLING 

595. The object of levelling is to determine the differ- 
ence between the true and apparent level at one place in 
reference to another, or the difference of true level of two 
places. 

596. A line of true level is such that all points in it are 
equally distant from the centre of the earth. 

597. A line of apparent level at any place is a horizontal 
line passing through that place. 

Let MN be an arc of the earth's surface, and LT another 
concentric with MN, and LP a tangent to 
the arc LT at L. Then L and P are in the 
same apparent level when P is seen from L ; 
also L and T are on the same true level ; and 
PT is the difference between the true and apparent level in 
reference to L at a distance from it equal to LT. 




LEVELLING 319 

The point P, on apparent level with L, is found by means 
of a level placed horizontally at L. 

598. The spirit-level (SL) consists of a glass tube nearly 
filled with spirit of wine, and enclosed in a brass tube, 
except the upper part. It is sometimes 

placed parallel to the axis of a tele- ' 

scope, and when brought to a level, ^ 

a point at a distance may be found on % ^5 ^ 

the same level with the axis of the 

telescope, by looking through it to a 

pole or other object at a distance, and finding the point on it 

that appears to coincide with the intersection of two very fine 

wires that cross each other within the telescope. 

The spirit-level is also sometimes attached to a bar of brass 
FG, with two upright pieces FE, GO, and small openings or 
sights at E and 0, so placed as to be on a horizontal line 
when the level SL is horizontal, which is the case when an 
air-bubble at B is at the middle point. 

The plumb-line level is furnished with sights like the spirit- 
level, or with a telescope. The horizontal 

position of the sights E, O is determined by E \ _j 

the vertical position of the plummet PW. 

The fluid-level consists of a tube EPO 
filled with some fluid to E and O, which are W w 

therefore on the same level. 

Square staffs are also used in levelling. They are wooden 
rods, divided into feet and parts of a foot, with movable 
vanes ; and when used, are fixed vertically in the ground. 

599. Problem I. To find the difference between the true 
and apparent level for any given distance. 

PT (fig. to Art. 597) is the difference between tnie and apparent 
level for the distance MN, and may be found by the formula 
h = %d?, where h is in feet and d in miles. 

But if refraction is taken into account, d must be previously 
diminished by T \ part, or the formula h = d? employed. In 



320 LEVELLING 

levelling, however, the distances are generally small, seldom more 
than 300 or 400 yards ; and this correction for so short a distance 
may generally be neglected. 

EXAMPLES. 1. A place at the distance of a mile from another' 
is on the same apparent level with the latter ; what is the height 
of the former above the point of true level with the latter? 

Here d=\ mile, and A = oJ 2 = x 1 = 8 inches. 

2. What is the difference between true and apparent level at 
the distance of 2022 feet ? 

Here d=fH miles =-383; hence A = d 2 =x -383 2 , or h ='0978 
foot = 1-176 inches. 

3. Required the difference between true and apparent level at a 
distance of 4 miles. 



EXERCISES 

1. Required the difference between true and apparent level at 
the distance of 2 miles ...... = 4 feet 2 inches. 

2. What is the difference between true and apparent level at 
the distance of 1240 feet ? ...... =0 '44 inch. 

3. Required the difference between true and apparent level at 
the distance of 1760 feet ....... - '888 inch. 

4. What is the difference between true and apparent level at 
the distance of 1 miles? ...... =12^ inches. 

5. If at a point in the surface of a canal it is found that for a 
distance of 3^ miles the surface of the earth is on an apparent 
level with it, required the depth of the surface of the canal below 
the surface of the earth at that distance. . . =8 feet 2 inches. 

600. It is convenient to have formulae when the distance is given 
in feet, yards, or chains, to find the difference of true and apparent 
level in inches. 

,, , . i * i h 2( d \ 

When a is yards and h inches, TH = O( 



When d is feet and A inches, then, instead of d? in the preced- 
ing formula, substitute (f) ==l<^> and h= -000000287^. 

When d is imperial chains and h inches, then, since 80 chains 
= 1 mile, 

, Set? d* , d* 



LEVELLING 321 

The two following formulae, in which logarithms are used, may 
sometimes be conveniently employed. Taking the formula when 
d and h are expressed in miles, or cP=Dh, it may be altered 
thus : 

When d and h are in feet, 



_, , 
" ' 01 '" 



___ 
\5280/ ~ 5280" 5280 ' '"7912x5280' 

and Lh = 2Lrf+ 8 '3790798. 

When h is feet and d imperial chains, 

d\ z 7912, , , 33d 2 



and LA = 2Ld+ 1 -0181675. 

Similarly, when h is inches and d chains, 
Lh = 2Ld+ 3 -0973487. 

EXAMPLES. 1. What is the difference between true and apparent 
level at the distance of 3540 feet ? 

h= -000000287^ = -000000287 x 3540 2 =3'59 inches. 

Or, LA=2Ld3540 + 8-3790798 = 2 x 3 '5490033 + 8 -3790798 
= 7-0980066 + 8 -3790798 = 1 -4770864 = LO -299976 feet, 
or 3-5997 inches. 

2. Find the difference between true and apparent level corre- 
sponding to a distance of 400 chains. 

, d 2 400 2 400 onn . . ., . 

h=- = =- = 1W> mches = 16 feet 8 inches. 

OvMJ oiX/ 

EXERCISES 

1. What is the difference between true and apparent level at 
the distance of 3100 feet ? . . . . . =276 inches. 

2. Required the difference between true and apparent level for 
a distance of 140 chains. ..... =2 -0437 feet. 

3. Required the difference of true and apparent level for a 
distance of 166 chains ....... =2 '8733 feet. 

601. Problem II. To find the difference of true level of 
two places on the surface of the earth not far distant. 

RULE. In each of the vertical lines passing through the 
two places, find a point on the same true level with some 



322 LEVELLING 

intermediate point, and the difference of the vertical heights 
of these two points above the given points is the difference of 
true level. 

Thus, let A and B be the two places on the earth's surface. 
Place a level at L, some intermediate position, and two square 
staffs at A and B, and find two points E and F on the same 
apparent level with L; and measure the heights AE, BF, and 
the distances MA, MB ; then calculate the 
distances EC and FD of true level below 
M ^ ^ apparent level by last problem. 

Were the instrument L placed in the 
middle between the two places, the points E and F that are on 
apparent level would evidently be also on a true level ; for then 
CE would be equal to DF, though the distances and refraction 
were considerable. 

Having found EC and FD, the points C and D of true level are 
then known ; and hence AC, BD are known, and their difference 
is the difference of true level. If AC exceed BD, then A is 
evidently lower than B. 

EXAMPLE. Let the distance of the level from the two stations 
A and B be = 240 yards and 300 yards; let AE and BF be = 10 
and 6 feet respectively ; what is the difference of true level of A 
andB? 

EC=A = -000002583e? 2 = '147 inch, 
DF=& = -000002583d 2 = '232 inch ; 

hence AC = AE -EC = 10 ft. - -148 in. =9 ft. 11 '852 in. 
BD = BF -DF= 6 ft. - '232 in. =5 ft. 11-768 in. 

Therefore AC-BD = 119'852 inches - 71 '768 inches = 48'084 inches 
= 4 feet '084 inch = the height of the point B above A. 

Were the instrument L in the middle between A and B, then E 
and F would be in the same true level, and the difference of level 
of A and B would be = 10-6 = 4 feet. 

EXERCISES 

1. Find the difference between the true level of two places A and 
B, having given the distances AM, MB, 1040 and 1820 feet, and 
the heights AE, BF, of apparent level with L, 5 feet and 6 feet 
respectively = 1 1 -36 inches. 

2. Let the distances AM, MB be = 12 and 18 chains, the heights 
AE, BE = 3 feet 2 inches and 5 feet 8 inches ; find the difference of 
true level of A and B. =2 feet 5*77 inches. 



LEVELLING 323 

602. Problem III. To find the difference of true level of 
two places at a considerable distance. 

Let A and E be the two places. 

Take intermediate places B, C, D, so that the distances of any 
two successive places may not exceed a quarter of a mile. By last 
problem, find the difference of true level of A and B by means 
of observations taken with a level 

at some convenient station between ^^ ^*g *c ^ *t 

A and B. Find in a similar man- 
ner the difference of true level of B and C, C and D, D and 
E ; then the difference of level of A and B is easily found 
thus : 

When one place is higher than the next succeeding, reckon 
the difference of level positive ; and when lower, negative ; find 
the sum of the positive and also of the negative, and then 
the difference of these sums is the difference of level of the 
first and last places. The first place is higher than the last, 
when the sura of the positive numbers exceeds that of the nega- 
tive, and lower when the contrary is the case. 

EXAMPLE. 

Let A be 4 feet 3 inches higher than B, or +4 feet 3 inches, 
B it 3 M 2 ii lower C, .. -3 2 .. 
C 2 6 ,, higher D, +2 6 
D 3 M 8 ii lower E, -3 8 M 
Sum of positive, . . . = 6 feet 9 inches, 

ti negative, . . . =6 n 10 ,, 

Difference, =0 feet 1 inch. 

Hence A is 1 inch lower than E. 

EXERCISES 

1. Let A be = 10 feet above B, B = 8 feet below C, and C = 12 feet 
above D ; find the difference of level of A and D. 

A is = 14 feet above D. 

2. Let A be = 12 feet 4 inches above B, B = 8 feet 3 inches below 
C, C = 10 feet 11 inches above D, and D = 3 feet 2 inches below E ; 
what is the difference of level of A and E ? 

A is = ll feet 10 inches above E, 



324 



LEVELLING 



603. Problem IV. To find the difference of level between 
two objects when the observations are taken nearly in the 
middle between every two successive stations. 

A back observation is one taken on a staff behind the station, 
and a fore observation is one taken on a staff before the station 
that is, in the direction in which the observer is advancing with 
his operations. 

In this method the effects of refraction and of the earth's curva- 
tare are the same for each pair of back and fore observations taken 
at the same station, so that the points of apparent level for these 
two observations are also points of true level ; and thus no correc- 
tion is necessary for either curvature or refraction. 

RULE. Find the sum of the back and fore observations sepa- 
rately ; the excess of the former above the latter will show the 
ascent from the first to the last station, or the excess of the latter 
above the former will show the descent. 

EXAMPLE. From stations at nearly equal distances between the 
points A and B, B and C, C and D, D and E, the observations 
were as in the following Table ; find the difference of level of A 
and E. 



Number of 
Station 


Distance of Station 


Back 
Observation 


Fore 
Observation 


from 


from 


1 


A 200 


B 200 


4-2 


T5 


2 


B 345 


C 342 


2-3 


5-7 


3 


C 500 


D 504 


2-1 


3-9 


4 


D 1285 


E 1280 


9-5 


4-2 




2330 


2326 


18-1 


15-3 




2326 




15-3 




4656 


2-8 



Hence the height of E above A is 2 '8 feet, and the distance is 
=4656 feet. 

EXERCISES 

1. What is the difference of level of A and D, and their distance, 
taking the data from the last example ? 

D is = 2-5 feet lower than A, and the distance is = 2091 feet. 

2, Required the height of the point A above E, and their dis- 



LEVELLING 



325 



tance from the data in the subjoined Table, arranged as in the 
preceding example. 



Number of 
Station 


Distance of Station 

A 


Back 
Observation 


Fore 
Observation 


from 


from 


1 


A 150 


B 150 


3'5 


2-5 


2 


B 542 


C 542 


4-3 


3-2 


3 


C 253 


D253 


2-7 


8-5 


4 


D751 


E 753 


7-4 


9'6 

i 



A above E = 5'9 feet, and their distance = 3394 feet. 



STRENGTH OF MATERIALS AND THEIR ESSENTIAL 
PROPERTIES. 

604. The properties of matter are almost innumerable, 
but they may be divided into two classes (1) Essential 
properties; (2) Contingent properties. The essential pro- 
perties are those without which matter cannot possibly exist 
The contingent properties are those which we find matter 
possessing, but without which we could conceive it to exist. 

Essential Properties. (l) Extension means that property by 
which every body must occupy a certain bulk or volume. When 
we say that one body has the same volume as another, we do not 
mean that it has an equal quantity of matter, but only that it 
occupies an equal space. 

(2) Impenetrability means that every body occupies space to 
the exclusion of every other body, or that two bodies cannot exist 
in the same space at the same time. 

Contingent Properties. (1) Divisibility means that matter 
may be divided into a great but not an infinite number of parts. 
The ultimate particles of matter are termed atoms, derived from a 
Greek word signifying indivisible. 

(2) Porosity signifies that every body contains throughout its 
mass minute spaces or interstices to a greater or less extent. 
This has been proved to be the case with many substances, and 
there is evidence that leads us to believe it to be true for all. 

Prac. V 



326 STRENGTH OF MATERIALS 

(3) Density is that property by which one body differs from 
another in respect of the quantity of matter which it contains in a 
given volume. The density of a substance is either the number of 
units of mass in a unit of volume, in which case it is equal to the 
heaviness (that is, weight of unit volume of substance in standard 
units of weight) ; or it is the ratio of the mass of a given volume 
of the substance to the mass of an equal volume of water, in 
which case it is equal to the specific gravity. 

(4) Cohesion is that property by which particles of matter 
mutually attract each other at insensible or indefinitely small 
distances. It is generally regarded as differing from gravitation, 
which acts at all distances. It is, however, Conceivable that 
the two kinds of attractive forces may be fundamentally the 
same. 

. (5). Compressibility and dilatability are properties common to 
all bodies, by which they are capable of being compressed like a 
sponge, or extended like a piece of india-rubber, in a greater or less 
degree. 

! ,(6) Uigidity signifies the stiffness to resist change of shape when 
acted on by external forces. Unpliable materials which possess 
this, property in a large degree are termed hard t 'whilst those which 
readily yield to pressure are called soft. Substances which cannot 
r resist a change of shape without breaking are termed brittle, whilst 
those that do resist, and at the same time change their form, are 
said to be tough. 

(7) .Tenacity is the resistance (due to cohesion) which a body 
offers to being pulled asunder, and it is measured by the ten^ 
site '"strength in Ib. per square inch of the cross section of the 
body. 

(8) Malleability is that property by Avhich certain solids may 
be rolled, pressed, or beaten out from one shape to another without 
fracture. It is therefore a property depending upon the softness; 
toughness, and tenacity of the material. 

(9) Ductility is that property by which some metals may .-be 
drawn through a die-plate into wires or tubes. A metal is said to 
c be homogeneous when it is of the same density and composition 
throughout its mass. It is isotropic when it has the same elastic 
'properties in all directions. 

(10) Elasticity is that property possessed by all substances in 
a greater or less degree of regaining their original size and shape 
after the removal of the force which caused a change of formi, 

7 



STRENGTH OF' MATERIALS 32? 

When a solid does not return to its- original form or shape after 
the force has been removed, it has been stretched beyond the 
elastic limit of the material; 

(11) Fusibility is that property whereby metals and many 
other substances, such as resins, tallows, &c., become liquid on 
being raised to a certain temperature. 

The following Table shows in round numbers the melting-points 
of a few of the commoner metals : 

MELTING-POINTS OF METALS IN DEGREES FAHRENHEIT 



Mercury, 
Tin, 
Bismuth, 
Lead, 
Zinc, 


: . - 3 

\ - +440 
. 500 
. . 600 
. 700 


Coppeiy. . . 
German silver, 
Gold, . 
Cast-iron, . . 
Steel, . ... 


2000 
2000 
2000 
2200 
2500 


Antimony, 
Brass, . 
Silver, . 


. 800 
. 1800 
..- ....- 1850 


Nickel, also Aluminium, 
Wrought-iron, 
Platinum,...,. -, *> ., . 


2800 
3300 
3500 



605. It is convenient to introduce here the definition and pro- 
perties of the moment of inertia and the radius of gyration. If the 
mass of eveiy particle of a body be multiplied by the square of 
its, distance from a given axis, the sum of the products is called the 
moment of inertia of the body about that axis. 
. - If M be the mass of a body, and k be such a quantity that M/f 2 
is the moment of inertia about a given axis, then k is called the 
radius of gyration of the body about that axis. 

Thus, M& 2 = I = moment of inertia ; 

or, %?== square of radius of gyration. 

A cylinder can be conceived as made up of a great number 
of circular discs threaded together on the same axis, and the 
moment of inertia will just be the sum of the moments of inertia 
of all the discs. Since the radius of gyration of each disc is 
independent of the thickness of the disc, it _ follows that the 
radius of gyration of the whole cylinder will be the same as that 
of one of the discs. 

The term ' moment of inertia ' has been defined above with 
respect to a solid body only, but it is easy to see that by a slight 
alteration in the wording of the definition it may be made to apply 
equally to an area or a section of a solid. Accordingly, we find the 



328 STRENGTH OP MATERIALS 

terms 'moment of inertia' and 'radius of gyration' applied to 
areas as well as solids. 

For instance, we speak about the moment of inertia and radius 
of gyration of a circle about a diameter, of a triangle about its 
base, and so on. 

The moment of inertia of a solid, or section of a solid, about a 
given axis is always proportional to the mass of the solid, or to 
the area of the section, as the case may be. 

The following rule has been stated by Routh, and will be 
found useful for finding the moments of inertia about an axis 
of symmetry : 

Moment of inertia = mass x (sum of the squares of the perpen- 
dicular semi-axes) -r (3, 4, or 5, according as the body is rectangular, 
elliptica,!, or r ellipsoidal). 

The Table on p. 329 gives the radius of gyration for certain 

sections, R = /V /TT- 

'Y M 

606. Load, Stress, and Strain. When force is applied to a 
body so as to produce either elongation or compression, bending, 
torsion, shearing, or a tendency to any of these, the force applied 
is 'termed the load; the corresponding resistance, or reaction in 
the material, is termed the stress due to the load. Any alteration 
produced in the length, volume, or shape of the body is termed the 
strain. 

Tensile Stress and Strain. If the line of action of a load be 
along the axis of a bar, tie-rod, or beam, so as to tend to elongate 
it, the reaction per square inch of cross section is termed the 
tensile stress, and the elongation per unit of length is called the 
tensile strain. 

607. Young's modulus of elasticity of any substance is the 
ratio of the tensile strength to the tensile strain. Thus Young's 
modulus 

stress ,, P / 1 m . JT? 
= -7 = E = -r/T-; orPL=AE, 
strain A/ L 

where P=pull, push, or load in Ib. on the bar, 
A = area of cross section of the bar, 
L=length of bar before the load was applied, 
1= length by which the bar is extended or compressed, 
or load per square inch of cross section =P/A. 



STRENGTH OF MATERIALS 



329 





to 



to 

CM 



a 

O) 
00 
OJ 






2 i* 



C 1 



- 
Tc S 



" 



3 II 



330 STRENGTH OF MATERIALS 

p 

Then, so long as -r- does not exceed the elastic limit, / varies 

I P 

directly as P for the same bar ; or =r- varies directly as -r for 

different bars of the same material} and subjected to the same 
conditions. In other words, so long as the stress does not 
exceed the elastic limit, the strain; will be proportional to the 
stress. 

Hooke's law holds good for metal bars under the action 
of forces tending to elongate or compress them. This law 
states : 

(1) The amount of extension or compression for the same bar is 
in direct proportion to the stress. 

(2) The extension or compression is inversely proportional to 
the cross sectional area ; consequently, if the area be doubled the 
extension or compression will be halved, or the resistance to the 
load will be doubled. 

(3) The extension or compression is directly proportional to 
the length. Since stress is reckoned by so many Ib. per square 
inch of cross section of a material, and strain is simply a ratio, 
it follows that the modulus of elasticity (E) must also be reckoned 
by so many Ib. per square inch. 

EXAMPLE. A steel bar 5 feet long and 2| square inches in cross 
section is suspended by one end ; what weight hung on the other 
end will lengthen it by -016 inch, if 'Young's modulus for steel is 
30000000 Ib. per square inch ? 

Now, the universal rule is, modulus ; of elasticity = '-,- , or stress 

= mod ulusx strain. For the strain is the elongation per unit of 
the length. 

Consequently, /=J-= = -00026. 

. ' . the stress = modulus x strain = 30000000 x '00026 

= 8000 Ib. per square inch ; 

and the total stress = 8000 Ib. x2'25 square inches = 18000 Ib. Or, 
we might have applied the formula previously deduced namely, 
PL = AIE, where P is the total pull required in Ib. 

A/E 2-25 square inches x -016" x (30 x 10 6 ) 1cnnnll 

. ' . r = ,= = = =75 J 8UUU 1 D. 

L 5x 12 

When the limit of elasticity is exceeded, the strain increases at a 
much greater rate than the stress producing it, 



STRENGTH OF MATERIALS 



331 



The constant E is termed by some writers the ' modulus ' or 
' coefficient of elasticity ; ' but such a term is inappropriate, for 
there are different coefficients or moduli of elasticity, according 
to the nature of the strain, and Young's modulus is but one among 
them. 

TABLE A 
YOUNG'S MODULUS OF ELASTICITY 

E= Young's modulus, in pounds weight per square inch. 

M=length corresponding with modulus. 

W= weight each square inch will bear without permanent alteration in length. 



Material 


M. (Feet) 


E. (Lb.) 


W. (Lb.) 


METALS 








Brass . 


2460000 


8930000 


6700 


Gun-metal . 


2790000 


9873000 


10000 


Iron, cast . ';'<' 


5750000 


18400000 


15300 


n wrought . 


7550000 


24920000 


. JL7800 


Lead . . ; 


146000 


720000 


1500 


(from 


8530000 


29000000 


45000 


Steel . { to 


12354000 


42000000 


65000 


Tin . . . 


1453000 


4608000 


2880 


Zinc . . -~ ' 


4480000 


13680000 


5700 


STONES 








Marble . , . '. 


2150000 


2520000 


4900 


Slate . . . ; 


13240000 


15800000 





Portland . ,. ' 


1672000 


1533000 


1500 


TIMBER 








Ash . ; '. 


4970000 


1640000 


3796 


Beech . . 


4600000 


1345000 


3113 


Elm . . .. 


5680000 


1340000 


3102 


Fir . L ',. . 


8330000 


2016000 


4667 


Larch -4 - 


4415000 


1074000 


2486 


Mahogany 


6570000 


1596000 


3694 


Oak . . ;,-j 


4730000 


1700000 


3935 



608. Limiting Stress, or Ultimate Strength. For every kind 
of material, and every way in which a load is applied, there must 



332 



STRENGTH OF MATERIALS 



be a value which, if exceeded, causes rupture or fracture of the 
body. The greatest stress which the material is capable of with- 
standing is called the limiting stress, or ultimate strength per 
square inch of cross section of the substance, for the particular 
way in which the load is applied. 

Factors of Safety. The ratio of the ultimate strength, or 
limiting stress, to the safe working load is called the factor of 
safety. This factor of necessity varies greatly with different 
materials, and even with the same material according to circum- 
stances. 

For materials which are subjected to oxidation, or to internal 
changes of any kind, the factor of safety must of necessity be 
larger than in those which are always kept dry, or are well painted 
and carefully handled. 

TABLE B 

ULTIMATE STRENGTH AND WORKING STRESS OF MATERIALS 
WHEN IN TENSION, COMPRESSION, AND SHEARING 



Material 


Ultimate Strength 
Tons per Square Inch 


Working Strain 
Tons per Square Inch 


Ten- 
sion 


Compres- 
sion 


Shear- 
ing 


Ten- 
sion 


Compres- 
sion 


Shear- 
ing 


Steel bars, 


45 


70 


30 


9 


9 


5 


ii plates, . 


40 








8 








Wrought-iron bars, 


25 


17 


20 


5 


34 


4 


ii M plates, 


224 


17 


20 


44 


34 


4 


Iron wire cables, . 


40 








8 








Cast-iron, 


?i 


48 


14 


14 


9 


3 


Copper bolts, 


15 


25 





3 


5 





Brass (sheet), 


14 








3 








Ash, . 


7i 


4 


1 


ii 








Beech, . 


5 


4 





i 








Elm, . 


6 


44 


| 


i 





i 


Fir, . . . 


5 


24 


4 


i 


4 


TV 


Oak, . 


6ft 


34 


i 


i 


4 


' i 


Teak, . 


64 


5 





i 


i 





Granite, 





34 








4 





Sandstone, . 





H 








i 





Brick in cement, . 


i 


k to A 





50 lb. 


180 lb. 






STRENGTH OF MATERIALS 333 

The breaking strain of iron and steel does not (as hitherto 
assumed) indicate the quality a high breaking strain may be due 
to hard, unyielding character, or a low one may be due to extreme 
softness. The contraction of area at the fracture forms an essential 
element in estimating the strength. 

609. Examples of Stress and Strain. What do you under- 
stand by stress and strain respectively ? If an iron rod 50 feet 
long is lengthened by J inch under the influence of a stress, what 
is the strain ? 

Stress is the reaction per unit area of cross section due to the 
load. 

Let P = the total tension acting on area A. 

p 
Then stress = -j- 

Strain is the ratio of the increase or diminution of length or 
volume to the original length or volume. 

Let L = original length of a bar of the material, 1= the amount 
by which the length is increased or diminished ; then when the bar 
is subjected to stress, 

the strain = e=Tp- 

Ju 

In the example given L=50' x 12" = 600 inches, and l=\ inch. 
.-. Strain, e==-^=: -00083. 



EXERCISE 

From the above question and answer, determine Young's 
modulus for the iron of which the rod is composed, if the load 
was 4366 lb., and the cross section of the rod 2 square inches. 

total load 

(1) Stress= ; 
cross area 

P 4366 lb. , 

or P = -r = - o - =2183 lb. 

A 2 

stress 
(2) Youngs modulus = 



E =?= =2500000 - 

.-.a load of 25000000 lb. would elongate a rod of the iron to 
double its length by tensile stress, 

EXAMPLE. A wire ^ square inch in cross section and 10 feet 



334 STRENGTH OF MATERIALS 

long is fixed at its upper end ; a load of 1000 Ib. is hung from the 
lower end, and then the wire is found to stretch 1 inch. (1) What 
is the stress ? (2) What is the strain ? 

(1) Here P = 1000 Ib., and A = T V square inch. 
Let p = stress, or pull per square inch in Ib. 

p 
.-. the stress, or P = -^ = 1000/ T V= 10000 Ib. per square inch. 

(2) Original length = L = 10' = 120", and the increase of length = 1 
= 1 inch. 

Let e= strain, or extension per unit of length that is, per inch 
in this case. 

increase of length I 1" 

. . the strain, or e = r-j fl = T = TBTT// = '0083. 

original length L 120 

610. Compressive Stress and Strain. If the line of action 
of a load be along the axis of a bar, shore, strut, or pillar, so 
as to tend to compress or shorten the same, the reaction per 
square inch of cross section is termed the compressive stress, 
and the diminution per unit of length is called the compressive 
strain. 

EXAMPLE. A vertical support in the form of a hollow pillar, 
having 2 square inches cross section of metal, is 10 feet long. 
With a load of 10000 Ib. resting on the top, it is found to be 
compressed -fa of an inch in length. (1) What is the stress? 
(2) What is the strain ? 

(1) Here P = 10000 Ib., and A=2 square inches. 

Let p = stress, or compression per square inch of cross section 
inlb. 

P 10000 
.-. the stress, or^= T = = 5000 Ib. per square inch. 

A 2i 

(2) Original length = L = 10' = 120", and the diminution of length 

!=A". 

Let e= strain, or compression per unit of length that is, per 
inch in this case. 

diminution in length !" 

. . the strain, or e = = ,-7^7. = '00083. 

original length 120 

611. Work done in Stretching a Bar: Resilience: If a 
load of gradually increasing amount be applied to a bar so as to 
stretch it, the amount of actual stretch, or elongation of the bar, 
will, within the limitations already specified, be directly propor- 
tional to the load producing it. 



STRENGTH OF MATERIALS 335 

When the bar is loaded to its elastic limit, or prooi stress, as it is 
sometimes called, then the work done in stretching it is termed the 

2 

resilience of the l>ar, and the ratio ^ ^- (where p is the direct tensile 

Hi , p 

stress) is its modulus, or coefficient of resilience.* 

EXAMPLE. If a wrought-iron tie-bar, 5 feet long and 3 inches in 
diameter, has a limit of elasticity of 15 tons per square inch, and 
a modulus of elasticity of 30000000 Ib. per square inch, what is its 
resil ience ? Take TT = *f. 

Formula W, or resilience =^ x -=- ; 

r.. 

w 2 
or ^ r= T? x i volume of the bar. 

A = area of the section (usually in inches), 

L = original length of bar in inches, 

p = direct stress = 15 tons per square inch in this case. 

Cx^= 665-28 foot-lb. 

Sit '2 

WORKING 

p=15 x 2240 Ib., E = 30000000 Ib. j)er square inch, 
x 3 2 square inches, and L=5 feet. 



(15x2240) 2 H , 

ReSlhenCe " 30000000 xI ~ = 665-28 foot-lb. 



612. The shearing force oi" load at any point or any trans- 
verse section of a beam is equal to the resultant or algebraical 
sum of all the parallel forces on either side of the point or 
section. 

When the section under consideration is in the same plane as 
the load, the only effect the load has at that section is a tendency 
to shear the beam, but in the more general case, where the load 
acts at a distance from the given section, we have, in addition, 
a tendency to curve or bend the beam at the section. Hence the 
name bending moment is given to this latter effect. 

The bending moment at any point in a beam is the algebraic 
sum of the moments with respect to that point of all the external 
forces acting on the portion of the beam on either side of that 
point. 

* Resilience is derived from the Latin re, back, and salio, I leap or spring. 



336 STRENGTH OF MATERIALS 

The resistance to bending depends only on Young's modulus and 
the form of the section, and has no reference to the direct resistance 
to crushing. 

613. A column or strut under pressure may fail in three ways : 
firstly, by the metal being absolutely crushed ; secondly, by the 
column bending and breaking near the centre of its length ; and 
thirdly, by the plates composing it wrinkling, owing to their 
breadth being out of proportion to their thickness. 

With timber, failure sometimes takes place by the fibres crush- 
ing into each other, and sometimes by their splitting apart. In 
the latter case, although it may be due to absolute crushing, the 
length has some influence on the resistance, for each fibre may 
be considered to be a column failing by cross breaking, assisted, 
however, by its adhesion to the adjacent fibres. 

The absolute resistance varies from 2 to 3 tons per square inch, 
and the limit of elasticity may be taken as about half this, and the 
safe resistance as 10 cwt. 

The following is the formula for failure by bending or cross 
breaking where I is the length of the beam, r the radius of gyration 
of the section (Art. 605), and E the coefficient tabulated below :- 

O 

Strain per square inch = -y-^ when the ends of the column are 



rounded, or of 1 strength. 



When the ends are flat and fixed the formula becomes .. x 3. 



Cast-iron, . . -00018 
Wrought-iron, . -00010 
Steel, . . . -00008 



Teak, . . . -001 
Oak and Pitch-pine, '002 



Fir, . . . . -003 
With timber, the value of E may be taken as '001 all round. 

Note. When the ratio of length to radius of gyration becomes 
so small that the column is on the point of failing by direct 
crushing, the resistance to bending will be less. 

A factor of safety of about J the absolute resistance of the 
material is commonly taken, but this is really only of the 
elastic resistance. Now, in a long column, until the breaking 
weight is actually reached, except in the case mentioned, there is, 
theoretically, no tendency to bend, and certainly the elastic limit 
is not exceeded ; hence so large a factor of safety is not required. 
Owing, however, to differences in the different parts of the same 



STRENGTH OP MATERIALS 337 

material in the value of E, and also to the possible divergence 
of the line of pressure from the neutral line, some margin must 
he allowed, and for all purposes and conditions the absolute 
resistance of the material is well within the limits of safety. 

The rigidity and strength of a column depends on the shape of 
its ends. 

In calculating the strength of a column by means of the formula 

o, three-fourths of the result so obtained will give the weight 



that will cause the column to deflect from the perpendicular that 
is, when the column or pillar is timber. In some cases, owing to 
different densities, seasoning, &c., it will require four-fifths of the 
breaking weight to cause the column to bend at all. 

In cast-iron columns, as already pointed out, there is theoretically 
no tendency to bend when the column's length exceeds fifteen times 
its diameter or thickness, and in such cases the breaking weight 
may be regarded as the bending weight. 

With timber, the formula for the safe weight or load in tons per 

. , . 4 1333 

square inch is r-^ or 

003 -^ 



The safe resistance for cast-iron is 7 tons per square inch, 
,i it ,i wrought-iron n 5 n n n 

,, ii n timber n 10 cwt. n 

n ii n steel 11 6 tons n n 

and a deduction for rivet-holes must be made in calculating the 

sectional area. 

- RELATIVE STRENGTH OF ROUND AND FLAT ENDS IN 
LONG COLUMNS 

Both ends rounded, 1 strength, . . . =1 

One end flat and firmly fixed, 1 strength, . . =2 
Both ends flat and firmly fixed, . . .- =3 

RELATIVE STRENGTH OF SECTION IN LONG SOLID 
COLUMNS 

Cylindrical 100 

Triangular, 110 

Square, ........ 93 



338 STRENGTH OP MATERIALS 

RELATIVE STRENGTH OF MATERIAL IN LONG COLUMNS, CAST- 

IRON BEING ASSUMED AS 1000 

Wrought-iron, .. .. ,. . '' '- . ' ^ . . 1745 
Cast steel, . .. -. . v . "1 /". . 2518 
Oak, . ..':.-'' -"-. -. ; -.. .';-'.'; 109 
Red deal, . . * . . , 78 

A further investigation of this problem will appear towards the 
end of this subject. 

614. Problem I. To find the weight that a rectangular 
cast-iron beam, supported at both ends, can sustain at its 
middle. 

RULE. Find the continued product of 850, the breadth and 
square of the depth, both in inches, and divide this product by the 
length in feet, and the quotient will be the required weight. 
That is, W=8506e? 2 -rZ; 

850M 2 ' IW. ' . IW 
and ^= 



For malleable iron, Use 950 instead of 850. The weight of the 
beam must always be added to the applied weight i the weight 
of the beam is equivalent to of its weight applied at the middle ; 
and any weight uniformly distributed is also equivalent to of 
itself applied at the middle. 

EXAMPLE. A bar of cast-iron is = 2 inches square and 15 feet 
long ; what weight will it be capable of supporting,? 



EXERCISES 

1. Find the weight that can be supported by a beam = 5 inches 
square and 10 feet long. .. .. , . . 7 . . . = 10625 Ib. 

2. A beam of cast-iron is =20 feet long and 2 inches broad, and 
it has to support a load of 10000 Ib. ; what must be its depth ? 

= 9*7 inches. 

3. A cast-iron joist is = 30 feet long, 10 inches deep, and 3 inches 
broad ; what weight, uniformly distributed, can it sustain ? 

: -T - i ;a>n =13812 -5 Ib. 

615. Problem II. To find the weight that a beam fixed 
at one end can sustain at its free end. 
The weight is of that found by the preceding problem. 



STRENGTH OP MATERIALS 339 

When the weight is uniformly distributed over the beam, take 
^ of that found by Problem I. 

EXERCISES 

1. A beam is = 30 feet long, 8 inches deep, and 2J broad ; what 
weight can it support at its extremity ? . . . = 1133Jlb. 

2. What load uniformly distributed over a beam = 32 feet long, 
4 inches deep, and 2 broad can it sustain ? . . = 425 Ib. 

3. A beam =20 feet long and 10 inches deep supports a load of 
17000 Ib. at its extremity ; what is its breadth ? . . = 16 inches. 

4. A beam = 24 feet long and 2 inches broad supports 1735 Ib. 
uniformly distributed ; required its depth. . . . =7 inches. 



STRENGTH OF SHAFTING TO RESIST VARIOUS 
STRESSES 

616. Problem III. To find the weight which a solid 
cylinder or square shaft of cast-iron, wrought - iron, or 
wood can sustain when the weight is applied at the 
centre, or distributed, and when the cylinder or shaft is 
supported at both ends. 

Let D = diameter in inches, or side if square ; 

L = length of shaft, supported at both ends, in feet; 
W = weight applied at the centre in Ib. 



Ihen , w, 



is weight distributed in Ib. 

Round Shafts Square Shafts 

For wood,,. -. K = 40 K = 70 

n cast-iron, . . K = 500 K = 850 

wrought-iron, . K = 700 K = 1200 

K.D 3 . ._, L.W , 
L= w , and D 3 = , for weight at centre; 

9~K Tft OTf T)3 

1 -tTTy ^XV . \J T ^CV . LJ . . .. . . 

and W= ^ , L= .... , n n distributed. 

EXERCISES 

1. What weight will a cylinder=10 feet long and 4 inches 
diameter support ? = 3200 Ib. 

2. What weight will a uniformly loaded cylinder support, its 
length being=24 feet, and diameter=10 inches? . =41666| Ib. 



340 STRENGTH OF MATERIALS 

3. What will be the diameter of a cylinder = 20 feet long, which 
is capable of supporting 3125 Ib. ? .... =5 inches. 

4. AVhat will be the limit of the length of a cylinder uniformly 
loaded by a weight of 100000 Ib., whose diameter is = 12 inches? 

= 17*28 feet. 

617. Problem IV. To find the weight which a solid 
cylinder or square shaft of cast-iron, wrought -iron, or 
wood fixed at one end can sustain at the free end. 

The weight is just the fourth of that found in the previous 
problem, and the formulas the same. All that is necessary is to 
take one-fourth of the value of K. 

RULE. Multiply the cube of the diameter, or side if square, 
in inches, by the value of K-r4, and divide the product by the 
length in feet, and the quotient will be the weight in Ib. 

- D 3 
W= V , or W = Jx KxD 3 -j-L; 

f-D 3 T w 

L = *vr- , and D 3 = 



EXERCISES 

1. What weight will a cylinder =10 feet long and 4 inches 
diameter support at its free end ? ..... =800 Ib. 

2. What will be the diameter of a cylinder =20 feet long that 
can support 781 '25 Ib. ? ..... '' =5 inches. 

3. What will be the length of a cylinder, which is = 12 inches 
diameter, that supports 12500 Ib. ?. . . . =17 '28 feet. 

618. Problem V. To find the exterior diameter of a 
hollow cylinder of cast-iron, supported at both ends, so 
as to sustain a weight applied at the middle, the ratio 
of the interior and exterior diameters being given. 

RULE. Let the ratio of the exterior to the interior diameter be 
that of 1 to n ; then take the difference between 1 and the fourth 
power of n, and multiply it by 500 ; find also the product of the 
length and the weight; divide the latter product by the former; 
then the quotient will be the cube of the diameter. 



When the exterior diameter d is found, the interior diameter 



STRENGTH OF MATERIALS 341 

will be obtained by multiplying d by n. If d' = the interior 
diameter, and t = the thickness of the metal, then 

d'=nd, and t = \(d-d') = %(\ -n)d. 

EXAMPLE. The weight supported by a hollow cylinder is 
32000 lb., its length is = 12 feet, and the ratio of the exterior and 
interior diameters = 10 to 1 ; what are its diameters? 

tP-lW' 500(1 n i- 12x32000 - 12x64 - 768 -7680000 

U '" 500(1 --1 4 )~1- -0001" -9999 ~ 9999 
= 768-08, and d= ^768-08 =9-15 inches; 
hence d'=nd='l x9'15= '915 inch, 

and <=(! -)d=i(l - '!) x 9'15= x -9 x 9-15=4-1175. 

EXERCISES 

1. A hollow cylinder=10 feet long supports 2500 lb., and the 
ratio of its diameters is =2 to 1 ; what are the diameters? 

= 3 76 and 1'88, and thickness of metal '94 inch. 

2. A hollow cylinder = 9 feet long is intended to support 15000 lb., 
and the thickness of the metal is to be = of the exterior diameter ; 
required its diameters =6 '769 and 4 '061 inches. 

619. Problem VI. To find the resistance to torsion (or 
torque) of solid and hollow shafts. 

RULE FOR SOLID SHAFTS. Multiply the cube of the diameter 
by the shearing stress in lb. per square inch permissible iu the 

material of the shaft, and the result by 
Or, putting this in formula form, 

T.R. (torsional resistance) = ^f> 3 f . ;. ' . [1], 

where D = outside diameter, and /= shearing stress in lb. per square 
inch. 

RULE FOR HOLLOW SHAFTS. From the 4th power of the outside 
diameter subtract the 4th power of the inside diameter, and divide 
by outside diameter. Multiply this quotient by the shearing stress 
in lb. per square inch permissible iu the material of the shaft, and 

, ,, ,3-1416 

by the quotient of ^ 
lb 

Or, putting this in formula form, 

7T/D 4 -rf 4 \ 

T.R. (torsional resistance) = TK( fT~)/ 

where D = outside diameter, d= inside diameter, both in inches, and 
/= shearing stress in lb. per square inch. 

Pnur. W 



342 STRENGTH OF MATERIALS 

It is instructive to compare the torsional resistances of solid and 
hollow shafts of the same weight and material. The result shows 
that, for the same length and weight, the hollow shaft having outer 
and inner diameters in the proportion of 2 to 1 will be 44 4 3 percent. 
stronger than the solid one. 

NOTE. The strength of shafts varies as the third power of their 
diameters, whilst their stiffness varies as the fourth power. 

EXAMPLE. Find the torsional resistance or ' twisting moment ' 
of a hollow shaft of cast-iron, the external and internal diameters 
of which are 20 inches and 8 inches respectively. Take the surface 
stress as 6000 Ib. per square inch. 

Here D-20 inches,/ 6000 Ib. per square inch, d=8 inches. 

Then, as T.R. =T.M. = . 5 ./, 



. 

. . T.M. = -F x ?r- x 6000 = 9183525 inch-lb. 
lo -< 

EXERCISES 

1. Find the resistance to torsion of a solid shaft of cast-iron 
whose diameter is 5 '25 inches, with a surface stress of 4500 per 
square inch. . . ..... = 127820'745 inch-lb. 

2. Find the torsional moment of resistance of a wrought-iron 
shaft (solid) whose diameter is 6 inches, the surface stress being 
8000 Ib. per square inch. . .' . . =339206 '5 inch-lb.* 

620. In order to transmit energy through a shaft, the driving force 
must be applied at some distance from its centre. The driving 
force and its effective leverage therefore constitute what is termed 
a turning or twisting moment (T.M.), which puts the shaft in a 
state of torsion. The tendency of a purely torsional moment ap- 
plied to a shaft is to cause the shaft to shear in planes normal to 
its axis, and this has to be met by the shearing resistance of the 
material, which resistance must, of course, be of the nature of a 
moment. The resistance the shaft offers to twisting we term its 
torsional resistance (T.R.), and as this balances the turning moment, 
we have 

T.M. = T.R. 

The turning moment driving a shaft may either be uniform or 
variable in amount. Shafts which are driven by means of gearing, 
and which revolve at a uniform speed, are generally considered as 
examples of uniform turning moment. 

* See Table, Art. 625, and compare the above answer. 



STRENGTH OP MATERIALS 343 

A typical example of variable turning moment may be recognised 
in the steam-engine crank-shaft, where both the driving force of 
the steam on the piston and its effective leverage are continually 
varying throughout the stroke. When the turning moment is 
uniform that is, when the shaft revolves uniformly at n revolutions 
a minute, and transmits energy at the rate of so many horse-power, 
we have all the data required in order to estimate T.M. The work 
done by a turning couple in one minute is equal to the magnitude 
of the turning couple multiplied by its angular displacement in the 
same time. Now, the turning couple, or turning moment, as it is 
termed, is T.M. inch-lb., or ^ T.M. foot-lb., and the angular velocity 
of the shaft is n x 2?r radians per minute. Therefore the work done= 

T M 

' x 2irn foot-lb. per minute, 

KM 

and the horse-power (H.P.) = 
T.M. 
12 XZim wxT.M. 



33000 " 63024 
T.M. =63024. 



n 

EXAMPLE 1. Supposing it was required to find the horse-power 
transmitted by the shaft in the first example, running, we will 
say, at the rate of 70 revolutions a minute, we proceed thus : 
T.M., as there found =9183525 inch-lb., 
T.M. x n 9183525 x 70 



63024 

EXAMPLE 2. If a steel shaft revolving at 60 revolutions per 
minute be required to transmit 220 horse-power, what should be 
its diameter so that the maximum stress produced in it may not 
exceed one-fifth of that at the elastic limit ? The elastic limit in 
torsion is 18 tons per square inch. 

Combining formulae [1] and [2], we have 
T.R.=T.M.; 

that is, jD 3 /= 63024 x 

lo ' n 

s/lTp". 

.-. D (outside diameter) = 68 -5 y nf [3]. 
Here H.P. = 220, n = 60, and /= i x 18 x 2240 = 8064 Ib. per 
square inch. 

s / 220 
.*. D = 68'5x A/RA v c<vu = 5'27 inches. 



344 STRENGTH OP MATERIALS 

621. Problem VII. To find the diameter of a shaft, the 
torsional moment of resistance being given, and with a 
shearing stress of not over 8000 Ib. per square inch. 

RULE. Divide the torsional moment of resistance in inch-lb. by 
the quotient of 3 '1416 -r 16 x 8000, and extract the cube root. 
Or, expressing this in formula form, 



D= 



The shearing stress for wrought-iron is from 8000 to 10000 Ib. 
per square inch, and for cast-iron from 4000 to 5000 Ib. 

The factor of safety is therefore the lowest value of / in each 
case. 

All shafts when in motion that is, rotating are subjected to 
a combined and simultaneous bending and twisting moment. 

EXAMPLE. Supposing it be required to find the diameter of a 
shaft whose bending and twisting moments =25000 inch-lb. 
We employ formula [1], Problem VI., making 




25000 __ 

= 2-51 inches. 



EXERCISES 

1. Find the diameters of the following wrought-iron shafts, 
whose combined bending and twisting moments are respectively 
in inch-lb. (take/=8000 Ib. per square inch) : 

42390; 82793; 120522; 196250. 
=3"; 3|"; 4J"; 5". 

2. Find the diameters of the following cast-iron shafts, whose 
combined bending and twisting moments are respectively in 
inch-lb. (take /= 4000 Ib. per square inch) : 

84780; 165586; 241044; 392500. 

=4", smallest; 6"; 6|"; 8". 

622. Stiffness of Shafts : Angle of Twist. The effect of a 
turning moment applied to a shaft is to twist one part relatively 
to another. So far we have been dealing only with the resistance 






STRENGTH OP MATERIALS 



345 



the shaft offers to being twisted that is to say, we have been 
concerned only with the strength of the shaft without regard to the 
question of stiffness. 

In many cases, especially in light machinery, the question of the 
stiffness of the shafting is of greater importance than that of the 
strength. 

The stiffness of a shaft is measured by the smallness of the angle 
of twist per unit length of the shaft. 

This figure illustrates strain in a shaft. A 

Let dl be the axial distance in inches 
between the two sections whose diameters 
are AB, ab, and let do be the circular 
measure of the angle between those 
diameters when the shaft is twisted ; then 
the torsional or shearing strain at the 
surface of the shaft is = 

D de 




D=as before, the extreme diameter of the shaft in inches. 

Let /= surface stress in the material of the shaft in Ib. per square 
inch. 

Let C = modulus or coefficient of shearing elasticity, or of rigidity 
in Ib. per square inch. 

TM r, stress f 

Then, as C=-r = Tf7\ 33> 

strain /D\ off 

\2j'dl 



Hence for a shaft L inches long, by a simple integration we have 
the angle of twist 



CD 







To express this result in terms of the twisting moment and the 
diameter of the shaft, we have 

f= ' for solid shafts, 



and 



re" 3 

T M 

f ^i t ' ,A for hollow shafts. 



346 STRENGTH OF MATERIALS 

Making these substitutions and simplifying, we get : 

T? IM i . 10-2(T.M.)L ,. 'j 
For solid shafts, 6= p,^ . radians, 

, 584(T.M.)L 



and for hollow shafts, 



CD 4 



10-2(T.M.)L 



radmns ' 



or 



a _ 

~ 



584(T.M.)L 
C(D 4 -d 4 ) 



[4]; 



[5]. 



EXAMPLE. The angle of twist or torsion of a shaft is limited 
to 1 for each 10 feet of length ; find the diameter of a solid 
round shaft to transmit 100 horse-power at 50 revolutions a 
minute, the modulus of resistance to torsion being 10000000 Ib. 
per square inch. 

Here 6 = 1 when L = 10 x 12 = 120 inches, 

C = 10000000; 

IT T> 1fu\ 

also, T.M.= 63024 x 



. = 6 3024 x = i 26 024 inch-lb. 
n 50 

Now, applying formula [4], the given conditions are : 
584 x 126048 x 120 



1=- 

and by solving for D, we get 



10000000 x D 4 



584 x 126048 x 120 
10000000 



= 5 '45 inches. 



EXERCISES 

1. What is the maximum horse-power which could be trans- 
mitted by a shaft 3 inches in diameter when making 150 revolutions 
per minute, and supposing the shearing stress in the material to 
be limited to 7500 Ib. per square inch ? . . =94 '5 horse-power. 

2. If a shaft of 3 inches diameter transmits safely 33 horse- 
power at 100 revolutions per minute, what size of shaft will 
transmit safely 20 horse-power at 150 revolutions per minute ? 

= 2 -22 inches. 

623. It may be accepted generally that the coefficient of tor- 
sion in substances is about of their Young's modulus. 
The limit of resistances of cast-iron, wrought-iron, and eteel to 



STRENGTH OP MATERIALS 347 

shearing may be stated as 9, 18, and 27 tons respectively, or in the 
ratio of 1, 2, 3. 

Some authorities, however, give these materials as high a value 
as 14, 20, and 30 tons. Cast-iron requires to be treated with 
greater caution than any other material used in structures, its 
elastic limit being only about half its breaking weight. A test- 
piece does not elongate perceptibly to ordinary observation, but 
breaks suddenly without giving warning. It is subject to flaws in 
process of manufacture, and these are generally carefully concealed, 
being beyond the range of detection. It therefore requires a large 
factor of safety. 

624. Problem VIII. To find the breaking strain of cast- 
iron shafts (solid cylindrical) when subjected to a tor- 
sional strain with a leverage (/), the diameter of the shaft 
being given. 

RULE. Multiply the cube of the diameter in inches by 2 "36, 
and divide by the leverage (I), also in inches, and the result is the 
breaking strain in tons. 

Or, expressing this in formula form 

2-36 x d 3 , . 

j =W in tons. 

EXAMPLE. Find the weight which must be applied at a leverage 
of 170 inches in order to break a solid cylindrical cast-iron shaft 
whose diameter is 2 inches. 

,,, 2-36 xeZ 3 2-36x8 18-88 

W = = rr = -^r- ='11 ton, or 246 '4 inch-lb. 



EXERCISE 

Find the weights which must be applied at a leverage of 
14 feet in order to break the under -mentioned solid cylindrical 
shafts of cast-iron, whose diameters are respectively 2^", 3J", 4". 

Answers, -22 ton, '48 ton, '88 ton. 
Actual experiment, '18 , '52 n , '86,11 

2-^fl v /73 lv-^~, : ' 

OU A {A TT -I*-' Ot>/ ITT 

w = = , or W= j , or W= 



W W W 

r= radius of shaft. 



348 



STRENGTH OF MATERIALS 



625. TORSIONAL MOMENT OF RESISTANCE FOR SHAFTS, CALCU- 
LATED FROM THE FORMULA T.M. =^./. rf 3 . 

M=inoment of resistance to torsion=a-=3'14159. 
/= stress per square inch ; d= diameter of shaft in inches. 
/=8000 to 10000 Ib. for wrought-iron, and 4000 to 5000 Ib. for cast-iron. 



Diameter 
Inches 


/=8000 Ib. 


/= 10000 Ib. 


Diameter 
Inches 


/=80001b. 


/= 10000 Ib. 


1 


1570 


1962 


7i 


598293 


747866 


u 


3066 


3832 


74 


662344 


827930 


tft 


5299 


6624 


71 


730810 


913512 


If 


8414 


10517 


8 


803840 


1004800 


2 


12560 


15700 


84 


964176 


1205220 


2i 


17883 


22354 


9 


1144530 


1430662 


24 


24531 


30664 


94 


1346079 


1682599 


2f 


32651 


40814 


10 


1570000 


1962500 


3 


42390 


52988 


104 


1817471 


2271839 


3i 


53895 . 


67369 


11 


2089670 


2612088 


3* 


67314 


84143 


"4 


2387774 


2984717 


3f 


82793 


103491 


12 


2712960 


3391200 


4 


100480 


125600 


13 


3449290 


4311612 


4i 


120522 


150652 


14 


4308080 


5385100 


44 


143066 


178833 


15 


5298750 


6623438 


4| 


168260 


210325 


16 


6430720 


8038400 


5 


196250 


245313 


17 


7713410 


9641762 


5i 


227184 


283980 


18 


9156240 


11445300 


54 


261209 


326511 


19 


10768630 


13460788 


6f 


298472 


373090 


20 


12560000 


15700000 


6 


339120 


423900 


21 


14539770 


18174710 


H 


383300 


479125 


22 


16717360 


20896700 


64 


431161 


538951 


23 


19102190 


23877738 


6| 


482848 


603560 


24 


21703680 


27129600 


7 


538510 


673138 









STRENGTH OP MATERIALS 349 

Note. The bending moment of resistance is half the numbers 
in the Table, as M = ^ ./. d 3 . 

Rl 

For cast-iron shafts half the numbers to be taken. 

EXAMPLE. Required to find a shaft for a drum having 2J tons 
pulling on it at 17-inch radius, and taking/=8000 Ib. 

The moment of weight = W. 1=2% x 2240 x 17 = 95200 Ib. ; the 
torsional moment of resistance must be equal to or greater than 
this amount. 

Find in the Table the number next higher, which in this case is 
100480, opposite 4-inch diameter, which will be the size of shaft 
required in wrought-iron. 

If for cast-iron shaft, and /= 4000 Ib., then 5-inch diameter is the 

IQfiOKA 

size, since - = 98125, or 95200x2=190400 Ib., and the next 

X 

higher number in the Table = 196250, opposite 5-inch diameter. 

626. Problem IX. To find the weight that could safely 
be supported by a column of cast-iron or other material, 
such as oak or deal, and resting on a horizontal plane. 
The column may be either square or cylindrical in shape. 

Before proceeding further it will be as well to state that the safe 
load in structures, including weight of structures, must be regarded 
as follows : 

In cast-iron columns, J breaking weight. 

ii wrought-iron structures, J 

ii cast-iron girders for tanks, J 

ii " ii bridges and floors, J 

M timber (live load), . . . ^ 

n (dead load), . 

In stone and bricks, ..... 
The shape of the ends of the column materially affects its strength, 
and at all times requires to be considered when computing its 
strength. 

Nature of Column Ends Rn<ledi Ends Flat i 

when L exceeds 15D when L exceeds SOD 



Solid cylinder of cast- \ w , 

V W=1TT=- re= 

iron, . . . J L 17 L" 

Hollow cylinder of) TV ._ 10 D" 6 -d'' 78 w_ u .o/ p *' 85 " 

. f W lo - T~T^T j * 44 "o4 T T 

cast-iron, . ) L 1 7 L 17 



350 



STRENGTH OF MATERIALS 



Nature of Column 
Solid square of Dantzic oak (dry), . 

Solid square of red deal (dry), . . 



Ends Plat, 
when L exceeds SOD 

W=10-95p 

T)4 

W= 7-81 



W= breaking weight in tons, 
L = length of column in feet, 
D = external diameter of column in inches, 
d= internal diameter in inches. 

Now, as it is required in the problem that the safe weight he 
stated in the answers, attention is directed to the factors of safety 
already afforded. 

In hollow columns the strength nearly equals the difference 
between that of two solid columns, the diameters of which are 
equal to the external and internal diameters of the hollow one. 

627. Strength of Short Columns in which L is less than 
SOD. 

w= breaking weight of short columns, 
W = breaking weight of long columns as found above, 
C = crushing force of material (expressed in tons per square 

inch) of which the column is formed x sectional area of 

column. 

WC 



To facilitate the working of the formulae, Tables of 3 '6 and 17 
power may be employed. 

3-6 POWER 



No. 


Power 


. No. 


Power 


No. 


Power 


3 


52 


10 


3982 


17 


26892 


4 


147 


11 


5611 


18 


33035 


5 


328 


12 


7674 


19 


40133 


6 


632 


13 


10233 


20 


48273 


7 


1102 


14 


13367 


21 


57543 


8 


1783 


15 


17136 


22 


68033 


9 


2723 


16 


21619 


24 


93058 



STRENGTH OP MATERIALS 
17 POWER 



351 



No. 


Power 


No. 


Power 


No. 


Power 


5 


15 


18 


136 


30 


325 


8 


34 


20 


163 


35 


421 


10 


50 


22 


191 


40 


529 


12 


68 


25 


238 


50 


773 


15 


100 


28 


288 







EXAMPLE. What weight can a solid cylindrical cast-iron 
column sustain safely when its ends are flat and its dimensions 
are : length 20 feet, diameter 6 inches ? 

We first determine the ratio of its length to the diameter, in 
order that we may know which formula has to be employed. 

Thus, L = 20 feet =240", D= 6"; .'. ^==40. 

We therefore use formula 'When L exceeds 30D,' which in this 
case is 

D3-65 

W = 44-16^ I? -; 

hence W=44'16x( S7 -^) = 44'16x T7 == 167 '808 tons. 
\ 20 1 '/ loo 



But as this is the breaking weight, we take of the same for 
the answer, neglecting, as will be seen, the weight of the column, 
which should also be determined. 

. . the safe weight which this column can sustain is 167 '808 4- 4 
=41 '952 tons, or, say, 42 tons, omitting weight of column. 



EXERCISES 

1. A cast-iron solid cylinder is 15 feet long, and 5 inches in 
diameter ; its ends are flat, and rest on a horizontal plane ; find 
its breaking and safe loads. 

Breaking load = 144 -84 tons; safe load, not including weight 
of cylinder \ of the above. 

2. What weight can a solid column of Dantzic oak (square in 
section) sustain safely, its ends being flat? Take L = 10 feet, and 
D (that is, a side of the section) = 10 inches, and let C (the crushing 
weight) = 2'61 tons per square inch. 



352 STRENGTH OP MATERIALS 

Safe weight for dead load=44'28 tons; safe weight for live 
load =22'14 tons, neglecting weight of column in the cal- 
culation. 

3. Find the safe weight that a hollow cast-iron cylinder, rounded 
at both ends, can sustain ; its external and internal diameters are 
respectively 6 and 5 inches, and its length = 10 feet. 

Safe weight = 1976 tons, neglecting weight of cylinder in the 
calculation. 

4. Find the breaking weight and safe strain of a solid cylin- 
drical column of cast-iron whose length is 33 feet, and diameter 
7 inches. 

Use the formula , Txo , and consider the ends flat and fixed. 
8 



square inch of cross sectional area=44'1787 square inches xl ton 
= 44-1787 tons for rounded ends ; but as the ends are flat and fixed 
in the question, we multiply by 3. 

.. the breaking weight = 132-5361 tons. 

Note. By referring to Art. 613 we find that -=4-r. 

T Cl 

Here = 33 feet = 396 inches, and d=7~5 inches; 

7 one 

. . 4^=4 x ^ = 4 x 52-8=211-2, which squared = 44605-44. 

5. Find the breaking weight of a solid cylindrical column of 
cast-iron whose ends are fixed, and length = 25 feet, diameter 
8 inches. 

Use formula 7^2, ..... =300 tons nearly. 



6. Find the breaking weight in tons per square inch of the 
following solid cylindrical columns of cast-iron, whose ends are 
flat and fixed : 

When ^=10 42 Ans. When ^= 40 5'2 Ans. 

d d 

=15 37 M it = 60 2-3 n 

=20 21 it =120 0-6 .. 

n =30 9-2 -, 



STRENGTH OP MATERIALS 



353 



628. TABLE OF CRUSHING WEIGHTS OF A FEW MATERIALS 



Material 


Lb. per 
Sq. Inch 


Material 


Lb. per 
Sq. Inch 


T ffrom 
Iron, cast, . j , 


80640 
143360 


Oak, English, | fro 


6400 
10000 


average, 


107520 


Pine, .... 


6000 


T . . ffrom 
Iron, wrought, i 


35840 
40320 


Teak, .... 
Basalt, Scotch, . 


12000 
8300 


ii average, 


37856 


ii greenstone, 


17200 


Lead, cast, 


6944 


Granite, Aberdeen, 


11000 


Steel, 


336000 


it Cornish, . 


14000 


Steel plates, 


201600 


ii Mt. Sorrel, 


12800 


Tin, cast, 


15008 


Marble, Italian, . 


9681 


Aluminium bronze, . 


129920 


it statuary, 


3216 


Ash, Ba . 


8600 


Sandstone, Arbroath, . 


7884 


Beech, 


7700 


ii Caithness, . 


6490 


Birch, 
Box, 


3300 
10300 


ffrom 
Slate, Anglesea, 1 


10000 
21000 


Cedar, 


5700 


Brick, red, . 


808 


Deal, 


5850 


n fire, . 


1717 


Ebony, 
Elm, 


19000 
10300 


, ffrom 
Portland cement, { 
' I to 


3795 
5984 


Fir, spruce, 


6500 


Glass, flint, . 


27500 


Larch, 


3200 


M crown, 


31000 


Lignum-vitae, . 


10000 


n common, . 


31876 


Mahogany, 


8000 







MISCELLANEOUS FORMULA AND TABLES 
629. WEIGHT AND STRENGTH OF ROPE AND CHAINS 

Rope 

C = circumference of rope in inches, 
L = working load n tons, 
S= breaking strain n 

W= weight of rope in Ib. per fathom. 



STRENGTH OP MATERIALS 



Chains 

D = diameter in eighths of an inch, 
W=safe load in tons ; 



, where 



D 2 

W = = 
y 



d= diameter of iron in inches ; 
85D 2 = weight of chain in Ib. per fathom. 

TABLE OF VALUES OF k, x, y, AND z 



Description of Rope 


fc 


X 


y 


2 


Common hemp, 


032 


18 


18 


6- 


Coir, hawser laid, .... 








131 





ii cable laid, 








117 





St Petersburg tarred hemp hawser, . 


037 


22 


235 


6-35 


ii ii ti cable, 


025 


15 


207 


8-28 


White Manilla hawser, 


045 


27 


177 


3'93 


H ii cable, 


033 


19 


155 


4-7 


Best hemp, ' cold register,' 


100 


60 








it warm, .... 


116 


70 








Iron wire rope, 


290 


1-8 


87 


2-9 


Steel wire rope, 


450 


2-8 


89 


1-91 



EXAMPLE. Find the breaking strain of a 4-inch common hemp 
rope. 



EXERCISES 

1. Find the breaking strain of the under-mentioned size ropes : 
5J" common hemp, 4" steel wire, 6" St Petersburg cable. 

=5-445, 44-8, 5'4 tons. 

2. Find the circumference of a white Manilla cable that will 
stand a strain of 7 tons without breaking. . . = 14'5 inches. 

3. Find the weight of 200 fathoms of 4" steel wire rope. 

= 1-27 tons. 

4. Find the safe load that may be put on the following chains : 
", $"i 1" =li 6$. and 13 tons. 



STRENGTH OF MATERtALS 



355 



630. Breaking Weight of Beams on the Slope. 

* 

L-^-r3 




W= breaking weight for horizontal beam, as found by rule, 

Problem VI. namely, W = j , 

l 

L = span on horizontal line, 
P=span on slope, 

w= breaking weight of beam on slope ; 
WP 



EXAMPLE. Suppose L in the above diagram = 10 feet and P = 12 
feet, and that a rectangular beam of female fir 3 inches square and 
12 feet long is laid on the span P ; what weight applied at the 
centre of the beam would break it ? 

As a horizontal beam, its breaking weight is found by the 
formula, 

4x3x3 2 xll40 4x3x9x1140 



W = 



But 



144 144 

WP_855lb. x!2 
: L ~ 10 



= 8551b. 



= 1026. 



.-. the weight required to break the beam on the slope would 
be 1026 Ib. 

631. Beams unequally loaded. 

Let W= breaking weight for load applied at the centre, as found 
by formula, 



I 

w= breaking weight for beam unequally loaded, 
P = length of the beam or span, 
x and y = distances of load from point of support ; 

WP 2 

w= - 

txy 

Now, supposing that the beam or span P, as shown in the follow- 
ing sketch = 12 feet, that its breadth and depth are each 3 inches, 



356 



STRENGTH OP MATERIALS 



that it is of precisely the same material as the beam mentioned in 
the preceding example, and that x and y are 9 feet and 3 feet 
respectively, find the breaking weight. 





r 


* 1 




'.' 

'.;'' 
'/ 


g 



We first find the breaking weight of the beam, supposing it to 
be strained by a load applied at its centre. 

4x3x9x1140 . , , , . 

= 855 lb., as already found. 



But 



144 

WP 2 855 x 144 2 



= 1140lb. 



4xy 4x108x36" 
632. Pressures on and Reactions from the Supports of 
Beams. If a beam is supported at its extremities and loaded 
at the middle, as shown by the following figure, then not only 
the weight of the beam, but also the load, produces pressures on 
and equal reactions from the supports A and B. 




633. Reactions at A and B, Load at Centre and Weight 
of Beam neglected. 

Let Rj be the reaction at A, and R 2 the reaction at B ; then, 
by taking moments about the point B, we have 
R 1 xAB=WxCB, 



WxL _W 

: 2xL ~ 2 



STRENGTH OF MATERIALS 357 

Also, by taking moments about the point A, we have 
R 2 xBA = WxCA, 

R _WxL = W 

' ' 2 ~ 2xL ~ 2 ' 

It will be at once seen that the upward reactions are each = \V, 
and as action and reaction are equal and opposite, the pressures 
downwards at A and B (due to the load W at the centre of the 
beam) must also be equal to |W. 

If we consider the beam as uniform throughout, and its weight 
as=?, then this force may be supposed to act at its centre of 
gravity, or at a distance = ^L from A to B. The load W also acts 
at a distance L from A to B. Consequently, by taking moments 
about B, we have 



L = 



In the same way, by taking moments about A, we should find 

W w 
that R2=-r- + 7j-; therefore the downward pressure at the points 



A and B must also be equal to 



W w 



EXAMPLE 1. A uniform beam of length L feet, and weight 
w lb., is supported at both ends, and carries a weight W at one- 




fourth of the distance between the supports from one end ; find 
pressures and reactions at each point of support. 



358 



STRENGTH OF MATERIALS 



634. Pressures and Reactions at Supports A and B, due 
to Weight of Beam and a Load at D. The above figure 
represents the data in the question ; for the distance between the 
supports A and B = L, the weight w of the uniform beam acts at 

its centre of gravity C, or at a distance - from each end, and the 



load W acts at D, or at a distance -r from one end. 

moments about the point B, we have 
RjxAB^Wx 

R,x L =Wx gL 



By taking 



x CB, 


(Divide both sides of the equation by L. ) 

.'. the upward reaction at A = R 1 = f\V + \w; and consequently 
the downward jiressure at A, being equal and opposite to the 
upward reaction at A, must also be = fW + ^w. 

In the same way, by taking moments about the point A, we 
have 

Rax BA = W x DA. + W x CA, 

R 2 x L =Wx ^ +iv xL. 

(Divide both sides of the equation by L. ) 

.*. the upward reaction at B = W + $w; and consequently the 
downward pressure at B, being equal and opposite to the upward 
reaction at B, must also be equal to 



EXAMPLE 2. A uniform beam 12 feet long, and weighing 100 lb., 
"^ 

HHI 

2l8Las;<----3n:-f5|^>4 9 FT- ->.10& US, 

\^m.< 




is supported at both ends, and carries a weight of 2 cwt. at a 
distance of 3 feet from one end ; find the pressure on each point of 
support. 



STRENGTH OP MATERIALS 359 

By taking moments round B, we have 



R! x 12' =224 x 9' + 100 x 6' ; . '. R^^IS Ib. 

US 

To find Rg we get 



.-^=224 + 100-218 = 

635. The following formulae will commend themselves as being 
concise and less difficult to remember. 

Taking the above example, and using the following formula 

P : W=CL :L; 

or, expressing this proportion in words as power is to weight, so 
is the counter-lever to the lever. 
P = power, W = weight, CL = counter-lever, L = lever. 
P: W = CL:L, 

P : 224 : : 3 : 12, 
fi79 
P=^f = 56 Ib. +50 Ib. = 106 Ib., 

\- 

and this is the power required to raise the beam from its support B. 
In other words, it is the pressure at the point B. 

By referring to the first figure it will be easily understood why 
50 is added to 56. 

In order to find the pressure at the point A we proceed thus : 
P: W = CL:L, 
P : 224 : : 9 : 12, 



Ifl 

Therefore the pressure at A =218 Ib., and at B 106 Ib. 

It will be observed that the beam is a lever of the third order 
that is, the weight is between the power and the fulcrum and that 
its entire length is a lever. The counter-lever is the distance 
between the weight and the fulcrum. 

Another solution is as follows : 



*---3 FT--*k --.--.-.9 F] 

W=224 Ib. 



360 STRENGTH OP MATERIALS 

Let AB represent the beam, C the point of application of the 
weight 224 lb., AC and BC = distances between supports and 
weight. 

The total distance between supports A and B = 12 feet; AC 
represents & or J of that distance, and BC = A or f of the same. 

Then the pressure at B, neglecting weight of beam = 

and the pressure at A = 

but to each of these results we must add half the weight of the 
beam, viz. 50 lb. ; 

. . 56 + 50 = 106 lb. = pressure at B, 
and 168 + 50 = 218 lb. A. 

EXERCISES 

1. A uniform beam 10 feet long, and weighing 1000 lb., is sup- 
ported at both ends ; a weight of 100 lb. is placed at a distance of 
2 feet from one end ; find the pressure and reaction at each point 
of support. i4: ., ; , ff _'. x =580 lb.; 520 lb. 

2. A 38-ton gun is being supported by a hydraulic jack at the 
breech and a tackle at the muzzle ; the length of the gun is 
16 feet ; the point of application of the jack is 6 feet from the 
centre of gravity of the gun, while that of the tackle is 10 feet ; 
find the pressure on the jack and the strain on the tackle. 

=23f tons on jack ; 14 tons on tackle. 

636. Stiffness of Beams (Tredgold). 

B = breadth of beam in inches, 
D = depth ii ii 

L= length n feet, 

W=load in lb. at the centre. 



IF' ' D 3 

a =-01 fir, 
= '01 ash, 
= -013 beech, 
= -008 teak, 
= -015 elm, 
= '02 mahogany, 
= -013 oak. 
When the beam is uniformly loaded, take '625W instead of W. 



STRENGTH OF MATERIALS 361 

637. Transverse Stress or Bending Moment of Beams. 
A transverse Stress is produced by a force or forces acting 
perpendicularly to the axis of a bar or beam. By axis we mean 
a line passing through the centres of gravity of all the transverse 
or cross sections of the bar or beam. 

(1) Take the case of a rectangular beam where the load is 
applied at the centre, the beam being supported at its ends A, B, 
and let it be required to find the transverse stress or bending 
moment. Then, neglecting the weight of the beam itself, and con- 
fining our attention solely to the load W, we see at once that an 

W . 

upward reaction = is produced at A and at B. Then, by taking 
2i 

moments about C (the centre of the beam), we have : 

W L \VL 

The bending moment, or B.M., at C = -^ x ~v~~I~' 

.. the bending moment of a beam loaded at the centre is -j-. 

Note. This is the maximum bending moment. 

(2) Should the load be uniformly distributed along its length, 

WL 

then the maximum bending moment is 5 

o 

This shows that the bending moment at C, when the load is 
uniformly distributed, is only half the magnitude that it would 
be if the load were concentrated at the centre C. Consequently a 
uniform beam of certain dimensions will bear double the load 
evenly distributed that it can support if the load be concentrated 
at or near its middle. 

(3) Should the beam be supported at both ends, and a con- 
centrated load be placed anywhere between the points of support, 

MMA 

the maximum bending moment is -fW, where m and n are the 

Li 

relative distances of the section from the ends, and the B.M. at 
mn 

L 

any section of the beam = - for a uniformly distributed load. 

EXERCISES 

1. A uniform beam 12 feet long weighs 400 lb., and is supported 
at its extremities ; find the bending moment tending to break 
the beam at a point 3 feet from one end, and the shearing force. 

Bending moment = 450 lb. 

As previously pointed out, the shearing force or load at any 
point or any transverse section of the beam is equal to the resultant 



362 STRENGTH OP MATERIALS 

or algebraical sum of all the parallel forces on either side of the 
point or section. Consequently the forces in this exercise on the 

W 

side of A, where the shearing force is asked for, are -^-, acting 

Z 

W 

vertically upwards at A, and downwards. 

W 400 
.. the shearing force to the left of the section = -j- = -j- = 

100 Ib. upwards. 

W 400 
The shearing force to the right of the section = -j- = = 100 Ib. 

downwards. 

2. A uniform beam 10 feet long weighs 500 Ib. , and is supported 
at its extremities ; find the bending moment tending to break the 
beam at a point 4 feet from one end. ; =600 Ib. 

638. In order to better understand the relation of the ' shearing ' 
and ' bending ' forces, an intimate acquaintance with the science of 
graphic statics is necessary, and although it does not fall within 
the province of this work to enter into the subject at any length, 
the reader's attention is nevertheless directed to its important 
application in connection with theoretical and applied mechanics. 
Graphic statics is the science and art of determining by scale 
drawings the total stresses in the various parts of a structure. 
The forces transmitted through each part of a structure may be 
ascertained in three ways namely, by calculation, the graphic 
method, and by the method of sections. The first method is 
extremely laborious, except in very simple problems, whereas the 
other methods are not only rapid, but at the same time afford 
self-evident means of checking the accuracy of the solution, and 
there is less chance of a grave error than there is in the purely 
analytical method. 

The following works commend themselves : Graphics, by Pro- 
fessor R. H. Smith, M.I.M.E. (Longmans, Green, & Co., London); 
Principles of Graphic Statics, by G. S. Clark (E. & F. N. Spon, 
London) ; Elements of Graphic Statics, by L. M. Hoskins (Mac- 
millan & Co., London). 

639. Problem X. Beam fixed at one end and loaded at 
the other. 

Let CD be a cross section anywhere within the length of the 
beam at a distance of x inches from the fixed end A. 
To find the shearing force at section CD. 






STRENGTH OP MATERIALS 



363 



It will be observed that the only force acting to the right of the 
section is W Ib. 
.*. the shearing force =W Ib. 




It is independent of x, and therefore the same for all such 
sections as CD. 

The bending moment at CD is Wxby its distance from the 
section in inches. 

.-. B.M.=WxBD=W(L-a;)inch-lb. 

The equation is true whatever may be the position of W on the 
beam, so long as L denotes its distance in inches from the fixed 
end, and CD is between W and the support. 

640. Problem XI. Beam fixed at one end and loaded 
uniformly. 




Regard the load on the beam as w Ib. per inch run, and let it be 
required to find the shearing force and bending moment at any 
section CD at x inches from the fixed end. Consider the part of 
the beam to the right of CD as before. The only force is the 
weight of that portion of the load carried by BD ; consequently, 

The shearing force (S.F.) = to x BD=;(L -a:) Ib. . [A]. 



364 



STRENGTH OP MATERIALS 



[C]. 



The moment of that portion of the load on BD with respect to 
CD is the same as if it were all concentrated at the middle point 
of BD. 

.-. the bending moment (B.M.) = 

wxBDx|BD = ^>xBD 2 =iw(L-o;) 2 inch-lb. , [B], 
and S.F. =wL lb., \ 

B.M.=iw>L 2 inch-lb.J ' 

Equations [A] and [B] demonstrate that both the S.F. and B.M. 
disappear when the quantity x = that of L ; and when x=0 we get 
Equations [C]. 

641. Problem XII, Beams supported at both ends and 
loaded at the middle. 







Here we measure x from the middle point of the beam. As W 
is equidistant from A and B, the reactions at those points, Rj and 
112, are equal to each other ; and as their sum is W, we have 

R 1 = R 2 = ^Wlb. 
The force to the right of CD is Rg, and its leverage is BD. 

.-. S.F.=R 2 = JWlb [D], 

and B.M. =R 2 x BD = iW(L-o:) inch-lb. . . [E]. 

Note in this case that the bending moment disappears when 
x=$L, and increases uniformly from this until 03=0; it then 
attains its maximum value, JWL ; or, 

Maximum bending moment = WL inch-lb.. . . [F]. 

EXAMPLE. Take a beam of length L, supported at both ends, 
and let it be loaded at the centre with any load W ; prove that the 
bending moment is greatest at the middle of the beam and equal 
to JWL ; then determine by graphic method the bending moment 
and shearing force at a point 6 feet from one support in a beam 
whose length is 25 feet between points of support, supposing it 
to be loaded with 5 tons at its centre. 



STRENGTH OP MATERIALS 



365 



From Equation [E] we find that, for a beam loaded as in this 
example, the bending moment at any distance x from its centre is 



This is obviously greatest when x0, that is, at the centre. 
Then, maximum bending moment = JWL, and shearing force =JW. 
Consequently, for the numerical values of W and L in the question 
before us, we have : 

Maximum B.M. = -25 x 5 x 25 = 31-25 foot-tons, 
and shearing force = -5 x 5 =2 '5 tons. 

S.F. 
30 



20 



SCALE 
10 




Diagram of B.M. and S.F., 
constructed for the Example. 



At 6 feet from one end the bending moment measures 15 foot- 
tons. This is easily verified by means of the formula for B.M., 
because x 12 '5 - 6 = 6 '5. Consequently the 

B.M. = i x 5 x (12-5 - 6-5) = 15 foot-tons. 



642. Beam supported at both ends and loaded any- 
where. The maximum bending moment with a single concen- 
trated load will always occur immediately under the load, whether 
it be at the middle of the beam or not. 

For the bending moment at any section at a distance x from one 
end is R x x, and this is greatest when x is largest that is, when 
the section is under the load. 

To find the reactions at the supports we take moments about A 
and B, and get R 2 x L = W x m. 



366 



STRENGTH OP MATERIALS 



Consequently R 2 =^W lb., and R X = ^W lb., and these are the 




values of the shearing force (S.F.) to the right and left of W 
respectively 

S.F. to the right = ?W lb., 

, JL 

S.F. totheleft = ^Wlb. 
Li 

Multiplying the first of these equations by n, or the latter by n, 
we get : 

Maximum bending moment = ^ W inch-lb. 



643. Beam supported at both ends, and loaded uni- 
formly. Let the weight, as before, per inch run be denoted by 
iv ; then the total load carried by the beam will be wL lb., and 




the reactions Rj and R 2 will each be \wL lb. Taking the forces 
to the right of the section CD, we have the S.F. = E^-wxBD 
-x) = wx lb. ; 



StttENGTH OF 



36? 



and 



=4;.BD(L-BD) 



' - a; 2 ) inch-lb. 
By plotting the diagrams of S.F. and B.M. we get this figure : 

f-~ 



S.F. curve is 
a straight 
line. 




B.M. curve is a 
parabola with 
vertex below 
the middle of 
the beam. 



The limit values of S.F. and B.M. are : 

Wheno;=L, then S.F. = iwL lb., and B.M. = 0; 

it a;=0, then S.F. 0, and maximum B.M. = |wL 2 inch-lb. 

644. When a beam carries more than one load, or is loaded in 
more ways than one, the simplest and safest way is to consider each 
load separately, without regard to the others, and then combine the 
separate effects so as to obtain the resultant action, as follows. 

EXAMPLE. Draw the bending moment and shearing force 
diagrams for a beam 12 feet long, supported at both ends, and 
loaded with weights of 4 and 6 tons at distances of 3 and 8 feet 
respectively from one end of the beam. 

Measuring distances from the left end of the beam, and considering 
each load separately, we have for the 4 tons to the left of the load 



and to the right of it 



1 =w = x 4=3 tons; 



1 = W = x4 = l ton. 



The maximum bending moment due to this load is : 

, , inn... 3x9 

B.M. 1 = ^-W = -T2~x4=9 foot-tons. 

It occurs immediately under the load. 



368 



Next, taking the 6 tons load, we have to the left of it 

S.F. 2 =w=^x 6=2 tons; 
and to the right of it 

S.F. 2 =yW = j2x6=4 tons. 

The maximum bending moment due to the 6 tons is : 
B.M. 2 =-y-W = -^-x6 = 16 foot-tons. 

By plotting these results we 
get this figure : 




S.F. and B.M. Curves for preceding Example. 

The thiii lines show the actions of the separate loads, and the 
full lines their combined results, obtained by taking the algebraic 
sum of the former. 

The reader should carefully note the necessity of attending to 
the sign of the shearing force. Thus, between the weights we 
have a shearing force of 2 tons, which, on account of its sign, is 
drawn below the base-line ; also a shearing force of 1 ton drawn 
above the base-line. The resultant shearing force between the 
loads is therefore the difference of these, and is drawn on the same 
side of the base-line as the greater of its components. 

The bending moments everywhere along the beam are of the 
same sign ; therefore, to obtain the combined bending moment 



STRENGTH OP MATERIALS 369 

diagram, we have simply to add the ordinates of each separate 
diagram. Thus, to get the total bending moment at the section 
under the 6 tons load, we add FG (viz. that due to the 4 tons at 
that point) to FH (that due to the 6 tons). The result FK is 
therefore the total bending moment at that point. 

It is quite sufficient to do this for the sections under each load, 
and then join each of the points so obtained with each other and 
with the ends of the beam by straight lines. If drawn to scale, 
the bending moment at any other point can then be obtained by 
measuring the corresponding ordinate. 

EXERCISES ON THE BENDING MOMENT AND SHEARING FORCE 
OF BEAMS. 

1. A beam 12 feet long is supported at its ends, and is loaded 
with a weight of 3 tons at a point 2 feet from one end ; find the 
bending moment at the centre of the beam, and also the shearing 
force B.M. = 36 inch-tons ; S.F. =0-5 ton. 

2. A beam is 20 feet in length, and is supported at both ends ; it 
is loaded with 1 ton evenly distributed along its length ; find the 
bending moment at a distance of 7 feet from one end (neglect 
its weight) =5096 foot-lb. 

3. A uniform beam, fixed at one end and free at the other, is 
10 feet long, and weighs 6 cwt. ; it carries two loads, one of 2 cwt. 
at the free end, and the other 4 cwt. at its middle point ; find 
the shearing forces at points 2 feet and 6 feet from the fixed end. 

= 10-5 cwt. ; 6'4 cwt. 

4. Find the bending moment and shearing force at a point 
8 feet from the same support in the beam here mentioned ; it is of 
uniform shape and weighs 15 cwt. , and rests on supports at its ends 
which are 20 feet apart ; it is loaded with three weights of 4, 6, 
and 10 cwt., at distances of 2, 7, and 12 feet respectively from one 
of the supports. . . . B.M. =98 foot-cwt. ; S.F. =3 cwt. 

645. The following formulae will be found useful for determining 
the strength of rectangular beams and girders : 
Let L = length of beam or span in inches, 
B= breadth ,, t> n 

D = depth it H it 

W= breaking weight in cwt., 
* K = coefficient of rupture, 
M = multiplier for deflection (see ' Deflection '). 

* The value of K is the transverse strength of the material expressed in cwt, 
or lb. (see Table). 



370 



STRENGTH OF MATERIALS 





W 


K 


M 


One end fixed, the other loaded, 


KBD 2 


LW 


33 


L 


BD 2 


One end fixed, weight distri- 
buted, ... 


2KBD 2 


LW 


125 


L 


2BD 2 


Ends supported, weight at 
centre, , . . 


4KBD 2 


LW 


02 


L 


4BD 2 


Ends supported, weight dis- 
tributed, .... 


8KBD 2 


LW 


013 


L 


8BD 2 


Ends fixed, weight distributed, 


12KBD 2 


LW 


0032 


L 


12BD 2 



' To find the breaking weight of beams of the following sections, 
use the formula for W given above, but substituting for BD 2 the 
values of V for the section required. I = moment of inertia. 




BD 3 
~ 12 
V=BD 2 



Eectangle 




1 = 



V= 



BD 3 

36 

BD 2 



^---_-\ 

""-"-"* 
Triangle 



BD 3 - bd 3 





= -7854CT 3 
= 4-7CT 2 



< B 

Hollow Rectangle 



Ellipse 



STRENGTH OF MATERIALS 



371 




Circle 

I=-7854R 4 
V = 4-7R 3 




S 4 
: 12 



Square 




Hollow Circle 
I=-7854(R 4 -r 4 ) 
V-W*^ 

v ~ 4 '\ R y 



I 



-6--0 



1 = 



12 




*---R---> 

Semicircle 



V=-38R 3 




e. The safe weight that may be put on beams in permanent 
structures is from \ to \ the breaking weight of the beam. 



372 



STRENGTH OP MATERIALS 



646. STRENGTH AND WEIGHT OF MATERIALS 
TABLE C 



Materials 


Weight of 
a Cubic 
Inch 


Weight of 
a Cubic 
Foot 


Tensile 
Strength 
per Square 
Inch 


Crushing 
Weight 
per Square 
Inch 


Transverse 
Strength 
per Square 
Inch 


METALS 


Lb. 


Lb. 


Tons 


Tons 


Tons 


Aluminium, sheet, . 


096 


166-6 


12 








it cast, 


092 


159-8 


8 








Antimony, cast,. 


242 


419-5 


47 








Bismuth, cast 


353 


613-1 


1-45 








Copper, bolts, . . 


318 


552-4 


17 








it cast, .... 


31 


537-3 


8-4 








it sheet, .... 


316 


548-1 


13-4 








ii wire, .... 


32 


555 


26 








Gold . 


665 


1150 


9'1 








( from 


252 


437 


6 


36 


2 


Iron, cast, . . 1 to 


273 


474-4 


13 


64 


3-4 


ii M average, . 


26 


451 


7'3 


48 


2-6 


( from 


273 


474-4 


16 


16 


3 


,, wrought, \ to 


281 


486-9 


29 


18 


5-5 


ii ii average, . 


28 


485-6 


22 


16-9 


3'8 


it wire, . 








40 










'408 


708-5 


8 


31 





ii sheet 


41 


711-6 


1-5 







Mercury, 


49117 


848-75 













775 


1343-9 











ii sheet, . . . 












_ 


Silver 


377 


653-8 


18-2 








Steel, . . . 


288 


499 


52 


150 





ii plates, .... 






35 


90 





Tin, cast, ..... 


262 


4551 


2 


6-7 







252 


437 


3-3 








ALLOYS 












Aluminium bronze, 90 to 95 per ) 


276 


478-4 


32 


58 


_ 


cent, copper, . . ) 












Bell metal (small bells), . 


29 


502-52 


1-4 








Brass, cast, .... 


3 


524-37 


8 








ii sheet, .... 


361 


526-86 


14 








ii wire, .... 


307 


533-109 


22 








ii 5 copper, 1 zinc, 


3 


525-09 


137 








ii 4 ii 1 ii . 


304 


527-36 


147 








3 ii 1 ti . . 


3 


524-18 


13-1 








ii 2 ti 1 ii 


299 


518-06 


12-5 








-, 1 ii l-i 


296 


513-75 


9-2 








.1 1 M 2 II . . 


298 


517-06 


19-3 








STRENGTH OF MATERIALS 



373 



Materials 


Weight of 
a Cubic 
Inch 


Weight of 
a Cubic 
Foot 


Tensile 
Strength 
per Square 
Inch 


Crashing 
Weight 
per Square 
Inch 


Transverse 
Strength 
per Square 
Inch 




Lb. 


Lb. 


Tons 


Tons 


Tons 


Brass, 1 copper, 4 zinc 


265 


400-13 


1-9 


- 





Gold (standard), 


638 


1106-42 











Gun metal, 10 copper, 1 tin, 


306 


528-36 


16-1 








.. 9 ii 1 ii 


305 


528-24 


15-2 








ii 8 ii 1 ii 


305 


528-05 


17-7 








7 1 


305 


527-89 


13-6 








Pewter, 

















Silver (standard), 


371 


643-72 











Speculum metal, 


264 


464-87 


3-1 








White metal (Babbett), 


263 


456-32 











TIMBER 


















Lb. 


Lb. 


Lb. 


i from 


025 


44 


16000 





1867 


Acacia, j to 


028 


49 




_ 





(from 


'025 


43 


12000 


8600 


2000 


Ash, . . . . { ; to 


027 


47 


17000 


9300 


3000 


( from 


025 


43 


11000 


7700 


1500 


Beech, { to 


025 


43 


22000 


9300 


2000 


(from 


026 


44 


15000 


3300 


1900 


Birch { to 


026 


45 





6000 


1930 


Box, 


046 


80 


20000 


10000 


2445 


Cedar, West Indian, . 


026 


47 


5000 


5700 


1443 


ii American, 


020 


35 








766 


ii Lebanon, . . 1 


017 


30 


11000 


5800 


1300 


Chestnut 


022 


38 


12000 





1770 


Cork 


008 


15 









Deal, Christiania, 


025 


43 


12000 


5850 


1562 


Ebony, 


043 


74 




19000 


2100 


Elm, English, . . - 


02 
021 


34 
36 


13200 
14000 


10300 


782 
1100 


ii Canadian, .... 


026 


45 


_ 





1920 


Fir, spruce, .... 


018 


32 


10100 


6500 


1490 




027 


47 


20000 


4600 




Ironwood, . 


041 


71 






3000 


Greenheart, .... 


041 


71 










Larch, .... 


019 


34 


8900 


3200 


1330 




02 


OE 


10200 


5500 


1660 


Lignum-vitse, .... 


048 


OO 

83 


11800 


10000 


3440 


Lime, ... 


'02 


35 








Mahogany, Nassau, . 


024 


42 








1719 


M Honduras, 


02 


35 


21000 


8000 


1910 


ii Spanish, . 


031 


53 





8200 


1300 


Maple, .... 


025 


42 


10600 




1694 


Oak, African 


035 


62 







2523 


M American, red, . 


03 


53 


10000 


6000 


1680 



374 



STRENGTH OF MATERIALS 



Materials 


Weight of 
a Cubic 
Inch 


Weight of 
a Cubic 
Foot 


Tensile 
Strength 
per Square 
Inch 


Crushing 
Weight 
per Square 
Inch 


Transverse 
Strength 
per Square 
Inch 




Lb. 


Lb. 


Lb. 


Lb. 


Lb. 


Oak, American, white, 


028 


49 


_ 








( from 
u English, t to 


028 
034 


48 
58 


10000 
19000 


6400 
10000 


1600 
1690 


j from 
Pine, red, j ^ 


021 
024 


36 
. 41 


12000 
14000 


5400 
7500 


1200 
1530 


( from 
M white, 1 to 


015 
02 


27 
34 








1229 


u yellow, .... 


018 


. 32 





5300 


1185 


ii Dantzic, .... 


023 


40 


8000 


5400 


1426 


J from 
M Memel, ( to 


02 
021 


34 
37 








1348 


( from 


017 


29 











R,ga,. - - { to 


023 


41 


14000 





1383 


Satinwood 


034 


60 








3200 


( from 

Tpjik ! 

eak, . ' \ to 


026 
031 


46 
' 54 


8000 
15000 


12000 


2110 


STONES, &c. 












Basalt, Scotch 


106 


184 


1469 


8300 





11 Greenstone, . 


104 


181 





17200 





it Welsh, .... 


099 


172 





16800 


_ 


( from 
Chalk,. . . . { to 

Firestone, 


084 
094 
065 


145 
162 
112 


- 


501 


- 


Granite, Aberdeen gray, . 


094 


163 





10900 





u u red, 


095 


165 





' 





u Cornish, . . , . 


096 


166 





14000 





M Mount Sorrel, 


096 


167 





12800 





Limestone, compact, 


093 


161 





7700 





u Purbeck, . 


093 


162 





9160 





M Anglesea, 











7579 





u Blue Lias, 


089 


154 


_ 








u Lithographic, . 


093 


162 











Marble, statuary, 


098 


170 


722 


3216 





u Italian, .... 


098 


170 





9681 





11 Brabant block, 


097 


108 





9219 





u Devonshire, . 











7428 





Oolite, Portland stone, 


087 


151 





4100 





11 Bath stone, . 


072 


123 











Sandstone, Arbroath pavement, 


089 


155 


1261 


7884 





u Bramley Fall, . 


09 


156 





6050 





u Caithness, 


095 


165 


1054 


6490 


857 . 


u Craigleith, 


OSS 


153 


453 


5287 





u Derby grit, 


086 


150 





3100 





u Rd, Cheshire, 


077 


133 





2185 






STRENGTH OF MATERIALS 



375 



Materials 


Weight of 
a Cubic 
Inch 


Weight of 
a Cubic 
Foot 


Tensile 
Strength 
per Square 
Inch 


Crushing 
Weight 
per Square 
Inch 


Transverse 
Strength 
per Square 
Inch 




Lb. 


Lb. 


Lb. 


Lb. 


Lb. 


Sandstone, Yorkshire paving, . 


09 


157 





5714 





Slate, Anglesea, 
ii Cornwall, 
ii Welsh 


103 
09 
104 


179 
157 
180 


| 9600 to 
1 12800 
it 


10000 to 
21000 
M 


j- 19(51 


ii Trap, .... 


098 


170 








- 


MISCELLANEOUS SUBSTANCES 












Asphalt 


09 


156 











i from 
Brick, common, ' i * 


057 
072 


100 
125 











ii London stock, 
n red, . . 


066 
077 


115 
134 





808 





ii Welsh fire, 


086 


150 










ii Stourbridge fire, 


079 


137 





1717 





Cement, Portland,) ( from 
in powder, > ' ( to 


05 
054 


86 
94 


400 
600 


3795 
5984 





Cement, Roman, 
Clay, .... 


057 
068 


100 
119 


185 








Coal, anthracite, 


055 


95 











ii runnel, .... 


046 


79 











ii Glasgow, .... 


046 


80 











n Newcastle, 
Coke 


045 
'026 


79 
46 











Concrete, ordinary, . 


068 


119 











n in cement, . 


079 


137 








_ 


( from 
Earth, - - - | to 

Glass, flint, 


054 
072 
111 


77 
125 
192 


2413 


27500 





ii crown, .... 


091 


157 


2546 


31000 





n common green, 


091 


158 


2896 


31876 





n plate 


099 


172 


_ 


_ 





Gutta-percha, .... 
Gypsum, . 


035 
082 


60 
143 


71 








India-rubber, 


033 


58 








Ivory 


065 


114 








Lime, quick, .... 


03 


53 





_ 


_ 


( from 
Mortar, . . . J to 


049 
068 


86 
119 











ii average, .... 
Pitch, ' 


061 
041 


106 
69 











Plumbago,. .... 


082 


140 











Sand, quartz, .... 
n river, 


099 
067 


171 
117 
























376 STRENGTH OP MATERIALS 

647. Deflection of Beams and Girders in terms of Weight. 
L= length of span in inches, 

W= weight on beam in lb., 

I = moment of inertia (see Table, ' Various Sections of Beams'), 
E= Young's modulus of elasticity (see Table A), 
S = stress in tons per square inch on material of beam or girder, 
d= deflection of beam or girder in inches, 
D = effective depth. 

Note. If W is in tons, the modulus of elasticity E is, say, 8000 
for cast-iron, 13000 for steel, and 11000 for wrought-iron ; but if 
W is in lb., the value of E must be taken from Table A. 

W.L 3 



One end fixed, the other loaded, d= 
n ii uniformly n d= 



3EI 
WL 8 



8EI 

WL 3 

Ends supported, load at centre, d= j^=^ 

4oHil 



ii load distributed, d= q^Tp-T ? 

SL 2 
M uniform stress, d= 

fixed, load at centre, d= 

WL 

M weight distributed, d= v ^ 




Note. The greatest deflection usually allowed in beams is 1 inch 
in 100 feet, or r^ of the span. 



EXAMPLE. Find the deflection of a cast-iron beam 18 feet in 
length, breadth 1 inch, and depth 12 inches, when loaded at the 
centre with a weight of 6000 lb. The ends of the beam are 
supported. 

Take modulus of elasticity = 18400000. 



_ _6000xl0077696 _ _ h 
~ 48EI ~ 48 x 18400000 x 144 



EXERCISES 

1. Find the deflection of a cast-iron beam supported at both 
ends, with a weight of 12000 lb. at its centre; its length = 18 feet, 
breadth 2 inches, and depth 12 inches. = '475 inch. 



STRENGTH OF MATERIALS 377 

2. What weight should be placed at the centre of a cast-iron 
beam of the following dimensions, length 18 feet, breadth 1 inch, 
depth 12 inches, in order to deflect the beam J inch, the ends 
being supported ? ....... =3155 -006 Ib. 

By integrating the formula we get 



3. A plate of steel 18 inches wide and inch thick sustains a 
weight of 1 ton at the centre of a 35-inch span ; find the deflec- 
tion (take =42000000) ....... =-458 inch. 

4. Find the deflection of a wrought-iron bar 2 inches square, 
25 inches span, with a load of 3 tons at the centre. = '065 inch. 

5. Find the breaking weight of a wrought-iron bar 1 inch square 
and 12 inches long, supported at the ends ; the bar is made of the 
best material ......... = 1 '83 tons. 

6. A wrought-iron solid beam of the following dimensions 
length 14 feet, breadth 6 inches, and depth 9 inches ; what weight 
uniformly distributed over it would be sufficient to break it, sup- 
posing its ends are fixed? Take the transverse strength = 3 '8 tons. 

= 131-9 tons. 

7. Suppose the beam in No. 6 exercise to be loaded up to 80 tons ; 
find its deflection ........ = -000008 inch. 

8. Find the weight that should be placed as a central load on 
this beam in order that the usual amount of deflection be not 
exceeded .......... =12873'21b. 

9. Find the breaking weight of a rectangular beam of ash, also 
its deflection, when its ends are fixed and it is loaded uniformly 
(distributed load) ; its dimensions are length 14 feet, breadth 
6 inches, and depth 9 inches. Transverse strength = 19 cwt. 

Breaking weight = 32-9 tons ; deflection = '00005 inch. 

10. Find what load placed at the centre of the beam in No. 9 
exercise would break it, and state the deflection just as the beam 
gave way. Let its ends be supported. = 10 '99 tons ; 4'03 inches. 

11. Find the breaking weight, safe load, and deflection under 
the breaking load of a square beam of pine whose length = 14 feet 
and side 12 inches, when the ends are supported and the weight is 
distributed. Take coefficient K = 13cwt. 

Breaking weight = 26 '74 tons, safe weight =5 -34 tons, deflec- 
tion = 1-3 inches. 

12. Find the breaking weight of a cast-iron beam whose length 



378 



STRENGTH OP MATERIALS 



is 20 feet, arid with a load at its centre ; the ends are supported, 
and the section as shown in the fig. Take the transverse strength 
= 52 cwt. 



D = 12 inches, 
B= 8 
d= 8 .1 
b= 4 

= 554 cwt. 




648. Breaking Weight of Cast-iron Girders. 

D = depth of girder in inches, 
A = area of bottom flange in inches, 
S = span in inches, 
W= breaking weight in tons. 

Supported at both ends, with load at 



centre, 



W = 



25AD 

S 



Supported at both ends, with load dis- 



tributed, 



W = 



50AD 

S 



If the depth = ] J 7 of the span, W = A4'17,\ where the weight is 
it ii =tV I' " W=Ax5, / distributed. 

A 

Area of the top flange if the load is applied on the top = 

o 

Area of the top flange if the load is applied on the bottom 

flange = - 

2D 

Depth at the ends may equal -5 

o 

Safe deflection, tV i ncn f r eac h foot of span, under a test load of 
J of the breaking weight. 

EXAMPLE. Find the breaking weight of a cast-iron girder of the 
above section from the following particulars : Length of span 
10 feet, depth of girder 10 inches, size of bottom flange 6" x 1", 
with a distributed load. 

,, T 50AD 50x7-5x10 . 

W= -S- = 10x12 =3" 



STRENGTH OP MATERIALS 



379 



EXERCISE 

Find the breaking weight of the following girders of the same 
section with a distributed load. 



Span in Feet 


Depth in Inches 


Size of Bottom 
Flange 


Answer 


15 

20 
25 


15 
20 
25 


8x1$ 
10x1^ 
13xl| 


50 tons. 
62-5 n 
94-79 


30 
35 


30 
35 


15x2 
17x2 


125-0 
141-6 



649. Deflection of Iron and Steel Girders, ends supported, 

The usual allowance in American bridges is T5 V<T after the girder 
is set. 

S = span in feet, 

P = stress on the metal by any load in tons per square inch, 

E = modulus of elasticity in tons = say 10000 for iron and 13000 

for steel, 

D = effective depth of girder in feet, 
d deflection of girder in inches ; 

SK. (For value of K, see Table, p. 380.) 



EXERCISES 

1. Find the deflection of a Avrought-iron girder whose effective 
depth = 3 feet and span 30 feet; the stress on the metal = 5 tons 
per square inch. . . ...... = -45 inch. 

2. What deflection would a steel girder have with a span of 
50 feet, its effective depth being 4 feet, and stress = 6 tons per 
square inch ? . ....... = -87 inch. 

It will be noticed that the deflection is too great, the girder 
being badly proportioned. 

3. Find the deflection of a steel girder under a stress of 8 tons 
per square inch, the span = 32 feet, and the effective depth of the 
girder = 4 feet ......... = -47 inch. 



380 



STRENGTH OF MATERIALS 







* 




GC * 






H: 


OOOOOOOOOOOO<^ 


^ G^ 

O O 














I 


-B 




11 




f Girder to 


* 


OO50OIT-CDIOJCOOOOOO5 
(NCOt i i US OS i iCOCDO5(NOt- 


O5 O5 




"S. 

Q 


c 


~~t CO 1T" O 1 ^ GO i~^ l *O GC "~H CC CO 


O3 (N 

F-H <N 






> 








02 
P 


3f Effect 


-B 


liillllliiill 





<3 


2 











a 

& 


-S 


lilililliiill 


CO GC 

O 


w 










H 




H. 


lilllllillill 


O 






* 


I i 1 1 i 1 i 1 1 i 1 1 1 


s 


















WMM 


KW 




1 

II 

OQ 


t. a 


ip ip 


.. 




. 

*: 


! 


c 13 









I 


"-i 02 





STRENGTH OP MATERIALS 381 

650. To find the deflection of a beam or girder of uniform 
section, we have the following formula: 

W I 3 W I 3 



where W= weight in tons, 
= span in inches, 
(MR) = moment of resistance of cross section in inches, 

d= depth of beam or girder, or, rather, twice the distance 

of the fibres most strained from the neutral axis, 
E = modulus of elasticity. 

E = -00018 for cast-iron, = -001 for teak. 

E= -00010 M wrought-iron, = '002 M oak and pitch-pine. 

E= -00008 ,, steel, ='003 fir. 

To be in a position to use the above formula, we must first 
determine the value of (MR) for each section. For a rectangular 
section, (MR), the moment of resistance or modulus of section, 

bd? 
= -=-. For other sections, see Table at the end of this subject. 

651. Determination of Moment of Inertia. 

I = moment of inertia, 

N = distance of neutral axis from lower edge of section, 
H = height of any particles from lower edge of section, 

d= distance of any particles from the neutral axis, 

B = breadth of section at any height H, 

S=sum, 

A = difference. 

1= - 7; - , if the neutral axis be in the centre and the figure 
be symmetrical ; if not, 



= 2BAH. 



The neutral axis, for all practical purposes, passes through the 
centre of gravity of any section. 

The following example demonstrates the application of the 
formula given. 

The more closely the section is divided into minute rectangles, 
the more accurate will be the result. 



382 



STRENGTH OP MATERIALS 



H 


H2 


H3 


AH 


AH2 


AH 


B 


BAH 


BAH2 


BAH3 

















l 










4 


16 


64 


4 


16 


64 / 


10 


40 


160 


640 


6 


36 


216 


2 


20 


152 


6 


12 


120 


912 


18 


324 


5832 


12 


288 


5616 


1 


12 


288 


5616 


20 


400 


8000 


2 


76 


2168 


3 


6 


228 


6504 


24 


576 


13824 


4 


176 


5824 


7 


28 


1232 


40768 
















98 


2028 


54440 
















= 2BAH 


SBAH 2 


ZBAH 3 
















= A 








SBA(tP)_2028 

2A ~ 196 ~ 4b ' 



= 18146-6-10489-8 
= 7656-7. 

.'. Height of neutral axis from lower edge 
of section = 10'346=N, and moment of inertia 
= 7656-7. 

652. Now, the moment of resistance (MR) or 
modulus of the section is found as follows : 
(MR) = moment of resistance, 
I = moment of inertia, 

N = height of neutral axis from farthest edge of section, 
M re = modulus of rupture, 
K= coefficient of fracture ; 

ft V T TUrreT 

(MR) = ^4^- (MR) = f 



N 

The modulus of rupture M re is found by multiplying the trans- 
verse strength of the material by 6. For transverse strength, see 
Table, 'Strength and Weight of Materials.' 

EXAMPLES. 1. To find the deflection of a 60-lb. double-headed 
rail, 4J" deep, under a load of 1 ton at the centre of a 33" span. 
The moment of resistance is 6'7. 



Here 



STRENGTH OP MATERIALS 

W = 1-0267 tons, Z = 33 3 = 35937 inches, 
(MR)=6'7, </=4-5 inches, 
W/ 3 1 -0267 x 35937 36896 '5 



383 



24x6-7x4-5 723'6 
50-9 x E = 50-9 x "0001 = -005 inch. 

If the neutral axis passes through centre of section, mean of 33 
experiments (Baker) = -005 inch. 




2. Find the deflection of the 84-lb. rail shown in the fig. when 
loaded with 2000 Ib. at the centre of a 60-inch span ; find also the 
strain on the extreme fibres, the depth being 4J inches. 

For this particular section (MR) 
= 2-49x3-5. 



Note. The strain on the ex- 
treme fibres is given by the 


" tl 

i 


1 


* 


---ly-io 
17-75" 


tons per square inch, W in tons, ^i~ 
I in inches. 
W/ 3 93 x 216000 


12-13* 


I 


1 


>' 


24(MR)d 24 x (2 -49 x 3 -5) x 4 -5 

OlQvlQv iim>] .AO1 > 


f NCU rA L 


| 




AX/5 


Neutral axis assumed passing 
through centre of section. x . 
\V7 93x60 10-95 


\ 

7 12 " 




1 


f i ' 1 T" 


' 4(MR) 4x8-715 8'715 


& n " 

diLIZ " 


i2 




iii^..o? 


= 1-25 tons per square inch. 
3. To determine the moment of 


resistance 


1 





bending of 



the section of the cast iron girder as shown in the above fig. 
The maximum safe tensile and compressive stresses are 2 



384 



STRENGTH OF MATERIALS 



and 7 tons per square inch respectively. Its dimensions are 
as follows : Top flange, 4" x 1J" ; bottom flange, 12" x If" ; web, 
16" x 1". 

Determine the moment of resistance if the girder is 20 feet long, 
and is supported at its two ends. Find the greatest safe load 
which it will carry when uniformly distributed along its length. 

We first find the position of the neutral axis thus : 



H 


HS 


H3 


AH 


AH2 


AH3 


B 


BAH 


BA(H2) 


BA(H3) 

















) 










1-75 


3-06 


5-35 


1-75 


3D6 


5-35 f 


12 


21 


3672 


64-2 


17-75 


315-06 


5692-31 


16-00 


812- 


5586-96 


1-5 


24 


468-00 


8380-44 


19-25 


370-56 


7133-28 


1-50 


55-5 


1540-97 


4 


6 


222-00 


6163-88 
















51 


726-72 


14608-52 














= 


2BAH 


2BAH2 


2BAH3 
















= A 







XT SBAH 2 726-72 _ __ . . 
N= -2^ ^02- = 7'12 inches, 



=2284-31. 

653. The neutral axis is of fundamental importance in the theory 
of beams and girders, because it is the fulcrum about which both 
the bending and resisting couples act. 

Should E not be the same for tensile and compressive stresses, 
then the neutral axis will not pass through the centre of the area, 
but will lie to the side having the greater value of E. 

The greatest stress comes on the fibres farthest from the neutral 
axis, and is the principal effect to be considered in the question of 
strength. 

Now, the moment of inertia for the whole section is found to be 
2284-31. 



For tension 



a. ^ i l 2284-31 oon 
the modulus = = _ = 320. 



2284-31 



= 188. 



it compression M ^^ 

a; 2 12-13 

Tensile stress =2*5 tons per square inch. 

Compressive n =7'5 n n 

We must therefore take the lower value of the two resisting 
moments in fixing the load to be carried by the girder. 



STRENGTH OF MATERIALS 



385 



These are 320 x 2'5 = 800 inch -tons, 
and 188x7-5 = 1410 

.. bending moment = resisting moment =800 inch-tons. 

The girder will therefore carry safely a uniformly distributed 
load given by the equation on bending moments, Art. 649 
namely, ^L 2 = 800. 

8x800 __, , 

: = 26| tons. 



W = 



20x12 



800 



This will make the maximum compressive stress r-^=4'255 tons, 

loo 

instead of 7 '5 as given; showing that' the girder is not well 

designed. 

In a properly proportioned girder we should have : 
Modulus for tension x tensile stress per square inch 

= M compression x compressive 



EXERCISES 

1. Find the breaking load (distributed) of the girder mentioned 
in the last example =84 - 2 tons. 

2. Find the breaking weight 
of the cast-iron girder in the 
accompanying fig. when loaded 
at the centre of a 30- foot span. 
What would its deflection be 
when carrying a load of 45 tons 
at its central point ? Top flange 
= 5" x 2", web = 26" x 1 -5", bottom ' 
flange = 15" x 2". Also state its 
moment of inertia, and the height 
of neutral axis from lower edge 
of section. Total depth of girder 
= 30". 

Moment of inertia = 9068 -275, 
Neutral axis =11 -45 inches, 
Breaking load =62-5 tons, 
Deflection = -58", or "76". ' 

For the deflection, see p. 387. 

3. If the tensile and compres- yE^^^^."*?; """^^3 p* 
sive stresses are limited to 1'5 

tons and 9 tons respectively in the girder mentioned in the second 
exercise, find the greatest safe load that the girder will carry 





I 


1 


r -'ica 




I 






18-55 










1 






. NEUTKAL 




2 


6' 
AXIS 










11-45 












V 


,,, - - 



386 



STRENGTH OF MATERIALS 



- // 

i*f"'"".::ft ." 



when loaded at its centre. What will the maximum compressive 
stress be? = 13-2 tons; 2 -4 tons. 

4. Determine the load that may safely 
be distributed on the section of wrought- 
iron as shown in the fig. if the span 
= 20 feet, the tensile and compressive 
stresses being limited to 5 tons and 3'5 
tons respectively. . . . =737 lb. 

5. If the position of the iron was re- 
versed, and the other conditions the same 

, as in the last question, what would be 
the safe load? . . . . =590 lb. 
6. Find the position of the neutral axis and moment of inertia 
for the following sections of f iron. 





W = width; D = depth; i=thickness, 

in inches 

when 4 4 I 

3 4 | I' . 

4 3 4 

3i 34 4 

34 34 i . 



I N.A. 
(Moment of Inertia) (Neutral Axis) 

. 5-5641 2-816 

. 5-0485 2-6731 

. 2-4234 2-1731 

. 3-635 2-4423 

2-865 2-487 



654. Should it be required to find the deflection when E and I 
are known, one of the formulae ('Deflection in Terms of Weight') 
may be employed when W has been ascertained. 

If M, the bending moment or moment of resistance, has been 
found, then the deflection may be determined by formula already 
given, or by one of those found in Table, ' Strength and Stiffness 
of Beams,' at the end of the subject. 

Any difference in the results will be due to the value of E, as 
already pointed out. Another point the student or reader will 
do well to notice is this : for the sections in the Table, the neutral 
axis passes through centre of gravity of each section. 



STRENGTH OF MATERIALS 387 

655. In the formula for deflection of beams and girders of uniform 

"W7 3 
section, namely, oZTM"' *^ ie momen * f resistance must 



first be ascertained. In the case of a cantilever of rectangular 
cross section loaded at the outer end, the moment of resistance 
(MR) = KbcP, where (MR) = resisting moment in inch-lb., and K = a 
constant number found by trial depending upon the nature of the 
material of which the beam is composed. It has been assumed 
that the beam or girder is of uniform section, so that I, the 
moment of inertia, is constant ; the more general cases where I 
varies being rather beyond the scope of this work. 

On the whole, it would be safer to adhere to the formulae 
containing I as a quantity ; but before closing the subject, the 
following examples will present the application of the formulae 
more fully. 

Taking the second exercise, let it be required to find the 
deflection of the girder in terms of the maximum bending moment, 
and also by formula as below. 

M.1 2 

(1) Deflection in terms of M (see Table) = T?^T, 

CjL 
W/ 
M =~ (see Prob. XII. p. 364), W in lb., I in inches, 

W = 45 tons = 100800 lb., 
/ = 30 feet =360 inches; 

...WlOOSOOxSK^ 

4 4 

lxMxJ2_ 1 x 9072000 x 129600 _ _ Q . 
12lTE^I ~ 12 x 18400000T9068 : 275 ~ 
This answer agrees with that already found in terms of W. 

Wx/ 3 

(2) The deflection (see formula ^-. /1t/rp . -,. E) 

24 x (MR) x d 

45 x 46656000 

= oi 1100 g nr~^ x '0001 8 = -71 inch. 
24 x 1186-5 x 18-55 

Here d= distance of fibres most strained from neutral axis = 
18'55 inches. 

We will now explain how (MR) has been obtained, for. in 
unsymmetrical sections there are two values of the modulus of 
the section to be considered. 

The ratio - is usually noted by z ; y is any distance above or 
below the neutral axis. 



388 STRENGTH OF MATERIALS 

The modulus for tension =z t = - . . . _ = 791 '9. 

y 11'45 

I 9068-275 
it it compression =z c =-= -.O.KK =488-8. 

Now, if the greatest permissible tensile and compressive stresses 
were limited to 1 '5 tons and 9 tons respectively per square inch 
and these, as already stated, are the working stresses for cast- 
iron then the tensile stress = 791 '9 x 1-5 = 1186 '5 inch-tons, and the 
compressive stress = 488 -8 x 9 = 4399 '2 inch-tons. We must there- 
fore take the lower value of the two resisting moments (MR) in 
order to determine the load to be carried by the girder. 
.-. BM = (MR) = 1186 -5 inch-tons. 

The girder will therefore carry a central load given by the 
equation WL (see Prob. XII. p. 364). 
r 4x1186-5 



This makes the maximum compressive stress -^ -^ = 2 '4 tons, the 

4oo o 

difference between the answers of the two formulae being T W of 
an inch. 
Note. Should the moment of resistance be calculated by 

6KI WP 

means of the formula (MR) = -^- (p. 382), then , . E 

3 
becomes 



-T^T . . 
4M.h 

EXERCISES 

1. Find the greatest load that may uniformly be distributed on a 
cast-iron girder, having top and bottom flanges united by a web of 
the following dimensions. Width of upper flange 3 inches, of 
lower flange 9 inches ; total depth 12 inches ; thickness of each 
flange and of the web 1 inch ; distance between the points of 
support 10 feet. The greatest admissible stress in the compression 
flange is 3 tons per square inch, and that in the tension flange is 
1 tons per square inch. ...... =8 '8 tons. 

'2. Find the deflection of this girder by means of formula 

MJ 2 
fa . =-, supposing it to be loaded at the centre with a weight of 

5 tons. Take 1 = 398. . . . . . . . ='05. 

W/ 3 
3. Find the deflection by formula T/' ^" ' ' = 6 ' 



6 KI 

4. Find the deflection when (MR) is =-. . . . ='06. 



STRENGTH OF MATERIALS 389 

5. A uniform beam of oak, 10 feet in length, 15 inches deep, and 
10 inches wide, sustains, in addition to its own weight, a load of 
5000 Ib. placed at its centre ; find the greatest bending moment 
and the greatest stress in the fibres. 

Take the specific gravity of oak as 0'934. 

Here the greatest bending moment takes place at the centre of 
the beam, and is made up of two parts (1) that due to the beam's 
own weight, which is uniformly distributed along its length ; and 
(2) that due to the 5000 Ib. concentrated at its middle. 

The greatest stress in the fibres is ascertained by formula 
,. B.M. or (MR) 
/=- j- 5 - -*y- 

/ stands for either the tensile or compressive stress, at any 
distance y above or below the neutral axis. 

B.M. = 159072 inch-lb. ; greatest stress = 424'l Ib. per square 
inch. 



656. To find the strength of thin wrought-iron girders. 
The formulae for the moment of resistance are very simple, for 
here the flanges are thin in comparison with their distance apart, 
the bending resistance of the web being disregarded as a provision 
against the shearing force acting at the section. 
Let A t = area of flange in tension, 

A<;= ii n compression, 

H = distance between centres of flanges, 
,/i=mean stress in tension flange, 
fe= 11 ii compression flange. 

Distance between centre of tension flange and the neutral axis is 



The moment of inertia of the flanges with respect to the neutral 
axis is 



, M 

f=j 

f 




390 STRENGTH OF MATERIALS 

Hence f t = 

Similarly, 

EXAMPLE. A wrought-iron girder of I section has a top flange 
of 9 square inches in sectional area, and a bottom flange of 8 square 
inches. The distance between the centres of gravity of the flanges 
is 12 inches, and the ends of the beam rest on abutments 16 feet 
apart. The girder is loaded uniformly with a load equal to 1 ton 
per lineal foot (including the weight of the girder). What would 
be the mean stress per square inch on the metal in each flange 
at the dangerous section ? 

The resistance of the web to bending is neglected. 

By ' dangerous section ' is here meant the middle section of the 
girder, where the maximum bending moment occurs. 

Maximum B.M. =i( T \) x (16 x 12) 2 = 32 x 12 inch-tons. 

. . mean stress in tension flange 

32 x 12 
ft rTr =4 tons per square inch ; 

O X 11^ 

and mean stress in compression flange 

32 x 12 

/ c = =3 '55 tons per square inch, 
y x \2t 

In compression, iron may be strained to 4 tons per square inch. 

In tension, iron may be strained to 5 tons per square inch. 

The regulations of the French department ' Fonts et Chaussees ' 
allow 3 '81 tons per square inch. 

Steel may be strained to 6 tons per square inch in tension and 
compression. 

657. Collision of Bodies. 

W = weight of one body, 

V = velocity of one body before impact, 

Y= ii ii ii after n 

K = coefficient of restitution of the one body, 

w = weight of the other body, 

v= velocity of the other body before impact, 

y= n n n after 

k coefficient of restitution of the other body, 

=0 for a non-elastic body, 1 for a perfectly elastic body. 
The ideal elastic body is one for which the coefficient of restitution 



STRENGTH OF MATERIALS 



391 



is unity, and should such a body strike a plane surface, it would 
rebound at an angle equal to that at which it struck the plane ; in 
other words, the angle of incidence () = the angle of reflection (6). 



a: 



\b 



Note. Practically this is never true, since no body is known 
which has its coefficient of restitution equal to unity. 
For notation, see ' Collision of Bodies. ' 



Conditions 


Non -elastic Bodies 


Elastic Bodies 


One body in 
motion, . 

Bodies moving 
in the same 
direction, 

Bodies moving 
in contrary 
directions, 


wv 


WV(I+*) 


y ~ W+w 
Y V(W-K>) 


f W + w 
WV + wv 


W+w 

WV(I + k) + v(w-JcW) 


W + w 
v V(W-Kw) + vw(l + K) 


y ~ W + w 
WV-wv 


W + w 
WV(I + k)-v(w-kW) 


W + w 
v V(W-Kw)-wXI + K) 


y W + w 


W + w 



When the bodies are inelastic their velocities after impact will 
be alike, or Y = y. 



392 



MOMENT OF INERT 
The plane of bending is supposed perpendici 



Form of Section 



Area of Section 










A 

S 2 



bh 



MODULUS, &c., OF SOME SECTIONS 
to plane of paper, and parallel to side of page 



393 





Moment of Inertia of Section 
about Axis through 
Centre of Gravity 


Square of Radius 
of Gyration of 
Section 
I 
A 


Modulus of Section 
I 

y 

2/=any dist. above or below N.A. 


I 








S 4 


S 2 


S 3 




12 


12 


6 


bh 3 


h? 


bh? 




12 


12 


6 


S 4 -s 4 


S2 + S 2 


i( 84 -*^ 


12 


12 


*\ S ) 











0491rf 4 


cP 
16 


0982^ 




i 
i ; 
[ 














0491(D*-rf) 


TP-cP 


.^(D 4 -^ 


16 


82 \ D / 



394 



MOMENT OF INERTIA, 

The plane of bending is supposed perpendicular 



Form of Section 



Area of Section 



H 




EU-bh 



-f- 





K----B --- * 



EH-bh 



MODULUS, &c., OF SOME SECTIONS 
to plane of paper, arid parallel to side of page 



395 





Moment of Inertia of Section 
about Axis through 
Centre of Gravity 


Square of Radius 
of Gyration of 
Section 

I 

A 


Modulus of Section 
I 

y 

y=any dist. above or below N.A. 




I 

BH 3 -6A 3 


1 /BH 3 -6A 3 \ 


BH 3 -6A 3 




12 




6H 




BA 3 + 6H 3 




BA 3 + 6H 3 




12 




6H 




(BH 2 - M 2 ) 2 - 4BH6A(H - A) 2 




(BH 2 - M 2 ) 2 - 4BH6A(H - A)- 




12(BH - bh) 




6(BH 2 + M 2 -26HA) 



396 



STRENGTH AND STIFFNESS OF 

/ stands for either the tensile or compressive stress 



Manner of Supporting and Loading 



Maximum 

Bending 

Moment 

M 




Cantilever Loaded at End 



W.I 




Cantilever Loaded Uniformly 



Wl 
"IT 




Supported at both Ends. Loaded at Centre 



Wl 



BEAMS UNDER A LOAD OF W LB. 

at any distance y above or below the neutral axis 



397 





Deflection in 
terms of 
W 


Deflection in 
tenns of 
M 


Deflection in 
terms of Stress 


Relative Stiffness 
under same Load 


i W/ 3 


, M/ 2 

*-^r 


j.2? 

i- % 


A 


l ' El 


, WP 
**ET 


, M^ 
*'ET 


^ 

*Ey 


i 


. WP 
"'El 


. M.P 
TV ET 


A > 

"'% 


i 



398 



STRENGTH AND STIFFNESS OF 
/ stands for either the tensile or compressive stress 



Manner of Supporting and Loading 



Maximum 

Bending 

Moment 

M 




Supported at both Ends. Loaded Uniformly 



Wl 




Ends Fixed. Loaded at the Centre 




Ends Fixed. Loaded Uniformly 



Wl 

8 



12 



BEAMS UNDER A LOAD OF W LB. 

at any distance y above or below the neutral axis 



399 





Deflection in 
terms of 
W 


Deflection iu 
terms of 
M 


Deflection in 

terms of Stress 


Relative Stiffness 
under same Load 


S W* 


v|g 


*f 


1 


***' El 


"'El 


. WP 
r ' El 


, Mr 8 
"ET 


,/P 
A % 


4 


t WP 

"^'ir 


. MP 

*'ir 


a/* 2 

*si 


8 



400 PROJECTILES AND GUNNERY 



PROJECTILES AND GUNNERY 

658. The subject of projectiles, considered in a practical 
point of view, treats of the methods of determining by 
calculation various circumstances belonging to the motions 
of bodies projected in the atmosphere. 

This subject is divided into two parts namely, the para- 
bolic and flat trajectory theories. 

I. THE PARABOLIC THEORY OF PROJECTILES 

In the parabolic theory several hypotheses not strictly 
correct are made ; but only one of them can lead to any 
sensible error in practice, though in some cases the error is 
comparatively small. This last hypothesis is, that there is 
no resistance from the atmosphere to the motion of a pro- 
jectile ; and the other two are, that gravity acts in parallel 
lines over a small extent of the earth's surface, and that its 
intensity is constant from its surface to a small height above 
it. The parabolic theory applies to all ordnance with high 
angle fire and low muzzle velocity, such as howitzers and 
mortars. 

659. Problem I. Of the height fallen through by a body, 
the velocity acquired, and the time of descent, any one 
being given, to find the other two. 

Let A=the height fallen through, 

v= n velocity acquired, 
t= n time of descent, 
#=32-2 feet; 

then h=W=W*=^> 

V=gt = *J2gh = , 

t-- =>J = 
~9 ~ ff ~ v' 



PROJECTILES AND GUNNERY 401 

These relations of h, v, and t are proved in treatises of theo- 
retical mechanics. Any two of these three quantities are said to 
be due to the other ; thus the velocity acquired by falling from a 
given height is said to be due to that height, and so of the other 
two quantities. The acquired velocity is also called the final 
velocity. The number 32'2 is the velocity in feet that a body 
acquires in falling during one second. The velocity with which a 
body is thrown upwards or downwards is called its initial velocity. 
Should the body be thrown upwards the force of gravity imparts 
a negative acceleration, and if thrown downwards it imparts a 
positive acceleration. 

Therefore the sign of g is + in the first case and - in the second 
case. 

In solving the following exercises, such a formula is to be chosen 
in each case as contains the elements concerned that is, the 
quantities given and sought. 

EXAMPLES. 1. What is the velocity acquired in falling 
10 seconds? 

v=gt = 32 -2x10 = 322. 

2. What is the height fallen through in 5 seconds ? 
h = lgt 2 =ls x 32-2 x 5 2 =402'5. 

EXERCISES 

1. What velocity would be acquired in falling 120 feet? 

= 87-9 feet. 

2. Kequired the height through which a body must fall to acquire 
the velocity of 1500 feet per second =34938 feet. 

3. In what time will a body acquire the velocity of 900 feet ? 

=27*95 seconds. 

4. In how many seconds would a body fall 27000 feet? 

= 40 '95 seconds. 

660. When a body is projected in any direction except that of a 
vertical line, it describes a parabola. 

Thus, if a body is projected in the direction PT it will describe 
a curvilineal path, as PVH, which will be a y *, 

parabola. \ / 

M X \i 

661. The velocity with which the body is 
projected is called the velocity of projection. 

During the time that the projectile would 
be carried, by the velocity of projection con- 
tinued uniform, to T, it would be carried by the force of gravity 




402 PROJECTILES AND GUNNERY 

from T to H. But the distance PT is evidently proportional to 
the time, whereas TH is proportional to the square of the time. 
Since (Art. 659) h = ^gt 2 !Q-\t-, therefore TH is proportional to the 
square of PT. And the same is true for any other line drawn, 
as TH, from a point in PT to the curve ; and this is a property 
of the parabola. 

662. The velocity of projection is that due to a height equal to 
the distance of the point of projection from the directrix of the 
parabola described by the projectile. 

Or, the velocity at P is that acquired in falling down MP, AM 
being the directrix. 

663. The velocity in the direction of the curve at any other 
point in it is equal to the velocity due to its distance from the 
directrix. 

The velocity at any point, as H, is that due to AH ; and if a 
body were projected with that velocity in the direction of the 
tangent HG, it would describe the same curve HVP, and on 
arriving at P, would have the velocity due to MP. 

664. The height due to the velocity of projection is called the 
impetus. 

Thus MP is the impetus. 

665. The distance between the point of projection and any body 
to be struck by the projectile is called the range, and sometimes 
the amplitude. When the range lies in a horizontal plane it is 
called the horizontal range. 

Thus, P being the point of projection and H the body struck, 
PH is the range, and PQ the horizontal range. 

666. The time during which a projectile is moving to the object 
is called the time of flight. 

667. The angle contained by the line of projection and the 
horizontal plane is called the angle of elevation. 

Thus TPQ is the angle of elevation. 

668. The inclination of the horizontal plane to the plane passing 
through the point of projection and the object is called the angle 
of inclination. 

Thus HPQ is the angle of inclination. 

The range of a projectile may be either on a horizontal or an 
oblique plane. 



PROJECTILES AND GUNNERY 403 

669. Projectiles on Horizontal Planes. The following for- 
mulae afford rules for calculating the impetus, range, velocity of 
projection, time of flight, and elevation : 
Let h = the impetus MP in feet, 

v= a velocity of projection in feet per second, 

t= ii time of flight in seconds, 

r= M horizontal range = PH, 

e ii angle of elevation = TPH, 

/= M greatest range, 

h'= height = VD; 

p 
then 7t=5- by Art. 659 ; r =2h sin 2e ; 

47 
v=\/2gh Art, 659; r'=2/t; 

2/4 
t = 2 sin e V ; h' h sin 2 e. 

9 

Let PT be the line of projection, and PVH the curve described. 
On PM describe a semicircle MBP, and from its 
intersection with the tangent PT in B, draw BC 
parallel to the axis, and BA perpendicular to 
the impetus MP. Then AB = PC = |PH = r, and 
BC = iTH, and VD = BC. Draw the radius OB, 
then (Eucl. III. 32) angle BPC or e = BMP 
= POB, or POB = 2e. Now, 

AB/OB = sin BOP, 
or i^/i^ = sin 2e ; 

hence ? = \h sin 2e, and r 2/t sin 2e. 

Again, the time of flight is just equal to the time of describing 
PT uniformly with the velocity of projection, or the time of falling 
through TH by gravity. Now, if TH = h", 

Hr/PH = tan TPH, or h"jr=t&n e ; 

...,2 sin e. cos e. sin e 

hence h = r tan e = 2/i, sin 2e tan, e 2/t - : 

cos e 

and therefore h" = <lh sin 2 e. 

But if t is the time due to h", then 

.2A" ,8h sin 2 e . ,2h 
=2 sin e v 




ff 9 9 

which is the expression above for t. 

Again, VD = BC = iTH, or h' = $h" = h sin 2 e. 

When e = 15 or 75, sin 2e = sin 30 = \, and r=h. 

The greatest value of r or 2/t sin 2e, for a given value of h, is 
when sin 2e is a maximum or 2e = 90, and e = 45; for then sin 2e 
= 1 , and r = 2k. 



404 PROJECTILES AND GUNNERY 

670. Two elevations, of which the one is as much greater than 
45 as the other is less, give the same horizontal range. 

For if these angles of elevation are 45 + d and 45 -d, then for 
these elevations 2e is 9Q + 2d and 90-2<f, which are each other's 
supplements; and hence sin (90 + 2e?) = sin (90 -2d), and the two 
values of r are 

r=2h sin (90 + 2d), and r=2k sin (90 -2d), 
which are equal. 

671. Problem II. Given the velocity of projection, or 
the impetus and the elevation, to find the range, the time 
of flight, and the greatest altitude of the projectile. 

v 2 
The formulae to be used are h ^-, r2h sin 2e, t 2 sin e 

ty 

.2k 2v . , ,, , . 

V = sin e, and h h sure. 
9 9 
Or, r, t, and h' may sometimes be more easily found by 

logarithms ; thus 

Lr = L2A + L sin 2e- 10. 

L* =L2v+ L sin e-(lQ + Lg). 

LA' = L& + 2Lsin e-20. 

EXAMPLE. A ball was discharged with a velocity of 300 feet 
at an elevation of 24 36' ; required the range, the time of flight, 
and the greatest altitude. 

* 300* _ 

- ' 



r=2h sin 2e = 2795 x -756995 = 2116, 

n*. (IdO 

t = sin e = -~x -4162808 ^7 '76 seconds. 
9 ' 



EXERCISES 

1. A shell being discharged at an elevation of 28 30', and with a 
velocity of 230 feet in a second, what is the impetus, the range, 
the time of flight, and the greatest elevation ? 

/t = 821, r=1378, h' = 187, and * = 6'82 seconds. 

2. The impetus with which a cannon-ball is fired is = 3600, and 
the elevation = 75, and the elevation of another fired with the 
same impetus was = 15 ; required the ranges. . . . =3600. 

3. Required the time of flight of a shell fired at an elevation of 
32, with an impetus of 1808 feet. . . . =1T23 seconds. 



PROJECTILES AND GUNNERY 405 

672. Problem III. Given the range and elevation, to find 
the velocity of projection. 

7* 

From r = 2h sin 2e is found h = =- -^-, the 1st formula; and 

2 sin 2e 

v* 
from h=Q~ is derived v=*J2gh, the 2nd formula. 

Or by logarithms 

L2h = 10 + Lr - L sin 2e, 
and Lv = |(L2gr + LA). 

The greatest altitude and time of flight are found as in last 
problem. 

EXAMPLE. A ball was projected at an elevation of 54 20', and 
was found to range 2000 feet ; required the initial velocity. 

*- r 2000 -1055.5 

l ~2 sin 2e~2x -9473966" 

and v = \J2gh = V2 x 32 -2 x 1055 -5 = V67974 -2 = 260 ?. 

EXERCISES 

1. A shell projected from a mortar at an elevation of 60 was 
found to range = 3520 feet ; required the impetus and velocity of 
projection h = 2032 -25, and v = 361 "77. 

2. A ball projected at an elevation of 15 or 75 was found 
to range over 5200 feet ; what was the impetus and velocity of 
discharge? h -5200, and v= 578 -69. 

3. The elevation being = 45, and range = 12000, what is the 
impetus? t =6000. 

673. Problem IV. Given the impetus or projectile velo- 
city and the range, to find the elevation. 

Since /t=^-, and r2h sin 2e, therefore sin 26=^=^, and the 
2g 2h v 2 

formulae are . _ r , . . 

sin 2e=-y when h is given, 

ZiiL 

and sin 2e=^ when v is given. 

Or, L sin 2e = Lr + 10 - L2h, 

and L sin 2e = L# + LT + 10 - 2Lv. 

EXAMPLES. 1. At what elevation must a piece of ordnance be 
fired so as to throw a ball =5600 feet, the initial velocity being 
= 800 feet? 

Pr*-. 2 A 



406 PROJECTILES AND GUNNERY 



Sin *== 



hence e = 8 10' 56", and 90-e=81 49' 4", 

which are the t\vo elevations. 

The greatest height and the time of flight can now be found as 
in the first problem. 

2. At what elevation will a mark at the distance of 5100 yards be 
hit with an impetus of 3000 yards ? 



hence e = 29 6' 30", and 90 - e = 60 53' 30". 

EXERCISES 

1. At what elevation must a shell be fired, with a velocity of 
420 feet, so as to range = 5400 feet? . . . =40 9', or 49 51'. 

2. Required the elevation necessary to hit an object = 4200 yards 
distant with an impetus of 4000 yards. . . = 15 50', or 74 10'. 

674. Problem V. Given the elevation and time of flight, 
to find the range and velocity of projection. 

The formulae are r = icr 2 cot e, v^-^-. 

2 sin e 

Or, L 2r = ~Lg + ZLt + L cot e - 10, 

Ltf + 10-Lsin e. 



EXAMPLE. A ball projected at an angle of 32 20' struck the 
horizontal plane 5 seconds after ; what was the range and projectile 
velocity ? 

r = \gP cot e = \ x 32 -2 x 25 x 1 -5798079 = 635 -87, 
, gt 32-2x5 161 _ 1KA( . 

2 sin e ~ 2 x -534844 ~ 1 "069688 ~ 

as v is known, h can now be found by Art. 673. 

t> 2 
The formulae are obtained thus : Since h=^-, and P= 

9 

4 sin 2 e ; hence h n ^ . , and therefore (Art. 669) r=2h sin 2e 
g 8 smV 

0/2 COS (5 

. . ., . sin 2e ; but sin 2e 2 sin e . cos e, and = cot e ; 
4 sin j e sm e 

hence r=\g& cote. 
Also, 



PROJECTILES AND GUNNERY 407 

EXERCISE 

The time of flight of a shell projected at an elevation of 60 was 
= 25 seconds ; what was the initial velocity and the range? 

r=5809'6, and v = 464'76. 

675. Besides the preceding theorems for projectiles on horizontal 
planes, many more might be given of less importance ; the two 
following are sometimes useful : 

676. For the same impetus, the ranges are proportional to the 
sines of twice the angles of elevation. 

Let r and r' be two ranges corresponding to the elevations e and e', 

then r : r' = sin 2e : sin 2e' ; and therefore r'=r -. ^r- ; 

sin 2e 

also, sin 2e'=- . sin 2e. 

r 

EXERCISES 

1. If a shell range 1000 yards at an elevation of 45, how far will 
it range at an elevation of 30 16'? . . . =870 '642 yards. 

2. If the range of a shell at an elevation of 45 is = 3750, what 
must be the elevation for a range of 2810 feet? =24 16', or 65 44'. 

3. A shell discharged at an elevation of 25 12' ranges = 3500 feet; 
how far will it range at an elevation of 36 15'? . . =4332 '2. 

677. The ranges are proportional to the impetus, or to the squares 
of the velocities. 

Or, r :r'=h : h', where h is the impetus corresponding to r, and 

Jt f f f 

h' to r' ; hence r'=r . 7-. and h' h-- 

h' r 

For rlh sin 2e=~ sin 2e ; hence roc/ice^ 2 when e is given. 

EXERCISE 

If a shell ranges 4000 feet with an impetus of 1800, how far 
will it range with an impetus of 1980? .... =4400. 

678. The square of the time is proportional to the tangent of the 

o r 

elevation ; also, t z = . tan e. 
ff 



For t=2 sin e 


fUt . . 9 Z'<< 

V , or t* = 4: sin e . , i 




9 ff 


therefore 


2A-- ? ' 

Alt' . pr- , 

sin 2e 


nnd hence / 2 <t sii 


~2* r * sin 2 e 



r 2r . 

. ; TT~ = -= tan e. 
g sin 2e 2 sin e . cos egg 



40,8 PROJECTILES AND GUNNERY- 

EXERCISES 

1. In what time will a shell range 3250 feet at an elevation of 
32? .. =11-23 seconds. 

2. What is the time of flight for the greatest range for any 

impetus? . ... . . . t= \/ ^\/r nearly. 

IL-PRACTICAL GUNNERY 

679. Although the parabolic theory of projectiles affords a 
tolerable approximation to fact in the case of smaller velocities 
not exceeding 300 or 400 feet per second for the larger kinds 
of shells, yet its results deviate so widely from truth for 
greater velocities that ranges which, calculated by this theory, 
exceed 20 or 30 miles are found in fact to be only 2 or 3 
miles. The cause of so great a difference is, that when the 
velocity of a projectile exceeds 1200 or 1300 feet there is a 
vacuum formed behind it, because air rushes into a vacuum 
with a velocity of only about 1300 feet in a second; and 
therefore there is not merely the ordinary resistance of the 
air retarding the motion in this case, but also the atmospheric 
pressure of the air on its anterior surface, with scarcely any 
pressure on its posterior surface to counteract it; and even 
with less velocities than this, the pressure of the rarefied air 
on the posterior surface is so small that the unbalanced 
pressure on the anterior surface causes a great retardation, 
far exceeding that produced by the ordinary resistance, which 
is nearly proportional to the square of the velocity. 

680. It has been found by experiment that the square 
of the initial velocity of a projectile varies as the charge of 
powder directly, and as the weight of the ball inversely. By 
experiments made by Dr Hutton and Sir Thomas Bloomfield, 

2c 
it was found that v= 1600 x /y, where v = the initial velocity, 

c = the charge of powder, and b = the weight of the ball ; but 
by more recent experiments performed by Dr Gregory and 
a select committee of artillery officers, it has been found that 



PROJECTILES AND GUNNERY 409 

the velocity is considerably greater on account of the im- 
proved manufacture of gunpowder, and that the formula 

3c 
v-lQQQJj- affords a near approximation to the initial 

velocity. (See ' Flat Trajectory Theory.') 

681. Experiments for determining the velocity of a pro- 
jectile are performed by means of wire screens placed in 
front of the gun. The projectile in flight passes through 
and cuts the wire of these screens, which are placed at 
known distances from each other, and by an ingenious 
electrical arrangement connected with the wires the actual 
velocity is definitely recorded that is, the 'muzzle' or 
' initial ' velocity. 

682. Problem VI. Of the charge of powder, the weight 
of the projectile, and the initial velocity, any two being 
given, to find the third. 

Let w = the initial velocity, 

c= ii weight of the charge in lb., 
and 6= ii n ball n 

3c 
then v = 1600\/T- for spherical projectiles, 

VO . r* Kfi 
p for elongated projectiles 

(see Flat Trajectory formula) ; 

b( v \ 2 
hence C =3ll60o)' 

and a 

Also, the velocities are proportional to the square roots of the 
charges directly, and of the weights of the projectiles inversely. 

For voc*^, 



when b is constant, 

and VQC>/- ,i c ii 

That is, if v, c, b are the velocity, charge, and weight of shot 
in one experiment, and v', c', b' the same quantities in another, 
then 



410 PROJECTILES AND GUNNERY 

,3c .3c' 
:= V T :V T - 

when b is constant, 



and v' = v\/-. 

c 

t; : i/= VT ' VTT when c is constant, 
or v:v'= \/b' : \/b, and V'=*V\/T/* 

EXAMPLES. 1. Find the initial velocity of a shell weighing= 
48 lb., the charge being = 3 Ib. 

v= 1600 v = 1600 VTS = 1600 v4 = 400 V3 
o 4o lb 

= 400x1-732 = 692-8. 

2. The weight of a ball is = 32 lb. ; what must be the charge of 
powder necessary to give it a velocity of 1500 feet ? 

bl v y 32/1500y_32 225 
'~3\1600/ ~ 3U600/ ~T 256"' 

3. The velocity of a ball, with a charge of 10 lb. of powder, is 
= 1200 feet ; what would be its velocity with a charge of 12 lb. ? 

v' = v\/- = 1200VTR = 120x7120 = 120 x 10 "95445 = 1314'534. 

C 1 U 

EXERCISES 

1. What is the velocity of a shell weighing = 36 lb. when dis- 
charged with 4 lb. of powder ? ..... =923-76. 

2. With what velocity will a 48-lb. ball be impelled by a charge 
of2lb. ? . . . ...... =632-456. 

3. The weight of a shell is = 100 lb. ; what charge of powder is 
necessary to project it with a velocity of 1000 feet ? . = 13'02 lb. 

4. A ball is discharged with a velocity of 900 feet by a charge of 
2 lb. of powder ; required its weight ..... =18 - 961b. 

5. The velocity of a ball of 24 lb. weight is = 800 feet; what 
would be the velocity of a ball of 18 lb. impelled with the same 
charge? .- ........ =923"76. 

683. Problem VII. Given the range for one charge, to 
find the range for another charge ; and conversely. 

The ranges are proportional to the charges that is, one charge 
is to another charge as the range corresponding to the former is 
to that corresponding to the latter. 



PROJECTILES AND GUNNERY 411 

Or, e : c' = r : r', 

c' r' 

and r' = r . -, also c'=c. . 

c r 

EXAMPLE. If a shell range 4000 feet Avhen discharged with 9 Ib. 
of powder, what will be the charge necessary to project it 3000 feet? 
, r' 3000 



It was found (Art. 677) that r is proportional to v 2 , or rccv 2 ; 

bf v \ 2 
and since (Art. 682) c = ^(-r^-:) , therefore c is proportional to v 2 

o \ 1 OUU / 

when b is given, or cocv 2 ; but rocv 2 ; therefore rocc, or r : r' = c : c'. 
It could be similarly shown that, when c is given or constant, 

1 11 ,, , , , b ,, r 

roc-, or r :r = T : T/> or r:r=b : o, and r=r. r ,,also b=b.-,; so 
bob or 

that the range is inversely as the weight of the ball, all other 
circumstances being the same. 

EXEKCISES 

1. If a shell range 2500 feet when projected Avith a charge of 
5 Ib., what will be its range when the charge is = 8 Ib. ? . =4000. 

2. If a charge of 6 Ib. is sufficient to impel a ball over a range 
of 3600 feet, what charge will be required that the range may be 
4500 feet? .......... =7'5 Ib. 

684. Some important problems in practical gunnery can be 
solved by means of the Table in Art. 689, calculated by Mi- 
Robins, in which the actual and potential ranges for the same 
elevation are given in terms of the terminal velocity. 

The actual range is the range in a resisting medium, the 

potential range is the range in a non-resisting medium or 

vacuum, and the potential random is the greatest range in 
a vacuum. 

685. The terminal velocity of a projectile is that velocity 
which it has in a resisting medium when the resistance against it 
is equal to its weight, or it is the greatest velocity it can acquire 
in falling by its own weight through that resisting medium. 

The resistance to a plane surface moving with a moderate 
velocity in a resisting medium is nearly equal to the weight of a 
column of the fluid, having the surface for its base, and a height 
equal to that due to the velocity in a vacuum. The resistance on 
a hemisphere or on the anterior surface of a ball is only half that 



412 PROJECTILES AND GUNNERY 

on a surface equal to the area of one of its great circles ; and 
hence the resistance to a ball moving with a small velocity in the 
atmosphere is nearly half the weight of a column of air having 
a great circle of the ball for its base, and a height equal to that 
due to the velocity ; for the resistance to a sphere is equal to only 
half the resistance to the end of a cylinder of the same diameter. 
When the velocity is not considerable the resistance is about ^ 
instead of of the above column, as appears by computing the 
example in Art. 689, but for great velocities it is considerably 
greater. 

Several formulae have been given for determining the terminal 
velocity of a ball. One of these, due to Hutton, is as follows : 

Let r=t\\e resistance in avoirdupois pounds, c?=the diameter of 
the ball in inches, and v ihe velocity in feet ; then 

=( -000007565v 2 - -00175^, or r= 0000044CW ; 
the former value referring to considerable, and the latter to 
smaller, velocities. 

In order to find the terminal velocity, for which r = w, the weight 
of the ball, 

V78 

w= -5236^ x ~ x 7-25= '137134^, 

and when r = iv, the terminal velocity v' will be found from the 
equation, -137134^= -0000044dV, 

and v' = 

The height due to this velocity is h' = ^ 1 J 1 d=487d; and for a 
shell, the weight of which is f of that of a ball of equal diameter, 

= 158 V^- 



686. Robins found that the resistance to a 12-lb. ball moving 
with a velocity of about 25 '5 feet in a second was \ ounce 
avoirdupois. Now, for velocities less than 1100 feet per second, 
the resistance is nearly proportional to the squares of the veloci- 
ties, and it is also as the squares of the diameter ; hence, if c is 
the constant to be determined, 

r = c(Pv*, or -fa Ib. = c x 4'45 2 x 25 '5 2 . 

It will be found from this equation that c is = -000002427 ; and the 
value of v' would be found as above to be 238 \/d, and /t' = 883rf. 
In the Table, p. 415, Robins has taken this quantity to be 900rf, 
and denotes it by F that is, F = 900o?. This appears to be the 
origin of this quantity F, which has not before been accounted for. 



PROJECTILES AND GUNNERY 



413 



Robins had probably found, by other experiments, that 900 would 
generally afford more correct results than 883. 

This quantity namely, the height due to the terminal velocity 
in a vacuum may be called the terminal height. 

687. Problem VIII. To find the terminal height. 

The terminal height is found by multiplying the diameter of the 
ball by 900. 

When the ball has a different specific gravity from iron, find 
the height for iron ; then the specific gravity of iron is to the 
specific gravity of the ball as the height for an iron ball is to 
the required height. 

For iron, F = 900rf. 

For a ball of other material, whose specific gravity is s, 



and 

For a shell, F = 

For lead, * = 1 1 '35, and F = 1409d 

688. The following Table gives the weight of a cast-iron ball 
when its diameter is known, and conversely. The weight is in 
avoirdupois pounds, and the diameter in inches : 



Weight 


Diameter 


Weight 


Diameter 


136 


1 


17-1 


5 


1 


1-94 


18 


5-09 


1-10 


2 


24 


5-61 


3 


2-8 


29-5 


6 


3-7 


3 


32 


6-21 


4 


3-08 


42 


6-75 


6 


3-52 


47 


7 


8-7 


4 


70 


8 


9 


4-04 


100 


9 


12 


4-45 




' 



The weight of any solid ball may be found by multiplying the 
cube of its diameter by '5236, and the result by the weight of a 
cubic inch of its material. Diameter to be in inches. 



414 



PROJECTILES AND GUNNERY 



II 

WEIGHT OF CAST-IRON SOLID CYLINDERS IN LB. 

Length of cylinder=l foot 



Weight 


Diameter 
Inches 


Weight 


Diameter- 
Inches 


2-4 


1 


89 


6 


9-9 


2 


120 


7 


21-9 


3 


156 


8 


39-0 


4 


198 


9 


61-0 


5 





EXERCISES 

1. Find the terminal height for an iron ball = 6 inches in diameter. 

= 5400. 

2. Find the terminal height for a 3-lb. iron ball. . . =2520. 

3. Find the terminal height for a shell = 12 inches in diameter. 

= 8640. 

4. Find the terminal height for a leaden ball = 2 inches in 
diameter =2818. 

689. Problem IX. Given the actual range of a given 
spherical projectile, at an angle not greater than 10, and 
its original velocity, to find its potential range, and the 
elevation to produce the actual range. 
CASE 1. When the potential random does not exceed 39000 feet. 
RULE. Divide the actual range by the terminal height, and find 
the quotient in one of the columns of actual ranges in the follow- 
ing Table; and opposite to it, in the next column of potential 
ranges, is a number which, multiplied by the preceding height, 
will give the potential range. The potential range and initial 
velocity being known, find the elevation by Art. 673. 
Let F = the terminal height in feet, 

r= i, actual range in feet, 
R= u potential range in feet, 
r'= u actual range in the Table, 
R'= n potential range in the Table, 
v M initial velocity, 
h= M impetus, 
e= M elevation, 
d= u diameter of the projectile in inches; 



PROJECTILES AND GUNNERY 



415 



then 2A = the potential random. 

If T1T 

Then F = 900, r = and R = tR, 

nearly, or Lh = 2(Lv - '903090). 

Or, LA = 2Lv- 1-806180. 

R_32R 

Or, L sin 2e = 10 + LR - L2A (Art. 673). 

In the following Table the first, third, and fifth columns con- 
tain the actual ranges of projectiles expressed in terms of F 
that is, the F for the ball in any particular case is the unit of 
measure ; and the second, fourth, and sixth columns contain the 
corresponding potential ranges that is, with the same elevation 
and initial velocity expressed in the same manner : 



Actual 
Range 


Potential 
Range 


Actual 
Range 


Potential 
Range 


Actual 
Range 


Potential 
Range 


01 


0100 


1-3 


2-1066 


3-3 


13-8258 


02 


0201 


1-4 


2-3646 


3-4 


15-0377 


04 


0405 


1-5 


2-6422 


3-5 


16-3517 


06 


0612 


1-6 


2-9413 


3-6 


17-7767 


08 


0822 


T7 


3-2635 


3-7 


19-3229 


1 


1034 


1-8 


3-6107 


3'8 


21-0006 


12 


1249 


1-9 


3-9851 


3-9 


22-8218 


14 


1468 


2-0 


4-3890 


4-0 


24-7991 


15 


1578 


2-1 


4-8249 


4-1 


26-9465 


2 


2140 


2-2 


5-2955 


4-2 


29-2792 


3 


3324 


2-3 


5-8036 


4-3 


31-8138 


4 


4591 


2-4 


6-3526 


4-4 


34-5686 


5 


5949 


2-5 


6-9460 


4-5 


37-5632 


6 


7404 


2-6 


7-5875 


4-6 


40-8193 


7 


8964 


2-7 


8-2813 


4-7 


44-3605 


8 


1-0638 


2-8 


9-0319 


4-8 


48-2127 


9 


1-2436 


2-9 


9-8442 


4-9 


52-4040 


1-0 


1-4366 


3-0 


10-7237 


5-0 


56-9653 


1-1 


1 -6439 


3-1 


11-6761 






1-2 


1-8669 


3-2 


12-7078 







In this case 2h does not exceed 39000, and v does not 
exceed 1112; for v=8\/^ = 8x 139, and e may be found without 



416 PROJECTILES AND GUNNERY 

previously calculating h ; by substituting in the last formula the 
value of L2A, it becomes 

L sin 2e= 11 '505150 + LR-2Li>. 

EXAMPLE. At what elevation must an 18-pounder be fired, with 
a velocity of 984 feet, in order that its actual range on a horizontal 
plane may be = 2925 feet? 

F = 900d=900 x 5-09=4581, 



7H57 = ' 64 and R = FR'=4581 x -8028 = 3678, 
.r 



LA = 2(Lv - -903090) = 2(2 -992995 - -903090) 
=2-089905x2 = 4-179810, and A = 15129. 
L sin 2e=10 + L3678-L2A = 13-565612-4-480840=9-084772, 
and 2e = 6 59', and e = 3 29'5'. 

EXERCISES 

1. At what elevation must a 12-lb. ball be fired, with a 
velocity of 700 feet, in order that it may reach an object = 2000 
feet distant? . . . ...... =4 28 -5'. 

2. Find the elevation at which a ball = 5 inches in diameter must 
be discharged, with a velocity of 800 feet, that its actual range 
may be = J of a mile ...... =2 53'. 

CASE 2. When the potential random exceeds 39000 feet. 
RULE. Find two mean proportionals between 39000 and the 
potential random ; then the less of these means is to the potential 
random as the potential range, found by the former case, is to the 
true potential range ; then the elevation is found as before. 
Find h as in the preceding case ; then, if 

R" = the potential range found by the preceding case, 
R = ii true potential range, 
then R =^00138R"\/7t 2 . 

Or, LR = 3-139977 + LR" + LA. 
Instead of LR", LF + LR' may be used. 
Then find e, as in the former case, or, by this formula, 

L sin 2e = 6 '8389469 + LR" - LA, 
which gives c at once, when h and R" are found. 

EXAMPLE. At what elevation must a 24-pounder be discharged, 
with a velocity of 1730 feet per second, in order that its actual 

range may be = 7500 feet? 

j, 

F = 900^=900x5 -61 = 5049, and -' = ^=x 

r 5049 



PROJECTILES AND GUNNERY 



41? 



hence R" = FR'= 5049x2-587 = 13060, 

LA = 2(Lv - -903090) = 2(3 '238046 - "903090) = 2x2 -334956 

= 4-669912, and h = 46764, 
and 2h = 93528, which exceeds 39000. 
Then L sin 2e = 6 -8389469 + LR" - JLA 

= 6-8389469 + 4-1159432 - 1-5566372 

= 9 -3982529 = L sin 14 29'; 
and therefore e = T 14'5'. 

Let a = 39000 ; then 2/t being the potential random, let x and y 
be two mean proportionals between a and 2h ; 
then a: x=x: y, and x: y=y : 2A ; 

yZ y2 y& 

hence y = , and 2A = = -gj 



and x 

also, a;:2A = R":R; 



therefore, R = = = -00138R"#A 2 . 

* _ \/2 2 A V 

Or, LR = 3 -1399769 + LR" + LA. 

Then e can be found for this potential range, and given initial 
velocity, as in the preceding case ; or, 

R 2/tR" R" 



therefore L sin 2e = 10 - L2a 2 + LR" - 

or L sin 2e = 6 -8389469 + LR" - JLA. 

EXERCISES 

1. At what elevation must a ball 4'5 inches in diameter be fired, 
with a velocity of 1200 feet per second, in order that its actual 
range may be = 4500 feet? ....... =4 45'. 

2. Required the elevation at which a 24-pounder must be fired, 
with a velocity of 1600 feet per second, that its actual range may 
be a mile .......... =4 2ff 30". 

690. Problem X. Given the elevation not exceeding 45, 
and the velocity with which a given projectile is dis- 
charged, to determine its actual range. 

CASE 1. When the potential random does not exceed 39000 feet. 

RULE. Reduce the terminal height F, corresponding to the 
given projectile in the ratio of radius to the cosine of f of the 
angle of elevation ; find the potential range by Art. 689 ; divide 
this range by the reduced F, and find the quotient in the tabular 



418 PROJECTILES AND GUNNERY 

column of potential ranges ; and opposite to it, in the preceding 
column of actual ranges, is a number, the product of which, by the 
reduced F, will give the actual range. 

Let F =the terminal height found by Art. 687, 

F' = it reduced height, 
the other letters denoting as before. 

Then, to find F', F'/F = cos f e, 

or LF' = LF + Leosfe-10, 

and h is to be found as in Art. 672. 
To find R, R/2/t = sin2e, 

or LR = L2A + L sin 2e - 10. 

Then r'=~, and r=FV, 

r 

or Lr=LF' + Lr'. 

EXAMPLE. What is the actual range of a musket-bullet, of the 
usual diameter of f of an inch, discharged at an elevation of 15, 
with a velocity of 900 feet ? 

F = 1409^= 1409 x | = 1057... (by Art. 687), 
LF' = L1057 + Lcosll 15'- 10 = 3024075 + 9 '991574- 10 
= 3-015649, and F' = 10367. 



Also, A= = = 12656, and2A = 25312, 

\o/ \ o / 

and (Art. 671 ), R = 2/t sin 2e = 2h x \ = h = 12656, 



and r=FV = 1036-7 x 3-15 = 3266 feet, the actual range. 

EXERCISES 

1. What is the actual range of a ball of 6 inches diameter, fired 
at an elevation of 25, with a velocity of 1000 feet ? = 10570 feet. 

2. What is the actual range of a shell = 10 inches in diameter, 
its Aveight being = * of that of a ball of the same diameter, when 
discharged at an elevation of 40, with a velocity of 400 feet ? 

= 3938 feet. 

CASE 2. When the potential random exceeds 39000, or the 
impetus exceeds 19500, or the velocity exceeds 1112 feet. 

RULE. Find two mean proportionals between 39000 and the 
potential random, and take the less of them for the reduced 
potential random ; then the true potential random is to the 
reduced potential random as the potential range to the reduced 
potential range. This reduced potential range, being divided by 



PROJECTILES AND GUNNERY 419 

the reduced terminal height F', will give the tabular potential 
range, from which the actual range is found as in the last case. 



and adding 10 to both sides, and substituting for LA its value 
2Lv- 1-806180 (Art. 689) ; then 

LR" = LR-Lv + 4-064143 .... (1), 
and LR = L2/t + L sin 2e - 10, 

or LR =2Lv+ L sin 2e- 11 -505150 . . . (2); 

hence LR" = Lv + L sin 2e- 7 -441007 . . (l) + (2). 

Find F and F', as in Art. 690. 

EXAMPLE. Required the range of the bullet in the example of 
the first case, discharged at the same elevation, with a velocity of 
2100 feet. 

In this case v>1112, or 2A>39000. 

As in the former example, F = 1057, and F'= 1036-7. 
And LR" = Lv + L sin 2e- 7-441007 = 2-214813 + 9-698970 -7'441007 

= 11-913783 - 7-441007 = 4-472776 ; 
hence R" =29701. 

LR' = LR" - LF' = 4 -472776 - 3 -015653 = 1 -457123, 
and R'= 28-65 ; hence, by Table, r' = 4-17303, 

and r = FV = 1036-7 x 4-17303 = 4326 feet, the actual range. 

Although the velocity in this example is more than double that 
in the preceding, yet the range is only 1060 feet greater. 

EXERCISES 

1. Find the actual range of a 42-lb. ball, discharged with a 
velocity of 1800 feet, at an elevation of 36. . . =15413 feet. 

2. What will be the actual range of a 24-lb. ball, fired at an 
elevation of 35, with a velocity of 1760 feet per second? 

= 13695 feet. 

691. It can be shown, by dynamical principles, that balls of the 
same density, projected at the same elevation, with velocities that 
are proportional to the square roots of their diameters, describe 
similar curves. The reason of this is, that the resistances are pro- 
portional to the masses or weights of the balls. Their velocities 
at their greatest height, which are horizontal, are proportional to 
their diameters ; and any corresponding lines of their trajectories 
that is, of the curves described by them are proportional to 
their diameters. Their actual ranges are therefore proportional to 
their diameters, or to the squares of their initial velocities, but their 
potential ranges are in the same proportion ; hence their actual 
and potential ranges are proportional. But the terminal heights, 



420 PROJECTILES .AND GUNNERY 

being QOOd, are proportional also to their diameters, or their ter- 
minal velocities are proportional to their initial velocities. The 
terminal heights are therefore also proportional to their ranges, 
both actual and potential. Hence the quotients of the actual and 
potential ranges of one ball by its terminal height are respectively 
equal to the corresponding quotients for another ball, both being 
projected under the conditions stated above that is, the tabular 
ranges, both actual and potential, are the same for all balls of the 
same density, discharged at the same elevation, with velocities 
proportional to the square roots of their diameters. Thus a 
comparatively limited set of experiments with a ball of given 
dimensions and density would be sufficient to determine the data 
for the construction of the preceding Table ; by means of which 
the ranges of balls, of an unlimited variety of density and size, 
could be computed. 

The weights of two balls being w, w', their diameters d, d', their 
velocities v, v', and the resistances to them r and r', then (Art. 686) 

r:r' = dV:d'V 2 nearly, 

if the velocities are both greater or both less than 1112 feet. 
And if v : v' = \/d : *Jd', then 

r:r' = d 2 d: d'W^d 3 : d' 3 =w : w'; 

so that in this case the resistances are as the weights. If v and v' 
are the terminal velocities, then r=w, and r' = iv'; 
hence r :r'w :w'. 

Or, d 2 v*:d' z v' 2 =d 3 :d' 3 , 

or v 2 :v'* = d:d',oi-v:v' = \/d:\/d'. 

THE FLAT TRAJECTORY THEORY 

692. This theory will be recognised as bearing especially 
on modern guns and rifles, for if the point-blank range of 
any gun is increased, its trajectory (or, more correctly speaking, 
the trajectory of its projectile) takes the form of a straight 
line, and less and less that of a parabola, 

All guns with a high muzzle velocity are affected by the 
investigations made in connection with this theory. 

To attain a high muzzle velocity various measures have from 
time to time been adopted. Careful consideration has been 
bestowed on the shape of projectiles, which in modern guns 
are elongated cylinders of iron or steel, with ogival heads, and 



PROJECTILES AND GUNNERY 421 

struck with a radius of 1^ the diameter of the projectile. In 
consequence of the rotatory motion imparted to the shell by 
the rifling of the gun, the shell on leaving the gun ' spins ' or 
rotates round its longer axis, thus only exposing its ogival 
head to the resistance of the atmosphere. Air-spacing in the 
powder-chamber also ensures a greater volume of gas being 
generated at the base of the shell, and consequently a greater 
pressure is set up than was the case in muzzle-loading guns with- 
out gas-checks. The twist of the rifling, and also windage, are 
also important factors connected with velocity, range trajec- 
tory, and time of flight. It will thus be seen that, in order to 
obtain a high initial velocity, the gun is strained to an extent 
far exceeding that demanded by the use of balls, on which 
the parabolic theory treats. The modern weapon and its pro- 
jectiles are therefore a distinct departure from the ancient 
cannon, and in consequence of the increased strain to which 
it is subjected, it is built up in coils, so distributed as to 
equalise the action of the combined stresses, and at the same 
time with a due regard to its minimum weight and mobility. 
Guns of modern design are made entirely of steel in the form 
of ribbon-wire, the ultimate tensile strength of which 100 
tons per square inch. By winding on with varying tension, 
any desired state of initial stress may be given ; and thus on 
firing, every part of the structure is made to take its due 
share of stress. For calculating the strength of a wire gun, 
the winding tensions must be known, as well as dimensions 
and strength of material employed. 

693. The height reached is given approximately by the formula, 



where T = total time of flight in seconds, 

=time of flight in seconds to a point where height of 

trajectory is h feet. 

Assuming the vertex to be reached at half-time, and putting 
t = f, and g=S2, we get the greatest height. H = 4T 2 is a useful 
approximation for comparing the flatness of trajectories of different 
guns. 

Prac. 2 5 



422 



PROJECTILES AND GUNNERY 



694. Velocity and Momentum of Recoil. 
v = muzzle velocity of projectile, 
W= weight of gun and carriage, 
V = velocity of recoil of gun and carriage, 
w= weight of projectile, 
w x = ,, powder charge, 
C = a constant deduced by experiment ; 



695. Gravimetric Density. 

To find gravimetric density GD of a charge. 

Let S = cubic space allotted per Ib. of powder in the chamber. 

GD=^- 3 - 



696. WORK DONE BY EXPLODING POWDER 



No. of 
Expansions 


Work 
per Lb. burned 
Foot-Tons 


No. of 
Expansions 


Work 
per Lb. burned 
Foot-Tons 


No. of 
Expansions 


Work 
per Lb. burned 
Foot-Tons 


1-25 


19-226 


5-5 


95-210 


11 


121-165 


1-5 


31 -986 


6 


98-638 


12 


124-239 


1-75 


41-494 


6-5 


101 -744 


13 


127-036 


2 


49-050 


7 


104-586 


14 


129-602 


2'5 


60-642 


7'5 


107-192 


15 


131 -970 


3 


69-347 


8 


109-600 


16 


134-108 


3'5 


76-315 


8-5 


111-840 


17 


136-218 


4 


82-107 


9 


113-937 


18 


138-138 


4-5 


87-064 


95 


115-905 


19 


139-944 


5 


91-385 


10 


117-757 


20 


141-647 



This Table is made out for charges of unit gravimetric density. 

Divide cubic contents of bore by cubic content of cartridge- 
chamber, which will give number of expansions. Multiply the 
number found opposite this in the Table by number of Ib. in the 
charge, and the result will be the Avork done. 

If the charge be not of unit gravimetric density. 

Suppose gravimetric density = -8, and number of expansions =5 ; 

Work done per Ib. of powder = work done in 5 expansions minus 

work done in , or in 1-25 expansions = (91 '385- 19 '226) foot-tons 
= 72-159 foot-tons. 






PROJECTILES AND GUNNERY 423 

In practice only a portion of this, called the factor of effect, 
varying from 0*7 to 0'9, is obtained. Thus, suppose factor of 
effect 0'8, the work realised is = 72'159x -8 foot- tons = 57 '727 foot- 
tons per Ib. of powder in the charge. 

wV 2 

p-r- foot-tons is also a measure of work contained in the 
2^x2240 

projectile, in which iv= weight of projectile in Ib., V=the niuzzle 
velocity in feet per second. 

By equating the two expressions, the probable muzzle velocity 
can be estimated before actual trial has taken place. 

697. Penetration of Armour. 

T = thickness of wrought-iron that can be penetrated by direct 

fire (inches), 

d= diameter of projectile (inches), 
v = striking velocity, feet per second; 

T = -^- J~,--Ud. 



For steel and compound or steel-faced armour, penetration 
= Tx'8 (approximately). Captain Orde Brown's rough rule: 



Various causes will modify the above, and when striking at an 
angle about 40 from the normal projectiles will be deflected. 

698. Penetration of Rifle - Bullets. Thickness of various 
materials proof against magazine-rifle fire at all ranges : 

Earth parapet free from stones, not rammed, 24 inches. 

Clay, ........ 24 

Fine loamy sand, ...... 20 n 

Wrought-iron or mild steel plate, ... inch. 

Fir, dry or green, ...... 38 inches. 

Elm, green, ....... 36 . n 

Oak, , ........ 24 

Sand-bags, filled, header, .... 1 bag. 

ii n stretcher, .... 2 bags. 

699. Use of Bashforth's Tables (A and B). The two 
following Tables (Bashforth's) give the relations between (A) the 
time of flight of a projectile and its velocities at the beginning 
and end of that time, and (B) the distance of flight of a projectile 
and its velocities at the beginning and end of that distance. 
Thus, the initial velocity of a projectile being given, the velocity 



424 



PROJECTILES AND GUNNERY 



13 







1 = -JI 

2 JH 

PH O Q) 

3 "1 1 1 

** _ irj s 



>H 

H 

i i 

O 
O 

*1 
w 

t> 

Q 
2 
! 

w 

S 



W8 
H 



> & 


1 = 



15 

^ r2 " 

CL,' g 

*O . J2 

= '5 i 

^^5 
?2l 

o ^- o 
= < tcsE 

.s'i 

-ta O 

II II II II 

"** S S 

EH S S 



I I 



222s-S:b 

i ii ii irti i^i^-ii (r ii ICMCMCMCMI 

OOiOOSCOiOCOCMCMCOOiCOI OCMlOOiCD 

coaoooocoiocct^-cocooot^ioocoLoi-^ 

I~~e'3Tt<p>pO5CMipOOpCM'*5paOO5p7H 

C-l'T'lCOCOCOCOCOCOCOCOCOCOCOCOCOCOCO 
0-1CMCMCNCMCMCMCMCMCMCMCMCM<M<MCMCM 

CM C^J CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO 
&CMCMCMCMCMCMCMCMCMCMCNCMCMCMCMCM 

6l Si CM CM CM CM CM CM Ol CM CM CM CM (M CM CM OJ 

00 00 IO O JM CM O iQ O> Si -* 

cSl 6l CM CM cVl CM CM CM CM C>5 CM CM C>5 CM CM \ O\ 

co co i 

CO CO i 

CM CM CM CN CN CM 

00 CM *^* P O- p Ol C^ O t^ CO p CO t^* *O r~* 

1O 00 ^^ 

y 9 s T 1 

iCMCMeocococococococococococococo 

CM Ch ell CM CM CM CM CM CM CM CM CM CM CM CM CM CM 

CMcoi^-oCMppcocposr-iroipcpocoip 

^1 ^l "T 1 ! CO CO CO CO CO CO CO CO CO CO CO CO CO CO 
CM 6l CM C>1 CM CM CM CM CM CM CM CM 'M CM CM CM CM 

S'MCNCOCOCOCOCOCOCOCOCOCOCOCOCOCO 
CM 5-1 (M CM CM CM (M (M C CM CM <M CN (M CM CM CM 

^ioioicocococorocofocococococococo 

<M CM <M (M CM C-J CM (M CM CN O) CM <N <N <N CN Ol 

SSSSSSS52: 



PROJECTILES AND GUNNERY 



425 





o = 



23 



-a G 5 

3 =.2'5> 
o 2 



* a S - 

H -5 w 5 

^ *3 -a I S 

L>H -in " *^ -* 

C .5 a = . 

^4 a ?> S iS ^ 

S 7 if ii H 

V >^ ^ r^ I 



fc 5 



HH H3 

Pa 
o 



- &J" 
3 s ? 



o *3 

w -*> ' S? 
^~ Sow 



H 

a 



a, o g 

a |~ 3 
^ -cS 
.2 ' g 






^ - - (N( 

igxosxccecceccppccpTicoipiffl' 
^^^~ 

xo5OiiioiiccccC'* ; *co'O>t5coc6i 
cc cc 

xxxoscccccp-T-ip 

O il X -* 6 c: : ~ i- rc x >h cb CO X 

r-cccc ~i^ -TCCI.--CCX-H oos 
m i~ -^ c >~ o i~ ~ cc t-- ** x i< -*i< co 
< 5^ ?i cc cc cc -t * Lt >c >-t cc cc co 

^cc^^ -^'Tir^pTJT"^-'*^?? 

i^ c*. O >! ?i cc re rj- ^h i?: i~ i- ;c ic cc 
O ^-l 71 O >p p * L- OS OS "* !N Tf Q-l t^ p CO 

cc cc * -^ * T^ * 

ccScc V:?:xrci~ irfc^icccs;?5o 
i^- x o ' ^H ->! o-i re cc ~t *& * i.t o c cc co 

i x M >-H t>- H x ?j i^ '-*5 ci cc cc c: ~i ic 

xo-*-^xxiSd;'- ; '?i5r~ ; >iiccc5i 

C- CC ^H O t^* Ol t~ ^1 CC ^ 'C C^ T) CC Ct ^1 i^ 
CC X O < "M >! CC CC -t -t *t i~ '~ L- CC CC 

-M 1^ i^ o cc 5- o '"^ t - x i^ x o c c; 

t^owaco'T'jr^O'icoO'^x'yi^^^i^t 

^ ^ ^ _ . re cc ^? -f -t 'rt uc i~ co cc 

O O O O O O O O O O O CO O O O 

> s 



426 PROJECTILES AND GUNNERY 

after any given interval of flight may be found, or vice versA; 
or, the initial and remaining velocities being known, the time or 
distance of flight may be found. 

The coefficient n in the formulae depends on the shape of head of 
the projectile, steadiness of flight, and density of the air. For 
ogival-headed projectiles, the heads of which are struck with a 
radius of 1J diameter, n = l; for more modern projectiles with 
heads of 2 to 2^ diameter, and under normal atmospheric condi- 
tions, n may be taken from about - 88 up to unity from the 
lightest to the heaviest. With the 0*303 magazine bullet, = 0'64. 

For high velocities, at all ordinary low angles of elevation, these 
Tables and formulae give results agreeing very closely with actual 
practice. 

EXAMPLE ON THE USE OF THE TABLES 

Suppose the muzzle velocity of a 10" gun to be 2040 feet per 
second, and that it has a remaining velocity of 1879 feet 
per second. It is required to find the distance of flight, or 
range of the projectile in yards, given the weight of projectile 
= 5001b. 

By referring to Table B we have the formula, 

s = C(Sv-Sv), 
where s range in feet. 

We first determine C ; and by taking the value of n as = -88 (see 
notes preceding the Tables), we get the ballistic coefficient C = w -"- nd?. 
.'. = 500 Ib. -f -88xlO" 2 =5-68. 

The initial velocity Sv=2040, and the number in the Table 
corresponding to this velocity is 45350 '3. 
.-. Sv=45350-3. 

The remaining velocity is 1879 feet per second, and the 
number in the Table corresponding to a velocity of 1870 feet per 
second (for 1870 is the nearest velocity in the Table to that given) 
is 44716-3. 

.-. Sv=44716-3. 

.-. s range = C(Sv-Sv) 

=5-68(45350-3 - 447 163 -3) 
= 5-68x634 

= 3601-12 feet, or 1200-37 yards. 

Further, let the time of flight of the above-named projectile be 
also required. 

The given muzzle velocity =2040 feet per second, and the number 
in Table A corresponding to this velocity is TV =233-567. 



fROJECtlLES AtfD GtTNNER? 427 

The given remaining velocity = 1879 feet per second, and as the 
nearest velocity in Table A is 1870, the number corresponding to 
this velocity is TV = 233 '242. 

The ballistic coefficient C is w-rnd?, and this has already been 
found =5 -68. 

.. = time of flight of projectile in seconds 
= C(Tv-Tv) 
=5 -68(233 -567 -233 -242) 
=5-68x -325 
= 1 '846 seconds. 

. -. the time of flight of a projectile whose weight=500 lb., and 
whose muzzle and remaining velocities are respectively 2040 and 
1879 feet per second, ranges over a distance of 1200 '37 yards in 
1 '846 seconds of time. 

It may be mentioned that the weight of powder in the charge of 
this gun, which affords the above results, is 252 lb., and that the 
projectile at this range will penetrate 20'5 inches of wrought-iron 
armour-plating. 

It will therefore prove interesting to learn what measure of 
work is stored up in a projectile whose weight is 500 lb., and 
which travels at the rate of 2040 feet per second. 
By the formula already afforded, we find that this measure of 

wV 2 

: ~ 20x2240' 

The symbol (g) is the acceleration due to gravity (see ' Parabolic 
Theory '). 

500 x 2040 2 



' 2gx 2240 "2x32x2240 
= 14515 -34 foot-tons. 

Should the muzzle velocity and time of flight be given, the 
remaining velocity can easily be found ; for, 

Ty= remaining velocity 

= Tv-/c. 

Let us suppose that it be required to find the remaining velocity 
of a shell whose weight is 500 lb. and diameter 10 inches ; let 
the value of (n) be assumed ='88, and the muzzle velocity 
of the shell =2040 feet per second, with time of flight = 1-846 
seconds. 

We first determine the value of C, the ballistic coefficient, vide 
formula C iv -f nd 2 , which in this case = 500 -^ '88 x 100. 

.'. C = 5-68. 



428 PROJECTILES AND GUNNERY 

The given time of flight = 1 '846 seconds. 

The muzzle velocity =2040 feet per second (vide Table) 233 '567. 
. . TV = remaining velocity 



= 233 -567 -'325 

= 233-242; 

and the velocity corresponding to this number in the Tables 
= 1870 feet per second. 

.. the shell has a remaining velocity of 1870 feet per second at 
the end of a time of flight = 1 '846 seconds. 

700. The reader's attention is directed to formula v = 1600. /_, 

\ b 

which appears in Art. 680. This formula can only be applied to 
guns which have more or less windage ; but in almost all modern 
weapons the system of obturation adopted is such as to totally 
exclude this factor. The object aimed at is to utilise to the very 
fullest extent the force set up by the expansion of the gas in 
the powder-chamber. The prevention of gas waste or escape in 
a breech-loading gun is technically termed 'obturation,' and is 
derived from the Latin obturo, I stop or close up. 

The ' windage ' of a gun (a term already used) is the difference 
between the diameter of the bore of the gun and that of its 
projectile. 

With muzzle-loading guns it was an inevitable necessity to 
have windage, otherwise it would have been impossible to load 
the gun. It has, however, its disadvantages namely, that a large 
volume of the gas generated by the ignition of the charge passed 
both over and under the projectile whilst being propelled through 
the bore of the gun, and thus escaped without having fulfilled its 
allotted work in respect to the projectile ; and consequently this 
loss of potential energy materially affected the muzzle or initial 
velocity of the projectile. 

The formula can, however, be modified and applied to modern 
guns by increasing the value of the coefficient 3 to 375 ; thus 

Initial velocity v=\QOG^J r --- 

It would be better, perhaps, to alter the symbol (b) to P where 
P weight of projectile ; this distinction would better characterise 
the formulae, and at the same time assist the memory. 



PROJECTILES AND GUNNERY 429 



Thus, for flat trajectories the initial velocity = 1600, 

where C = weight of charge in lb., and P = weight of projectile also 
in lb. The formula affords a close approximation to the initial 
velocity, but should not be preferred to that by which the Tables 
are calculated. 

EXERCISES 

1. A 5" breech-loading gun, whose shell weighs 16 lb., has a 
muzzle velocity of 1800 feet per second, and a remaining velocity 
of 1200 feet per second ; find the distance of flight, or range in 
yards, of the projectile, and state the measure of work contained 
in the shell ; given n = l. 

Distance of flight = 1823'296 yards; measure of work = 361-6 
foot- tons. 

2. Taking the previous exercise, let it be required to find the 
time of flight of the projectile in seconds, with the velocities there 
mentioned ......... = 1 '250 seconds. 

3. Supposing the shell mentioned in the first exercise had a 
striking velocity of 1200 feet per second, determine its penetration 
by formula for direct fire, in both wrought-iron and steel-faced 
armour. 

Wrought-iron, 3^002 inches ; steel-faced armour, 2'401 inches. 

4. The muzzle velocity of a 3-pounder Hotchkiss quick-firing 
gun is 1873 feet per second ; its remaining velocity is required, 
given time of flight of projectile = 3 seconds, weight of projectile 
= 3 lb., n= '88, rf=l'85 inches. . . . = 1060 feet per second. 

5. With the data before you in the above question and answer, 
state the distance of flight, or range in yards, of the shell. 

= 4141 feet, or 1380'3 yards. 

6. What would be the greatest height the projectile would 
attain to in a given time of flight = 3 seconds? 

= 36 feet, approximately. 

701. Strength of Guns. In calculating the circumferential 
strength of a gun built up by shrinking on successive layers of 
metal, the general formula employed is 



where n is the number of layers. Thus, for a 6 inch breech- 
loading, steel gun, having over the powder-chamber a tube, 



430 PROJECTILES AND GUNNERY 

breech-piece, and jacket, commence from the exterior and put n 

= 3, 2, 1 successively ; thus 

r z_ r 2 

P_'3 '2m 
9 o . o J- >, 



_ 
PO = r 2 + ? ,~2( T + Pi 

in which r is the radius of the powder-chamber ; r lt r 2 , and r 3 
the outer radii of the tube, breech-piece, and jacket respectively ; 
P , P 15 P 2 , P 3 are the radial pressures in tons per square inch at 
the surfaces, where the radii are r Q , r 1} and r 2 respectively. 

Note. In the case of the outside layer, or jacket, P 3 =0, as the 
pressure of the atmosphere may be neglected. 

T , Tj, and T 2 are the maximum allowable hoop tensions in tons 
per square inch at r u , r\, and r 2 respectively. 

Practically, with modern gun-steel, the values for strength 
calculations may be taken at T =15, T a and T 2 , &c. 18. A large 
margin of safety is thus provided, as 40 tons per square inch would 
be an average ultimate tensile strength. 

For wrought-iron coils, T , T 1( &c. =9. 

If F is taken as the circumferential factor of safety, usually 
about 1'5, and P the safe working pressure in the chamber, 

PF-PO- 

For the longitudinal strength of the above gun (where the 
breech-screw gears into the breech-piece), with the same system of 
notation, p Q denoting longitudinal pressure, 



The longitudinal factor of safety f generally equals 6 or 7, and 



702. Concluding Remarks, and Momentum of Eecoil. 
From what has already been said, the reader will at once per- 
ceive that the science of gunnery aims at the further development 
of the last theory. We have but touched the fringe of this 
interesting science, as space will permit us to do no more ; but 
before closing the subject we can safely predict that the high 
initial velocities of projectiles fired from breech-loading rifled guns, 
such as the 16 '25-inch and 13 '5-inch, with charges of 1800 Ib. 
and 1250 Ib. of powder respectively, will shortly be eclipsed by the 
introduction of electricity as an agent of propulsion. 



PROJECTILES AND GUNNERY 431 

Experiments have already been made in this direction with 
lighter projectiles, and the marvellous results obtained therefrom, 
as regards range, velocity, and time of flight, are such as will 
necessitate all future formulae being expressed in terms intimately 
associated with this prime-motor. 

In order to obtain a high initial velocity for any gun, various 
complex considerations present themselves, and these require to be 
regarded as the sum of so many positive and negative quantities. 

The construction of any gun depends on the total stress the 
material will be subjected to in the different parts of the gun, 
modified, of course, by the particular circumstances connected 
with its manipulation, which may either be field, naval, moun- 
tain, siege, or position (that is, coast defence). 

Having decided upon its calibre and determined its employ- 
ment, we have to consider the questions of the charge and weight 
of the projectile ; and in arriving at these we are governed by 
the all-important matter of initial velocity, which in turn regulates 
the momentum of the shell, and consequently its penetrative 
work. 

The twist of rifling (expressed in so many turns or revolutions of 
the projectile in a certain number of calibres) produces frictional 
resistance. 

For instance, let the diameter of the bore of a gun be 3 inches, 
and let the twist of rifling be expressed as 1 turn in 30 calibres. 
It will be readily understood that the projectile, in passing through 
the bore of the gun, makes one complete turn round its longer axis 
in a distance = 90 inches or 30 calibres. 

Then, again, there is another factor which cannot be overlooked 
namely, the ' momentum of recoil.' 

When a gun fires a projectile, the force of the explosion pro- 
duces momentum in the gun equal in amount but opposite to 
that of the projectile, and causes recoil. The other effects pro- 
duced in the gun and the projectile are not, however, numerically 
equal. 

According to a well-known law in dynamics, we are told ' that 
when two bodies mutually act upon each other, the momenta 
developed in the same time are equal, but opposite in direction ; ' 
or, every action is accompanied by an equal and opposite reaction. 

EXAMPLE. The 5-inch B.L. gun whose weight is 2 tons fires a 
projectile weighing 50 Ib. Avith an initial velocity of 1800 feet per 
second. Find the velocity of the gun's recoil and the mean force of 
the explosion, supposing the bore of the gun to = 25'l calibres. 



432 PROJECTILES AND GUNNERY 

Let W, w = weight of gun and projectile respectively. 

V, v velocity .1 .1 ,, 

By the above law, momentum of gun = momentum of projectile. 

.-. WV=wv; 

that is, 2 x 2240 x V = 50 x 1800. 

Kf\ Y i son 

'" V 2x2240 =2 ' 089 feet Pei ' second ' 

Now, to find the mean effort executed during the explosion of 
the powder, we must first ascertain the acceleration of the pro- 
jectile along the bore of the gun. Since the bore is 25*1 calibres, 
or 10 '458 feet, in length, and the initial velocity of the projectile 
as it leaves the gun = 1800 feet per second, we have 

v*=2as, 

a formula deduced from uniformly accelerated motion, where 
a = acceleration per unit time, 
*=distance described during interval (t z -tj. 

.'. 1800 2 = 2xxlO-458. 
1800 2 
. '. tt= oo^qTft = 154905-335 feet per second per second. 

But P=-xa. 

9 

This equation expresses the force P in the same units as (w), and 
if w be stated in Ib. weight, this will be in what is termed 
gravitation units. 

en 

,:. P=px 154905-335=242039-595, &c., Ib. 

Large charges of powder alone will not produce a high velocity, 
although in a great measure they assist it. The object to which 
gunnery is rapidly trending is minimum charges and higher 
velocities. 

Although the introduction of electricity will revolutionise this 
science to a very great extent, still, under circumstances where 
high angle fire appears necessary and advantageous, there can be 
but little doubt that the theory on which the general problems rest 
will still be found the pangenesis of formulae connected with this 
science. 

Its renaissance is dependent on a recognition of the theories 
already treated, for they embody certain fundamental laws of 
natural science inseparable from any speculation or experiment 
connected with gunnery. 

These laws cannot therefore be affected in any way by a mere 
change of an agent representing force. The force, if electricity, is 



PROJECTILES AND GUNNERY 433 

still a force, and only differs from other forces by virtue of its 
highly subtile character and the magnitude of its power. 

The word power is very frequently misapplied by writers and 
students, for they often call the mere pull, pressure, or force 
exercised on or by an agent the power. 

It should never be employed in any other sense than as expressing 
a rate of doing work, or activity. 

In electrical engineering the unit of power is called the watt, 
and it equals 10 7 ergs per second, or 746 watts = 1 horse-power. 



PROJECTIONS 

GENERAL DEFINITIONS 

703. The representation on a plane of the important points 
and lines of an object as they appear to the eye when situated 
in a particular position is called the projection of the object. 

704. The plane on which the delineation is made is called 
the plane of projection. 

705. The point where the eye is situated is called the point 
of sight, or the projecting point. 

706. The point on the plane of projection where a perpen- 
dicular to it from the point of sight meets the plane is called 
its centre. 

707. The line joining the point of sight and the centre is 
called the axis of the plane of projection. 

708. Any point, line, or other object to be projected is 
called the original, in reference to its projection. 

709. A straight line drawn from the point of sight to any 
original point is called a projecting line. 

710. The surface which contains the projecting lines of all 
the points of any original line is called a projecting surface. 
When the original line is straight, the projecting surface will 
be a projecting plane. 

COR. The projection of any point is the intersection of its 
projecting line with the primitive. 



434 



PROJECTIONS 



STEREOGRAPHIC PROJECTION OF THE SPHERE 

DEFINITIONS 

711. The stereographic projection of the sphere is that in 
which a great circle is assumed as the plane of projection, and one 
of its poles as the projecting point. 

712. The great circle upon whose plane the projection is made 
is called the primitive. 

713. By the semi-tangent of an arc is meant the tangent of half 
that arc. 

714. By the line of measures of any circle of the sphere is meant 
that diameter of the primitive, produced indefinitely, which is perpen- 
dicular to the line of common section of the circle and the primitive. 

715. Let A be the pole of the primitive BD, and MN a circle 

to be projected ; MN. being 
in the first figure a small 
circle, and in the second a 
great circle ; then the point 
M has for its projection the 
point m, and n is the projec- 
tion of N, and the circle mn 
is the projection of the circle 

MN. The line AM is the projecting line of the point M, and the 

plane AMN is the projecting plane of the diameter MN, whose 

projection is the line mn. 
In the stereographic projection, the projection of every circle of 

the sphere is a circle. 

716. Problem I. To find the locus of the centres of the 
projections of all the great circles that pass through a 
given point. 

Let F be any given point within the primitive ABCM. 

Through F draw the diameter BM and AC 
perpendicular to it ; draw AF, and produce it 
to D ; draw the diameter DL ; draw AL, and 
produce it to meet BM in G ; bisect FG per- 
pendicularly by II', and IF is the required 
locus. Thus any circle, PFN, passing 
through F, and having its centre in any 
point as I in IHF, is the projection of a 

great circle, and hence it cuts the primitive in two points, 

P, N, diametrically opposite. 





PROJECTIONS 



435 




717. Problem II. Through any two points in the plane 
of the primitive, to describe the projection of a great 
circle. 

1. When one of the points is in the centre of the primitive. 
Draw a diameter passing through the other point, and it will be 

the required projection. For the great circle passes through the 
pole of the primitive. 

2. When one of the points is in the circumference, and the 
other is neither in the circumference nor in the 

centre. 

Let A and P be the two points, and ACBD 
the primitive. 

Draw the diameter AB, and describe the circle 
APB through the three points A, P, B ; and it 
is the required circle. 

3. When neither of the points is in the centre or circumference. 
Let F, G be the given points, and ABC the . 

primitive. 

Find IH the locus of the centres of all the pro- 
jections of great circles passing through one of 
the points, as F (Art. 716) ; join F, G, and bisect 
FG perpendicularly by KH ; and the centre of 
every circle through F and G is in KH ; but the 
centre of the required circle is in IH ; hence H 
is its centre ; and a circle, DFG, through the two given points, 
described from the centre H, is the circle required. 

718. Problem III. About some given point, as a pole, to 
describe the projection of a great circle. 

1. When the given point is the centre of the primitive. 
The required projection is evidently the primitive itself. 

2. When the given point is in the circumference of the primitive. 
Draw a diameter through the given point, and another diameter 

perpendicular to the former ; the latter diameter is the required 
projection. 

For, since the primitive passes through the pole of the required 
projection, its original circle must pass through the pole of the 
primitive, and its projection is a diameter. 

3. When the given point is neither in the centre nor the cir- 
cumference of the primitive. 

Let P be the given point, and ADBC the primitive. 

Through P draw the diameter AB, and another CD perpendicular 




436 



PROJECTIONS 




to it. Draw DP, and produce it to E ; make the arc EF equal to 

a quadrant ; draw DF, cutting AB in G ; and the circle CGD, 
through the points C, G, D, is the required 
circle. 

For, considering APB as the primitive, and 
D its pole, PG is evidently the projection of a 
quadrant EF. Now, if ADBC be the primitive, 
since APB passes through P, the pole of the 
required circle, it must pass through C, D, the 
poles of AB. Hence the required circle must 

pass through C, G, and D. 

COR. Hence the method of finding the pole of a projected great 

circle is evident. 

1. When the projection is a diameter of the primitive. The 
extremities of the diameter perpendicular to it are evidently its 
poles. 

2. When the given projection is inclined to the primitive, as 
CGD. 

Join C, D, and draw the diameter AB perpendicular to CD. 
Draw DG, and produce it to F ; make the arc FE a quadrant ; 
draw DE, cutting AB in P, and P is the pole of the given circle. 

719. Problem IV. To describe the projection of a small 
circle about some given point as a pole. 

1. When the pole is in the centre of the primitive, or the original 
small circle parallel .to the primitive. 

Let AB, CD be two perpendicular diameters of the primitive. 
L Make CE equal to the distance of the small 

circle from its pole as, for example, 34. Draw 
DE, cutting AB in F ; from P as a centre, with 
the radius PF, describe the circle FGK, which 
will be the required projection. 

For PF is evidently the projection of CE, and 
the centre of the required circle is evidently 
in P. 

2. When the given pole is in the circumference of the primitive, 
or the original circle is perpendicular to the primitive. 

Let C be the given pole ; AB, CD two perpendicular diameters. 
Make CE equal to the distance of the circle from its pole. Draw 
EL a tangent to the primitive at E, and let it meet DC produced 
in L. A circle described from the centre L, with the radius LE 
namely, MNE is the required circle. 




PROJECTIONS 



437 




3. When the pole is neither in the centre nor the circumference 
of the primitive. 

Let P be the given point, and AB, CD two perpendicular 
diameters of the primitive. Draw CP, and produce it to E ; lay 
off EF, EG, each equal to the distance of the 
circle from its pole for instance, 62 ; draw 
CF, CG, cutting AB in H and I, and on 
HI, as a diameter, descril>e the circle HKI, 
and it is the required projection. For if AB 
be the primitive, and C its pole ; E the pole 
of a small circle, and F, G two points in its 
circumference, then HI is the diameter of its 
projection. Hence, if ACBD be the primitive, HI is evidently 
the diameter of the projected small circle, whose pole is P. 

COR. The method of finding the projected pole of a given 
projected small circle is manifest from this problem. 

1. When the small circle is concentric with the primitive, the 
centre of the latter is the projected pole of the former. 

2. When the small circle is perpendicular to the primitive, as 
MNE, its pole is in C, the middle of the arc MCE. 

3. When the circle is inclined to the primitive, as HKI, draw a 
diameter AB through its centre, and CD perpendicular to it ; draw 
CH, CI, cutting the primitive in F, G ; bisect FEG in E ; draw CE, 
and P is the required pole. 

720. Problem V. To measure any given arc of a pro- 
jected circle. 

1. If the given arc be a part of the primitive, 
it may be measured as the arc of any other circle 
(Art. 130 or 162). 

2. When the given arc is a part of a circle 
projected into a straight line. 

Let KL be any given arc of the projected 
circle AKB ; find C its pole, and draw CK, CL, 
cutting the primitive in F and G, and FG is the measure of KL, 
and is in the present instance 32. 

3. When the given circle is inclined to the primitive. 

Let HI be the given arc of the projected circle AIB. Find 
P its pole ; draw PH, PI, cutting the primitive in D, E, and 
DE is the measure of HI, which is therefore, in the present 
example, 45. 

?wc, 2 C 




438 



PROJECTIONS 




721. Problem VI. To measure the projection of a 
spherical angle. 

1. When the circles containing the given angle are the primitive 
and a diameter of it. 

The angle is a light angle. 

2. When one of the circles is the primitive, and the other is 
a circle inclined to it. 

Let AEB be the primitive, and AIB the other 
circle, and IAD the angle. Find F and C their 
centres ; draw AC, AF, and the angle CAP 
measures the given angle. Or, find F and P 
their poles ; draw AP, AF, cutting the primi- 
tive in G and B, and GB measures the given 
angle, which is in the present instance 40. 

3. When one of the circles is a diameter of the primitive, and 
the other is inclined to the latter. 

Let AFB and AIB be the two circles, and FAI the given angle. 

Draw the radius AC of the circle AIB, and AH perpendicular to 
AFB, and the angle HAG measures the given angle. Or, find P 
and E the poles of the circles ; draw AE, AP ; then GE measures 
the given angle, which is 50. 

4. When both the circles are inclined to the primitive. 

Let ABD, A'BD' be the two circles, and ABA' 
the given angle. Find C, C', the centres of the 
circles, then the two radii drawn from these to 
B will contain an angle CBC' equal to that at B. 
Or, find P, P', the poles of the circles, and lines 
drawn from B through these points will intercept 
on the primitive an arc which measures the given 
angle. The angle in this instance is 32. 

722. Problem VII. Through a given point in a given pro- 
jected great circle, to describe the projection of another great 

circle cutting the former at a given angle. 
Let ABCD be the primitive, and Z the given 
angle. 

1. When the given circle is the primitive. 
Let A be the given point ; draw the perpen- 
dicular diameters AC, BD ; make angle EAF 
= Z = 32, suppose; and from F as a centre, 
with a radius FA, describe the circle AGC ; it is the required 
projection, and angle GAD = 32, 





PROJECTIONS 



439 




When the angle is a right angle, the diameter AC is evidently 
the required projection. 

2. When the given projected circle is a diameter of the primitive. 
Let BD be the given projection, and F the given point. Find 

GH the locus of all the great circles passing 
through F ; draw FL perpendicular to BD, and 
FH, making an angle LFH = Z=46, for instance ; 
from the centre H, with the radius HF, describe 
the circle IFK ; it is the required projection, and 
angle DFK=46. 

If the angle be a right angle, G is the centre, 
and AFC the required projection, for angle LFG 
=a right angle. Or, since the required circle is 
in this case perpendicular to BFD, it must pass through its poles 
A and C. Hence the circle AFC, passing through the three points 
A, F, C, is the required projection. 

3. When the given circle is inclined to the 
primitive. 

Let AFC be the given circle, and F the given 
point in it. Find EG the locus of the centres of 
all the great circles passing through F. Draw 
FH a radius of the given circle, and draw FG, 
making the angle GFH = Z=23, suppose; from 
the centre G, with the radius GF, describe IFE ; and it is the 
required projection, and angle IFC = 23. 

When the angle Z is a right angle, draw from F a line perpen- 
dicular to FH, and it will cut EG in the centre of the required 
circle. Or since in this case the required projection must pass 
through the pole of AFC, find its pole, and describe the projec- 
tion of a great circle passing through this pole and the point F 
(Art. 717), and it will be the required circle. 

723. Problem VIII. Through a given point in the plane 
of the primitive, to describe the projection of a great circle 
cutting that of another great circle at a given angle. 

Let AKB be the given circle, Z the given angle, and C the given 
point in the plane of the primitive AMB. 

Find F the pole of AKB, and about it describe a small circle 
IGN, at a distance from its pole equal to the measure of angle 
Z = 44, for example. About the given point C, as a pole, describe 
a great circle LHM, intersecting the small circle in L and G. 
About either of these points, as G, for a pole, describe a great 





440 PROJECTIONS 

circle DCE, and it is the required projection. For the circle DCE 
must pass through C, since C is at the distance of a quadrant from 
G, a point of the circle LGM. Also, the distance 
between F and G, the poles of AKB and DCE, is 
the measure of the given angle, and hence the in' 
clination of the circles is equal to that angle = 44. 
SCHOL. 1. Let an arc of a great circle FCK be 
described through F and C ; then, FK and CH 
being quadrants, FH = CK. Now, FH must not 
exceed FN, the measure of the .angle, otherwise 
the circle LHM would not meet IGN, and the problem would be im- 
possible. ButCK = FH; therefore the distance of the given point 
from the given circle must not exceed the measure of the angle. 

SCHOL. 2. If the point C were in the centre of the primitive, 
the circle LGM would coincide with the primitive. If C were in 
the circumference of the primitive, the circle LGM would be a 
diameter perpendicular to that passing through C. 

724. Problem IX. To describe the projection of a great 
circle that shall cut the primitive and a given great circle 
at given angles. 

Let ADB be the primitive, AEB the given circle, and X, Y the 
given angles which the required circle makes respectively with 
these circles, and let these angles be respectively 47 and 45. 
About F, the pole of the primitive, describe a small circle at a 
distance of 47, the measure of angle X, and 
about G, the pole of AEB, describe another 
small circle at a distance of 45, the measure of 
angle Y. Then from either of the points of 
intersection H, I, as I for a pole, describe the 
great circle CED, and it is the required circle. 
For the distances of its pole I from F and G, 
the poles of the given circles, are equal to the 
measures of the angles X and Y ; and therefore the inclinations of 
CED to the given circles are equal to these angles that is, angle 
ACE = 47, and AEC = 45. 

SCHOL. When any of the angles exceeds a right angle, the 
distance of the small circle from its pole is greater than a quadrant. 
The same small circle will be determined by finding the more 
remote pole that is, the projection of the pole nearest to the 
projecting point and then describing a small circle about it at 
a distance equal to the supplement of the measure of the angle, 




8TEREOGRAPHIC PROJECTION 



441 




STEREOGRAPHIC PROJECTION OF THE CASES OF 
TRIGONOMETRY 

PROJECTION OF THE CASES OF RIGHT-ANGLED 
TRIGONOMETRY 

725. CASE 1. Given the hypotenuse AC = 64, and the angle 
C=46, to construct the triangle, and to measure its other parts. 

Let ECFD be the primitive ; draw the circle 
CAD, making angle C = 46 (Art. 722) ; about C, 
as a pole, describe the small circle I AH at a 
distance = 64 from C (Art. 719) ; then through A 
draw the diameter BK ; and ABC is the given 
triangle. 

Measure the sides AB, BC, and angle A (Arts. 
720 and 721) ; and it will be found that AB = 40 17', BC = 54 55', 
and A = 65 35'. 

726. CASE 2. Given the hypotenuse AC = 70 24', and the side 
BC = 65 10', to construct the triangle. 

Make the arc BC 65 10', and describe the small circle I AH at a 
distance from its pole C equal to 70 24' (Art. 
719) ; draw the diameter BAG, and then through 
A and C describe the great circle CAD ; and 
ABC is the required triangle. 

Measure the side AB, and angles A and C, as 
in the preceding problem. 

Angle C = 39 42', A = 74 26', and AB = 37. 

727. CASE 3. Given the side AB = 37, and BC = 65 10', to 
construct the triangle. 

Make BC = 65 10'; draw the diameter BAR 
as a pole, describe the small circle AIH at a 
distance from G=the complement of AB=53 
(Art. 719), then is AB = 37; through A and C 
describe the great circle CAD (Art. 717); and 
ABC is the required triangle. 

Measure AC, and angle A and C as before. 

AC = 70 24', A =74 26', and C = 39 42'. 

728. CASE 4. Given angle A = 32 30', and C = 106 
construct the triangle. 

Draw a diameter BL, and find its pole P (Art. 718, Cor.) ; about 
the pole P describe the small circle KI'I at a distance from Pof 




and about G, 

c 




442 




32 30' ; and about G, the pole of the primitive, describe a small 
circle 1'IQ at a distance from it = 73 36', the supplement of angle C 
(Art. 719) 5 and about I, the intersection of these 
small circles, describe the great circle CAD (Art. 
718) ; and ABC is the required triangle. 
Measure AB, BC, AC as before. 
AC = 117 31', AB = 126 42', and BC = 28 28'. 

729. CASfi 5. Given the side BC = 140 53', 
and angle C = 105 53', to construct the triangle. 

Make BGC = 140 53'; draw the diameter 
BAD, and through C describe the circle CAE, 
making angle FCE = 74 7', the supplement of 
105 53' ; and ABC is the required triangle. 

Measure AB, AC, and angle A. 

AC = 70 24', AB = 114 17', and A = 138 16'. 

730. CASE 6. Given AB = 40 25', and angle 
C = 44 56', to construct the triangle. 

Describe the circle CAD, making angle ACB = 44 56'; and 
about G, as a pole, describe the small circle 
AA' at a distance from G = 49 35', tlie com- 
plement of AB ; then through A and A' draw 
the diameters BH, B'H', and ABC, A'B'C are 
two triangles, constructed from the same data 
that is, having their sides AB, A'B' of the given 
magnitude, and the angle C common. 

Measure AC, BC, and angle A ; also A'C and B'C, and angle A'. 

AC = 66 38', BC = 58 36', and A = 68 25'; and A'C = 113 22', 
B'C = 121 24', and A' = 111 35'; the three latter parts are the 
supplements of the three former. 

PROJECTION OP CASES OF OBLIQUE-ANGLED 

SPHERICAL TRIGONOMETRY 

731. CASE 1. Given the side AB = 132 11', BC = 143 46', and 
AC = 67 24', to construct the triangle. 

Make ADB=132 11'; about A, as a pole, de- 
scribe the small circle DCE at a distance AD of 
67 24' ; and about B', the small circle FCG at 
a distance B'F = 36 14', the supplement of BC; 
then through A, C, and B, C, describe the 
great circles AC A', BCB'; and ABC is the 
required triangle. 
By measurement, angle A = 143 18', B = lll 4', and C = 131 30'. 





STEREOGRAPHIC PROJECTION 



443 




and 



732. CASE 2. Given the angle A = 114 30', B = 83 12', and C 
= 123 20', to construct the triangle. 

Describe the great circle ACA', making angle BAG = 114 30'; 
then about G, as a pole, describe a small circle 
PP'R at a distance from it=83 12' (Art. 719) ; 
and about the remote pole of ACA' describe the 
small circle P'PS at a distance from it = 56 40' 
=the supplement of 123 20'; then about either 
of the points of intersection P, P', as P, describe 
the great circle B'CB ; and ABC is the required 
triangle. 

It will be found by measurement that the side BC = 125 24'. 

733. CASE 3. Given the side AC =44 14', BC = 84 14', 
angle C = 36 45', to construct the triangle. 

Make AC =44 14'; make angle ACB = 36 45' 
(Art. 722) ; draw the small circle IBH about C, 
as a pole, at a distance = 84 14'; and through the 
points A, B describe the circle ABK (Art. 717) ; 
and ABC is the required triangle. 

By measurement, AB = 51 6', angle A = 130 5', 
and B= 30 26'. 

734. CASE 4. Given angle A = 130 5', -B= 
30 26', and the side AB = 51 6', to construct 
the triangle. 

Make AB = 51 6', angle BAG = 130 5', or 
EAC=49 55' (Art. 722), and ABC = 30 26'; and 
ABC is the required triangle. 

By measurement, AC =44 14', BC = 84 
and angle C=3645'. 

735. CASE 5. Given the side AC = 80 
angle A =51 30', to construct the triangle. 

Make AC = 80 19', and angle BAC = 51 30' (Art. 722); about 
C, as a pole, describe B'B at a distance =63 50'; 
and through B and C describe the circle EEC ; 
or through B' and C describe EB'C ; and either 
ABC or AB'C is the required triangle. 

By measurement, in the triangle ABC, AB = 
120 46', angle B = 59 16', and angle C = 131 32'. 
In the triangle AB'C, angle B' is the supplement 
of B = 180-59 16' = 120 44'; bnt AB' is not the 
supplement of AB, nor angle ACB' of ACB. It is found that 
AB'=28 34', and ACB' =24 36'. 




14', 
19', BC = 63 50', and 




444 STEREOGRAPHIC PROJECTION 

736. CASE 6. Given angle A=31 34', B = 3028', and the side 
BC = 40, to construct the triangle. 

Make BC = 40, and angle ABC = 30 28' (Art. 722); about the pole 
of BAD, and at a distance = 31 34', describe 
a small circle PP'G, cutting the diameter 
PP', which is perpendicular to CK in P and 
P'; about P. as a pole, describe the great 

e j[._c/ _\!-/ ,/T j circle CAK, and ABC is the required tri- 

/ angle. 

The great circle described about P' as a 
pole would cut the circle BAD at the given 
angle ; but it would be an exterior angle of 
the triangle, to which the side BC belongs. 

But if A were<B, there would then be two triangles ; in this case 
the two poles, P and P', would lie on the same side of CK. 
By measurement, AC = 38 30', AB = 70, and C = 130 3'. 




SPHERICAL TRIGONOMETEY 

737. Spherical Trigonometry treats of the methods of 
computing the sides and angles of spherical triangles. 

DEFINITIONS 

738. A sphere is a solid every point in whose surface is equi- 
distant from a certain point within it. 

This point is called the centre. A sphere may be conceived to 
be formed by the revolution of a semicircle about its diameter as 
an axis. 

739. A line drawn from the centre to the surface of a sphere is 
called its radius ; and a line passing through the centre of the 
sphere, and terminated at both extremities by its surface, is called 
a diameter. 

740. Circles whose planes pass through the centre of the sphere 
are called great circles ; and all others, small circles. 

741. A line limited by the spherical surface, perpendicular to the 
plane of a circle of the sphere, and passing through the centre of 
the circle, is called the axis of that circle ; and the extremities of 
the axis are the poles of the circle. 

742. The distance of two points on the surface of the sphere 
means the arc of a great circle intercepted between them. 



SPHERICAL TRIGONOMETRY 



445 



743. A spherical angle is an angle at a point on the surface of 
the sphere, formed by arcs of two great circles passing through the 
point, and is measured by the inclination of the planes of the circles, 
or by the inclination of their tangents at the angular point. 

744. A spherical triangle is a triangular figure formed on the 
spherical surface by arcs of three great circles, each of which is 
less than a semicircle. 

When one of the sides of a spherical triangle is a quadrant, it is 
called a quadrantal triangle. 

745. The sides of a spherical triangle being arcs of great circles 
of the same sphere, their lengths are proportional to the number 
of degrees contained in them ; and hence the sides of spherical 
triangles are usually estimated by the number of degrees they 
contain. 

The definitions of trigonometrical ratios given in ' Plane Trigono- 
metry ' are employed in reference to the sides and angles of spherical 
triangles. 

746. A spherical angle is measured by that arc of a great circle 
whose pole is the angular point which is intercepted by the sides 
of the angle. 

Thus, the spherical angle ABC, which is 
the same as the angle contained by the planes 
ABF, CBF of the two arcs AB, BC that 
contain the angle, is measured by the arc AC 
of a great circle ACD, whose pole is the angu- 
lar point B ; or by the angle MEN contained 
by the tangents MB, NB to the arcs BA, BC. 
For angle MEN = angle AEG, and AEC is 
measured by AC. 

747. Two arcs are said to Be of the same species, affection, or 
kind when both are less or both greater than a quadrant ; and 
consequently the same term is applied to angles in reference to 
a right angle. 

The species of the sides and angles of spherical triangles can 
generally be easily determined by means of the algebraical signs 
of their cosines, cotangents, &c. 

748. To find the relations between the trigonometrical 
functions of the three sides and the three angles of any 
spherical triangle. 

Let ABC be a spherical triangle, and let O be the centre of 
the sphere on which it is described; then OA = OB = OC, and let 





446 SPHERICAL TRIGONOMETRY 

AD be a tangent to the arc AB, produced to meet OB in D, and 

AE a tangent to the arc AC, produced to meet OC in E and join 
DE. Then, if A, B, and C represent the 
\ D three angles, and a, b, and c the sides 
opposite them ; since AD and AE are tan- 
gents to the arc AB and AC, the angle 
DAE is the measure of the spherical angle 
BAG ; also, c is the measure of the angle 

AOD, and b is the measure of the angle AOE, and a is the 

measure of the angle BOG or DOE ; hence 

AO AD OA . AE 

^pT = cos c, -pyjf^sin c, ;^pr =cos o, and ^pp=sm b. 

Therefore in the triangle DAE, 

DE 2 = AD 2 + AE 2 - 2AD . AE cos A ; . . [.] ; 
and from triangle DOE, 

DE 2 = OD 2 + OE 2 -20D.OEcosa. . . . [6]. 
Subtracting [a] from [6], and observing that OD 2 -AD 2 and 
OE 2 -AE 2 are each equal to OA 2 (Eucl. I. 47), since the angles 
OAD and OAE are right angles, we obtain 

0=2OA 2 + 2AD . AE cos A - 2OD . OE cos a ; 
transposing and dividing by 2OD . OE, 

coaa = OA x OA AD x AE cosA 

or cos a=cos c . cos & + sin c . sin b . cos A. 

Similarly, cos b =cos a . cos c + sin a . sin c . cos B, 
and cos c cos a . cos 6 + sin a . sin b . cos C. . 

Again, transposing and dividing by the coefficients of the cosines 
of the angles, cos a - cos b cos c "\ 



cosB = 

and cos C = 



sin b sin c 

cos b - cos a cos c 
sin a sin c 

cos c - cos a cos b 



[rf]; 



c 

and reducing to a common denominator, and putting 1 - cos 2 & for 
sin 2 6, and 1 - cos 2 c for sin 2 c in the numerator, there results 

1 - cos 2 6 - cos 2 c - cos 2 + 2 cos a cos b cos c 
sm 2 A= 



sin a sin b 

. (cos a - cos b cos c) 2 
whence, 1 - cos-'A, or sin^A = 1 TZT =y ; 



SPHERICAL TRIGONOMETRY 447 

Taking the root of this, and dividing the two sides by sin a, the 
second side will be a symmetrical function of a, b, c, which we 
shall call M namely, 

sin A _ VI - cos 2 - cos 2 6 - cos 2 c + 2 cos a cos b cos c _ , ^ 
sin a sin a sin b sin c 

But if A and a be now changed into B and 6, or into C and c, 
the second side will remain the same ; hence the first side must 
continue constant, and we shall have 

-, sin A sin B sin C , 

M= = r= -; hence . . . [el 

sin a sin b sin c 



749. In every spherical triangle the sines of the angles 
are proportional to the sines of the opposite sides. 

According to the property of the supplemental triangle,* change, 
in [c], a into 180- A, &c., and we shall have 

- cos A = cos B cos C - sin B sin C cos a. ~ 
Similarly, - cos B =cos A cos C - sin A sin C cos b, 
and - cos C = cos A cos B - sin A sin B cos c. 

cos A + cos B . cos C 

COR. Whence, cosa= -. ~ 

sin B . sin C 

cos B + cos A . cos C 

COS b = ; 7 ; ~ -> 

sin A . sin C 

cos C + cos A . cos B 

cos c = j T -. 5 

sin A . sin B. 

To' eliminate b from equation 1 of [c], put sin b= B [el, 

sin A 

and cos b = cos a cos c + sin a sin c cos B [c] ; substituting in the 
result 1 - sin 2 c for cos 2 c, and dividing the whole by the common 
factor sin a sin c, we have 

sin c cot a = cos c cos B + sin B cot A. " 

Similarly, sin c cot b = cos c cos A + sin A cot B, 

sin a cot c=cos cos B + sin B cot C. 

(sin a cot 6 = cos a cos C + sin C cot B,"j 
sin b cot a cos b cos C + sin C cot A, J- . . [/]. 
and sin b cot c=cos b cos A + sin A cot C. J 

* If the angular points of a spherical triangle be made the poles of three great 
circles, these three circles by their intersections will form a triangle which is said 
to be supplemental to the former ; and the two triangles are such that the sides of 
the one are the supplements of the arcs which measure the angles of the other. 



448 SPHERICAL TRIGONOMETRY 

The equations [c], [e], [/], [A] are the foundation of the whole of 
Spherical Trigonometry, and serve for the solution of all triangles ; 
but as they are not suited to logarithmic calculation, we proceed to 
deduce from them more convenient formulae. 

SOLUTION OF RIGHT-ANGLED SPHERICAL TRIANGLES 

750. In the preceding formulae, if one of the angles, as B, be a 
right angle, then sin B = l, and cos B = 0, and we at once have, 
by making these substitutions in the above formulae, [c], [e], [/], 
and [A]. 

From [c], . . cos 6 = cos a. cos c . . . . (I). 

. 1 fsin a=sin A . sin b . . . . (m). 

\sin c=sin C . sin 6 . . . . (n). 

(cos 6 = cot A . cot C . . . . (o). 
cos A = cos a . sin C . . . (p). 

cos C = cos c . sin A . . . . (q), 
/-sin c = tan a . cot A . . . . (r). 
.-,, | sin = tan c.cot C . . . . (s). 

1cosA=tan c.cot b . . . . (t). 
vcos C = tan a. cot b . . . . (). 
Collecting the values of each quantity into one line, and multi- 
plying the first side by R, to make it true for any radius, we have 
R . cos b = cot A . cot C = cos . cos c . . (o) and (I), 
R . sin a = tan c . cot C = sin 6. sin A . . (*) n (m), 
R . sin c = tan a . cot A = sin b . sin C . . (r) (n), 
R. cos A = tan c.cot b cos a. sin C . . (t) n (p), 
R . cos C = tan .cot 6 = cos c . sin A . . (u) (q). 
The above ten equations are all included in two rules, called 
Napier's Rules, for the circular parts ; they are the following : 

751. If in a right-angled spherical triangle the right angle be 
omitted, there remain other five parts. Napier observed that if 
the two sides which contain the right angle, the complements 
of the other two angles, and the complement of the side opposite 
the right angle be called the five circular parts, then any three 
of these being taken, they will either be adjacent, or one of them 
will be separated from each of the other two by another of the 
circular parts. Let now that part which lies between the other 
two, or which is separated from the other two, be called the 
middle part ; and the remaining two, when they all lie together, 
the adjacent parts ; and when they are separated from it, the 
opposite parts ; then, 



SPHERICAL TRIGONOMETRY 449 

752. R. xsine of the middle ( tangents of the adjacent parts, 

part = prod act of the \cosines opposite n 

It will, in fact, easily be seen that these two conditions contain 
all the ten preceding equations, which are true as first given, for 
radius = 1 ; and in the second form and in the rule, are true for any 
radius. 

These equations, taken in connection with the signs of the 
trigonometrical ratios, demonstrate various general properties 
which it will be of use to observe in all right-angled spherical 
triangles. 

1st. From equation (I) we conclude that any one of the three 
sides is < or > 90, according as the other two sides are of the same 
or different species. 

2nd. Equation (o) shows that if the hypotenuse be compared 
with the two adjacent angles A and C, any one of these three 
arcs is < or > 90, according as the two others are of the same or 
different species. 

3rd. The equations (p, q) prove that each of the angles A and C 
is always of the same species as the opposite sides and c ; and 
conversely. 

4th. Equations (t, u) prove that the hypotenuse and a side 
are of the same species when the included angle is acute, and of 
different species when the included angle is obtuse. 

5th. Equation (/) proves that if cos a = 0, or = 90, cos 6 = 0, 
and .'. 6 = 90; hence cot 6 = 0, and from (t) cos A=0, or A = 90; 
so that the sides a and b are both quadrants, and perpendicular to 
the third side c ; and C is the pole of the arc c, consequently c is 
the measure of the angle C ; the triangle is then isosceles, and has 
two right angles, and the third side and third angle contain the 
same number of degrees. 

The above five theorems will be found useful when any triangle 
is divided into two right-angled triangles by an arc drawn from 
one of its angles perpendicular to the opposite side. 

753. To find expressions suited to logarithmic calculation for the 
three angles of a spherical triangle, in terms of the three sides. 

cos a - cos b . cos c . 

By Art. 748 (d), cos A = - : =- = - - ; hence 



1+COS A = 



: =- = 
sin b . sin c 

cos a - cos 6 . cos c 



sin b . sin c 
cos a - (cos b . cos c - sin b . sin c) 
sin b . sin c 



450 



SPHERICAL TRIGONOMETRY 



cos a -cos (b + c) 



and hence, 

Similarly, 
and 
Again, 

and hence 

Similarly, 
and 
Also, 


sin b . sin c 
_2 sin ^(a + b + c) sin ^(b + c-a) 


sin b sin c 


/sin s . sin (s - a) * 

(*r\c\ 1 \ / . v ' 


'V sin o . sin c 


1T> /sin s . sin (5-6) 


^ sin a . sin c 


/sin s . sin (s - c) 


^ sin . sin b J 
A , cos a - cos 6 . cos c 


sin b . sin c 
cos b . cos c + sin b . sin c - cos a 


sin b . sin c 
cos (6 - c) - cos a 


sin o. sin c 
2 sin ^(a + & - c) sin (a + c - 6) ^ 


sin 6 . sin c 


: i A /sin ^(a + 6-c) sin ^(a + c-6) 


V sin o . sin c 


-in 31 A /sin(.9-6).sin(s-cn 


'y sin o . sin c 


. n 1T} /sin (s - rt) . sin (s - c) 


V sin a . sin c 


in /sin (,?-). sin (s- &) 


biUjjC-y S in a. sin 6 J 




Vsin (s - 6) . sin (s - c) sin b . sin c 


sin b . sin c " sin s . sin (s - a) 


Vsin (s - b) sin (s - c) 


sin s . sin (s - a) 


/sin (s - a) . sin (s - b) . sin (s - c) __ 



sin s . sin 2 (s - a) 



SPHERICAL TRIGONOMETRY 



451 



and therefore tan 



sm (s- 
Similarly, tan | 



and tan 



L 


f 1 


. 


. 


/o 


b) , sin 


i n 


/>\ 


sin 






\s 


(> 


-C). 


1 


. 


- a) . sin 


iy 


b) . sin 


tm 


f,\ 


sin 


s 


\ s 


\ S 


:jj 


1 1 


. 


- a) . sin 


(*- 


b) . sin 


(s-c). 




s 



The three angles of a spherical triangle can be calculated 
logarithmically from either of the three sets of formulae given 
above ; but the last, which gives the tangent of the semi-angle, 
will be found the most convenient in practice, as all the angles can 
be found in terms of four arcs, while those formulae which give the 
semi-angles in terms of the cosine or sine require the use of seven 
arcs ; besides, the angles can be found with greater accuracy from 
the tangents than from the sine or cosine, as the tangent varies 
more rapidly than either the sine or cosine. 

754. When the three angles are given to find the sides, the 
supplements of the given angles may be taken for the sides 
of a new triangle, and the angles of this triangle, found from 
the formulae of last article, will be the supplements of the 
sides of the given triangle, from which the sides can easily be 
found. 

Formulae similar to the above, expressing the sides in terms of 
the angles, may be deduced from [g] in the same manner as those 
in the last article ; they are the following : s= 



and 



and 



- cos s . cos (s - A) 
sin B . sin c 



a _ /cos (s - B) . cos (s - C) 
sin B . sin C 



. b I 
in = =./ 

2 AJ 

b /cos (s 

os H=*/ '*-> 
2 A/ si 



- cos s . cos (s - B) 
sin A . sin C 

-A).cos(s-C) 



c I 
m 2 = Ay 



-> = 
sin A . sin 



- cos s . cos (s - C) 
sin A . sin B ' 



452 SPHERICAL TRIGONOMETRY 



ix /cos (s- A) . cos (s-B) 

and cos ^= A / 4i 

sin A . sin B 



cot s = - - - T- . / -- . cos (s - A) . cos (s - B) . cos (s - C), 
2 cos (s- Aw cos s 

COt 7; = - ; - =7. / --- . COS (S- A) . COS (S - B) . COS (S-C), 

2 cos (s-B)Ay coss 

1 / I 

; j=r . / . cos (s - A) . cos (s - B) . cos (s - C). 
(s - C)Ay cos s 



c 
cot 7; = 

2 cos ( 



755. When the parts given are either two sides and the contained 
angle, or two angles and the side lying between, the other parts 
are most conveniently found by a set of formulae called Napier's 
Analogies, which may be established as follows : 



Let M = . / . sin (s - a) . sin (s-b) sin (s - c), then 
V sin s 

AM B M 
tan 7r = r.tan-T^ : rr> 

2 sin (s -a) 2 sin (s - b) 

and tan -^ = . ; ; hence 

2 sin (s - c) 

M M 



A + B_sin (s-ffl) sin (s-b) 

0~~ M2 

1- 



sin (s - a) . sin (s - 6) 



5 . sin (s-a) . sin (s-b) . sin (s - c){sin (s - b) + sin (s - a)} 
sin (s - a) . sin (5 - 6){sin s - sin (s - c)} 

sin . sin (s - c) sin (s - b) + sin (s - a) 



sin (s - a) . sin (s - b) sin * - sin (s - c) 

, c (a-b) 

2 sin-, cos - 77 p, ,, 

, C . C cos *(a - 1 

= COt . 1 =COt 77 . - 

2 , /a + b\ c 2 c< 



In a similar manner, after some reduction, we find 

A-B / sin s . sin (s - c) sin (s-b)- sin (s - a) 
tan ^ = . /^ : . , ' x - 



J_ / sin s . sin 
A/ sin (*-). si 



sin(s-6) sin s + sin (s-c) 
c 



n 2 cos p: sin *( - b) n 

C 2 , C sin 



TT- -^- -. 

2 ... c 2 sin 4(a + 6) 

2 sin J(a + 6) . cos 5 



SPHERICAL TRIGONOMETRY 453 

Therefore, 




and 
In a similar manner, it may be shown from Art. 753 (k) that 




and tan i(a-6) = tan ^- sin 

The above four equations, which can easily be converted into 
proportions, are called Napier's Analogies. 

756. Rule for determining the sign of the answer in a proportion. 
If the fourth term is a cosine, tangent, or cotangent, and of the arcs 
whose cosines, tangents, or cotangents enter in the first three terms, 
if one or three are greater than a quadrant, so is the fourth term. 

757. CASE 1. Given the hypotenuse and one of the angles of a 
right-angled triangle, to find the other parts. 

EXAMPLE. In the spherical triangle ABC right-angled at B, the 
hypotenuse AC is 64, and the angle C 46 3 ; what 
are the remaining parts ? 

1. To find BC 

When angle C is the middle part, BC and the 
complement of AC are the adjacent parts ; there- 
fore R . cos C = cot AC . tan BC ; and as BC is 
wanted, the proportion must be (Art. 750) 

Cot AC : R = cos C : tan BC. 
+ Cot AC 64 . . . . = 9-6881818 

+ Radius ..... = 10- 

+ CosC46 ..... = 9-8417713 

+ Tan BC 54 55' 35-8" . . = 10 '1535895 

Since the signs of the first three terms are +, for radius is always 
positive, that of the fourth must be so, and +tan B shows that 
Bis<90. 

2. To find AB 

AB being the middle part, AC and C are opposite parts ; therefore 
(Art. 750, n) R . sin AB-sin C . sin AC ; or, since AB is required, 

R : sin AC = sin C : sin AB. 
Radius ...... =10- 

Sin AC 64 . . . . , = 9-9536602 

Sin C 46 ..... = 9-8569341 

Sin AB 40 16' 52" . 9-8105943 

Prac 2 D 



454 SPHERICAL TRIGONOMETRY 

The sine for 9-8105943 may be either that of 40 16' 52", or its 
supplement 139 43' 8" ; but in the given triangle, the angle C 
opposite to the side AB is acute ; hence AB is<90 (Art. 742). 

3. To find angle A 

When AC is the middle part, angles A and C are the adjacent 
parts, and (Art. 750, o) R . cos AC = cot A . cot C ; hence 

CotC:R = cos AC: cot A. 
Cot C 46 . . ... . = 9-9848372 

Radius =10- 

Cos AC 64 = 9-6418420 

Cot A 65 35' 4" . 9 "6570048 

EXERCISES 

1. The hypotenuse is = 75 20', and one of the oblique angles 
=57 16' ; what are the other parts? 

The two sides = 64 10' 20" and 54 28' 3", and the other angle 
= 68 30' 4". 

2. The hypotenuse is = 64 40', and an angle = 64 38' 11" ; find the 
other parts. 

The other angle =47 55' 50", its opposite side = 42 8' 24 -5", 
and the other side = 54 45' 25". 

758. CASE 2. Given the hypotenuse and a side. 

EXAMPLE. Let the hypotenuse AC and the side BC of the tri- 
angle ABC be given equal to 70 24' and 65 10' respectively, to 
find the other parts. 

1. To find angle C 

By Art. 750 (), R . cos C = cot AC . tan BC ; 
hence R : cot AC tan BC : cos C. 

Radius =10- 

Cot AC 70 24' .... = 9-5515524 
Tan BC 65 10' .... = 10-3346338 
Cos C 39 41' 40" . 9-8861862 

2. To find AB 

By Art. 750 (1), R . cos AC = cos BC . cos AB ; 
hence Cos BC : R = cos AC : cos AB. 

Cos BC 65 10' . . . . = 9-6232287 

Radius =10' 

Cos AC 70 24' .... = 9-5256298 
Cos AB 36 59' 27" = 9-9024011 



SPHERICAL TRIGONOMETRY 



455 



3. To find angle A 

By Art. 750 (m), R . sin BC = sin AC . sin A ; 
hence Sin AC : R=siu BC : sin A. 



Sin AC 70 24' 
Radius . 
Sin BC 65 10' 
Sin A 74 26' 26" 



9-9740774 
= 10- 
= 9-9578626 

9-9837852 



EXERCISES 

1. The hypotenuse is = 75 20', and a side is = 64 10'; required 
the other parts. 

The other side = 54 28' 32", its opposite angle = 57 16' 32", and 
the other angle = 68 29' 40". 

2. The hypotenuse AC is = 50, and the side BC=44 18' 39"; 
what are the other parts ? 

AB=26> 3' 53", angle A =65 46' 6", and angle C = 35 e . 

759. CASE 3. Given the two sides. 

EXAMPLE. The side AB is=37, and BC is=65 10 7 ; find the 
other parts. 

1. To find AC 

By Art. 750 (1), R . cos AC = cos AB . cos BC ; 
hence R : cos AB = cos BC : cos AC. 

Radius =10- 

CosAB37 = 9-9023486 

Cos BC 65 10' . . . . = 9-6232287 
Cos AC 70 24' 9" . . . . = 9 '5255773 

2. To find angle A 

By Art. 750 (r), R . sin AB = cot A . tan BC ; 
hence Tan BC : R = sin AB : cot A. 

Tan BC 65 10* . . . = 10-3346338 

Radius =10' 

SinAB37 ..... = 97794630 
Cot A 74 26' 14-5" . 9-4448292 

3. To find angle C 
By Art. 750 (s), R . sin BC = cot C . tan AB ; 

hence Tan AB : R = sin BC : cot C. 

TanAB37 = 9-8771144 

Radius = 10- 

Sin BC 65 10' 9-9578626 

Cot C 39 42' 14" . . = 10-0807482 



456 SPHERICAL TRIGONOMETRY 

EXERCISES 

1. The two sides are = 54 28' and 64 10' ; find the other parts. 
The angles are = 57 16' 1 "4" and 68 29' 48", and the hypotenuse 

= 75 19' 48". 

2. The two sides are =42 12' and 54 41' 28" ; what are the other 
parts ? 

The angles are = 48 0' 49" and 64 33' 24", and the hypotenuse 
= 64 38' 54". 

760. CASE 4. Given the two oblique angles. 

EXAMPLE. The angle C is = 106 24', and angle A = 32 30'; 
required the other parts. 

1. To find AC 

By Art. 750 (o), R . cos AC = cot A . cot C ; 
hence R : cot A = cot C : cos AC. 

+ Radius =10' 

+ Cot A 32 30' . . . . = 10-1958127 

- Cot C 106 24' (73 36') . . = 9-4688139 
-Cos AC 11 7 30' 55" . 9-6646266 

In the Tables the cosine here belongs to an arc of 62 29' 5" ; 
but since the sign of cot C, one of the terms, is negative, that of 
the fourth term, cos AC, must also be negative (Art. 756) ; and 

hence AC > 90. 

2. To find AB 

Angle C being M, A and com p. AB are and o. 
By Art. 750 (q), R . cos C = sin A . cos AB ; 
hence Sin A : R = cos C : cos AB. 

+ Sin A 32 30' . . . . = 9'7302165 
+ Radius =10' 

- Cos C 106 24' . . . . = 9-4507747 

- Cos AB 121 42' 3" . 97205582 

AB is also < 90, for angle C is so (Art. 756). 

3. To find BC 

By Art. 750 (p), R . cos A = sin C . cos BC ; 
hence SinC : R=cos A : cos BC. 

+ Sin C 106 24' . 9-9819608 

+ Radius =10- 

+ CosA3230' . 9-9260292 

+ Cos BC 28 27' 31" . . . 9-9440684 

BC is <90, for angle A is so. 



SPHERICAL TRIGONOMETRY 
EXERCISES 



457 



1. The two angles are = 39 42' and 74 26' ; find the other parts. 
The sides are = 36 59' 39" and 65 9' 28", and the hypotenuse 

= 70 23' 39". 

2. The angles A and C are respectively = 138 15' 45" and 
105 52' 39" ; what are the other parts ? 

The sides AB and BC are = 114 15' 54 -2" and 140 52' 39'6", 
and AC = 71 24' 30-3". 

761. CASE 5. Given a side about the right angle and its 
adjacent angle. 

EXAMPLE. The side BC is = 140 53', and angle C is = 105 53'; 
find the other parts. 

1. To find AC 

By Art. 750 (), R . cos C = cot AC . tan BC ; 
hence Tan BC : R = cos C : cot AC. 

- Tan BC 140 53' . . . = 9-9101766 
+ Radius . . . . = 10- 

- Cos C 105 53' . . . . = 9-4372422 
+ Cot AC 71 23' 55-3" . . = 9-5270656 

Angle A is of the same species with BC, and hence A and C are 
of the same species ; therefore (Art. 752) AC<90. 

2. To find AB 

By Art. 750 (*), R . sin BC = cot C . ten AB ; 

hence Cot C : R =sin BC : tan AB. 

-CotC 105 53' . . . . = 9-4541479 

+ Radius =10- 

+ Sin BC 140 53' . 9-7999616 

-Tan AB 114 16' 33" = 10-3458137 

The side AB and angle C are of the same species (Art. 752). 



3. To find angle A 
R . cos A = sin C . cos BC ; 
R : sin C=cos BC : cos A. 



By Art 750 (p), 
hence 

+ Radius % . = 10- 

+ Sin C 105 53' . 9*9830942 

- Cos BC 140 53' . 9-8897850 

- Cos A 138 15' 57" . 9'8728792 
Angle A is of the same species as BC, 



458 SPHERICAL TRIGONOMETRY 

EXERCISES 

1. Aside and its adjacent angle are respectively = 119 11' and 
126 54' ; find the other parts. 

The hypotenuse = 71 27' 43", the other side = 130' 41' 42", and 
the other angle = 112 57' 0'7". 

2. The side AB is = 54 28' 10", and angle A = 68 29' 48" ; what 
are the other parts ? 

AC = 75 19' 54-3", BC = 64 10' 3'2", and 0=57 16' 10'3". 

762. CASE 6. When a side ab ut the right angle and its 
opposite angle are given. 

EXAMPLE. Given AB=40 25', and angle C=44 56', to find the 
other parts. 

1. To find AC 

By Art. 750 (n), R . sin AB = sin AC . sin C ; 

hence Sin C : R = sin AB : sin AC. 

Sin C 44 56' . . . . . = 9-8489791 

Radius =10' 

Sin AB 40 25' . 9-8118038 

Sin AC 66 37' 48" . . . = 9-9628247 

Or (Art. 752), AC is also 180 -66 37' 48" = 113 22' 12". 

2. To find angle A 

By Art. 750 (q), R . cos C = sin A. cos AB ; 

hence Cos AB : R = cos C : sin A. 

Cos AB 40 25' . 9-8815842 

Radius =10" 

Cos C 44 56' = 9-8499897 

Sin A 68 24' 30" . 9 "9684055 

Or, A is also 180 - 68 24' 30" = 111 35' 30". 

3. To find BC 

By Art. 750 (s), R . sin BC = cot C . tan AB ; 

hence R : cot C = tan AB : sin BC. 

Radius = 10- 

CotC 44 56' . 10-0010107 

Tan AB 40 25' .... = 9-9302195 
Sin BC 58 36' 0'6" . 9-9312302 

Or, BC is also 180 -58 36' 0'6" = 121 23' 59'4", 



SPHERICAL TRIGONOMETRY 



459 



As either of the triangles ABC, A'B'C (fig. to Art. 730), fulfils 
the conditions given in this case, it is hence called the ambiguous 
case. In practical applications of this subject it is generally easily 
known which of the two triangles is to be taken. If, for example, 
it were known that the side BC or the angle A is less than 90, 
the triangle ABC alone would satisfy the given conditions, and the 
triangle A'B'C would be excluded. 

EXERCISES 

1. Given AB or A'B' (fig. to Art. 730) = 26 4', and the opposite 
angle C = 35, to find the other parts. 

AC = 50 0' 18", or AC' = 129 59' 42"; angle A = 65 46' 13", 
or angle A' = 114 13' 47"; and BC=44 18' 57", or B'C = 
135 41' 3". 

2. Given (fig. to Art. 730) AH or A'H' 115, and angle C 
= 114 14', to find the other parts. 

AC = 83 39' 43", or A'C = 96 20' 17"; angle A = 103 46' 50", 
or A'=76 13' 10" ; also, CH = 105 8' 33", or CH' = 7451' 27". 

3. Proof triangle in which all the parts are given, and in which, 
if any two of the five parts be taken as the given parts, the other 
three will be found by the previous rules ; it will therefore afford 
an exercise in each of the ten cases of right-angled trigonometry ; 
the right angle is A. 



Elements 


Log. Sine 


Log. Cosine 


Log. Tangent 


a = 71 24' 30" 


9-9767235 


9-5035475 + 


10-4731759 + 


b =140 52 40 


9-8000134 


9-8897507- 


9-9102626- 


c =114 15 54 


9-9598303 


9-6137969- 


10-3460333 - 


B = 138 15 45 


9-8232909 


9-8728568- 


9-9504341- 


C = 105 52 39 


9-9831068 


9-4370867 - 


10-5460201 - 



QUADRANTAL SPHERICAL TRIANGLES 

763. If the supplements of the sides and angles of a quadrantal 
triangle be taken, they will be the angles and sides respectively of 
a right-angled spherical triangle ; the supplement of the quadrantal 
side being the right angle. 

This is evident from the properties of the polar triangle, which 
are the following : 

764. If the angular points of a spherical triangle are made the 
poles of three great circles, another spherical triangle will be 




460 SPHERICAL TRIGONOMETRY 

formed, such that the sides of each triangle are the supplements 
of the angles opposite to them in the other triangle. 
Thus, if the angular points A, B, C of the triangle ABC are 
respectively the poles of the sides EF, DE, DF 
opposite to them in the triangle DEF, then EF 
is the supplement of angle A, DE of B, DF of C, 
BC of D, AB of E, and AC of F. 

Hence, if ABC were a quadrantal triangle, 
AB being the quadrantal side, then DEF would 
be a right-angled triangle, E being the right 
angle. 

Quadrantal triangles can therefore be solved by the rules for 
right-angled triangles. It will also be sufficient to know two parts 
of such a triangle besides its quadrantal side ; for then two parts of 
the supplemental right-angled triangle are known ; and if its other 
parts are then calculated, the supplements of its sides and angles 
will be respectively the angles and sides of the given triangle. 

EXERCISE 

Aside and the angle opposite to the quadrantal side are = 136 8' 
and 61 37' ; find the other parts. 

The other side = 65 26' 35", and the other two angles =53 9' 
and 142 26' 2". 

OBLIQUE-ANGLED SPHERICAL TRIGONOMETRY 

765. The number of cases in oblique-angled trigonometry 
formed in reference to the given parts is six, as in the former 
section. 

These cases, except when the three sides or three angles are 
given, can be solved by the method used in the preceding section, 
as explained afterwards under the next head. The solutions may, 
however, frequently be more conveniently effected by means of 
other methods, which are here employed for that purpose. 

The rules used in the first four cases to determine the species of 
the part sought are : 

766. The half of a side or angle of a spherical triangle is less 
than a quadrant. 

For a side or an angle of a spherical triangle is less than two 
right angles. 

Other two rules are given under the fifth and sixth cases, to be 
employed in their solution. 






SPHERICAL TRIGONOMETRY 461 

The rule in Art. 759 is also applicable to oblique angled 
spherical triangles. 

767. Half the difference of any two parts of a triangle is less 
than a quadrant. 

For each part is less than 180". 

768. It is to be observed in forming examples in spherical 
trigonometry that the sum of the three sides of a spherical 
triangle is less than the circumference of a circle, and the sum of 
any two sides is greater than the third ; also, the greater angle is 
opposite to the greater side, and conversely. 

769. CASE 1. When the three sides are given. 

This case can be conveniently solved by any of the three 
following rules : 

RULE I. From half the sum of the three sides subtract the side 
opposite to the required angle ; then add together the logarithmic 
sines of the half-sum and of this difference and the logarithmic 
cosecants of the other two sides ; and half the sum, diminished by 
10 in the index, will be the logarithmic cosine of half the required 
angle. 

Let the three sides be denoted by a, b, c ; the angles respectively 
opposite to them being A, B, and C ; and half the sum of the sides 
by s; 

/sin s . sin (-a)\J 

then (Art. 753. z cos AA= - ' r. 

V sin b . sin c J 

And for B and C the formulae are exactly analogous ; that for B, 
for instance, being formed from the above by changing A into B, 
a into 6, and b into a. 

RULE II. From half the sum of the three sides subtract sepa- 
rately the sides containing the angle ; add together the logarithmic 
sines of the two remainders and the logarithmic cosecants of these 
two sides ; and half the sum, diminished by 10 in the index, will 
be the logarithmic sine of half the required angle. 

If A is the required angle, 

.. .. . , /sin (s-b) sin (s-c)\i 
then (Art. 7o3, j) sin 4A = ( - -^ 5^ - I . 

\ sin b . sin c J 

RULE III. From half the sum of the three sides subtract the 
side opposite to the given angle, and also each of the sides con- 
taining it ; then add together the logarithmic cosecant of the half- 
sum and the logarithmic sines of the three remainders ; and half 
the sum will be a constant, which, being diminished by the siue of 



462 SPHERICAL TRIGONOMETRY 

the half-sum, minus the side opposite the angle sought, will be 
the logarithmic tangent of half the required angle. 
Let A be required, then (Art. 753, k) 

tan iA= - ; .( cosec s , sin (s-a) . sin (s - b) . sin (s - c) } . 

sin (*-a)\ / 

770. The third rule is given in a new form, and is both more 
accurate in particular cases and more easy in practice than either 
of the other two when all the three angles are sought. 

EXAMPLE. The sides of a spherical triangle are = 143 46', 67 24', 
and 132 11'; find the angles (see fig. in Art. 731). 
Let a = 143 46' 

b= 67 24 B y Rule m 

c = 132 11 L cosec s . = 10'8392676 

2)343~21 L sin (s-a). = 9 '6703000 

then ,=171 40 30" Lsin(*-&) . = 9-9863791 

-= 27 54 30 Lsin(s-c) . = 9-8034339 

s-6 = 104 16 30 2)40-2993806 

s-c= 39 29 30 20-1496903 = 

C - L sin (s - a) = L tan A = 10-4793903 
C-Lsin (,-6) = L tan |B = 10-1633112 
- L sin (s - c) = L tan C = 10'3462564 
Hence angle A = 143 18' 34" 

B = lll 3 18 
and C = 131 29 32 

EXERCISES 

1. Find the three angles of the spherical triangle whose three 
sides are a = 33 4', 6 = 74 16', and c = 94 18'. 

A = 26 34' 54-6", B = 52 7' 47 '6", and C = 125 7' 57 '2". 

2. The sides of a spherical triangle are = 62 54' 4", 125 20', and 
131 30'; what are its angles? 

= 83 12' 10", 114 30', and 123 20' 32". 

771. CASE 2. When the three angles of a spherical triangle are 
given. 

RULE I. From half the sum of the three angles subtract the 
angle opposite to the required side ; then add together the loga- 
rithmic cosines of the half-sum and of this remainder and the 
logarithmic cosecants of the other two angles ; and half the sum, 
diminished by 10 in the index, will be the logarithmic sine of half 
the required side. 



SPHERICAL TRIGONOMETRY 463 

Let be the required side, and S half the sum of the angles ; 

/ - cos S . cos (S - ) \i 

then (Art. 7o4) sin ia = ( = R ^-^ ) . 

\ sin B . sin C / 

RULE II. From half the sum of the three angles subtract 
separately the angles adjacent to the required side ; then add 
together the logarithmic cosines of the two remainders and the 
logarithmic cosecants of the other two angles ; and half the sum, 
diminished by 10 in the index, will be the logarithmic cosine of 
half the required side. 

Let be the required side ; 

/cos(S-B)cos(S-C)U 
then (Art.7o4) cos fr = ( ^ B '. rin * c ') - 

RULE III. From half the sum of the three angles subtract 
each angle separately ; then add together the logarithmic secant 
of the half sum and the logarithmic cosines of the three re- 
mainders; and half the sum will be a constant, from which, if 
the logarithmic cosine of the half-sum, diminished by any angle, 
be subtracted, the remainder will be the logarithmic cotangent of 
half the side opposite to that side. (See Art. 754.) 

EXAMPLE 

The angles of a spherical triangle are = 114 30', 83 12', and 
123 20' ; find the sides. 

To find the side a by Rule III 

Here A = 11430' 

B = 83 12 

C = 123 20 

2)321 2 



S =160 31 
S-A= 46 1 
S-B= 77 19 
S-C= 37 11 


L sec = 10-0256087 
Lcos = 9-8416404 
L cos= 9-3415580 
L cos= 9-9012980 



2)39-1101051 
= 19-5550525 
Log. cot = C-L cos (S-A)= 9'7134121; .-. Ja= 62 39' 55 

2 

.-. = 125 19 50 

In the same manner, the other sides may be found to b 
= 62 54' 16", and c = 131 23' 32". 



464 SPHERICAL TRIGONOMETRY 

EXERCISES 

1. The three angles are = 111 4', 143 18', and 31 30'; find the 
sides. The sides are = 67 25' 35", 143 44' 46", and 132 10' 26". 

2. The three angles A, B, C of a spherical triangle are 
respectively = 70 39', 48 36', and 119 15'; what are the sides? 

The side a = 89 16' 53 "4", 6 = 52 39' 4*5", and c = 112 22' 58 -6". 

772. The two following cases can be solved by means of the 
analogies of the circular parts,* which are expressed in the 
following manner : 

Let one of the six parts of a triangle be omitted, and let the 
part opposite to it, or its supplement when it happens to be an 
angle, be called the middle part (M) ; the two parts next it, 
the adjacent parts (A, a) ; and the two remaining parts, the 
opposite parts (0, o) ; then 

sin %(A +a) : sin \(A -a) = tan \M : tan \(0 -o), 
and cos \(A +a) : cos \(A -) = tan \M : tan %(0 + o). 

By means of these two analogies, half the sum and half the 
difference of and o are found, and each of them is then easily 
found. 

773. When A, a, 0, and o are given, M can be found from the 
first of these analogies by placing it for the last term, and 
sin \(A - a) for the first, and the other two indifferently for the 
second and third ; thus, 

sin \(A - a) : sin \(A + ) = tan %(0 - o) : tan \M. 
Or, M can be similarly found from the second analogy. 

774. CASE 3. When two sides and the contained angle are 
given, as a, b, and C. 

Omit the side c, and make the supplement (Art. 772) of C the 
middle part M; then the sides a, b are the adjacent parts A, a ; 
and the angles A, B the opposite parts 0, o ; hence (Art. 755, v) 
sin %(a + b) : sin (~6) = cot C : tan (A~B), 
cos !( + &) : cos J(a~6) = cot C : tan $(A + B). 
The half-sum and half-difference of A and B being found by 
these two analogies, each of them is then easily found. 

To find the side c 

Reject C, and make c the middle part; then c is M ; angles 
A, B are adjacent parts ; and the sides a, b are opposite parts ; 
hence (Art. 773), 

sin (A~B) -. sin ^(A + B) = tan \(a~b) : tan c. 



These are called Napier's Analogies, as they were discovered by him. 



SPHERICAL TRIGONOMETRY 



465 



EXAMPLE. In a spherical triangle two sides are = 84 14' 29" 
and 44 13' 45", and the contained angle = 36 45' 28"; required 
the remaining parts. 

a =84 14' 29" 
6 = 44 13 45 
C = 36 45 28 

i( + 6) = 64 14' 7", \(a- 6) =20 0' 22", 
iC = 1822'44". 



Here 



and 



Sec 
Cos 
Cot 



1. To find i(A+B) 



Hence i(A + B) = 81 15' 44-4". 



2. To find J(A~B) 



Sin 
Cot 
Tan 



i(a-6) 



= 10-3618336 

= 9-9729690 

= 10-4785395 

10-8133421 



10-0454745 
= 9-5341789 
' = 10-4785395 
= 10'0581929 



J(A~B) . . . 
Hence |( A ~ B) = 48 49' 38". 

Since a>b, therefore A>B; hence 

A =81 15' 44 -4" + 48 49' 38" = 130 5' 22-4", 
and B = 81 15' 44 "4" -48 49' 38"= 32 26' 6 "4". 

3. To find the side c 

Cosec|(A~B)4849'38" . . = iO'1233621 

Sin i(A + B)81 15'44-4", . . = 9-9949302 

Tan J(a-6) 20 0' 22" . . = 9-5612100 

Tan c2533' 5 -8" . 9 '6795023 
And c=51 6' 11 -6". 

EXERCISES 

1. Given two sides = 89 17', 52 39', and the contained angle 
= 119 15', to find the other parts. 

The other side is = 112 23' 2", and the other angles = 70 39' 3" 
and 48 35' 58 -5". 

2. The sides a and b are = 109 21' and 60 45', and angle C is 
= 127 20' 55-5" ; find the other parts. 

Angles A and B are = 90 43' 6 -6" and 67 37' 1'4", and the 
side c is = 131 24'. 



466 



SPHERICAL TRIGONOMETRY 



775. CASE 4. When two angles and the interjacent side are 
given. 
Let the angles A and B and the interjacent side c be given. 

1. To find the sides a and b 

Omit C, and let c be the middle part ; then A and B are the 
adjacent parts, and a and b the opposite parts ; hence 
sin |( A + B) : sin J( A ~ B) = tan \c : tan \(a ~ b), 
cos \( A + B) : cos (A ~ B) = tan \c : tan %(a + b). 

2. To find angle C 

Omit c, and make the supplement of C the middle part ; then 
the sides a, b are adjacent parts ; and the angles A, B are opposite 
parts; hence (Art. 772), since tan M = cot JC, 

sin \(a~b) : sin $(a + b) = tan ^(A~B) : cot ^C. 
EXAMPLE. The angles A and B are = 130 5' 22-4" and 32 26' 6'4", 
and the side c is = 51 6' 1T6" ; required the other parts. 

1. To find a and b 

(A + B)=81 15' 44-4", 

(A~B) = 48 49' 38", and c = 25 33' 5 -8". 

. 9-1815881 
. 9-8184449 
. 9-6795022 
19-4979471 
. 10-3163590 
14' 7". 



Sin 

Sin (A~B) 

Tan \c . 



9-9949302 
9-8766379 
9-6795022 
19-5561401 

Tan|(a~6) . 9 '5612099 
Andi(a~6)=200'22". 



Cos 

Cos i(A~B) 

Tan \c . 



And !( + &) = 64 



And since A>B, therefore a>b ; whence 

a =84 14' 29", 6 = 44 13' 45". 

2. To find angle C 

Cosec J( - b) 20 0' 22" .. = 10-4658211 

Sin }(a + b) 64 14' 7" . = 9'9545255 

Tan i(A~B) 48 49' 38" . . = 1Q-Q581929 

Cot |C 18 22' 44" . 10-4785395 
And C = 36 45' 28". 

EXERCISES 

1. The angles A and B are = 82 27' and 57 30', and side c is 
= 126 37' ; what are the other parts ? 

The angle C is = 124 42', and the sides and 6 = 104 34' 28" 
and 55 25' 32". 



SPHERICAL TRIGONOMETRY 467 

2. Given A=66 57' 36", B = 97 20' 31'6", and the side c 
=41 9' 46", to find the third angle and other two sides. 

C=42 30' 55", a = 63 39' 58", and 6 = 75 0' 51 -6". 

776. CASE 5. When two sides and the angle opposite to one of 
them are given. 

Let a, b, and A be given ; then B can be found by the analogy, 
sin a : sin 6 = sin A : sin B. 

When B is found, there are then two sides and their opposite 
angles known ; and hence c and C can be found as in the third 
and fourth cases ; thus 

sin (A~B) : sin J(A + B) = tan J( ~ 6) : tan \c, 
sin \(a <~ b) :sin \(a + 6) = tan ^(A~B) :cot C. 

There will, however, be sometimes two values of B, as in the 
analogous case of plane trigonometry, and consequently two 
triangles can be formed from the same data; hence this is an 
ambiguous case, as is also the next case, for a similar reason (see 
fig. to Art. 735). 

When B has two values, so have c and C. The values of B are 
supplementary ; and by using first one of its values, the correspond- 
ing values of c and C will be found by the last two analogies, and 
then all the parts of one of the triangles are known. When the 
other value of B is taken, and the corresponding values of c and C 
are computed in the same manner, all the parts of the other 
triangle will then be known. 

Whenever differs from 90 in excess or defect less than 6 does, 
there will be only one triangle, and therefore only one value of B, 
which will be of the same species as b ; in other cases B has two 
values that are supplementary. 

When the difference of a from 90 is less than that of 6, then 
it is evident that sin >sin b; that is, if (Jir~a)<(^T'~6), then 
sin a>sin b. 

EXAMPLE. The sides a, b are = 38 30' and 40, and angle 
A = 30 28' ; required the other parts. 

To find angle B 
Sin a : sin 6 = sin A : sin B. 

Cosec a 38 30' .... = 10-2058504 
Sin b 40 . . . . . = 9-8080675 
Sin A 30 28' . . . . = 9*7050397 
Sin B 31 34' 14", . 9 '7189576 

Or, B = 148 25' 46". 



468 SPHERICAL TRIGONOMETRY 

B has two values, for (^ir~>a)>(%ir~b), since 

90 -38 30' = 51 30', and 90 -40 = 50, or sin <sin 6. 

Taking the triangle that has B acute, then B = 31 34' 14". 
There are now known , b, A, and B, to find c and C, which are 
calculated exactly as in the third and fourth cases ; and when this 
is done all the parts of this triangle are known. 

Taking next the triangle that has B obtuse, then B = 148 25' 46" ; 
hence in this triangle are known a, b, A, and B ; and consequently 
c and C in it are found also as in the preceding triangle. 

It will be found in the triangle in which B is acute that 
C = 130 3' 50", and c = 70 0' 29". 

EXERCISES 

1. Given a =24 4', & = 30, and A = 36 8', to find the other parts. 
B = 46 18' 6", or 133 41' 54"; C = 103 59' 50", or 11 23' 33"; 

and c=42 8' 49", or 7 51' 5 -4". 

2. Given = 76 35' 36", 6 = 50 10' 30", and A =121 36' 19 "8", to 
find the other parts. 

B = 42 15' 13-5", = 34 15'2'8", and c=40 0' 10". 

777. CASE 6. When two angles and a side opposite to one of 
them are given. 

Let A, B, and a be given ; then b will be found by the analogy, 
sin A : sin B = sin a : sin b. 

When b is found, there are then two sides and their opposite 
angles known ; and hence c and C can be found as in the preceding 
case ; thus 

sin J(A~B) : sin (A + B) = tan ( ~ b) :tan |c, 
sin \(a ~ b) : sin %(a + b) = tan (A~B) : cot |C. 

There will sometimes be two values of b admissible, as there were 
of B in the preceding case, and consequently also two triangles 
(see fig. in Art. 736). 

When b has two values, so have c and C. The values of b 
are supplementary, and by taking one of them there will then 
be known in one of the triangles the parts A, B, a, b ; hence 
c and C can now be found, and all the parts of this triangle 
will be known. Taking then the second value of b, the remaining 
parts c, C of the other triangle can similarly be found. 

Whenever A differs from 90 in excess or defect by less than B 
does, there will be only one triangle, and therefore only one value 
of b, which will be of the same species as B ; in other cases b has 
two values that are supplementary. 

When (iT~A)<(j7r~B), then sin A>sin B. 



SPHERICAL TRIGONOMETRY 469 

EXAMPLE. The angles A and B are = 31 34' and 30 28', and the 
side is =40 ; required the other parts. 

To find the side b 

Sin A : sin B = sin a : sin b. 

Cosec A 31 34' . . . . = 10-2810914 
Sin B 30 28' . . . . = 97050397 

Sin a 40 = 9*8080675 

Sin b 38 30' 18-5" . 9-7941986 

The side b has only one value, for sin A>sin B. 
In the triangle are now known the parts A, B, a, b, and the 
remaining parts c and C may be computed in the same manner 
as in the third and fourth cases. 

EXERCISES 

1. Given A = 51 30', B=59 16', and = 63 50', to find the other 
parts. 

6 = 80 19' 9", or 99 40' 51" ; C = 131 29' 53", or 155 22' 19" ; 
and e = 120 48' 5", or 151 27' 3". 

2. Given A = 97 20' 31 '6", B=66 57' 3 "6", and = 75 0' 51'3", to 
find the other parts. 

C=42 30' 54-7", 6 = 63 39' 57 -8", and c=41 9' 45 -6". 

Besides determining the species of the parts of oblique spherical 

triangles by means of the algebraical sines of the required parts, 

they can also be ascertained by certain theorems in spherical 

geometry. 

OTHER SOLUTIONS 

The preceding methods of solution .are generally the most con- 
venient when all the parts of a spherical triangle are required ; but 
when only one part is required, it will be more concise and simple 
to use some of the following methods : 

778. The third, fourth, fifth and sixth cases can be solved by 
dividing the given triangle into two right-angled triangles by 
means of & perpendicular from one of the angles upon the opposite 
side, so that one of the right-angled triangles shall contain two of 
the given parts. 

By the method, however, of right-angled trigonometry alone, it 
would be necessary always to calculate the perpendicular ; but this 
unnecessary calculation is avoided by eliminating the perpendicular 
from two equations. 

P. 2 E 



470 SPHERICAL TRIGONOMETRY 

779. THE THIRD CASE. Let the given parts be a, b, and C ; and 
let a perpendicular BD be drawn from 

^j\ /T\ angle B upon the side b ; let be the 

jr I \ y I \ segment of b that is nearest to C, 

A ^ --- D "~Ac A<C__i_ J" reckoning from C towards A when 

C is acute, but from C in AC pro- 

duced when C is obtuse; then AD = 6-0 when C is acute, and 
AD = b + when C is obtuse. 

1. To find and A 

From the triangle BCD, by making C, or its supplement when 
it is obtuse, the middle part, BC and CD are the adjacent parts ; 
therefore, 

cos C = tan 6 cot , or tan = tan a. cos C . . [1]. 

Again, from the triangles ABD and BDC, by making AD and 
DC the middle parts, we have 

sin = cot C tan BD, and sin (b + 0) = cot A tan BD ; 
hence, by division, 

sin 6 cot C . tan BD cot C tan A 



sin (6 + 0)~cot A. tan BD~cot A~tan C' 
or sin (6+0): sin = tan C : tan A . . . [2], 

2. To find the side c 

In the triangles ABD and CBD, we have by right-angled 
trigonometry, 

cos c=cos (b + 6) cos BD, and cos = cos cos BD ; 

cos c cos (b + 6). cos BD_cos (b + 0) f 
' ' cos a~ cos 6 . cos BD cos 6 

cos a . cos (b + 0) 

hence cos c= - ^ - > 

cos 6 

or . cos 6: cos (6 + 0) = cos a : cos c . . . [3], 

The angle B can be found exactly in the same manner as A by 
supposing the perpendicular to be drawn from A upon the side a. 
The formulae for this purpose are easily obtained from those for A 
by merely changing A into B, a into b, and b into a. 

EXERCISE 

Given a = 89 17', 6 = 52 39', and C = 119 15', to find A, B, and c. 
A = 70 39' 3", B=48 35' 57", and c = 112 22' 60". 



SPHERICAL TRIGONOMETRY 471 

780. THE FOURTH CASE. Let the given parts be A, B, c, and 
let a perpendicular be drawn from B upon b ; let angle ABD 
= <t>, then, 

To find the side " and the angle C 

From the triangle ABD, cos c=cot <f> . cot A ; 
or cos c . tan A = cot <p ; . . R : cos c = tan A : cot (f> . [4]. 

From the triangles ABD and CBD we have 

cos = cot c. tan BD, and cos (B~0)=cot a. tan BD ; . 
cos (B~ft)cot. tan BDcot qtan c 



cos (f> cot c . tan BD cot c tan a ' 

hence cos (B<~0) : cos c/> = tan c : tan a . . . [5J, 

Again, from the same triangles we have 

cos A=sin <f> . cos BD, and cos C = sin (B~</>). cos BD; 

cos A_ sin <(> cos BD sin <f> 

' cos C ~sin (B~0) cosBD~sin (B~#) ' 
and hence sin (f> : sin (B~#) = cos A : cos C . . . [6J. 

EXERCISE 

Given A = 82 27', B = 57 SO 7 , and c = 126 37', to find a, b, and C. 
a = 104 34' 30", 6=55 25' 32", and C = 124 42? 7". 

781. THE FIFTH CASE. Let a, b, and A be given, and let a 
perpendicular CD be drawn from C upon c ; let the segment of c 
next to A be denoted by 0, and the opposite angle ACD by <f> ; then, 

1. To find the angle B 
. sin b . , T> 

bin B = -* . sin A, or sin a : sin b = sin A : sin B. 

sin a 

When a is nearer to 90 than b, B has only one value, which is 
of the same species as b. When a differs more from 90 than b, 
then B has two supplementary values (Art. 777). 

2. To find the side c 

Let AD = 0, then BD = (c~0) ; and 
cos A=cot b . tan 0, or 
tan = tan b. cos A . . [7]. 
Also, cos 6 = cos cos CD, 
and cos a = cos (c ~ 0) cos CD ; 

cos b cos 6 . cos CD cos 




cos a cos (c~6) cos CD cos (c~6) ' 
hence cos b : cos a = cos 6 : cos (c~6) , . . [7]. 



472 SPHERICAL TRIGONOMETRY 

782. The value of 6 found above does not show whether the 
perpendicular is within or without the triangle, as c is not known ; 
but the species of A and B determine this circumstance, for the 
perpendicular falls within or without the triangle according as 
angles A and B are of the same or of different affection. 

3. To find the angle C 

Let ACD = #, then BCD = C~c/>; and from the triangle ACD we 
have cos b = cot A. cot <f>, or cot </>= cos b tan A . . [8]. 

Also, from the triangles ACD and BCD, 

cos 0=cot b tan CD, and cos (C~<j>) = cot a tan CD. 

cos <j> _cot b . tan CD_cot &_tan a 
' cos (C~<p)~ cot a tan CD ~cot a~~tan b ' 
hence tan a : tan 6 = cos <j> : cos (C~0) . . . [9]. 

783. When B has two values, one of them for instance, its acute 
value should first be taken, and the side c and angle C of the 
triangle to which it belongs are then to be calculated ; then, its 
other value being taken, the side c and angle C of the triangle to 
which it belongs are to be calculated. 

EXERCISE 

'Given a = 76 35' 30", 6 = 50 10' 30", and A = 121 36', to find 
B, C, and c. B = 42 15' 26", C = 34 15' 15", and c = 40 0' 14". 

784. THE SIXTH CASE. Let A, B, and a be given, and let a 
perpendicular be drawn, as in the preceding case ; then, 

1. To find the side b 

Sin A : sin B = sin a : sin b . . . [10]. 
When A is nearer to 90 than B, b has only one value, which 
is of the same species as B ; in any other case b has two supple- 
mentary values (Art. 777). 

2. To find the angle C 

By last case [8], cot = cos b. tan A, and from the triangles 
ACD and BCD we have 

cos A=sin . cos CD, 
and cos B = sin (C~<j>) cos CD 

cos A _ sin <f> . cos CD sin <f> 

' ' cos B~sin (C~0) cos CD~sTnTC~0) ' 

hence cos A : cos B = sin <j> : sin (C-~</>) . . [11]. 

It is known, as in Art. 782, whether the perpendicular falls 
within or without the triangle. 






SPHERICAL TRIGONOMETRY 473 

The species of C - <f> is not thus determined, as its sine is the 
fourth term ; but those of B and a being known in triangle BCD, 
that of DCB = C-< is known, if the formula (Art. 750, o) cos a 
= cot B. cot (C~0) be used; or, R : cos a = tan B : cot (C -<f>). [12]. 

3. To find the side c 

By Art. 781, tan = tan 6. cos A, and from the triangles ACD 
and BCD we have 

siu = cot A . tan CD, and sin (c~0) = cot B tan CD ; 

sin _cot A . tan CD_cot A_tan B 
' sin (e~0)~cot B. tan CD~cot B~tan A' 

hence tan B : tan A = sin 6 : sin (c~0) . . [13]. 

The species of c - 6 is not determined, as its sine is the last term ; 
but in triangle BCD, the species of (c - 0) is known from those of 
a and B. 

785. When b bas two values, one of them for instance, its acute 
value can first be taken, and the angle C and side c of the triangle 
to which it belongs are then to be calculated ; its other value being 
next taken, the angle C and side c of the triangle to which it 
belongs are to be computed. 

EXERCISE 

Given A = 97 W 30", B = 66 57', and a =75 1', to find C, b, andc. 
C=42 31' 16", 6 = 63 39' 59", and c=41 Iff 8". 

786. All the forms of a triangle that can exist when two sides 
and an angle opposite to one of them are given are contained in 
the following diagram : 

Let MM', PIT be two perpendicular diameters of the circle 
MPM', which is the base of a hemisphere ; and let C be any point 
in its surface, and the arcs passing through C be all the halves 
of great circles, except CB", CjS", which are 
portions of great circles. Let all these be 
symmetrically situated in the semicircles PMir, 
PMV, so that every two corresponding arcs 
as, for instance, AC, A'C are equally in- 
clined to PC*-. 

Hence every two corresponding arcs, 
reckoning from C, as CB and CB' or C/3 and 
C/3', are equal. Also, of all these arcs, Cir is 
the greatest, CP the least ; and any arc, as Ca, nearer to the 
greatest is greater than any other, as CA' that is, more remote 




474 SPHEBICAL TRIGONOMETRY 

(Solid Geoni., p. 54). Also, the arcs CM, CM' are quadrants; and 
therefore all the arcs, reckoning from C to the semicircle MPM', 
are less than quadrants, andthose terminating in the semicircle 
MirM' are greater than quadrants. 

Let PC = /i, and denote the parts of the triangle ABC as usual ; 
then R . sin A = sin B sin a, 

. R . sin h 

and sm B = : 

sin a 

It is evident that when h is constant, sin B is least when sin a 
is greatest that is, when a is a quadrant and equal to CM' or 
CM. Hence, of all the angles subtended by CP at the points 
B, A', B", M', a, ...that at M' is the least, and that at a point 
nearer to M', either in the quadrant PM' or irM', is less than one 
more remote ; and these angles, therefore, are all less than right 
angles, for h<a; and their adjacent angles are greater than right 
angles. Also, when two arcs CA' and Ca are supplementary, the 
acute angles at A' and o are equal. 

When each of the sides and b is less than a quadrant, and 
angle A is acute, and sin >sin b, only one triangle can be 
formed, and the unknown angle B is of the same affection as b. 
This appears from the triangle ACB", where B"C = , AC = b, and 
angle CAP = A ; for B"C> AC, and <aC. 

When each of the sides a and b is greater than a quadrant, and 
sin >sin b, and A is obtuse, there can be only one triangle, 
and angle B will be of the same affection as b. This appears from 
the triangle aC/3", where )3"C = a, aC = b, and angle Cair = A ; where 
/3"C is intermediate between aC and its supplement AC, and there- 
fore sin >sin 6. 

When sin <sin b there will be two triangles. This appears 
when A is acute from the triangles ACB, ACB', in which CB 
= CB' ; and when A is obtuse, from the triangles aC/3, aC/3'. And 
in this case the two values of B are supplementary. 

By examining all the possible cases, it will be found that they 
are comprehended in the rules for the signs of the trigonometrical 
ratios (Art. 195). 

Let A, B, and a be given. When each of the given parts is 
less than a quadrant, and sin A>sin B, there can be only one 
triangle, and >the unknown side b will be of the same species 
as B. This appears from triangle ACB", where B"C = a. Were 
sin A<sin B, there could then be two triangles as BCA, BCa'. 
And by examining in the same way all the possible cases, the 
theorem stated in Art. 776 is easily established. 



ASTRONOMICAL PROBLEMS 475 



ASTRONOMICAL PROBLEMS 

CIRCLES AND OTHER PARTS OF THE CELESTIAL 
SPHERE 

787. To an observer placed on the surface of the earth 
the heavenly bodies appear to be situated on the surface of 
a concave sphere, of which the place of the observer is the 
centre; for the magnitude of the earth is a mere point in 
reference to the distance of all celestial bodies, except those 
belonging to the solar system, and it becomes sensible in 
regard to the distances of the latter only when accurate 
observations are taken with proper instruments. 

The apparent diurnal revolution of these bodies from east 
to west, caused by the real daily rotation of the earth on its 
axis in the opposite direction from west to east, and the 
apparent annual motion of the sun in the heavens, arising 
from the earth's annual revolution in its orbit in an opposite 
direction, are, for convenience, in the practice of astronomy 
and navigation, considered as real motions ; and the positions 
of these bodies are determined accordingly, for any given 
time, with the aid of the principles of spherical trigonometry. 

DEFINITIONS 

788. The celestial sphere is the apparent concave sphere on 
the surface of which the heavenly bodies appear to be situated. 

789. The axis of the celestial sphere is a straight line passing 
through the earth's centre, terminated at both extremities by the 
celestial sphere. About this axis the heavenly bodies appear to 
revolve. 

790. The poles of the celestial sphere are the extremities of 
its axis, one of them being called the north, the other the south 
pole. 

The poles appear as fixed points in the heavens, without any 
diurnal rotation, the bodies near them appearing to revolve round 
them as centres. 



476 ASTRONOMICAL PROBLEMS 

791. The equinoctial or celestial equator is a great circle in 
the heavens equidistant from the poles ; and it divides the celestial 
sphere into the northern and southern hemispheres.* 

This circle referred to the earth is the equator ; also the axis of 
the earth is a portion of that of the celestial sphere. 

792. The ecliptic is a great circle that intersects the equator 
obliquely, and is that in which the sun appears to perform its 
annual motion round the earth. 

793. The two points in which the ecliptic and equinoctial inter- 
sect are called the equinoxes, or equinoctial points; that at 
which the sun crosses the equator towards the north is called the 
vernal equinox, and the other the autumnal equinox. 

794. The zodiac is a zone extending about 8 on each side of the 
ecliptic ; and it is divided into twelve equal parts, called the signs 
of the zodiac. 

The names and characters of these signs are : 

Cp Aries, 50 Cancer, ^ Libra, V5 Capricornus, 

y Taurus, Q Leo, tl^ Scorpio, %& Aquarius, 

IE Gemini, H Virgo, / Sagittarius, X Pisces. 

The first six lie on the north of the equinoctial, and are called 
northern signs ; the other six, on the south of that circle, are 
called southern signs. Each sign contains 30. 

The signs are reckoned from west to east according to the 
apparent annual motion of the sun. The first, Aries, is near the 
vernal equinox ; and Libra, near the autumnal equinox. t 

795. The solstitial points are the middle points of the northern 
and southern halves of the ecliptic ; the northern is called the 
summer, and the southern the winter, solstice, t 

796. The horizon is the name of three circles : one, the true 
or rational ; another, the sensible ; and a third, the visible or 
apparent horizon. 

The first is the intersection of a horizontal plane passing through 

* This circle is called the equinoctial because when the sun is in it the nights, 
and consequently the days, are equal everywhere on the earth's surface. 

t About three thousand years ago the western part of Aries nearly coincided 
with the vernal equinox ; but from the slow westerly recession of this point, 
called the precession of the equinoxes, it is nearly a whole sign to the west of Aries. 

J These points are so named because when the sun (sol) has arrived at either 
of them it appears to stop (sto), in reference to its motion north and south, and 
then to return ; and hence, also, the origin of the term tropics, from a Greek word 
(TJT) which means a turn. 



ASTRONOMICAL, PROBLEMS 477 

the earth's centre with the celestial sphere ; the second is the 
intersection with this sphere of a plane parallel to the former 
touching the earth's surface at the place of the observer ; and the 
third is the intersection with the same sphere of the conic surface, 
of which the vertex is at the eye of the observer, and the surface of 
which touches on every side the surface of the earth, considered as 
a sphere. 

797. A vertical line passing through the earth's centre and the 
place of the observer may be called the axis of the horizon ; and 
the extremities of this axis, where it meets the celestial sphere, 
the poles of the horizon ; the upper pole being called the zenith, 
and the lower the nadir. 

798. The north, east, south, and west points of the horizon are 
called the cardinal points. 

799. Meridians are great circles passing through the poles of the 
celestial sphere ; they are also called hour circles. 

These circles correspond to meridians on the earth. 

800. A meridian passing through the equinoctial points is called 
the equinoctial colure ; and that passing through the solstitial 
points, the solstitial colure. 

801. Circles passing through both poles of the ecliptic are called 
circles of celestial longitude. 

802. Vertical circles are great circles passing through both the 
poles of the horizon. 

803. A vertical circle passing through the east and west points 
of the horizon is called the prime vertical. 

804. Small circles parallel to the equinoctial are called parallels 
of declination. 

805. Small circles parallel to the ecliptic are called parallels of 
celestial latitude. 

806. Small circles parallel to the horizon are called parallels 
of altitude. 

807. The right ascension of a heavenly body is an arc of the 
equinoctial intercepted between the vernal equinox and a meridian 
passing through the body. 

808. The longitude of a heavenly body is an arc of the ecliptic 
intercepted between the vernal equinox and a circle of longitude 
passing through the body. 



478 ASTRONOMICAL PROBLEMS 

809. The azimuth of a body is an arc of the horizon intercepted 
between the north or south point and a vertical circle passing 
through the body.* 

810. The amplitude of a body is an arc of the horizon inter- 
cepted between the east or west point and a vertical circle passing 
through the body.t 

811. The declination of a body is its distance from the equi- 
noctial, measured by the arc of the meridian passing through it 
which is intercepted between the body and the equinoctial. 

812. The latitude of a body is the arc of a circle of longitude 
intercepted between the body and the ecliptic. 

813. The altitude of a body is the arc of a vertical circle passing 
through the body, intercepted between it and the true horizon. 

814. The dip or depression of the horizon is the angle of de- 
pression of the visible horizon below the sensible, in consequence 
of the eye of the observer being situated above the surface of the 
earth. 

815. The observed altitude is the altitude indicated by the 
instrument, the apparent altitude is the result after correcting 
the observed altitude for the error of the instrument and the dip, 
and the true altitude is the result after correcting the apparent 
altitude for refraction and parallax. The meridian altitude of a 
body is its altitude when on the meridian. When a body is on the 
meridian it is said to culminate ; and its culmination is said to 
be upper or lower according as it is then in its highest or lowest 
position. 

816. The polar distance or codeclination of a body is its 
distance from the pole of the equinoctial, measured by the arc 
of a meridian intercepted between the body and the pole. 

817. The zenith distance or coaltitude of a body is its dis- 
tance from the zenith, measured by the arc of a vertical circle 
intercepted between the body and the zenith. 

818. The obliquity of the ecliptic is the inclination of the 
ecliptic to the equator. This inclination is nearly 23 274'. 

819. The horary angle of a body at any instant is an angle at 
the pole of the equator, contained by the meridian passing through 

* The azimuths may be named according to the quadrants in which they lie ; 
thus, N.E. when between the north and east points, S.W. between the south and 
west points, and so on. 

t Amplitudes may be named, also, according to the quadrants in which they lie, 
as the azimuths are named. 



ASTRONOMICAL PROBLEMS 479 

the body and the meridian of the place of observation. It measures 
the time between the instant of observation and the instant of the 
body's passage over the meridian of the observer. 

820. The rising or setting of a body is the time when its centre 
is apparently in the horizon when rising or setting. 

821. The diurnal arc of any body is that portion of its parallel 
of declination which is situated above the horizon, and its noc- 
turnal arc that portion of the same parallel which is below the 
horizon. 

The diurnal arc, reckoned at the rate of 15 to 1 hour, will 
express the interval of continuance of a body above the horizon, 
for a star in sidereal time, for the sun in solar time, for the moon 
in lunar time, and for a planet in planetary time (Art. 831). 

822. The precession of the equinoxes is a small motion of the 
equinoxes towards the west.* 

823. A tropical year is the time in which the sun moves from 
the vernal equinox to that point again. 

. 824. A sidereal year is the time in which the sun moves from 
a fixed star to the same star again, or the time in which it 
performs an absolute revolution.t 

825. Apparent time is that which depends on the position of the 
sun, and is also called solar time. This is the time shown by a 
sun-dial, the days of which are unequal. 

826. Mean time is the time shown by a well-regulated clock, 
the days of which are equal. 

827. An apparent solar day is the time between two successive 
transits of the sun's centre over the meridian, and is of variable 
length. I 

828. A mean solar day is a constant interval of time, and is 
the mean of all the apparent solar days in a year ; or it is what 
an apparent solar day would be were the sun's motion in the 
equinoctial uniform. 

* This retrograde motion is about 1 in 72 years, or 50'2" annually ; and in conse- 
quence of it the sun returns to the vernal equinox sooner than it would do were 
this point at rest ; hence the origin of the term. 

t The tropical year, in consequence of the precession of the equinoxes, is shorter 
than the sidereal year by the time the sun takes to move over 50'2", or 20 m. 19'9 s. 
The length of the former is 365 d. 5 h. 48 in. 51-6 s., and that of the latter 365 d. 
6h. 9m. 11-5 s. 

t In consequence of the obliquity of the ecliptic and the sun's unequal motion 
in its orbit, its motion referred to the equinoctial is not uniform, and consequently 
the intervals between its successive transits are variable. 



480 ASTRONOMICAL PROBLEMS 

829. The equation of time is the difference between mean and 
apparent time. 

It is just the difference between the time shown by a regulated 
timepiece and a snn-dial ; at mean noon it is the difference 
between twelve o'clock mean time and the mean time of the 
sun's passing the meridian. 

830. A sidereal day is the interval between two successive tran- 
sits of the same star over the meridian. It is the time in which 
the earth performs an absolute rotation on its axis ; and as this 
motion is uniform, the sidereal day is always of the same length. 

The sidereal day begins when the vernal equinox that is, the 
first point of Aries arrives at the meridian ; and its length is 
23 h. 56 m. 4'056 s. in mean solar time, or 24 sidereal hours. 

A meridian of the earth returns to the same star in a shorter 
interval than it does to the sun ; the difference, expressed in mean 
time, is called the retardation of mean on sidereal time ; and 
when expressed in sidereal time, it is called the acceleration of 
sidereal on mean time.* 

831. Generally, the interval of time between the departure of a 
given meridian from a celestial body and its return to that body 
is called a day in reference to the body. If the body is a star, the 
interval is called a sidereal day ; if the sun, a solar day ; if the 
moon, a lunar day ; each day consisting of 24 hours, the hours 
for these days being of course of different magnitudes. 

The astronomical day begins at noon, and is reckoned till next 
noon ; and it is thus twelve hours later than the civil day. 

832. The refraction of the atmosphere causes the altitude of a 
celestial body to appear greater than it would be were there 
no atmosphere ; the increase of altitude from this cause is the 
refraction of the body (see Art. 849). 

When the body is in the horizon its refraction is greatest, and 
when in the zenith it is nothing ; at other altitudes the refraction 
is intermediate. 

* The retardation for 24 hours of mean time is=3 m. 55-9094 s., or 24 sidereal hours 
= 23 h. 56 m. 4'0906 s. of mean time. The acceleration for 24 hours of sidereal time 
is=3 m. 56 '5554s., or 24 hours of mean solar time are=24h. 3m. 56 -5554s. of sidereal 
time. These equivalents are obtained from the fact that the sun's mean increase 
of right ascension in a mean solar day is 59' 8'3", or 3 m. 55-9 s. of mean time ; so that 
a meridian of the earth moves over 360 in a sidereal day, and 360 59' 8'3" in a mean 
solar day ; the former motion, which is just the time of the earth's rotation on its 
axis, is performed in 24 sidereal hours, and the latter in 24 h. 3 in. 56'5554 s. of 
sidereal time ; or the former is performed in 23 h. 56 m. 4 - 09 a. mean time, and the 
latter in 24 hours of mean time. 



ASTRONOMICAL PROBLEMS 



481 



833. The parallax of a celestial body is the quantity by which 
its altitude, when seen from the surface of the earth, is diminished, 
compared with its altitude seen from the earth's centre. 

The parallax, like the refraction, is greatest when the body is in 
the horizon, and is nothing when it is in the zenith. 

When the body is in the horizon its parallax is called horizontal 
parallax ; its parallax at any altitude is called its parallax in alti- 
tude ; and its parallax supposed to be subtended by the greatest 
or equatorial ratlins of the earth is called its equatorial parallax. 

834. Most of these definitions will be readily understood from 
the two following diagrams : 

Let PQM be a meridian passing through the pole of the ecliptic ; 
MQ the equator, N and S its north and south poles ; CC' the 
ecliptic, P and p its north and south poles. Then A is the first 
point of Aries, C that of Cancer, and C' 
that of Capricornus ; and angle CAQ, 
measured by CQ, is the obliquity of the 
ecliptic. The parallels of declination CB, 
C'T are the tropical circles, the former 
being the tropic of Cancer, and the latter 
that of Capricorn ; and PR, Ep are the 
polar circles, the former being the arctic, 
and the latter the antarctic, circle. Also, 
if PO/) and NOS are respectively a circle 

of celestial longitude and a meridian passing through any celestial 
body O, then AL is its longitude, OL its latitude, AH its right 
ascension, and OH its declination. The meridian NQSM is the 
solstitial colure, and NAS the equinoctial colure. 

Again, let RH be the horizon, Z and N its poles, the former 
being the zenith and H the south point; EQ the equator, Pandjo 
its north and south poles ; also, let B be 
any celestial body, and ZBN a vertical 
circle through it ; then BL is its altitude, 
HL its azimuth ; and if O is its position 
when rising, OW is its amplitude. Let A 
be the first point of Aries, and POp a meri- 
dian through O ; then the distance between 
A and F is the right ascension of O, the 
distance between A and W its oblique ascen- 
sion, and WF its ascensional difference. The small circle GBT, 
parallel to RH, is a parallel of altitude, and ZWN is the prime 
vertical. 





482 ASTRONOMICAL PROBLEMS 

835. Problem I. To convert degrees of right ascension 
or of terrestrial longitude into the corresponding time; 
and conversely. 

RULE. To convert degrees into time, multiply by 4, and con- 
sider the product of the degrees by 4 as minutes of time, the 
product of the minutes of space as seconds of time, and so on. 

To convert time into degrees, reduce the hours to minutes, and 
consider the number of minutes of time as degrees, the seconds 
of time as minutes of space, and so on ; divide by 4, and the 
quotient will be the required number of degrees. 
EXAMPLES. 1. Convert 36 12' 40" to time. 

(36 12' 40") x 4 = 144 m. 50 s. 40 t. =2 h. 24 m. 50 s. 40 t. 
2. Convert 2 h. 24 m. 50 s. 40 t. to degrees. 

i(2 h. 24 m. 50 s. 40 t.) = |(144 m. 50 s. 40 t.) = 36 12' 40". 
If the sun moved uniformly it would pass over 360 of the 
equator or equinoctial in 24 hours of mean time that is, 15 in 
1 hour ; hence, if c?=the number of degrees, and A = the number of 
hours corresponding, 

1 h. : h = 15 : d, and therefore 
, d 4 , , , .,, 60A 
A = 15 = 60 rf ' a d=15h = ~^- ' 
and from these expressions the rules are easily obtained. 

EXERCISES 

1. 80 6 32' 40" are equivalent to 5 h. 22 m. 10 s. 40 t. 

2. 161 5 20 i, 10 44 21 20 

3. 98 14 48 ii i, 6 32 59 12 

4. 5 h. 22 m. 10 s. 40 t. are equivalent to 80 32' 40" 

5. 28 6 H 71 30 

6. 14 1 12 M 210 18 

836. Problem II. To express civil time in astronomical 
time; and conversely. 

RULE. When the given time is P.M., the civil time and astrono- 
mical time are the same; and when the civil time is A.M. add 12 
hours to it, and the sum will be the astronomical time, reckoning 
from the noon of the preceding day. The rule for the converse 
problem is evident. 

EXAMPLES. 1. April 6 at 3 h. 12 m. P.M. civil time is in 
astronomical time also 6th April 3 h. 12 m. 

2. June 1 at 10 h. 15 m. A.M. of civil time is 31st May 22 h. 15 m. 
of astronomical time. 



ASTRONOMICAL PROBLEMS 483 

EXERCISES 

Civil Time Astronomical Time 

1. Feb. 10 d. 4 h. 20 ni. P.M. is Feb. 10 d. 4 h. 20 m. 

2. July 13 2 12 A.M. ,. July 12 14 12 

3. Aug. 1 11 40 A.M. July 31 23 40 

4. Oct. 9 10 1 P.M. M Oct. 9 10 1 

837. Problem III. To reduce the time under any given 
meridian to the corresponding astronomical time at that 
instant at Greenwich ; and conversely. 

RULE. To find the time at Greenwich corresponding to that at 
another place : to the given time expressed astronomically apply 
the longitude in time by addition when it is W., and by subtraction 
when it is E. 

To find the time at a given place corresponding to a given time 
at Greenwich : to the given time expressed astronomically apply 
the longitude in time by addition when it is E., and by subtraction 
when it is W. 

EXAMPLES. 1. Find the time at Greenwich corresponding to 
20th June, at 9 h. 12 m. A.M., at a place in longitude = 14 2' 30" W. 
Given time June, . . . . 19 d. 21 h. 12 m. s. 

Longitude in time W. , + 56 10 

The reduced time at Greenwich, . 19 22 8 10 

2. Find the time at Greenwich corresponding to 30th August, at 
2 h. 40 m. 10 9. P.M., at a place in longitude = 75 34' 45" E. 

Given time August, . . . . 30 d. 2 h. 40 jii. 10 s. 
Longitude in time, ... 5 2 19 

Astronomical time at Greenwich, . 29 21 37 51 

From these examples the converse problem is evident. 

EXERCISES 

1. Find the time at Greenwich corresponding to 18th July, at 
5 h. 24 m. A.M., at a place in longitude = 40 20' W. 

July = 17 d. 20 h. 5 m. 20 s. 

2. Find the time at Greenwich corresponding to 19th June, at 
1 h. 12 m. 40 s. P.M., in longitude = 90 37' 30" E. 

June = 18 d. 19 h. 10 m. 10 s. 

3. Find the time at a place in longitude = 40 20' W. correspond- 
ing to 18th July, at 8 h. 5 m. 20 s. A.M., at Greenwich. 

July = 17 d. 17 h. 24 in. 



484 ASTRONOMICAL PROBLEMS 

4. Find the time in longitude = 90 37' 30" E. corresponding to 
19th June, at 7 h. 10 m. 10 s. A.M., at Greenwich. 

June = 19 d. 1 h. 12 m. 40 s. 

838. Problem IV. To reduce the registered* declination 
or right ascension of the sun to any given meridian and 
to any time of the day. 

RULE. As 24 hours is to the given time, so is the change of 
declination for 24 hours to its change for the given time. When 
the declination is increasing add this proportional part to it, 
and when diminishing subtract it, and the result will be the 
declination required. 

By the same method, the sun's right ascension can be found for 
any time at a given place, only it is always increasing. 

Or, if t = reduced time at Greenwich past the previous noon, 
v' = variation of declination in 24 hours, 
v = H ii given time, 

D' = declination at noon at Greenwich, 
D = H required ; 

then 24 : t v' : v, and v = ^v't, D = T)'v. 

Or, if P.L. denote proportional logarithms, 
P.L., v=P.L., t + P.L., v'. 

EXAMPLE. Find the sun's declination in 1854, 30th August, at 
2 h. 40 m. 10 s. P.M., at a place in longitude = 75 34' 45" E. 
Time at Greenwich (Art. 837), 29th August =21 h. 37 m. 51 s. =t 
Sun's declination at noon on 29th . . = 9 24' 1" =D 

, 30th . . =92 35 

Variation of declination in 24 h. . . = 21 26 = v' 
And 24 h.: 21 h. 37m. 51 s.=21'26":v, and v . . =0 19' 19" 
Hence the required declination D'-v = D . . . =9 4 42 

Or, by proportional logarithms 

P.L., t 21 h. 37 m. 51 s. = 4515 
P.L., vf 21' 26" = 4912 

hence P.L., v 19 19 = 9427 

and D'= 9 24 1 

therefore D = 9 4 42 = required declination. 

By changing declination to right ascension in the preceding rule, 
it will be adapted to the finding of the sun's right ascension. 

* These elements of the sun's place are registered in the Nautical Almanac lot 
noon of every day at Greenwich. 



ASTRONOMICAL PROBLEMS 485 

EXERCISES 

1. Find the sun's declination in 1854, 12th October, at noon, at 
a place in longitude 4 15' W., the declination at Greenwich at 
noon, 12th October, being = 7 22' 52" S. and increasing, and its 
variation in 24 hours = 22' 32" =7 23' 8". 

2. Find the sun's declination in 1854, 20th June, at 9 h. 12 m. 
P.M., at a place in longitude = 14 2' 30" W. ; having given D 7 
=23 27' 13" N., and t/ = 20", and the declination increasing. 

=23 27' 21". 

3. Find the sun's declination in 1854, 29th May, at 2 h. 37 m. 20 s. 
A.M., in longitude = 32 4ff W. ; having given D'=21 27' 33" N., 
andi/ = 9'31" =21 34' 13". 

4. What is the sun's right ascension in 1854, 4th September, at 
4 h. 45 m. 39 s. P.M., at a place in longitude = 72 35' 15" W., when 
jy = 10 h. 52 m. 4-65 s., and i/ = 3 m. 37 s.? =10 h. 53 m. 31'45 s. 

PROPORTIONAL LOGARITHMS 

839. These logarithms, which are useful in calculating small 
quantities, such as minutes of time or space, as they generally 
require to be carried out only to four decimal places, are obtained 
in this manner : 

Let a, b, c,...be any quantities ; assume another quantity q, such 
that it exceeds any of the quantities a, b, c,...then Lg-La is the 
proportional logarithm of a, Lq - Lb that of b, and so on ; also, 
P. Lq Lq-Lq = o. The quantity q so assumed is 3 hours or 
3 degrees, and sometimes 24 hours. 

840. Problem V. To reduce the registered declination 
and right ascension of the moon to any given meridian 
and to any time of the day.* 

RULE. Find the reduced time, and the declination for the pre- 
ceding hour ; then, as 10 minutes is to the time past that hour, so 
is the variation in 10 minutes to the variation in the past time, 
which being applied to the given declination by addition or sub- 
traction, according as the declination is increasing or diminishing, 
will give the declination required. 

The same rule applies for finding the right ascension, only 1 hour 
or 60 minutes must be used for 10 minutes ; and as the right ascen- 
sion is always increasing, the variation is always to be added. 

* The moon's declination and right ascension are given for every hour in the 
Nautical Almanac, and the variation of the former for every 10 minutes. 

Prac. 2 F 



486 ASTRONOMICAL PROBLEMS 

Let D', D = the earlier given and required declination, 

R', R= M M it right ascension, . 

t = time past the hour preceding the reduced time, 
v' ihe variation for 60 ra. of right ascension, or 10 m. of 

declination, 

v = the correction sought ; then 
for the declination, 10 m. : t m. =v' : v, 

and v = stv' ; hence D = D' + w; 

and for the right ascension, 60 m. : t m. =v r : v, 
and v = -fatv' ; hence R=R' + v. 



Right Ascension 
R'=16h. 53m. 37 -35 s., 
16 55 57-69 


Declination 
D'=26 18' 47 -5" 
26 20 25-9 


v' = 2 20-34 


1 38-4 



EXAMPLE. What will be the declination and right ascension of 
the moon on 15th November 1841, at h. 30 m. A.M., at a place in 
longitude = 36 45' W.? 

Given time on 14th November . = 12 h. 30 m. 

Longitude in time . . . . = + 2 36 

Reduced time (Art. 837) . . . = 15 6 and t = Q m. 

On 14th, at 15 h., 

M 16 h., 

Change in 60 m., 

Diff. dec. in 10 in. 16'4". 

For declination, v' in 10 m. = 16*4" ; 

hence v = &tv' = & x ! 6 '4 = 9 -8", 

and D = D' + v = 26 18' 57 "3". 

For the right ascension, 

v=j s ti/ = f- s x 140-34 s. = 14-03 s., 
and R = R' + v = 16 h. 53 m. 51 -38 s. 

From the variation of declination 1' 38'4" in 60 m., its change in 
the past time 6 m. could be found in the same manner as that for 
right ascension. 

EXERCISES 

1. What was the moon's right ascension and declination at 
Greenwich on the 18th of October 1841, at 10 h. 40 m. P.M., from 
these data ? 

October 1841 Right Ascension Declination 

On 18th, at 10 h., R' = 17 h. 2 m. 18'88 s., D' = 26 34' 19'7" 
i. ,, 11 h., 17 4 38-13 26 35 31 '9 

R =17 h. 3 m. 51-71 s., and D =26 35' 7'8" 



ASTRONOMICAL PROBLEMS 487 

2. Required the moon's right ascension and declination on the 
27th of October 1854, at 10 h. 43 m. A.M., at a place in longitude 
=40 15' E., from these data : 

October 1854 Right Ascension Declination 

On 26th, at 20 h., R' = 19 h. 4 m. 5 -50 s., D'=26 49' 5'1" 
21 h., 19 6 44-40 26 47 4'8 

Reduced time = 26 d. 20 h. 2 m., R=19 h. 4 m. 10-79 s., 
and D= 26 49' 1'09". 

841. Problem VI. To reduce the registered semi-diameter 
or horizontal parallax of the moon to any meridian and 
any time of the day.* 

RULE. Find the civil time at Greenwich ; then, as 12 hours is 
to the reduced time, so is the change in either of these elements in 
12 hours to its change for the intermediate time, which is to be 
applied by addition or subtraction to the earlier given element 
according as it is increasing or decreasing. 
Let s', s =the earlier given and required semi-diameter, 
p', p = ii H ii horizontal parallax, 

t = M time in hours past noon or midnight, 
u 7 H change in either of these elements for 12 h., 
and v = M ii for the reduced time ; then 
for the semi-diameter, 

12 h. : t \\. = v r : v, v=^tv', and s = s' + v ; 
and for the horizontal parallax, 

12 h. : t h. =tf : v, v=^tv', a.ndp=p'v. 

EXAMPLE. Find the semi-diameter and horizontal parallax of 
the moon at a place in longitude = 4 20' 15" W., on 20th March 
1854, at 7 h. 42 m. 39 s. P.M.; having given the registered elements 
for the preceding noon and midnight. 

The reduced time is 20th, 8 h. P.M. 

20th March 1854 Semi-diameter Horizontal Parallax 

At noon, .... J = l& 9'3" p' = 59' 10'3" 

midnight, . . . 16 IQ-2 59 13 '5 

Change in 12 h., . . 0'9 3'2 

Computed change in 8 h., 0'6 2'1 

Required elements, . . 16 9-9 59 12-4 

For to find s, v = ^tv' = ^ x -9" = -6", and s = s' + v ; 
and n p, w= T Vv' = T^x3-2"=2-l", ., p=p' + v. 

* These elements are registered for every noon and midnight. 



488 ASTRONOMICAL PROBLEMS 

EXERCISES 

1. Find the semi-diameter and horizontal parallax of the moon 
on 17th November 1854, at 10 h. 30 m. P.M., at Greenwich from 
these elements : 

17th Nov. 1854 Semi-diameter Horizontal Parallax 

At noon, .... s=15'39'2" p = 5T 19'8" 

midnight, ... 15 46 57 44'7 

s=15' 45-2", andj9=57' 41 "6" 

2. Find the semi-diameter and horizontal parallax of the moon 
on the 10th of November 1854, at 4 h. 7 m. 10 s. P.M., at St Helena, 
in longitude = 5 42' 30" W., from these elements : 

10th Nov. 1854 Semi-diameter Horizontal Parallax 

At noon, .... sf = U' 48 "9" y = 54' 15'8" 

H midnight, ... 14 48'3 54 13'5 

Reduced time = 4 h. 30 m., 5 = 14' 48'7", and j9 = 54' 14-9" 



AUGMENTATION OF THE MOON'S SEMI-DIAMETER 

842. Since when the moon is in the zenith it is nearer to the 
observer than when in the horizon by the radius of the earth, its 
apparent magnitude is consequently increased, and at intermediate 
altitudes its augmentation will be intermediate. The amount of 
this augmentation for any given altitude is sensibly constant for 
the same diameter, and is given in a Table, and can be easily 
applied. 

For the altitude of 15, and semi-diameter 14' 43 '7", this aug- 
mentation is 4", so that the semi-diameter found above must be 
augmented by this quantity for this altitude, and would then be 
= 14' 43-7" + 4" = 14' 47'7". 

The semi-diameter of the moon given in the Nautical Almanac 
is that which it would appear to have when seen from the centre of 
the earth. If this semi-diameter be denoted by s, and its apparent 
semi-diameter at the given place by *', and a its altitude, then s' 
can be calculated from the equation, 

s'=s + ms 2 sin a. 

Where m = k sin 1", &=3'6697, and k = -, where h and s are the 

s 

registered horizontal parallax and semi-diameter. The value of m 
is -00001779, for the ratio of h to s is constant. When 
a=0, then s'=s. 



ASTRONOMICAL PROBLEMS 489 



CONTRACTION OF THE MOON'S SEMI-DIAMETER 

843. The lower limb of the moon is apparently more elevated 
by refraction than its upper limb, as its altitude is less ; and con- 
sequently every diameter of the moon except the horizontal one 
is contracted, and the vertical one is subject to the greatest con- 
traction. This contraction is greater the less the altitude, and is 
sensibly constant for the same diameter and for a given altitude ; 
and is therefore conveniently applied by means of a Table. 

The contraction for an altitude of 15 is 4" for the vertical 
diameter ; so that the semi-diameter 14' 48", previously found, now 
becomes = 14' 47 "7" - 4" = 14' 43'7". 

This semi-diameter, neglecting these corrections, Avas found to be 
14' 43 - 7" ; so that in this instance these two corrections exactly 
compensate each other. 

THE SUN'S SEMI-DIAMETER 

844. The sun's daily change of distance from the earth is so small 
compared with its distance that its semi-diameter does not sensibly 
change in apparent magnitude in the course of a day ; so that its 
registered semi-diameter may be considered as constant for at least 
one day. Its distance also is so great compared with the earth's 
radius that its semi-diameter is not subject to an apparent aug- 
mentation dependent upon altitude. 

The sun's diameter, however, like that of the moon, is subject to 
an apparent contraction by the unequal refraction of its upper and 
lower limbs, and its amount is sensibly the same as for the moon. 

845. Problem VII. Given the horizontal parallax of a 
celestial body, and its altitude, to find its parallax in 
altitude. 

RULE. Radius is to the cosine of the apparent altitude as the 
sine of the horizontal parallax to the sine of the parallax in 
altitude. 

Let p' = the horizontal parallax, 

p = ii parallax in altitude, 
a = ii apparent altitude ; 
then R : cos a=sinj0' : smp, 

or sin p = sin p' cos when rad. = 1. 

Or, by proportional logarithms, 

P.L, p = P.L, p' + L sec -10. 



490 



ASTRONOMICAL PROBLEMS 




EXAMPLE. When the horizontal parallax of the moon is = 54' 
20", and its altitude = 36 45', what is its parallax in altitude? 

Here = 36 45', and ^'=54' 20". 

By Logarithms By Proportional Logarithms 

L, radius . . = 10' L, radius . . = 10' 

L, cos a . . = 9-9037701 L, sec a . . = 10-0962 

L, siny . . = 8-1987581 P.L, p' . . = -5202 

L, smp . . = 8-1025282 P.L, p . . = -6164 

Hence p . . = 43' 32". Hence p . . = 43' 32" 

846. The principle on which the rule is founded is very simple. 

Let PEA be the earth, O its centre, M' the 
moon in the horizon, M its position at any 
altitude, OZ a vertical line ; and let 
angle OM'P =p' the horizontal parallax, 
IT OMP p H parallax in altitude, 
ii MPM'=a M apparent altitude 

OP r it earth's semi-diameter ; and 
OM or OM.' d .. moon's distance ; 
then in triangle OPM, angle P=90 + a, and sin P = cos a ; 
and ii OPM, sin P or cos a : sin pd : r ; 

OPM', R :sinp' = d :r; 
hence R : cos a = sin p' : sin p. 

847. The sun's horizontal parallax varies only about of a 
second, and may in practice generally be considered as invariable. 
The parallaxes in altitude for the sun at any given time may 
therefore be considered the same for any other time ; and thus, 
being constant, they are given in a Table. 

EXERCISES 

1. When the horizontal parallax is = 54' 16", and altitude = 
24 29' 30", what is the parallax in altitude ? . . =49' 23". 

2. When the horizontal parallax is = 57' 32", and the altitude 
= 50 40', what is the parallax in altitude? . . . =36' 28". 

REDUCTION OF THE EQUATORIAL PARALLAX 

848. The horizontal parallax given in the Nautical Almanac is 
calculated for the equatorial radius of the earth, and is the true 
horizontal parallax only at the equator ; for, the earth's radius 
being less the greater the latitude, the horizontal parallax will be 
less at any other place. If I denote the latitude, and e the ellip- 
ticity of the earth, the value of which is nearly ^, and if p' and 



ASTRONOMICAL PROBLEMS 491 

p" denote the horizontal parallax at the given place and at the 
equator, then is p'=P" (1 ~ e sin 2 /). 

For if a, r are the radii of the earth at the equator and the 
given place, it is proved in the theory of the figure of the earth 
that r = a (l-e sin 2 /). Also (Art. 846), r = d sin p' t and a=d 

sin "; therefore, since ^ -. *% ver Y nearly, =^7 = -= (l-e sin 2 /); 
p' sin fr J ' p " a 

hence P'=P" (1 - e sin 2 /). 

A Table contains the corrections, calculated by this formula, 
that must be deducted from the equatorial horizontal parallax in 
order to reduce it to the horizontal parallax for any given latitude. 

Thus, for the equatorial horizontal parallax in the preceding 
example, the reduction in the Table under 54', and opposite to 
latitude 36, is 3'7" ; and the correct horizontal parallax for this 
latitude is =54' 20" - 3'7"=54' 16'3". 

849. Problem VHL Given the observed altitude of a 
heavenly body, to find the altitude when corrected for 
refraction. 

The refraction for the observed altitude is given in a Table, and 
is always to be subtracted from the observed altitude. 

EXAMPLE. Let the apparent altitude be 32 10', to find the true 
altitude. 

In the Table the refraction for this altitude is 1' 30", 
and the apparent altitude . . . = 32 10' 0" 

The refraction = - 1 30 

Hence the true altitude . . . . = 32 8 30 

Let ER be a part of the earth's surface, and ZP a portion of the 
upper limit of the atmosphere ; B the real 
place of a heavenly body, B' its apparent 
place ; O the eye of the observer ; OH a 
horizontal and OZ a vertical line. When 
a ray of light BO from the body B 
enters the atmosphere at P, which in- 
creases in density downwards, the direc- 
tion of the ray approaches always nearer 
to that of the vertical line OZ, and thus 

it moves in a curved path BPO ; but the direction of the body is 
referred to the direction of the ray when entering the eye at O 
that is, to the direction OB' of the tangent to the curved path at 
O and the body thus appears at B' higher than its real position. 




402 ASTRONOMICAL PROBLEMS 

The greater the altitude of the body the less is its refraction, 
and in the zenith it vanishes. 

850. The mean refraction of a body is its true refraction when 
the barometer stands at 29 '6 inches, and Fahrenheit's thermometer 
at 50. Braclley's formula for calculating the mean refraction is 

r'-57" tan (z-3r'), 
where r' = the mean refraction, and z = the zenith distance. 

A Table of mean refractions can thus be calculated ; and to 
find the true refraction r when the pressure of the atmosphere 
is h, and the temperature t, multiply the mean refraction 1 J by 

400 h ,. . 400A , _ , ,, . , 

6 ; that is, r=^ r'. But a Table is also calcu- 



350 + t' 29-6' ~ 29 -6(350 + *)' 

lated containing the corrections that must be applied to the mean 
refractions when the pressure and temperature differ from 29 - 6 
and 50. 

Thus, the refraction in the preceding example namely, 1' 30" 
is the mean refraction ; but if the temperature and pressure were 
69 and 30 '35, then the correction 

For altitude 32 10', and temperature 69, is = - 4" 

And n 32 10', pressure 30 '35, is . . = + 2 

Hence the correction for both is . . . = - 2 

And the mean refraction was found . . . =1' 30" 

Therefore the true refraction . . . = 1 28 

Apparent altitude =32 10' 0" 

And the true altitude . ." . . . =32 8 32 

851. Unless when great accuracy is required, or when the alti- 
tude is small, the corrections for change of pressure and temperature 
are unnecessary. 

852. Problem IX. Given the height of the eye of the 
observer above the surface of the earth, to find the depres- 
sion of the visible horizon. 

The depression of the horizon HOR (fig. to preceding problem) 
can be calculated when the height OE of the eye and the diameter 
of the earth are known ; for it is just the angle at the earth's 
centre, subtended by the arc ER, because (fig. to Art. 585) angle 
RBH = BCH, and the latter angle can be calculated in the same 
manner as angle E (fig. to Art. 587). 

The real depression, however, will be this angle diminished by 
TV of itself on account of refraction. 



ASTRONOMICAL PROBLEMS 493 

EXAMPLE. Find the depression of the horizon when the height 
of the eye is = 30 feet. 
Opposite to 30 in the Table is 5' 18", the dip. 

853. Problem X. Given the observed altitude of a fixed 
star, to find its true altitude. 

RULE. Correct the observed altitude by applying to it the index 
error of the instrument with its proper sign, and subtract the dip 
from the result, and the remainder will be the apparent altitude. 

From the apparent altitude subtract the refraction, and the 
remainder will be the true altitude. 

When the observed altitude is taken by a back observation, the 
dip must be added to it. When great accuracy is required, the 
corrections for the temperature and pressure of the atmosphere 
must be applied to the refraction. 

EXAMPLE. The observed altitude of a star was =40 20' 34", the 
height of the eye = 12 feet, and the index error 2' 25" in excess; 
find the true altitude. 

Observed altitude . . . . = 40 20* 34" 

Index error = - 2 25 

40 18 9 

Dip = 3 21 

Apparent altitude . . . . = 40 14 48 
Refraction . = - 1 8 



True altitude = 40 13 40 

EXERCISES 

1. When the observed altitude of a star is =25 36 40", the index 
error = 1' 54" in defect, and the height of the eye = 20 feet, what is 
the true altitude ? =25 32' 15". 

2. What is the true altitude of a star when its observed altitude 
is = 38 2' 20", the height of the eye=18 feet, and the temperature 
and pressure = 45 and 30'6? =37 57' 4". 

854. Problem XI. -Given the observed altitude of the 
upper or lower limb of the sun, to find the true altitude 
of its centre. 

RULE. Apply the sun's registered semi-diameter to the observed 
altitude by addition or subtraction, according as the lower or upper 
limb was observed, and subtract the dip, and the result will be the 



494 ASTRONOMICAL PROBLEMS 

apparent altitude of its centre ; from which subtract the refraction 
corresponding to it, and add the parallax in altitude, and the sum 
will be the required altitude. 

EXAMPLE. If the observed altitude of the sun's upper limb on 
the 6th of November 1854 should be = 28 21' 24", and the height 
of the eye = 13, what would be the true altitude? 

Observed altitude of sun's upper limb = 28 21' 24" 

Semi-diameter = -. 16 11 

28 5 13 
Dip = - 3 29 



Apparent altitude of centre . = 28 1 44 

Refraction = - 1 47 



27 59 57 
Parallax in altitude = +8 



28 5 

EXERCISES 

1. If the observed altitude of the sun's lower limb on the 
19th of April 1854 was = 42 10' 15", the height of the eye being 
=25 feet, what was the true altitude of its centre, its semi- 
diameter being = 15' 57" ? =42 20' 25". 

2. If the observed altitude of the sun's upper limb on the llth 
of June 1854 was = 20 40' 15", and the height of the eye = 16 feet, 
what was the true altitude of its centre, its semi-diameter being 
= 15' 47"? - .. . . . =20 18' 12". 

855. Problem XII. Given the observed altitude of the 
upper or lower limb of the moon, to find the true altitude 
of its centre. 

RULE. To the observed altitude apply the semi-diameter by 
addition or subtraction according as the altitude of the lower or 
upper limb is given ; from this result subtract the depression, and 
the remainder is the apparent altitude of the moon's centre ; and 
to this altitude apply the refraction and parallax in altitude, as in 
the preceding problem. 

EXAMPLE. If on the 12th of July 1841, in latitude 56 40', the 
observed altitude of the moon's upper limb was =57 14' 20", 
the height of the eye =22 feet, the semi-diameter = 15' 35", and the 
horizontal parallax = 57' 14"; required the true altitude of the 
moon's centre. 



ASTRONOMICAL PROBLEMS 



495 



Ob. alt. moon's upper limb 
Moon's semi-diameter 
Augmentation . 
Aug. semi-diameter 
Depression 
Ap. alt. of centre a! 
Moon's par. in alt. p 
Ref. to ap. altitude . 
True altitude a 


= 57 14' 20" 
15 35 
13 


Hor. par. . 
Reduction 
Tr. hor. par. 

P.L., 57' 6", 
Secant ' . 
P.L., p, . 


= 57' 14" 

= - 8 


= 57 6 

= '4986 
'= '2627 


= - 15 48 
4 32 


= 56 54 
= + 31 11 
= - 37 


= -7613 


= 57 24 34 



856. The longitude of the place of observation and the time of 
observation must be known in order to determine the reduced time 
(Art. 837), and then the semi-diameter and horizontal parallax are 
found for the reduced time according to the rule in Art. 841. 

EXERCISES 

1. At a place in latitude =36 50' the observed altitude of the 
moon's lower limb was =24 18' 40", the height of the eye = 17 '3 
feet, the moon's semi-diameter at the time of observation = 15', and 
its horizontal equatorial parallax =55' 2" ; what was the true alti- 
tude, the corrected semi-diameter, and parallax in altitude ? 

The true altitude of moon's centre = 25 17' 41", semi-diameter 
= 15' 6", and parallax = 50' 1". 

2. If, at a place in latitude = 24 30', and longitude = 23 E., on 
31st May 1796, at 5 h. 36 m. P.M., the observed altitude of the 
moon's lower limb was = 23 48' 15", the height of the eye = 17 '3 
feet, the moon's semi-diameter and horizontal parallax at the 
preceding noon and following midnight being = 15' 49", 15' 56", 
and 58' 1", 58' 29" ; required the true altitude of the moon's centre. 

The reduced time = 4 h. 4 m. P.M.; corrected semi-diameter 
= 15' 57", parallax in altitude = 53' 6", and altitude 
=24 51' 9". 

3. On 10th September 1841, in latitude = 28 40' N., longitude 
24 45' W., at 5 h. 51 m. P.M., the observed altitude of the moon's 
lower limb was 32 40' 15", height of eye = 16 feet; required the 
true altitude of the moon's centre, having also given the moon's 

Semi-diameter Horizontal Parallax 

At noon preceding, . . .16' 15" 59' 37" 

., midnight following, . . 16 19 59 51 

Reduced time = 7 h. 30 m. P.M., corrected semi-diameter 
= 16' 26", parallax in altitude=50' 8", and altitude 
= 33 41' 29". 



496 ASTRONOMICAL PROBLEMS 

857. Problem XIII To find the polar distance of a 
celestial object. 

RULE. When the declination and the latitude of the place are 
of the same name, subtract the declination from 90 ; and when 
of different names, add the declination to 90 ; then the difference 
in the former or the sum in the latter case is the polar distance. 

Let D = the declination of the body, P = the polar distance ; 
then P = 90 + D. 

EXAMPLE. What will be the moon's north polar distance on 
the 12th of November 1854 at noon at Greenwich, its declination 
being then =21 18' 4 -4" N. ? 

P=90-D = 90-21 18' 4 -4" =68 41' 55 -6". 

EXERCISES 

1. Find the moon's north polar distance on the llth of Septem- 
ber 1854 at 11 h. P.M. at Greenwich, its declination being then 
= 18 21' 15" N =71 38' 45". 

2. Find the north polar distance of Mars on the 10th of Decem- 
ber 1841 when on the meridian of Greenwicli, its declination being 
at that time = 19 15' 16" S =109 15' 16". 

858. Problem XIV. To convert intervals of mean solar 
time to intervals of sidereal time ; and conversely. 

RULE. As 1 h. is to 1 h. m. 9 '8565 s., so is the given interval 
of mean solar time to the required interval of sidereal time ; and 
1 h. is to h. 59 m. 50-1704 s. as the given interval of sidereal time 
to the required interval of mean time. 

859. Or find the equivalents by means of a Table of time equiva- 
lents, or by means of a Table of accelerations and retardations. 

Let m = the mean time, 

s = ii equivalent sidereal interval, 
a = ii acceleration, 
r = M retardation ; 
then s m + a, m=s-r. 

P'.L. a = P.L. m + 1-65949, 

P'.L. r = P. L. s +1-66068, 

where P'.L. stands for 3 h. proportional logarithms, and P.L. for 
24 h. ones. 
EXAMPLE. Convert 10 h. 20 m. 40 s. of sidereal time to mean 

time. 

By the First Method 

1 h. : h. 59 m. 50-1704 s. = 10 h. 20 m. 40 s. : 10 h. 18 m. 58'32 s. 



ASTRONOMICAL PROBLEMS 497 

By the Second Method 

P. L., 10 h. 20m. 40s = '36550 

Constant =1-66068 

F.L., 1m. 41-68s. . . . =2'02618 

Sidereal time, 10 h. 20 40 
Required 10 18 58 "32 

The rules depend on the facts that a meridian describes 360 in 
24 sidereal hours, and 360 59' 8'3" in 24 hours of mean solar 
time. Hence it describes 59' 8'3" in 3 m. 55 '91 s. mean time, 
and in 3 m. 56 '56 s. sidereal time. 

Hence 24 h. mean time =24 h. 3 m. 56 '56 s. sidereal time, 
and 24 sidereal t, =23 56 4 '09 mean M 
Or, 1 mean ,, = 1 9 '856 sidereal M 

and 1 sidereal .1 = 59 50*170 mean n 

The acceleration of sidereal on mean time in 24 sidereal hours 
is 3 m. 56 '56 s., and the retardation of mean on sidereal time in 
24 J mean hours is 3 m. 55 - 91 s. A Table of accelerations and 
retardations for any number of hours, of minutes, &c. can easily 
be calculated. 

The rule by proportional logarithms is obtained thus : The 
first two terms of the proportion are either 3600 s. and 3609'85 s., 
or 3600s. and 3590 - 17 s., and the difference of their logarithms is 
00119; then proportional logarithms maybe taken for the other 
two terms ; for if a, b, c, d are the terms of a proportion, then 
(Art. 839), 

La~L6 = P.L. c~P.L. d, 
L6-La = P.L. c-P.L. d, 
and P.L. d=P.L. c + (La-Lb). 

EXERCISES 

1. Convert 7 h. 40 m. 15 s. of sidereal time to mean time. 

= 7 h. 38 m. 59-6 s. 

2. Convert 7 h. 38 m. 59 '6 s. of mean time to sidereal time. 

= 7 h. 40m. 15 s. 

860. Problem XV, Given the sun's registered mean right 
ascension at mean noon, to find its mean right ascension 
at any place and at any time of the day.* 

RULE. Find the reduced time ; then, as 24 hours is to the 
reduced time, so is 3 m. 56 '555 s. to a proportional part, which, 

* In the Nautical Almanac the mean right ascension is called the sidereal time. 



498 ASTRONOMICAL PROBLEMS 

when added to the given right ascension at the preceding mean 
noon, will give that required. 

Let A' = the registered mean right ascension at mean noon, 
A = ii required mean right ascension, 
d' = H increase of A' in 24 mean hours =3 m. 56 '555 s., 
d = n proportional part for t, 
t M reduced time ; 

then 24 : t=d' : d, and d=^d't, 

and A=A' + d. 

EXAMPLES.1. Find the sun's mean right ascension at mean 
noon on the llth of April 1854 at a place in longitude = 36 15' W. 

t = longitude in time = 2 h. 25 m. 

Sun's given mean right ascension, or A' = l h. 17 m. 29 "86 s. 
Increase in time t, or . t>- T . . V . d =0 23'82 
Right ascension required, or . .A =1 17 53*68 
2. What is the sun's mean right ascension at 2 h. 40 m. P.M. on 
the 30th of April 1841 at a place in longitude =50 20' 30" W.? 
Given time . . = 2 h. 40 m. s. 
Longitude in time = + 3 21 22 
Reduced time t . = 6 1 22 
Sun's given right ascension, or . A' = 2 h. 33 m. 1'27 s. 
Increase in time t, or . . d =0 59 '36 
Right ascension required, or . A =2 34 - 63 

861. The principle of the rule is evident, for 3 m. 56'555 s. is the 
increase of the sun's mean right ascension in 24 hours mean time, 
and terrestrial longitude reduced to time by the usual rule is mean 
time. 

EXERCISES 

1. Find the sun's mean right ascension at mean noon at a place 
in longitude = 45 25' W. on the 25th of June 1841, its registered 
mean R.A. at mean noon being=6 h. 13 m. 48 '5 s. 

= 6 h. 14 in. 18 -34s. 

2. Required the sun's mean R.A. on the 20th of July 1841, at 
3 h. 20 m. P.M., at a place in longitude = 56 15' W. ; its registered 
mean R.A. at mean noon on the same day being =7 h. 52 m. 22 '45 s. 

= 7 h. 53m. 32-26 s. 

3. What will be the sun's mean R.A. on the 14th of November 
1854, at 10 h. 40 m. A.M., at a place in longitude = 36 24' 15" E. ; 
its registered mean R.A. at mean noon on the 13th being = 15 h. 
29 m. 5-9 s. ? 15 h. 32 m. 25 '4 s. 



ASTRONOMICAL PROBLEMS 499 

862. Problem XVI. To convert any given mean time on 
any given day to the corresponding sidereal time; and 
conversely. 

RULE. When mean time is given, express it astronomically 
and convert it into the equivalent sidereal time ; then to this 
result add the sidereal time at the preceding mean noon, and the 
sum will be the required sidereal time. 

When sidereal time is given, subtract from it the sidereal time 
at the preceding noon, and convert the remainder into its equiva- 
lent mean time, and it will be the required time. 

The sidereal time at the preceding noon that is, the sun's mean 
right ascension is found by the preceding problem. 

Let m = the mean astronomical time, 

s = H equivalent interval of sidereal time, 
a = ii acceleration for m, 
r = H retardation for s, 
S = ii sidereal time, 

S'= ii registered sidereal time or sun's mean R.A. at pre- 
ceding mean noon at the given place. 
When m is given, s = m + a, and S = S' + s ; 
and when S is given, s=S-S', and m=s-r. 

EXAMPLES. 1. Find the sidereal time corresponding to 2 h. 
22 m. 25 '62 s. mean time at Greenwich, 2nd January 1854. 

Here m= 2 h. 22 m. 25'62 s. 

a = 23-39 
s = 2 22 49-01 

S' = 18 47 10-92 at noon, 2nd Jan. 
S =21 9 59-93 required time. 

2. Find the mean time corresponding to 21 h. 9 m. 59 '93 s. 
sidereal time at Greenwich, 2nd January 1854. 

Here S =21 h. 9m. 59 '93s. 

S' = 18 47 10-92 at noon, 2nd Jan. 

s 

r 

m= 2 22 25-61 2nd Jan., required time. 

3. Find the sidereal time corresponding to 3 h. 40 m. P.M. 
mean time on the llth of April 1854 at a place in longitude 
= 36 15' W. 



= 2 
= 


22 




49-01 
23-40 



500 ASTRONOMICAL PROBLEMS 

The sun's mean R.A. at mean noon that is, the sidereal time 
at the preceding noon at the given place, according to the first 
example of the preceding problem is = l h. 17 m. 53 '68 s. 

m = 3 h. 40 m. s. 
a = 36-14 
s =3 40 36-14 
and S' = l 17 53 '68 



hence S =4 58 29-82 

863. The sidereal time at mean noon at any place is just the 
right ascension of its meridian at that time that is, the sidereal 
interval since the transit of the first point of Aries and this is 
just the right ascension of the mean sun at the mean noon. This 
time is given in the Nautical Almanac for Greenwich, and is 
easily found from the sun's right ascension at mean noon, by 
applying to it the equation of time ; it could also be deduced 
from the sun's right ascension at apparent noon. 

EXERCISES 

1. Convert 2 h. 21 m. 13-08 s. mean solar time on the 2nd of 
January 1854 at Greenwich into the corresponding sidereal time, 
the sidereal time at mean noon being = 18 h. 47 m. 10 '92 s. 

= 21 h. 8 m. 47-2 s. 

2. Convert 21 h. 8 m. 47 '2 s. of sidereal time on the 2nd of 
January 1854 at Greenwich into the corresponding mean solar 
time, the sidereal time at mean noon being = 18 h. 47 m. 10 - 92 s. 

=2 h. 21 m. 13-08 s. 

3. Find the sidereal time corresponding to 8 h. 20 m. A.M. mean 
time on the 26th of June 1841 at a place in longitude = 45 25' W., 
the registered sidereal time at the preceding mean noon being 
= 6 h. 13 m. 48-5 s =2 h. 37 m. 38'75 s. 



- 864. Problem XVII. To find the mean time of the sun's 
transit over the meridian of any place. 

RULE. Find the equation of time for the reduced time corre- 
sponding to the longitude, and it will be the time from mean noon, 
either before or after, at which the transit of the centre happens. 

To the time of the meridian passage of the centre apply the 
time of the semi-diameter's passing the meridian, by subtraction 
or addition, according as the time of transit of the first or second 
limb is required. 



ASTRONOMICAL PROBLEMS 501 

EXAMPLES. 1. Find the mean time of the transit of the sun's 
centre, and that of its first limb, over the meridian of Greenwich 
on the 10th of January 1854. 

Equation of time at apparent noon to 

be added to apparent time . . =0 h. 7 m. 50 '43 s. 
Time of semi-diameter's passage . = - 1 10 '52 

6 39-91 

Hence the time of transit of the centre is=0 h. 7 m. 50*43 s., 
nnd of the first limb=0 h. 6. m. 39-91 s. 

2. Find the mean time of the transit of the sun's centre, and 
of its second limb, at a place in longitude = 54 30' E. on the 
28th of April 1854. 

Longitude in time = 3 h. 38 m. s. 

Reg. equation of time on the 28th . = + 2 m. 36*28 e, 

27th . =0 2 26*87 

Increase of equation of time in 24 h. . =0 9 -41 

3h. 38m. = - 1'43 



Equation of time for reduced time . =0 2 34 -85 
Time of passage of semi-diameter . =0 1 5'80 

3 40-65 

Hence the time of the passage of the centre is=0 h. 2 m. 34*85 s., 
and of the second limb = 3 m. 40'65 s. 

EXERCISES 

1. Find the time of the meridian passage of the sun's centre, 
and of its second limb, at Greenwich on the 30th of April 1854, 
the equation of time at apparent noon being =2 m. 53-58 s., to 
be subtracted from apparent time, and the mean time of the semi- 
diameter's passing the meridian = 1 m. 5 -96 s. 

For centre =29 d. 23 h. 57 m. 6 '42 s., and for the second limb 
=29 d. 23 h. 58 m. 12-38 s. 

2. Required the time of the meridian passage of the sun's centre, 
and that of its first limb, at Edinburgh, in longitude = 12 m. 44 s. 
W., on the 16th of November 1854, the equation of time at apparent 
noon (to be subtracted from apparent time) on the 15th and 16th 
being=15 m. 15*56 s., and 15 m. 4'77 s., and the mean time of the 
semi-diameter's passage being 1 m. 8 '71 s. 

For centre = 15 d. 23 h. 44 m. 55 '32 s., and for the first limb 
= 15d. 23 h. 43m. 46-61 s. 

rw- 2 Q 



502 ASTRONOMICAL PROBLEMS 

865. Problem XVIII. To find the mean time of a star's 
culmination at any given meridian. 

BULK Find the sun's mean right ascension at mean noon at 
the given place, and subtract it from the star's right ascension, 
increased if necessary by 24 hours ; and the remainder, which is 
a sidereal interval, being converted into mean time, will be the 
mean time of culmination. 

Let A'= the star's apparent right ascension at given time, 

A = IT sun's mean right ascension at mean noon preceding 

the transit at given place (Art. 860), 
T'= ti sidereal time of transit after mean noon, 
T = 11 mean \< \< 

then T' = A' - A, and T = T' - r, by Art. 859. 

EXAMPLES. 1. When will Arcturus culminate at Greenwich on 
the 1st of April 1854 ? 

R.A. of Arcturus, or . . A' = + 14 h. 9 m. 0'82 s. 
R. A. of sun at mean noon, . A = - 42 4 '21 
Sidereal time of cul. after noon, T" = 13 26 56 '61 
Retardation, . . . r = - 2 12 '20 



Mean time of transit, . . T = 13 24 44-41 

2. Find the time of the passage of Arcturus over the meridian of 
a place in longitude = 62 15' W. on the 7th of December 1854. 
Longitude in time 62 15' W. =4 h. 9 m. 
Sun's registered mean R.A. on 7th = 16 h. 55 m. 21 '88 s. 
Increase or accel. for 4 h. 9 m., or a = + 40-90 

Hence A = - 16 56 2'78 

And A' = + 14 9 2'69 



Therefore . . . . T' = 21 12 59-91 

Retardation, or . . r = - 3 28 '55 



Mean time of transit, . . T = 21 9 31-36 
Or, in civil time, on the 8th, at 9 h. 9 m. 31 '36 s. A.M. 

EXERCISES 

1. At what time will Sinus culminate at Greenwich on the 17th 
of December 1854, its right ascension being = 6 h. 38 m. 45'46 s., 
and sun's mean right ascension, or the sidereal time, at mean noon 
being = 17 h. 39 m. 2773 s. ? . . . At 12 h. 57 m. 10-06 s. 

2. When did Aldebaran culminate at New York, in longitude 
;= 73 59' W., on the 17th of November 1841, its right ascension 



ASTRONOMICAL PROBLEMS 503 

being 4 h. 26 m. 51 '19 s., and the registered sidereal time at mean 
noon being = 15 h. 45 m. 29-03 s. ? . . At 12 h. 38 m. 28-96 s. 

866. Problem XIX. To find the mean time of the moon's 
culmination at any given meridian. 

RULE I. Find the sun's mean right ascension for the reduced 
time corresponding to the longitude, and find also the moon's right 
ascension for the same time ; subtract the former from the latter, 
increased if necessary by 24 hours, and the remainder will be an 
approximate time. 

Then as 1 hour diminished by the difference between the hourly 
variations in right ascension of the sun and moon is to the 
approximate time, so is this difference to a fourth term, which, 
added to the approximate time, will give the true time. 

Let A' = the moon's R.A. at the reduced time, 
A = n sun's ii ii ii M 

T" = n approximate time of culmination, 
T = n true time of culmination, 
vf = n hourly variation in R.A. of the moon, 
v = it n n n n sun, 

r = fourth term. 
Then T'=A'-A. 

\-(vf-v) :T' = t/-v : r, andT = T' + r. 

867. The interval of time between two successive meridian 
passages of the moon is called the moon's daily retardation. 

If the transit is required only to about a minute of accuracy, 
it can easily be found by the following rule : 

RULE II. Find the difference between the times of the pre- 
ceding and succeeding meridian passages that is, the moon's daily 
retardation ; then find the proportional part for the longitude in 
time ; add this part to the first registered time, and the sum will 
be the required time. 

Let P' = registered time of preceding passage, 
P = required time of meridian passage, 
rf =the moon's daily retardation, 
v = n variation for the given longitude in time, 
t = n longitude in time. 
Then 24 : t=v' : v, and v=^t. And P=P' + v. 

EXAMPLE. Find the time of the moon's culmination in longi- 
tude =40 45' W. on the 2nd of May 1854, 



504 ASTRONOMICAL PROBLEMS 

By the Second Method. 
Longitude 40 45' =2 h. 43 m. 

Time of reg. meridian passage on 2nd May= + 4 h. 13 m. Os. 
ii 3rd = 53 12 

Eetardation in 24 h = 50 12 

Hence retardation in 2 h. 43 m. . . = + 5 41 



Required time of transit . . = 4 18 41 

By the First Method 

Sun's m. R.A. at m. noon on 2nd May = 2 h. 40 m. 17'50 s. 

Increase or acceleration in 2 h. 43 m. . = + 26 '77 

Sun's R.A. at noon at given place, A = - 2 40 44 '27 

Moon's reg. R. A. on 2nd May at 2 h. . = 6 49 6'80 

3h. 6 51 18-74 



Increase in 1 h., or . . . i>'= 2 1T94 
.1 43 m. =+01 34-56 



Moon's R. A. at noon at place, . A' = + 6 50 41 -36 
Approximate time . = A'-A = T' = 4 9 57 '09 
Also, by Table of accelerations, . v = m. 9*858 s. 
And by Nautical Almanac, . . v'= 2 11-94 

Hence v'-v= 2 2'08 

And 1 -(v'-v): T'=v'-v : r. 

Or, 57 m. 57 '92 s. : 4 h. 9 m. 57'09 s. =2 m. 2'08 s. : r. 

P.L., 57 m. 57-92 s. . . . = 1 '39520 
P.L., 4 h. 9 m. 57-09 s. . . = '76050 
F.L., 2m. 2-08 s. . . . = 1-94701 

2-70751 

P.L., h. 8 m. 47-16 s. . . =1-31231 
And T' = 4 9 57 "09 

Hence T = 4 18 44-25 = mean time of transit. 

Let the meridian be at S at mean noon at the 
given place, and the moon then at M' ; then, if M 
be the position of the same meridian of the earth 
at the culmination of the moon, the meridian will 
have moved over the arc SM, while the moon has 
moved over M'M. Now, if arc SM' = T' in sidereal 
time = A' -A, 
SM =T" in sidereal time, 

T =the mean time corresponding to T", 
h, h' a mean and sidereal hour respectively, 
v, i/ = the variations in R.A. of sun and moon in 24A, 



ASTRONOMICAL PROBLEMS 505 

Then 24A' + v : v' = SM : M'M = T" : T" - T'. 

Or, 24A' - ( vf - v) : 24A' + v = T' : T". 

But 24A' + v : 24A' = T" : T. 

Therefore 24A' -(tf-v): 24/i/ = T' : T. 

By this proportion the required time T can be found. Or if 
v, vf refer to 1 instead of 24 hours, then 

l-(i/-v): 1 = T':T. 

Or, l-(i/-v):i/-v=T: T-T'. 

Or, l-(tf-v):T'=v'-v.r, if r=T-T. 

This proportion is the rule, and the calculation can be made by 
proportional logarithms, taking those for 24 hours for the first two 
terms, and for three hours for the third and fourth. 

EXERCISES 

1. Find the time of the moon's meridian passage at a place in 
longitude = 168 30' W. on the 27th of November 1854, the sidereal 
time or sun's mean light ascension at noon on the 27th being 
= 16 h. 24 m. 17'70 s., and the moon's right ascension on the same 
day at 11 h. =23 h. 17 m. 17'8 s., and at 12 h. = 23 h. 19 m. 23-67 s. 

At7h. 5m. 21 -15s. 

2. Find the time of the moon's meridian passage at a place in 
longitude=68 30' W. on the 5th of November 1841, the registered 
sidereal time on the 5th being=14 h. 58 m. 10-35 s., and the moon's 
registered right ascension on the same day at 4 h. being 8 h. 35 m. 
1-64 s., and at 5 h. =8 h. 37 m. 22-13 s. . At 18 h. 17 m. 15'17 s. 

868. Problem XX. To find the time of culmination of a 
planet at any given meridian. 

The rule is exactly the same as that in last problem ; only, as 
the increase in right ascension of a planet for 24 hours is small, the 
right ascension is given, not for every hour, but only for every 
noon ; and the meridian passages are given to the tenth of a 
minute. In the first formula of last problem, therefore, when 
adapted to this one, v and v' are the daily variations in right 
ascension of the sun and planet. Hence 

24-{v'-v) :T=v' -v :r, and T = T' + r. 

When v'< v, r is negative, and T = T'-r. When v' is negative- 
that is, when the motion of the planet is retrograde then r is also 
negative, and 

24-(t/ + v) :T'=<y' + y : r, and T = T'-r. 

EXAMPLE. Find the time of the meridian passage of Mars, at a 
place in longitude =45 30' W., on the 6th of April 1854, 



506 ASTRONOMICAL PROBLEMS 

By the Second Method 
Longitude in time = 3 h. 2 m. 

Time of registered meridian passage on 6th , = + 9 h. 6 m. 

. ,i ., . 7th . = + 9 1-9 

Retardation in 24 h. = 04-1 

i. M 3 h. 2 m. = - 0-5 



Time of rner. passage at given place on 6th . = 9 5 '5 

By the First Method 

Registered R. A. of Mars on 6th . . = + 10 h. 5 m. 12-72 s. 
ii it 7th . .= 10 5 5-28 

Decrease in 24 h = 00 7'44 

3 h. 2 m. =-00 0-94 



R, A. at noon at given place, . A'= 10 5 11 '78 
Registered R. A. of sun on 6th . . = 57 47 '09 
Increase or acceleration in 3 h. 2 m. = 29 '9 



R.A. of sun at noon at given place, A = 58 16-99 
Approximate time . =T' = A'-A= 9 6 54-79 
The motion of the planet being retrograde, 

v' - v = - m. 7'4 s. - 3 m. 56'6 s. = - 4 m. 4 s., 
ftnd 24 - (v 1 -v)=24 h. 4 m. 4 s.* 

P.L., 24 h. 4 m. 4 s. = - '00123= - P.L., 23 h. 55 m. 56 a, 
P.L., 9 6 54-79 =+ -42045 
P'.L., 04 4 =+1-64603 

F.L., 1 32-41 = 2-06771 
T'= 9 6 54-79 
T=9 5 22 -38 = T-r, the required time. 

EXERCISES 

1. Find the time of Jupiter's transit over the meridian in longi- 
tude =160 E. on the 10th of January 1854, the registered times of 
meridian passage on the 9th and 10th being 23 h. 20 -8 m. and 

23 h. 17-9 in. At 23 h. 19 '5 m. 

2. Find the time of the meridian passage of Jupiter on the 10th 
of September 1854 at a place in longitude ==160 E., its registered 
right ascension on the 9th and 10th being=19 h. 17 m. 6'93 s. and 
19 h. 17 m. 3-55 s., and that of the sun on the 10th being=ll h. 
16m. 46-43 s At 8 h. m. 43'63 s. 

* When if \ v, the first term, 24 (i/ v), exceeds 24 hours. But it will never exceed 

24 hours by more than 10 minutes, and L1440 L1430=L1450 L1440, when carried 
only to 5 figures. Hence L[24 (tf+v)] may be taken instead of 

and it must be added to the logarithms of the second and third terms. 



ASTRONOMICAL PROBLEMS 507 

869. Problem XXI. To find the meridian altitude of a 
celestial body at a given place, the declination of the 
body and the latitude of the place being given. 

RULE. Find the declination of the hody at its meridian passage 
at the given place ; then take the sum or difference of the colati- 
tude and declination, according as they are of the same or of 
different names, and the result will be the meridian altitude. 

Let L, C = the latitude and colatitude, 

D, P n declination and polar distance, 
A, A'= ii meridian altitudes at upper and lower culmi- 
nation ; 
then A=CD. 

When the declination exceeds the latitude, the altitude then 
would exceed 90, and the supplement is to be taken, which is the 
altitude from the opposite point of. the horizon below the pole ; 
or in this case A=L + P. 

When the declination exceeds the colatitude, the lower meridian 
passage will be above the horizon, and the altitude then is 
A' = D-C, or A' = L-P. 

In all these formulae the latitude and declination are of the 
same name, except in A=C-D. 

EXAMPLE. Required the meridian altitude of the moon at a 
place in latitude = 56 20' 10" N., and longitude =40 45' W., on 
the 12th of May 1854. 

Longitude in time = 2 h. 43 m. 

Time of registered passage on 12th . = +12 h. 16'8 m. 
,i 13th . = 13 16-6 

Retardation in 24 h = 59'8 

M 2 h. 43 m. = + 6-7 



Time of meridian passage on 12th . = 12 23 '5 



Declination on 12th at 12 h. . . = 19 O' 39'5" S. 

u n n 13 h. . . = 19 12 24-4 S. 

Increase in 1 h. . . . . = 11 45 '9 

.. ,i 23-5 m. =+04 36'5 



Hence declination at transit, . D= 19 5 16-0 S. 

Colatitude, C= 33 39 50 "0 N. 

Meridian altitude, . . A = C-D= 143434-0 

If the apparent altitude were required, this altitude just found 
Nvould require to be corrected for refraction and parallax. 



508 



ASTRONOMICAL PROBLEMS 




Let NS be the horizon, EQ the equator, NZSR a meridian, 
and B, B', B" celestial bodies in the meridian, 
and B6, B'6', and B"6" parallels of declina- 
tion ; then ZE L. ; and hence ES = C, and 
A=BS = C + D. So when B" is south of the 
equator, B"E = D, and B"S = A ; hence 
A=C-D. For the body B', D>L., and 
B'S = C + D is>90, and A = B'N = 180-(C + D) 
= L + P. When D>L., and of the same name, 
then A' = 6'N = D-C. 

EXERCISES 

1. Find the meridian altitude of Castor on the llth of May 
1854 at a place in latitude = 28 30' 25", its declination being 
= 32 12' 10-7" N =86 16' 30T'. 

2. Find the meridian altitude of Jupiter on the 10th of Septem- 
ber 1854 at a place in latitude = 46 35' 28" N., and longitude 
= 160 E., its registered declination on the 9th and 10th being 
= 22 45' 6-0" and 22 45' 14 -2" S. (See 2nd exercise, Art. 868, for 
time of transit.) =20 39' 19'7". 

870. Problem XXII. Of the obliquity of the ecliptic, the 
sun's longitude, declination, and right ascension, any two 
being given, to find the other two. 

Let PEQ be the solstitial colure, EQ the equator, CC' the 
ecliptic, PRP' a meridian through the sun's centre S. Then A is 
the first point of Aries, C of Cancer, C' of 
Capricorn, and the point diametrically oppo- 
site to A is the first point of Libra ; also, 
AS = L is the sun's longitude, 
AR = A it it right ascension, 
SR=D M it declination; 
angle SAR = O is the obliquity of the ecliptic. 

Now, the triangle ARS is right-angled at R, 
and any two parts of it, except the right angle, being given, the 
other two can be found by the rules of right-angled spherical 
trigonometry. 

EXAMPLE. The sun's registered longitude on the llth of 
May 1854 was = 50 25' 39 "4", and the obliquity of the ecliptic 
= 23 27' 34-26"; find the sun's declination and light ascension 
at mean noon at Greenwich. 




ASTRONOMICAL PROBLEMS 509 

In the triangle ARS are given AS = L = 50 25' 39 '4", and angle 
= O = 2327'34-26". 

To find AR=A To find RS-D 



L, cot L . . = 9-9172221 

L, radius . . =10* 

L, cos O . . = 9-9625311 



L, tan A . . =10-0453090 



L, radius . . =10' 

L, sin O . . = 9-5999732 

L, sinL . . = 9-8869532 

L, sin D . . = 9-4869264 



And A=47 59' 0'12" = 3 h. 11 m. 56-1 s., and D = 17 52' 10-2". 

When the longitude exceeds 90, so will the right ascension. 
Since right ascension is reckoned from the vernal equinox, and 
since the equator and ecliptic intersect at two points diametrically 
opposite, it is evident that to any particular declination there 
belong four different light ascensions, and of these the one re- 
quired must he determined by the time of the year. 

EXERCISES 

1. The sun's longitude on the 10th of June 1854 will be 
= 79 13' 1'8"; what will be its declination and right ascension, 
the obliquity of the ecliptic being = 23 27' 34 -04" ? 

D=23 1' 16-1" N., and A=5 h. 13 m. 5 s. 

2. The obliquity of the ecliptic on the 20th of July 1854 being 
=23 27' 34-47", and the sun's declination =20 42' 11-7" N., re- 
quired its right ascension and longitude. 

A=7 h. 57 m. 46-25 s., and L = 117 22' 23'3". 

3. The sun's right ascension and declination on the 8th of 
Septemher 1854 will be = ll h. 6 m. 30'79 s., and 5 43' 52'4" N. ; 
what will be its longitude and the obliquity of the ecliptic ? 

L = 165 28' 20-7", and O = 23 27' 34-1". 

4. On the 27th of December 1854 the sun's longitude and right 
ascension will be =275 28' 44-8", and 18 h. 23 m. 52-67 s. ; required 
its declination and the obliquity of the ecliptic. 

D = 23 20' 49-3", and O = 23 27' 37 T. 

871. Problem XXIH Having given the longitude and 
latitude of a celestial body, to find its right ascension and 
declination; and conversely, the obliquity of the ecliptic 
being supposed known in both cases. 

Let M be the moon or any celestial body (last fig.), and TMT' 
an ecliptic meridian ; then 

AL = L' is its longitude, reckoning from the nearest pre 
ceding equinox, 



510 



ASTRONOMICAL PROBLEMS 



L = the true longitude, 
ML I is its latitude, 
AR=A' is its right ascension, reckoned from the nearest 

preceding equinox, 
A = true right ascension, 
MR=D is its declination, 
angle SAR=O the obliquity of the ecliptic. 

Let AM = H its distance from preceding equinox, 
angle MAR = E, and MAL = C ; 
then E = O + C, according as M is without or within thd 

angle of the equator and ecliptic ; 
and C = E~O. 

EXAMPLE. If the moon's longitude on the 2nd of August 1841, 
at noon at Greenwich, be = 310 50' 1", its latitude=0 10' 1" N., 
and the obliquity of the ecliptic =23 27' 42", required its right 
ascension and declination. 

Here L' = 130 50' 1", 1 = 0" 10' 1" N., and O=23 27' 42", and M 
is between the equator and ecliptic. 

In triangle MAL 
To find H 

L, radius . . =10' 

L, cos L' . . = 9-8154878 

L, cos I . ."' = 9-9999982 
L, cos H 



. = 9-8154860 

And H = 130 50' 0'2". And C=0 13' 14-4" 

Hence, also, E = O - C = 23 14' 27 '6". 



L, tan I 
L, radius 
L, sin L' 
L, cot C 
And C = 



To find C 

. = 7-4644506 
. =10- 

= 9-8788730 



= 12-4144224 





To find D 


L, radius 


. =10- 


L, sin H 


. = 9-8788749 


L, sin E 


. = 9-5961565 


L, sin D 


. = 9-4750314 



In triangle MAR 



L, cot H 
L, radius 
L, cos E 
L, tan A' 



To find A 

. = 9-9366113 
. =10- 
. = 9-9632462 
= 10-0266359 



And D = 17 22' 16". I And A' = 133 14' 38-7". 

Hence A = A' + 180 = 313 14' 38 -7" =20 h. 52 m. 58 -58 s. 

EXERCISES 

1. On the 19th of May 1854, at noon at Greenwich, the moon's 
longitude was = 331 3' 6'7", and its latitude=5 17' 22'3" S. ; 
required its right ascension and declination, the obliquity of the 
ecliptic being =23 27' 34". 

A=22 h. 20 m. 11-52 s., and D = 16 2' 51 -3". 



ASTRONOMICAL PROBLEMS 



511 



2. On the 17th of September 1854 the registered right ascension 
of the moon at noon will be 8 h. 10 m. 31 '05 s. , and its declination 
=24 47' 45 '7" N. ; what will he its longitude and latitude at the 
same time, the obliquity of the ecliptic being =23 27' 35 '65" ? 

L = 119 24' 41-8", and J=4 36' 32-45". 

872. Problem XXIV. The right ascensions and declina- 
tions, or the longitudes and latitudes, of two stars being 
given, to find their arcual distance apart. 

Let PNP' be the solstitial colure, A the vernal equinox, MN 
the equator, P its pole ; D, E two celestial bodies, of which AB, 
AC are the right ascensions, and DB, EC the 
declinations. Then in triangle DPE are given 
the codeclinations PD, PE, and angle P = the 
difference of their right ascensions ; hence there 
are known two sides and the contained angle, 
and therefore the distance DE can be found. 

When the latitudes and longitudes of two 
bodies are given, their distance is found exactly 
in the same way. When one of the bodies, as E', is on the opposite 
side of MN, then PE' = 90 + CE'. 

Let C, C' = the complements of their declinations, 
P = M difference of their right ascensions, 
D = their distance ; 

then (Art. 779), R : cos P=tan C : tan . 

and cos : cos (C' - 0)=cos C : cos D 

Or D can be found, though not so concisely, by the method in 
Art. 774. 

EXAMPLE. Find the distance between Capella and Procyon on 
the 21st of January 1841, their right ascensions being=5 h. 4 m. 
59-77 s. and 7 h. 31 m. 0'79 s., and their declinations =45 49' 58'2" 
N. and 5 37' 37 -9" N. 

Here P = 2 h. 26 m. 1-02 s. =36 30' 15'3". 

C=44 10' 1-8", and C' = 84 22 22-1". 




[1], 
[3]- 



To find the arc 9 
L, radius . . =10' 



L, cos P 
L, tan C 
L, tan 

And = 37 58' 55". 



= 9-9051549 
= 9-9873727 
= 9-8925276 



To find the distance D 
L, cos . . = 9-8966390 



L, cos (C' - 0) 
L, cos C 

L, cos D 

And D = 51 7' 10-5". 



= 9-8386823 
= 9-8557069 
19-6943892 
= 9-7977502 



512 



ASTRONOMICAL PfcOBLEMS 



When the difference of the right ascensions exceeds 12 hours, 
add 24 hours to the less, and from the sum subtract the greater, 
and the difference will be the included angle at the pole. 

EXERCISES 

1. When the latitudes of Sirius and Procyon were = 39 34' S. 
and 15 58' S., and their longitudes = 101 14' and = 112 56', what 
was their distance? =25 42' 52". 

2. Find the distance of Rigel and Regulus on the 1st of August 
1854, their right ascensions being 5 h. 7 m. 33 s. and 10 h. m. 37 s., 
and their declinations = 8 22' 30" S. and 12 40' 34" N. 

= 75 45' 44 7". 

873. Problem XXV. Given the latitude of the place, 
the declination and altitude of a celestial body, to find 
its azimuth. 

Let HZR be the meridian of the place, HR the horizon, and Z 
the zenith ; EQ the equator, and P its pole ; 
S the body ; then 

ZS = Z = zenith distance or coaltitude, 
PS = P=body's polar distance or codeclina- 

tion, 

PZ = C =colatitude, and angle 
PZS=A=supplement of azimuth AZR from 

south. 

Here P, C, Z are given ; and hence A can be found by Art. 769 ; 
thus, if S = |(P + C + Z), then 

2LcosJA = Lsin S + Lsin (S-P) + Lcosec C + L cosec Z - 20. 
If the body's declination is south, as at S', while the given 
latitude is north, the polar distance PS' = 90 + DS'. 

EXAMPLE. When, in latitude =44 12 7 N., the sun's altitude 
was = 36 30', its declination being =15 4' N., what was its 
azimuth ? 




Here 



and 



P= 74 56' 
C= 45 48 
Z= 53 30 

2)174 14 



L, sin S 
L, sin (S - P) 
L, cosec C . 
L, cosec Z . 



= 9-9994498 

= 9-3243657 

= '1445350 

= -0948213 



Hence S= 87 7 2)19-5631718 

AndS-P= 12 11 L, cos ^A, . = 9-7815859 

Hence A=52 47' 17'3", and A = 105 34' 34'6" = the azimuth 
from the north. 



ASTRONOMICAL PROBLEMS 513 

EXERCISES 

1. When, in latitude=48 51' N., the sun's declination is = 18 
30' N., and its altitude 52 35', what is its azimuth from the 
north? =134 36' 7". 

2. If, in latitude =51 32' N., the altitude of Arcturus was found 
to be = 44 30', when its declination was =20 16' N., what was its 
azimuth from the north ?. . . . . =117 8' 28". 

3. When, in latitude = 51 32* N., the sun's altitude was=;25 , 
and its declination =4 47' S., what was its azimuth from the 
north? 137 17' 37". 

METHODS OF DETERMINING TIME 

874. Problem XXVI Given the latitude of the place, 
the sun's declination and altitude, to find the hour of the 
day in apparent time. 

The angle P (last fig. ) in triangle SPZ is evidently the horary 
angle. If this angle be denoted by H, then (Art. 769) 

2Lcos JH = Lsin S + Lsin (S-Z) + L cosec P + L cosec C-20. 

EXAMPLE. On the 8th of May 1854, at 5 h. 30 m. 32 s. P.M. 
per watch, in latitude = 39 54' N., and longitude = 80 39' 45" W., 
the altitude of the sun's lower limb was observed to be 15 40' 57" ; 
required the error of the watch. 

Given time = 5 h. 30 m. 32 s. 

Longitude in time .... = 5 22 39 
Greenwich time . . . . =10 53 11 

1. To find the sun's declination 

Sun's registered declination on 8th . . = 17 4' 36" 

9th . . = 17 20 45 

Increase in 24 h. = 16 9 

Hence increase in 10 h. 53 m. 11 s. =07 19 



Required declination = 17 11 55 

2. To find the sun's true altitude 

Observed altitude 15 40' 57" 

Refraction = - 3 21 

Semi-diameter = + 15 52 

Contraction . ...... . . .=-003 

Parallax = + 008 

True altitude of centre , , , . = 15 53 33 



514 ASTRONOMICAL PROBLEMS 

3. To find the horary angle 

Z = 74 6' 27" L, sin S . . . = 9-9951982 

P = 7248 5 L, sin(S-Z) . . = 9-6160088 

C = 50 6 L, cosec P . . . = -0198668 

2)197 32 L, cosec C . . . = -1151111 

S = 98 30 16 2)197461849 

S^Z = 24 23 49 L, cos |H . . . = 9*8730924 

Hence H = 41 42' 9 -5", 

and H=5 h. 33 m. 37 "28 s. = apparent time. 

Time by watch = 5 30 32 

Watch slow by 3 5 -28 for .. H 

EXERCISES 

1. In latitude=52 12' 42" N., in the afternoon, the true alti- 
tude of the sun's centre was = 39 5' 28", when its declination was 
= 15 8' 10" N. ; what was the apparent time of observation ? 

= 2 h. 56 m. 42-7 s. 

2. In latitude = 24 30' N., in the forenoon, the true altitude of 
the sun's centre was found to be 33 20', when its declination was 
= 6 47' 50" S. ; required the apparent time of observation. 

= 8 h. 45 m. 57-4 s. A.M. 

875. Problem XXVII. Given the latitude and longitude 
of the place, the right ascension and declination of a fixed 
star and its altitude, to find the mean time. 

RULE. Find the horary distance of the star from the meridian ; 
then find the sun's mean right ascension at the preceding mean 
noon at the given place, and subtract it from the star's right 
ascension, increased if necessary by 24 hours ; to this interval 
apply the horary angle by addition or subtraction, according as 
the star is west or east of the meridian, and the result is the 
sidereal interval from mean noon, and its corresponding interval 
of mean time will be the required time. 
As in last problem, 

2Lcos JH = Lsin S + L sin (S-Z) + L cosec P + L cosec C- 20. 
Let A = sun's mean R.A. at preceding noon at place, 
A' = star's right ascension, 
d =the difference of A and A', 
s = sidereal interval from mean noon, and m, r the 

corresponding mean time and retardation ; 
then d = A'- A, s = d H, and m=$-r. 



ASTRONOMICAL PROBLEMS 515 

EXAMPLE. At a place in latitude = 48' 56' N., and longitude 
=66 12' W., the true altitude of Aldebaran, which was west of 
the meridian, was = 22 24' on the 10th of February 1854; what 
was the mean time of observation ? 

Star's dec. = 16 12' 43" N, and R.A. =4 h. 27 m. 33'3 s. 
Z= 67 36' 0" L, sinS . = 9-9999004 

P= 734717 L, sin(S-Z) . . = 9-6029108 

C = 41 4 L, cosec P . . . = -0176222 

2)182 27 17 L, cosec C . . = -1824765 

S = 91 13 38 2)19-8029099 

S-Z = 23 37 38 L, cos JH, . = 9'9014549 

Hence |H=37 9' 21'7", and H = 4 h. 57 m. 14-9 s. 

Sun's reg. RA. or sidereal time on 10th = 21 h. 20 m. 56'6 s. 
Acceleration for long. 4 h. 24 m. 48 s. W. = + 43 -5 

Sun's R.A. for noon at place, . A = + 21 21 40'1 
Star's I. A'= 4 27 33 '3 



Hence d=A'-A= 1 5 53'2 



and S=rf + H = 12 3 8-1 

and retardation = - 1 58 -5 




Mean time required . . . . = 12 1 9 '6 

Let A be the first point of Aries, MP the meridian of the place, 
B the place of the star, and S of the mean sun at 
time of observation. Then, if since noon the sun's 
increase of mean right ascension be SS', the sidereal 
interval at noon between the sun and star is = BS', 
and as B has passed the meridian by the horary 
arc BM, the sidereal time from noon till the obser- 
vation is expressed by S'M = S'B + BM = d + H 
=A'-A + H. And when the star is east of the meridian when 
observed, as at B', the sidereal time from noon = S'M = S'B' - B'M 
=A'-A-H. And this interval, reduced to mean time, gives that 
required. 

EXERCISES 

1. If, at a place in latitude = 53 24' N., and longitude=25 18' W., 
the altitude of Coronse Boreal is when east of the meridian was 
found to be =42 8' 0" cm the 31st of January 1841, its right ascen- 



516 ASTRONOMICAL PROBLEMS 

sion being=15 h, 27 m. 58 s., its declination =27 15' 12" N., and 
the registered mean right ascension of the sun = 20 h. 42 ra. 8 s., 
find the mean time of observation. 

H=3 h. 40 m. 19 '3 s., and the mean time = 15 h. 2 m. 45 '8 s. 
2. Find the hour of observation in mean time at which the 
altitude of Procyon was = 28 10' 13", when east of the meridian 
in latitude = 7 45' N., its declination being =5 41' 52" S., its 
right ascension = 7 h. 29 m. 30 s., and that of the mean sun at 
mean noon = 11 h. 4 m. 40 s. 

H = 4 h. 2 m. 0-5 s., and time = 16 h. 20 m. 8-5 s. 

876. The equation of equal altitudes is a correction, generally 
of a few seconds (and seldom exceeding half-a-minute), that must 
be applied to the middle time between the instants of two obser- 
vations at which the sun has equal altitudes in the forenoon and 
afternoon. It depends on the change of the sun's declination in 
the interval between the observations. 

877. Problem XXVIII. To find the equation of equal 
altitudes. 

Let L = the latitude of the place, 

I = H interval of time expressed in degrees, &c., 
P= M sun's polar distance, 

t/ = variation of declination in 24 hours in seconds, 
v = H H interval I in seconds, 

E = the equation of equal altitudes in seconds. 
Then an arc 0, called arc first, is such that 

L, tan = L, cot L + L, cos 1 - 10 ; 
and if is another arc, called arc second, then </> = P - 6. 
And L, E = L, cot 41 + L, cosec + L, cosec P + L, sin 

+ L, I + L, v' + 45-3645, 

in which the quantity I in L, I is expressed in minutes. The 
logarithms require to be carried only to four places. This rule is 
approximate, but it will give the result correct to a small fraction 
of a second. The polar distance at the nearest noon may be used, 
as any small change in it or in the latitude produces a very small 
effect on the equation. 

EXAMPLE. Find the equation of equal altitudes for an interval 

of 7 h. 45 m. 30 s., and latitude = 46 30' S., on the meridian of 

Greenwich, the sun's declination being = 7 10' N. 

Here L = 46 30' S., 1 = 7 h. 45 m. 30 s., I = 58 11'. 

P=97 10' N., I m. =465-5 m., v' = 22' 14" = 1334". 



ASTRONOMICAL PROBLEMS 517 

L, cot L . . = 9-97725 Constant . . = 45 -3645 

L, cos |I . . = 9-72198 L, cot 1 . = 9'7927 

L, tan 6, 26 35' = 9'69923 L > cosec 6 . . = 10-3492 

p = 97 10 L, cosec P . . = 10-0034 

, = ^r L, sin * . . = 9-9746 

L, I m. . . = 2-6679 

L, v' . . . = 3-1251 

L, E 18-9 s. . . = 1-2774 
Hence the equation of equal altitudes is 18-9 s. 

It is evident that when the declination of the sun has varied 
in one direction during the interval between two equal altitudes, 
the intervals between the meridian passage, or apparent noon, 
and the instants of the two observations are different. When 
it increases, the interval in the afternoon will exceed that of the 
forenoon, and conversely when it diminishes. For a demonstration 
of the rule, see Riddle's treatise on Navigation and Nautical 
Astronomy. A slight alteration has been made here which 
improves it a little. Instead of L, ^v, where v is the variation 
due for the interval I, and which requires v to be separately 
calculated, there has been introduced above the constant 5-3645, 
L, I m. and L, v'. 

For 24 : I=i/ : v ; and hence L, v = L, I + L, i/ - L, 24. 
And if v', v are in seconds of space, and 24 and I m. in minutes 
of time, then, since 24 h. = 1440 m., and L, ^v = ~L, v-L, 30; 
therefore, L, ,V = L, I m. + L, i/ + 5'3645. 

EXERCISE 

If at a given place, when the sun's declination at noon was 
= 17 54' N., the sun had equal altitudes at an interval of 5 h. 
40 m. 6 s., the latitude of the place being = 57 10', what was the 
equation of equal altitudes ? = 14-36 s. 

878. The middle time for the times of observation of two equal 
altitudes of the sun is half the sum of the times. 

879. Problem XXIX. To find the time by equal alti- 
tudes of the sun. 

RULE. Apply the equation of equal altitudes to the middle 
time by addition or subtraction, according as the polar distance is 
increasing or diminishing, and the result is the time shown by the 

fne. 2 H 



518 ASTRONOMICAL PROBLEMS 

clock at apparent noon ; find the mean time at apparent noon, and 
the difference between it and the preceding time will be the error 
of the clock. 

When a chronometer is used for the times of observation, apply 
the longitude in time to the mean time at apparent noon, and the 
result is the mean time at Greenwich at that instant ; and the 
difference between it and the time found by the chronometer for 
the same instant will be the error of the chronometer. 

EXAMPLE. At a given place the altitude of the sun was the 
same at 9 h. 34 m. 20 s. A.M., and 2 h. 32 m. 26 s. P.M.; required 
the error of the clock, the polar distance being decreasing, the 
equation of equal altitudes = 8'4 s., and the equation of time 
*= 1 m. 57'6 s. to be added to apparent time. 

Time of first observation . . . = 21 h. 34 m. 20 s. 

ii second . = 2 32 26 

24 6 46 

Middle time of observation . . . = 12 3 43 

Equation of equal altitudes . . = - 8 '4 

Time by clock at apparent noon . = 12 3 34 '6 

Meantime ,, n H -' 12 1 57 '6 



Clock is fast MM n . . = 1 37 

Instead of only two observations being taken, several corre- 
sponding pairs may be taken, and the sum of the times of observa- 
tion in the forenoon being divided by their number, and also the 
sum of the afternoon observations being similarly divided, the 
quotients are the mean times of observation, which are then to 
be treated as the two times of observation. The times of observa- 
tion ought to be more than two hours distant from noon. 

EXERCISES 

1. At a given place the altitude of the sun was the same at 
8 h. 4 m. 54 s. and 4 h. 2 m. 36 s. ; required the error of the clock, 
the polar distance being increasing, the equation of equal altitudes 
= 12-4 s., and the equation of time = 4 m. 16 '7 s. to be subtracted 
from apparent time. ........ Clock fast 8 m. 14'1 s. 

2. Suppose that at a given place the altitude of the sun was the 
same at 9 h. 40 m. 2 s. A.M. and 2 h. 10 m. 25 s. P.M. ; required the 
error of the clock, the polar distance being decreasing, the equation 
of equal altitudes = 14 '5 s., and the equation of time = 3 m. 50'2 s., 
to be added to apparent time. . . . Clock slow 8 m. 5J/2 s. 



ASTRONOMICAL PROBLEMS 



519 




880. Problem XXX. Given the latitude of the place 
and the declination of a celestial body, to find its ampli- 
tude and ascensional difference. 

Let HZR be the meridian of the place ; EQ the equator, P its 
pole ; HR the horizon, Z the zenith, and B the 
body in the horizon when rising ; then in the 
triangle ODB, D is the right angle, O is the 
colatitude, BD the declination, OB the am- 
plitude, OD the ascensional difference, and 
EO + OD the semi-diurnal arc. When the 
declination is south, OB' is the amplitude, OD' . 
the ascensional difference, and EO - OD 7 the 
semi-diurnal arc. When the body is setting, the figure is exactly 
similar. 

It is evident that if any two parts of the right-angled triangle 
OBD are given, the other parts can be found. 

Let angle BOD = C the colatitude, 
BD =D H declination, 
OB =M H amplitude, 
OD =N H ascensional difference, 
BW =1 ii semi-diurnal arc. 

To find the amplitude OB 
R. sin BD = sin O. sin OB, or sin C : R=sin D : sin M. 

To find the ascensional difference OD 
R. sin OD=cot O . tan DB, or R : tan D = cot C : sin N. 
Then 1=6 h. N. 

EXAMPLE. When the declination of a celestial body is = 14 15' 
N., what is its amplitude and ascensional difference in latitude 
= 3645'N.? 

Here C=53 15', D = 14 15'. 



To find the amplitude M 
L, sin C . . = 9-9037701 
L, radius . . = 10* 
L, sin D . . = 9-3912057 
L, sin M . . = 9-4874356 
And M = 17 53' 28" N. 



To find N 
L, radius . 
L, tan D . 
L, cot C . 

L, sin N 

= 10 55' 56". 



= 10- 

= 9-4047784 

= 9-8731668 

= 9-2779452 



The semi-diurnal arc 1 = 6 h. + N = 6 h. 43 m. 43-7 s. 



52Q ASTRONOMICAL PROBLEMS 

EXERCISES 

1. The declination of a celestial body is = 37'34'N. ; what is its 
amplitude and ascensional difference in latitude = 46 8' N. ? 

M = 61 37' 4-5" N., and N = 3 h. 32 m. 36'4 s. 

2. The declination of a celestial body is = 26 3' 53" S. ; what is 
its amplitude and semi-diurnal arc in latitude = 55 N. ? 

M = 50 S., and 1 = 3 h. 2 m. 45'4 s. 

881. Problem XXXI. To find the apparent time at which 
the sun's centre rises or sets at a given place. 

Find the sun's zenith distance when its centre appears on the 
horizon ; then, its polar distance and the colatitude being known, 
find the corresponding semi-diurnal arc, and this arc, converted 
into time, will be the time of rising or setting before or after 
apparent noon. 

The sun's zenith distance, when the apparent altitude of its 
centre is zero, is found by subtracting its parallax from the sum 
of the dip and horizontal refraction, and adding the remainder 
to 90. 

The declination to be used is of course that at rising and 
setting, which can be found by first determining the semi-diurnal 
arc, as in last problem, supposing the declination to be that 
at noon at the given place ; and then the approximate times 
of rising and setting are known, and the longitude being also 
known, the reduced time, and hence also the reduced declina- 
tion, can be found. 

EXAMPLE, Find the mean time of the apparent rising of the 
sun's centre on the 24th of May 1841 at a place in latitude = 55 
57' N., and longitude =25 30' W., the observer's eye being at the 
height of 24 feet. 

Approximate apparent time of rising on 23rd = 15 h. 38 m. 

Longitude in time = 1 42 

Reduced time of rising on 23rd . . = 17 20 

Hence reduced declination . . . . 20 44' 29" N. 

Horizontal refraction 33 51 

Depression . . ... . . = 4 47 

Horizontal parallax . . . . .=-009 

Depression of centre . . . . = . 38 29 



ASTRONOMICAL PROBLEMS 521 

By (Art. 769), 2 L cos H = L sin S + L sin (S - Z) 
+ L cosec P + L cosec C - 20. 

Here Z = 90 38' 29" L, sin S . . . = 9'9967739 

P = 9 15 31 L, sin(S-Z) . . = 9O426439 

C = 34 3 L, cosec P . . . = 10'0291009 

2)193 57 L, cosec C . . . = 1Q-251877Q 

S = 96 58 30 2)19-3203957 

S - Z = 6 20' 1" L, cos H . . . = 9*6601978 

And H= 62 47 8 And H = 8 h. 22 m. 17'8 s. 

Hence apparent time of rising on 24th is 3 h. 37 m. 42'2 s. A.M. 
Equation of time . . . . = - 3 30 -9 

Mean time of rising . . . = 3 34 11-3 

EXERCISE 

Find the mean time of the setting of the sun on the 20th of July 
1841 in longitude = 35 45' E., and latitude = 55 57' N., the eye of 
the observer being 20 feet high, its registered declination on the 
20th and 21st being = 20 40' 38" and 20 29' 12", and the equations 
of time = 5m. 58 -7s. and 6m. 2-1 s. . . At 8 h. 27 in. 9'3 s. 

882. The time of a star's rising or setting may be found thus : 
Compute the star's semi-diurnal arc, and it will be the sidereal 
interval from its rising to its culmination, which is to be reduced 
to the mean solar interval by Art. 858 ; then find the mean time of 
the star's culmination by Art. 865, and apply to it the preceding 
interval by subtraction for the mean time of rising, and by addition 
for the time of setting. 

The time of the moon's rising or setting may be found thus : 
Find approximately its semi-diurnal arc, considering its declination 
and horizontal parallax to be that at the nearest noon ; and find 
the time of its meridian passage ; then the approximate time of its 
rising or setting is known. Compute its declination and parallax 
for the reduced approximate time of rising, and find again its 
semi-diurnal arc ; then 24 hours is to the semi-diurnal arc found 
as the daily retardation to a fourth term, which, added to the 
preceding arc, will give the interval between the rising or setting 
and the meridian passage in mean time. For a sidereal day is to 
any sidereal arc as a lunar day (expressed in mean time) is to the 
corresponding lunar arc (expressed also in mean time). Then the 
sum or difference of this interval and the time of transit will be 
the time of rising or setting. 



522 ASTRONOMICAL PROBLEMS 

Let H = semi-diurnal arc in sidereal time, 

H' = corresponding lunar arc in mean time, 

R = moon's daily retardation n n 

r = n retardation for arc H' in mean time, 

t' the mean time of transit, 

t = it it rising or setting ; 

then 24:R = H:r, or P.L, r=P.L, R + P.L, H, 

and H' = H + r; then = <'H', 

where the upper sign refers to the time of setting, and the lower 
to the time of rising. 

The same method applies in finding the rising or setting of the 
planets ; but when v' < v, or when v' is negative (Art. 868), r is 
negative, and H' = H - r. 

EXERCISE 

Find the mean time of the rising of the moon for the data of the 
example in Art. 867 ; having also given the moon's declination on 
the 2nd of May at noon = 26 21' 52'9" N., its horizontal parallax 
on the 2nd at noon and midnight =54' 9 - 5" and 54' 10 "8", and its 
declination on the 1st at 21 h.=26 21' 49", and at 22 h. = 
26 21' 58", the horizontal refraction being =33' 50", and the 
latitude of the place = 54 30' S. . . =7 h. 1 m. 46 -7 s. A.M. 

METHODS OP FINDING THE LATITUDE 

883. Problem XXXII. Given the declination of a celes- 
tial body, and its meridian altitude, to find the latitude 
of the place of observation. 

Call the true zenith distance of the object north or south, accord- 
ing as the zenith is north or south of the body ; then, when the 
zenith distance and declination are of the same name, their sum is 
the latitude also of the same name ; but when of different names, 
their difference is the latitude of the same name as the greater. 

When the body is at the lower culmination, the latitude is equal 
to the sum of the altitude and polar distance, and is of the same 
name as the latter. 

That is, L = ZD, or L = A' + P, where A' is the altitude for the 
lower culmination. 

EXAMPLE. On the 2nd of May 1854, in longitude = 50 15' AY., 
the observed meridian altitude of the sun's lower limb was 
70 31' 18" S., the height of the eye beiug = 15 feet; what is the 
latitude ? 



ASTRONOMICAL PROBLEMS 523 

Observed altitude A"= 70 31' 18" S. 

Depression, d - 3 45 

70 27 33 

Refraction, r = - 20 

True altitude of lower limb . . . = 70 27 13 

Sun's semi-diameter . . . . = + 15 53 

Parallax = + 003 

True altitude of centre, . . A = 70 43 9 

Longitude 50 15' W. = 3 h. 21 in. 

Registered declination on 2nd May . = 15 21' 52 -2" N. 
Increase in 3 h. 21 m. . _, . . = 2 28'9 
Declination at given time, . . . D = 15 24 21 ! N. 
Zenith distance, . . . . . P = 19 16 51 

Latitude, L = 34 41 12-1 N. 

The principle of the rule is easily explained by a reference to the 
figure in Art. 869. Let B be the body, then the zenith distance 
ZB and the declination BE are of the same name, whether P be 
the south or the north pole ; and the latitude EZ is their sum, and 
of the same name. If B" be the body, then the zenith distance 
B"Z and declination B"E are of different names, and the latitude 
EZ = B"Z - B"E. So when B' is the body, Z and D are of different 
names, and EZ = EB'-ZB'. Let b' be the body at the lower 
culmination, A' = 6'N, and P=6'P; then L = A' + P. 

EXERCISES 

1. If the true altitude of Aldebaran, at a place in longitude = 
48 3tf W., on the 20th of May 1854, was = 54 20' 35" S., required 
the latitude, the star's declination being =16 12' 45" N. 

= 51 52' 10" N. 

2. If the meridian altitude of the moon's centre on the 2nd of 
May 1841, in longitude = 40 45' W., was=25 13' 45'2" S., when its 
declination was -S" 26' 3 -8" S., what was the latitude of the place? 

=56 20' 10" N. 

884. Problem XXXIII. Given the sun's declination and 
altitude, and the hour of the day, to find the latitude of 
the place. 

Let the parts of the triangle PZS in Art. 873 represent the same 
quantities as in that problem ; then the polar distance P = PS, the 
zenith distance Z = ZS, and the horary angle H = SPZ are given, to 
tind the colatitude C = PZ, 



524 



ASTRONOMICAL PROBLEMS 



The side PZ can be obtained by the method in Art. 776, or, 
more concisely, by that in Art. 781. In it the quantities a, b, 
A, and c are respectively the same as Z, P, H, and C in this 
problem. Hence R . tan 6 = tan P . cos H, 
and cos P : cos Z=cos : cos (c~ 0). 

The algebraic signs of the terms indicate whether these arcs are 
greater or less than quadrants. 

EXAMPLE. On the 8th of May 1854, at 5 h. 33 m. 33 '4 s. apparent 
time, the altitude of the sun's lower limb was observed to be = 
15 40' 57", the longitude of the place being=80 39' 45" W. ; what 
was the latitude ? 

The declination of the sun at the time of observation is found to 
have been = 17 12', and the true altitude of its centre = 15 53' 37" ; 
hence P = 72 48', Z = 74 6' 23", and H = 83 23' 21". 



To find the segment 6 
L, radius . . = 10* 
L, cos H . . = 9-0611695 
L, tan P . . = 10-5092668 


To find the segment (c 9) 
L, cos P . . = 9-4708631 
L, cos Z . . = 9-4375159 
L, cos . . = 9-9718687 


L, tan . . = 9-5704363 


19-4093846 


Hence = 20 24' 2". 


L, cos(e-0) . = 9-9385215 
Hence (c - 0) . = 29 46' 21" 
= 20 24 2 


Therefore the colatitude, or . 
And therefore the latitude, . 


. C = 50 10 23 
. L = 39 49 37 



EXERCISES 

1. At a given place in south latitude, when the sun's declination 
was = 15 8' 10" S., its true altitude was = 39 5' 28" at 2 h. 56 m. 
42'7 s. P.M. ; required the latitude of the place. =52 12' 42". 

2. At a place in north latitude, when the sun's declination was 
= 6 47' 50" S., its true altitude was = 33 20' at 8 h. 46 m. A.M. ; 
required the latitude . =24 31' 10 -3" 

885. Problem XXXIV. Given two altitudes of the sun or 
of a star, and the interval of time between the observa- 
tions ; or the altitudes of two known stars, taken at the 
same instant, to find the latitude of the place. 

Let P be the pole, Z the zenith, and B, B' the body in two 
different positions, or two different bodies. Then PB, PB' are 
the polar distances, and ZB', ZB the zenith distances ; these four 
quantities being given. Also, when B and B' are the sun in two 




ASTRONOMICAL PROBLEMS 525 

different positions on the same day, or of a star on the same night, 
or of two different stare at the same instant, the angle BPB', which 
measures the elapsed time, or the difference of 
right ascensions of two different stars, is known. 
But in the second case, when B, B' are two 
different stars, and the elapsed time between 
the observations is measured in mean time, it H | 
must be reduced to sidereal time. Hence the 
latitude may be found thus : 
Let P, P' = the polar distances PB, PB', 
Z, Z' = i, zenith ,. BZ, B'Z, 

H = ii angle BPB', 

E = side BB', 
and L, C = latitude and colatitude ZE and ZP. 

1. To find angle B' in triangle PBB' 

By Art. 774, sin (P + P') : sin |(P~P')=cot JH : tan i(B~B'), 
and cos i(P + P) : cos ^(P~P') = cot H : tan 

From which B and B' can be found. 

2. To find E in triangle PBB' 
Sin (B~B') : sin (B + B') = tan 4(P-P') : tan 



3. To find angle B' in triangle BZB' 

By Art. 769, 2 L cos B' = L sin S + L sin (S - Z) + L cosec 
E + LcosecZ'-20. 

4. To find angle B in triangle PB'Z 

B' = BB'Z~PB'B. 

5. To find C in triangle PB Z 
By Art. 779, R : cos B' = tan Z' : tan 0, 

and cos : cos (P'-#) = cos Z' : cos C=sin L. 

When B and B' are the same star at the times of the two 
observations, P is = P'; and when they represent the sun, if its 
declination for the middle time between the observations be taken, 
PB and PB' may be considered equal to this declination P. 
The solution may then be simplified, for PBB' will be an isosceles 
triangle, and a perpendicular from P on BB' will bisect it, and 
will form two equal right-angled triangles. Hence, instead of the 
preceding formulae at No. 1 and 2, take these two : 

1. To find B' in one of the right-angled triangles 
CotiH:R = cosP:cotB'. 



526 ASTRONOMICAL PROBLEMS 

2. To find E in the same triangle 
R : sin P = sin |H : sin JE. 

EXAMPLE. If in the forenoon, when the sun's declination was 
= 19 39' N., at the middle time between two observations of its 
altitude, these altitudes corrected were = 38 19' and 50 25', what 
was the latitude, the place of observation being north, and the 
interval between the observations one hour and a half ? 

Here P = 70 21', Z = 51 41', and Z' = 39 35'. 

H = l h. 30 m., and 4H = 11 15', and PZ = C. 



1. To find angle B' 


in PBB' 2. To find E in 


PBB' 


L, cot H . . = 


10-7013382 L, radius . . = 


= 10- 


L, radius . . = 


10- L, sin P . . = 


= 9-9739422 


L, cos P . . = 


9-5266927 L, sin H . . = 


= 9-2902357 


L, cot B' . . = 


8-8253545 L, sin |E . . = 


- 9-2641779 


And B' = 86 10' 24". 


And E = 21 10' 26". 




3. To find angle B' in BZB and PB'Z 


Z 


= 51 41' 0" 




L, cosec Z' 


= 39 35 . . . = 


10-1957243 


L, cosec E 


= 21 10 26 . . . = 


10-4422527 




2)112 26 26 




L, sin S . 


= 56 13 13 . . . = 


9-9196958 


L, sin (S - Z) . 


= 4 32 13 . . . = 


8-8981869 






2)19-4558597 


L, cos ^B', 


= 57 41' 29-4" . = 


9-7279298 




2 




B' 


= 115 22 59 




PB'B 


= 86 10 24 




4. PB'Z 


= 29 12 35 






5. To find C in PB'Z 




L, radius . . = 


10- L, cos (a . c)* . 


= 0-0910331 


L, cos B' . = 


9-9409342 L, cos (P - 0) 


= 9-9158171 


L, tan Z' . . = 


9 9173911 L, cos Z' 


= 9-8868846 


L, tan . = 


9-8583253 L, sin L 


= 9-8937348 


And = 35 48' 58". 


And L = 51 31' 54". 




P - = 34 32 2. 







* Here (a. c) means the arithmetical complement of L. cos 6 (see Table of ' Num- 
bers of Frequent Use in Calculation,' p. 620). 



ASTRONOMICAL PROBLEMS 



527 



The following method is somewhat simpler when only one star is 
observed, or when the mean declination of the sun is used, as in the 
last example.* 

Let P, H, and L denote the same quantities as in the preceding 
rule ; let S = A + A', the sum of the true altitudes, 

D=A~ A' ti difference of the true altitudes ; 
and let M, N, O, Q, R denote what are called the first, second, 
third, fourth, and fifth arcs ; then, 

1. Lsin M = Lsin P + Lsin ^H-10. 

2. L cos N = L cos P + L sec M - 10. 
And N is of the same species as P. 

3. L sin O = L sin ^D + L cos S + L cosec M - 10. 

4. L cos Q - L cos D + L sin S + L sec M + L sec O - 10. 

5. R=NiQ. 

When the zenith and elevated pole are on the same side of the 
great circle that passes through the two positions of the sun or 
star, R=N~Q; otherwise R = N + Q. 

6. L siu L = L cos O + L cos R - 10. 

The preceding example, computed by this rule, is added here : 

P =70 21' A =38 19' iS =44 22' 

H = 11 15 A'=50 25 D= 6 3 



1. To find M 


2. To find N 


L, sin P . . = 9-9739422 


L, cos P . . = 9-5266927 


L, sin H . . = 9-2902357 


L, sec M . . = 10-0074565 


L, sin M . . = 9-2641779 


L, cos N . . = 9-5341492 


M = 1035' 13". 


N = 69 59' 44". 


3. To find O 


4. To find Q 


L, sin D . . = 9-0228254 


L, cos JD . . = 9 9975743 


L, cos |S . .van 9-8542329 


L, sin iS . . = 9 8446310 


L, cosec M . = 10-7358221 


L, sec M . . = 10-0074565 


L, sin O . . = 9-6128804 


L, sec O . . = 10-0399840 


O=24 12' 38". 


L, cos Q . . = 9-8896458 


5. R = 3051'22" = N~Q. Q=398'22". 



6. To find L, 

L, cos O = 9-9600160 

L, cos R = 99337191 

L, sin L = 9-8937351 

And L = 5P 31' 54", as before. 

* For the demonstration of this rule, consult any good treatise on Navigation 
ud Nautical Astronomy. 



528 ASTRONOMICAL PROBLEMS 

In this problem it is not necessary to know the times of obser- 
vation, but merely the interval between them ; but the latitude 
being found, the time of either observation could be calculated, 
and if the time was taken with a chronometer, the longitude could 
also be found. 

EXERCISES 

1. At a place in north latitude, when the sun's declination was 
= 23 29' N., its true altitude at 8 h. 54 m. A.M. was 48 42', and at 
9 h. 46 m. A.M. it was = 55 48' ; required the latitude. 

= 49 49' 23" N. 

2. At a place in north latitude, when the sun's declination was 
=2 46' S., its true altitude -was = 33 11' at 9 h. 20 m. A.M. and 
42 44' at 1 h. 20 m. P.M. ; required the latitude. =40 50' 7" N. 

3. Find the time at which the first altitude was taken in the 
example to this problem, and the azimuth. 

The time=8 h. 30 m. 2 s., and the azimuth from N. = 107 47' 42". 

4. On the 6th of October 1830, at a place in north latitude, the 
true altitude of the sun at 7 h. 5 m. 49 s. A.M. mean time was 
= 8 37' 42-6", and at 1 h. 2 m. 47'8 s. it was = 33 43' 46'1", and the 
sun's declination for the middle time was = 5 9' 48'1" S. ; required 
the correct mean time of the first observation, and the latitude, 
the equation of time to be subtracted from apparent time being 
= 11 m. 42-1 s. 

The latitude=48 42' 42-9" N. ; the time = 7 h. 5 m. 39'8 s. 

In t,\te following example the first method must be employed, as 
the polar distances are different. Also, as Atair is to the east of 
Arcturus, and the altitude of the former was taken some minutes 
later than that of the latter, this elapsed time, converted to 
sidereal time, must be subtracted from the difference of their 
right ascensions in order to obtain the angle- H. 

On the 19th of September 1830 the zenith distance of Arcturus 
was found to be = 73 19' 26'5" at 8 h. 2 m. 47'8 s. mean time, and 
that of Atair was = 40 53' 56 "3" at 8 h. 22 m. 3 s. ; the polar 
distance of the former was = 69 55' 36 "4", and that of the latter 
= 81 34'; required the latitude, and the correct time of the first 
observation, the sun's mean right ascension at mean noon being 
= 11 h. 51 m. 29-76s. 

The latitude = 48 42' 12", and the time = 8 h. 4 m. 19 '6 s. 




ASTRONOMICAL PROBLEMS 529 

LUNAR DISTANCES 

886. Problem XXXV. To find the true angular distance 
between the moon and the sun or a star, having given 
their altitudes and apparent distances. 

Let M', S' be the apparent places of the centres of the moon and 
the sun, or a star, or a planet, and M, S their true 
places, and Z the zenith ; then M will be above M', 
because the moon's parallax exceeds the refraction 
clue to its height ; but S will be below S', in conse- 
quence of the refraction exceeding the parallax of the 
body. M'S' is the apparent distance, and MS the 
true distance. 

Let h = the apparent height of the moon's centre = the comple- 
ment of M'Z, 
h' = the apparent height of the centre of sun or star = the 

complement of S'Z, 
H = the true height of the moon's centre = the complement 

of MZ, 

H'= the true height of the centre of sun or star = the com- 
plement of SZ, 

d the apparent distance of the centres = S'M', 
D = the true distance of the centres SM, 
s = h + h', and S = H + H'. 

Then, by Spherical Trigonometry (Art. 748, d), we have 
cos d - sin h . sin h' cos D - sin H . sin H' 



cos TA-- 



cos h . cos h' cos H . cos H' 



hence 



cos d - sin h . sin h' _ cos D - sin H . sin H' 
cos h . cos A' cos H . cos H' ' 

cos d+cos s_cos D + cos S ^ 
cos h . cos h' cos H . cos H' ' 

2 cos %(s + d) cos %(s ~ d) _ cos D + cos S' 
- - - - 



- 5 - ^-, - - ^f. . 
.cos h . cos n cos H . cos J 

But l+cosS=2cos ! 4S, 

and 1 - cos D = 2 sin 2 D ; 

. . cos D + cos S =2 cos 2 JS - 2 sin 2 JD. 

Substituting in (a) the above value of cos D + cos S, and 'dividing 
both sides by 2, we have 

cos^(s+c?}cos \ (s ~ d) _ cos 2 S - sin 2 D _ 
cos h . cos h' cos H . cos H' 



530 

hence sin 2 iD = < 



ASTRONOMICAL PROBLEMS 
, cos H . cos H' . cos i(s + d} . cos 



COS 



cos h . cos h' 
cos H . cos H' cos Ms + d) . cos 



~" ~ i ~~ iT o 1 ci 

cos A . cos n! cos 2 b 

cos H . cos H' . cos Ms + d] . cos 

- 



, 

assume sin 2 = j - - ~^, 

cos h . cos n . cos-' $S 

then sin D = cos 6 . cos S. 

Or logarithmically 

T . , /L cos H + L cos H' + L cos 

Ll Sin t'= : Wl-r TT 7T 

\. +L sec /i + Lsec A +Lcos 



and then 



L sin JD=L cos ^ + L cos 4S-10 . 



[11 

. [2]. 



EXAMPLE. On the 14th of December 1818, at 12 h. 10 m. nearly, 
latitude = 36 7' N., longitude by account = 11 h. 52 m. W., the 
following observations were made, the height of the observer's eye 
being = 19'5 feet, in order to find the distance between the moon 
and Regulus. 

Ob. dis. of moon's nearest 1. and Regulus, d' = 33 15' 25". 

Ob. alt. moon's 1. 1. 

Depression 

Moon's semi-diameter 
Refraction . ,. . . f 

Hor. par. 53' 59", 
Par. in alt. 



61 26' 12" 
- 4 18 



61 




21 54 
14 56 



h = 



61 36 50 
00 3M 



61 36 18-9 



= 25 40-5 



L. sec. 

P.L. 

P.L. 



= -32274 
= -52301 

= -84575 



H = 62 2 nearly. 



Observed altitude of Regulus. 
Depression .... 

Refraction 



= 28 29' 17" 
= -0 4 18 
h' = 28 24 59 
= 1 45 



H'= 28 23 14 



d' = 

Moon's semi-diameter ..... = 
Hence ... . d - 33 30 21 



33 




15 25 
14 56 



ASTRONOMICAL PROBLEMS 531 

Then L, sec h .... 61 36' 50" = 10-3229308 
L, sec h' . . . . 28 24 59 = 10O557580 
L, cos%(s + d) . . .6146 5 = ' 9'6748997 
L, cos(s-rf) . . . 281544= 9-9448723 
L, cos H . . . . 62 2 = 9-6711338 
L, cos H' . . . . 28 23 14 = 9-9443616 

2)59-6139562 
29-8069781 

L, cos^S + 10 . . . . . = - 19-8478853 
L, sin 0=65 31' 13" . = 9-9590928 

L, cos0 = 9-6173895^ 

L, cos }S = 9-8478853) 

L, sin D = 16 58' 24 -2" . = 9-4652748 

Hence D = 33 56 48 '4. 

The calculation of the time of observation, and of the longitude 

of the place, is performed in the example to the next problem. 

EXERCISES 

1. Given the apparent altitudes of the centres of the sun and 
moon = 32 0' 1" and 24 0' 8", their true altitudes = 31 58' 38" and 
24 51' 48", and their apparent distance = 68 42' 15" ; required their 
true distance =68 19' 34". 

2. Suppose that on the 6th of April 1821, in latitude = 47 39' N., 
and longitude = 57 16' W., by account, at 3 h. 56 m. P.M. per watch, 
it was found that the apparent altitudes of the centres of the 
sun and moon were =26 9' 7" and 46 34' 44", their true altitudes 
=26 7' 19" and 47 14' 19", and the apparent distance of their centres 
= 76 0' 7"; required the true distance and the true apparent time of 
observation, the sun's declination at the time being = 6 32' 12" N. 

The distance = 75 45' 43", and time = 3 h. 51 m. 24 s. 

3. If in longitude = ll 15' W. by account, at 3 h. 45 in. A.M. per 
watch, the apparent altitude of the centre of the moon was = 24 29' 
33", and that of Regulus = 45 9' 12", and their true altitudes = 25 
16' 50" and 45 8' 15", and the apparent distance of Regulus from the 
moon's centre = 63 35' 4" ; required the true distance. =63 4' 54". 

4. On the 9th of April 1837, at 5 h. 29 m. 36-8 s. mean time, 
suppose that the nearest limbs of the sun and moon were = 54 30' 
12" distant ; that the apparent height of the sun's centre was =21 
50' 14", and that of the moon's = 61 10' 10"; the moon being east 
of the sun, and botli west of the meridian ; the latitude by account 
was = 41 47' N., and the longitude =2 h. 10 m. W. ; the horizontal 



532 ASTRONOMICAL PROBLEMS 

equatorial parallax was = 55' 31'1", and the semi-diameters of the 
sun and moon = 15' 59" and 15' 7 "8" ; required the true distance. 

=55 KK 43". 

5. On the 6th of May 1840, at a place in latitude = 36 40' N., 
and longitude by account = 39 W., at 7 h. 40 m. A.M., the apparent 
altitudes of the centres of the sun and moon were = 30 33' 0'2" and 
53 15' 40-9", their true altitudes = 30 31' 27 '5" and 53 50' 31 "3", 
and the apparent distance of their centres = 62 0' 9'1"; required 
their true distance. ........ . =61 52' 40'8". 

6. At a place in latitude = 10 1' 50" N., and longitude by account 
= 30 5' W. of Paris, on the 17th of December 1823 at 14 h. 59 in. 
48 '8 s. P.M., the apparent altitudes of the moon's centre and of 
Regulus were = 48 0' 49" and 70 34' 9", their true altitudes = 48 
40' 38" and 70 33' 49", and their apparent distance was = 58 25' 
36" ; what was their true distance ? . . . =57 47' 12 -6". 

7. Required the distance between the centres of the sun and 
moon from these data : 

Distance of nearest limbs of the two bodies . = 83 26' 46" 

Altitude of lower limb of sun . . . . =48 16 10 

ii ii upper n moon . . . . =27 53 30 

Semi-diameter of sun = 15 46 

n n moon, including augmen. . = 15 1 

Correction of sun's altitude, including clip . = 5 27 

n M moon's M M = 46 43 

The dip being = 4' 24", the latitude = 10 16' 40", and longitude 

by account = 149 E., and the observations taken on the 5th of 

June 1793, about 1 h. 30 m. P.M. . . . . =83 20' 55". 

THE LONGITUDE BY LUNAR DISTANCES 

887. Problem XXXVI. Given the true angular distance 
between the moon and the sun or a star, and the time of 
observation, to find the longitude. 

The time, when not previously known, can be calculated by 
means of the altitude of one of the bodies, as in Art. 874. 

The time at Greenwich can be found thus : Take from the 
Nautical Almanac the two distances to which the given distance is 
intermediate ; then the difference between the registered distances is 
to that between the first registered distance and the given distance 
as three hours is to a fourth term, which is to be added to the time 
of the first registered distance, to give the required time. 



ASTRONOMICAL PROBLEMS 533 

Then the difference between the time at the place and that 
found at Greenwich will be the longitude. 
Let f =time of the first registered distance, 

rf' = the difference between the two registered distances 

that is, for intervals of three hours, 
d = the difference between the first registered distance and 

the given distance, 

t = the interval of time corresponding to d, 
T = I. time required at Greenwich j 

then d' : d=3 h. :t; hence t = ~jf, and T = f + t. 

Or, L t=L 3 + L d-L d', or P.L t=P.L rf-P.L d'. 

Since the moon moves over 360 in about 30 days, therefore an 
error of 10" in measuring the lunar distance will cause an error of 
about 5' on the longitude. For 

360 : 10" = 30 d. : x, and x=^^=~=5'. 

o o 

When only the first differences of the lunar distances are taken, 
the result will be a few seconds of time wrong. Thus, in the 
fourth of the following exercises, the correction found, when the 
second differences are used, is 5 '8 s., or about 1'4'; the correct 
longitude being 2 h. m. 16*9 s. 

EXAMPLE. Find the time of observation and the longitude of 
the place of observation from the data of the example to the 
preceding problem. 

1. To find the time at Greenwich 

Distance at h. . . =33 58' 7" 33 58' 7" 

n 3h.. . =32 30 3 D = 33 56 48'4 

d' . . = 1 28 4 d= 1 18-6 

and d' : d=3 h. : t, or 1 28' 4" : 1' 18'6"=3 h. : 2 m. 40-6 s., 
and T=f + t=Q h. +2m. 40'6s.=0h. 2 m. 40 "6 s. 



2. To find the time at the place 

By Art. 875 (the declination of star being= 12 50' 53'4"), 

Z= 61 36' 46" L, sinS . . . =9 "9973493 

P=77 9 6-6 L, sin(S-Z) . . =97554481 

= 53 53 L, cosecP . . = -0110120 

2)192 38 52-6 L, cosec C . . = -0926862 

S=96 19 26-3 2)19-8564956 

S - Z = 34 42 40-3 L, cos H . . . =9-9282478 

Prc. 2 I 



534 ASTRONOMICAL PROBLEMS 



And H = 32 2' 11-1", and . H= 4 h. 16 m. 17'5 s. 
And star's right ascension . = 9 58 43 '6 

Right ascension of meridian . = 5 42 26 '1 

Sun's R.A. at noon at place . . =17 27 147 

12 15 11-4 
Acceleration . . . . =0 2 - 4 



Time at place on 14th . . . =12 13 11-0 
., Greenwich on 15th =0 2 40 '0 



Longitude of place . . . . =11 49 29'OW. 

The time at the place could also be found from the observed 
altitude of the moon. 

The principle on which the rule is founded is so simple as to 
require no explanation. It proceeds, however, on the hypothesis 
that the moon's motion is uniform, which it is so nearly for three 
hours that the error arising from this assumption amounts at 
most only to a few seconds. When extreme accuracy is required, 
what is called the equation of second differences, which depends on 
the differences of the first differences, is used as a correction.* 

EXERCISES 

1. Find the true longitude for the data in the third exercise of 
last problem, supposing the true time of observation to be the 24th 
of January 1813, at 3 h. 45 m. A.M., the true distance = 63 4' 54", 
and that the distance of the centre of the moon from Regulus on 
the 23rd at 15 h. = 62 20' 8", and at 18 h. = 63 48' 54". 

= 11 26' 45" W. 

2. Find the longitude from Paris for the data in the fourth 
exercise of the last problem, the exact mean time of observation 
nt the place being = 5 h. 29 m. 36'8 s., the true lunar distance 
= 55 17' 20", and the registered distance at 6 h. in the Connais- 
sance des Temps being = 54 11' 36", and the difference for 3 li. 
= 125'55" =2 h. 48 m. 6-2 s. W. 

3. Required the longitude west of Paris for the fifth exercise in 
last problem, the exact mean time of observation at the place being 
= 7 h. 40 m. A.M., the true lunar distance = 61 52' 35'4", and the 
distances registered in the Connaissance des Temps being on the 
5th at 21 h. = 61 6' 22", and on the 6th at h. = 62 45' 51". 

= 2 h. 43 m. 38 s. W. 

4. Find the longitude for the data in the sixth example of the 
preceding problem, the true distance of the moon's centre from 

* See Nautical Almanac. 



ASTRONOMICAL PROBLEMS 535 

Regulus being = 57 47' 12-6", and their registered distances in 
the Connaissance des Temps being on the 17th at 15 h. = 59 2' 7", 
and at 18 h.= 57 9' 45". .... =2 h. m. 11-2 s. 

5. Find the true apparent time at the place, and the longitude 
for the true lunar distance = 83 20' 55" in the seventh exercise 
of the preceding problem, the latitude being = 10 16' 40" S., and 
the sun's declination = 23 22' 48" N. ; also the next less and greater 
registered lunar distances being at 15 h. apparent time = 83 6' 1", 
and at 18 h. = 84 28' 26". 

Time = l h. 39 m. 38 '5 s. ; longitude = 10 h. 7 m. 6 s. 



NAVIGATION 

888. The department of navigation that belongs to Prac- 
tical Mathematics consists in the solution of the problems of 
determining the direction and distance of the intended port 
from the port left, or from the place of the ship at any time, 
and also the determining of the ship's place at any instant 
during the voyage. The principles of plane trigonometry, 
modified in their application according to circumstances, are 
sufficient for the solution of these problems. 

The ship is navigated, as nearly as possible, by the path 
which is the shortest distance between the two places, but, 
from contrary winds and intervening land, it is generally 
necessary to sail in a track of a zigzag form ; the distance 
sailed in each direction being known, as also the direction, 
the ship's place can always be found, as will be afterwards 
explained. 

DEFINITIONS OF TERMS 

889. When a vessel is obliged to sail to the right or left of the 
direction of the intended port, she is said to tack. When the ship 
is tacking towards the left, and the wind consequently on the 
right, she is said to be on the starboard tack ; and when she is 
tacking towards the right, she is said to be on the larboard tack. 

890. A ship does not sail exactly in the direction of her keel 
or longitudinal axis, but deviates towards the side that is opposite 
to the wind ; ami the angle contained between the apparent and 



536 NAVIGATION 

real direction is called leeway. The real direction is observable 
by the track of the vessel in the water, called the ship's wake, 
or by the direction of the log-line ; and the leeway can therefore 
be estimated. 

891. The angle formed by the meridian and the direction of the 
ship's track is called the course. 

892. A line cutting all the meridians at the same angle is called 
a rhumb-line, which when continued approaches nearer and 
nearer to the pole, in a spiral form, but without ever reaching it ; 
it is also called a loxodrome; whereas the arc of a great circle, 
which is the shortest distance between two places, is called the 
orthodrome. 

893. The portion of a rhumb-line intercepted between two 
places is called their nautical distance. 

894. The distance of a ship from the meridian left, reckoned on 
the parallel of latitude of the ship's place, is called her meridional 
distance. 

895. If the nautical distance is supposed to be divided into an in- 
definite number of minute equal parts, the sum of all the meridional 
distances belonging to these parts is called the departure. 

896. The difference of latitude of two places is an arc of a 
meridian, intercepted between the parallels of latitude passing 
through these places. 

897. The difference of longitude of two places is an arc of the 
equator intercepted between their meridians. 

INSTRUMENTS USED IN NAVIGATION 

898. The mariner's compass is the instrument by which the 
course is measured. This compass consists of a circular card 
suspended horizontally on a point, and having for one of its 
diameters a small magnetised bar of steel, called the needle. The 
circumference of the card is divided into 32 equal parts, called 
points of the compass ; and each point is divided into 4 equal 
parts, called quarter points. The point of the card which coin- 
cides with the north end of the needle is called the magnetic 
north ; the opposite point, the magnetic south ; and the middle 
points between these, on the extremities of the diameter perpen- 
dicular to the needle, are called the magnetic east and west. 
These are called the cardinal magnetic points, and the other 




NAVIGATION 537 

points are named from their situation in reference to these points. 
The true cardinal points are consequently the north, south, east, 
and west. Since there are 8 points in each quadrant, therefore 
a point is = an angle of 11 15'. 

At the same place the needle points nearly in the same direc- 
tion for many years, but in different places its direction is 
not towards the same part of the 
horizon. The angular difference 
between the magnetic and true 
north is called the variation of 
the compass, being west or east 
according as the magnetic north is 
towards the left or right of the 
true north. 

The compass needle may lie affected 
sensibly by the attraction of iron 
placed near it, and even by a great 
mass of iron at a considerable dis- 
tance, as in a ship-of-war by the 

guns. When the metal is symmetrically distributed in refer- 
ence to the longitudinal axis, the needle is not affected when the 
direction of this axis coincides with the magnetic meridian or 
vertical plane passing through the needle'; and its local attraction 
produces the greatest error in the true variation when the direction 
of the axis of the ship is perpendicular to the former direction. 
The variation of the needle at London is at present about 24- 

The points of the compass are seen in the foregoing figure. The 
middle point between N. and E. is called NE. ; that between N. 
and NE. is called NNE. ; and so on. 

899. The log is a piece of wood, of the form of a circular sector, 
which is nearly quadrantal ; and the arc of it is loaded with lead, 
so that it floats vertically with the central point uppermost. The 
line called the log-line is so attached to the log that when the 
line is drawn gently the log turns its flat side towards the ship, so 
that it remains nearly immovable while the line is unwound from 
the reel. 

The log-line is about 100 fathoms long, and is divided into equal 
parts called knots, each of which is generally subdivided into 
fathoms. A knot is the 120th part of a nautical mile, or of 6079 
feet, and ought therefore to be 50 feet 8 inches. In practice, how- 
ever, 50 feet is usually made the length of a knot, for the log being 
drawn a small way towards the vessel during the operation of 



538 



NAVIGATION 



estimating the ship's rate, or, as it is called, of heaving the log, the 
distance given by this line is nearer the truth ; and, besides, it 
is safer that the reckoning should be in advance of the ship, or 
ahead of it, as it is termed. 

The time, when observing the ship's rate by the log-line, is 
estimated by a sand-glass, which measures half-minutes that is, 
it runs out in 30 seconds. 

Since 30 seconds is the same part of an hour that a knot is of a 
mile, the number of knots run out in 30 seconds shows that the 
rate of the vessel is just the same number of miles per hour. 

Sometimes the sand-glass and log-line, from various causes, 
become incorrect, and therefore the rate measured by them, or 
the distance sailed, must be corrected. 

900. The angular instruments used in navigation are Hadley's 
quadrant and sextant. The principles on which these instru- 
ments are constructed will be under- 
stood from the adjoining figure. 

The graduated arc AB is the limb of 
the instrument, CM an index, movable 
about an axis at M, with a vernier at 
its extremity C. M is a small mirror 
attached to the index CM, and placed 
perpendicularly to the plane ABM of 
the instrument ; N is a similar small 
plate of glass, called the fore horizon 
glass, one half of which is a mirror ; 
and it is placed parallel to the mirror 
M when the index coincides with MB, 
or rather with the zero point at B, and 
is fixed in this position. When the angular distance between two 
objects, as two stars, at S and I is to be measured, the plane of 
the instrument is first placed in the same plane with the objects, 
and in such a position that one of them, I, is visible through the 
glass N to the eye situated at E, and then the index CD is moved 
till the image of S, after two reflections from M and N, appears 
to coincide with I, seen directly through the plate ; and the angle 
subtended by their distance namely, angle E is then measured 
by double the arc BC. 

The ray SM proceeding from S is reflected in the direction MN 
by the mirror M, and then at N by the mirror N, in the direction 
NE, so that, PM being perpendicular to MD, the angle of reflec- 
tion PMN is equal to that of incidence PMS, or the inclination 




NAVIGATION 539 

of the incident ray SMD is equal to that of the reflected ray 
NMC, and also angle GNM is equal to FNE. From these 
relations of the angles, it is easily proved that the angle E of 
the triangle MNE is equal to twice the angle F of the triangle 
NMF. But angle F is equal to FMB, as GN is parallel to MB ; 
hence the double of angle CMB, which is measured by twice the 
arc BC, is the measure of angle E. 

In Hadley's quadrant the arc AB is an octant that is, the 
eighth part of a circle and therefore it contains only 45 ; the 
sextant differs from the quadrant merely in having its limb AB 
a sextant, or the sixth part of a circle. The sextant is furnished 
with a small telescope, to show with more precision when the 
image of one of the objects coincides with the other. The arcs of 
the sextant and quadrant are both graduated, so as to give the 
reading of the true angle, though they are only the measure of 
half that angle. 

. PRELIMINARY PROBLEMS 

901. Problem I. Given the distance sailed as determined 
by the reckoning, and the error of the log-line and sand- 
glass, to find the true distance. 

I. When only the log-line is incorrect as the correct length of 
the knot or 50 feet is to the incorrect length, so is the incorrect 
distance to the true distance. 

II. When only the sand-glass is incorrect the number of 
seconds run by the glass is to 30 seconds as the incorrect distance 
to the true distance. 

III. When both the log-line and sand-glass are incorrect, mul- 
tiply six times the measured length of the log-line by the observed 
distance, and divide the product by ten times the seconds the 
glass takes to run out. 

Let fc, &' = the true and incorrect lengths of a knot, 
s, s' = ii M number of seconds, 

d t d'= it n distances; 

k' 1 
then, for I. k:K = d' : d, and d-j-d' = ^lc'd' J 

jt jt 

II. s'-.s=d':d, and c?=s-r = 30 , 
s s 

, K s , 3 V , , 6 k' , 

III. d= T . -,.d' = -=--' d= Tn'~' c ? 
K s 5s 10 s 



540 NAVIGATION 

k' 
For if k:k' = d' : d", then d" = -r-d', and s' : s=d" : d ; 

hence d= -,d" = -j- -d' = T^ d', 

s k s 10 * 

EXAMPLE. The distance by reckoning is = 92 miles, the length 
of the knot=51 feet, the seconds by the sand-glass = 28 ; what is 
the true distance ? 



EXERCISES 

The true distance is required from the data in the first three 
columns : 

Distance Length of Seconds Answer : 

by Log a Knot by Glass True Distances 

1. 245 miles 48 feet 30 235-2 

2. 156 it 50 ii 32 146-2 

3. 126 M 46 ii 27 128 '8 

4. 164 .. 49 33 146'1 



902. Problem II. Given the magnetic course that is, 
the course per compass and the variation, to find the 
true course. 

RULE. Apply the variation to the magnetic course towards the 
left when the variation is W., and towards the right when E. 

EXAMPLE. What is the true course when the compass course 
is NW., and the variation 2 points W. ? 

The variation, being W., must be applied to the left of the course, 
which will therefore increase it by 2 points, and the true course is 
therefore WNW. 

EXERCISES 

Find the true courses from the magnetic courses and variations 
given in these exercises : 

Magnetic Course Variation T nm Course 

1. NE. 1 W. NE. b N. 

2. SW. b W. 2 W. SW. b S. 

3. N. 6 E. 3 E. NE. 

4. SSW. \ W. 2 i E. SW. | W. 

5. WNW. \ W. 1 \ W. W. 

6. SE. S. 1 i E. SSE. E. 



NAVIGATION 541 

903. Problem III. Given the true course and the varia- 
tion, to find the magnetic course. 

Tliis problem is solved exactly as the last, only the variation is 
applied in the opposite direction to the true course. The true 
courses and variations in the exercises to the preceding problem 
may be taken as data for exercises to this problem, and the cor- 
responding magnetic courses will be the answers. 

904. Problem IV. Given the compass course, the varia- 
tion, and leeway, to find the true course. 

Apply the variation, then apply the leeway in a direction from 
the wind that is, to the left when the vessel is on the starboard 
tack, and to the right when on the larboard tack. 

EXAMPLE. The magnetic course is NE. b E. on the larboard 
tack ; required the true course, the variation being 2 points W. , 
and the leeway 5 points. 

EXERCISES 

Find the true course in the following exercises, the compass 
course, leeway, and variation being given : 

Compass Tar-k Variation Leeway Answer : 

Course Points Points True Course 

1. NE. b N. Larboard 2 W. 2 NE. b N. 

2. SE. b E. 2 W. 1J ESE. } S. 

3. WNW. Starboard 3 W. 2 SW. } W. 

4. N. | E. 5 E. 3J NNE" \ E. 

905. Problem V. Given the latitudes and longitudes of 
two places, to find their difference of latitude and 
longitude. 

RULE. When the latitudes are of the same denomination find 
their difference, but when they are of different names take their 
sum ; and the remainder in the former case, or the sum in the 
latter, will be the difference of latitude. 

Find the difference" of longitude in the same manner as that 
of latitude, observing that when the longitudes are of different 
names, and their sum exceeds 180, it must be subtracted from 
360, and the remainder will be the difference of longitude. 

EXAMPLE. What is the difference of latitude and longitude of 
Quito and Canton ? 



542 NAVIGATION 

Canton,. . Lat. = 23 8' 9" N. Long. = 113 16' 54" E. 

Quito, . . = 14 S. i, = 78 45 6 W. 

Difference of lat. . = 23 22 9 192 2 

60 360 



Dif. of long. . . = 1402-15 miles 167 58 

60 

= 10078 miles. 
The difference of longitude in miles is estimated on the equator. 

EXERCISES 

Find the difference of latitude and longitude of the places stated 
in each of the following exercises : 

1. Liverpool, lat. =53 24' 40" N., long. =2 58' 55" W. ; and 
New York, lat. =40 42' 6" N., long. =73 59' W. 

Dif. of lat. =762-57 miles ; dif. of long. =4260*08 miles. 

2. Valparaiso, lat. =33 1' 55" S., long. =71 41' 15" W. ; and 
Manila Cathedral, lat. = 14 35' 26" N., long. = 120 59' 3" E. 

Dif. of lat. = 2857 '35 miles ; dif. of long. = 100397 miles. 

906. Problem VI. Given the latitude and longitude of 
the place left, and the difference of latitude and longitude 
made by the ship, to find the latitude and longitude of 
the place reached. 

RULE. Apply the difference of latitude and longitude respec- 
tively to the latitude and longitude left by addition or subtraction, 
according as they are of the same or different denominations. 

When the longitude and difference of longitude are of the same 
name, and their sum exceeds 180, subtract it from 360, and the 
remainder is the longitude of a contrary denomination from that 
left. 

EXAMPLE. The latitude and longitude of the place left are 
= 24 36' N. and 174 40' W. respectively; and after sailing SW. 
for some time, the differences of latitude and longitude made 
were found to be =245 miles and 384 miles ; what are the latitude 
and longitude in ? 
Lat. left . = 24 36' N. Long, left . =174 40' W. 

Dif. lat. 245 . = jt 5_ S. Dif. long. 384 . = 6 24 W. 

Lat. in . . = 20 31 N. ~181 T W. 

360 
Long, in . = 178 56 E. 



NAVIGATION 543 

EXERCISES 

Find the latitude and longitude arrived at in the following 
exercises : 

1. Lat. left = 34 4' S., long, left = 12 5' E. ; dif. of lat. = 145 
miles S., dif. of long. =365 miles W. 

Lat. in =36 29' S. ; long, in = 6 0' E. 

2. Lat. left=20 40' N., long left = 178 14' W. ; dif. of lat. 
=216 miles S., dif. of long. = 420 miles W. 

Lat. in = 17 4' N. ; long, in = 174 46' E. 

Navigation is divided into different branches, according to the 
methods of calculation employed. 

PLANE SAILING 

907. In plane sailing the surface of the earth is considered to be 
a plane, the meridians being equidistant lines, and the parallels of 
latitude also equidistant, cutting the meridians perpendicularly. 
This supposition, though incorrect, will lead to no error, so far 
as the nautical distance, difference of latitude, and departure are 
concerned ; for, as appears from the explanation following the 
example given below, these elements will be the same whether 
they are lines drawn on a plane or equal lines similarly related 
drawn on a sphere. As the north is on the upper side of the 
figure of the mariner's compass, and the upper side of maps, the 
top of a page is considered to be directed towards the north ; 
therefore the upper parts of diagrams in navigation are considered 
to be the northern parts of the figure. 

Hence a vertical line, BC, will denote the difference of latitude ; 
a horizontal line, AB, the departure ; the oblique line or hypo- 
tenuse, AC, the nautical distance ; angle C the c 
course, and A the complement of the course. 
Hence 

908. If any two of the four parts namely, the 
nautical distance, departure, difference of latitude, 

and course are given, the other two can be found ^/ 

by the rules of right-angled trigonometry. A '^ fur 

There will therefore be six cases, of which the 
first, however, is the most important. These cases may also be 
calculated very easily, and with sufficient accuracy, by means of 
the Table of the difference of latitude and departure, or, 
as it is sometimes called, a Traverse Table ; this method of 
solution is called inspection. They can also be solved by 




544 NAVIGATION 

construction, as in the problems from Art. 136 to 139, or by means 
of logarithmic lines, as explained from Art. 154 to 157. 

909. Problem VII. Of the course, distance, difference of 
latitude, and departure, any two being given, to find the 
other two. 

EXAMPLE. A ship from a place in latitude = 56 14' N. sails 
SW. ^ W. 425 miles ; required the latitude in and the departure. 

The proportions are the same as in the second case of right- 
angled trigonometry, only the nautical terms are used for the 
angles and sides of the triangle. 

Construction 

Let BC be the meridian, and make angle C = 4 points = 50 37', 
and CA = 425, and draAV AB perpendicular to BC. Then measure 
AB and BC. 

By Calculation 

1. To find the departure AB 

Rad. : sin C = AC : AB, or 

Radius . . . . . . = 10' 

Is to sin course 4| points . = 9-888185 

As distance 425 = 2*628389 

To departure 328 -53 . . . . = 2-516574 

2. To find the difference of latitude BC 
Rad. :cosC = AC:BC, or 

Radius = 10- 

Is to cos course = 9-802359 

As distance 425 = 2-628389 

To difference of latitude 269 -6 . . = 2-430748 

Latitude left = 56 14' N. 

Dif. of lat. 269-6 = 4 30 S. 

Latitude in = 51 44 N. 

By Gunter's Logarithmic Lines 

When the course is given in points, use sine rhumbs or tangent 
rhumbs instead of the lines of sines and tangents. 

1. To find the departure 

The distance from radius, or 90 on the line S. Rhumb, to 
4| points will extend on the line of numbers from 425 to 328, the 
departure. 



NAVIGATION 545 

2. To find the difference of latitude 

The distance from 90 to the complement of the course 3J points 
(as sine 3 points is = cosine 4 points) on the line S. Rhumb will 
extend on the line of numbers from 425 to 270, the difference of 
latitude. 

By Inspection 

In the Traverse Table in the page containing the course 4^ points, 
and opposite to the distance 425, is the departure 328 - 5 and the 
difference of latitude 269 -6. 

As the distance in the Table is not greater than 300, take out 
first the difference of latitude and departure for 300, and then for 
125, and their sum will give the above ; or take the difference of 
latitude and departure corresponding to one-fifth of the distance, 
and multiply them by 5. 

When the course is not given, the problem cannot be con- 
veniently solved by inspection. 

Let AC, BD be the parallels of the latitude left and reached, 
BC, DA their meridians, and AGB their nautical distance, 
which therefore is at every point equally 
inclined to the meridian. Let the dis- 
tance AB be divided into a great number 
of minute equal parts AG, GH,...and let 
Gg, HA, ...be portions of parallels of lati- 
tude, and Ag, Gh,... portions of meridians 
passing through the points A, G, H. 
Then, since these parts differ insensibly 
from straight lines, and the angles GAg, 
HGA,...are equal, therefore the parts AG, 
GH,...are proportional to Ag, GA;...and 
hence AG: Agr^AG + GH-f ... :Ag + Gh+... 
or as AB : AD. But AG : Ag... =rad. : cos course ; hence 
AB : AD = rad. : cosine course. 

It is similarly shown that AG :G#=AG + GH + ... : Gg + Hh+... 
= distance : departure ; and hence 

rad. : sin course = distance : departure. 

910. The distance, difference of latitude, departure, and course 
are therefore related as the sides and angles of a plane right- 
angled triangle, and their various relations are therefore determin- 
able in the same manner as those of the sides and angles of the 
triangle. 

The following exercises, which illustrate the six cases, are to be 




546 NAVIGATION 

performed by construction, calculation, and logarithmic lines, and 
by inspection : 

EXERCISES 

1. A ship from a place in latitude = 49 57' N. sails SW. b W. 
244 miles ; required the departure and latitude. 

Departure = 203; latitude =47 41'4' N. 

2. A ship sails SE. b E. from a place in 1 45' north latitude, and 
is then found by observation to be in 2 46' south latitude ; required 
the departure and distance. Departure = 405 '6 ; distance =487 '8. 

3. A ship sails NE. b E. f E. from a port in latitude = 3 15' S., 
till her departure is 406 miles ; what is the distance sailed and the 
latitude in? .... Distance = 449 ; latitude = 3' S. 

4. A ship sails between the south and east 488 miles from a port 
in latitude = 2 52' S., and then by observation she is found to be in 
latitude T 23' S. ; what course has she steered, and what departure 
has she made? The course = 56 16' or SE. b E. ; departure =405 '8. 

5. A ship has sailed between the north and west from the island 
of Bermuda, in latitude = 32 25' N., till her distance is 488 miles 
and departure 405 miles ; what has been her course, and what is 
the latitude ? . The course N. =56 6' W. ; latitude = 36 57' N. 

6. A ship sails between the north and west till her difference of 
latitude is 271 miles, and departure 406 miles ; what is the course 
and distance sailed ? 

Course N.=56 17' W. or NW. b W., and distance = 488 '2. 

TRAVERSE SAILING 

911. Problem VIII. Given several successive courses and 
distances sailed by a ship between two places, to find 
the single course and distance by which she would have 
arrived at the same place. 

Find the difference of latitude and departure for each course and 
distance, and then the whole difference of latitude and departure, 
and the course and distance corresponding to these two elements. 

The difference of latitude and departure for each course and 
distance are to be found by the last problem, the method by the 
Traverse Tables being the most expeditious ; then these are ar- 
ranged in a table called a Traverse Table, the courses being in 
the first column, the distances in the second, the north and south 
differences of latitude, marked N. and S., in the third and fourth, 
and the east and west departure, marked E. and W., in the iiftli 
and sixth columns. 



NAVIGATION 



54? 



The difference between the sums of the columns N. and S., or of 
the northings and southings, will be the whole difference of lati- 
tude of the same name as the greater ; and the difference between 
the sums of the columns E. and W., or of the eastings and 
westings, will be the whole departure of the same name as the 
greater. 

EXAMPLE. A ship from Cape Clear, latitude = 51 25' N., sails 
S. b W. 20 miles, SE. 12 miles, SW. b S. 18 miles, WNW. \ N. 14 
miles, and SSW. 24 miles ; required the equivalent course and 
distance and the latitude in. 

TRAVERSE TABLE 







Difference of Lat. 


Departure 1 








i 






N. 


S. 


E. 


W. 


S. b W 


20 




19-6 




3-9 


SE. 


12 




8-5 


8-5 




SW. b S. 


18 


... 


15- 




10- 


WNW. \ N. 


14 


6-6 






12-3 


SSW. 


24 




22-2 




9-2 






6-6 


65-3 


8-5 


35-4 








6-6 




8-5 








58-7 




26-9 



Latitude left 
Difference of latitude . 
Latitude in . . 



= 51 25' N. 
= 58-7 S. 
= 50 26 



The whole difference of latitude and departure being now known 
namely, 587 S. and 26'9 W. the corresponding course and dis- 
tance can be found, as in the sixth example of the last problem. 



1. To find the course 



Dif. of lat. 58-7 
Is to dep. 26-9 

As radius 



= 1-768638 
= 1-429752 
= 10- 



2. To find the distance 



To tan. course 24 37'= 9-661114 

The course is therefore S. =24 
= 64 6 miles. 



Sin course 
Is to radius 
As dep. 26-9 
To dist. 64-6 
37' W., 



and 



= 9-619662 
= 10- 

= T429752 

= 1-810090 

the distance 



548 



NAVIGATION 




These two proportions can also be performed by Gunter's scale 
as formerly. 

Construction 

The different courses and distances may also be drawn as in the 
annexed diagram, and the equivalent course and distance measured. 

Describe a circle bed, and let OL represent the meridian ; draw 
the radii Oa, Ob, Oc, Od, Oe, making angles 
with OL equal to the given courses ; then on 
Oa lay off the corresponding distance OA=20 ; 
draw AB parallel to Ob, and = 12 ; BC parallel 
to Oc, and = 18 ; CD parallel to Od, and = 14 ; 
and DE parallel to Oe, and =24. Draw EL 
perpendicular to OL, then OL is the whole 
difference of latitude, EL the whole depar- 
ture, and (supposing O and E joined) angle 
EOL is the equivalent course, and OE the 
equivalent distance. 

EXERCISES 

1. A ship takes her departure from the Lizard W. light in 
latitude = 49 58' N., which then bears NNW., its distance being 
= 15 miles, and sails SE. 34 miles, W. b S. 16, WNW. 39, and 
S. b E. 40 ; what is the latitude in, and the bearing and distance 
of the Lizard ? 

Latitude in = 48 53' N. ; bearing of Lizard N. = 12 16' E. ; and 
its distance = 66 '8. 

2. A ship's place is in north latitude =50 36', and she sails during 
24 hours in the following manner : SSW. 54 miles, W. b S. 39, 
NW. b N. 40, NE. b E. 69, and NNW. 60 ; what is the latitude in, 
and the equivalent course and distance from the former place ? 

Lat. in = 51 45'; course N. = 33 57' W. or NW. b W. ; and 
distance = 83 '8. 

912. If the ship has sailed in a current during any time, its 
effect for that time is allowed for as a separate course and distance. 
For instance, if the ship has been sailing for 10 hours under the 
influence of a current setting NE. at the rate of 2 miles per hour, 
the effect is the same as if the ship had sailed NE. 25 miles, and 
should be entered as an additional course. 

GLOBULAR SAILING 

913. In globular sailing the methods of calculation are derived 
on the supposition that the earth is of a spherical form, and they 



NAVIGATION 549 

apply with sufficient accuracy for the determination of the ship's 
place at any time, and the bearing and distance of the port bound 
for or of that left. 

CASE 1. When the ship sails between two places on the same 
meridian. 

The difference of latitude is just the distance sailed, and the 
course is due north or south, and there is no difference of longitude. 

CASE 2. When the ship sails on the equator. 

The distance sailed is the difference of longitude, the course is 
due east or west, and there is no difference of latitude. 

CASE 3. When the ship sails on the same parallel of latitude. 

914. To find the distance when the latitude is given, and the 
longitudes of the two places. 

Radius is to the cosine of the latitude as the difference of longi- 
tude to the distance. 

Rad. : cos lat. = dif. long. : distance. 

915. To find the difference of longitude when the latitude and 
distance on the same parallel are given. 

Radius is to the secant of the latitude as the distance to the 
difference of longitude. 

Rad. : sec lat. = distance : dif. long. 

916. To find the latitude when the distance and difference of 
longitude are given. 

The difference of longitude is to the distance as radius to the 
cosine of the latitude. 

Dif. long. : distance = radius : cosine lat. 

This case is sometimes called parallel sailing. 

The proportions in this case can be represented by this con- 
struction : 

ABC is a right-angled triangle, of which B is the 
right angle, AB the distance, AC the difference of 
longitude, and angle A the latitude. Then, when 
AC is radius,... 

Dif. long. : distance = radius : cos lat. 
And when AB is radius, 

Rad. : sec lat. = distance : dif. long. 

EXAMPLE. The longitudes of two places in the latitude of 56 S. 
are=140 20' and 148 45' ; find the distance. 

Dif. of long. = 8 25' = 505 miles. 

Prac. 2 J 




550 NAVIGATION 

To find the distance 

L, radius =10' 

L, cos lat. 56 = 9747562 

L, dif. long. 505 . . . . . = 2 "703291 

L, distance 282 -4 = 2-450853 

The proportion can be derived from the figure in Art. 909. Let 
O be the centre of the earth's equator, and OQ its radius ; P the 
centre of the parallel of latitude at B, and PB its radius ; then 
the distance between two meridians, measured on the equator, is 
to their distance on the parallel at B as OQ : PB that is, dif. 
long. : dist. = radius : cos lat. 

EXERCISES 

1. A ship in latitude=49 30' sails due E. till her difference of 
longitude is = 3 30'; what is the distance sailed? . =136 '4 miles. 

2. A ship sails 136'4 miles due W. on the parallel of latitude 
= 49 30'; required the difference of longitude made. =210 miles. 

3. A ship sails 136'4 miles due E., and her difference of longitude 
is then = 3 30'; on what parallel of latitude did she sail ? 

Latitude = 49 30'. 

CASE 4. When the course is compound, to find the difference of 
latitude and longitude. 

917. METHOD I. The first method of solution is by middle 
latitude sailing. 

This method combines plane and parallel sailing ; and in it, it 
is supposed that the departure made by a ship is equal to the 
meridional distance on the middle parallel that is, the meridional 
distance EF (fig. to Art. 909) on the parallel of latitude in the 
middle between the latitudes of A and B, the latitude left and 
that arrived at, is equal to the sum of the elementary meridian 
distances Gg, HA,... which it nearly is. There are two proportions 
used namely, 

918. The difference of latitude is to the difference of longitude 
as the cosine of the middle latitude to the tangent of the course. 

Diff. lat. : dif. long. =cos mid. lat. : tan course. 

919. The cosine of the middle latitude is to the sine of the 
course as the distance to the difference of longitude. 

Cos mid. lat. : sin course = distance : dif. long. 

920. These two proportions can be obtained by means of two 
right-angled triangles, ABC, DBC, having their right angles at C. 




NAVIGATION 551 

In the triangle ABC angle A is the course, AB the distance, 
BC the departure, and AC the difference of latitude ; and in the 
triangle BCD, BC is the departure, angle B the 
middle latitude, and BD the difference of longitude. 
ABC is a triangle in plane sailing, and BCD in 
parallel sailing; and from the triangle ABD the 
proportion in Art. 919 is easily derived, while the 
two proportions 

Dif. lat. : dep. =radius : tan course, 
Dep. : dif. long. =cos mid. lat. : rad., 
being compounded, give Art. 918, 

Dif. lat. : dif. long. = cos mid. lat. : tan course. 

921. METHOD II. The second method of solution is by Mer- 
cator's sailing. 

In this method the surface of the earth is considered to be 
plane, the meridians being parallel lines, and also the parallels 
of latitude, as in plane sailing ; and since by this hypothesis the 
distance between the meridians is increased, except at the equator, 
the lengths of the arcs of the meridians are increased in the same 
proportion, so that the distances between the parallels of latitude 
for every successive minute are continually increasing with the 
latitude ; and the relative bearings of places are thus preserved. 
This method is so accurate that it may be used without sensible 
error for any distance on the earth's surface. 

The lengths of the meridians from the equator to any latitude 
are thus increased, and the increase is greater the higher the 
latitude. For instance, the increased distance of the parallel of 
10, instead of being 600 miles, is found to be 603 miles ; and that 
of the latitude of 50, instead of 50x60=3000 miles, is 4:527 miles. 
The increased lengths of the meridians, from the equator to any 
latitude, are called the meridional parts, from the manner in 
which they are computed; and their numerical values are con- 
tained in tables. 

Q22. The difference of the meridional parts for any two lati- 
tudes is called the meridional difference of latitude ; and the 
true difference of latitude is sometimes, for distinction, called 
the proper difference of latitude. 

The analogies peculiar to this method are : 

923. The difference of latitude is to the departure as the 
meridional difference of latitude to the difference of longitude. 
Dif. lat. : dep. =mer. dif. lat. : dif. long. 



552 



NAVIGATION 



Dif.Lony. 




924. The meridional difference of latitude is to the difference of 

longitude as radius to the tangent of the course. 

Mer. dif. lat. : dif. long. = radius : tan course. 

These proportions can be obtained from two 
right-angled triangles, * ABC, ADE, in which 
AB and AD are the proper and the meridional 
differences of latitude, BC the departure, DE 
the difference of longitude, AC the distance, and 
A the course. 

925. Problem IX. Given the place left and that bound 
for, to find the course and distance. 

1. By Middle Latitude Sailing 

To find the course 
Dif. lat. : dif. long. =cos mid. lat. : tan course. 

To find the distance 
Radius : sec course = dif. lat. : distance. 

2. By Mercator's Sailing 

To find the course 
Mer. dif. lat. : dif. long. = radius : tan course. 

To find the distance 
Radius: sec course = dif. lat. : distance. 
The second proportion in the two methods is the same. 

EXAMPLE. Required the bearing and distance of New York 
from Liverpool. 

By Middle Latitude Sailing By Mercator's Sailing 



Liverpool, lat. 
New York, lat. . 
Dif. lat. 
Mid. lat. . 
Liverpool, long. . 
New York, long. 


= 53 25' N. Mer. parts . 
= 40 42 N. Mer. parts . 


. = 3806 
. = 2678 


= 12 43 Mer. dif. lat. 
= 47 4 P. dif. lat. = 12 43' = 

= 259'W.\ TV , 
= 73 59 W .j Dif. long. =71 = 


. = 1128 
763 miles. 

tU)U it 



1. To find the course 



Dif. lat. 763 (a . c) 
Dif. long. 4260 
Cos mid. lat. 47 4' 
Tan course 75 16' 


= 7-117475 
= 3-629410 
= 9-833241 


Mer. dif. lat. 1128 
Dif. long. 4260 
Radius . 
Tan course 75 10' . 


= 3-052309 
= 3-629410 

= 10- 


= 10-580126 


= 10-577101 



NAVIGATION 553 



2. To find the distance 



Radius . . . =10- 

Sec course 75 16' . = 10'594618 

Dif. lat. 763 . . = 2-882525 

Distance 3000 . . = 3-477143 



Radius . . . =10' 

Sec course 75 10' . =10'591746 

Dif. lat. 763 . . = 2 -882525 

Distance 2980 . = 3-474271 



EXERCISES 

1. What is the bearing and distance of a place in latitude 
= 71 10' N., longitude =26 3' E., from another place in latitude 
=60 9' N., and longitude =0 58' W. ? 

Course N. =45 18' E., distance = 940 miles, by mid. lat. 

sailing. 
Course N=44 49' E., distance = 931 '8 miles, by Mercator's 

sailing. 

2. A ship having arrived at a place in latitude = 49 57' N., 
longitude = 5 14' W., is bound for another place in latitude 
= 37 N., longitude =25 6' W. ; required the bearing and distance 
of the latter place from the former. 

Course S. =48 4' W., distance =1162 '7 miles, by mid. lat. 

sailing. 
Course S. =47 54' W., distance = 1159 miles, by Mercator's 

sailing. 

926. Problem X. Given the place sailed from, the course 
and distance, to find the place arrived at. 

1. By Middle Latitude Sailing 

To find the difference of latitude 

Radius : cos course = distance : dif. lat. 

To find the difference of longitude 
Cos mid. lat.: sin course = distance :dif. long. 

2. By Mercator's Sailing 
To find the difference of latitude 
The analogy is the same as in the preceding method. 
Or, Radius : cos course = distance : dif. lat. 

To find the difference of longitude 
Radius : tan course = mer. dif. lat. : dif. long. 
By the first proportion in these two methods the difference of 



554 



NAVIGATION 



latitude is found, and by the second the difference of longitude ; 
and hence the latitude and longitude of the place in are known. 

EXAMPLE. A ship from a place in latitude = 25 40' S. and 
longitude =35 12' W. sails SW. 6 S. 246 miles; required the 
latitude and longitude in. 

To find the difference of latitude by both methods. 



Radius . 

Cos course 3 points 

Distance 246 ... 

Dif. lat. 204-6 

Lat. left . . = 25 40' S. 

Dif. lat. . . = 3 25 S. 

Lat. in . . = ~29 5~ S. 

Mid. lat. . . = 27 22 

By Middle Latitude Sailing 
To find the dif. long. 



Mer. parts . 
Mer. parts . 
Mer. dif. lat. 



10- 

9-919846 

2-390935 

2-310781 

. = 1594-3 

. = 1825-2 

= 230-9 



Cos mid. lat. (a . c) . 
Sin course 3 points . 
Distance 246 , 
Dif. long. 153-9 . 

Longitude left 



= 0-051547 
=9-744739 
= 2-390935 
= 2-187221 



Different longitude 
Longitude in 



By Mercator's Sailing 
To find the dif. long. 
Radius . . . =10' 
Tan course 3 points = 9-824893 
Mer. dif. lat. 231 . = 2-363612 
Dif. long. 154-4 . = 2-188505 

. = 35 12' W. 

. = 2 34 W. 

= 37 46 W. 



The two methods give the differences of longitude to within less 
than a mile of each other. The place arrived at is in latitude 
=29 5' S., and longitude = 37 46' W. 

EXERCISES 

1. A ship from a place in latitude =50 30' N. and longitude 
= 145 20' W. sails 450 miles SSW. ; find the latitude and 
longitude in. 

Latitude = 43 34' N., longitude = 149 33' W., both by mid. lat. 
and Mercator's sailing. 

2. A ship from latitude = 51 15' N. and longitude = 9 50' W. 
sails SW. b S. till the distance run is 1022 miles; what are the 
latitude and longitude in ? 

Latitude = 37 5' N., longitude =23 2' W., by mid. lat. sailing, 
and =23 8' W. by Mercator's sailing. 



NAVIGATION 555 

927. Problem XI. Given the latitude left, the differ- 
ence of latitude and departure, to find the difference of 

longitude. 

By Middle Latitude Sailing 

Cos mid. lat. : radius = dep. : dif. long. 

By Mercator's Sailing 
Dif. lat : dep. = mer. dif. lat. : dif. long. 

EXAMPLE. A ship on a course between the south and west from 
latitude = 54 24' N. and longitude =36 45' "W. has made 346 miles 
of difference of latitude and 243 miles of departure ; what is the 
latitude and longitude in ? 

Lat. left . . = 54 24' N. Mer. parts . . = 3905 '7 
Dif. lat. 346 . = 5 46 S. Mer. parts . . = 3348*7 
Lat. in . . = 48 38 N. Mer. dif. lat. . = 557 
Mid. lat. . . = 51 31 

To find the difference of longitude 



By Middle Latitude Sailing 
Cos mid. lat. 51 31' = 9-793991 
Radius . . . =10' 
Dep. 243 . = 2-385606 

Dif. long. 390-5 . = 2-591615 



By Mercator's Sailing 
Dif. lat. 346 (a . c) . = 7 '460924 
Dep. 243 . . = 2-385606 
Mer. dif. lat. 557 . = 2-745855 
Dif. long. 391-2 . = 2 "592385 



Long, left . . = 3645'W. Long, left . . =3645'W. 
Dif. long. 390-5 . = 6 31 W. Dif. long. 391'2 . = 6 31 W. 
Long, in . . =43 16 W. Long, in . . =43 16 W. 

The place arrived at is therefore in latitude = 48 38' N. and 
longitude = 43 16' W. 

EXERCISE 

A ship from latitude = 37 N., longitude = 48 20' W., sails 
between the north and east till her difference of latitude and 
departure are 855 and 564 miles ; required the latitude and 
longitude in. 

Latitude=51 15' N., longitude in 35 14' W., by mid. lat. 
sailing, and 35 8' W. by Mercator's. 

928. Problem XII. To perform a traverse or compound 
course by middle latitude and Mercator's sailing. 

RULE. Form a Traverse Table, and find by it the whole differ- 
ence of latitude and the departure ; then find the latitude in and 



556 NAVIGATION 

the course made good, as in Art. 911. Find then the middle lati- 
tude between that left and that arrived at, or find the meridional 
difference of latitude for these two latitudes ; then, to find the 
difference of longitude, 

Cos mid. lat. : radius = dep. : dif. long, by mid. lat. sailing. 
Or, rad. : tan course = Mer. dif. lat. :dif. long, by Mer. sailing. 

EXAMPLE. Find the longitude and latitude of the place of the 
ship arrived at, after sailing the various courses and distances given 
in the example of a traverse in plane sailing in Art. 911, supposing 
the longitude left to be =23 40' W. 

Construct the traverse as in that example, and it will be found 
that the difference of latitude is = 58'7 S., and the departure 
= 26-9 W. Hence 

Lat. left . . = 51 25' N. Mer. parts . . - 3608-7 

Dif. lat. . . = 59 S. Mer. parts . . = 3515-1 

Lat. in . . = 50 26 Mer. dif. lat. . = 93 '6 

Mid. lat. . . = 50 55 

Then dif. lat. : dep. = radius : tan course, and, as found in that 
example, the course is S. 24 37' W. 

. To find the difference of longitude 



By Middle Latitude Sailing 

Cos mid. lat. 50 55' = 9799651 
Radius . . . =10* 
Dep. 26-9 . . = 1-429752 


By Mer color's Sailing 

Radius . . . =10- 
Tan course 24 37' . = 9-661043 
Mer. dif. lat. 93 "6 . = 1-971276 


Dif. long. 42 7 
Long, left 
Dif. long. 
Long, in 


. = 1-630101 
. =23 40' W. 
. = 43 W. 


Dif. long. 42-9 
Long, left 
Dif. long. 


. = 1-632319 
. =2340'W. 
. = 43 W. 


. =24 23 W. Long, in 


. =24 23 W. 



929. Since the difference of latitude, distance, and course are 
the same parts of a right-angled triangle in plane sailing that 
the departure, difference of longitude, and middle latitude are in 
middle latitude sailing ; and the difference of latitude, departure, 
and course are the same parts of the triangle in plane sailing 
that meridional difference of latitude, difference of longitude, and 
course are in Mel-eater's sailing ; therefore the difference of longi- 
tude for the last two proportions can be found, by inspection, in 
the Table of the difference of latitude and departure. 



NAVIGATION 557 

In Plane Sailing In Mid. Lat, Sailing 

Course corresponds to . . . . Mid. lat. 

Dif. lat. ii it .... Departure. 

Distance n u Dif. long. 

In Plane Sailing In Mercator's Sailing 

Course corresponds to . . . . Course. 
Dif. lat. . .... Mer. dif. lat. 

Departure H n . Dif. long. 

Thus, for the proportion above by middle latitude sailing, in the 
Table of difference of latitude and departure in the page for course 
=51, and departure 26'9 in the difference of latitude column, 
there is 42 in the distance column for the difference of longitude, 
as above ; and for the proportion by Mercator's sailing, in the same 
Table for course = 25 (for 24 37'), and meridional difference of 
latitude 93 '6 in the difference of latitude column, there is 44 for 
the difference of longitude in the departure column. 

930. When great accuracy is required, or when sailing in high 
latitudes, it is necessary to calculate the difference of longitude for 
each course and distance, supposing the distances not to exceed a 
few miles, instead of merely finding the difference of longitude on 
a whole day's sailing. This method is called a globular traverse. 

EXERCISES 

1. A ship from a place in latitude = 50 6' N. and longitude 
=5 55' W. is bound to a port in the island of St Mary's in 
latitude = 36 58' N., and longitude = 25 12' W., and steers the 
following courses : S. b W. 24 miles, WSW. 32, NW. \ W. 41, 
SSE. i E. 49, ENE. f E. 19, W. 21, NE. \ E. 36, S. 41, SSW. 92, 
and N. 36 ; what is the latitude and longitude in, and also the 
direct course and distance to the intended port ? 

Latitude in = 48 9', and longitude in = 7 19', by mid. lat. and 

by Mercator's sailing. 

Course =42 26', and dist. = 990'4, by mid. lat. sailing. 
Or, =42 19', =988-6, .1 Mer. sailing. 

In the following exercise the difference of longitude is found on 
each course, as explained in Art. 930 that is, by the globular 
traverse : 

2. A ship from latitude = 68 38' N. and longitude = 8 40' E. is 
bound for the North Cape in latitude = 71 10' N., and longitude 
=26 3' E., and sails the courses and distances in the subjoined 



NAVIGATION 



Table ; what is the latitude and longitude in, and the direct course 
and distance of the Cape ? 



Courses 


Dis- 
tance 


Latitude 


Departure 


Lat. in 


Dif. Long. 


N. 


S. 


E. 


W. 


E. 


W. 














68 38' 






NE. b N. 


63 


52-4 




35-0 




69 30 


97-2 




NE. 


38 


26-9 




26-9 




69 57 


77-6 




NNE. 


56 


51-7 




21-4 




70 49 


64-2 




N. 


30 


30-0 








71 19 






NW. b N. 


25 


20-8 






13-9 


71 40 




43-8 


NNW. J W. 


36 


31-8 






17-0 


72 12 




55-2 


N. &E. 


40 


39-2 




7-8 




72 51 


25-8 




NE. b E. } E. 


72 


33-9 




63-5 




73 25 


219-1 




SE. 


50 




35-4 


35-4 




72 50 


120-5 




ENE. 


65 


24-9 




60-1 




73 15 


206-9 








311-6 


35-4 


250-1 


30-9 




811-3 


99-0 






35-4 




30-9 






99-0 








276-2 




219-2 






712-2 





Latitude in = 73 14', longitude in = 20 32' E. 

The course required is S. =38 1' E., distance = 157 '4, by mid. 

lat. sailing. 
Or, The course is S. = 37 59' E., distance = 157 '3, by Mercator's 

sailing. 

3. A ship in latitude = 67 30' N., longitude = 8 46' W., sails NE. 

64 miles, NNE. 50, NW. b N. 58, WNW. 72, W. 48, SSW. 38, S. b 

E. 45, and ESE. 40 ; what is the latitude and longitude in ? 

By the plane traverse, the lat. in is = 68 43' N., and longitude 

= 11 3' W. ; and by the globular traverse, the long, in is 

= 11 37' W. by mid. lat., and = ll 42' by Mercator's sailing. 

Departure of a Ship 

The place of departure of a ship that is, the place from which 
the beginning of a voyage is reckoned is generally some promon- 
tory or other convenient object whose latitude and longitude are 
known ; and as the vessel is usually some miles distant from it, 
observations must be taken to determine this distance. 




NAVIGATION 559 

931. Problem XIII. Given the bearing of a headland 
from a ship at two places, and the distance and direction 
sailed between them, to find the distance of the promon- 
tory from the ship. 

Let P be the promontory, S and H the two places of the ship, 
mn and ab parts of the meridians through H 
and S ; then angle mSP, the bearing of P from 
S, is given ; and angle mSH or 6HS, the ship's , H _ 
course ; and also angle PH6, the bearing of P 
from H. 

Therefore, in the triangle PHS, the angle 
at S = iSP + iSH is known, and that at H 
= PH6-6HS is also known, and the side HS ; 
therefore the distances PH and PS can be 
found (Art. 186). 

EXERCISES 

1. A headland was observed from a ship to bear NE. b N., and 
after sailing 7 '5 miles on a NNW. course, the headland then bore 
ESE. ; required the distance of the headland from both places of 
the ship. =5'4 and 6 '36 miles. 

2. A lighthouse was observed to bear from a ship NNE., and 
after sailing 15 miles on a WNW. course its bearing was found to 
be NE. b E. ; required its distance from the last place of the ship. 

=27 miles. 

3. A cape was observed to bear E. b S. from a ship, and after 
sailing NE. 18 miles, its bearing was SE. b E. ; what was the dis- 
tance of the cape from the second place of the ship ? =391 miles. 

932. After taking the departure, the next important problem is 
to find the bearing and distance of the port bound for, which 
is solved by Art. 925. After performing a day's sailing, as nearly 
as possible in the proper direction, the place of the ship is then 
to be determined by Art. 928 ; and then again, if necessaiy, the 
bearing and distance of the intended port ; and these problems are 
to be successively repeated during the voyage. The latitude and 
longitude of the ship, determined in this manner, are said to be 
the latitude and longitude by account. As the place of a ship 
determined in this manner cannot be depended upon on a long 
voyage, on account of the errors occasioned by unknown currents, 
storms, and the unavoidably imperfect means of measuring the 
courses and distances, it becomes necessary to employ the prin- 



560 NAVIGATION 

ciples of practical astronomy to determine the latitude and lon- 
gitude with greater accuracy. This method of determining the 
various elements in navigation is called nautical astronomy. 

MISCELLANEOUS EXERCISES ON NAVIGATION 

1. A gunboat of a blockading squadron lies 4 miles to the south 
of a harbour, and observes that a ship leaves the harbour in a 
direction E. 30 S. If the blockading ship sails 12 miles an hour, 
find in what direction she must go so as to cross the course of the 
other ship in three-quarters of an hour. . . =E. 7 21' 45". 

2. From a given point the position of two gunboats is found to 
be 29 east and 56 west. Supposing them to occupy a position in 
line distant from the given point 80 yards, and that their line is 
at right angles to one produced from the given point, find the 
distance between them = 163 yards nearly. 

3. A yacht which is known to be sailing due east at the rate of 
12 miles an hour was observed at noon to be 150 to the east of 
south at 1 h. 30 m. after noon. She was seen in the south-east. 
Determine the distance of the ship at noon. . . =25 '45 miles. 

4. From a ship a rock and a headland are observed to bear 18 E. 
of N. The ship sails 6 miles NW., and then the rock is due east, 
and the headland NE. What is the distance between the rock and 
the headland? =8'755 miles. 

5. A ship sailing out of harbour is watched by an observer from 
the shore, and at the instant she disappears below the horizon he 
ascends to a height of 20 feet, and thus retains her in sight 40 
minutes longer. Find the rate at which the ship is sailing, 
assuming the earth to be a sphere of 4000 miles radius and 
neglecting the height of the observer. . =8 '257 miles per hour. 

6. An observer from the deck of a ship 20 feet above the level of 
the sea can just see the top of a distant lighthouse, and on ascend- 
ing to the masthead, which is 60 feet above the deck, he sees the 
door, which he knows to be one-fourth of the height of the light- 
house above the level of the sea. Find his distance from the light- 
house, and its height, assuming the earth to be a sphere of 4000 
miles radius. . . . = Height, 80 ft. ; distance, 87196 '32 ft. 

7. Two ships sail at the same time from the same port, and sail 
for 5 hours at the respective rates of 8 and 10 knots an hour in 
straight lines inclined to each other at an angle of 60. They then 
sail directly towards each other. Find the inclination of their new 
course to their original courses, . . . 70 55' 36"; 49 6' 24". 

8. A buoy is moored 9 miles north of a port from which a yacht 



NAVIGATION 561 

sails in a direction ENE. She tacks and sails towards the buoy 
until the port is SW. of her, when she tacks and sails into port. 
Prove that the length of the course is about 16 miles. 

NAUTICAL ASTRONOMY 

933. By the principles of nautical astronomy the time at the 
ship's place, the variation of the compass, the latitude and longi- 
tude, and various other elements used in navigation can be 
determined. As the complete solutions of these problems have 
already been given in the problems in practical astronomy, except- 
ing the circumstances peculiar to navigation, by which the solutions 
are in some cases modified, it will be necessary here merely to add 
the methods of calculating the effect of these circumstances. 

934. Problem XIV. To find the variation of the compass. 
RULE. Find the azimuth or amplitude of some celestial object 

by the methods formerly given in Art. 873 and 880 ; and find 
also its bearing per compass, and the difference between the 
azimuth and bearing will give the variation of the compass. 

EXERCISES 

1. The azimuth of the sun was found to be S. =48 54' E. when 
its true bearing was S. =77 1' E. ; what was the variation of the 
compass ? = 28 7', or 2^ points E. 

2. The amplitude of a star was found to be E. = 10 15' N. when 
its true bearing was S. =84 12' E. ; what was the declination of 
the needle ? = 16 3' W. 

3. The azimuth of a star was found to be N. =68 10' E. when 
its true bearing was NE. b E. ; what was the variation of the 
compass ? =11 55' E. 

935. Problem XV. Having given two altitudes of a 
celestial body, the ship having sailed for several hours 
during the interval, to reduce the first altitude to the 
place at which the second was taken. 

RULE. Find the angle of inclination between the ship's course 
and the bearing of the body at the first place of observation, or its 
supplement when greater than a right angle ; then 

The radius is to the cosine of this angle as the distance run to 
the correction in minutes ; which is to be applied by addition or 
subtraction to the first altitude, according as the inclination is less 
or greater than a right angle. 



562 NAVIGATION 

If d=the distance, =the inclination, c=the correction, a' = the 
first altitude, and a = the first altitude reduced to the second place, 
then Rad. : cos id :c, and a=a' + c. 

It is evident that c can be found by inspection of the Table of 
difference of latitude and departure, by considering i as the course, 
and d the distance ; then c will be found in the latitude column. 

EXERCISES 

1. The altitude of a star when east of the meridian was observed 
to be 20 40', and its bearing at the time was SE. b S. ; and after 
sailing 40 miles W. b S. its altitude was again observed when it 
was west of the meridian ; what would have been its altitude at 
the time of the first observation if it had been taken at the place 
of the second observation ? 

Here i = 3 + 7 = 10 pts. = 112 30', d = 40, c = 15'3', and = 20 
24-7'. 

2. The sun's altitude was observed to be = 30 41 '5', and its 
bearing was SE. b E. ; and after sailing 48 miles E. b S. its 
altitude was again taken ; required the sun's altitude at the latter 
place of the ship at the time of the first observation. 

Here i=2 pts. =22 30', d=48, c=44'3', and = 31 25 '8'. 
The principle of the rule may be proved thus : Let S be the 
zenith of the place of the first observation, and S' that of the 
second, B and B' the positions of the body at these two instants 
of time, SS' the intermediate dis- 
tance sailed by the ship in minutes 
of space, HRn' and H'R?i the hori- 
zons of S and S' ; then BH and B'H' 
are the two altitudes. Now, to find 
the altitude BA of the body when at 
B, supposing the altitude to be taken 
then at S', produce BS to m, and 
from S' draw the perpendicular mS' 
from S' on Sm ; then, since SS', and consequently mS, is a small 
distance, mn may be considered as differing insensibly from S'A, at 
least for ordinary nautical purposes ; but S'A is a quadrant, as also 
SH ; hence mn = SH nearly, therefore mS = nH nearly. Consequently 
BA, the altitude of B, when taken at S', which is nearly = Bn, is 
less than BH by mS. Now, angle BSS' is evidently = i, SS' = c?, 
mS = c, and radius : cos i = d : c, which is the rule. If the ship had 
sailed from S to S" instead of S', it could in the same manner be 




NAVIGATION 563 

shown, by drawing S"m' perpendicularly to SH, that 8m,' would 
require to be added to the altitude of B, taken at S, in order to 
obtain its altitude at the same time if it were taken at S". 



CONSTRUCTION OF MAPS AND CHARTS 

936. Maps and charts are representations of portions or of 
the whole of the surface of the earth, with meridians and 
parallels of latitude at some convenient distance from each 
other, as at 5 or 10 degrees. The principal kinds of con- 
struction are the plane construction, the method of conical 
projection, the stereographic projection, and Mercator's 
projection. 

PLANE CONSTRUCTION 

937. In the first method of plane construction the meridians 
are parallel straight lines, as are also the parallels of latitude. It 
is used for a very small portion of the earth's surface, extending 
only a few degrees in length and breadth, as for a portion of a 
kingdom. 

The breadth from north to south, AC, is the 
number of degrees of latitude, each of which 
is equal to 60 geographical or 69 '02 imperial 
miles ; and the length from east to west, AB, is 
just the length of the number of degrees of 
longitude contained in it, estimated on the 
parallel EF of middle latitude by the proportion 
in Art. 914. 

Let L', L = the lengths of a degree of longitude at the equator 

and at the middle latitude, 
=the middle latitude ; 

then rad. :cos l = L' : L, and L = L' cos I, if R = l ; 

and since L' = 60 geographic miles, L = 60 cos I. 

If / = 56, then L = 60x -5592 = 33'552 miles. 

938. In the second method of plane construction the parallels 
of latitude are parallel straight lines, and the meridians are con- 
verging straight lines. This method is used for projecting larger 
portions of the earth's surface, as for a kingdom. 




564 



CONSTRUCTION OF MAPS AND CHARTS 



rr 



The breadth of the map in this case is AB, which is equal to 
the length of the degrees in the latitude con- 
tained in the map, as in the preceding method. 
The parallels of latitude are perpendicular to AB, 
which is the middle meridian. The lengths of 
the degrees of longitude in the parallel CD are 
the lengths for the latitude of B found as those in 
EF in the preceding method ; and the lengths for 
F the parallel EF are also those corresponding to 
the latitude of A, and the corresponding divisions of EF and CD 
are joined by straight lines, which are the meridians. 

CONICAL, PROJECTION 

939. The method of conical projection is used for still larger 
portions of the earth's surface, as for a continent. This method is 
thus derived : A conic surface is supposed to touch the earth's 
surface along the parallel of the middle latitude, and the former 
surface is supposed to coincide with that of the earth 
for a few degrees of latitude on both sides of the 
middle parallel. 

Let APB be the earth, C the middle latitude, 
and CE a tangent, meeting the axis DP produced ; 
then E is the vertex of the cone, CE its slant side, 
and ED its axis. When this conic surface is de- 
veloped on a plane it will be represented by E'MN, 
in which the points E', C' correspond to E and C in 
the above figure. The breadth of the map MQ, therefore, is just 
the length of the number of degrees of latitude contained in the 
map, as in those of the last two articles ; and the 
length of the middle parallel C'P is just the same 
length as on the earth, which is