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 curvetracing, 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 sevenplace 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
fourplace 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 RIGHTANGLED TRIANGLES . 59
OBLIQUEANGLED 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
LANDSURVEYING
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 PLANETABLE . . .
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 SLIDINGRULE
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 GAUGEPOINTS . 268
PROBLEMS AND EXERCISES ..... 270
CASKGAUGING ....... 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
MALTGAUGING . . . . . . . 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
MELTINGPOINTS 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 CASTIRON 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 RIFLEBULLETS .... 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 RIGHTANGLED TRIGO
NOMETRY . . . . . . .441
PROJECTION OF CASES OF OBLIQUEANGLED SPHERICAL
TRIGONOMETRY ...... 442
SPHERICAL TRIGONOMETRY
DEFINITIONS ....... 444
RIGHTANGLED SPHERICAL TRIANGLES . . . 448
QUADRANTAL SPHERICAL TRIANGLES .... 459
OBLIQUEANGLED SPHERICAL TRIGONOMETRY . . 460
OTHER SOLUTIONS OF THE CASES OF OBLIQUEANGLED
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 SEMIDIAMETER . . 488
CONTRACTION OF THE SAME . . . . .489
THE SUN'S SEMIDIAMETER .. . . . .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 EQUALSIDED
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
CURVETRACING
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
FOURPLACE 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 threesided 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 rightangled triangle has one
right angle, as R ; an obtuseangled tri
angle has one obtuse angle, as O ; and
an acuteangled 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 reentrant, as B.
50. Any side of a rectilineal figure may be
called its base. In a rightangled 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 + 2DBBE, 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 onesixth 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 tenthparts 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 halfdegrees, 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 (n1) 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 rightangled 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 rightangled 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 rightangled 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 rightangled 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=90C = 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.^Given 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 = L0001,
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 10L20,
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. Rlog. sin A, and log. sec
A =2 log. Rlog. 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=906 = 9034 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 (45A)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 onefourth
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
= 25. 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 onefourth
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 = onefourth 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 onefourth 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
twentyeight 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 rightangled 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
to1
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 (90A) = ^ = = 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 ' Q80A)   ' 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 (180A)=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(180A) =  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 rightangled 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 ABFC 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 = (AB).
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(AB) = 2 sin A.cos B [].
Sin (A + B)sin(AB)=2cos A. sin B [&].
Cos(AB) + cos(A + B) = 2cos A.cosB [c].
Cos(AB)cos(A + B)=2sin A.sinB [d].
191. If A + B = S, and AB=D, then A = 4(S + D),
and B = J(SD); and substituting these values in the last four
expressions, they become
SinS +sinD=2 sin (S + D) cos (SD) [a].
SinS sinD=2cosi(S + D)sin i(SD) [6].
CosD + cosS=2cos(S + D)cos(SD) [c].
CosDcos S=2sin^(S + D) sin (SD) [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(AB) [a].
Sin Asin B=2cos4(A + B) sin i(AB) ..". [&].
CosB + cos A=2cos(A + B)cosJ(AB) [c].
CosBcos A=2sini(A + B) sini(AB) [efj.
52 PLANE TRIGONOMETRY
194. In Arts. 186189 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].
CosOcos2A = 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 = 12 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 2Asin 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 A3 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 Bsm 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(AB)= ^ ...... [61.
1 + tan A tan B
PLANE TRIGONOMETRY 53
If in (a) B be made equal to A, we obtain
. 2 tan A
Tan2A=  57 ............ [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 %(AB) = tan j(A + B)
Sin Asin B~2 cos (A + B) sin (AB)~tan (AB) '" '
Sin A + sin B_2 sin j(A + B) cos j(AB)_, 1/A . W
Cos A + cosB~2 cos i(A + B) cos 4(AB)
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(AB)~ (
Sin Asin B^ 2 cos j(A + B) sin j(AB) _,,*_,>. r 71
CosA + cosB 2cosi(A + B)cos(AB)
Sin Asin B_2 cos j(A + B) sin j(AB)
Cos B cos A~2 sin 4(A + B) sin i(AB)~'
Cos B + cos A_2cos(A + B)cos ^(AB)_co
~ ~
CosBcosA~2sin^(A + B)sin i(AB)~tan 4(AB)
^cot^(AB)
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= fr = cot 2 A ...... [].
1  cos A tan \ A
Also by inverting, we have
1cosA tan ^A r7l
  r = 4r= 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 Bl ...... [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 (AB)
Sin (A + B) s ,
Sin Asin B~sm (AB)
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(AB)
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 rightangled 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 Asin B~tan (AB)
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 Al, 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 + ca)=sa, %(a + cb)=sb ;
and ^(a + b c)=sc ; and inserting these values in the above,
it becomes
21 . s(sa) . .. , , 1T) s(sb) , s(sc)
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_ ksb)\
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 + bc)(a + cb)
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 =
(sa)(sc)
(sa)(si
ab
206. Now, since tan A = fr, by dividing the value of
COS 2^*
sin JA by that of cos ^A, we obtain
tan JA =
Similarly,
tan 4B =
(sl>)(sc)_ /I (sa)(sb)(sc)
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
sb^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 rightangled triangle having AC
= BC; then /B=45.
Denote AC and BC by 1 ; then (Eucl. I. 47)
. ,_ AC 1 ._ BC 1
Sin 45 = js = = 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 Asin 2 B
4. Prove that   .  ^ = tan A . tan B ; and . .  ^^5
cot A + cot B smfA . sinrB
= cot 2 A . cot 2 Bl.
5. Prove that cos 2A + cos 2B = 2cos (A + B) . cos (AB).
6. Prove that cos (A + B) . cos (AB) = 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 40 = 180, prove (1) that sin A + sin B + sin C =
4 cos A . cos B . cos C ; (2) that cos Atcos 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 rightangled 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 (AB) = ^ ;
<2ah ffl _
(2) cos (AB) = ; (3) tan (AB)=
(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 rightangled 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 rightangled
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' 484".
Or thus :
. AC 415
sec A= r5 = ;
therefore
AB~249
L sec A = L 415  L 249 + 10
= 2618048123961993+10
= 102218488
= L sec 53 7' 484".
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
=23961993 + 10124938810
=25211381
= 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)(ACAB) = (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, . 28221681
L(ACAB) 166, . 22201081
LBC 2 , . . . . , . . = 50422762
hence (I. T.*), L BC 332, . . . = 25211381
* 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 rightangled triangle ABC, given
the hypotenuse AC 324 feet, and angle A 48 17', to
find the other parts.
1. To find angle C
C = 90A=904817'
=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
=25105450 + 98729976  10
=23835426
= 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
= 25105450 + 9823113810
=23336588
= 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
= 241848.
2. To find AB
rp=cos A; therefore AB = AC cos A
= 324 cos 48 17'
= 324x 6654475
= 215605.
62 PLANE TRIGONOMETRY
EXERCISES
1. In a rightangled 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 1124621.
2. The hypotenuse is 645, and an oblique angle 39 10' ; find the
other sides =500076, and 407 '368.
3. In a triangle rightangled at B, given the side AB 125, and
angle A 51 19', to find the other parts.
C = 3841', BC = 1561186, AC = 199'9949.
4. In a rightangled 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 rightangled 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 rightangled 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 rightangled
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. =854284, 51 39' 16", 38 20' 44".
COMPUTATION OF THE SIDES AND ANGLES OF
OBLIQUEANGLED 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', . . . ' . = 100470825
L sin B 49 25', . 98805052
LBC275, = 24393327
LAC 232 7665, .
23669204
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, = 24393327
LAB 28167,. . . . _. = 24497405
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',
76345120
25378191
97827958
99551269
L sin C 64 24' 1", . , ; ,
2. To find angle B
B=180(A + C) = 18010144' 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, . . _ . .
= 102172042
= 99908287
= 23654880
25735209
L AC 374559, ....
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 = 3259183, 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 = 243978, 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 = 133639 or 69992, 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 (AB).
_. (ab) . tan (A + B)
By natural sines, tan $( A  B) =  ^ ;
and by logarithms,
L tan (AB) = L tan i(A + B) + L(a6)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, . 75100415
L(ACBC)43, . 16334685
L tan(A + B)5330', . . = 101307911
Ltan^(BA) 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, . 99805963
LAC 176, . . ... = 22455127
L AB 187022, . 22718930
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 + 3097646816 x 2923717 = 48665 1368767=3497733, and c=
V3497733 = 187022.
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 = 110114.
3. Two sides of a triangle are respectively = 3754 and 3226 "4,
and the contained angle = 58 53' ; iequired 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
rightangled 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 201205.
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
CosAAD 84 
~AC~140'
L cos A = 10 + L 84  L 140
= 11 9242793 21461280
= 97781513
=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
= 11819543921139434
= 97056005
= L cos 59 29' 23".
Pnc. p
70
PLANE TRIGONOMETRY
4. To find angle ACB
= 180(A + B) = 180112 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,
= 23222193
= 17781513
78860566
= 78538720
L cos
C 33 41' 24" 2
2
2)198402992
= 99201496
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 =
176777807
*= 80
L(sa) =
19030900
sb= 70
L(s6) =
18450980
sc= 60
L(s c) =
17781513
Tan
Tan
Tan
Hence
2)232041200
Constant = 116020600
A = 26 33' 54 2" = 9 6989700 {cont.  L(s  a)}.
B =29 44 41 6 =9 '7569620 {cont.  L(s  6)}.
C=33 41 242 = 9 8239087 {con t.L(sc)}.
A= 53 7' 484"
B= 59 29' 23 2"
C = 67 22' 484" ; and adding
= 180
PLANE TRIGONOMETRY 71
2 1302+1402150 2
By Rule V.,... Nat. cos C ^ 2 x 130x140
3650022500 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 rightangled 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
=2505068 and 3508176.
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 plumbline.
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 LANDSURVEYING.
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 100039,
Height of eye = 5166
it tower =105 205
99209898
= 20791812
120001710
= 20001710
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=mn (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 = (mn) ; 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 baseline, were =40 12' and 50 42'; required
the height of the lull.
Here C=mn=5Q 42'40 12' = 10 30 7 .
LAB 384, = 25843312
L sin n 40 12*, . . . . = 9 '8098678
L sin m 50 42', = 9'8886513
L cosec (mn) 10 30 7 , = 10'7393670
L . CD, = 30222173
.. CD =1052488.
76 MENSURATION OP HEIGHTS AND DISTANCES
The two proportions may also be wrought separately.
In the following exercises the baseline is measured as in the
above example :
EXERCISES
1. In order to find the height of a hill, a baseline 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 =179276 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 = CBDA, 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', . . . . = 104063406
L sin A 38 19', . . ... = 9 '7923968
L AB 360, = 25563025
.'. LBC, . . . . . . = 27550399
MENSURATION OP HEIGHTS AND DISTANCES
77
2. To find CE in triangle BOB .
E = 90 + EBD = 90 + 48 20' = 138 20'.
B = CBDDBE = 6125'48 20' = 13 5'.
L cosec E 138 20' (41 40'), . . = 101773117
L sin B 13 5', . . . . = Q'3548150
LBC, ...... = 27550399
LCE, ...... = 22871666
.. CE = 1937165.
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. . . =1119978.
2. In order to determine the height of a lighthouse, situated
on the top of an inaccessible eminence, the following data were
obtained : A baseline = 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
?=DBFt, = DAE', and r = mn;
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', = 104818934
L sin n 29 5', . = 96867088
L AB 112, . = 20492180
L BD . 16512, . = 22178202
To find CD
L cosec v 105 20', = 100157411
L sin m 48 20', = 9 8733352
L BD 16512, . = 22178202
LCD,
= 21068965
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 baseline 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 = FBAFBD, and A=180ABF; 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=FBAFBD=
47 42* 17 52 / =29 50', and A=180 ABF = 18047 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', . = 105131405
L sin A 132 18' (47 42'), . . = 9'8690152
LAB 204, ..... = 23096302
BD 491 797, . ' . '. . . = 26917859
2. To find DE in triangle DBE
L cosec E 101 40' (78 20'), . . = 100090662
L sin B 6 12', = 90334212
LBD491797, . 26917859
DE 542342, ...; . , .. = 17342733
3. To find AD in triangle ADB
L cosec D 17 52', . 105131405
L sin B 29 50', = 9 '6967745
LAB 204, ..... = 23096302
AD 3307846, . 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 B30.
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
) = 180159 55'=205'.
31085650
104642168
99833805
96868981
L AB 1284, .
L cosec B 20 5',
L sin D 74 15',
L tan A 25 56',
LBC,
To find BC
32430604
BC = 175009.
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, = 100112761
L sin B 43, = 98337833
L AB 1140, = 30569049
AD 797 929, ..... = 29019643
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? =673624 and 6524 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? =680323 and 56257 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(ABBC)242, .
2558', .
Tan(A~C) 7 36' 40",
.*. Angles C= 33 34' 40"
n A= 18 21 20
7 '0545314
23838154
96875402
= 91258870
2. To find AC
L cosec C 33 34' 40"
L sin B 128 4', .
L AB 562, .
L AC 800008,
= 102572213
98961369
27497363
29030945
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, . . . = 107962026
L sin B 106 23' (73 37'), . . = 9 "9819979
LAB 520, = 27160033
LAK 312035, = 34942038
84 MENSURATION OF HEIGHTS AND DISTANCES
2. To find AM in triangle AMB
L cosec M 19 C 20', ... = 104800888
L an B 56 15', . ' . . . = 99198464
LAB 520, . . . . . = 27160033
LAM 130538, . . . . . = 31159385
3. To find the angles M and K in triangle AMK
M + K=180 A=18040=140.
AT. oa L(AK + AM) 4426133, . ,., . = 63539563
HAK AM) 181437, . ' '. fc . = 32587259
Ltan J(M + K)70, . 104389341
Tan KM K) 48 23' 48", . . = 100516163
Hence K= 21 36' 12"
4. To find MK in triangle AMK
L eosec K 21 36' 12", ... = 104339415
Lain A 40, . . . . = 98080675
L AM 130598, = 31159385
LMK 228006, . 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 =14104;
ADB = 45 1' 3",
ADC = 302'0",
..CD =208088.
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 615797,
= 101307551
99997174
26589648
27894373
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 816318,
= 101925354
99999927
27193313
29118594
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, . 03010300
L AC . BC = L AC + L BC, = 5 7012967
L cos C 88 50', . . = 83087941
 10
431 1 1208 = L 20470 14
Hence AB 2 , = 1025110'86
And AB = \/102511086=101248.
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 =11261, AC=519685, 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, . .'... = 34031205
L(s)737, . . . 28674675
10 L b 1540, . 68124793
10Lcl727, . 67627077
2)198457750
LcosJA33 9 8' 33", . . . = 99228875
..A=6617' 6"
2. To find AD in triangle ADB
Angle D = 180 (m' + ') = 180 53 24' = 126 36'.
L cosec D 126 36', . 100953832
L sin B 27 44', . 96677863
L AB 1727, = 32372923
L AD 100106, = 30004618
MENSURATION OF HEIGHTS AND DISTANCES 89
3. To find angle C in triangle ADC
r=An' = 66 17' 6"  25 40' = 40 37' 6".
C + D = 180  r = 139 22' 54".
AT. co. L( AC + AD) 2541 06, . 65949850
L( AC AD) 538 94, . 27315404
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', = 103322137
L sin v 39 52' 29", . 9 '8069333
L AC 1540, . 31875207
LAE 212162, . 33266677
Angle A = 180  (m + v) = 180  67 36' 29" = 1 12 23' 31".
L cosec m 27 44', . 103322137
Lsin A 112 23' 31", . 9'9659537
L AC 1540, . 31875207
L CE 305976 = 34856881
5. To find BE in triangle ABE
Angle A = CAECAB = 112 23' 31" 66 17' 6"=46 6' 25".
L cosec E 53 24', = 10'0953832
L sin A 46 6' 25", . 98577155
L AB 1727, = 32372923
L BE 155021, = 31903910
The distances, therefore, are AE = 212162, 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 = 165357, 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 = 5325, angle BEC= = 13 30', and AEC = m 29 50'; what
are the distances of E from A, B, and C ?
= 13106, 151428, 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 = CABCAE.
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 = 35771, CE = 37862, 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 ? ..... =221548 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=89818 feet, and distance = 74 '9666.
5. From the top of a tower 1365 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.
=529647 and 251319 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 = 35686, 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 = 5001 6, and DC = 48908 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 = 10663, BD = 15641, and CD = 140107 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 shrapnelshell 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 = 76501 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 seawall
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 seawall.
= 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" = 38194 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 = 2175 x li> '5 = 383625 sq. ft. =383 sq. ft. 90 sq. in.
=42 sq. yd. 5 sq. ft. 90 sq. in.
Or L^l = L6 + LA = L2475 + L 15 5 = 13935752 + 1190331 7
= 25839069 ; . . ^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=245x85=20825 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}(db).
Then JR = bh = b\/(d+b)(db) ;
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 + (23521825 + 13979400)
= 2 + x 37501225 = 38750612 = 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 12152 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 v10 ......... [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 + Lsm2w10 ...... [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 = 8229955 sq. feet.
Or by [2], L^l = 2L6 + L tan v  10 = 2L 36 + L tan 32 25'  10
= 31126050 + 9802792510 = 29153975;
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], ^BX 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
= 53351242 = 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 = bh24: 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=26314438 + 25670264 = 51984702;
hence .51=157932 sq. links = l acre 2 roods 126912 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 383488 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 e10 [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 = 16283889 + 13273589
+ 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' + ~Lsmi10 ... [2].
EXAMPLES. 1. Find the area of a square whose diagonal is
= 20 feet 3 inches.
Here d=d' =2025 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
=26128500 = 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 = 30951694 + 30170333
+ 96821349 10 =5 '7943376 = L 622784,
and yR = 311392 sq. links = 3 acres 1823 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 ? . .* . . * ' .. =1948557 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 = 30962146 + 2 9754318
= 60716464 = 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 =1025 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 rightangled 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 + LsinilO.
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=27958800 + 28061800 + 98118038
 10=54138638 = 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 =8347788 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". =174284 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?
= 122088 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 15416 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 halfsum and each of the sides ; then find the
continued product of the halfsum and the three differences, and
the square root of the product will be the area ; or,
Add together the logarithms of the halfsum 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(sa)(sb)(sc)}.
Or L, JR = %{Ls + L(s  a) + L(sb) + 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 sa = 979
c = 3650 sb = 2671
2s = 10430 S ~ C =
And &, = \Js(s a)(s b)(s c) = \J5215 x 979 x 2671 x 1565
= V21341514430775 = 4619687.
Or Ls5215, = 37172543
L(sa)979, = 29907827
L(s6)2671 = 34266739
L(sc)1565, = 31945143
2)133292252
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(sa)(sb)(sc)}.
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 = 110848.
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 = 32697464 + 28129134
= 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 = 64302, 42848, and perpendicular distance 34232?
= 18339795 square feet.
4. Find the area of a trapezium whose parallel sides are = 41 and
24'5 feet, and perpendicular distance = 43. =140825 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 332 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?' + Lsinz10.
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 = 442493 sq. feet.
Or L 2M = Ld + LeP + L sin t  10 = 2 6232493 + 2 51 18834
+ 98118038 10 =4 9469365 = L 88498 "6 ;
and .51 = 44249 3 sq. feet = 4916 sq. yards 53 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 i10.
EXERCISES
1. What is the area of a quadrilateral whose c
diagonals are = 50 and 40 feet, and the included
angle = 60? . . . . . . =8660254 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 2355 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 3384 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 = \/{(sa)(sb)(sc)(sd)};
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 = 7003125 sq. Ik.
= 30951694 + 3'0511525 = 61463219 = L 1400625,
and 51 = 7003125 sq. links = 7 acres 05 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 114 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 onefourth
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 ) = 405244 = 161,
and JR = x 161 tan 84 25' = i x 161 x 10229428 = J x 1646 9379
= 4117345.
Or L 4^l = L 161 + L tan 84 25'  10=22068259 + 11 "0098513  10
= 32166772 = L 164694;
and 1 = 411735 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 = ABCDBC 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 ^= (2n4) = (n2), and c=
n ^ n v n
EXAMPLE. Find the interior and central angles of a regular
pentagon.
i=^(n2) = "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 rightangled 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
= 17781513 + 102385606 10=20167119;
hence j9 = 103'923.
EXERCISES
1. Find the apothem of a regular pentagon whose side is = 10.
= 68819.
2. What is the length of the apothem of a regular heptagon
whose side is 80? ..... . . . =830608
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 = 14485 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 z10 = L6 + 2L5 + L tan 60 10= 7781513
+ 1 '3979400 + 102385606  10 = 2 '414651 9 ;
hence ^1 = 259808.
EXERCISES
1. Find the area of a regular pentagon whose side is = 30 feet.
= 15484275 square feet.
2. What is the number of square yards in a regular heptagon
whose side is = 20 yards? ... =1453564 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
02886751
04330127
60 0'
120 0'
Square
4
05
1
90
90
Pentagon
5
06881910
17204774
108
72
Hexagon
6
08660254
25980762
120
60
Heptagon
7
1 0382607
36339124
128 34?
51 25f
Octagon
8
12071068
48284271
135
45
Nonagon
9
13737387
61818242
140
40
Decagon
10
15388418
76942088
144
36
Undecagon
11
17028436
93656399
147 16 T * T
32 43 T \
Dodecagon
12
18660254
111961524
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 36339=225 x 3 6339 = 817 '6275 sq. feet.
Or L.5l = L 36339 + 2 L 15=05603730 + 23521826 = 29125556, and
51 =817 '628 square feet.
EXERCISES
1. How many square yards are contained in a regular hexagon
whose side is = 50 feet ? . . . =721688 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? =34839667 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.
= 619371864 square feet.
6. Find the area of a regular octagon whose side is = 80 yards.
= 3090193 square yards.
7. How many square yards are contained in a regular decagon
whose side is = 12 feet? . . . =12310734 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 31416, 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=31416e?=W, or c=2x31416r = 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 = 7854 feet.
EXERCISES
1. Find the circumference of a circle whose diameter is = 28 feet.
= 879648 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 = 143235 = 14 feet 3'882 inches.
EXERCISES
1. Find the diameter of a circle whose circumference is = 177
feet =563391 feet.
MENSURATION OF SURFACES 117
2. What is the diameter of a circle whose circumference is = 32
feet? ......... =101856 feet.
3. What is the diameter of a wheel whose rim is = ll feet?
= 35013 feet.
4. What is the diameter of a circular pond whose circumference
is=200feet? ........ =6366 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 12732 = 12732 sq. feet.
EXERCISES
1. Find the area of a circle whose diameter is =21, and circum
ference = 65 973 ......... =346358.
2. What is the area of a circle whose diameter is = 20, and
circumference = 628318? ...... =314159.
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 3471 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 =31416x 60 2 = 3'1416x 3600 = 1130976 sq. feet.
Frac. I
118 MENSURATION OF SURFACES
For by last problem, M = cd; and by Prob. XXIII., c = 3141Gd;
therefore JR= x 3'1416e^= 7854^.
And since d=2r, or cP^r" ; therefore .1 = 31416^= srr 2 .
EXERCISES
1. What is the area of a circle whose diameter is = 60 feet?
= 282744 square feet.
2. Find the area of a circle whose diameter is = 35 feet.
= 962115 square feet.
3. Find the area of a circle whose diameter is = 397 "885 feet.
= 1243384 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 145 square poles.
6. How many square yards are contained in a circle whose
diameter is = 350 feet? . . . =1069016 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^ = 3263 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.
= 4973594 square feet.
2. What is the area of a circle whose circumference is = 15 708?
= 19635.
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.
= 31831 square yards.
MENSURATION OF SURFACES 119
5. What is the number of square yards in a circle whose circum
ference is 251328 yards? .... =502656 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 onefourth 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')(cc')
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 = 345576.
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 188496.
l& = (c + c')(dd')=x 345576x10=86394.
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.
=201063.
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 ? =120264 square feet.
5. The circumferences of two circles are = 62'832 and 94248 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 2rh = 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 + 35=122.
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 = 156204994.
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 = 84? =14.
2. What is the radius of a circle, the chord of an arc of which
is = 12, and the apothem = 10? =116619.
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 =425.
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 = QPPR, orp=rh ; 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 + 10Lr
= 13010300 + 1017781513 = 95228787;
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 = 31416rf, and 360 : =3'1416rf : I ;
, 314l6nd
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 = 130905.
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 = 11126.
EXERCISES
1. What is the length of a circular arc of 45, the diameter being
= 12? =471258.
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'.
= 59559.
5. Find the length of an arc whose chord is = 12, and radius = 18.
= 12234.
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? =32461.
7. Find the length of an arc whose chord is=4'8, and that of its
half=2443. . ... . . . /. . = 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 = 1308997.
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 = 1112648.
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 = 1308997,
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. . . =314172.
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? =110106.
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 =22344.
10. What is the area of a sector whose chord is = 30, and height
= 4? =473015.
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? =102188.
5. Required the area of a segment whose chord is = 16, and
diameter = 20 =447293.
6. The chord of a segment is = 24, and the radius =20; what is
its area? =65401.
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. .... =733966 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 65 ; what is its area? . =216597 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 = 1006416.
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=rp.
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 Maa' ...... [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'=rp' = 51438 45 = 6 '478.
Also h =b + h' = 2Q+ 6 478 = 26 '478.
The tabular height for h' is== = 06292.
d 10^ 'Doo
h 26478
Tabular area for 0629 is = 020642
H for 2572 is =159811
Difference, = 139169
Hence JR =139169^= 139169 x 10600 = 147519.
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. . . . =53419.
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?
= 15817475.
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. =44343.
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? =17205.
2. The chord is = 30, and the heights = 3 and
15 ; find the area of the luue . . =292954.
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 1846 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 2127 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 onethird 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.
LANDSURVEYING 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.
LANDSUEVEYING
293. Landsurveying 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 surveyingcross, 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 LANDSURVEYING
at the end of each chainlength when measuring in the
field.
298. Offsetstaffs are wooden rods ten links long divided
into links for measuring offsets (Art. 305).
299. Other staffs, about six feet long, called picketstaffs
or stationstaffs, 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 surveyingcross, or crossstaff, 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 crossstaff 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
LANDSURVEYING 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 chainlines or stationlines.
305. Lines measured perpendicularly to chainlines, to the
angular points, and other points of the boundary of a field
are called offsets.
306. The crossstaff is used for finding the position of
offsets. The point in the chainline from which an offset is
to be measured to any point in the boundary is found by
fixing the staff in the chainline 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 spiritlevel, ~N the tele
scope moves vertically about a horizontal axis which
136 LANDSURVEYING
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 halfdegrees, 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 spiritlevels, 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 chainlines. Measure the base and height, or else the three
sides of each of the principal triangles, then calculate their
LANDSURVEYING
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 chainline
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/,;=ABA = 145.
= ABAi = 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 fieldbook, as
follows. The beginning of the fieldbook 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 chainlines, 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 fieldbook 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 Oj.
138
LANDSURVEYING
EXERCISES
1. Find the area of a field, the dimensions of which are given in
the following fieldbook :
Left Offsets
Chainline
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 :
Chainlines
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 fieldbook.
= 1 acre 1 '3184 square poles.
= 9000
= 24150
= 2750
= 1950
= 7975
= 2875
LANDSURVEYv^G
139
Left Offsets
Chainline
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 fieldbook ' From O 3
on L of Go' means that a chainline 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 chainline,
AC, is measured ; so ' From Oj on R of O 2 ' means that the next
chainline 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 fieldbook, 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 LANDSURVEYING
the measurements in the following fieldbook, A, B, C, D, E
being respectively the 1st, 2nd, 3rd, 4th, and 5th stations.
Offsets on Left
Chainlines
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
LANDSURVEYING
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
LANDSURVEYING
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,
1515
66
Yards into links, .
Square feet into acres, . ,
Square yards into acres,
Feet into miles,
4545
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
LANDSURVEYING 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
LANDSURVEYING
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.
LANDSURVEYING
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 118 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 PLANETABLE
320. By means of the planetable, 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
tripodstand, having at the top a ball
andsocket 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 spiritlevels 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 compassbox, C, is fixed to one side of the table for
determining the bearings of stations and other objects, and for
146 LANDSURVEYING
the purpose of fixing the table in the same relative position in
different stations.
There is also an indexrule 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 planetable 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 indexrule 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 planetable 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 magneticneedle, as in last problem. Apply
the nearer end of the indexrule 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
LANDSDRVEYINO 147
point of the compasscard as before. If the indexrule 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 planetable.
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 indexrule 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 LANDStJRVEtING
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 landsurveying 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.
LANDSURVEYING 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 LANDSURVEYING
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.
LANDSURVEYINa 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 = 9804 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
LANDSURVEYING
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
LANDSURVEYING 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
LANDSURVEYING
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 fieldbook and following sketch :
Plan from following Fieldbook
Scale of Chains
155
Left Offsets
Chainlines
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 LANDSURVEYING
The chainlines in this fieldbook 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 chainline
that is, its inclination to the meridian ; and as the inclination of
each of the successive chainlines 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 chainline, 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 chainline lies to the left of the first ; and hence the
angle of the polygon is here a reentrant angle, and = 360  156 15'
= 203 45'.
When any boundaryline crosses a chainline, 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 chainline ; and when the
line is straight to the. end, these lines are marked S, as in the
fieldbook 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 fieldbook. Area = 66 ac. 3 ro. 13'46 sq. po.
337. In the following fieldbook, the expression 'in line with
fence,' with dots ... on the left or right of the chainlines, means that
the point arrived at in the chainline is in the same line with some
straight fence on the left or right, which does not extend so far as
the chainline. 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
Chainlines
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 14ft
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
LANDSURVEYING
Plan from preceding Fieldbook
EXTENSIVE SURVEYS WITH THE THEODOLITE
338. In large surveys with the theodolite, as that of an estate or
the mapping of a district, extensive chainlines 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 stationline 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 chainline.
Let ABCD (fig. to Art. 335) represent a portion of a chainline,
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 polestar 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
clampingscrew, 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
LANDSURVEYING 159
picket placed at N'. The head is now fixed in this position by
means of the lockingscrew 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 eyeglass of the telescope, is noted, as it measures the bearing
of the second station from the first. The first stationline, AB, is
now measured, and the instrument removed to the second station
at B ; then, after being levelled, and the lockingscrew 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 clampingscrew 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
stationline, BC, from the meridian drawn through B towards the
south. The upper part of the head is now clamped, and the second
stationline 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 stationlines 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 LANDSURVEYING
this direction ; the length of this first stationline 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 chainlines 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 bearinglines (or rhumblines) 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
objectend 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 rightangled 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 ?
=459375 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.
=20125 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 = (%)*= ^ = 15625 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.
=2072671875 cubic feet.
4. Find the contents of a cube whose edge is = 6*5 yards.
=274625 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?
= 765625 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. . . . =128125 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, =83138 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(l5) 2 = 6x 225 = 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. 10d.
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?
= 1150037 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. =25566 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?
= 61491 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 ? . . =27056818 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 =2x31416x5 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?
= 15146439 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 onethird of the
product is the solidity.
ViW,
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 ? =7624 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 onethird 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 ? . . =282744 cubic feet.
2. The diameter of the base of a cone is = 10, and its altitude
= 12 ; what is its solidity ? =31416.
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 ? =462856 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= 188496,
and s = z + b = 188496 + '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 ? . =6291054 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.
= 9809375 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 onethird 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 onethird
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
10/36x616x4\ 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 VBV& = 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 EhEbhe_fi/'BEbe\
~ * Ee ~3\ Ee )'
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? ....... =15796303 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.
= 788643 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 ..... .;_.', . =155981 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=80629 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.
= 1309x19=24871.
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 ?
= 73304 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.
=288634 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 ?
= 329868 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.
= 1914072 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?
= 132112 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=188496,
B + b = 7854(D 2 + d 2 ) = 7854 x 20 = 15 "708,
and s = ip(C + c) + B + 6 = x20x 188496 + 15708
= 204204 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"Cp"c} = l{pC+p"(Cc)} ...... [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. . =745875 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?
= 19854912 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 onesixth 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 = 4132 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={& + $(le)}b
Were the edge longer than the base, the formula would be
the same; for the expression \e + ^(le) would then become
), as before.
EXERCISES
1. Find the contents of a wedge whose base is = 16 inches long
and 2J broad, its height being = 7 inches, and its edge = 10J inches.
= 1115625 cubic inches.
2. The length and breadth of the base of a wedge are = 5 feet
10 inches and 2^ feet, the length of the edge is =9 feet 2 inches,
and the height = 34 '29016 inches ; what is its solidity ?
=248048 cubic feet.
389. Problem XVI. To find the solidity of a prismoid.
RULE. To the sum of the areas of the two ends add four times
the area of the middle section parallel to the ends ; multiply this
sum by the height, and take onesixth of the product.
Let L =the length of the base AB,
B = M breadth of the base BD,
I = M length of the top GH,
b = breadth of the top GH,
M= n length of middle section,
m = M breadth of middle section,
h = ii height of the prismoid ;
then M = J(L + ), and wi=^(B + 6),
and V = i(BL + bl + 4Mm)A.
EXAMPLE. Find the solidity of a prismoid, the length and
bread tli of its base being = 10 and 8, those of the top = 6 and 5,
and the height = 40 feet.
Here M = (L + /) = ^(10 + 6) = x 16 = 8,
i = ^(B + 6) = ^( 8f 5) = x 13 = 6*5,
=$(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 twothirds 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=9929 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
twothirds 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 ........ =3215558 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 twothirds 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 sheetcopper are contained
in a hollow copper globe =25 inches in diameter?
s = 31416d 2 =3'1416 x 25 2 =31416 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 goldleaf 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=31416c#t = 31416x 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 )
= 41888 x 652 = 27310976 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)
= 20944 x 11 =230384 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 , = 31416rfA,
w 5236^x31416^/1
hence V =
For the cone, volume = V"=rA5236r 2 (rf2A) ; hence
volume of segment,
V = V  V"= 52361^ r 2 (d2A)} ...... [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(dh)(d2h)}= 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.
=4356352 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 onethird 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 twothirds of the square of the height,
or find the difference between the square of the diameter of the
sphere, and onethird 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 ) = 15708x2(4 +  + f)
=31416 x f = 238238 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.
= 47124 x 40= 188496 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 )
= 47124 x 40 = 188496 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?
= 1958264 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 ..... =316338176 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?
= 4942784 cubic feet.
6. Find the volume of a spherical zone whose height is =4 feet,
and the end diameters = 6 feet. = 146608 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 = onefourth
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 ^=^=^^=972, 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 fourthirds 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 ? .......  . =1847.
2. The abscissa and ordinate of a parabola are = 10 and 8; what
is the length of the curve ? . . . . . =2809.
419. Problem VI. To find the area of a parabola.
RULE. Multiply the base by the height, and twothirds 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 twothirds 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
56 feet. ... ....... . . =1344 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 twothirds 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 twothirds 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.
30318 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 =(ah)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^ = 3024 = 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
semiaxis minor and the ordinate to the distance of the ordinate
from the centre.
This distance being added to the semiaxis 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
semiaxes, and h AM, and dPM ;
MENSURATION OF CONIC SECTIONS 195
then b*:a?=(b + d)(bd) re 2 , or c 2 =p(& + rf)(&rf),
and h=a + c, 2ah=ac.
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=ac=159 = 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 semiaxis minor and the ordinate, and, according as the less
or greater abscissa is given, add this root to or subtract it from the
semiaxis 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 semiaxes, 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=:2516 = 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 =(2ah)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 semiaxes, 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 31416, 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 ?
= 31416 x 223607 = 702484.
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? ...... . . . =347001.
2. The axes are = 6 and 8 ; what is the length of the curve ?
= 22214.
3. The axes are =4 and 6 ; what is the length of the curve?
= 16019.
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=23562 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(ah) 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 ?
= 1507968.
4. What is the area of an ellipse whose axes are = 6 and 7 ?
=329868.
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=16200975.
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 =335472.
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.
= 1565536.
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 semiaxis minor and the
ordinate to the square of the distance between the ordinate and
the centre.
The sum of this distance and the semiaxis 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
semiaxis minor and the ordinate ; and, according as the less or
greater abscissa is given, find the sum or difference of this root and
the semiaxis 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 semiaxes ;
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 semiaxes, 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? . . . . =109641.
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(hh') = 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
= 35343x369 = 13041567.
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?
=52166268.
2. Find the solidity of a frustum of a paraboloid, the diameters
of its ends being = 29 and 15, and its height = 18. . =75351276.
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(32A)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 = 408408.
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? . . . ...... =879648.
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.
=306306.
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
= 10472 x 3540 = 3707088.
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, W1 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? =22242528.
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. . . =463386.
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
= 1309 x 1832=2398088.
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 semiaxes ; 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 ?
= 165876375.
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.
= 111265.
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 onetwelfth 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 1732 = 3897.
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? . . . . ..... =254558 and 62352.
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 onethird 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 = 124704;
or s= 2e 2 V3 =2 x6 2 x 1732= 124704.
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.
= 193087.
2. What are the volume and surface of an octahedron whose
edgeis = 3? =127279 and 311769.
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 76631.
Also the surface s=15e 2 V * ^ = 15e 2 x 1376382.
5
EXAMPLE. What are the solidity and surface of a dodecahedron
whose edge is =2?
V=e3x 76631 = 8 x 7'663 1=6 13048,
and s= 15e 2 x 1 376382 = 15x4x1 376382 = 8258292.
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 rightangled
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 rightangled 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(102\/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.
= 16552296.
REGULAR SOLIDS 217
2. What are the solidity and surface of a dodecahedron whose
side is =4? ...... =490 4384 and 33033168.
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 fivesixths of this product.
Or, V = f6 3 V 7 + 2 X/ . or V = fe 3 x261803=218169e 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 + 3x223606
=^V6 85409 =Y x 261803 = 174535,
and s= 5e 2 V3 = 5 x 2 2 x 173205 = 34641 ;
or s=20e 2 A' =20x22 x 433 = 3464.
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.
= 47124504.
2. What are the volume and surface of an icosahedron whose
edgeis=5? ...... =272711 and 216506.
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
17320508
01178513
6
Hexahedron
6
1
8
Octahedron
34641016
04714045
12
Dodecahedron
206457288
76631189
20
Icosahedron
86602540
21816950
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 8660254 = 34641016.
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=24674eft 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 31416tf=24674eft 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 =2442726.
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=3'1416x31416d=98696cft,
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. =34740992.
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 onetwelfth 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' = $dr2r; 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 = 62832( T y 3  Me) = 6'2832( x 16000  441 '9288 x *f)
= 62832 x 2755415 = 173128235.
EXERCISES
1. The length of a circular spindle is = 8, and its middle
diameter = 6 ; what is its solidity ? .... =138503.
2. The length of a circular spindle is = 24, and its middle
diameter = 18; find its solidity. .... =37395696.
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 onetwenty
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 = (Drf), 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 = 2687776.
= 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
= (31866 2643 8)62832 = 542 866 x 6 '2832 =341093984.
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 eightfifteenths 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? .... =36316896.
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. =75649728.
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 onefourth 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 onefourth
of the quotient will be the central distance.
To the central distance add half the middle diameter, and the
sum will be the semiaxis 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 semiaxis minor. '
Then, if =semiaxis 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^ 10890. 1624 _
'  = *' 18.990818 ~
and
*'=46 = 30 =  2 ' andA ' = ' 111824 '
s=46A'=4x 125x7'5x 111824=41934.
'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?=(2bh)h : (&)*, 6 2 : a?=C2bh')h' : (J/) 2 . Hence (2bh)h
4^,^ _ ^2
= <l(2bh')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.
= 25784748.
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 onefourth of the quotient will be the central
distance.
To the central distance add half the middle diameter, and the
sum will be the semiaxis 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
4681836108
= 15l, and /t = (D
x14 2114
~
072
6;
_
~'
= 25'08;
' = 4 x 2508 x 15'1 x 003712=5623,
= 2618{288 + 1 1664 
 1 2
Vl}14
= 2618(40464  357)14= 2618 x 404283 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
= 1507878; 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 or^ >
a=cJD =
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 = ACFE, and IK = IM + MK
= EF + MK.
Or, if AL = D, MK=e/, and A' = HA, 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,
, ,,, .(REbe\ u , 600x20384x16 _
and y= l(^^r^ ~ 2016
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 246825.
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 50353776.
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 = 6989,
= i{Jx20 3  6989(25 10)}Y = (13333 104835)12= 17099.
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? . . =1094334.
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 ? =734218 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 !
159'6
2618 x 1098 x 300
= 159698.
54
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? . . =5110336.
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
,,,_hD + 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 96, 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  9e 3 x 371872 x 21739V27139) x
90
90
=J(1877 '2628 1054 537)^ = 8227258
O *4
= 1015701.
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 ?
= 4062787.
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= Dd (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 ^ = Dd=1596 = 5'4,
and h' = ~ . = ~ = 36, and A' = '254551 ;
U 15
hence A = cPA' = 15 2 x 254551 = 57 '273975,
and V
= (1590943  9216)20=446229.
EXERCISES
1. Let a vessel in the form of a conic frustum, the diameters of
its bottom and top being =30 and 192 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. . =17849166 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 ah' 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 Dd :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 : ah' = d:D h; hence a : h' = d:h (Dd) ;
hence a=
Also, h' :a = h:d' ;
, ,. ah dh L j , h
hence d' =  T7 = j ^ T. ; hence bd\/, ^ r..
h h(Dd) h(Dd)
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(hD + 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
Ddh Ddh
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 oaktree 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 ?
=623 feet.
3. The area of an equilateral triangle is = 720 ; required its side.
= 40784.
4. What must be the length of the radius of a circle which
contains an acre ? =117*752 feet.
5. A circular fishpond 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? =54 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 churchspire,
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 3inch 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?
= 977846 yards.
16. A rectangular bowlinggreen, 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? =7feet.
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 ?
= 3269, 4559, and 10172.
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 ?
= 5057, 7209, 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 489898.
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 ? .... =1259921 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 423 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 onefifth 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 onefifth
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 flatbottomed 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 oneeighth 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? =4899 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?
=428567 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 SLIDINGRULE 241
THE COMMON SLIDINGRULE
486. The common or carpenter's slidingnile consists of
two pieces, each a foot long, connected by a foldingjoint. 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 onehalf 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 SLIDINGRULE
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 = ; .'. L6Ll = LpLa;
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; ' L6L = LoLl;
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 SLIDINGRULE 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
=.=.=; .. LaU=2La2Lx.
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 Slidingrule. 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 105 on A is 7 on B.
The first rule depends on Art. 247.
The reason of the method by the slidingrule 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. 10d.
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 Slidingmle. 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 LVL6=2Lm2L12;
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 Slidingrule. 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
25067 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 ?
= 7495 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?
=41278 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, onefourth of
the mean circumference and multiply its square by the
length.
By the Slidingrule. 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 onefifth of the girt, and multiply its square by
twice the length.
By the Slidingrule. Set twice the length on C to 12
on D, and opposite to onefifth 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 . ^=58074 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
onefifth 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 = 943, 7 '92, 6'15, 4 "74, and 3'16 feet, and its length is = 17 feet
3 inches ; what is its content ?
= 425195 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
0690
Ueecli j
Aluminium, sheet, . . 2670
\ to
0696
ii cast, . . 2560
Birch / fl ' m
0711
Antimony, . . 6720
\ to
0730
Bismuth, n . . 9 "822
Box, ....
1 280
Copper bolts, . ' ..   . 8850
Cedar, West Indian,
0748
H wire, . . . 8900
n American, .
0554
Gold, . . . .18417
n Lebanon,
0486
Iron, cast, average, . 7 '248
Chestnut, ;. . .
0606
M wrought, average, . 7780
Cork,
0240
Lead, cast, . . . 11360
Deal, Christiania, . ;>.'
0689
sheet, . . . 11400
Ebony, . . ^ . ,/ .
1187
Mercury, .' : . 13596
Elm, English, .
0553
Platinum, . . . 21531
n n . . .
0579
n sheet, . . 23000
n Canadian,
0725
Silver, .... 10474
Fir, spruce,
0512
Steel, .... 8000
n male,
0550
Tin, cast, . . . 7 '291
M female,
0498
Zinc, .. . . .7000
Hornbeam, . .
0760
Iron wood,
1150
ALLOYS
Greenheart,
1143
Aluminium bronze, 90 to
Larch, . .., . .
0543
95 per cent, copper, . 7 "680
n ....
0556
Bellmetal (small bells), . 8050
Lignumvitae, . . '. ;..
1333
Brass, cast, . . . 8400
Lime, . . }\
0564
ii sheet, . ..;., 8440
Mahogany, Nassau,
0668
t . Avire, . .."" .; 8 '540
M Honduras, .
0560
Gold (standard), .. .17724
n Spanish,
0852
Gun metal (10 copper,
Maple, . . v .
0675
1 tin) 8464
Oak, African, ' .
0988
Silver (standard), . . 10312
M American, red,
0850
Speculum (metal), . . 7'447
M ii white, .
0779
Whitemetal (Babbett), . 7'310
ii English,
0777
n ii .
0934
TIMBER
Pine red !
0576
ffrom 0710
Acacia, . . { ^ ^
\ to
... ffrom
0657
0432
. , ffrom 0690
n white, .
I to
0553
SI1 ' * ' I to 0760
n yellow, . ....
0508
254
RELATIONS OP WEIGHT AND VOLUME OF BODIES
Pine, Dantzic,
0649
MISCELLANEOUS SUBSTANCES
TIT i (from
n Memel, .
0550
Asphalt, .
2500
I. to
0601
. , ffrom
1600
. /from
0466
Brick, common, c
2000
" Ri s a ' i to
0654
n London stock,
1840
Satinwood,
0960
n Red,
2160
/from
0740
n Welsh fire, .
2400
Teak, . .  to
0860
M Stourbridge fire, .
2200
Cement, Portland, . from
3100
STONES, &c.
ii n in powder,
3155
Agate, ....
2590
M Roman,
1600
Amethyst, common,
2750
Clay, ....
1900
Basalt, Scotch,
2950
Coal, Anthracite, .
1530
ii Greenstone,
2900
i, Cannel,
1272
Welsh,
2750
M Glasgow,
1290
. .. ffrom
2330
n Newcastle,
1269
Chalk, . .  to
2620
Coke, . . . ,'rtjj.i
0744
Firestone,
1800
Concrete, ordinary,
1900
Granite, Aberdeen gray,
2620
u in cement,
2200
n M red,
2620
tf 1.1 ffrom
Earth, 1
1520
n Cornish, .
2660
\ to
2000
n Mount Sorrel, .
2670
Glass, flint,
3078
Limestone, compact,
2580
M crown, .
2520
M Purbeck,
2600
M common green,
2528
n Blue Lias,
2467
M plate, .
2760
M Lithographic,
2600
Guttapercha, .
0966
Marble, Statuary, .
2718
Gypsum, ....
2286
M Italian,
2726
Ice, if from water purged
n Brabant block, .
2697
from air,
0954
Oolite, Portland Stone, .
2423
Indiarubber, .
0930
u Bath n
1978
Ivory, ....
1820
Sandstone (Arbroath
Lime, quick, . .
0843
pavement),
2477
,, . ffrom
Mortar.
1380
n Bramley Fall,
2500
\ to
1900
ti Caithness,
2638
M average,
1700
ii Craigleith,
2450
Pitch, . . . V
1150
n Derby grit,
2150
Plumbago,
2267
M Red (Cheshire),
2'510
Snow, . . .
0083
Slate, Anglesea,
2870
Sand, quartz, .
2750
ii Cornwall,
2510
n river,
1880
n Welsh, .
2880
n pit (coarse), .
1610
Trap,
2720
n (fine),
1520
RELATIONS OF WEIGHT AND VOLUME OP BODIES
255
Sand, pit, Thames, .
1640
Shingle, .
1420
Tallow, .
0940
Tar, .
1016
Tile, common,
1810
il ii '"'i^ ^ 1<
1850
LIQUIDS, &c.
Water, distilled,
1000
ii sea, . . .
1027
Acetic acid,
1060
Alcohol, absolute, .
0792
it of commerce, .
0800
n proof, . i'i
0916
Chloroform,
1490
Citric acid, . .,
1034
Ether, . . . .
0716
Fluoric acid, .
1060
Hydrochloric acid, .
1200
Milk, . . , l
1032
Nitric acid,
1420
Oil, Linseed, . k ..
0940
it Olive,
0915
Whale, . . . .
0923
Petroleum, crude, .
0885
n refined, '
0910
Sulphuric acid, , : .
1843
Oil, Amber, . \ .
0868
n Cinnamon,
1043
1640
Oil, Lavender,
0894
1420
ii Turpentine,
0864
0940
it Sweet almonds,
0932
1016
n Codfish, .
0923
1810
ii Hempseed, . i
0926
1850
Porter, brown stout,
1011
Proof spirit,
0922
Strong ale, . . ' i
1050
1000
Wine, Port, . " : *
0997
1027
it Champagne, j
0997
1060
Brandy, French, . .
0941
0792
Rectified spirit,
0838
0800
0916
GASES
1490
Atmospheric air, . .
1000
1034
Ammoniacal gas,
0590
0716
Carbonic acid gas, . *
1527
1060
it oxide gas,
0972
1200
Carburetted hydrogen
1032
gas, ....
0972
1420
Chlorine gas, . . i
2500
0940
Cyanogen gas,
1805
0915
Hydriodic Acid gas,
4340
0923
Hydrogen gas, ,
0069
0885
Iodine, vapour of, .
8716
0910
Nitrous oxide gas, . j
1527
1843
Oxygen gas, .
1111
0868
Prussic acid gas, . .
0937
1043
Steam of water at 212, .
0623
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 999278 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 = 624 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. =3323 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
2956 ounces in water ; what is its specific gravity?
* For ordinary calculations the weight Of a cubic foot of fresh water is assumed
=to 625 Ib. or 1000 ounces.
RELATIONS OP WEIGHT AND VOLUME OF BODIES 257
ws' 33 6s' 33 '6s' .
* = ;^7 = 3362956 = ^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" = ww'', and hence w" : w =s':s,
or ww' : 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 + 4025 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=& = xj 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 firplanks 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 = 12288 cubic inches.
lUl/US liO U
EXERCISES
1. How many cubic feet are in a tonweight of male fir ?
= 651636.
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 ;
(Ss')s,, r , , (sS)s'
then w=] jTrrW, andw/ = ; ~ W.
(ss)S (ss)S
EXAMPLE. A composition weighing 56 Ib., having a specific
gravity 8784, 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 1464x9x56 13'176 _ _.
4 76712
and hence w'=Ww=5Q50 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 6154 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 sidewalls
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 = 15708,
and V=JRl = 157 '08.x 60 = 94248 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 ?
= 170281218 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 = 31416x 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 ? =21631353 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 = 46764 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=(2016) a xlx 2= 2133
C = ah. . = 16x12x2 . =384
Hence V = A + B + C . . =58126 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.
= 3077946 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=62484 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 ......... =21696 cubic feet.
523. Problem VIII. To find the surface of a groin.
RULE. Multiply the area of the base by 11416.
This rule will give very nearly the surface for circular and
elliptical groins of small eccentricity.
s = 114166.
EXAMPLE. Find the surface of a circular groin with a square
base, whose side is =12 feet
5 = 114166 = 11416 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
gaugepoints.
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 gaugepoints.
The gaugepoints 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 gaugepoint ;
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
gaugepoints.
The gaugepoints 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 gaugepoints.
The gaugepoints 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 gaugepoint, 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 gaugepoints 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 gaugepoint 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 gaugepoints 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 gaugepoint 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 gaugepoints.
TABLES OF MULTIPLIERS, DIVISORS, AND
GAUGEPOINTS
I. FOR PRISMATIC VESSELS WITH SQUARE BASES.
Measures
Divisors
Multipliers
Gaugepoints
Inches in the area of unity,
1
1
1
Superficial foot,
144
006944
12
A solid foot,
1728
000578
4157
Imperial gallon, .  .'
277274
003607
1665
ii bushel,
2218192
000451
471
A pound of hard soap,
2714
036845
521
ti ii dry starch, .
4030
024813
635
n ii green glass, .
1218
082102
348
GAUGING
269
II. FOR CYLINDRIC VESSELS
Measures
Divisors
Multipliers
Gaugepoints
Inches in the area of unity,
127324
785398
1128
A superficial foot, .
18334
005454
1354
A solid foot,
220016
000454
4691
Imperial gallon,
35304
002833
1879
ii bushel,
282429
000356
5314
A pound of hard soap, .
3565
02805
597
ii it dry starch, .
513
019491
716
H M green glass, .
155
064516
394
III. FOR REGULAR POLYGONAL PRISMATIC VESSELS
Measures
Divisors
Multipliers
Gaugepoints
PENTAGONAL BASE
Imperial gallons,
ii bushels, .
161161
1289288
006205
000776
1269
3591
HEXAGONAL BASE
Imperial gallons,
it bushels, . .
106723
853782
00937
001171
1033
2922
HEPTAGONAL BASE
Imperial gallons,
ii bushels, .
76302
610414
016106
001638
873
2471
OCTAGONAL BASE
Imperial gallons,
it bushels,
57425
459403
017414
002177
758
2143
IV. FOR CONICAL VESSELS.
Measures
Divisors
Multipliers
Gauge points
Imperial gallons,
ii bushels, . .
1059109
847287
000944
000118
3254
92049
270
GAUGING
V. FOE SPHERICAL VESSELS
Measures
Divisors
Multipliers
Gauge points
Imperial gallons,
it bushels,
529554
4236434
001888
000236
2301
6509
VI. PRISMOIDAL VESSELS, FRUSTUMS, OR CYLINDROIDS
Measures
Divisors
Multipliers
Gaugepoints
WITH SQUARE ENDS
Imperial gallons, .
it bushels, . .
1663644
1330915
000601
000075
4079
11536
WITH CIRCULAR ENDS
Imperial gallons,
ii bushels,
2118217
1694574
000472
000059
4602
13017
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~ 277274
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 _
m2218192
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. =14426.
3. Required the content of a parallelogram in pounds of hard
soap, the length being = 84 inches, and the perpendicular breadth
= 32 inches. . = 9904.
4. What is the area in imperial bushels of a trapezoid, the parallel
sides being=60 and 145, and the perpendicular breadth = 80 inches?
= 3698.
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. . . . '.... .... =22947.
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 ^ = 5877 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 16236.
2. What is the content in imperial bushels of a cylindric vessel
whose diameter is =48 inches, and depth = 64 inches?
=522 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.
=413.
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? . . =4496.
6. Find the content of a conical vessel in imperial gallons, the
diameter of its base being = 60 inches, and its height = 60 inches.
= 2039.
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? =9605.
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? . ', . . . =1041.
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 onesixth 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 onesixth 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= = = 8024 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 = 246109 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? ..... =9229.
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.
= 12812.
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? .......... =10939.
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 ? . = 14255.
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 mashtuns, 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
688
694
10
13
72
722
721
10
23
736
735
735
10
33
74
738
739
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 ,. =1473x10=1473
ii 3rd =153 x 10 =1530
4th n =1547x10=1547
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 mashtun or still, for every inch in depth.
The contents for the first inch of depth from the bottom of the
preceding vessel is 1547; 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 1547 ; 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 flatbottomed 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
12 544
10 519
10 496
10 473
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 \
\552 548/
9 598 602 9177
9 638 644 10474
9 64 646 1054
105 62 624 11507
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 1059108,
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 )Lr 1059 108 = (2 x 32*+ 24 2 ) x 40 r 1059*108 :, >,\ '
= (2048 + 576) x 40j 1059'108 = 991 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 1059108, will give the
content in imperial gallons.
C = {2B 2 + H 2  (B  H) 2 }L =1059108.
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 1059108
= (2x32 2 + 24 2 f x8 2 )x 40 f 1059108
= (2624256)40r 1059108 = 981 imperial gallons.
EXERCISES
1. Find the content of a cask whose Lung and end diameters
are =48 and 36, and length = 60 inches. . . . . =33124.
2. What is the content of a cask whose bung and head dia
meters are = 36 and 20, and its length =40 inches? . . =10914.
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 7060724 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 7060724 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 10591086 for the content.
C = (B 2 + BH + H 2 )L f 10591086.
For this is the formula of Art. 386, except that, instead of the
factor 2618 or Jx7854, 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
10000
10000
10000
10000
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
353036, or multiplied by 0028325, is the content in imperial
gallons; or,
C = D 2 Lf 353036, 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 = 2814,
and C= 0028325D 2 L= 0028325 x 2814 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, 8944.
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
=0^353036, 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 353036 = '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
100
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
= 949, 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)L43177325.
EXAMPLE. Find the content of a cask whose diameters are
= 32 and 24, and length = 40 inches.
C = (39B 2 + 25H 2 + 26BH)Lf 3177325
= (39 x 32 s + 25 x 24 2 + 26 x 32 x 24) x 40 r 31773 25
= 935 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(Dd+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 = WfB,
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=WfB = ^='3125, D = 5 3125= 1875, D=0469;
hence U = (R  JD)C = ( '3125  0469) x 98 = "2656 x 98 =2603.
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=WrL = !=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.
MALTGAUGING
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 couchframe, 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 floorgauge, to determine
which is the greatest or duty gauge.
RULE. Multiply the greatest cistern or couch gauge by 12 if
the floorgauge 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 floorgauge, it is to be taken for the
dutygauge, otherwise the floorgauge is to be taken.
Since the cistern and couch gauges are each to be multiplied
by the same number 12 to obtain the floorgauge, 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 floorgauge = 205*2 bushels, which will afford the greatest
or duty gauge ?
The couchgauge exceeds the cisterngauge ;
hence 132 x 16 = 2112.
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 floorgauge 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, 59, 6'4, and 5'5.
and c = 48x32x6x 000451 =4156 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
=468 inches? ......... =3647.
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 = 216, 223, 229, 234, and 2355? .... =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.
=4236.
THE DIAGONAL ROD
563. The diagonal gaugingrod 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 bunghole to the opposite side of the
head (BH, fig. to Art. 557) that is, to the part where the start'
opposite to the bunghole meets the head ; then the number on the
rod at the bunghole 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 narrowbored 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 freezingpoint is marked 32, and
the boilingpoint 212, the scale is called Fahrenheit's ; when
the freezingpoint is marked 0, and the boilingpoint 100, it is
called centigrade ; and when the freezing point is marked 0, and
the boilingpoint 80, it is called Reaumur's.
Reaumur's thermometer is not in use in any Englishspeaking
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 32808, 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 32809 = 1 1358 "47.
EXERCISES
1. How many imperial feet and fathoms are contained in 6254 '6
metres? =205207 feet, or 34201 fathoms.
2. In 7645 metres how many imperial feet ? . = 2508248 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 604524 = 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 sealevel 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 + (f32)*'
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 (tf) 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 + if fs (t32)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 ? ..... =528125.
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' + iQ&rijdp', or 1 L(p l p') = Ld + Lp'  4 ;
h" = l(mO(LpLp 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 =31725, t =6575, T =60 '05,
p' = 2784, 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 . . =15014016
L/J! . . =14450746
0563270, or 563 27= h"
563 = 3547
59874 = ^
Then 6^ . = 3592 '44= h'
. = 599
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 s32, 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 (LpLpj), 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 = LpLj'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
476943
75
478465
105
479890
135
481268
45
993
76
513
106
936
136
314
46
477042
77
562
107
983
137
359
47
092
78
610
108
480030
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
479040
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
482031
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
478029
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
481042
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
999996
59
999944
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
999899
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=
25
52
107
167
23
298
369
444
522
605
692
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
19
29
3'8
48
58
67
77
8'6
96
EXAMPLE. Find the altitude for the observations in the example
of the first method, supposing the latitude to be = 55.
p =31725, t =6575, T =6005,
2? &1 "o4 * 54 *o , A i)U * / o y
Hence . . s =120, d = 9 '3, and Z=55.
L.^> . . =150140 And . L.R= 275074
L.y . . =144467 A =480584
05673 = 999960
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, AJ = 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 13.
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 = s32, 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 hmk 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 hmk 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 freezingpoint 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=TT". 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 ^Jj 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 =3265 lines, T =7 '6, t =7 '8,
p' = 3178 T' = 64, i!' = 62, 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 =30295 inches, T =75 '5, t =75 '5,
/>' = 27064 T' = 60l, 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 =73558 centimetres, T =18625, t = 19'125,
y = 53'72 T'= 975, t' = 4, and Z=43.
= 85799 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 =76315, T =t =253,
2/ = 60'095, T = t' =213, and 1 = 21.
The height is 6846 '2 feet ; but if only  of the correction c + c' is
taken (Art. 579), the height is = 68384 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 =27267, T =t = 226,
p'= 16042, T = t'=  23, 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 =762, T =253, t = 25 '3,
j' = 37727, T' = 10, t'=  16, 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 twohundredth 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 fieldguns, . . . 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 + 114^.
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+ 114 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 + 228 = 11128,
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 thundercloud 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 = 1358616 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 = 90HBR is known.
In the triangle ABT, rightangled 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 5633284, .
= 114497317
= 03010300
L, sec B 87 58', .
L, AB 2, .
L, BT 5636834,
= 114500052
= 03010300
117507617
10
117510352
10
= 17507617
= T7510352
and BH = AT + BT = 1 127012.
2. To find CH in triangle BCH
L, tan B 87 58',
L, BH 1127012,
= 114497317
= 20519285
135016602
10
35016602
L, CH 317439, ....
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. =7914826 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, rightangled 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 39576015, . = 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 39576, . = 35974321
L, AS 11257, . = 20514349
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 = 11257.
The tangential distance of the horizon AS = 11257
S= 1023
The actual distance .... =1228
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
= 18015 = 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' 1812",
= 35972563
= 22174839
10
L, sin E,
L, AS 165, .
L, ES 3959437
BE = 3956
BS= 3437
. = 86198504
10
. = 22174839
= 86202276
= 35976335
miles = 18147 ft.
EXERCISES
1. The distance at which a mountain is visible at sea is = 142
miles; required its height ...... = 2143 miles.
2. The distance at which a mountain is visible at sea is = 120
miles ; required its height ...... =1852 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(ee') = 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
6905 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 halfaminute 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 =2017 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=2017 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)=18090 33' 29" = 89 26' 31".
L, cosec C 89 26' 31", . 10 0000206
L, sin A 23' 24", = 7 '8329386
L, AD 2017, = 13047059
L, CD in nautical miles, . . = T1376651
L, 6076 feet (in a mile), . . = 37836178
L, CD in feet 29212829
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
= 8342 + 240 = 10742.
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 = 432, 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
CR
D
C
CR
D
C
CR
1
66
57
6
24
2057
12
96
82
2
267
229
7
3267
2800
14
130
112
3
6
514
8
4267
3657
16
170
146
4
1067
914
9
54
4630
18
216
185
5
1667
1429
10
6667
5714
20
2667
2286
D= distance in statute miles, C = curvature in feet = D 2 approxi
mately, CR= 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?
=98 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 spiritlevel (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 spiritlevel 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
airbubble at B is at the middle point.
The plumbline 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 fluidlevel 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 = 1176 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 + 83790798 = 2 x 3 '5490033 + 8 3790798
= 70980066 + 8 3790798 = 1 4770864 = LO 299976 feet,
or 35997 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. 11768 in.
Therefore ACBD = 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 = 106 = 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
42
T5
2
B 345
C 342
23
57
3
C 500
D 504
21
39
4
D 1285
E 1280
95
42
2330
2326
181
153
2326
153
4656
28
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 = 25 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
25
2
B 542
C 542
43
32
3
C 253
D253
27
85
4
D751
E 753
74
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 indiarubber, 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 dieplate 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 meltingpoints
of a few of the commoner metals :
MELTINGPOINTS OF METALS IN DEGREES FAHRENHEIT
Mercury,
Tin,
Bismuth,
Lead,
Zinc,
: .  3
\  +440
. 500
. . 600
. 700
Coppeiy. . .
German silver,
Gold, .
Castiron, . .
Steel, . ...
2000
2000
2000
2200
2500
Antimony,
Brass, .
Silver, .
. 800
. 1800
.. .... 1850
Nickel, also Aluminium,
Wroughtiron,
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 semiaxes) 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, tierod, 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 225 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
Gunmetal .
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
Wroughtiron bars,
25
17
20
5
34
4
ii M plates,
224
17
20
44
34
4
Iron wire cables, .
40
8
Castiron,
?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 = rj 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 wroughtiron tiebar, 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^= 66528 footlb.
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 ~ = 66528 footlb.
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.
Castiron, . . 00018
Wroughtiron, . 00010
Steel, . . . 00008
Teak, . . . 001
Oak and Pitchpine, '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, threefourths 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 fourfifths of the
breaking weight to cause the column to bend at all.
In castiron 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 castiron is 7 tons per square inch,
,i it ,i wroughtiron 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 rivetholes 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
Wroughtiron, .. .. ,. . '' ' . ' ^ . . 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
castiron 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 castiron 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 castiron 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 castiron 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 castiron, 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 castiron, . . K = 500 K = 850
wroughtiron, . 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 castiron, 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 onefourth of the value of K.
RULE. Multiply the cube of the diameter, or side if square,
in inches, by the value of Kr4, 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 jL;
fD 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 castiron, 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 = \(dd') = %(\ 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?
tPlW' 500(1 n i 12x32000  12x64  768 7680000
U '" 500(1 1 4 )~1 0001" 9999 ~ 9999
= 76808, and d= ^76808 =915 inches;
hence d'=nd='l x9'15= '915 inch,
and <=(! )d=i(l  '!) x 9'15= x 9 x 915=41175.
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
, ,, ,31416
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 castiron, 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 D20 inches,/ 6000 Ib. per square inch, d=8 inches.
Then, as T.R. =T.M. = . 5 ./,
.
. . T.M. = F x ?r x 6000 = 9183525 inchlb.
lo <
EXERCISES
1. Find the resistance to torsion of a solid shaft of castiron
whose diameter is 5 '25 inches, with a surface stress of 4500 per
square inch. . . ..... = 127820'745 inchlb.
2. Find the torsional moment of resistance of a wroughtiron
shaft (solid) whose diameter is 6 inches, the surface stress being
8000 Ib. per square inch. . .' . . =339206 '5 inchlb.*
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 steamengine crankshaft, 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 horsepower,
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. inchlb., or ^ T.M. footlb., and the angular velocity
of the shaft is n x 2?r radians per minute. Therefore the work done=
T M
' x 2irn footlb. per minute,
KM
and the horsepower (H.P.) =
T.M.
12 XZim wxT.M.
33000 " 63024
T.M. =63024.
n
EXAMPLE 1. Supposing it was required to find the horsepower
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 inchlb.,
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 horsepower, what should be
its diameter so that the maximum stress produced in it may not
exceed onefifth 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 inchlb. 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 wroughtiron is from 8000 to 10000 Ib.
per square inch, and for castiron 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 inchlb.
We employ formula [1], Problem VI., making
25000 __
= 251 inches.
EXERCISES
1. Find the diameters of the following wroughtiron shafts,
whose combined bending and twisting moments are respectively
in inchlb. (take/=8000 Ib. per square inch) :
42390; 82793; 120522; 196250.
=3"; 3"; 4J"; 5".
2. Find the diameters of the following castiron shafts, whose
combined bending and twisting moments are respectively in
inchlb. (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 . 102(T.M.)L ,. 'j
For solid shafts, 6= p,^ . radians,
, 584(T.M.)L
and for hollow shafts,
CD 4
102(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 horsepower 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 inchlb.
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 horsepower 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 horsepower.
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 horsepower 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 castiron, wroughtiron, 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. Castiron 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
236 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 castiron shaft
whose diameter is 2 inches.
,,, 236 xeZ 3 236x8 1888
W = = rr = ^r ='11 ton, or 246 '4 inchlb.
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 castiron, 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 wroughtiron, and 4000 to 5000 Ib. for castiron.
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 castiron shafts half the numbers to be taken.
EXAMPLE. Required to find a shaft for a drum having 2J tons
pulling on it at 17inch 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 4inch diameter, which will be the size of shaft
required in wroughtiron.
If for castiron shaft, and /= 4000 Ib., then 5inch diameter is the
IQfiOKA
size, since  = 98125, or 95200x2=190400 Ib., and the next
X
higher number in the Table = 196250, opposite 5inch diameter.
626. Problem IX. To find the weight that could safely
be supported by a column of castiron 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 castiron columns, J breaking weight.
ii wroughtiron structures, J
ii castiron 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
castiron, . ) 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=1095p
T)4
W= 781
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.
36 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 castiron
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
D365
W = 4416^ 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 castiron 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 castiron 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 castiron 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
= 441787 tons for rounded ends ; but as the ends are flat and fixed
in the question, we multiply by 3.
.. the breaking weight = 1325361 tons.
Note. By referring to Art. 613 we find that =4r.
T Cl
Here = 33 feet = 396 inches, and d=7~5 inches;
7 one
. . 4^=4 x ^ = 4 x 528=2112, which squared = 4460544.
5. Find the breaking weight of a solid cylindrical column of
castiron 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 castiron, whose ends are
flat and fixed :
When ^=10 42 Ans. When ^= 40 5'2 Ans.
d d
=15 37 M it = 60 23 n
=20 21 it =120 06 ..
n =30 92 ,
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
Lignumvitae, .
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
635
ii ii ti cable,
025
15
207
828
White Manilla hawser,
045
27
177
3'93
H ii cable,
033
19
155
47
Best hemp, ' cold register,'
100
60
it warm, ....
116
70
Iron wire rope,
290
18
87
29
Steel wire rope,
450
28
89
191
EXAMPLE. Find the breaking strain of a 4inch common hemp
rope.
EXERCISES
1. Find the breaking strain of the undermentioned size ropes :
5J" common hemp, 4" steel wire, 6" St Petersburg cable.
=5445, 448, 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.
= 127 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 + 100218 =
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 counterlever to the lever.
P = power, W = weight, CL = counterlever, 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 counterlever 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 38ton 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
selfevident 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(La;)inchlb.
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.) =
wxBDxBD = ^>xBD 2 =iw(Lo;) 2 inchlb. , [B],
and S.F. =wL lb., \
B.M.=iw>L 2 inchlb.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(Lo:) inchlb. . . [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 inchlb.. . . [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 = 3125 foottons,
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 (125  65) = 15 foottons.
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 inchlb.
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(LBD)
'  a; 2 ) inchlb.
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 inchlb.
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 foottons.
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 =yW = ^x6 = 16 foottons.
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 baseline ; also a shearing force of 1 ton drawn
above the baseline. The resultant shearing force between the
loads is therefore the difference of these, and is drawn on the same
side of the baseline 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 inchtons ; S.F. =05 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 footlb.
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.
= 105 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 footcwt. ; 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
= 47CT 2
< B
Hollow Rectangle
Ellipse
STRENGTH OF MATERIALS
371
Circle
I=7854R 4
V = 47R 3
S 4
: 12
Square
Hollow Circle
I=7854(R 4 r 4 )
VW*^
v ~ 4 '\ R y
I
60
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
1666
12
it cast,
092
1598
8
Antimony, cast,.
242
4195
47
Bismuth, cast
353
6131
145
Copper, bolts, . .
318
5524
17
it cast, ....
31
5373
84
it sheet, ....
316
5481
134
ii wire, ....
32
555
26
Gold .
665
1150
9'1
( from
252
437
6
36
2
Iron, cast, . . 1 to
273
4744
13
64
34
ii M average, .
26
451
7'3
48
26
( from
273
4744
16
16
3
,, wrought, \ to
281
4869
29
18
55
ii ii average, .
28
4856
22
169
3'8
it wire, .
40
'408
7085
8
31
ii sheet
41
7116
15
Mercury,
49117
84875
775
13439
ii sheet, . . .
_
Silver
377
6538
182
Steel, . . .
288
499
52
150
ii plates, ....
35
90
Tin, cast, .....
262
4551
2
67
252
437
33
ALLOYS
Aluminium bronze, 90 to 95 per )
276
4784
32
58
_
cent, copper, . . )
Bell metal (small bells), .
29
50252
14
Brass, cast, ....
3
52437
8
ii sheet, ....
361
52686
14
ii wire, ....
307
533109
22
ii 5 copper, 1 zinc,
3
52509
137
ii 4 ii 1 ii .
304
52736
147
3 ii 1 ti . .
3
52418
131
ii 2 ti 1 ii
299
51806
125
, 1 ii li
296
51375
92
.1 1 M 2 II . .
298
51706
193
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
40013
19

Gold (standard),
638
110642
Gun metal, 10 copper, 1 tin,
306
52836
161
.. 9 ii 1 ii
305
52824
152
ii 8 ii 1 ii
305
52805
177
7 1
305
52789
136
Pewter,
Silver (standard),
371
64372
Speculum metal,
264
46487
31
White metal (Babbett),
263
45632
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
Lignumvitse, ....
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
_
_
Guttapercha, ....
Gypsum, .
035
082
60
143
71
Indiarubber,
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 castiron, 13000 for steel, and 11000 for wroughtiron ; 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^TpT ?
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 castiron 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 castiron 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 castiron
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 35inch span ; find the deflec
tion (take =42000000) ....... =458 inch.
4. Find the deflection of a wroughtiron 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 wroughtiron 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 wroughtiron 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.
= 1319 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 = 329 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 = 13 inches.
12. Find the breaking weight of a castiron 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 Castiron 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 castiron 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 50x75x10 .
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.
625 n
9479
30
35
30
35
15x2
17x2
1250
1416
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 Avroughtiron 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
*
OO50OITCDIOJCOOOOOO5
(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
FH <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 castiron, = 001 for teak.
E= 00010 M wroughtiron, = '002 M oak and pitchpine.
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 '
= 181466104898
= 76567.
.'. Height of neutral axis from lower edge
of section = 10'346=N, and moment of inertia
= 76567.
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 60lb. doubleheaded
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 = 10267 tons, Z = 33 3 = 35937 inches,
(MR)=6'7, </=45 inches,
W/ 3 1 0267 x 35937 36896 '5
383
24x67x45 723'6
509 x E = 509 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 84lb. rail shown in the fig. when
loaded with 2000 Ib. at the centre of a 60inch span ; find also the
strain on the extreme fibres, the depth being 4J inches.
For this particular section (MR)
= 249x35.
Note. The strain on the ex
treme fibres is given by the
" tl
i
1
*
lyio
1775"
tons per square inch, W in tons, ^i~
I in inches.
W/ 3 93 x 216000
1213*
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 1095
\
7 12 "
1
f i ' 1 T"
' 4(MR) 4x8715 8'715
& n "
diLIZ "
i2
iii^..o?
= 125 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)
)
175
306
535
175
3D6
535 f
12
21
3672
642
1775
31506
569231
1600
812
558696
15
24
46800
838044
1925
37056
713328
150
555
154097
4
6
22200
616388
51
72672
1460852
=
2BAH
2BAH2
2BAH3
= A
XT SBAH 2 72672 _ __ . .
N= 2^ ^02 = 7'12 inches,
=228431.
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
228431.
For tension
a. ^ i l 228431 oon
the modulus = = _ = 320.
228431
= 188.
it compression M ^^
a; 2 1213
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 188x75 = 1410
.. bending moment = resisting moment =800 inchtons.
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 castiron 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 =625 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
1855
1
. NEUTKAL
2
6'
AXIS
1145
V
,,,  
386
STRENGTH OF MATERIALS
 //
i*f"'"".::ft ."
when loaded at its centre. What will the maximum compressive
stress be? = 132 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)
. 55641 2816
. 50485 26731
. 24234 21731
. 3635 24423
2865 2487
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 inchlb., 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 11865 x 1855
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 9068275
it it compression =z c == .O.KK =4888.
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 15 = 1186 '5 inchtons, and the
compressive stress = 488 8 x 9 = 4399 '2 inchtons. 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 inchtons.
The girder will therefore carry a central load given by the
equation WL (see Prob. XII. p. 364).
r 4x11865
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
castiron 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 inchlb. ; greatest stress = 424'l Ib. per square
inch.
656. To find the strength of thin wroughtiron 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 wroughtiron 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 inchtons.
. . 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 nonelastic 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(WK>)
f W + w
WV + wv
W+w
WV(I + k) + v(wJcW)
W + w
v V(WKw) + vw(l + K)
y ~ W + w
WVwv
W + w
WV(I + k)v(wkW)
W + w
v V(WKw)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)
TPcP
.^(D 4 ^
16
82 \ D /
394
MOMENT OF INERTIA,
The plane of bending is supposed perpendicular
Form of Section
Area of Section
H
EUbh
f
KB  *
EHbh
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*
vg
*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,
#=322 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 322 x 5 2 =402'5.
EXERCISES
1. What velocity would be acquired in falling 120 feet?
= 879 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 902<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 e20.
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 cannonball 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 90e=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 + 10Lsin 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 322x5 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? .. =1123 seconds.
2. What is the time of flight for the greatest range for any
impetus? . ... . . . t= \/ ^\/r nearly.
ILPRACTICAL 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
vlQQQJj 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
= 400x1732 = 6928.
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 ? ..... =92376.
2. With what velocity will a 48lb. ball be impelled by a charge
of2lb. ? . . . ...... =632456.
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 nonresisting 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 725= '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 12lb. 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 castiron 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
171
5
1
194
18
509
110
2
24
561
3
28
295
6
37
3
32
621
4
308
42
675
6
352
47
7
87
4
70
8
9
404
100
9
12
445
'
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 CASTIRON SOLID CYLINDERS IN LB.
Length of cylinder=l foot
Weight
Diameter
Inches
Weight
Diameter
Inches
24
1
89
6
99
2
120
7
219
3
156
8
390
4
198
9
610
5
EXERCISES
1. Find the terminal height for an iron ball = 6 inches in diameter.
= 5400.
2. Find the terminal height for a 3lb. 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 1806180.
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
13
21066
33
138258
02
0201
14
23646
34
150377
04
0405
15
26422
35
163517
06
0612
16
29413
36
177767
08
0822
T7
32635
37
193229
1
1034
18
36107
3'8
210006
12
1249
19
39851
39
228218
14
1468
20
43890
40
247991
15
1578
21
48249
41
269465
2
2140
22
52955
42
292792
3
3324
23
58036
43
318138
4
4591
24
63526
44
345686
5
5949
25
69460
45
375632
6
7404
26
75875
46
408193
7
8964
27
82813
47
443605
8
10638
28
90319
48
482127
9
12436
29
98442
49
524040
10
14366
30
107237
50
569653
11
1 6439
31
116761
12
18669
32
127078
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 + LR2Li>.
EXAMPLE. At what elevation must an 18pounder 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 509=4581,
7H57 = ' 64 and R = FR'=4581 x 8028 = 3678,
.r
LA = 2(Lv  903090) = 2(2 992995  903090)
=2089905x2 = 4179810, and A = 15129.
L sin 2e=10 + L3678L2A = 135656124480840=9084772,
and 2e = 6 59', and e = 3 29'5'.
EXERCISES
1. At what elevation must a 12lb. 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 = 3139977 + 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 24pounder 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'= 5049x2587 = 13060,
LA = 2(Lv  903090) = 2(3 '238046  "903090) = 2x2 334956
= 4669912, and h = 46764,
and 2h = 93528, which exceeds 39000.
Then L sin 2e = 6 8389469 + LR"  JLA
= 68389469 + 41159432  15566372
= 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 24pounder 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 + Leosfe10,
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 musketbullet, 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
= 3015649, 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 = 10367 x 315 = 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 1806180 (Art. 689) ; then
LR" = LRLv + 4064143 .... (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'= 10367.
And LR" = Lv + L sin 2e 7441007 = 2214813 + 9698970 7'441007
= 11913783  7441007 = 4472776 ;
hence R" =29701.
LR' = LR"  LF' = 4 472776  3 015653 = 1 457123,
and R'= 2865 ; hence, by Table, r' = 417303,
and r = FV = 10367 x 417303 = 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 42lb. ball, discharged with a
velocity of 1800 feet, at an elevation of 36. . . =15413 feet.
2. What will be the actual range of a 24lb. 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',oiv:v' = \/d:\/d'.
THE FLAT TRAJECTORY THEORY
692. This theory will be recognised as bearing especially
on modern guns and rifles, for if the pointblank 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. Airspacing in the
powderchamber 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 muzzleloading guns with
out gaschecks. 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 ribbonwire, 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 halftime, 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
FootTons
No. of
Expansions
Work
per Lb. burned
FootTons
No. of
Expansions
Work
per Lb. burned
FootTons
125
19226
55
95210
11
121165
15
31 986
6
98638
12
124239
175
41494
65
101 744
13
127036
2
49050
7
104586
14
129602
2'5
60642
7'5
107192
15
131 970
3
69347
8
109600
16
134108
3'5
76315
85
111840
17
136218
4
82107
9
113937
18
138138
45
87064
95
115905
19
139944
5
91385
10
117757
20
141647
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 125 expansions = (91 '385 19 '226) foottons
= 72159 foottons.
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
pr foottons 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 wroughtiron 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 steelfaced 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 magazinerifle fire at all ranges :
Earth parapet free from stones, not rammed, 24 inches.
Clay, ........ 24
Fine loamy sand, ...... 20 n
Wroughtiron or mild steel plate, ... inch.
Fir, dry or green, ...... 38 inches.
Elm, green, ....... 36 . n
Oak, , ........ 24
Sandbags, 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
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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
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222sS:b
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Cl'T'lCOCOCOCOCOCOCOCOCOCOCOCOCOCOCO
01CMCMCNCMCMCMCMCMCMCMCMCM<M<MCMCM
CM C^J CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO
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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 *
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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
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CM 6l CM C>1 CM CM CM CM CM CM CM CM 'M CM CM CM CM
S'MCNCOCOCOCOCOCOCOCOCOCOCOCOCOCO
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^ioioicocococorocofocococococococo
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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?
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a
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^   (N(
igxosxccecceccppccpTicoipiffl'
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xo5OiiioiiccccC'* ; *co'O>t5coc6i
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xxxoscccccpTip
O il X * 6 c: : ~ i rc x >h cb CO X
rcccc ~i^ TCCI.CCXH 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 Ql t^ p CO
cc cc * ^ * T^ *
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i x M >H t> H x ?j i^ '*5 ci cc cc c: ~i ic
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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
ogivalheaded 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(SvSv),
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 =568.
The initial velocity Sv=2040, and the number in the Table
corresponding to this velocity is 45350 '3.
.. Sv=453503.
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 447163.
.. Sv=447163.
.. s range = C(SvSv)
=568(453503  447 163 3)
= 568x634
= 360112 feet, or 120037 yards.
Further, let the time of flight of the abovenamed projectile be
also required.
The given muzzle velocity =2040 feet per second, and the number
in Table A corresponding to this velocity is TV =233567.
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 wrnd?, and this has already been
found =5 68.
.. = time of flight of projectile in seconds
= C(TvTv)
=5 68(233 567 233 242)
=568x 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 wroughtiron
armourplating.
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 foottons.
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 = 1846
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 = 568.
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
= 233242;
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 powderchamber. The prevention of gas waste or escape in
a breechloading 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 muzzleloading 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" breechloading 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 = 3616
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 wroughtiron and steelfaced
armour.
Wroughtiron, 3^002 inches ; steelfaced armour, 2'401 inches.
4. The muzzle velocity of a 3pounder Hotchkiss quickfiring
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 powderchamber a tube,
430 PROJECTILES AND GUNNERY
breechpiece, 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 powderchamber ; r lt r 2 , and r 3
the outer radii of the tube, breechpiece, 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 gunsteel, 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 wroughtiron 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,
PFPO
For the longitudinal strength of the above gun (where the
breechscrew gears into the breechpiece), 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 breechloading rifled guns,
such as the 16 '25inch and 13 '5inch, 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 primemotor.
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 allimportant 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 wellknown 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 5inch 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 = 2xxlO458.
1800 2
. '. tt= oo^qTft = 154905335 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 154905335=242039595, &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 horsepower.
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 semitangent 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 RIGHTANGLED
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 OBLIQUEANGLED
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 = 18059 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 RIGHTANGLED 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 rightangled 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 rightangled 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 rightangled 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 + ca)
sin b sin c
/sin s . sin (s  a) *
(*r\c\ 1 \ / . v '
'V sin o . sin c
1T> /sin s . sin (56)
^ 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 + 6c) sin ^(a + c6)
V sin o . sin c
in 31 A /sin(.96).sin(scn
'y sin o . sin c
. n 1T} /sin (s  rt) . sin (s  c)
V sin a . sin c
in /sin (,?). sin (s &)
biUjjCy 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
(sc).
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 semiangle,
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
semiangles 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(sC)
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 (sB)
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 (SC),
2 cos (sB)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 (sb) sin (s  c), then
V sin s
AM B M
tan 7r = r.tanT^ : rr>
2 sin (s a) 2 sin (s  b)
and tan ^ = . ; ; hence
2 sin (s  c)
M M
A + B_sin (sffl) sin (sb)
0~~ M2
1
sin (s  a) . sin (s  6)
5 . sin (sa) . sin (sb) . 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 (ab)
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
AB / sin s . sin (s  c) sin (sb) sin (s  a)
tan ^ = . /^ : . , ' x 
J_ / sin s . sin
A/ sin (*). si
sin(s6) sin s + sin (sc)
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(a6) = 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
rightangled triangle, to find the other parts.
EXAMPLE. In the spherical triangle ABC rightangled 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 . . . . = 96881818
+ Radius ..... = 10
+ CosC46 ..... = 98417713
+ Tan BC 54 55' 358" . . = 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 ABsin C . sin AC ; or, since AB is required,
R : sin AC = sin C : sin AB.
Radius ...... =10
Sin AC 64 . . . . , = 99536602
Sin C 46 ..... = 98569341
Sin AB 40 16' 52" . 98105943
Prac 2 D
454 SPHERICAL TRIGONOMETRY
The sine for 98105943 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 . . ... . = 99848372
Radius =10
Cos AC 64 = 96418420
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' .... = 95515524
Tan BC 65 10' .... = 103346338
Cos C 39 41' 40" . 98861862
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' . . . . = 96232287
Radius =10'
Cos AC 70 24' .... = 95256298
Cos AB 36 59' 27" = 99024011
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"
99740774
= 10
= 99578626
99837852
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 = 99023486
Cos BC 65 10' . . . . = 96232287
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* . . . = 103346338
Radius =10'
SinAB37 ..... = 97794630
Cot A 74 26' 145" . 94448292
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 = 98771144
Radius = 10
Sin BC 65 10' 99578626
Cot C 39 42' 14" . . = 100807482
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' . . . . = 101958127
 Cot C 106 24' (73 36') . . = 94688139
Cos AC 11 7 30' 55" . 96646266
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' . . . . = 94507747
 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' . 99819608
+ Radius =10
+ CosA3230' . 99260292
+ Cos BC 28 27' 31" . . . 99440684
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' 303".
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' . . . = 99101766
+ Radius . . . . = 10
 Cos C 105 53' . . . . = 94372422
+ Cot AC 71 23' 553" . . = 95270656
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' . . . . = 94541479
+ Radius =10
+ Sin BC 140 53' . 97999616
Tan AB 114 16' 33" = 103458137
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' . 98897850
 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' 543", 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' . . . . . = 98489791
Radius =10'
Sin AB 40 25' . 98118038
Sin AC 66 37' 48" . . . = 99628247
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' . 98815842
Radius =10"
Cos C 44 56' = 98499897
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' . 100010107
Tan AB 40 25' .... = 99302195
Sin BC 58 36' 0'6" . 99312302
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 rightangled trigonometry ;
the right angle is A.
Elements
Log. Sine
Log. Cosine
Log. Tangent
a = 71 24' 30"
99767235
95035475 +
104731759 +
b =140 52 40
98000134
98897507
99102626
c =114 15 54
99598303
96137969
103460333 
B = 138 15 45
98232909
98728568
99504341
C = 105 52 39
99831068
94370867 
105460201 
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 rightangled 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 rightangled triangle, E being the right
angle.
Quadrantal triangles can therefore be solved by the rules for
rightangled 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 rightangled 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".
OBLIQUEANGLED SPHERICAL TRIGONOMETRY
765. The number of cases in obliqueangled 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 halfsum 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 (sb) sin (sc)\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 halfsum, 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 (sa) . 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 (sa). = 9 '6703000
then ,=171 40 30" Lsin(*&) . = 99863791
= 27 54 30 Lsin(sc) . = 98034339
s6 = 104 16 30 2)402993806
sc= 39 29 30 201496903 =
C  L sin (s  a) = L tan A = 104793903
CLsin (,6) = L tan B = 101633112
 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' 546", 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 halfsum 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(SB)cos(SC)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 halfsum, 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
SA= 46 1
SB= 77 19
SC= 37 11
L sec = 100256087
Lcos = 98416404
L cos= 93415580
L cos= 99012980
2)391101051
= 195550525
Log. cot = CL cos (SA)= 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 halfsum and halfdifference 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' 444".
2. To find J(A~B)
Sin
Cot
Tan
i(a6)
= 103618336
= 99729690
= 104785395
108133421
100454745
= 95341789
' = 104785395
= 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' 224",
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 15'444", . . = 99949302
Tan J(a6) 20 0' 22" . . = 95612100
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' 555" ; 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' 224" 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' 444",
(A~B) = 48 49' 38", and c = 25 33' 5 8".
. 91815881
. 98184449
. 96795022
194979471
. 103163590
14' 7".
Sin
Sin (A~B)
Tan \c .
99949302
98766379
96795022
195561401
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" .. = 104658211
Sin }(a + b) 64 14' 7" . = 9'9545255
Tan i(A~B) 48 49' 38" . . = 1QQ581929
Cot C 18 22' 44" . 104785395
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' .... = 102058504
Sin b 40 . . . . . = 98080675
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' 135", = 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' . . . . = 102810914
Sin B 30 28' . . . . = 97050397
Sin a 40 = 9*8080675
Sin b 38 30' 185" . 97941986
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' 547", 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 rightangled triangles by
means of & perpendicular from one of the angles upon the opposite
side, so that one of the rightangled triangles shall contain two of
the given parts.
By the method, however, of rightangled 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 = 60 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 rightangled
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
sundial, the days of which are unequal.
826. Mean time is the time shown by a wellregulated 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. 516 s., and that of the latter 365 d.
6h. 9m. 115 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 snndial ; 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. 559094 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. 559 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. 465 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 LgLa is the
proportional logarithm of a, Lq  Lb that of b, and so on ; also,
P. Lq LqLq = 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 5769
Declination
D'=26 18' 47 5"
26 20 259
v' = 2 2034
1 384
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 14034 s. = 1403 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 3813 26 35 31 '9
R =17 h. 3 m. 5171 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 4440 26 47 4'8
Reduced time = 26 d. 20 h. 2 m., R=19 h. 4 m. 1079 s.,
and D= 26 49' 1'09".
841. Problem VI. To reduce the registered semidiameter
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 semidiameter,
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 semidiameter,
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 semidiameter 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 Semidiameter Horizontal Parallax
At noon, .... J = l& 9'3" p' = 59' 10'3"
midnight, . . . 16 IQ2 59 13 '5
Change in 12 h., . . 0'9 3'2
Computed change in 8 h., 0'6 2'1
Required elements, . . 16 99 59 124
For to find s, v = ^tv' = ^ x 9" = 6", and s = s' + v ;
and n p, w= T Vv' = T^x32"=2l", ., p=p' + v.
* These elements are registered for every noon and midnight.
488 ASTRONOMICAL PROBLEMS
EXERCISES
1. Find the semidiameter 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 Semidiameter Horizontal Parallax
At noon, .... s=15'39'2" p = 5T 19'8"
midnight, ... 15 46 57 44'7
s=15' 452", andj9=57' 41 "6"
2. Find the semidiameter 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 Semidiameter 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' 149"
AUGMENTATION OF THE MOON'S SEMIDIAMETER
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 semidiameter 14' 43 '7", this aug
mentation is 4", so that the semidiameter found above must be
augmented by this quantity for this altitude, and would then be
= 14' 437" + 4" = 14' 47'7".
The semidiameter 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 semidiameter be denoted by s, and its apparent
semidiameter 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 semidiameter. 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 SEMIDIAMETER
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 semidiameter 14' 48", previously found, now
becomes = 14' 47 "7"  4" = 14' 43'7".
This semidiameter, 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 SEMIDIAMETER
844. The sun's daily change of distance from the earth is so small
compared with its distance that its semidiameter does not sensibly
change in apparent magnitude in the course of a day ; so that its
registered semidiameter may be considered as constant for at least
one day. Its distance also is so great compared with the earth's
radius that its semidiameter 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 . . = 99037701 L, sec a . . = 100962
L, siny . . = 81987581 P.L, p' . . = 5202
L, smp . . = 81025282 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 semidiameter ; 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 (le sin 2 /). Also (Art. 846), r = d sin p' t and a=d
sin "; therefore, since ^ . *% ver Y nearly, =^7 = = (le 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 (z3r'),
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' 296' ~ 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 semidiameter 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"
Semidiameter = . 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 semidiameter 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 semidiameter 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 semidiameter = 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 semidiameter
Augmentation .
Aug. semidiameter
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 semidiameter 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 semidiameter at the time of observation = 15', and
its horizontal equatorial parallax =55' 2" ; what was the true alti
tude, the corrected semidiameter, and parallax in altitude ?
The true altitude of moon's centre = 25 17' 41", semidiameter
= 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 semidiameter 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 semidiameter
= 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
Semidiameter Horizontal Parallax
At noon preceding, . . .16' 15" 59' 37"
., midnight following, . . 16 19 59 51
Reduced time = 7 h. 30 m. P.M., corrected semidiameter
= 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=90D = 9021 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. 501704 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=sr.
P'.L. a = P.L. m + 165949,
P'.L. r = P. L. s +166068,
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. 501704 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 =166068
F.L., 1m. 4168s. . . . =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,
L6La = P.L. cP.L. d,
and P.L. d=P.L. c + (LaLb).
EXERCISES
1. Convert 7 h. 40 m. 15 s. of sidereal time to mean time.
= 7 h. 38 m. 596 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. 3226 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. 59 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=SS', and m=sr.
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 = 2339
s = 2 22 4901
S' = 18 47 1092 at noon, 2nd Jan.
S =21 9 5993 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 1092 at noon, 2nd Jan.
s
r
m= 2 22 2561 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
4901
2340
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 = 3614
s =3 40 3614
and S' = l 17 53 '68
hence S =4 58 2982
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. 1308 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. 472 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. 1308 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. 485 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 semidiameter'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 semidiameter's passage . =  1 10 '52
6 3991
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. 3991 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 semidiameter . =0 1 5'80
3 4065
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. 5358 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. 1238 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
semidiameter'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. 4661 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 4441
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 = + 4090
Hence A =  16 56 2'78
And A' = + 14 9 2'69
Therefore . . . . T' = 21 12 5991
Retardation, or . . r =  3 28 '55
Mean time of transit, . . T = 21 9 3136
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. 1006 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. 2903 s. ? . . At 12 h. 38 m. 2896 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.
\(vfv) :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 1874
Increase in 1 h., or . . . i>'= 2 1T94
.1 43 m. =+01 3456
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 1194
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. 5792 s. . . . = 1 '39520
P.L., 4 h. 9 m. 5709 s. . . = '76050
F.L., 2m. 208 s. . . . = 194701
270751
P.L., h. 8 m. 4716 s. . . =131231
And T' = 4 9 57 "09
Hence T = 4 18 4425 = 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' (tfv): 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: TT'.
Or, l(tfv):T'=v'v.r, if r=TT.
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. 2367 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. 1035 s., and the moon's
registered right ascension on the same day at 4 h. being 8 h. 35 m.
164 s., and at 5 h. =8 h. 37 m. 2213 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 19
Retardation in 24 h. = 041
i. M 3 h. 2 m. =  05
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. 1272 s.
ii it 7th . .= 10 5 528
Decrease in 24 h = 00 7'44
3 h. 2 m. =00 094
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 1699
Approximate time . =T' = A'A= 9 6 5479
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 5479 =+ 42045
P'.L., 04 4 =+164603
F.L., 1 3241 = 206771
T'= 9 6 5479
T=9 5 22 38 = Tr, 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. 179 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. 355 s., and that of the sun on the 10th being=ll h.
16m. 4643 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' = DC, or A' = LP.
In all these formulae the latitude and declination are of the
same name, except in A=CD.
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 . = 13 166
Retardation in 24 h = 59'8
M 2 h. 43 m. = + 67
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 12 244 S.
Increase in 1 h. . . . . = 11 45 '9
.. ,i 235 m. =+04 36'5
Hence declination at transit, . D= 19 5 160 S.
Colatitude, C= 33 39 50 "0 N.
Meridian altitude, . . A = CD= 1434340
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=CD. 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 = DC.
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' 107" 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' 60" 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 rightangled at R,
and any two parts of it, except the right angle, being given, the
other two can be found by the rules of rightangled 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' 3426"; 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 = 2327'3426".
To find AR=A To find RSD
L, cot L . . = 99172221
L, radius . . =10*
L, cos O . . = 99625311
L, tan A . . =100453090
L, radius . . =10'
L, sin O . . = 95999732
L, sinL . . = 98869532
L, sin D . . = 94869264
And A=47 59' 0'12" = 3 h. 11 m. 561 s., and D = 17 52' 102".
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' 161" N., and A=5 h. 13 m. 5 s.
2. The obliquity of the ecliptic on the 20th of July 1854 being
=23 27' 3447", and the sun's declination =20 42' 117" N., re
quired its right ascension and longitude.
A=7 h. 57 m. 4625 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' 207", and O = 23 27' 341".
4. On the 27th of December 1854 the sun's longitude and right
ascension will be =275 28' 448", and 18 h. 23 m. 5267 s. ; required
its declination and the obliquity of the ecliptic.
D = 23 20' 493", 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' . . = 98154878
L, cos I . ."' = 99999982
L, cos H
. = 98154860
And H = 130 50' 0'2". And C=0 13' 144"
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
. = 74644506
. =10
= 98788730
= 124144224
To find D
L, radius
. =10
L, sin H
. = 98788749
L, sin E
. = 95961565
L, sin D
. = 94750314
In triangle MAR
L, cot H
L, radius
L, cos E
L, tan A'
To find A
. = 99366113
. =10
. = 99632462
= 100266359
And D = 17 22' 16". I And A' = 133 14' 387".
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. 1152 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' 418", and J=4 36' 3245".
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.
5977 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. 102 s. =36 30' 15'3".
C=44 10' 18", and C' = 84 22 221".
[1],
[3]
To find the arc 9
L, radius . . =10'
L, cos P
L, tan C
L, tan
And = 37 58' 55".
= 99051549
= 99873727
= 98925276
To find the distance D
L, cos . . = 98966390
L, cos (C'  0)
L, cos C
L, cos D
And D = 51 7' 105".
= 98386823
= 98557069
196943892
= 97977502
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 (SP) + 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 .
= 99994498
= 93243657
= '1445350
= 0948213
Hence S= 87 7 2)195631718
AndSP= 12 11 L, cos ^A, . = 97815859
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 (SZ) + L cosec P + L cosec C20.
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
Semidiameter = + 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 . . . = 99951982
P = 7248 5 L, sin(SZ) . . = 96160088
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. 427 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. 574 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 (SZ) + 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 . = 99999004
P= 734717 L, sin(SZ) . . = 96029108
C = 41 4 L, cosec P . . . = 0176222
2)182 27 17 L, cosec C . . = 1824765
S = 91 13 38 2)198029099
SZ = 23 37 38 L, cos JH, . = 9'9014549
Hence H=37 9' 21'7", and H = 4 h. 57 m. 149 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 81
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'AH. 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. 05 s., and time = 16 h. 20 m. 85 s.
876. The equation of equal altitudes is a correction, generally
of a few seconds (and seldom exceeding halfaminute), 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' + 453645,
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. =4655 m., v' = 22' 14" = 1334".
ASTRONOMICAL PROBLEMS 517
L, cot L . . = 997725 Constant . . = 45 3645
L, cos I . . = 972198 L, cot 1 . = 9'7927
L, tan 6, 26 35' = 9'69923 L > cosec 6 . . = 103492
p = 97 10 L, cosec P . . = 100034
, = ^r L, sin * . . = 99746
L, I m. . . = 26679
L, v' . . . = 31251
L, E 189 s. . . = 12774
Hence the equation of equal altitudes is 189 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 53645,
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, vL, 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 ? = 1436 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
= 124 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 semidiurnal arc. When the
declination is south, OB' is the amplitude, OD' .
the ascensional difference, and EO  OD 7 the
semidiurnal arc. When the body is setting, the figure is exactly
similar.
It is evident that if any two parts of the rightangled 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 semidiurnal 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 . . = 99037701
L, radius . . = 10*
L, sin D . . = 93912057
L, sin M . . = 94874356
And M = 17 53' 28" N.
To find N
L, radius .
L, tan D .
L, cot C .
L, sin N
= 10 55' 56".
= 10
= 94047784
= 98731668
= 92779452
The semidiurnal arc 1 = 6 h. + N = 6 h. 43 m. 437 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' 45" 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 semidiurnal 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 semidiurnal 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 semidiurnal
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(SZ) . . = 9O426439
C = 34 3 L, cosec P . . . = 10'0291009
2)193 57 L, cosec C . . . = 1Q251877Q
S = 96 58 30 2)193203957
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 34 113
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. 21 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 semidiurnal 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 semidiurnal 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
semidiurnal arc ; then 24 hours is to the semidiurnal 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 = semidiurnal 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 semidiameter . . . . = + 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 41 121 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 . . = 90611695
L, tan P . . = 105092668
To find the segment (c 9)
L, cos P . . = 94708631
L, cos Z . . = 94375159
L, cos . . = 99718687
L, tan . . = 95704363
194093846
Hence = 20 24' 2".
L, cos(e0) . = 99385215
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(PP') : 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 rightangled triangles. Hence, instead of the
preceding formulae at No. 1 and 2, take these two :
1. To find B' in one of the rightangled 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 . . =
107013382 L, radius . . =
= 10
L, radius . . =
10 L, sin P . . =
= 99739422
L, cos P . . =
95266927 L, sin H . . =
= 92902357
L, cot B' . . =
88253545 L, sin E . . =
 92641779
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 . . . =
101957243
L, cosec E
= 21 10 26 . . . =
104422527
2)112 26 26
L, sin S .
= 56 13 13 . . . =
99196958
L, sin (S  Z) .
= 4 32 13 . . . =
88981869
2)194558597
L, cos ^B',
= 57 41' 294" . =
97279298
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)* .
= 00910331
L, cos B' . =
99409342 L, cos (P  0)
= 99158171
L, tan Z' . . =
9 9173911 L, cos Z'
= 98868846
L, tan . =
98583253 L, sin L
= 98937348
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 ^H10.
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 . . = 99739422
L, cos P . . = 95266927
L, sin H . . = 92902357
L, sec M . . = 100074565
L, sin M . . = 92641779
L, cos N . . = 95341492
M = 1035' 13".
N = 69 59' 44".
3. To find O
4. To find Q
L, sin D . . = 90228254
L, cos JD . . = 9 9975743
L, cos S . .van 98542329
L, sin iS . . = 9 8446310
L, cosec M . = 107358221
L, sec M . . = 100074565
L, sin O . . = 96128804
L, sec O . . = 100399840
O=24 12' 38".
L, cos Q . . = 98896458
5. R = 3051'22" = N~Q. Q=398'22".
6. To find L,
L, cos O = 99600160
L, cos R = 99337191
L, sin L = 98937351
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' 426", 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. 421 s.
The latitude=48 42' 429" 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. 2976s.
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'= : Wlr TT 7T
\. +L sec /i + Lsec A +Lcos
and then
L sin JD=L cos ^ + L cos 4S10 .
[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 semidiameter
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 36 189
= 25 405
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 semidiameter ..... =
Hence ... . d  33 30 21
33
15 25
14 56
ASTRONOMICAL PROBLEMS 531
Then L, sec h .... 61 36' 50" = 103229308
L, sec h' . . . . 28 24 59 = 10O557580
L, cos%(s + d) . . .6146 5 = ' 9'6748997
L, cos(srf) . . . 281544= 99448723
L, cos H . . . . 62 2 = 96711338
L, cos H' . . . . 28 23 14 = 99443616
2)596139562
298069781
L, cos^S + 10 . . . . . =  198478853
L, sin 0=65 31' 13" . = 99590928
L, cos0 = 96173895^
L, cos }S = 98478853)
L, sin D = 16 58' 24 2" . = 94652748
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. 368 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 semidiameters 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' 409", 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
Semidiameter 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 dL d', or P.L t=P.L rfP.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= 1 186
and d' : d=3 h. : t, or 1 28' 4" : 1' 18'6"=3 h. : 2 m. 406 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 66 L, sin(SZ) . . =97554481
= 53 53 L, cosecP . . = 0110120
2)192 38 526 L, cosec C . . = 0926862
S=96 19 263 2)198564956
S  Z = 34 42 403 L, cos H . . . =99282478
Prc. 2 I
534 ASTRONOMICAL PROBLEMS
And H = 32 2' 111", 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 15 114
Acceleration . . . . =0 2  4
Time at place on 14th . . . =12 13 110
., 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. 62 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' 126", 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. 112 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 logline ; 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 rhumbline, 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 rhumbline 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 shipofwar 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 logline 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 logline 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 logline, is
estimated by a sandglass, which measures halfminutes 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 sandglass and logline, 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 logline and sand
glass, to find the true distance.
I. When only the logline 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 sandglass 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 logline and sandglass are incorrect, mul
tiply six times the measured length of the logline 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 djd' = ^lc'd' J
jt jt
II. s'.s=d':d, and c?=sr = 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" = rd', 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 sandglass = 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 2352
2. 156 it 50 ii 32 1462
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 Tark 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. . . = 140215 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. =76257 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 rightangled 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 . = 9888185
As distance 425 = 2*628389
To departure 328 53 . . . . = 2516574
2. To find the difference of latitude BC
Rad. :cosC = AC:BC, or
Radius = 10
Is to cos course = 9802359
As distance 425 = 2628389
To difference of latitude 269 6 . . = 2430748
Latitude left = 56 14' N.
Dif. of lat. 2696 = 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 onefifth 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 + GHf ... :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
196
39
SE.
12
85
85
SW. b S.
18
...
15
10
WNW. \ N.
14
66
123
SSW.
24
222
92
66
653
85
354
66
85
587
269
Latitude left
Difference of latitude .
Latitude in . .
= 51 25' N.
= 587 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. 587
Is to dep. 269
As radius
= 1768638
= 1429752
= 10
2. To find the distance
To tan. course 24 37'= 9661114
The course is therefore S. =24
= 64 6 miles.
Sin course
Is to radius
As dep. 269
To dist. 646
37' W.,
and
= 9619662
= 10
= T429752
= 1810090
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 rightangled 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 = 2450853
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
rightangled 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
rightangled 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'
= 7117475
= 3629410
= 9833241
Mer. dif. lat. 1128
Dif. long. 4260
Radius .
Tan course 75 10' .
= 3052309
= 3629410
= 10
= 10580126
= 10577101
NAVIGATION 553
2. To find the distance
Radius . . . =10
Sec course 75 16' . = 10'594618
Dif. lat. 763 . . = 2882525
Distance 3000 . . = 3477143
Radius . . . =10'
Sec course 75 10' . =10'591746
Dif. lat. 763 . . = 2 882525
Distance 2980 . = 3474271
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. 2046
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
9919846
2390935
2310781
. = 15943
. = 18252
= 2309
Cos mid. lat. (a . c) .
Sin course 3 points .
Distance 246 ,
Dif. long. 1539 .
Longitude left
= 0051547
=9744739
= 2390935
= 2187221
Different longitude
Longitude in
By Mercator's Sailing
To find the dif. long.
Radius . . . =10'
Tan course 3 points = 9824893
Mer. dif. lat. 231 . = 2363612
Dif. long. 1544 . = 2188505
. = 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' = 9793991
Radius . . . =10'
Dep. 243 . = 2385606
Dif. long. 3905 . = 2591615
By Mercator's Sailing
Dif. lat. 346 (a . c) . = 7 '460924
Dep. 243 . . = 2385606
Mer. dif. lat. 557 . = 2745855
Dif. long. 3912 . = 2 "592385
Long, left . . = 3645'W. Long, left . . =3645'W.
Dif. long. 3905 . = 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
= 269 W. Hence
Lat. left . . = 51 25' N. Mer. parts . .  36087
Dif. lat. . . = 59 S. Mer. parts . . = 35151
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. 269 . . = 1429752
By Mer color's Sailing
Radius . . . =10
Tan course 24 37' . = 9661043
Mer. dif. lat. 93 "6 . = 1971276
Dif. long. 42 7
Long, left
Dif. long.
Long, in
. = 1630101
. =23 40' W.
. = 43 W.
Dif. long. 429
Long, left
Dif. long.
. = 1632319
. =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 rightangled 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 Meleater'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', =9886, .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
524
350
69 30
972
NE.
38
269
269
69 57
776
NNE.
56
517
214
70 49
642
N.
30
300
71 19
NW. b N.
25
208
139
71 40
438
NNW. J W.
36
318
170
72 12
552
N. &E.
40
392
78
72 51
258
NE. b E. } E.
72
339
635
73 25
2191
SE.
50
354
354
72 50
1205
ENE.
65
249
601
73 15
2069
3116
354
2501
309
8113
990
354
309
990
2762
2192
7122
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
= PH66HS 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 threequarters 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 southeast.
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 onefourth 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
247'.
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