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William Collins, Sons, & Co.'s Educational Works.
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IN MEMORIAM
FLORIAN CAJORl
Old Testament History. By Rgv. C. Ivexs, -
Nbw Tjestament Histojiy. By Ilov. C- Iyexs, •
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MENSUEATION,
FOR
THE USE OF SCHOOL&
v^/
GLASGOW Aiq'D LONDON:
WILLIAM COLLINS, SONS, & OOMPANT.
PREFACE TO SECOKD EDITIOS.
To this Edition there is an AppendL"^, printed in a separate
form, for the use of Teachers, containing the leading pro-
perties of the Conic Sections, and the Demonstrations of
the Eules of Mensuration. These were in the First Edition
interspersed through the work, partly interwoven with the
text, and partly in the shape of notes. It is hoped that
the present arrangement will better suit the convenience of
both Teachers and Pupils. Several other alterations have
been made, which, it is hoped, will be found to be improve-
ments.
Teachers should direct their Pupils to learn only such
portions of the work as may be necessary for their intended
occupations ; for most pupils, the first and second sections,
and a few problems in the fourth and sixth, will be quitf
j^iifficJent
CONTENTS.
Mensuration of Sv^EnviciES— {continued).
Given the side of a square, to find the diameter of a circle
equal in area to the square . . , .
Given the side of a square, to find the circumference of a
circle whose area is equal to the square whose side ia
given ......
To find the area of a sector of a circle . .
To find the area of a segment of a circle .
To find the area of a zone of a circle
To find the area of a circular ring .
To find the area of a part of a ring, or of the segment of a
sector .......
To find the area of a lune . . . ,
To measure long irregular figures . . ,
Exercises in Mensuration of Superficies . •
36
3G
37
89
.40
40
40
41
42
SECTION III.
CoNio Sections—
Of the Ellipsis
Of the Parabola
Of the Hyperbola . • •
SECTION IV.
Mensuration of Solids—
Definitions .
To find the solidity of a cube
Of a parallelopipedon .
Of a prism
Of a cylinder ,
To find the content of a solid, formed by a plane
parallel to the axis of a cylinder
To find the solidity of a pyramid
Of a cone . . ,
Of the frustum of a pyramid ,
Of the frustum of a cone .
Of a wedge ....
Of a prismoid ....
Of a cylindroid
Of a sphere ....
Of the segment of a sphere . •
Of the frustum of a sphere . •
Of a circular spindle .
Of the middle frustum of a circular spindle
Of a spheroid ....
Of the segment of a spheroid .
Of the middle zone of a spheroid
Of a parabolic conoid .
Of the frustum of a parabolic conoid .
Of a parabolic spindle
Of the middle frustum of a parabolic spindle
Of a hyperbolic conoid
Of a frustum of a hyperbolic conoid
Of a frustum of an elliptical spindle
Of a circular ring
passing
45
49
52
57
69
59
60
50
61
61
62
62
63
63
64
64
65
66
66
67
68
^9
69
70
70
71
71
72
72
73
73
74
CONTENTS.
SECTION V.
'J'nB Five REOtiLAR Bodies—
Definitions ....
To find the solid contents of the regular bodies
To find their superficial contents •
75
76
r
SECTION VI.
Surfaces op Solids —
To find the surface of a prism . . . . 80
Of a pyramid ....
80
Of a cone
81
Of a frustum of a pyramid
81
Of a frustum of a cone
82
Of a wedge . . . ,
82
Of the frustum of a wedge
82
Of a globe
83
Of a segment or zone of a sphere
83
Of a cylinder .
84
Of a circular cylinder .
84
Of a parallelopipedon .
1 84
SECTION VII.
Mensuration of Timber and of Artificsbs* Work—
Description of the carpenter's rule ... 85
Use of the sliding rule
86
Timber measure
88
Carpenters' and joiners' work—
Of flooring ....
96
Of partitioning
97
Of wainscotting .
99
Bricklayers' work —
Of tiling or slating . •
101
Of walling
102
Of chimneys
103
Masons' work ,
104
Plasterers' work
105
Plumbers' work . . . ,
106
Painters' work
107
Glaziers' work ....
108
Pavers' work
109
Vaulted and arched roofs .
\ . 109
SECTION VIII.
Specific Gravity—
To find the specific gravity of a body • . 115
A table of specific gravities .... 117
To find the tonnage of ships .... 122
Floating bodies
123
rl C0NTEN1:3.
SECTION IX.
Weights and dimensions of balls and shells
Piling of balls and shells
Determining distances by sounds
125
128
131
SECTION X
. I Gauging—
^
Ofthe Ganging rule ..... 132
Verie's sliding rule . . . . , 134
A table of multipliers, divisors, and gaugo -points, for
squares and circles . . • , ,136
The gauging or diagonal rod . . . .141
Ullaging . 149
SECTION XL
LAND-SURVEYING—
Form of a field-book , ,
To measure land with the chain only
To survey a field by means of the theodoliti
To survey a field with crooked hedges
To survey any piece of land by two stations
To survey a large estate
To survey a town or city .
To compute the content of any survey
153
158
160
160
161
162
164
165
Miscellaneous Problems
167
A Table of the Areas 09 thu Ssomrnts ov a Circle, whose
diameter is 1 • • . . •
171
MENSUEATION.
SECTION I.
\
\
A
\
\
PEACTICAL GEOMETET.
DEFINITIONS.
1. Geometry teaches and demonstrates the properties of all kinds
of magnitudes, or extension ; as solids, surfaces, lines, and angles.
2. Geometry is divided into two parts, theoretical and practical.
Theoretical Geometry treats of the various properties of extension
abstractedly ; and Practical Geometry applies these theoretical pro-
perties to the various purposes of life. When length and breadth
only are considered, the science which treats of them is called Plane
Geometry ; but when length, breadth, and thickness are considered,
the science which treats of them is called Solid Geometry.
3. A Solid is a figure, or a body, having three dimen-
Bions, viz., length, breadth, and thickness ; as A.
The boundaries of a solid are surfaces or superficies.
4. A Superjicies^ or surface, has length and breadth i ~ <
only ; as B. ^ I 1
The boundaries of a superficies are lines. ^
5. A Line is length without breadth, and is q ^
formed by the motion of a point ; as C B.
The extremities of a line are points.
6. A Straight or Right Line is the shortest distance between two
points, and lies evenly* between these two points.
7. A Point is that which has no parts or magnitude ; it is indi-
visible ; it has no length, breadth, or thickness. If it had length, it
would then be a line ; were it possessed of length and breadth, it
would be a superficies ; and had it length, breadth, and thickness,
it would be a solid. Hence a point is void of length, breadth, and
thickness, and only marks the position of their origm or termination
in every instance, or of the direction of a line.
8. A Plane rectilineal Angle is the inclination
of two right lines, which meet in a point, but are S-=
Qot in the same direction ; as S.
8
PRACTICAL GEOMETRY.
9. One angle is said to be less than another,
when the lines which form that angle are nearer
to each other than those which form the other, B <
measuring at equal distances from the points in
which the lines meet. Take Bw, Bm, Ex, and
Ew, equal to one another ; then if w n be greater
than X w, the angle A B C is greater than the E -
angle FED. By conceiving the point A to move
towards C, till m n becomes equal to x w, the
angles at B and E would then be equal ; or by conceiving the point
F to recede from D, till x n becomes equal to m w, then the angles
at B and E would be equal.
Hence it appears that the nearer the extremities of the lines forming
an angle approach each other, while the point at which they meet
remains fixed, the less the angle; and the farther the extreme points
recede from each other, the vertical point remaining fixed, as before^
the greater the angle.
10. A Circle is a plane figure contained by one line L^
called the circumference, which is everywhere equally
distant from a point within it, called its centre, as o:
and an arc of a circle is any part of its circumference ;
as A B.
11. The magnitude of an angle does not
consist in the length of the lines which form
it : the angle C B G is less than the angle ABE, B<
though the lines C B, G B are longer than AB,
EB.
12. When an angle is expressed by three letters, as ABE, the
middle letter always stands at the angular point, and the other two
anywhere along the sides ; thus the angle A B E is formed by A B
and B E. The angle A B G by A B and G B.
13. In equal circles, angles have the same ratio to each other as
the arcs on which they stand (33. VI.) Hence also, in the same, or
equal circles, the angles vary as the arcs on which they stand ; and,
therefore, the arcs may be assumed as proper measures of angles.
Every angle then is measured by an arc of a circle, described about
the angular point as a centre ; thus the angle A B E is measured by
the arc A E ; the angle A B G by the arc A F.
14. The circumference of every circle is generally divided into
360 equal parts, called degrees ; and every degree into 60 equal parts,
called minutes; and each minute into 60 equal parts, called seconds.
The angles are measured by the number of degrees contained in the
arcs which subtend them ; thus, if the arc A E contain 4(>^e^rees,
or the ninth part of the circumference, the angle A B E is said to
measure 40 degrees.
15. When a straight line H 0, standing
on another A B, makes the angle H 0 A
equal to the angle HOB; each of these
angles is called a right angle ; and the line
H 0 is said to be a pei'pendicular to A B.
PRACTICAL GEOMETRY.
Tlie measure of the angle H 0 A is 90 decrees, or the fourth part
of 360 degrees. Hence a right angle is 90 degrees.
16. An acute angle is less than a right angle ; as A 0 G, or G 0 H.
17. An obtuse angle is greater than a right angle ; as GOB.
IS. k plane Triangle is the space enclosed by three
straight lines, and has three angles ; as A.
19. A right angled Triangle is that which has
one of its angles right ; as A B C. The side B C,
opposite the right angle, is called the hypothenuse ;
the side AC is called the perpendicular; and the
Bide A B is called the base.
20. An oltuse angled Triangle has one of its
angles obtuse; as the triangle B, which has the
obtuse angle A.
21. An acute angled Triangle has all its three angles acute, as
m figure A, annexed to definition 18.
22. An equilateral Triangle has its three sides
equal, and also its three angles ; as C.
23. An isosceles Triangle is that which has two of
its sides equal ; as D.
24. A scalene Triangle is that which has all its / b
sides unequal ; as E.
25.\ A quadrilateral figure is a space included by four straight
lines. , If its four angles be right, it is called a rectangular parallelo-
gramr«
26. A Parallelogram is a plane figure bounded by four straight
lines, the opposite ones being parallel; that is, if produced ever so
far, would never meet.
27. A Square ia a four sided-figure, having all its
sides equal, and aJ^ its angles right angles ; as H.
10
PRACTICAL GEOMETRY.
28. An Ohlong^ or rectangle, is a right angled paral-
lelogram, whose length exceeds its breadth ; as I.
_J
29. A Rhomhus is a parallelogram having all
its sides equal, but its angles not right angles;
as K.
30. A Rhomboid is a parallelogram having ,
its opposite sides equal, but its angles are not
right angles, and its lengtti exceuds its breadtli ;
asM.
31. A 7Vapezium is a figure included by four
straight lines, no two of which are parallel to each
other ; as N.
A line connecting any two of its opposite angles, is called a
diagonal.
32. A Trapezoid is a four-sided figure having two / F \
of its opposite sides parallel ; as F. ^ i
33. Multilateral Figures, or Polygons, are those which have more
than four sides. They receive particular names from the number ol
their sides. Thus, a Pentagon has five sides ; a Hexagon has six
sides; a Heptagon^ seven; an Octagon^ eight; a Nonagon, nine;
a Decagon, ten; an Undecagon, eleven; and a Dodecagon has
twelve sides.
If all the sides and angles of each figure be equal, it is called a
regular polygon ; but if either or both be unequal, an irregular
polygon.
34. The Diameter of a circle is a straight line passing through
the centre, and terminated both ways by the circumference ; thus
A B is the diameter of the circle. The diameter divides the circle
into two equal parts, each of which is called a
semicircle ; the diameter also divides the circum-
ference into two equal parts, each containing 180
degrees. Any line drawn from the centre to the ^\
circumference is called the radius, as A 0, 0 B, or
OS. If 0 S be drawn from the centre perpen-
dicular to A B, it divides the semicircle into two
equal parts, A 0 S and BOS, each of which is called a quadrant,
or one-fourth of the circle ; and the arcs A S and B S contain each
90 degrees, and they are said to be the measure of the angles A 0 S
and BOS.
35. A Sector of a circle is a part of the cncle comprehended under
PRACTICAL GEOMETRY.
11
two Radiu not forming one line, and the part
of the circumference between them. From this
definition it appears that a sector may be either
greater or less than a semicircle ; thus A 0 B is a
sector, and is less than a semicircle; and the
remaining part of the circle is a sector also, but
is greater than a semicircle.
36. A Chord of an arc is a straight line joining its extremities,
and is less than the diameter ; T S is the chord of the arc T H S, or
ofthe arc TABS.
37. A Segment of a circle is that part of the circle contained between
the chord and the circumference, and maybe either greater or less than
a semicircle ; thus T S H T and T A B S T are segments, the latter being
greater than a semicircle and the former less.
38. Concentric circles are those having the
same centre, and the space included between their
circumference is called a ring ; as F E.
PROBLEM I.
To bisect a given straight line A B ; that is, to divide it into two
equal parts, O
From the centres A and B, with any radius,
greater than half the given line A B, describe two
arcs intersecting each other at 0 and S, then
the line joining 0 S will bisect A B.
PROBLEM XL
Through a given point x to draw a straight line CD paralla
to a given straight line AB.
In A B take any point 5, and with the C %^ ? -D
centre s and radius s x describe the arc
0 x; with X as a centre and the same ra-
dius s a;, describe the arc s y. Lay the ^
extent ox taken with the compasses from s to y ; through xy draw
C D, which will be parallel to A B.
PROBLEM III.
To draw a straight line C D parallel to A B, and at a given
distance F from it.
In A B take any two points x^f; ^ r ^
and from the two points as centres V^' ^^\ /-""^ ^\
with the extent F taken with the
compasses, describe two arcs s, r;
then draw a line C D touching these A * a jb
aies at r and 5, and it will be at *
the given distance fTOTn A B, and parallel to it. ^
12
PRACTICAL GEOMETRT.
PROBLEM IV.
To divide a straight line AB into any numher oj equal parts.
Draw A K making any angle with A B ; and through B draw B T
parallel to AK; take any part A E and
repeat it as often as there are parts to
be in AB, and from the point B on the
line BT, take BI, IS, SV, and VT
equal to the parts taken on the line
AK; then join AT, EV, GS, HI, and
K B, which will divide the line A B into
the number of equal parts required, as ^
AC, CD, DF, FB.
PROBLEM V
From a given point P in a straight li?ie AB to erect a perpendicular,
1. When the given point is in^ or near the middle of the line,
D
On each side of the point P take equal por-
tions, P a;, V f; and from the centres, x, /,
ivith any radius greater than P a;, describe two
arcs, cutting each other at D ; then the line
joining D P will be perpendicular to A B.
Or thus:
From the centre P, with any radius P w,
describe an arc n x y ; set off the distance,
P n from n to a:, and from x to y ; then from
the points x and y with the same or any
other radius, describe two arcs intersecting
each other at D ; then the line joining the
points D and P will be perpendicular to A B.
2. When the point is at the end of the line.
From any centre q out of the line, and with
the distance ^ B as radius, describe a circle,
cutting AB in p; draw pa 0; and the line
joining the points 0, B, will be perpendicular
to AB.
—3
A-
Or thus:
Set one leg of the cornpasses on B, and with
any extent fi p describe an arc p x ; set off
the same extent from p to q ; ]oinpq ; from
7 as a centre, with the extent jt? q as radius,
describe an arc r ; produce p q to r^ and
the line joining r B will be perpendicular to
/
PUACTICAL GEOMETRT.
lb
PROBLEM VI.
From a given point D to let fall a perpendicular upon a given
line A B.
1. When fhe point is nearly opposite the middle ofjhe given line.
From the centre D, with any radius,
describe an arc x y^ cutting A B in a: and
y / from x and y as centres, and with the
same distance as radius, describe two arcs
cutting each other at S ; then the line join-
ing D and S will be perpendicular to A B.
-'•r-
When the point is nearly opposite the end of the given line^ and
when vie given line cannot be conveniently produced.
Draw any line D x, which bisect in o; from o
as a centre with the radius ox describe an arc
cutting A B in 2/ ; then the line joining D y will
be perpendicular to A B.
PROBLEM VIL
To draw a perpendicular^ from any angle
of a triangle ABC, to its opposite side.
Bisect either of the sides containing the
angle from which the perpendicular is to be
drawn, as BC in the point r; then with the
radius r C, and from the centre r, describe
an arc cutting A B (or A B produced if
necessary, as m the second figure), in the
point P; the line joining CP will be
pendicular to A B, or to A B produced.
per- A
PROBLEM VilL
Upon a given right line k^ to describe an equilateral triangle.
From the centres A and B, with the given line
A B as radius, describe two arcs cutting each other
at C ; then the lines drawn from the point C to the
points A and B will form, with the given line A B,
an equilateral triangle, as A B C*
u
PRACTICAL GEOMETTir.
PROBLEM IX.
To make a triangle whose sides shall be equal to three given right
lines A B, AD, and B D, any two oj which are greater tJian
the third.
From the centre A with the extent A D, describe
an arc, and from the centre B with the radius
B D describe another arc cutting the former at D;
tiien join DA, D B, and the sides of the triangle »
A B D will be respectively equal to the three given /
right lines. f — ^
PROBLEM X. B ^
Two sides A B and BC of a right angled
triangle being given, to Jind the hypo-
thenuse.
Place B C at right angles to A B ; draw A C, ^
and it will be the hypothenuse required. ^
PROBLEM XL
The hypothenuse AB, and one side AC, of a riyht angled triangle
being given, to find the other side.
Bisect A B in a; ; with the centre x, and a; A as
radius, describe an arc; and with A as a centre,
and A C as radius, describe another arc cutting
the former at C ; then ioin A C and C B ; and A B C a ^
ivill be a right angled triangle, and BC the re- ^_
quired side.
PROBLEM XIL
To bisect a given angle; that is, to divide it into two equal part:'
c
Let A C B be the angle to be bisected.
From C as a centre, with any radius C x, de-
scribe the arc xy; from the points x and y as cen-
tres, wi^h the same radius, describe two arcs
cutting each other at 0 ; join 0 C, and it will ^
bisect the angle A C B.
PROBLEM XIIL
At a given point A in a given right line A B, to make an angle
equal to ihe given angle C.
From the centre C with any radius C y, de-
scribe an arc xy ; and from the centre A, with
the same radius, describe another arc, on which
take the distance mn equal to xy; then a line
drawn from A through m will make the angleC
w A n equal to the angle x C y.
"o""^
PRACTICAi:. GEOMETRY.
15
M XV. \^
I a scale ofdhqri
'rt'nfni'yxy \
PROBLEM XIV.
To make an angle containing any proposed number of degrees.
1. Vt^nthe required angle is less than a quadrant^ as 40 degrees,
TakeHn the compasses the extent of 60 de-
grees fromithe line of chords, marked ciio. on
the scale ; aiad with this chord of 60 degrees as
radius, and trh» centre A, describe an arc ^yt\^
take from the limiof chords 40 degrees, which set
off from n to m ; nsmi A draw a line through m;
and the angle m A n>^i\\ contain 40 degrees.
2. When the required^angle is greater than a quadrant^ as 120
degrees. /
From the centre o, with ihe chord of 60 de- n
greesas radius, describe the semicircle warwB;
set off the chord of 90 degrees from. B to n,
and the remaining 30 degrees from n to x ;
join ox ; and the angle Boa; wiU contain 120^^-
degrees ; or subtract 120 froijar 180 degrees,
and set off the remainder (6^^ degrees) taken from the line of chorda
from y to x: then join a: o^ and Box will contain 120 degreea as
before. /
PROBLEM
An angle being gUfen, tojind^ by a scale ofehqrds^ how many
degrees it contains.
From the verWSc A as centre, with the chord
of 60 degrees a/ radius, describe an arc xy; take
the extent :i^y with the compasses, and setting
one foot at'the beginning of the line of chords, "^
the others leg will reach to the number of degrees which the angle con-
tains :iut if the extent a:?/ should reach beyond the scale, find the num-
ber oldegrees in xy^ which deducted from 180, will leave the degrees
iuyme angle B o x. See figure to the second case of the last Problem.
/ PROBLEM XVI.
Upon a given right line A B, to construct a square.
With the distance A B as radius, and A
as a centre, describe the arc E D B ; and
with the distance A B as radius, and B as
a centre, describe the arc AFC, cutting
the former in x; make x E equal to a; B ;
join E B ; make x C and x D each equal to
A F, or Fa:; then join AD, DO, OB, and
A D C B will be the required square.
Or thus.
Draw B C at right angles to A B, and equal to
it; then from the centres A and C, with the
radius A B and C B, describe two arcs cutting
each other at D ; join D A and D C, which will
complete the square.
PRACTICAL GEOMETRY.
PROBLEM XVII.
To make a rectangular parallelogram of a given length and
breadth. ,
A"
Let A B be the length, and B C the breadth.
Erect B C at right angles to A B ; through C a
and A draw C D and A D, parallel to A B and B
BC.
PROBLEM XVIII.
To find the centre of a given circle.
Draw any two chords AC, C B ; from the
points, A, C, B, as centres, with any radius
greater than half the lines, describe four
arcs cutting in ra; and wv, draw rx and
y V, and produce them till they meet in 0,
>vhich will be the centre.
PROBLEM XIX.
Vpon a given right line A B, to describe a rhombus having an
angle equal to a given angle A.
c
B. y^
Make the angle CAB equal to the angle at A ; make AC equal
to A B ; then from C and B as centres, with the radius A B describe
two arcs crossing each other at D ; join D C and D B, which will
complete the rhombus.
PROBLEM XX.
To find a mean proportional between two given right lines A B
and B C.
Place A B and B C in one straight line ;
bisect A C in o; froiiu o as a centre, with A o
or 0 C as radius, describe a semicircle A S C ;
erect the perpendicular B S, and it will be a
mean proportional between AB and BC;
that is, A B : B s : : B S : B c.
PROBLEM XXL
To divide a given right line A B into two such parts^ as shall be to
each other as xo to of.
X ?i f
From the point A draw A S equal to x o, and ^
produce it till F S becomes equal to o f; Jom
F B, and draw S T parallel to F B ; tfien will
A T : T B : : ar 0 : (?/•
TRACTICAL GEOMETRT.
17
PROBLEM XXII.
To find a third proportional to two given right lines A B, A S.
Place AB and AS so as to make any angle
at A : from the centre A, with the distance
A S, describe the arc S D ; then draw D x
parallel to BS, and kx will be the third
proportional required ; that is, A B : A S : :
A S • Aa:
PROBLEM XXIII.
To find a fourth proportional to three given right lines^ A B, AC,
and A D.
Place the right lines A B and AC so as to
make any angle at A ; on AB set off AD ;
join B C ; and draw D S parallel to it ; then
A S will be the fourth proportional required,
viz. AB: AC:: AD: AS.
PROBLEM XXIV.
In a given circle to inscribe a square.
Draw any two diameters A C, D B at right
angles to each other ; then join their ex-
tremities, and the figure ABCD will be a
square inscribed in the given circle.
If a line be drawn from the centre o to the
middle of AB, and produced to/; the line
joining /B will be the side of ah octagon
mscribed in the circle.
PROBLEM XXV.
To make a regular polygon on a given right line^ A B.
Divide 360 degrees by the number of sides
contained in the polygon ; deduct the quotient
from 180 degrees, and the remainder will be
the number of degrees in each angle of the
polygon. At the points A and B make the
tingles oABandoBA each equal to half the
angle of the polygon ; then from o as a centre,
and with o A or o B as radius, describe a circle,
in which place A B continually.*
Or thus:
Take the given line A B from the scale of equal parts, and multiply
the number of equal parts in it by the number in the third column of
the following table, answering to the given number of sides; the
product will give the number of equal parts in the radius A o, or o B,
which taken from the scale of equal parts in the compasses, will give
* See Appendix, Demonstration 1.
18
TRACTICAL GEOMETRY.
the radius, with which describe a circle, and place in it the line A B
continually, as shewn in the first method.*
TABLE I.
When the side of the polygon is 1.
No. of
Name of the
Radius of the circumscrib-
Angle OAB, or
sides.
Polygon.
ing circle.
OB A.
3
Trigon
•5773503
30
4
Tetragon
•7071068
45
5
Pentagon
•8506508
64
6
Hexagon
1, Side = radius.
60
7
Heptagon
1-1523825
64^
8
Octagon
1-3065630
67^
9
Nonagon
1-4619022
70
10
Decagon
1-6186340
72
11
Undecagon
1-7747329
73i^
12
Dodecagon
1-9318516
75
PROBLEM XXVI.
In a given circle to inscribe any regular polygon ; or^ to divide
the circumference of a given circle into any number of equal
parts.
Divide the diameter A B into as many
equal parts as the figure has sides ; erect
the perpendicular o x, from the centre o ;
divide the radius oy into four equal
parts, and set off three of these parts
from y to x; draw a line from x to the
second division z, of the diameter A B,
and produce it to cut the circumference
at C ; join A C, and it will be the side of
the required polygon. f
PROBLEM XXVII.
To draw a straight line equal to any given arc, of a circle^ AB.
Divide the chord AB into four equal parts ;
and set off one of these parts from B to D ;
then join D C, and it will be equal to the
length of half the given arc nearly.J
* Set Api«ucL.v, £>duiousoniuoii 2. t Ibid. 3 X Ibid. 4.
PRACTICAL GEOMETKY.
19
Or thus:
From the extremity of the arc AB, whose length
is required to be found, draw kom^ passing through
the centre ; divide o n into four equal parts, and set
off three of these parts from n to m ; draw m B, and
produce it to meet A C drawn at right angles to
A m ; then will A C be nearly equal in length to the
arc AB.*
PROBLEM XXVIII.
To make a square equal in area to a given circle.
First divide the diameter AB into fourteen
equal parts, and set off eleven of them from
A to S ; from S erect the perpendicular S C,
and join A C, the square of which will be very
nearly equal to the area of the given circle, f
PROBLEM XXIX.
To construct a diagonal scale.
Draw an indefinite straight line ; set off any distance A E according
to the intended length of the scale ; repeat A E any number of times,
EG, G B, &c. ; draw C D parallel to A B at any convenient distance ;
then draw the perpendiculars A C, E F, G H, B D, &c. Divide A E
and A C each into ten equal parts; through 1, 2, 3, &c., draw lines
parallel to AB, and through xy^ &c., draw a:F, ^Z, &c., as in the
annexed figure.
8 r y z E
wv
V
9
8
7
6
5
W \ M ' 1
I 1 \ '■
p M ^1
MM
1
i
2
-L-UJuU- -i
2
1
11 11 M n 1 1
C
'987654324]
F
]
S
J)
The principal use of this scale is, to lay down anj line from a given
measure ; or to measure any line and compare it with others. — What-
ever number C F represents, F Z will be the tenth of it, and the sub-^
divisions in the vertical direction FE v/ill be each one-hundredth
part. Thus, if C F be a unit, the small divisions in C F, viz. F 2,
Kc, will be lOths, and the divisions in the altitude will be the 100th
* See Apjaendix, DemonstratioL. 6.
t Ibid. Q.
20
PRACTICAL GEOMETRY.
parts of a unit. If C F be ten, the small divisions F Z, &c., will be
units, and those in the vertical line, tenths ; if C F be a hundred, the
others will be tens and units.*
To take any number off the scale, as suppose 2^^^^, that is, 2-38:
place one foot of the compasses at D, and extend the other to the
division marked 3 ; then move the compasses upward, keeping one
foot on the line D B, and the other on the line 3 5, till you arrive at
the eighth interval, marked 88, and the extent on the compasses will
be that required. This, however, may express 2 "38, 23-8, or 288,
according to the magnitude of the assumed unit.
Note. — If C F were divided into 12 equal parts, each division would be 1 inch,
and each vertical division 1-1 0th of an inch, by making C F one foot.
PROBLEM XXX.
To reduce a rectilinear figure to a similar one upon either a
smaller or a larger scale.
Take any point P in the figure
ABODE, and from this as-
sumed point draw lines to all the
angles of the figure ; upon one
of wiiich P A take P a agreeably
to the proposed scale; then draw
a h parallel to A B, 5 c to B C,
&c., then shall the figure ahcde
be similar to the original one,
and upon the required scale.
Or, measure all the sides and
diagonals of the figure by a scale, and lay down the same measures
respectively from another scale, in the required proportion.
When the figure is complex, the reduction to a different scale ia
best accomplished by means of the Eidograph, an instrument invented
by Professor Wallace, or by means of the improved Pentograph.
PROBLEM XXXr.
2^0 divide a circle into any number of equal parts^ having their
perimeters equal also.
Divide the diameter AB into the re-
quired number of equal parts, at the points
C, D, E, &c. ; then on one side describe
the semicircles 1, 2, 3, 4, &c., and on the
other side of the diameter describe the
semicircles 7, 8, 9, 10, &c. on the diame-
ters B F, B E, B D, B C, &c. ; so shall the
parts 111, 2 10, 3 9, 4 8, &c. be equal
both in area and perimeter. — Leslie's
Geometry.
• ft3ee Appendix, Demonstration 7.
MENSURATION OF SUPERFICIES. 21
MENSUEATION OF SUPERFICIES.
SECTIOISr II.
I.
Long Measure,
12 Inches . . 1 Foot.
3 Feet ... 1 Yard.
6 Feet ... 1 Fathom.^
16^ Feet Eng. \ (1 Pole or
6 J Yards )* ( Perch.
40 Perches . . 1 Furlong.
8 Furlongs . 1 Mile.
1 Yard.
1 Fathom,
f 1 Pole or
\ Perch.
1 Furlong.
1 Mile.
The area of any plane figure is tlie space contained within its
boundaries, and is estimated by the number of square miles, square
yards, square feet, &c. which it contains.
II.
Square Measure,
144 Inches ... 1 Foot.
9 Feet . .
36 Feet . .
272i Feet Eng. 1
30i Yards i
1600 Perches ,
64 Furlongs .
In Ireland 21 feet make 1 pole or perch, and 7 yards therefore
will make a pole or perch. There are other measures used, for which
eee Arithmetical Tables.
Land is generally measured by a Cliain^ of 4 poles, or 22 yards ;
it consists of 100 links, each link being -22 of a yard. See Section
XL Surveying.
Duodecimals are calculations by feet, inches, and parts, which
decrease by twelves : hence they take their name.
Multiplication of feet, inches, and parts, is sometimes called Cross
Multiplication, from the factors being multiplied crosswise. It is
used m finding the contents of work done by .artificers, where the
dimensions are taken in feet, inches, and parts.
Rule.
I. Write the multiplier under the multiplicand in such a manner,
that feet shall be under feet, inches under inches, &c.
II. Multiply each term of the multiplicand by the number of feet
in the multiplier, proceeding from right to left ; carry 1 for every 12,
in each product, and set down the remainder under the term multi-
plied.
III. Next multiply the terms of the multiplicand by the number
under the denomination inches, in the multiplier ; carry 1 for every
12, as before, but set down each remainder one place farther to the
right than if multiplying by a number under the denomination feet.
ly. In like manner proceed with the number in the multipliei
22
MENSURATION OF SUPERFICIES.
under the denomination parts or lines, remembering to set down each
remainder one place farther to the right than if multiplying by a
number under the denomination inches. And so on with numbers of
inferior denominations.
V. Add the partial products thus placed, and their sum will be tho
whole product.
IN CROSS MULTIPLICATION IT IS USUAL TO SAY,
Feet multiplied by feet, give feet.
Feet by inches, give inciies.
Feet by parts, give parts.
Inches by inches, give parts.
Inches by parts, give thirds.
Inches by thirds, give fourths.
Parts by parts, give fourths.
Parts by thirds, give fifths.
Parts by fourths, give sixths, &c *
1. Multiply 7 feet 9 inches by 3 feet 6 inches.
F. I.
7.9
3 . 6
23
3
10 . 6
27
1 . 6Ans.
F. I. P. F. I. P.
2. Multiply 240 . 10 . 8 by 9 . 4 . 6
9.4.6
2168 .
0 ,
. 0
80 .
3 ,
. 6 .
. 8
10 .
0 ,
. 5 .
. 4
2258
4.0. 0 Ans.
9.
10.
11.
Multiply 8
Multiply 9
Multiply 7
Multiply 4
Multiply 7
Multiplv 10
Multiply 75
Multiply 57
Multiply 75
I. P. '"
6 . 11.
6.
1
11
• 7
6.
10.
6.2.3.
0.6.6
8.
9 . 9.
11 . 3.
* In multiplication, the multiplier must always be a number of times ; to
talk of multiplying feet by feet, &c., is absurd, for what notion can be formed
of 7 feet taken 3 times? However, since the above easily suggests the correct
meaning, and is a concise method of expressing the rule, it haa been thought
proper to retain it. See Appendix, Demonsti'&tiou «
MENSURATION OF SUPERFICIES.
23
12. Multiply 321
13. Multiply 4
U. Multiply 39
16.
6'=|
2'=i
I.
p. F. I. P. F. I. p.'"""
7 . 3 by 9 . 3.6. Ans. 2988 . 2 . 10.4.6.
7 . 8by 9 . 6 . 44 . 0 . 10.
10 . 7byl8 . 8 . 4. 745 . 4 . 10.2.4.
be solved by the method of aliquot parts, thus :—
F. ' " F. ' "
Multiply 368 . 7 . 5 by 137 . 8 . 4
137 . 8 . 4
2576
1104
368
184 . 3 . 8 . 6
61 . 6 . 2 . 10
10 . 2 . 10 . 5,8
68 . 6
11 . 5
3.9.8
0 . 11 . 5
Ans. 50756 . 7 . 10 . 9.8
6
D
C
J
PROBLEM I.
To find the area of a square.
Rule. Multiply the length of the side by
Itself, and the product will be the area.*
1. Let the side of the square ABCD be 6:
what is its area?
Ans. 6x6 = 36, the area.
2. What is the area of a square whose side is
15 chains? Ans. 22^.
3. What is the area of a square whose side is
7 feet 9 inches? Ans. 60^.
4. What is the area of a square whose side is 4769 links?
Ans. 22743361.
PROBLEM II.
To find the area of a rectangle.
Rule. Multiply the length of the rectangle by its breadth, and the
product will be the area.* * '" ^
1. Let the sides of the rectangle
A B C D be 12 and 9, what is its
^ea ? Ans. 12 x 9 = 108, the area.
2. What is the superficial content
if a plank, whose length is 5 feet 6
inches, and breadth 7 feet 8 inches?
Ans. 42 feet 2 inches.
3. What is the area of a field
whose boundaries form a rectangle,
its length being 176 links and
Dread th 154 links ? Ans. -27104 of an acre.
* See Appendix Demonstration*
n
24 MENSURATION OF b'UPERFICIES.
4. What 18 the superficial content of a floor, whose length is 40
feet 6 inches, and breadth 28 feet 9 inches ?
Ans, 1164 feet, 4 inches, f> parts.
PROBLEM III.
To find the area of a rhombus.
Rule. Multiply the length by the per-
pendicular breadth, and the product will be
the area.*
^ 1. What is the area of a rhombus, whose
side is 16 feet, and perpendicular breadth
10 feet?
Ans, 16 X 10 = 160 feet, the area.
2. What is the content of a field in the
form of a rhombus, whose length is 7*6
chains, and perpendicular height 5*7 chains? Ans. 43'32 chain*!,
- 3. What is the area of a rhombus whose side is 7 feet 6 inches,
and perpendicular height 3 feet 4 inches ? Ans. 25 feet.
4. What is the area of a rhombus whose length is 3 yards, and
perpendicular height 2 feet 3 inches ? Aiis. 20 feet 3 inches.
PROBLEM IV.
To find the area of a triangle*
Rule. Multiply the base by the perpendicular height, and divide
the product by two for the area.f
1. The base of a triangle is 76*5 feet, and perpendicular 92*2 feet;
what is its area ?
Ans. 76-6 x 92-2 -^ 2=3526-65 square feet, the area.
2. The base of a triangle is 72*7 yards, and the perpendicular
height of 36*5 yards ; what is its area ? Ans. 1326*775 yards.
3. The base of a triangular field is 1276 links; and perpendiculai
976 links; how many acres in it? Ans. 6 acres 36*3008 perches.
4. The base of a triangle measures 15 feet 6 inches, and the per-
pendicular 12 feet 7 inches ; what is its area ?
Ans. 97 feet 6J inches.
* See Appendix, Demonstration 0. t Ibid. 10.
l^ENSURATION OF SUPERFICIES.
25
PROBLEM V.
Having the three sides of any triangle given ^ to find its area.
Rule I. From half the sum of the three
Bides subtract each side separately, then multi-
ply the half sum and the three remainders
together, and the square root of the last pro-
duct will be the area of the triangle.*
Rule II. Divide the difference between the
squares of two sides of the triangle by the
third side ; to half this third side add half the
quotient, and deduct the square of this sum
from the square of the greater side, the re-
mainder will be the square of the perpendicular,
the square root of which, multiplied by half
the base, will give the area of the triangle.f
1. Given the side AB = 9-2, BC = 7*5,
and AC = 5*5 ; required the area of the
triangle?
9-2
7-5
6-5
Sum 22-2
JSumIl-1— 9'2 = l-9) : then V (lMxl-9x 3-6x5-6) =
)M— 7-5 = 3-6V V 425'1744 = 20-619 the area by
ll-l— 5-5 = 5-6 ) Rule I.
Again, 9-2^— 7-52 = 84-64— 56-25 = 28'39 ; then 28•39-^5•5 =
B- 161818, quotient.
Now (5-161818^2) + (5•5-^-2) = 2*580909 + 2-75 = 5-3309 =
half quot. plus half third side: then 84-64 — 28-41849481 =
56-22150519, and V 56*22150519 = 7*498 = perpendicular; thep
7-498 X 2-75 = 20-619 the area as before.
2. What is the area of a triangle whose sides are 50, 40, and 30 ?
Ans. 600.
3. The sides of a triangular field are 4900, 5025, and 2569 links;
how many acres does it contain ?
Arcs. 61 acres, 1 rood, 39*68 perches.
4. What is the area of an isosceles triangle, whose base is 20, and
each of its equal sides 15? Ans. 111*803.
5. How many acres are there in a triangle, w^hose three sides are
380, 420, and 765 yards? Ans. 9 acres 38 poles.
6. How many square yards in a triangle, whose three sides are 13,
14, and 15 feet ? Ans. 9^ square yards.
* See Appendix, Demonstration 11.
t Ibid, 12.
26 MENSURATION OF SUPERFICIES.
7. How many acres, &c., in a triangle, whose three sides are 49
60-25, and 25*69 chains? Ans, 61 acres, 1 rood, 39*68 perches.
PROBLEM VI.
To find the area of an equilateral triangle.
Rule. Square the side, and from this square deduct its fourth
part ; then multiply the remainder by the fourth part of the square
of the side, and the square root of the product will give the area.*
Or multiply ab'^ by V ^ fo^ the area.f
4
1. Each side of a triangular field, ABC, measures 4 perches, what
is its area?
42 = 16, then 16 -^ 4 = 4, and 16 — 4 = 12: then 12 x Y =
12x4 = 48, and V48 = 6-928, the area.
2. How many acres in a field of a triangular form, each of whose
Bides measures 70 perches? Ans. 13 acres, 1 rood, 1 perch.
3. The perimeter of an equilateral triangle is 27 yards, what is its
area? Ans. 35-074:.
jjjote. — "When the triangle is isosceles, the perpendicular is equal to th«
Bquare root of the difference between the squares of either of the equal sides,
ftud half the base.
PROBLEM yn.
Given the area and altitude of a triangle^ to find the base.
Rule. Divide the area by the altitude or per-
pendicular, and double the quotient will give the
base.J
1. Given the area of a triangle = 12 yards,
and altitude = 4 ; what is its base?
Ans. 12 H- 4 = 3 ; then 3x2 = 6 yards, the
base, AB.
2. A surveyor having lost his field book, and requiring the base
of a triangular field, wiiose content he knew from recollection was
14 acres, and altitude 7 yards, how much is the base ?
Ans. 19360 yards.
PROBLEM VIIL
Given the area of a triangle, and its base, to find its altitude.
Rule. Divide the area by the given base, and double the quo-
^tient will give the perpendicular.
'" The reason of this rule is manifest, from the last.
1. Given the area of a triangle = 12, and its base = 6 ; what ia
•Us perpendicular height ?
' Ans. 12-i-6 = 2 ; then 2 x 2 = 4 the altitude.
• Soe Appendix, Demonstration 13. f Ibid. 14, I Ibid. 15
MENSURATION OF SUPERFICIES. 27
PROBLEM IX.
Given any two sides of a right angled triangle^ to find the third
side, and thence its area.
Rule.
I. To the square of tlie perpendicular add the square of the base,
and the square root of the sum will give the hypothenuse.
II. The square root of the difference of the squares of the hypo-
thenuse and either side, will give the other.
III. Or multiply the sum of the hypothenuse, and either side, by
their difference ; and the square root of the product will give the
other.*
1. Given the base ACS, the perpendicular C B 4 ;
required the hypothenuse 'A B'?"
32 + 423^25; then V 25 = 5, the hypothenuse A B.
2. Given A B 5, A C 3 ; required C B ?
53 — 33 = 16 ; then V 16 = 4 the side B C ; or,,
(5 + 3) X (5 — 3J = 8 X 2 = 16; then Vl6 = 4, as
before.
3. Given A B 5, B C 4 ; required AC?
53_ 43 = 9 ; then V 9 = 3 the side AC; or (5 + 4) x (5 — 4)
= 9x1 = 9; then V 9 = 3, as before. And 3x4-^2 = 6 the
area of the triangle.
4. The wall of a building on the brink of a river is 120 feet, and
the breadth of the river is 70 yards ; what is the length of the chord
in feet that will reach from the top of the building across the river ?
Ans. 241-86 feet.
5. A ladder 60 feet long, will reach to a window 40 feet from the
dags on one side of a street, and by turning the ladder over to the
other side of the street, it will reach a window 50 feet from the flags;
required the breadth of the street ? Ans. 77*8875 feet.
6. The roof of a house, the side walls of which are the same
height, forms a right angle at the top, the length of one rafter being
10 feet, and its opposite one 14 feet; what is the breadth of the
house? Ans. 17*204.
PROBLEM X.
Given the base and perpendicidar of a right angled triangle^ to
find the perpendicidar let fall on the hypothenuse from the right
angle ; and also the segments into vAich the hypothenuse is
divided by this perpendicidar.
Rule. Find the hypothenuBe by Problem IX. Then
divide the square of the greater side by the hypothenuse,
and the quotient will give the greater segment, which
deducted from the entire will give the less. Having
found the segments, multiply them together, and the
square root of the product will give the perpendicular.f
* tsce Appendix, Demonstration 16. f IWd. IT.
W MENSURATION OF SUPERFICIES.
1. Given AC 3 yards, and CB 4 yards; required the segments
B D, DA, and the perpendicular D C.
33 + 42 = 25 ; then V 25 = 5 = A B.
42^5 = 16-T-5 = 3-2 =BD; then 5 — 3*2 = 1*8 = AD.
Again, 32 x I'S = 576; then ^5*76 = 2*4 = D C.
2. The roof of a house whose side walls are each 30 feet high,
forms a right angle at the top ; now if one of the rafters be 10 feet
long, and its opposite yoke-fellow 12, required the breadth of the
building, the length of the j)rop set upright to support the ridge of
the root, and the part of the floor at which it must be placed?
Ans. Breadth of the building 15*6204 feet, greater segment 9*2186
feet, lesser segment 6'4018 feet, and length of the prop 37*68 feet.
PROBLEM XI.
To find the area of a trapezium.
Rule. Divide the trapezium into two triangles, by joining two
of its opposite angles ; find the area of each triangle, and the sum of
both areas will give the area of the trapezium.
Or,
Draw two perpendiculars from the opposite angles to the diagonal ;
then multiply the sum of these perpendiculars by the diagonal, and
half the product will give the area.*
1. In the trapezium ABCD, the diagonal AC is 100 yards, the
perpendicular D E 35, and B F 30 ; what is its area?
DE = 35 D
BF=30
A<=
65
100
2)6500
3250 the area.
2. What is the area of a field, whose south side is 2740 links, easi
side 3575 links, north side 3755 links, west side 4105 links, and the
diagonal from south-west to north-east 4835 links ?
Ans. 123 acres 11*8633 perches.
3. In the trapezium ABCD, the side AD is 15, DC 13, CB 14,
and AB 12 ; also the diagonal A C 16 ; what is its area?
A71S. 172-5247.
4. In the trapezium ABCD, there are given A B 220 yards, D C
265 yards, and AC 378 yards; also AF 100 yards, and E C 70 yards;
what is its area ?
Ans. 85342*2885 yards=17 acres, 2 roods, 21 perches.
5. In the trapezium ABCD, there are given A B 220 yards, D C
265 yards, BF 195*959 yards, DE 255*5875 yards; also FE 208
yards; required the area of the trapezium? Ans. 85342*2885 yards.
6. Suppose in the trapezium ABCD, on account of obstacles, I can
only measure AB, DC, BF, DE, and FD, which are respectively 22
yards, 26 yards, 19 yards, 25 yards, and 32 yards ; required the area ?
Ans. 840*55 square yards.
* See Appendix, Demonstration 18.
MENSURATION OF SUPERFICIEa
PROBLEM XII.
To find the area of a trapezium inscribed in a circle^ or of any
one whose opposite angles are together equal to two right angles.
Rule. Add the four sides together, and take half the sum, from
this half sum deduct each side separately ; and the square root of the
product of the four remainders will give the area of the trapezium.*
1. What is the area of a four-sided field, whose opposite angles
are together equal to two right angles, the length of the four sides
being as follows, viz. AB 12-5, AD 17, DC 17-6, and BC 8 yards?
12-5
17 A
17-6
8
2)55
27-5
27-5 27-6 27-5
12-5 17 17-6
15 X 10-5 X 10 X 19-5=30712-50; then
30712-50= 175-26, the area in yards.
2. There is a trapezium whose opposite angles are together equal
to two right angles; the sides are as follows, viz. A B 25, AD 34,
D C 35, and B C 16 ; required its area? Ans, 70099.
PROBLEM XIII.
To find the area of a trapezoid,
Rqle. Multiply half the sum of the two parallel sides by the
perpendicular distance between them, and the product will give the
1. Let AB C D be a trapezoid, the
side AB = 40, DO=25, CP=18; re- \/
quired the area ?
40
G
25 A F PB
65-j-2 = 82-6x 18 = 585, area.
2. What is the area of a trapezoid, whose parallel sides are 750
and 1225 links, and the perpendicular height 1540 links?
Ans. 15 acres 33-2 perches.
3. What is the area of a trapezoid whose parallel sides are 4 feet
f> inches, and 8 feet 3 inches ; and the perpendicular height 5 feet 8
inches? Ans. 36 feet 1^ inches.
4. What is the area of a trapezoid whose parallel sides are 1476
and 2073 yards, and perpendicular height 976 yards?
Ans. 220 acres, 3 roods, 25 perches, 7 yards Irish.
* Boe Appendix, Demonstration 19l t I^id. 20.
30
MENSURATION OF SUPERFICIES.
PROBLEM XIV.
To find the area of an irregular polygon.
Rule. Divide the figure into triangles and trapeziums, and find
the area of each separately, by Problem IV. or XI. Add these areas
together, and the sum will be the area of the polygon.*
1. What is the area of the irregular polygon ABCDEFGA,
the following lines being given ?
A0= 9
GB = 29
Cw =11
GC=28-4
Fa;=U-5
'w/\
Cw =id
FD=36
E 2 = 7-4
G F
A0= 9
Cn =11
2)20 sum
10 half
29diag.-GB
290=areaof ABCGA,
C2/=13
E2;= 7.4
2)20-4 sum.
10^
35 FD
357 0 area of FCDEF.
Far = 14-5
i GC=14-2
205-9 areaofGFC.
290 =areaofABCGA
357 =areaofF CDE F
205-9 = area of GF C
Ans. 852-9 = area ofABCDEFGA.
2. In a five-sided field G C D E F G there is G C = 28 perches, F x
— 14 perches, C ?/ = 1 3 perches, 2 E = 7 perches, and F D = 35 perches ;
required its area? Ans. 3 acres, 1 rood, 26 perches.
* In finding the area of an irregular figure, draw a line through the extrema
angles of the figure, on which let fall perpendiculars from all the other anglesr
of the polygon, which will divide it into triangles and trapeasoids; then find
tho area of these by Problems IV. and XIIL
MENSURATION OF SUPERFICIES.
81
3. In the annexed figure, there are given in perches,
AX = 15
XR ■
RT :
. TD
AP
PS :
SD :
GX
FR :
ET
BP :
CS =
Required the area ?
14
6
17
14
12
6
10
12
20
14
Ans, 4 acres, 3 roods, 19J perches.
PROBLEM XY.
To find the area of a regular polygon.
Rule I. Add all the sides together, and multiply half the sum by
tlie perpendicular drawn from the centre of the polygon t<\) the middle
of one of the sides, and the product will give the area. This perpen-
dicular is the radius of the inscribed circle.
Rule II. Multiply the square of the side of the polygon by the
number standing opposite to its name in the following table, under
ihe word area, and the product will give the area of the polygon.
Rule III. Multiply the side of the polygon by the number standing
opposite to its name in the column of the following table, headed
'^ Radius of Inscribed Circle," and the product will be the perpendicu-
lar from the centre of the polygon to the middle of one of its sides ;
tiien multiply half the sum of the sides by this perpendicular, and the
product will give the area.*
TABLE II.
When the side of the polygon is 1.
I
[No. of
Sitles.
3
Radius of inscrib-
ed Circle.
Area of Tolygon.
0-2886751
0-4330127 -
f tan. 30° = ^V3
4
0-6000000
1-0000000 =
1 tan. 45°= 1 x 1
5
0-6881910
1-7204774 =
f tan. 54°=|V(l+iV5)
6
0-8660254
2-5980762 =
f tan. 60° = f V3
7
1-0382617
3-6339124 =
1 Um. 64°f
8
1-2071068
4-8284271 =
f tan. 67°| = 2x(l+ V2)
9
1-3737387
6-1818242 =
1 tan. 70°
10
1-5388418
7-6942088 =
V tan. 72°=:|V(3 + 2V^'i)
11
1-7028437
9-3656404 =
V tan. 73VV
12
1-8660254
11-1961524 =
V.tan. 75° = 3x(2+ V3)
♦ Eee Appeodix, Demonstration 21.
C
32 MENSURATION OF SUPEKFICIES.
Note. — ^Tlie radius of the circumscribed cii'cle, when the side of the polygon
is 1, may be seen in Table I.
The expressions in the fourth column may be seen in Trigonometry, to
ivhich the pupil is referred for a full investigation of them. The tangents of
the angle O a G iu the heptagon, nonagon, and undecagon, are extremely
difficult to be found without a table of tangents.
1 . The side of a pentagon is 20 yards, and the perpendicular from
the centre to the middle of one of the sides is 13*76382 ; required the
area ?
By Rule I. 20x5x 13'76382-^2 = 1376-3824-2 = 688-191. Ans,
By Rule II. 20 x 20 x 1-720477 = 688*19, the area as before.
2. The side of a hexagon is 14, and the perpendicular from the
centre 12*1243556 ; required the area? Ans. 509*2229352.
3. The side of an octagon is 5*7, required its area?
Ans. 156*875596479.
4. The side of a heptagon is 19*38 yards, what is its area ?
Ans. 1364*84.
5. The side of an octagon is 10 feet, what is its area ?
Ans. 482*84271.
f>. The side of a nonagon is 50 inches, what is its area ?
Ans. 15454*5605.
7. The side of an undecagon is 20, what is its area ?
Ans. 3746-25616.
8. The side of a dodecagon is 40 yards, what is its area ?
Ans. 17913-84384.
PROBLEM XYI.
Given the diameter of a circle^ to find the circumference; or^ t'^^
circumference to find the diameter, and thence the area.
Rule.*
I. Say as 7 : 22 : : the given diameter : cir-
cumference.
Or, as 113 : 355 : : the diameter : the circum-
^ference.
Or, as 1 : 3*1416 : : the diameter : the circum-
ference.
II. Say as 22 : 7 : : the given circumference :
the diameter.
Or, as 355 : 113 :: the circumference : the
diameter.
Or, as 3*1416 : 1 : : the circumference : the diameter.
1. The diameter of a circle. is 15, what is its circumference?
7 : 22 : : 15 : 22 X 15-r-7 = 330-T-7 = 47*142857.
Or, 113 : 355 : : 15 : 355 X 15-j-113 = 5325-^113 = 47'124.
Or, 1 : 3*1416 :: 15 : -3*1416x15 = 47*124.
2. The circumference of a circle is 80, what is its diameter ?
22 : 7 : : 80 : 7 X 80^22=25*45.
355 : 113 :: 80 : 113 X 80-^355 = 25*4647.
8*1416 ; 1 : : 80 : 80^3*1416 = 25-4647.
♦ See Appendix, Deraonstraticn 7lSi.
MENSURATION OF SUPERFICIES. 33
8. What is the circumference of a circle whose diameter is 10 ?
Ans. 31-4285.
4. What is the diameter of a circle whose circumference is 50 ?
s Ans. 15-909.
5. The diameter of the earth is 7958 miles, what is its circum-
terence? ^W5. 25000*8528 miles.
6. The circumference of the earth being 25000-8528 miles, what
is its diameter V Ans, 7958 miles.
PROBLEM XVII.
To find the length of an arc of a circle.
Rule I. Multiply tlie radius of the circle by the number of de-
grees in the given arc, and that product by -01745329, and the last
product will be the length of the arc*
Rule II. From eight times the chord of half the arc, subtract the
thord of the whole arc, one- third of the remainder will give the
length o( the arc, nearly.f
1. If the arc AB contain 30 degrees, the
radius being 9 feet, what is the length of the
arc?
30x9 = 270, and 270 x -01745329 = 4-7124.
2. If the chord A D of half the arc A D B be
20 feet, and the chord A B of the whole arc 38 ;
what is the length of the arc ?
20x8— 38= 122; then 122^3 = 401 feet. ^?2«.
3. The chord of an arc is 6 feet, and the chord of half the arc
is 3J ; required the length of the whole arc ? Ans. 7^.
4. The chord of the whole arc is 40, and the versed sinej or
height of the segment 15 ; what is the length of the arc?
Ans. 53^.
5. The chord AB of the whole arc is 48*74, and the chord AD of
half the arc 30-25 ; required the length of the arc?
6. AB = 30, DP=8; required the length of the arc?
Ans, 35 J.
PROBLEM XVIII.
To find the area of a circle.
Rule I. Multiply half the circumference by half the diameter,
for the area.§
Rule II. Multiply the square of the diameter by -7854, for the
area. II
Rule III. Multiply the square of the circumference by -07958.1
Rum IV. As 14 to 11, so is the square of the diameter to the
area.
Rule V. As 88 to 7, so is the square of the circumference to the
area.
* See Appendix, Demons tJ-ation 23. f Ibid. 24.
t By "versed sine," in works on mensuration, is not meant the trigono
»netrical versed sine of the whole arc, hut of half the arc.
g See Appendix, Demonstration 22, Jl Ibid. '28. % Ibid. 25.
84 MENSURATION OF SUPERFICIES.
1. To find the area of a circle whose diameter is 100 and circunr
fereiice3U-J6.
By Rule I. By Rule II. By Rule III.
314-16 -7854 98696*5 sq. cir.
100 100'^= 10000 -07958
4)31416 Area 7854 7854- Area.
Area 7854
By Rule IV. By Rule V.
100^ = 10000 98696-5 sq. cir.
11 7
2)110000 8)090875-5
7)55000 11)86359-4
Area 7857 7850-85
2. What is the area of a circle whose diameter is 7 ?
Ans. 38 J nearly.
3. How many square yards are in a circle whose diameter is
l^yard? Ans. 1-069.
4. The surveying wheel turns twice in the length of 16| feet ; in
going round a circular bowling-green it turns exactly 200 times (
how many acres, roods, and perches in it ?
Ans. 4 acres, 3 roods, 35*8 perches.
6. The circumference of a fish-pond is 56 chains, what is its area!
A71S. 249-56288.
6. What is the area of a quadrant, the radius being 100 ?
Ans. 7854.
?. Required the length of a chord fastened to a stake at one end,
and to a cow's horns at the other, so as to allow her to feed on an
acre of grass and no more? . Aiis. Sd^ yards.
8. The circumference of a circle is 91, what is its area?
Ans. 659-00198.
9. The diameter of a circle is 15 perches, what is its area?
Ans. 176-715.
10. What is the area of the semicircle of which 20 is the radius ?
Ans. 628-32.
PROBLEM XIX.
Given the diameter of a circle., to find the side of a square equal m
area to the circle.
Rule. Multiply the diameter by -8862269, and the product will
be the side of a square equal in area to the circle.*
1. If the diameter of a circle be 100, what is the side of a square
equal in area to the circle? Ans, 88-62269.
2. The diameter of a circular fish- pond is 200 feet, what is the side
of a Rquare fish-pond equal in area to the circular one ?
Ans. 177-24538.
• See Appendix, Demonstration 26.
MENSURATION OF SUPERFICIES. 36
PROBLEM XX.
Given the circumference of a circle^ to find the side of a square
^ equal in area to the circle.
Rule. Multiply the circumference by -2820948, and the product
will be the side of the square.*
1. The circumference of a circle is 100, what is the side of a square
equal in area to the circle? Ans. 28*20948.
2. The circumference of a round fish-pond is 200 yards, what is
the side of a square fish-pond equal in area to the round one?
Ans. 56-41896.
PROBLEM XXL
Given the diameter., to find the side of the inscribed square.
Rule. Multiply the diameter by -7071068^^ t-^^ ^'^
and the product will give the side of the in-
scribed square, f
1. The diameter of a circle is 100, what is
the side of the inscribed square ?
Ans. 70-71068.
2. The diameter of a circle is 200, what is
the side of the inscribed square ?
Ans. 141-42136.
PROBLEM XXIL
Given the area of a circle^ to find the side of the inscribed squa/ c
Rule. Multiply the area by -6366197, and extract the square
i'oot of the product, which will give the side of the inscribed square. J
1. The area of a circle is 100, what is the side of the inscribed
square? Ans. 7-97884.
2. The area of a circle is 200, what is the side of the inscribed
square ?
200 X -6366197= 127-3239400; then V127-3239400 = 11-2837. Ans.
PROBLEM XXIIL
Given the side of a square., to find the diameter of the
circumscribea circle.
Rule. Multiply the side of the square by 1-4142136, and the pro >
duct will give the diameter of the circumscribed circle.§
1. If the side of a square be 10, what is the diameter of the cir-
cumscribed circle ? Ans. 14-142136.
2. If the side of a square be 20, find the diameter of the circum-
scribed circle ? ^n5. 28-284272.
PROBLEM XXIV.
Given the side of a square., to find the circumference of the
circumscrined circle.
Rule/ Multiply the side of the square by 4*4428934, and the
product will be the circumscribed circle ? 1|
* See Appendix, Demonstration 27. t Ibid. 28. X Ibid. 29.
§ Ibid. 30. il Ibid. 31.
86 MENSURATION OF SUPERFICIES.
1. If the side of a square be 100, what is the circumference of tho
circumscribed circle ? Ans. 444*28934.
2. If the side of the square be 30, what is the circumference of the
circumscribed circle? Ans. 133'286802.
PROBLEM XXV.
Given the side of a square^ to find the diameter of a circle equal
in area to the square.
Rule. Multiply the side of the square by 1-1283791, and the
product will be the diameter of a circle equal in area to the square
whose side is given.*
1. If the side of a square be 100, what is the diameter of the
circle whose area is equal to the square whose side is 100?
Ans. 112-83791.
2. What is the diameter of a circle equal in area to a square whose
side is 200 ? Ans. 225-67582.
PROBLEM XXYI.
Given the side of a square^ to find the circumference of a circle
whose area is equal to the square whose side is given.
Rule. Multiply the side of the square by 3-5449076, and the
product will give the circumference of a circle equal in area to the
given square.f
1. What is the circumference of a circle, whose area may be equal
to a square whose side is 100? Ans. 354*49076.
2. Find the circumference of a circle equal in area to a square
whose side is 300? Ans. 1063-47228.
PROBLEM XXVII.
To find the area of a sector of a circle.
^ Rule I. Multiply half the length of the arc by the radius of the
circle, and the product is the area of the sector. J
Rule II. As 360 is to the degrees in the arc of the sector, so is
the area of the whole circle to the area of the sector.§
1. Let A C B 0 be a sector less than a semi-
circle whose radius A 0 is 20 feet, and chord
A B 30 feet, what is the area ?
Eirst, V (A 02 - A D2) = V 400 - 225) =
13-228 = OD; then 0 C-0 D = 20- 13-228
= 6-772 = CD.
Again, V (A Ds + C Da) = V 225 +
45-859984) = 16-4578 = AC, the chord of half
the arc.
Hence, by Problem XVII. the arc A B is 33-8874 ; then
qo.OQ'T'J.
■ — 2 ^ ^^ = 338-874, the area required.
* See Appendix, Demonstration a ; f Ibid. 33. t Ibid. 34. f Ibid. 35
MENSURATION OF SUPERFICIES.
2. Let aEFBOA be a sector greater
than a semicircle, whose radius A 0 is 20,
the chord E B, 38, and chord B F of half
E F B, ^23 ; required the area ?
23 = chord BF
8
87
184
38 = chord BE
3)146
48-666 &c. = arc BFE
20
973J area.
3. What is the area of a sector whose arc contains 18 degrees, the
diameter being 3 feet ?
•7854
18 : : 7*0686 : the area of the sector ;
1 : : 7-0686 : -35343. Ans.
Then 360 ;
Or, 20 :
4. What is the area of a sector whose arc contains 147 degrees
29 minutes, and radius 25 ? Ans. 804-3986.
5. What is the area of a sector whose arc contains 18 degrees, the
radius being 3 feet? Ans, 1-41372/
PROBLEM XXVIII.
Tofina the area of the segment of a circle.
Rule I. Find the area of the sector having the same arc with
tlie segment, by the last problem; find also the area of the triangle,
formed by the chord of the segment and the two radii of the sector.
Tlien add these two areas together, when the segment is greater
than a semicircle, but find their difference when it is less than a
semicircle, the result will evidently be the answer.
1. What is the area of the segment A C B D A,
its chord A B being 24, and radius A E or E C
20?
V(AE2 — AD2) = V (400 — 144) = 16
= DE; EC— ED = 20 — 16 = 4 = CD;
V (AD2 + DC2) = V (144 4- 16) =
12-64911 = AC; then^^^ x 8) - 24 ^
= AC;
25-7309 = arc A CB,
And 12-8654 = half arc
20 = radius
3
257-308 = area of sector E B CA.
192 =areaofAABE
192 = areaof AABB
65-308 = area of segment A B C A .
58 MENSURATION OF SUPERFICIES.
2. Let AG F B A be a segment greater than a semicircle, there are
given the chord AB 205, FD 17-17, AF 20, FG 11-5, and AE
11*64 ; required the area of the segment?
(FGx8)— AF (11-5x8) — 20 , , , . ,
^ = ^^ ^ = 24 the length of the art
AGF (Problem XYII.) ; then 24 x 11-64 = 279-36, area of sector
AEBFGA (Problem XXVII.) Again, FD - EF = 17-17 —
,1^.1 K -o vT. .u ABxED 20-5x5-53 ^^ ^oo^
Il-64 = 5-o3 = LD; then ^ = o = 56 6825
the area of the triangle ABE, which being added to the area of the
sector before found will give the area of the segment, viz., 279-36 +
56-6825 = 336-0425 the area of the segment A G F B A.
Rule II. To two-thirds of the product of the chord and versed
sine of the segment, add the cube of the versed sine divided by twice
the chord, and the sum will give the area of the segment, nearly.
When the segment is greater than a semicircle, find the area of the
remaining segment, and deduct it from the area of the whole circle,
the remainder will give the area of the segment.*
3. What is the area of the segment AC B, less than a semicircle,
Us chord being 18*9, and height or versed sine DC 2-4?
AB X DC = 18-9 x 2-4 = 45*36, and | AB x DC = | x
45-36 = 30-24; then --?^^^ = -36571 ; hence 30-24 + •36571 =
^ X lO V
30-60571 the area.
Note. — If two cliords of a circle cut one another, the rectangle contained by
the segments of one of them is equal to the rectangle contained by the seg-
ments of the other. This is the 35th Proposition of Book III. of Euchd.
4. Required the area of the segment AG FB whose height FD is
20, and chord AB 20?
^?=^= 10 = AD, and AD2= 100; but AD2=FD x DC
'•^^ - FD-20-^-
The area of the segment ACB is, by the last case, 69 7916; and
the area of the whole circle, by Problem XYIIL, is 490*875; then
490-875— 69-7916 = 421-0834 = area of the segment AGFB.
5. W'hat is the area of the segment AGFB, greater than a semi-
circle, whose chord AB is 12, and versed sine 18? Ans. 297*81034.
Rule III. Divide the height of the segment by the diameter of the
civcle, to three places of decimals. Find the quotient in the column
Height of the table at the end of the practical part of this treatise,
and take out the corresponding Area Seg., which multiply by the
square of the diameter, and the product will be the area of the seg-
ment required.^
Note. I. — If the quotient of the height by the diameter be greater than '5 sub-
tract it from 1, and find the Area Seg. corresponding to the remainder, whicli
subtract from •7864 for the correct Area Seg.
* Se<9 Appendix, Demonstration 36. t Ibid. 37.
MENSURATION OF SUPERFICIES.
89
Kote 11.— If the quotient of the height by the diameter does not terminato
In three figures, find the Area Seg. corresponding to the first three decimal
figures of the quotient, subtract it from tlie next greater Area Seg., multiply
the remainder \^ the fractional part of the quotient, and add the product to tho
area segment first, taken out of the table. When great accuracy is not required,
the fractional part may be omitted.
6. Let the diameter be 20, and the versed sine 2, required the area
of the segment ?
^2^= -1, to which answers -040875
Square of diameter, 400 «
16-35 area.
7. What is the area of a segment, whose diameter is 52, and
versed sine 2 ?
^2^=-038y\ which is the tabular versed sine. Then to '038 an-
swers '009763, and the difference between this area and the next is
•000385, which multiplied by j\ gives -000177, which added to
•009763 gives -009940, which is the area corresponding to the
versed sine •038xV Then 52^ x 009940 = 26-87776 is the area
required.
PROBLEM XXIX.
Tojind the area of a zone^ or the space included by two parallel
chords and the arcs contained between them.
Rule. Join the extremities of the parallel chords towards the same
parts, and these connecting lines will cut off two equal segments, the
areas of which, added to the area of the trapezoid then formed, wiU
give the area of the zone.
1. Suppose the greater chord A B = 30, the
less CD 20, and tlie perpendicular distance
D a: = 25, required the area of the zone
ABDC?
i (AB-CD) = a:B=J (30-20) = 5: then
V (x D2 + X B'O = D B = V (25^ + 52) =
25-49. A B— B a: = A a; = 30—5 = 25, and
( A a: X B a:) -^ D a; = F a; = (25 X 5) ~ 25
= 5. Da: + Fa;=DF = 25+5 = 30;i
V(CD2 + iyY-2) =^CF=G2; = J V(20'
+ 30'-^) = 18-027, the radius of the circle; (DBxAa:)-r-2Da:=:
G 2/* = (25-49 X 25) ~- {2 x 25) = 12-745 ; G z - G y = z y =z
18-027 - 12-745 = 5-282, the height of the segment A 2; C.
36 -05)5 '28(- 146, the tabular area seument answering to which is
•071033, then '071033 x (36-05)^ = 92*315 = the area of the seg-
ment AzC.
J (A B + C D) X D a; = H30 + 20) X 25 = 625 the area of tho
trapezoid ABDC: then 625 + 92.315 x 2 = 809-63 = the area of
the zone.
2. Let the chord AB be 48, the chord CD 30, the chord AC
15*81 14 ; what is the area of the zone ABDC?
Ans. Ti> diameter C F = 50, height of the segment A 2; 0=3
* Se© Appendix, Demonstration 33.
40 MENSURATION OF SUPERFICIES.
1-2829, area by the table of segments = 13-595. Area of the zone
ABDC = 534-19?
3. Let AB = 20, C D = 15, and their distance = 17j: required
the area? Ans. 395*4369.
4. Let AB = 96, CD = 60, and their distance = 26; required
the area? Ans. 213Q'7627.
PROBLEM XXX.
Tojind the area of a circular ring^ or of the space included
between two concentric circles.
Rule. Multiply the sum of the two diameters by their difference,
and the product arising by '7854 for the area of the ring.*
1. The diameter AB is 30, and CD 20;
what is the area of the ring XX?
30
20
60 sum
10 difference
600
•7854
392-7000 area of the ring X X.
2. "What is the area of the circular ring, when the diameters are
40 and 30? Ans. 549-78.
3. What is the area of the circular ring, when the diameters are
60 and 45? Ans. 373-065.
PROBLEM XXXI.
To find the area of apart of a ring^ or of the segment of a sector
Rule. Multiply half the sum of the bounding arcs by their dis*
tance asunder, and the product will give the area.f
1. Let AB be 50, and ah 30, and the distance
a A 10 ; what is the area cf the space a & B A ?
. 50 + 30 ,^ ,^^
Ans. ^ — X 10 = 400
2. Let A B = 60, a & = 40, and a A = 2 ; re-
quired the area of the space a 5 B A ? Ans. 100.
3. Let AB = 25, a5 = 15, and aA = 6; re-
quired the area of the segment of the sector ?
Ans. 120.
PROBLEM XXXII.
To find the area of a lune, or the space included between the
intersecting arcs of two eccentric circles.
Rule. Find the areas of both segments which form the lune, and
deduct the less from the greater ; the remainder will evidently be
the area required.
• See Appendix, Demonstration 39* i Ibid. 40.
MENSURATION OF SUPERFICIES.
41
1. Let the chord AB = 40, EC = 12,
and E D = 4 ; what is the area of the
lune ADBCA?
By note, page 38, (A E^ ^ E C) + E C
= diameter of the circle of which A C B
is an arc ; and (AE'^ -4- E D) + E D =
the diameter of the circle of which A D B is an arc ; hence (20" -^ 12)
+ 12 = 45-3; and (202-^4) + 4=104; are the two diameters.
12 +- 45-3 = -264. 4 -^ 104 = '038.
The Area Seg. answering to -264 is -165780, and (45-3)2 x
•165780 = 340-1954802 = area of the segment AE B C A.
The Area Seg. answering to '038 is -009763, and (104)2 x -009763
= 105-596608 = area of the segment A E B D A ; then 340-1954802
— 105-596608 = 234-5988722 = the area of the lune.
2. Let the chord A B be 40, and the heights of the segments E C
and E D 15 and 2 ; required the area of the lune? Ans. 388*5
PROBLEM XXXIIL
TO MEASURE LONG IRREGULAR FIGURES.
When irregular figures^ not reducible to any known figure^ present
themselves^ their contents are best found by the method oj
equi-distant ordinates.
Rule. Take the breadths in several places, at equal distances, and
divide the sum of the first and last of them by 2 for the arithmetical
mean between those two. Add together this mean and all the other
breadths, omitting the first and last, and divide their sum by the
number of parts so added, the. quotient will give the mean breadth
of the whole, which being multiplied by the given length will give
the area of the figure, very nearly.
It is not necessary sometimes to take the breadths at equal dis-
tances, but to compute each trapezoid separately, and the sum of all
the separate areas thus found will give the area of the entire nearly.
Or, add all the breadths together and divide by the number of
them for a mean breadth, which being multiplied by the length, as
before, will give the area nearly.
1. Let the ordinate AD be 9-2, bf 7, eg 9, dh 10, BC S'^
and the length AB 30 ; required the area?
9-2 AD
8-8 BO
2)18
9 mean breadth of first and last.
7 bf
9 eg
10 dh
A b
4)35 sum
8-75 mean
30
262-50 area of the whole figure
8-75 mean breadth of all.
30
42 EXERCISES IN
2. The length of an irregular figure is 39 yards, and its breiidths,
in five equi-distant places, are 4-8, 5*2, 4*1, 7*3, and 7'2; what is
its area? Ans. 220*35 square yards.
3. The length of an irregular figure is 50 yards, and its breadths,
at seven equi-dlstant places, are 5*5, 6*2, 7*3, 6, 7*5, 7, and 8'8 ;
what is its area? Ans. 342*9150 square yards.
4. The length of an irregular figure being 37*6, and the breadths,
at nine equi-distant places. 0, 4*4, 6*5, 7*6, 5*4, 8, 5*2, 6 5, 6-1;
what is the area? Ans. 218*315.
EXEECISES.
1. Find the area of a square whose side is 35*25 chains.
Ans. 124 acres, 1 rood, 1 perch.
2. Find the area of a rectangular board, whose length is 12J feet,
and breadth, 9 inches. Ans. 9| feet.
3. The sides of three^uares being 4, 5, and 6 feet, what is the
length of the side ofjj(™fe which is equal to all three?
'^^ \ Ans. 8*7749 feet.
4. Required the area of a rtomboid whose length is 4^1, chains,
and breadth, 4*28 chains? ^A7is. 4 acres, 1 rood, 39']^erclies.
5. There is a triangle whose base is 12*6 chains, and altitude 6*4
chains, wliat is its area? Ans. 40-32.
6. Find the area of a triangle whose sides are 30, 40, and 50 yards.
Ans. 600 square yards.
7. There is a triangular corn-field whose sides are 150, 200, and
250 yards, determine the number of acres contained in the field, and
the expense of reaping the corn at 9s. 6d. per acre.
Alls. Content of the field, 3 acres 15 perches ; expense of reaping,
£1, 9s. 5d.
8. What must the base of a triangle be to contain 36 square feet,
whose vertex is to be 9 feet from the base? Ans. 8 feet.
9. What must be the altitude of a triangle equal in area to the last,
whose base is 12 feet? Ans. 6 feet.
10. The height of a precipice standing close by the side of a river
is 103 feet, and a line of 320 feet will reach from the top of it to the
opposite bank; required the breadth of the river? Ans. 302 97 feet.
11. A ladder 12|- feet in length stands upright against a wall, how
far must the bottom of it be pulled out from the wall so as to lower
the top 6 inches? Ans. 3 J feet.
12. A person wish -
:*ng to measure the
distance from a point
A, at one side of a
canal, to an object 0,
at the other, and
having no mstrument
but a book, placed a
MENSURATION OF SUPERFICIES. 43
comer of it on the point A, and directed an edge of it, as in tha
figure, in a straight line with the object 0, and drew the straight
lines A B, A C ; he then placed the book so that a comer of it rested
on the point B, at the distance of eight times its length from the
point A, and directed an edge of it, as before, to the object 0, and
drew the straight line BC which met AC at the distance of three
times the length of the book from A ; how many times the length of
the book is the object 0 tVom the points A and 6 ?
Ans. 21 i and 22*78 times.
13. What is the area of a trapezium whose diagonal is 70*5 feet,
and the two perpendiculars 26*5 and 30*2 feet?
Ans, 1998*675 square feet.
14. "What is the area of a trapezium whose diagonal is 108 feet 6
inches, and the perpendiculars 56 feet 3 inches, and 60 feet 9 inches?
Ans. 6347 feet 36 inches.
15. What is the area of a trapezoid whose two parallel sides are
75 and 122 links, and the perpendicular distance 154 links?
Ans. 15169 square links.
16. A field in the form of a trapezoid, whose parallel sides are
6340 and 4380 yards, and the perpendicular distance between thena
121 yards, lets for £207, 14s. per annum; what is that per acre?
Ans. £1, lis.
17. Two opposite angles of a four-sided field are together equal to
t^70 right angles,, and the sides are 24, 26, 28, and 30 yards ; what
is its area? Ans. 723*99 square yards, nearly.
18. Kequired the area of a figure similar to that annexed to the
first question under Problem XIV., whose dimensions are double of
those there given ? Ans. 3411-6.
19. What is the side of an equilateral triangle equal in area to &
Bquare, whose side is 10 feet? Ans. 15*196 feet, nearly.
20. Required the area of a regular nonagon, one of whose sides is
8 feet, and the perpendicular from the centre = 10*99 feet?
Ans. 395*64 square feet.
21. Required the area of a regular decagon, one of whose sides is
20*5 yards? Ans. 3233*491125 square yards.
22. A wheel of a car turns round 4400 times in a distance of 10
jniles; what is its diameter? Ans. 3*819708 feet.
23. If the diameter of a circle be 9 feet, what is the length of the
circumference? Ans. 28f feet, nearly.
24. Required the length of an arc of 60°, the radius of the circle
being 14 feet? Ans. 14*660772 feet.
25. The chord of an arc is 30 feet and the height is 8 feet, what is
the length of the arc? Ans. 35^ feet, nearly.
26. The diameter of a circle is 200, what is the area of the quad-
rant? Ans. 7854.
27. The diameters of two concentric circles are 15 and 10, what is
the area of the ring formed by those circles ? Ans. 98*175.
28. The circumference of a circle is 628*32 yards, what is the
radius of a concentric circle of half the area ? Ans. 70*71.
29. What is the side of a square equal in area to the circle whose
diameter is 3? Ans 2*6586807
44 EXERCISES IN MENSURATION OF SUPERFICIES.
30. The two parallel chords of a zone are 16 and 12, and theif
perpendicular distance is 2, what is the ai'ea of the zone?
Ans. 28-376.
31. The length of a chord is 15, and the heights of two segments of
circles on the same side of it are 7 and 4 ; wliat is the area of the
lune formed by those segments? Ans. 38, nearly.
32. The base and perpendicular of a right-angled triangle are each
1, what is the area of a circle having the hypothenuse for its
diameter? Ans. 1*5708.
33. If the area of a circle be .^00, what is the area of the inscribed
SQuaie? dns. 63-66.
CONIC SECTIONS.
45
OOmO SECTIONS.
SECTION III.
(or THE ELLIPSIS*^
PROBLEM I.
llie transverse and conjugate diameters of an ellipsis being given ^
to find the area.
Rule. Multiply the transverse and conjugate diameters together,
and the product arising by '7854:, and the result will be the area.f
1. Let the transverse axis be 35, and the conjugate axis 25;
required the area? . 35 x 25 x •7854 = 687'225. Ans.
2. The longer diameter of an ellipse is 70, and the shorter 50 ;
what is the area? ^W5. 2748-.9.
3. What is the area of an ellipse whose longer axis is 80, and
shorter axis is 60? Ans. 3709-92.
4. What is the area of an ellipse, whose diameters are 50 and 45?
Ans, 1767-15.
PROBLEM IL
To find the area of an elliptical ring.
Rule. Find the area of each ellipse separately, and their difference
will be the area of the ring.
Or, From the product of the two diameters of the greater ellipse
deduct the product of the two diameters of the less, and multiply
the remainder by -7854: for the area of the ring. J
Q
1. The transverse diameter A B is 70, and
the conjugate CD 50; and the transverse
diameter E F of another ellipse having the same
centre 0, is 35, and the conjugate GH is 25;
required the area of the elliptical space between
their circumferences ?
^"^or definitions of the ellipsis (or, as it is frequently written, ellipse) and
k. lEier Conic Sections, see Appendix, Properties of the Conic Sections.
t JSeti Appendix, Demonstration 41. J Ibid. 42.
4C CONIC SECTIONS.
70 X 50 X -7854 = 27489 ; and 35 x 25 x -7854 = G87-226 ; tbcn
2748'9 — 687*225=: 2061-675 = area of the elliptical ring.
70x50 = 3500
35x25= 875
2625 X -7854 = 2061 '675 = area
2. The transverse and conjugate diameters of an elh'pse are 60 and
40, and of another 30 and 10 ; required the area of the space between
their cu'cumferences? Ans. 1649 •34.
3. A gentleman has an elliptical flower garden, whose greater
diameter is 30, and less 24 feet ; and has ordered a gravel walk to be
made round it of 5 feet 6 inches in width ; required the area of the
walk? Ans. 371-4942.
PROBLEM III.
Given the Tieiglit of an elliptical segment., whose base is parallel to
either of the axes of the ellipse] and the two axes of the ellipse^
to find the area.
Rule. Divide the height of the segment by that diameter of which
it is a part, to three places of decimals, find the quotient in the
column Height of the table referred to in page 38, and take out the
correspondent Area Seg. Multiply the Area Seg. thus found and
both the axes of the ellipsis together, and the result will give the
area required.*
1. Required the area of an elliptical seg-
ment R A Q, whose height A P is 20 ; the
transverse axis AB being 70, and the
tonjugate axis CD 50 ?
20 -7- 70 = •285f = the tabular versed
sine, the corresponding segment answering
.^0 which is -185166; then •]85166x70x
50=648-081, the area.
2. What is the area of an elliptical segment cut off by a chord
parallel to the shorter axis, the height of the segment being 10, and
the two diameters 35 and 25 ? Ans. 162-0202.
3. What is the area of an elliptical segment cut off by a chord
parallel to the longer axis, the height of the segment being 10, and
the two diameters 40 and 30? Ans. 275-0064.
4. What is the area of an elliptical segment cut off by a chord
parallel to the shorter diameter, the height being 10, and the two
diameters 70 and 50 ? ^?is. 240-884.
PROBLEM IV.
To find the circumference of an ellipse., by having the two
diameters given.
Rule. Multiply the sum of the two diameters by 1*5708, and the
product will give the circumference nearly; that is, putting t for the
transverse, c for the conjugate, andjo for 3*1416 j the circumference
will be {t+c}x^p.f
* Soe Apftendis, Demonstration 43. ♦ TbM. 44.
OF THE ELLIPSIS. 47
1. Let the transverse axis be 24, and the conjugate 18; required
the ^B^ circvwierence-
(24 + 18) X 1-5708 = 42 xl'5708 = 65'9736 is the circumference
Dearly.
2. Required the circumference of an ellipse whose transverse axis
fs 30, and conjugate 20? Ans. 78*54.
3. Required the circumference of an ellipse whose diameters are
60 and 40? Ans. 157-08.
4. What is the circumference of an ellipse whose diameters are
6 and 4? Ans, 15-708.
5. What is the circumference of an ellipse whose diameters are
3 and 2? Ans. 7*854.
PROBLEM V.
To find the length of any arc of an ellipse.
Rule. Find the length of the circular arc xy^ intercepted by 0 C,
0 B, and whose radius is half the sum of 0 C, OB: and it will be
equal to the elliptical arc B C nearly.*
Note. — ^The nearer the axes of the ellipse apin-oach towards equality, the
uiore exact the result of the operation by this R\ile ; and the less the elliptical
arc, the nearer its exact length will approach the arc x y.
1. Let the axis AD be 24, CK 18, and 0 T 3 ; required the
length of the arc B C ?
Here we have TD = 9, and AT=15;
then from the property of the ellipsis, we
have A02 : 0 C2 : : ATxT D : T Ba=
9^x9x15 9x9x15 , ^ ^j //rkT2.
-i2^rT2-=— 16-'^"^^^=^^^^ +
T B2)= V (9 + ^^^f^) =9-21616, the radius
of the circle of which G B is an arc ; but 0 C is the radius of the
circle of which C V is an arc; therefore the radius of the circle of
which a; V is an arc, is \ 0 C + ^-O B = 9-10808. But by Trigono-
me^r?/,tHB-^OB = 3-4-9-21616 = -325515, is the sine of the angle
COB, or arc xy^ to the radius 1, answering to 18*9968 degrees.
Therefore, by Problem XYII. Rule I., the length of the arc xy is
•01745 X 18-9968 x 9-10808 = 3-0192, which is also equal to the length
of the elliptical arc CB, nearly.
2. Given A D 30, C K 20, and 0 T 5 ; required the length of
the arc B C ? Ans. 5-03917786255.
3. Given AD 40, CD 30, and 0 T 5 ; required the length of
the arc B C? Ans. 5-033880786.
* See Appendix, Demonstration 45.
t It may be done without Trigonometry, by first finding the lencrth of tho
aic B by Rule XL Prob. XVII. Sec. 2, tbeu OG:Oy::GB:YZ.
D
48 CONIC SECTIONS.
PROBLEM VI.
Given the diameter and abscissas^ tojind the ordinate.
Rule. Say, as the transverse is to the conjugate, so is the square
root of the rectangle of the two abscissas, to the ordinate.*
1. In the ellipse A C D K, the transverse diameter A D is 100,
the conjugate diameter C K 80, and the abscissa DT 10; required
the length of the ordinate T B?
100 : 80 : : V (90 x lO) : T B = 24. (See the last figure.)
2. Let the transverse axis be 35, the conjugate 25, and the
abscissa 7 ; required the ordinate. Ans. 10.
3. Given the two diameters 70 and 60, and the abscissa 10 ; re-
quired the ordinate? Ans. 209956.
PROBLEM VII.
Given the transverse axis^ conjugate and ordinate^ to find the
abscissas.
Rule. As the conjugate is to the transverse diameter, so is the
equare root of the difference of the squares of the ordinate and semi-
conjugate, to the distance between the ordinate and centre. Then
this distance being added to, and subtracted from, the semi- diameter,
will give the two abscissas, f
1. Let the diameters be 35 and 25, and the ordinate 10 ; required
the abscissas?
By the Bale f + |f V ([f ]' - 10«) = — ^^ = 28 ami 7,
the two abscissas.
2. Let the diameters be 120 and 40, and the ordinate 16; re-
quired the abscissas ? i4?i5. 96 and 24.
PROBLEM VIII.
Given the conjugate axis^ ordinate^ and abscissas^ to find the
transverse axis.
Rule. Find the square root of the difference of the squares of the
semi- conjugate axis and the ordinate, which add to, or subtract from,
the semi-conjugate, according as the less abscissa or greater is given.
Then say, as the square of the ordinate is to the rectangle of the
conjugate, and the abscissa, so is the sum or difference found above
to the transverse required. J
1. Let the ordinate be 10, and the less abscissa 7; what is the
diameter, allowing the conjugate to be 25?
V( [yT"~^^^ ) = 7-5 ; then 7*5 + 12-5 = 20 ; then 10^ : 25
X 7 : : 20 : 35, the transverse required.
2. Let the ordinate be 10, the greater abscissa 28, and the con-
jugate 25 ; required the transverse diameter ? Ans. 35.
* S©8 Appendix, DomoaHtiation -Id. \ Ibid. 47. \ J bid. 48-
OF THE PARABOLA. 40
PROBLEM IX.
Given the transverse axis, ordinate, and abscissa, to find the
conjugate. 't
Rule. The square root of the product of the two afflpissas is to
the ordinate, as the transverse axis is to the conjugate. mB ':
1. Let the transverse axis be 35, the ordinate 10, anl^p ^scissaa .
28 and 7; required the conjugate ?
//oo rrx 1A n^ 35x10 35x10 ^. ,, *!
V (28 X 7) : 10 : : 35 : ^.^g^y. = ~~II~ =^^' ^^® conjugate.
2. Let the transverse diameter be 120, the ordinate 16, and the
abscissas 24 and 96; required the conjugate? * Ans. 40*
OF THE PAEABOLii.
PROBLEM X.
Given the base and height of a parabola, to find its area.
Note. — Any double ordinate AB, to the axis of a parabola, may be callecJ. ita
base, and the abscissa O D, to that ordinate, its height. q
Rule. Multiply the base by the height, and § of
the product will be the area.f
1 . Required the area of a parabola, whose height qI
is 6 and base 12 ? /
6 X 12 X 1 = 48 the area. A d b
2. What is the area of a parabola, whose base is 24, and height 4?
Ans. 64.
3. What is the area of a parabola, whose base is 12, and height 2 ?
Ans. 16.
PROBLEM XI.
To find the area of the zone of a parabola, or the space between
two parallel double ordinates.
Rule I. When the two double ordinates, their distance, and the
altitude of the whole parabola are given ; find the area of the whole
parabola, and find also the area of the upper segment, their diflference
will be the area of the zone.
II. When the two double ordinates and their distance are given;
to the sum of the squares of the two double ordinates, add their pro-
duct, divide the sum by the sum of the two double ordinates, multip'y
the quotient by | of the altitude of the zone, and the product will be
the area of the zone. J
1. Given A B 20, S T 12, and D j; 8 ; what is the area of the zone
ASTB, the altitude D 0 being 12-5?
« See Appendix, Demonstration 49. f Ibid. 50. X Ibid. 51.
%
BO CONIC SECTIONS.
(20 X 12-6) x| = 166| = area of the parabola ABO, and (12-5—8)
X 12 = 64, and 54x1 = 36; hence 166|— 36 = 130|, the area.
III. When the altitude of the whole parabola is not given.
2. Suppose the double ordinate A B = 10, the double ordinate
ST =6, and their distance Da;=4: what is the area of the zone
ASTB?
^^'^1^0 + 6^"^^^ 12i ; then 12i x 4 x | = 32|, the area as before.
3. Let the double ordinate AB = 30, CP = 25, and their distance
D G= 6 ; required the area of the zone A B P C? Ans. 165j\
PROBLEM XIL
7'o Jind the length of the curve^ or arc of a ftarahola^ cut off by
a double ordinate to the axis.
Rule.
L Divide the double ordinate by the parameter, and call the
quotient q.
IL Add 1 to the square of the quotient ^, and call the square root
ot the sum s.
in. To the product of q and 5, add the hyperbolic logarithm of
their sum, then the last sum multiplied by half the parameter, will
give the length of the whole curve on both sides of the axis.
Putting c for the curve, q for the quotient of the double ordinate
divided by the parameter, s for V (1 + 9'^) ^^^ ^ for half the para-
meter; then
c^ay. {75 + hyp. log. of (<7 + s.)}*
Note.— The common logarithm of anv number multiplied by 2*302586093
gives the hyperbolical logarithm of the same number.
1. What is the length of the curve of a parabola, cut off by a
double ordinate to the axis, whose length is 12, the abscissa being 2 ?
y^ y
x—2 and y — Q\ then a = — = 3^« = 9, and g = - = f = |^
f Iso 5 = V (1 + ^^) = V (1 + t) = V (¥) = k V (13) =
1-2018504=5. Then | + 1-2018504 = 1-868517, whose common
logarithm is -271497, which being multiplied by 2-302585093,
produces -6251449 for its hyperbolic logarithm; and also |x
1-2018504= -8012836; the sum of these two is 1 4263785, there-
fore 9x1-4263785 = 12-8374065, is the length of the curve re-
quired.
Rule II. Put y equal to the ordinate, and q equal the quotient
arising from the division of the double ordinate by the parameter,
or from the division of doubio the abscissa by the ordinate ; then the
length of the double curve will be expressed by the infinite series.
"^^^^V 2.3 2.4.5^2.4.6.7'^^ )
* See Appendix* Demons 52.
OF THE TARABOLA. 51
Note. — This series 'will converge no longer than till 3=1. For when q is
greater than 1, the series will diverge.
Let the last example be resumed, in wliich the abscissa is 2, and
the ordinate 6.
Hence, 2 x 2-7-6 = f =5'; then employing f instead of q in the last
series, we get
12x (l+QI"— 114 +3xh^?^) = 12'837 the length of the curve
as before.
Rule III. To the square of the ordinate, add ^ of the square of
the abscissa, and the square root of the sum will be the length of the
single curve, the double of which will be the length of the double
tourve, nearly.*
Note. — The two first rules are not recommended in practice. The practical
application of this is much simpler, and is therefore to be employed in prefer-
ence to either.
Eetaining the same example, in which a: =2, and 2/ = 6, we shall
geti;=V(2/^ + #^')= V(36+V) = 6-1291, and C = 12-8582, nearly.
2. Required the length of the parabolic curve, whose abscissa is 3,
and ordinate 8 ? Ans. 17*4:35.
PROBLEM XIIL
Grtven any two abscissas and the ordinate to one ofthem^ to find
the corresponding ordinate to the second abscissa.
Rule. Say, as the abscissa, whose ordinate is given, is to the
square of the given ordinate, so is the other given abscissa to the
square of its corresponding ordinate.f
1. If the abscissa a;0 = lO, and the ordinate x S = 8, what is the
ordinate AD, whose abscissa D 0 is 20 ?
a; 0 : X S2 : : D 0 : A D2, viz. 10 : 64 : : 20 : 128, the square root
of which is 11-313, &c.,=AD.
2. If 6 be the ordinate corresponding to the abscissa 9, required
the ordinate corresponding to the abscissa 16? Ans, 8.
PROBLEM XIV.
Given two ordinates^ and the abscissa corresponding to one of them,
tojind the abscissa corresponding to the other.
Rule. Say, as the square of the ordinate whose abscissa is given,
is to the given abscissa, so is the square of the other ordinate to its
corresponding abscissa.J
1. Given Sa;=6, a; 0 = 9, and AD- 8; required the abscissa
CD? 36:9: : 64: 16=0D.
2. GivenSa;=8, a;0=10, and AD-^9; required OD?
Ans. 12-656.
* See Appendix, Demonstration 63. f Ibid. 54- t Ibid. 64,
52 CONIC SECTIONS.
PROBLEM XV.
Jiven two ordinates perpendicular to the axis and their distance^
tojina the corresponding abscissas.
Rule. Say, as the difference of the squares of the ordinates is to
their distance, so is the square of either of them to the corresponding
abscissa.*
1. Given Sa:=6, A D = 8, and a;D = 7; required the abscissas?
(64 — 36) : 7 : : 64
28 : 7: :64: 16 = 0D, and
28 :7: :36: 9=0 x.
2. Given Sa; = 3, AD = 4, and a;D = 2; required the abscissas?
Ans, 4f and 2^.
OF THE HYPERBOLA.
PROBLEM XVI.
Given the transverse and conjugate diameters^ and any alscissa^
to find the corresponding ordinate.
Rule. As the transverse is to the conjugate, so is the mean pro^
portional between the abscissas to the ordinate.^
1. If the transverse be 24, the conjugate 21, and the less abscissa
ADS; required the ordinate ?
Note. — The less abscissa added to ttie transverse gives the
greater.
24:2i::vr32x8):y^^^ip^ = i4the ._
ordinate.
2. If the transverse axis of an hyperbola be 120, the less abecissa
40, the conjugate 72 ; required the ordinate? Ans. 48.
3. The transverse axis being 60, the conjugate 36, and the less
abscissa 20 ; what is the ordinate ? Ans. 24.
PROBLEM XVII.
Given the transverse.^ conjugate^ and ordinate^ to find the abscissa/
Rule. To the square of half the conjugate, add the square of the
ordinate, and extract the square root of the sum. Then say,
As the conjugate is to the transverse, so is that square root to half
the sum of the abscissas.
* See Appendix, Demonstration 56. T Ibid 56.
OF THE HYPERBOLA. 53
Then to tliis half sum, add half the transverse, for the greater
ahscissa ; and from the half sum take half the transverse for tlie less
abscissa.*
1. If the transverse be 24, and the conjugate 21 ; required the
abscissas to the ordinate 14 V
10-5=. J conjugate 14 = ordinate
10-5 14
110-25 196
196
306*25, the square root of which is 17*5 ; then 21 : 24 : : 17*5 :
20 = half sum, 20 H- 12 = 32 the greater abscissa, and 20 — 12=8 the
less abscissa.
2. The transverse is 120, the ordinate 48, and the conjugate 72;
required the abscissas ? Ans. 40 and 160.
PROBLEM XVIII.
(riven the conjugate^ ordinate^ and abscissas, tojind the transverse.
Rule. To or from the square root of the sum of the squares of the
ordinate and semi-conjugate, add or subtract the semi-conjugate,
according as the less or greater abscissa is used; then, as the square
of the ordinate is to the product of the abscissa and conjugate, so is
the sum or difference, above found, to the transverse. f
1. Let the conjugate be 21, the less abscissa 8, and its ordinate 24 »
required the transverse ?
21x8x V(142 + ^) + 104
IP ^
1 X V(3* + 42) + 3) = 3 X (5 + 8) = 24 the transverse.
2. The conjugate axis is 72, the less abscissa 40, the ordinate 48 i
-,equired the transverse? Ans. 120.
3. The conjugate is 36, the less abscissa 20, and its ordinate 24 ;
required the transverse? Ans. 60.
PROBLEM XIX.
Given the ahscissa, ordinate, and transverse diameter, to find the
conjugate.
Rule. As the mean proportional between the abscissas is to the
ordinate, so is the transverse to its conjugate %
1. What is the conjugate to the transverse 24, the less abcissa
being 8, and its ordinate 14 ?
24 X 14
;^,^3^3^^ = 21 the conjugate.
2. The transverse diameter is 60, the ordinate 24, and the lesa
abscissa 20 ; what is the conjugate ? Ans. 36.
* See Appendix, Demonstration 57. f Ibid. 58. | ibid. 59.
^4 CONIC SECTIONS.
PROBLEM XX.
divert any two abscissas^ X, x, and their ordinates^ F, y^ to find
the transverse to which they belong.
Rule. Multiply each abscissa by the square of the ordinate belong-
ing to the other; multiply also the square of each abscissa by tlie
square of the other's ordinate ; then divide the diffepence of the latter
products by the difference of the former ; and the quotient will be
the transverse diameter to which the ordinates belong.*
1. If two abscissas be 1 and 8, and their corresponding ordinates
4r| and 14, required the transverse to which they belong?
^ S'^x 41x41—1^x14^^35x35-14x14 5x5 — 2x2
® Ixl42-8x4|x4| ~14xl4-35x4|~2x2-5x|
= — = — =24, the transverse.
PROBLEM XXL
To find the area of a space AN OB, bounded on one side by thi
curve of a hyperbola, by means of equi- distant ordinates.
Let AN be divided into any given number of equal parts, AG,
CE, EG, &c., and let perpendicular ordinates AB, CD, EF, &c.,
b« erected, and let these ordinates be terminated by any hyperbolic
curve BDF, &c. ; and let A=AB
+ N0, B = CD+GH4-LM,&c.,and
C = EF+IK, &c. ; then the com-
mon distance A C, of the ordinates,
being multiplied by the sum arising ^ ^
from the addition of A, 4 B, and 2 C, A C' E G~
and one-third of the product taken will be the area, very nearly..
^, ,. A + 4B + 2C -p, ,, ... T^ * n ,
That IS, ^ X D = the area, puttmg D = A C.f
1. Given the lengths of 9 equi-distant ordinates, viz., 14, 15, IC,
17, 18, 20, 22, 23, 25 feet, and the common distance 2 feet ; required
the area? Ans. 300| feet.
2. Given the lengths of 3 equi-distant ordinates, viz., AB = 5,
CD = 7, and EF=8, also the length of the base AE 10; what is
the area of the figure A B F E ? Ans. 68^ feet.
3. If the length of the asymptote of a hyperbola be 1, and there
be 11 equi-distant ordinates between it and the curve, the common
distance of the ordinates will then be ^, and from the nature of the
curve their lengths will be ^, ^, 4-f, ^, ii, ih \h ^, H, rh U;
Mhat is the area of the'curved figure? Ans. -69315021.
This formula will answer for finding the area of all curves by
using the sections peq)endicular to the axis. The greater the num-
ber of ordinates employed, the more accurate the result ; but in real
practice three or five are in most cases sufficient.
* See Appendix, Demonstraticai 6Q» f Ibid. 6L
OF THE HYPERBOLA. 65
PROBLEM XXII.
To find tjit length of any arc of an hyperbola beginning at
the vertex.
Rule.
I. To 19 times the square of the transverse, add 21 times the
iquare of the conjugate ; also to 9 times the square of the transverse
add, as before, 21 times the square of the conjugate, and multiply
each of these sums by the abscissa.
II. To each of these two products, thus found, add 15 times the
product of the transverse and the square of the conjugate.
III. Then, as the less of these results is to the greater, so is the
ordinate to the length of the curve, nearly.*
1. In the hyperbola BAG, the transverse diameter is 80, the
conjugate 60, the ordinate BD 10, and the abscissa AD 2 ; required
the length of the arc B A C ? (Fig. p. 52.)
Here 2 (19x802 + 21 x 602) = 2 (121600 + 75600) = 394400.
And 2 (9x802 + 21x602) = 2 (57600 + 75600) = 266400.
Whence 15 x 80 x 60« +394400 = 4320000 + 314400 = 4714400.
And 1 5 X 80 X 602 + 266400 = 4320000 + 266400 = 4586400.
Then- 4586400 : 4714400 : : 10 : ^^^^ = 10-279 = AB.
458641^
Hence A FC = 10-279 x 2 = 20*558.
2. In the hyperbola BAG, the transverse diameter is 80, the
conjugate 60, the ordinate BD 10, and the abscissa AD 2-1637;
required the length of the arc A B ? Ans. 10-3005.
PROBLEM XXIII.
Given the transverse axis of a hyperbola^ the conjugate^ and the
abscissa, to find the area.
Rule.
I. To the product of the transverse and abscissa, add f of the
square of the abscissa, and multiply the square root of the sum by 21.
II. Add 4 times the square root of the product of the transverse
and abscissa, to the preceding product, and divide the sum by 75.
III. Divide 4 times the product of the conjugate and abscissa by
the transverse ; this quotient, multiplied by the tormer quotient, will
give the area of the hyperbola, nearly.f
1. In the hyperbola BAG (see figure, page 52), the transverse
axis is 30, the conjugate 18, and the abscissa A D is 10 ; what is the
area?
* See Appendix, Demonstration 62. t Ibid. 63.
56 CONIC SECTIONS.
Hpre 21 V(30 X 10 + ^ X 10^) = 21 V (300 + 71-42857) = 21
V (371-42857) = 21 x 19272 = 404-712 ;
And^ V (30 X 10) + 404-712 ^ 4 x 17-3205 + 404-712 _^
75 ~ 75 ^
C9-282 + 404-712 473994 _ ^^^^
~-T5 = ~75- = ^■^^^^•
Whence -i^^^^^^x 6-3199 = 24 x 6-3199= 151-6776, the area re-
quired.
2. What is the area of an h}^erbola whose abscissa is 25, the
transverse and conjugate being 60 and 30? Ans. 805 0909.
3. The transverse axis is 100, the conjugate 60, and abscissa 60 j
rti<iuired the area V -4w5. 322 3633684
MENSUEATION OF SOLIDS.
SECTION ly.
DEFINITIONS.
1. A solid is that which has length, breadth, and thickness.
2. The solid content of any body is the number of cubic inc
feet, yards, &c., it contains.
3. A cube is a solid, having six equal sides at right
angles to one another.
4. A prism is a solid whose ends are plane
figures, which are parallel, equal, and similar.
Its sides are parallelograms.
It is called a triangular prism, when its ends
tre triangles; a square prism, when its ends are squares; a penta*
gonal prism, when its ends are pentagons ; and so on.
5 A parallelopipedon is a solid having six
rectangular sides, every opposite pair of which
are equal and parallel.
6. A cylinder is a round solid, having
cn-cular ends, and may be conceived to be
described by the revolution of a rectangle
about one of its sides, which remains fixed.
7. A pyramid is a solid, having a plane figure for its
base ; and whose sides are triangles meeting in a point,
called the vertex.
Pyramids have their names from their bases, like
prisms. *
When the base is a triangle, the solid is called a
triangular pyramid ; when the base is a square., it is
called a square pyramid ; and so on.
8. A cone is a round pyramid, having a circle for its m
base. ML
68
MENSURATION OF SOLIDS.
9. A sphere is a round solid, which may be conceived
to be formed by the revolution of a semicircle about its
diameter, which remains fixed.
10. The axis of a solid is a line joining the middle of both ends.
11. When the axis is perpendicular to the base, the solid is called
a right prism or pyramid, otiierwise it is oblique.
12. The height or altitude of a solid, is a line drawn from its
vertex, perpendicular to its base, and is equal to the axis of a right
prism or pyramid ; but in an oblique one, the altitude is the perpen-
dicular of a right-angled triangle, whose hypothenuse is the axis.
13. When the base is a regular figure, it is called a regular prism
or pyramid ; but when the base is an irregular figure, the solid on it
is called irregular.
14. The segment of any solid, is a part cut off from the top by a
plane parallel to its base.
15. A frustum is the part remaining at the bottom, after the seg^
ment is cut off.
16. A zone of a sphere is a part intercepted between two planes
which^^re parallel to each other.
17. k' circular spindle is a solid generated by
the revolution of a segment of a circle about its
chord, which remains fixed.
18. A wedge is a solid, having a rectangular
base, and two of its opposite sides meeting in an
cdga.
19. A prismoid is a solid, having for its.
,'WO ends two right-angled parallelograms,
parallel to each other, and its upright sides
are four trapezoids.
fiO. A spheroid is a solid, generated by the
rotation of a semi-ellipsis about one of its
axis, which remains fixed.
When the ellipsis revolves round the trans-
verse axis, the figure is called a prolate, or
oblong spheroid ; but when the ellipsis re-
volves round the shorter axis, the figure is
called an oblate splieroid.
MENSURATION OF SOLIDS.
59
21. An elliptical spindle is a solid, generated
by the rotation of a segment of an ellipsis about
its chord.
22. A parabolic conoid^ or paraboloid, is a solid
generated by the rotation of a semi-parabola about
Us axis.
23. An ungula or hoof^ is a part cut off a solid by a plane oblique
to the base.
PROBLEM I.
To find the solidity of a cube.
Rule. Multiply the side of a cube by itself, and that product again
by the side, for the sohdity required.*
1. If the side of a cube be 4 inches, required
its solidity?
Here, 4 x 4 = 16, the number ot cubes of 1
inch deep in the square E F G D, and as the
entire solid consists of four such dimensions,
Us content is 16 x 4=64 cubic inches.
2. What is the solidity of a cubical piece of
marble, each side being 6 feet 7 inches?
Ans. 174 feet, nearly.
3. A cellar is to be dug, whose length,
breadth, and depth, are each 12 feet 3 inches ; required the numbei
of solid feet in it? Ans, 1838 feet 3 inches, nearly,
PROBLEM II.
To find the solidity of a parallelopipedon.
Rule. Multiply continually the length, breadth, and depth to-
gether for the solidity.f
1. What is the solidity of the
parallelopipedon ABCDEFG,
the length A B being 10 feet,
the breadth A G 4 feet, and
thickness A D 5 feet ?
ABxAGxAD = 10x
4x5 = 200 feet.
2. A piece of timber is 26
feet long, 10 inches broad,
and 8 inches deep ; required its solid content? Ans. 14| feet.
3. A piece of timber is 10 inches square at the end.s and 40 feet
long ; required its content ? Ans. 27J feet.
4. A piece of timber 15 inches, square at each end, and 18 feet
• See Appendix. Demonstration 6* t Ibid, <J4
60
MENSURATION OF SOLIDS.
long, is to be measured ; required its content, and how far fiom tho
end must it be cut across, so that the piece cut off may contain 1
solid foot ?
Ans. The solidity is 28'125 feet; and 7*68 in length will make
one foot.
5. What length of a piece of square timber will make one solid
foot, being 2 feet 9 inches deep, and 1 foot 7 inches broad?
Ans. 2*756 inches in length will make one solid foot.
PROBLEM III.
To Jind the solidity of a prism.
Rule. Multiply the area of the base by the perpendicular height,
and the product will be the solidity.*
1. What is the solidity of a prism,
A B C F I E, whose base C A is a pentagon,
ejvdi side of which being 3*75, and height
15 feet?
When the side of a pentagon is 1, its
area is 1*720477 (Table II.); therefore
1 720477 x3-753=24*1942 = the area of
the base in square feet; hence 24*1942 x
15 = 362*913 solid feet, the content.
2. What is the solidity of a square prism,
whose length is 5J feet, and each side of
its base 1 1 foot ? Ans. 9|- solid feet.
3. W^hat is the solidity of a prism, whose
base is an equilateral triangle, each side
being 4 feet, and height 10 feet?
Ans. 69*282 teet.
4. What quantity of water will a prismatic vessel contain, its base
being a square, each side of which is 3 feet, and height 7 feet?
Ans. 63 feet.
PROBLEM IV.
To find the solidity of a cylinder.
Rule. Multiply the area ©f the base by its height, and the product
will be the solid content.f
1. What is the capacity of a
fight cylinder A B G C, whose
height, and the circumference of
its base, are each 20 feet ?
20
First — r J— 7 = the diameter,
lialf of which multiplied by half
the circumference will give the area of the base (Prob. XYIII. Sec.
XL), that is, 10 x q.t.i/. = .n^-A =^^^^ ^^'®^ ^^ ^^® ^"^ » *^^^
25
--r^T- X 20 = 636*61828, the content.
♦ See Appendix, Demonstration 64, f Ibid. 64.
MENSURATION OF SOLIDS. 61
2 What is the content of the oblique cylmder A B F E, the circum*
ference of whose base is 20 feet, and altitude A C 20 feet ?
As before, the area of the base is ; then ^ ^' x 20 =
€36-61828, the solid content as before.
3. The length of a cylindrical piece of timber is 18 feet, and its
circumference 96 inches ; how many solid feet in it ?
Ans. 91-676 feet. ^
4. Three cubic feet are to be cut off a rolling stone 44 inches in
circumference ; what distance from the end must the section be
made? Ans. 33*64 inches.
PROBLEM V.
Tojlnd the content of a solid formed by a plane passing parallel
to the axis of a cylinder.
Rule. Find by Prob. XXVIII., Sec. II., the area of the base,
which, multiplied by the height, will give the solidity.*
1. In the cylinder A B G C, whose diameter is 3, and height 20
feet; let a plane LN pass parallel to the axis, and 1 foot from it;
what is the solidity of each of the two prisms into which the cylinder
vs divided? — (See the last figure.)
i-5 = (|_l)^3 = | = i= -1661 the tabular versed sine, to
which, in the Table of Circular Segments, corresponds the area
•08604117
which taken from . . . ... . . -78539816
leaves the other segment '69935699
Then 3^ = 9 which x -08604117 = 7-7437053 = seg. DC N.
Also 9x •69935699 = 6-29421291 = seg. D G N.
Hence 20 x 7*7437053= 15-4874 = the slice LKACND; and
20 x 6-29421699 = 125-88434 = the slice L K B G N D.
2. Suppose the right cylinder, whose length is 20 feet, and diameter
50 feet, is cut by a plane parallel to, and at the distance of, 21*75
feet from its axis ; required the solidity of the smaller slice ?
Ans. 1082-95 feet.
PROBLEM VI.
To find the solidity of a pyramid.
Rule. Multiply the area of the base by the one-third of the
height, and the product will be the solidity.f
1. What is the solidity of a square pyramid, each side of its base
being 4 feet, and height 12 feet?
4x4 = 16 the area of the base :
Then 16 x V = 64 feet, the solidity.
2. Each side of the base of a triangular pyramid is 3, and height
30; required its solidity ? Ans. 38-97117.
3. The spire of a church is an octagonal pyramid, each side at
the base being 5 feet 10 inches, and its perpendicular height 45 feet j
* See Appendix, Demoustrtttion 64. t Ibid. 65.
62 MENSURATION OF SOLIDS*
also each side of the cavity, or hollow part, at the base is 4 feet 1 1
inches, and its perpendicular height 41 feet; it is required to know
how may solid yards of stone the spire contains.
Ans. 32-19738 yards.
■ 4. The height of a hexagonal pyramid is 45 feet, each side of the
hexagon of the base being 10 ; required its solidity?
Ans, 3897-1143.
PROBLEM VII.
To find the solidity of a cone.
Rule. Multiply the area of the base by one-third of the height,
and the product will be the solidity.*
1. The diameter of the base of a cone is 10 feet, and its perpen-
dicular height 42 feet ; what is its solidity ?
102= 100 X -7854 = 78-54; then 78'54 x V = i099-56 feet.
2. The diameter of the base of a cone is 12 feet, and its perpen-
dicular height 100 ; required its solidity? Ans. 3769-92 feet.
3. The spire of a church of a conical form measures 37-6992 feet
round its base ; what is its solidity, its perpendicular height being
100 feet? ^ ^ ^ Ans. 3769-92.
4. How many cubic yards in an upright cone, the circumference
of the base being 70 feet, and the slant height 30? Ans. 134-09.
5. How many cubic feet in an oblique cone, the greatest slant
height being 20 feet, the least 16, and the diameter of the base 8
feet? Ans. 254-656588 feet.
PROBLEM VIII.
To find the solidity of the frustum of a pyramid.
Rule. Add the areas of two ends and the mean proportional
between them together ; then multiply the sum by one-third of the
perpendicular height, and the product will give the solidity.f
V
1. In a square pyramid, let A 0 = 7, PD = 5, and
the height 0Q=6; the solidity of the frustum is re-
quired.
7 X 7 = 49 = the area of the base.
6x5 = 25 = the area of the section S D
7 X 5 = 35 = the mean proportional between 49
and 25.
^, „ 49 + 35 + 25 ^ „-Q ,, , ,
Therefore, ^ x 6 = 218 = the content
o
of the frustum.
2. What is the content of a pentagonal frus-tum, whose height is
5 feet, each side of the base 1 foot 6 inches, and each side of the lesh
end 6 inches? Ans. 9-31925 cubic feet.
3. What is the content of a hexagonal frustum, whose height is 6
feet, and the side of the greater end 18 inches, and of the less 12
inches? Alls. 24-681724.
* See Appendix, Demonstration, 66. * Ibid. 67.
MENSURATION OF SOLIDS. 03
4 How many cubic feet in a squared piece of timber, the areas of
t'ne two ends being 504 and 372 inches, and its length 3l|- feet?
Ans. 95-447 feet.
6. What is the solidity of a squared piece of timber, its length
being 18 feet, each side of the greater base 18 inches, and each side
of the small end 12 inches? Ans. 28'6.
PROBLEM IX.
Tojind the solidity of the frustum of a cone*
Rule. Add the two ends, and the mean proportional between them
together, then multiply one-third of the sum by the perpendicular
height, and the product will be the content.*
1. How many solid feet in a tapering round piece of timber, whose
length is 26 feet, and the diameters of the ends 22 and 18 inches re-
spectively ?
Here 22^ x •7854 = 380-134 inches, the area of the greater end, and
IS^x •1854=1 254-47 inches = the area of the less end, (380-134 x
.?54-47)^ = 311-018=the mean proportional between the areas of the
ends ; then by the rule
.854-47 + 380-134 + 311-018 ^^ ,„ ^.„,^ u- - u «./. a i •
^ X 26 X 12 = 98345 cubic mches= 56-9 cubio
o
feet, the answer.
2. How many cubic feet in a round piece of timber, the diameter
of the greater end being 18 inches, and that of the less 9 inches, and
length 14-25 feet? Ans. 1468943 feet.
3. What is the solid content of the frustum of a cone, whose
height is 1 foot 8 inches, and the diameters of the ends 2 feet 4
inches, and 1 foot 8 inches? Ans, 5*284.
PROBLEM X.
V Tojind the solidity of a wedge.
Rule \. Add the three parallel edges together, and multiply one-
third of the sum by the area of that section of the wedge which is
perpendicular to these three edges, and the product will give the
content.f
Note. — When the quadrangular sides are parallelograms,
the wedge is a triangular prism, having for its base the
triangle B O C ; when the quadrangles are rectangular, R
A. O is the height of the prism, and the area of the trianpfle
1^ O C multiplied by A O will give its content ; when the ^
triangle B O G is isosceles and perpendicular to the plane
A C, the wedge is of the common kind ; C G is its QdgGy
and A R B O its back.
Rule II. To twice the length of the base, add the length of the
edge, multiply the sum by the breadth of tlie base, and the product
by the height of the wedge, and one-sixth of the last product will be
the solidity, that is, (2 L + /) x| 5 ^, by putting L=R B, the length
• See Appendix. Demonstration tW. f Ibid. 69.
E
64 MENSURATION OF SOLIDS.
of the base, Z=GC, tlie length of the edp^e, & = AIl, the breadth of
the base, A = the perpendicular height of the wedge.*
1. Let A0 = 4, GC = 3, RB = 2i, the perpendicular D T = 12, and
p the perpendicular distance of B R from the plane of the face A C =
3| feet; required the solid content?
^"^^/^^ X 12 X ^= 661 cubic feet.
2. The perpendicular height from the point T to the middle of the
back AB is 24-8, the length of the edge C G 110 inches, the base
R B 70 inches, and its breadth A R 30 inches ; required the solidity?
Ans. 31000 cubic inches.
3. How many cubic inches in a wedge whose altitude is 14 inches,
its edge 21 inches, the length of its base 32 inches, and its breadth
44 inches? Ans. 892*5 cubic inches.
PROBLEM XL
To find the solidity of a prismoid^ which is the frustum of a wedge.
Rule. By either of the foregoing rules, find the solidity of two
wedges whose bases are the two ends of the frustum, and height the
distance between them, and the sum of both will be the solidity of
the prismoid or frustum. +
1. In the prismoid A B P Q, there is given R B= 18,
AO = 27, PD = 21, SQ=24, B0 = 12, DQ=4, and
B 1 = 30 ; what is its solidity?
18 + 27 + 21 30x12 ^„_^^ ,, ^ ^ r .u
^ X — 2 — = 3960 = the content of t^e
, , 24 + 27 + 21 30x4 ...^ ,,
greater wedge, and ^ x — ,^ — =1440, the
content of the other; then 3960+1440 = 5400, the \ ^
content of the frustum.
2. What is the solidity of a piece of wood in the form of a pris-
moid, whose ends are rectangles, tlie length and breadth of one being
1 foot 2 inches and 1 foot respectively, and the corresponding sides
of the other 6 and 4 inches respectively; the perpendicular height
being 30j feet? Ans. 18*074 cubic feet.
Note. — The following rule will answer for any prismoid, of whatever figure
each end may be.
Rule. — If the bases be dissimilar rectangles, take two corresponding dimen-
sions, and multiply each by the sum of double the other dimension of the same
end, and the dimension of the other end corresponding to this last dimension;
then multiply the sum of the products by the height, and one-Bixth of the last
product will be the solidity. J
PROBLEM XII.
To find the solidity of a cylindroid; or the frustum of an
elliptical cone.
Rule.
L To the longer diameter of the gieater end, add half the longef
» E€9 Appendix, Demonstration 70. t Ibid. 71 X Ibid. 72.
MENSURATION OF SOLIDS.
C5
diameter of the less end, and multiply the sum by the shorter diameter
of the greater end.
II. To the longer diameter of the less end, add half the longer
diameter of the greater end, and multiply the sum by the shorter
diameter of the less end.
III, Add the two preceding products together,
and multiply the sum by -2618 (one-third of -7854)
and then by the height; the last product will be
the solidity.*
1. Let A B C D be a cylindroid, the base of which
is an ellipsis, whose two diameters are 40 and 20
inches, the top a circle, whose diameter is 30 inches ;
what is its solidity, allowing the height to be 10
feet?
(AB + 4 CD)xGH = (40 + 15)x20=1100
(CD + i AB)xmr=C30 + 20)x30 = 1500
sum = 2600
Then (2600 x '2018 x 10) 6806-8, which, divided by 144, gives
47-27 feet, the answer.
2. The transverse diameter of the greater base of a cylindroid is
13, and conjugate 8 ; the transverse diameter of the less base 10,
and conjugate 5-2; what is the solidity of the cylindroid, its height
being 12? Ans. 721-ff8968.
3. The transverse diameter at the top of the cylindroid is- 1 2 inches,
and conjugate 7; the longer diameter at the bottom is 14 inches,
and shorter 12, and its height 10 feet; required its solidity?
Ans. 6-78 feet.
PROBLEM XIIL
To find the solidity of a sphere.
Rule I. Multiply the cube of the diameter by '5236, and the
product will be the content.
Rule II. Multiply the diameter by the circumference of the
sphere, and the product multiplied by one- sixth part of the diameter
will be the solidity.f
1. Suppose the earth to be a perfect sphere,
and its diameter 7957 1 miles, how many solid
miles does it contain ?
7957f x3-1416 = the circumference of the
earth (Prob. XVI., Sec. II.); then
79571 X 3-1416 x 7957f = 198943750 = the sur-
face of the sphere; then
198943750 x 79571 X 1^ = 263857437760 miles,
the solidity by Rule II.
Again,-5236 x 6/3= -5236 x (7957|)3 = 263858149120miles, thesoli-
dity by Rule I., which gives the result too great on account of taking
•6236 a little too great.
■ See Appendix, Demonstration 73.
t Ibid 74-
66 MENSURATION OF SOLIDS.
2. What is the solidity of a sphere, whose diameter is 24 inches?
Ans. 7238-2464 cubic inches.
3. What is the solid content of the eartli, allowing its circumference
to be 25000 miles? Ans. 263858149120 miles.
4. Kequired the solidity of a globe whose diameter is 30 feet V
Ans. 14137-2.
PROBLEM XIV.
Tojind the solidity of the segment of a sphere.
Rule I. From three times the diameter of the sphere deduct
twice the height of the segment ; multiply the remainder by the
square of the height, and that product by -6236 ; the last product
will be the solidity.*
Rule II. To three times the square of the radius of the segment's
base add the square of its height ; multiply this sum by the heig^.t,
and the product by '5286 ; the last result will be the solidity.
1. What is the solidity of each of the frigid
zones, the diameter of the earth being 7957f
miles, and half the breadth, or arc of the
meridian intercepted between the polar circle
and the pole 234 degrees; that is, AD = 234
degrees, supposing A B to represent the polar
circle.
E
By Rule I.
As 1 ( = tabular radius) : 3978|^ ( = radius of the earth) : :
•0829399 ( = tabular versed sine of 23^ degrees) : 330*0074946, the
versed sine, or height of the segment
Then -5236 7^2= (3 d—2 h) = -6236 x 330'0074946a x 23213-2350108
= 1323679710, the solid content.
By Mule II.
As 1 : 39781- : : -3987491 (=the tabular sine of 23^ degrees;
: 1586-57282526, the radius of the base.
Then -5236 /ix (3 r^ + /i2) = -5236x3300074946 x 7660544936=
1323680299-69, the solidity.
2. Let ABDO be the segment of the sphere whose solidity is
required. The diameter AB of the base is 16 inches, and the height
OD 4 inches. Ans. 435*6352 cubic inches.
3. Required the solidity of the segment of a sphere, whose
diameter is 20 feet, and the height of the segment 5 feet ?
Ajis. 654-5 feet.
PROBLEM XV.
To find the solidity of the frustum or zone of a sphere.
RiTLE.
i. To the sum of the squares of the radii of the two ends, add i o/
* See Appendix, Dem'^ustraiion 7&.
MENSURATION OF SOLIDS. 67
t!ie square of their distance, or of the height of the zone ; tl.is sum
niLiiLiplied by the height of the zone, and the product again by
1 5708, will be the solidity.
II. For the middle zone of a sphere. To the square of the
diameter of the end add two-thirds of the square of the height;
multiply this sum by the height, and then by -7854:, the last result
will be the solidity.
(9r, From the square of the diameter of the sphere, deduct one-
third of the square of the height of the middle zone ; multiply the
remainder by the height, and then by '7854, the last result will be
the solidity.*
1. Required the solidity of the frustum of
a sphere, the diameter of whose greater end is
4 feet, the diameter of the less end 3 feet, and
the height 2^ feet?
(22 + 1-53 + ^ X 2-52) X 1-5708 X 2-5 = Six
3-927=32-725, the solidity of the frustum.
2. What is the solidity of the temperate zone,
its breadth being 43 degrees, the radius of the
top being 1586*57282526, and the radius of the base 3648*86750538,
and height 2062-2655?
(3648-867505382+ 1586-572825262 + J x 2062-26552) x 2062-2655
X 1-5708= 17249136x2062-2955x1-5708 = 55877778668, the soli-
dity of each temperate zone.
3. Required the solidity of the torrid zone, which extends
23|- degrees on each side of the equator, the diameter being
7957f miles, and the height 3173-14565052?
(7957-752— i X 3173-145650522) x 3173-14565052 x -7854 =
149455081137, the answer.
4. What is the solidity of the middle zone of a sphere, whose top
and bottom diameters are each 3 inches, and height 4 inches ?
Ans. 61-7848. ^
5. Wliat is the solid content of a zone, whose greater diameter ia
20 feet, less diameter 15 feet, and the height 1 0 feet ?
Ans. 189-58. ^
6. How many solid feet in a zone, whose greater diameter ia
12 feet, and less diameter 10 : the height being 2 ?
Ans, 195-8264.
PROBLEM XYI.
To find the solidity of a circular spindle.
Rule. Find the distance of the chord of the generating circular
segment from the centre of the circle, and also the area of this
fcegment.
Then, from one-third of the cube of half the length of the spindle
or half chord of the segment, subtract the product of the central
distance, and half the area of the segment ; the remainder, multi-
plied by 12-5664, will give the solidity.f
* See Appendix, Demonstration 76. Ibid. 77.
68 MENSURATION OF SOLIDS.
1. Let the axis AC of a circular spindle be
40 inches, and its greater diameter 1> L 30
inches ; what is its solidity?
202-^15 = 261, then 26| + 15 = 41|, the dia-
meter of the circle. Again, -~ — -= 5|, the
central distance.
Now 15-^4:l| = •36, the area segment corres-
ponding to V hich is '254550, which multiplied
by the squaie of 41|, produces 441-92708 the
area of the generating segment ABC, the half of which is 220*96354.
Lastly, (20'-^ 3)— (5|x 220-96354) = 1377-7 1268, and this multi-
plied by 12-5664 produces 17312-88862 cubic inches, the solidity
required.
2. The axis of a circular spindle is 48, and the middle diameter 36;
required the solidity of the spindle? Ans. 29916-6714.
PROBLEM XVIL
7^0 find the solidity of the middle frustum of a circular spindle.
Rule.
I. Find the distance of the centre of the middle frustum, from the
centre of the circle.
IL Find the area of a segment of a circle, the chord of which is
equal to the length of the frustum, and height half the difference
l)etween its greatest and least diameters ; to which add the rectangle
of the length of the frustum and half its least diameter ; the result
will be the generating surface.
in. From the square of the radius subtract the square of the
central distance, the square root of the remainder will give half the
length of the spindle.
IV. From the square of half the length of the spindle take one-
third of the square of half the length of the middle frustum, and
multiply the remainder by the said half length.
V. Multiply the central distance by the generating surface, and
subtract this product from the preceding ; the remainder, multiplied
by 6-2832, will give the solidity.*
1. Required the solidity of the middle frustum of a circular
spindle, the length D E being 40, the greatest diameter Q F 32, and
the least diameter P S 24 ?
First, 202-^4=100, and 100 + 4=104, the diameter of the circle.
Again, 52—16 = 36, the central distance. Also, ^ (32— 24) = 4,
and 44- 104 = -0381^2 ^^^ ^'^^^ segment corresponding to which ia
•009940, which, multiplied by the square of 104, produces 107*51104,
the area of PLQ ; and 40 x 12 = 480 the area ot the rectangle P D E L.
Hence 107*51104 + 480=587*51104 the area of the generating
surface P D L E.
* See Appendix, Demonstration 78.
MENSURATION OF S0LID3. 69
Next y (522 _ 362) = ^ (1408) = 8 V (22) = B 0 half the length
of the spindle ;
And (U08 — ^)x 20 = 25493^.
Then 36 x 587-51104 = 21150-39744, and
(25493J — 21150-39744) X 6-2832 ==27287-5347, the required so-
lidity.
2. What is the solidity of the middle frustum P S R L of a circular
spindle, whose middle diameter F Q is 36, the diameter P S of the
end 16, and its length D E 40? A72S. 29257-2904.
PROBLEM XYIII.
To find the solidity of a spheroid.
Rule. Multiply the square of the revolving axis by the fixed axis,
and this product again by -5236 for the solidity.*
1. What is the solidity of a prolate spheroid
whose longer axis A B is 55 inches, and shorter
axis CD 33?
Here 33^ x 55 x -5236 = 31361-022 cubic
uiches, the answer.
2. What is the solidity of an oblate spheroid,
whose longer axis is 100 feet, and shorter axis 6?
Ans. 31416 cubic feet.
3. What is the solidity of a prolate spheroid, whose axes are 40
and 50? ^ Ans. USSS.
4. What is the solidity of an oblate spheroid, whose axes are 20
and 10? Ans. 2094-4.
PROBLEM XIX.
To find the solidity of the segment of a spheroid^ the base of the
segment being parallel to the revolving axis of the sj)heroid.
CASE I.
Rule. From three times the fixed axis, deduct twice the height of
the segment, multiply the remainder by the square of the height, and
that product by -5236.
Then say, as the square of the fixed axis is to the square of the
revolving axis, so is the product found above to the solidity of the
spheroidal segment. f
1. What is the content of the segment of a prolate spheroid, the
height 0 C being 5, the fixed axis 60, and the revolving axis 30? —
See last Jgure.
50x3-5x2 = 150-10 = 140; then
140x52=3500, and 3500 X -5236= 1832-6; then
25 : 9 :: 1832-6 : 659-736, the answer.
CASE n.
When the base is elliptical.) or perpendicular to the revolving axis.
Rule. From three times the revolving axis, take double the
* See Appendix, Demonstration 79. * 'bid. 80.
70
MENSURATION OF SULIDS.
height; multiply that difference by the square of the height, and the
product again by •5236.
Then as the revolving axis is to the fixed axis, so is the last pro-
duct to the content.*
2. What is the content of the segment of a sphe-
roid, whose fixed axis is 50, revolving axis 30, and
height 6?
30x3 — 2x6 = 90 — 12 = 78;
Then 78x62=2808; and 2808 x •5236 = 1470-2688:
Then 30 : 50 : : 1470-2688 : 2450-448, the answer.
3. In a prolate spheroid, the transverse or fixed
axis is 100, the conjugate or revolving axis is 60,
and the height of the segment, whose base is parallel
to the revolving axis, is 10: required the solidity? Ans. 6277-888
4. If the axes of a prolate spheroid be 10 and 6, required the con-
tent of the segment, whose height is 1, its base being parallel to tlie
revolving axis? Ans, 6-2"7888.
PROBLEM XX.
To find the solidity of the middle zone of a spheroid^ the diameter
of the ends being perpendicular to the fixed axis ^ the middle dia^
meter ^ and that of either end being given^ together with the Ungth
of the zone.
Rule. To twice the square of the middle diameter, add the square
of the diameter of the end; multiply the sum by the length of tha
zone, and the product again by -2618 for the solidity. f
1. What is the solidity of the middle
zone of an oblate spheroid, the middle
diameter being 100, the diameter of the
end 80, and the length 36 ?
1002 X 2 + 802 ^ 26400 ; then 26400 x
36 = 950400, and 950400 x ^2618 =
248814-72, the answer.
2. What is the solidity of the middle
frustum of a spheroid, the greater dia-
meter being 30, the diameter of the end
18, and the length 40? Ans, 22242-528.
PROBLEM XXI.
To find the solidity of a parabolic conoid.
Rule. Multiply the square of the diameter of its base by ;3927,
and that product by the height; the last product will be the solidity.^
1. What is the solidity of the parabolic conoid, whose height
Is 10 feet, and the diameter of its base 10 feet?
102 X -3927 = 39-27; then 39-27x 10 = 392-7, the solidity required.
* See Appendix, Demonstratiou 81.
t Ibid. 82.
X Ibid. 83.
MENSURATION OF SOLIDS.
71
2. WhAt is the solidity of a parabolic conoid,
^hose height is 30, and the diameter of its
base 40? Ans. 18849-6.
3. What is the content of the parabolic
conoid, whose altitude is 40, and the diameter
of its base 12 ? Ans. 2261-952. ^
4. Required the solidity of a parabolic
conoid, whose height is 30, and the diameter
of its base 8 ? Ans. 753 -984.
PROBLEM XXII.
To find the solidity of the frustum of a parabolic conoid.
Rule. Multiply the sum of the squares of the diameters of the
two ends by tlie height, and that product by '3927 ; the last product
will be the solidity.*
1 . The greater diameter of the frustum is 10, T
and the less diameter 5 ; what is the solidity, the
length being 12?
103=100
63= 25
"125. Then 125x12 = 1500, and ^.1
1500 X •3927 = 589-05, the solidity.
2. The greater diameter of the frustum of a parabolic conoid is 20,
the less 10, and the height 12 ; what is the solidity? Ans. 2357*4.
3. The greater diameter of the frustum of a parabolic conoid is 30,
the less 10, and the height 50; required the solidity? Ans. 19635.
4. The greater diameter of the frustum of a parabolic conoid is 15,
the less 12, and the height 8 ; required the solidity ?
Ans. 1159-8408.
PROBLEM XXIII.
To find the solidity of a parabolic spindle.
Rule. Multiply the square of the middle diameter by -7854, and
that product by the length; then ^ of this product will be \he
solidity, t
1. The middle diameter CD, of a parabolic ^^-^"^ ^"-^^
spindle is 10 feet, and the length A B is 40 ; re- A«^— — 4— — ^B
quired its solidity ? D
103x •7854x40 = 3141-6 feet.
Then ^^x 3141*6 = 1675-52 feet, the answer.
2. The middle diameter C D, of a parabolic spindle is 12 feet, and
the length AB is 30; required the sohdity? Am. 1809-5616.
3. The middle diameter of a parabolic spindle is 3 feet, and the
length 9 feet; required its solidity? Ans, 3392928.
* See Appendix, Demonstration 84. t IbicL 85*
72
MENSURATION OF SOLIDS.
4. The middle diameter of a parabolic spindle is 6 feet, and the
length 10; required its solidity ? Ans. 150'7908.
5. The middle diameter of a parabolic spindle is 30 feet, and the
length 60; required its solidity ? Ans. 18849-6.
PROBLEM XXIY.
Tojind the solidity of the middle frustum of a parabolic spindle.
Rule. To double the square of the middle diameter, add the
square of the diameter of the end ; and from the sum subtract y*^ of
the square of the difference between these diameters ; the remainder
multiplied by the length, and that product by '2618, will be the
solidity.*
1. In a parabolic spindle, the middle dia-
meter of the middle frustum is 16, the least
diameter 12, and the length 20 ; required the
solidity of the frustum?
Here 2 x 16^ + 122- 3*5- x 42= 512 + 144—
6-4 = 649-6 ; hence 649*6 x 20 x '2618 =
3401-3056, the solidity.
2. The bung diameter of a cask is 30 inches, the head diam.eter
20 inches, and the length 40 inches ; required its content in ale
gallons, allowing 282 cubic inches to be equal to one gallon?
Ans. 80-211 gallons.
8. The bung diameter of a cask is 40 inches, the head diametei
BO inches, and tlie length 60 ; how many Mnne gallons does it con-
lain, 231 cubic inches being equal to one gallon ?
Ans, 276-08 gallons.
PROBLEM XXV.
Tofnd the solidity of a hyperbolic conoid.
Rule. To double the height of the solid add three times the trans-
verse axis, multiply the sum by the square of
the radius of the base, and that product by the E
height, and this last product by -5236; the
result divided by the sum of the height and
transverse axis, will give the solidity.'!'
1. Required the solidity of an hyperbolic
conoid, whose height Y m is 50, the diameter
A B 103-923048, and the transverse axis
VEIOO?
H«re(2x 50 + 3 X 100) X
(103-923048)^
= 400x
2700 = 1080000; and
1080000 X 50 X -5236
150
: 188496, the solidity.
2. What is the content of an hyperboloid, whose altitude is 10|
the radius of its base 12, and the transverse 30 ?
Ans. 2073-451151369.
• See AiBDendix, Demonstration dfi.
t Ibid. 87.
MENSURA'nON OF SOLIDS.
73
PROBLEM XXVI.
To find the solidity of the frustum of an hyperholoid, or
hyperbolic conoid.
Rule. To four times the square of the middle diameter, add the
Bum of the squares of the greatest and least diameters; multiply the
result by the altitude, and that product by -1309, for the solidity.*
1- Required the solidity of the frustum
A C E H I) B of an hyperbolic conoid, whose
f»;reatest diameter A B is 96, least diameter E II
54, middle diameter C D 76*4264352, and the
altitude m n 25?
Here 4 C D^ + A B^ + E H^ = (5841 x 4) +
9216 + 2916 = 35496, and 35496 x 25 x -1309=
116160-66, the answer.
2. What is the solidity of an hyperboloidal cask, its bung diameter
being 32 inches, its head diameter 24, and the diameter in the middle
between the bung and head f V^IO, and its length 40 inches?
Ans. 24998-69994216 inches.
PROBLEM XXVII.
To find the solidity oj a frustum of an elliptical spindle^ or awj
other solid formed hy the revolution of^a conic section about
Rule. Add together the squares of the greatest and least diameters,
and the square of double tlie diameter in the middle between the
two ; multiply the sum by the length, and the last product by '1309
tor the solidity.f
1. What is the content
of the middle frustum
C D I H of any spindle,
the length OP being 40,
the greatest, or middle
iliameter E F 32, the least,
or diameter at either end
C D 24, and the diameter
GK 30-157568?
Here 322 +(2x30-157568)^+ 242 = 5237-89 sum;
Then 5237-89x40 = 209515-6, and
209515-6 X -1309 = 27425-7, the answer.
2. What is the content of the segment of any spindle, the length
being 20, the greatest diameter 10, the least diameter at either end 5,
jaud the diameter in the middle between these 8? Ans. 997-458.
^ * See Appendix, Demonstration 88.
t Ibid, m
74
MENSURATION OF SOLIDS.
PROBLEM XXVIII.
To find the solidity of a circular ring.
Rule. To the thickness of the ring add the inner diameter;
multiply the sum by the square of the thickness, and the product by
2-4674, for the solidity.*
1. The thickness of a cylindrical ring is
2 inches, and the diameter CDS inches ;
required its solidity ?
(2 + 5)x4=:28; then 28x2-4674 =
69-0872 cubic inches, the answer.
2. Required the solidity of an iron ring
whose axis forms the circumference of a
circle ; the diameter of a section of the ring
2 inches, and the inner diameter, from side
to side, 18 inches ?
Ans. 197-3925 cubic inches.
3. The thickness of a cylindrical ring is 7 inches, and the innev
diameter 20 inches; required its solidity ? Ans. 3264-3702.
4. What is the solidity of a circular ring, whose thickness is 2
inches, and its diameter 12 inches ? Ans. 138*1744 cubic inches.
* Bkso Appendix. Domonetration 90.
THE FIVE REGULAR BODIES.
75
THE FIYE EEGULAE BODIES.
SECTION V.
DEFINITIONS.
A regular body is a solid contained under a certain number of
similar and equal plane figures.
Only five regular bodies can possibly be formed. Because it is
proved in Solid Geometry that only three kinds of equilateral and
equiangular plane figures joined together can make a solid angle.
1. The tetraedron^ or equilateral pyramid, is a solid
having four triangular faces.*
2. The Tiexaedron^ or cube,
six square faces.
is a solid having
3. The Qctaedron is a regular solid having
eight tiiangular faces.
* If figures similar to those annexed to the definitions, be drawn on paste-
board, and cut out by cutting through the bounding liues, and if the othex
lines be cut half through, and then the parts be turned up and glued together,
the bodies defined will be formed.
76 THE FIVE REGULAR BODIES.
4r. The dodecaedron has twelve pentagonal faces.
6. The icosaedron has twenty triangular faces.
PROBLEM I.
To find the solidity of a tetra&dron.
Rule I. Multiply -^ of the cube of the lineal side by the sqaare
root of 2, and the product will be the solidity.
Rule II. Multiply the cube of the length of a side of the body by
tlie tabular solidity, and the product will be the solidity of the body.*
This rule is general for all the regular bodies.
1. If the side of each face of a tetraedron be 1 ; C
required its solidity?
Here ^x l»x V2 = tVx V2=*11785113, the
solidity.
2. The side of a tetraedron is 12; what is its / fL
solidity? ^W5. 203*6467. /C^^^ \
PROBLEM II.
To find the solidity of a hexaedron^ or a cube.
Rule. Cube the side for its solidity.f
1. If the linear side of a hexaedron be 3, what is its content?
Ans. 3x3x3 = 27.
PROBLEM III.
To find the solidity of an octaedron.
Rule. Multiply the cube of the side by the square root of J?, and
^ of the product will be the content. J
1. What is the solidity of an octaedron, when the linear side
iol?
♦ See Appendix, Demonstration 91. t Ibid. 61 t Ibid, 92.
THE FIVE REGULAR BODIES.
It^
1' X V2 X J = ^2 = -4714045.
2. What is the solidity of the octaedron,
whose linear side is 2 ? Ans, 3 "7712.
PROBLEM IV.
To find the solidity of a dodecaedron.
Rule. To 21 times the square root of 5 add 47, and divide the
gum by 40; multiply the root of the quotient by 5 times the cube of
the lineal side, and the product will be the solidity.*
1. If the lineal side of the dodecaedron be
1, what is its solidity?
Here A=l, consequently 5 A'V Tri"
= 7-66311896, the content.
2. The side of a regular dodecaedron is 12
inches ; how many cubic inches does it con-
tain? Ans, 13241-8694592.
PROBLEM y.
To find the solidity of an icosaedron.
Rule. To 7 add three times the square root of 5, take half the
sum, multiply the square root of this half sum by f of the cube oi
the lineal side, and the product will be the solidity.+
1. What is the solidity of an icosaedron,
whose lineal side is 1?
Let the side be denoted by A. Then A= 1,
and consequently
I AV^^ = iV^^t|V5 = 2-18169499,
the content.
2, What is the solidity of an icosaedron, whose lineal side is 12
fuet? Ans, 3769-9689 feet.
iVo/e. — The following table may be collected from the examples given in the
forep:oing rules, each of which has been demonstrated under its particular
head. It has also been demonstrated that the cube of the lineal side of any
regular solid multiplied by the tabular number corresponding to the figure,
will give its content. It is particularly recommended to the pupil to employ
the general rule given in Problem I. whenever the content of any of the five
rogular bodies is required.
* See Appendix Demonstration 93.
t Ibid. 91
THE FIVE REGULAR BODIES.
TABLE IIL
Shewing the solidity of the Jive regular bodies^ the length
in each being 1.
No of
Bides.
Names.
Solidity.
4
6
8
20
12
Tetraedron
Hexaedron
Octaedron
Icosaedron
Dodecaedron
•1178511
10000000
•4714045
2-1816950
7-6631189
PROBLEM VI.
To find the surface * of a tetraedron.
Rule I. Multiply the square of the linear side by the square root
of 3, and the product will be the whole surface.f
Rule II. Multiply the square of the length of a side of the body,
by the tabular area corresponding to the figure, and the product will
be the surface of the body. This is a general rule for finding the
surfaces of the regular bodies.
1. If the side of a tetraedron be 1, what is its surface?
Here, Px V3= V3 = 1*7320508 = the whole surface.
2. The side of a tetraedron is 12 ; what is its surface ?
Ans, 249-415316^.
PROBLEM VII.
To find the surface of a hexaedron^ or cube.
Rule. Square the side and multiply it by 6, and the product will
be the surface.^
1. If the side be 1, what is the surface of a hexaedron?
12x6 = 6 the whole surface.
2. If the side be 4, what is the surface of a hexaedron?
Am. 9G.
PROBLEM VIII.
To find the surface of an octaedron.
Rule. Multiply the square of the side by the square root of 3, and
double the product will be the surface. §
♦ Though the next section treats exclusively of the surfaces of solids, and
would therefore seum to be the proper place for this problem and the following
ones in this section, yet it has been thouglit more convenient to place together
the rules both for finding the solidities and surfaces of those curious bodies.
t Sco Appendix, Demonstration 95. J Ibid. 96. § Ibid. Vt
THE FIVE REGULAR BODIES.
/9
1. If the side of an octaedron be 1, what is its surface?
2x 12V3 = 2 V3 = 3-4641016 = the wliole surface.
2. If the side of an octaedron be 12, what is its superficies ?
Ans. 498-8300304.
3. If the side of an octaedron be 4, what is its surface ?
Ans, 55-4256256.
PROBLEM IX.
7 0 find the superficies of a dodecaedron.
KULE. To 1 add \ of the root of 5 ; multiply tlie root of the sum
by 15 times the square of the lineal side, and tlie product will be the
surface.*
1. If the lineal side be 1, what is the surface of a regular dode-
caedron V
Herel2xl5V(l + IV5) = 15V(l + iV5) = 20-645728807,thesur-
ace.
2. What is the surface of a dodecaedron, whose lineal side is 2?
Ans. 82-58292.
PROBLEM X.
To find the superficies of an icosaedrou.
Rule. Multiply five times the square of the lineal side by the
equare root of 3, and the product will be thu surface.f
1. The side of an icosaedron is 1, what is its surface?
SxPx V3 = 5V3 = 8-66025403.
2. What is the surface of an icosaedron whose side is 2?
Ans. 34-641.
3. What is the surface of an icosaedron whose side is 3 ?
Ans. 77-9423.
j^ote. — In finding the superficial content of the regular bodies, it is particu-
larly recommended to employ the general rule given in Problem VI. in prac-
tice, in preference to any other. The particular rules given for each solid are
introduced merely to find the tabular numbers by which the pupil is to work.
From the examples given in the preceding rules, in which the lineal side of
each regular solid is 1, the following tabular numbers may be collected.
TABLE IV.
Shewing the surfaces oj the five regular todies., when the linear
side is 1.
Number
of sides.
Names.
Surface.
4
6
8
12
20
Tetraedron
Hexaedron
Octaedron
Dodecaedron ...
Icosaedron
1-7320508
6-0000000
3-4641016
20-6457288
8-6602540
" tJee Ai'pQudix, Demonstration 98.
t ibid. 9d.
80 SLTIFACKS OF SOLIDS.
SURFACES OF SOLIDS.
SECTION VL
H I
PROBLEM I.
To find the surface of a prism.
Rule. Multiply the perimeter of the end of the solid by its length,
ko the product add the area of the two ends, and the sum will be the
surface.*
1. If the side H I of the pentagon be 25 feet, and /'Ax.
height I D 10, what is its surface? A|<^/ \^'^
25 X 5 = 125, the perimeter ;
Then 125 x 10=1250 = the upright surface;
252 X 1-720477= 1075-298125 = tlie area of one end;
And 1075-298125 x2 = 2150-596250=the area of
both ends ;
Then 2150-596250 + 1250 = 3400-59625 = the entire
surface.
2. If the side of a cubical piece of timber be 3 feet
6 inches, what is the upright surface and whole super- ^
ficial content? ^ ^
A (49 feet upright surface.
* (73 feet 6 in. whole superficial content.
3. If a stone in the form of a parallelopipedon be 12 feet 9 inches
lon^, 2 feet 3 inches d«ep, and 4 feet 8 inches broad, what is the
upright surface and whole superficial content ?
A (176 feet 4 in. 6 sec. upright surface.
* (197 feet 4 in. 6 sec. whole sup. content.
PROBLEM n.
To find the surface of a pyramid.
Rule. Multiply the slant height by half the circumference of the
base, and the product will be the surface of the sides, to which add
the area of the base for the whole surface.f
Note. — The slant height of a pyramid is the perpendicular distance from the
vertex to the middle of one of the sides, and the perpendicular height is a
Btraight line drawn from the vertex to the middle of the base,
* See Appendix, Demonstration 100. t itld. lOL
SURFACES OF SOLIDS.
81
1. The slant height of* a triangular pyramid is 10 feet,
and each side of the base is 1 ; what is its surface ?
Half circumference = f
Slant height =10
Upright surface = 15
Area of the base = -433013
The entire surface = 15-433013 B
2. The perpendicular height of a heptagonal pyramid
is 13*5 feet, and each side of the base 15 inches ; required its surface?
Ans. 65-0128 feet.
PROBLEM III.
To find the surface of a cone.
Rule. Multiply the slant height by half the circumference of the
base, and the product, with the area of the base, will be the whole
surface.*
1 . What is the surface of a cone whose side is 20,
and the circumference of its base 9 ?
Here 20x|=90=the convex surface.
92 X -07958 = 6-44598 = the area of the base.
Then 90 + 6-44598 = 96-44598 = the whole sur-
face,
2. The perpendicular height of a cone is 10-5
feet, and the circumference of its base is 9 feet;
what is its superficies? Ans, 54-1336 feet.
PROBLEM IV.
To find the superficies of the frustum of a rights regular pyramid.
Rule. Add the perimeters of the two ends together, and multiply
half the sum by the slant height, the product will be the upright
surface ; to which add the areas of both ends, and the sum will be
the whole surface. f
1. What is the superficies of the frustum of a square pyra-
mid, each side of the greater base A B being 10 inches, and
each side of the less base C D 4 inches, and slant height
20 inches?
Here 10 x 4 = 40 the perimeter of the greater base.
And 4 X 4 = 16 the perimeter of the less end.
Sum 56, the half of which is 28.
Then 28 x 20 = 560 = the upright surface.
10 X 10= 100 = the area of the greater base.
4 X 4= 16 = the area of the less end.
Hence 560+ 100+ 16 = 676 = the whole surface.
2. What is the superficies of the frustum of an octagonal pyramid,
each side of the greater base being 9 inches, each side of the less base
5 inches, and the height 10'5 feet? Ans. 52-59 feet.
* See Appendix, Demonstratiou 10i2.
f Ibid. 103.
82 SURFACES OF SOLIDS.
PROBLEM V.
To find the superficies of the frustum of a cone.
Rule. Add the perimeters of both ends together, and multiply
half the sum by the slant height, to which add the areas of both ends,
for the whole superficies.* ^ ^^rr^^ a
1. If the diameters of the two ends CD and AB
are 7 and 3, and the slant height D B 9, what is the
whole surface of the frustum A B C D ?
^^x 3-1416x9 = 141-372, the convex surface, d
7x7x -7854 = 38-4846, the area of the base CD
3x3x-7854= 7-0686, the area of tlie end A B.
Then 141-372 + 45-6532= 186-9252 = the whole surface of the
frustum.
2. What is the superficies of the frustum of a cone, whose greater
diameter is 18 inches, and less diameter 9 inches, and the slant
height 171-0592 inches? Ans, 7672-981.
PROBLEM VL
To find the superficies of a wedge.
Rule. Find the area of the back, which is a right*angled paral-
lelogram; find the areas of both ends, which are triangles; and also
of both sides, which are trapezoids ; all these areas added together
will evidently be the whole surface.f ^ ^
1. The back of a wedge is 10 inches long,
dnd 2 inches broad, each of its faces is 10
inches from the edge to the back ; required its
whole surface?
10x2 = 20= the area of the back.
10 x 10 X 2 = 200 the areas of both faces.
V(A£2-Ex2)z=V(100--l) = 9-949 = Aic;then J
9-949 X 2= 19-898 = areas of both ends. *
Hence 200 + 20 + 19-898 = 239-898 = the
whole surface of the wedge.
2. The back of a wedge is 20 inches long,
and 2 inches broad ; each of its faces is 10 inches from the back to
the edge; what is its whole surface? Ans. 459*898.
PROBLEM VIL
To find the area of the frustum of a wedge.
Rule. Find the areas of the back and top sections ; of the two
faces ; and of the two ends ; the sums of all the separate results will
evidently be the whole surface.
* See Appendix, Demonstration 104. t Ibid* 106.
SURFACES OF SOLIDS.
83
1. The length and breadth of the back are
10 and 2 inches, the length and breadth of the
upper section are 10 and 1 inches, the length
of the edge from the back to the upper section
is 10 inches; required the whole surface?
10x2 = 20 = the area of the back.
10 X 1 = 10 = the area of the upper section.
10 X 10 X 2 = 200= the areas of both faces.
?^=i = -5, andV(100--25) = 9-98 = B2^.
Then(2 + l)x9-98 = 29-94 = areasofbothends.
Hence 20 + 10+ 200+ 2994 = 259-94 inches, the answer.
2. The length and breadth of the back are 10 and 4, the length
and breadth of the upper section are 5 and 2, and the length of each
of the faces is 20; required the whole superficies? Ans, 470* 78»
PROBLEM VIII.
To find the surface of a glohe or sphere.
Rule. Multiply the diameter of the sphere by its circumference,
and the product will be its convex surface.*
1. What is the surface of a globe, whose diameter is 24 inches ?
24x3-1416 = 75-3984, the circumference:
75-3984x24=1809-5616 inches, the answer.
2. What is the surface of the earth, its diameter being 7957|, and
circumference 25000 miles? Ans. 198943750 square miles.
PROBLEM IX.
To find the convex surface of any segment, or zone of a sphere.
Rule. Multiply the circumference of the whole sphere by the
neight of the segment, or zone, and the product will be the convex
surface.f
1. If the diameter of the earth be 7970 miles, the height of the
frigid zone will be 252-361283 miles, what is its surface?
Here 7970 x 3-141 6 = the circumference ; then
7970 X 3-1416 x 252-361283 = 6318761-107182216 miles.
2. If the diameter of the earth be 7970 miles, the height of the
temperate zone will be 2143*6235535 miles ; what is its surface?
Ans. 53673229-812734532 miles.
3. If the diameter of the earth be 7970 miles, the height of the
torrid zone will be 3178 030327 miles; what is its surface?
A?is. 79573277-600166504 miles.
Note. — By adding the surfaces of both frigid zones and both temperate zonea^
to the surface of the torrid zone, the sum 199557259 -44, is the surface of the
earth in square miles.
4. The diameter of a sphere is 3, the height of the segment 1 ;
what is its convex surface ? Ans. 9 4248.
5. The circumference of a sphere is 33, the height of the segment
la 4 ; what is its convex surface? Ans. 132.
* Bee Appendix, Demonstration 107, t Ibid. 108.
84 SURFACES OF SOLIDS.
PROBLEM X.
To find the surface of a cylinder.
Rule. Multiply the circumference by the length, and the product
will be the convex surface ; to which add the area of the two ends,
and the sum will be the surface of the entire solid.*
1. What is the entire surface of a cylinder, whose length is 10 feet,
and its diameter 5 feet ?
3-1416
5
15-7080, then 15-708 x 10=157'08 the convex surface.
5 X 5 X -7854: = the area of the base ; then
2 X 5 X 5 X -7854 = 50 x -7854 = 39 -2700 = the area of both bases; then
157*08 + 39-27= 196-35, the answer.
2. Required the superficial content of a cylinder, whose diameter
is 21-5 inches, and height 16 feet? Ans. 95-1 feet.
3. What is the surface of a cylinder whose diameter is 20-75 inches,
and its length 65 inches? Ans. 29*595 feet.
PROBLEM XL
2^0 find the superficies of a circidar cylinder.
Rule. Add the inner diameter to the thickness of the ring, multi-
ply the sum by the thickness, and that product by 9*8696 for the
superficies, t
1. The thickness AC of a cylindrical ring is 2 inches, the inner
diameter CD 5 inches ; required its superficial content ?
Here (2 + 5) x 2=14 ; then 14 x 9-8696 = 138-1744 square inches.
PROBLEM XIL
To find the surface of a parallelopipedon.
Rule. Find the area of the sides and ends, and their sum will be
the surface.
1. What is the surface of a parallelopipedon, whose length is
10 feet, breadth 4, and depth 2 ? Ans. 136 teet.
10 X 4 = 40=the area of one face.
10x4=40 = the area of its opposite face.
10x2 = 20 = the area of one face.
10 X 2 =20= the area of its opposite face.
4x2= 8 = the area of one end.
4x2= 8 = the area of its opposite end.
136=the surface of the whole solid.
2. The length of a parallelopipedon is 5, breadth 4, and depth 3 ,
v/hat is its S'arface ? Ans. d^.
* &QQ Appendix, Demonstration 109. % I^id. 110.
MENSURATION OP TIMBER AND OF ARTIFICERS' WORK, 85
MENSUEATION OF TIMBER AND OF
AETIFIOERS' WORK.
SECTIOlSr VII,
DESCEIPTION OF THE CARPENTERS' RULE.
^
This instrument is sometimes called the sliding rule, and is usea
in measuring timber and artificers' work. By it dimensions are
taken, and contents computed.
It consists of two eq^ual pieces of box-wood, each one foot long,
connected by a folding joint..
One face of the rule is divided into inches and half-qnarters or
eighths. On the same side or face are several plane scales divided by
diagonal lines into twelfths; these are chiefly used in planning
dimensions which are taken in feet and inches. The edge of the |
rule is divided decimally ; that is, each foot is divided into 10 equal I
parts, and each of these again into 10 equal parts. By means of this I
last scale, dimensions are taken in feet, tenths, and hundredths ; and I
then multiplied as common decimal numbers.
In one of these equal pieces, there is a slider on which are marked
the two letters B, C; on the same face are marked the letters A, D.
The same numbers serve for both these two middle lines, the one
being above the numbers, and the other below.
Three of these lines, viz.. A, B, C, are called double lines, as they
proceed from I to 10 twice over. These three lines are exactly alike
l30th in division and numbers, and are numbered from the left hand
towards the right, 1, 2, 3, 4, 5, 6, 7, 8, 9 to 1, which stands in the
middle ; the numbers then go on, 2, 3, 4, 5, 6, 7, 8, 9 to 10, which
stands at the right-hand end of the rule.
These four lines are logarithmic ones ; the lower line D, is a single
one, proceeding from 4 to 40, and is called the girt line, from its use
m finding the content of timber.
Upon it are also marked W G at 17*15, A G at 18*95, and I G at
18*8. These are the wine, ale, and imperial gauge points.
On this face is a table of the value of a load, or 50 cubic feet, of
timber, at all prices from 6 pence to 2 shillings per foot.
To ascertain the values of the figures on the rule, which have no
determinate value of their own, but depend upon the value set on the
anit at the left hand of that part of the rule marked 1, 2, 3, &c. : if
86 MENSURATION OF TIMBER AND OF ARTIFICERS WORK.
»
the first unit be called 1, the 1 in the middle will be 10, the other
figures that follow will be 20, 30, 40, &c., and the 10 at the right-
hand end will be 100. If the left-hand unit be called 10. the 1 in
the middle will be 100, and the following figures will be 200, 300,
400, 500, &c. : and the 10 at the right-hand end mil be 1000. L
the 1 at the left-hand end be called 100, the middle 1 will be 1000,
and the following figures will be 2000, 3000, 4000, &c., and the 10
at the right hand will be 10,000. From this it appears that the
values of all the figures depend upon the value set on the first unit.
The use of the double line A, B, is to find a fourth proportional,
and also to find the areas of plane figures.
The use of the several lines described here is best learned in
practice.
If the rule be unfolded, and the slider moved out of the grove, the
back part of it will be seen divided like the edge of the rule, all
measuring 3 feet in length.
Some rules have other scales and tables delineated upon them;
such as a table of board measure, one of timber measure, another foi
Bhewing what length for any breadth will make a square foot. There
is also a hue shewing whal length for any thickness will make a soli^
foot. — ^--- '" " ' '^"^'— ' .,^«,«.*.«*«!*,...*«K.
THE USE OF THE SLIDING EULE.
PROBLEM I.
To multiply numbers together.
Set 1 on B to the multiplier on A ; then against the multiplicanci
on B, stands the product on A.
1. Multiply 12 and 18 together.
Set 1 on B, to 12 on A ; then against 18 on B, stands the product
216 on A.
2. Multiply 36 by 22.
Set 1 on B, to 36 on A ; then as 22 on B goes beyond the rule,
look for 2-2 on B, and against it on A stands 79*2 ; but as the real
multiplier was divided by 10, the product 79*2 must be multij)lied by
10, which is effected by taking away the decimal point, leaving the
product 792.
PROBLEM II.
To divide one number by another.
Set the divisor on A, to 1 on B ; then against the dividend on A,
stands the quotient on B.
1. Divide 11 into 330.
Set the divisor 11 on A, to 1 on B ; then against the dividend 330
on A, stands the quotient 30 on B
2. Divide 7680 by 24.
THE USE OF THE SLIDING RULE. 87
Set 24 on A, to 1 on B ; then because 7680 goes beyond the rule
on A, look for 768 (the tenth of 7680), on A, and against it stands
82 on B ; but as the tenth of the dividend was taken that the number
should fall within the compass of the scale A, the quotient 32 must
be multiplied by 10, which gives 320 for the answer.
PROBLEM III.
To square ariy number.
Set 1 upon C, to 10 upon D ; then if you call the 10 upon D 1,
the 1 on C will be 10 ; if you call the 10 on D, 10, then the 1 on
C will be 100; if you call the 10 on D, 100, then the 1 on C will
be 1000 ; this being understood, you will observe that against every
number on D, stands its square on C.
1. What are the squares of 25, 30, 12, and 20?
Proceeding according to the above direction, 625 stands against
25, 900 against 30, U4 against 12, 400 against 20.
PROBLEM IV.
To extract the square root of a number.
Set 1 or 100, &c., on C, to 1 or 10, &c., on D ; then against every
number found on C, stands its root on D.
1. What are the square roots of 529 and 1600?
Proceeding according to the above directions, opposite 529 stands
23 J opposite 1600 stands 40, and so on.
PROBLEM V.
To find a mean proportional between two numbers^ as 9 aiid 25.
Set the number 9 on C, to the same 9 on D ; then against 25 on
V, stands 15 on D, the required mean proportional.
The reason of this may be seen from the proportion, viz.,
9: 15 :: 15:25.
1. What is the mean proportional between 29 and 430 ?
Set one number 29 on C, to the same on D ; then against the other
number 430 on C, stands 112 on D, which is the mean proportional,
nearly.
PROBLEM VL
To find a third proportional to two numbers^ as 21 and 32.
Set the first number 21, on B, to the second, 32, on A; then
against the second, 32, on B, stands 48 8 on A, which is the required
third proportional.
PROBLEM VII.
To find a fourth proportional to three given numbers.
Set the first term on B, to the second on A ; then against the third
term on B, stands the fourth on A.
If either of the middle numbers fidl beyond the line, take one-tenth
part of that number, and increase the fourth number found, ten times.
1. Find a fourth proportional to 12, 28, and 114.
Set the first term, 12, on B, to the second term, 28, on A; then
against the third term, 114, on B, stands 266 on A, which is the
answer.
HENSURATION OF TIMBER AND OF ARTIFICERS' WORK.
TIMBEE MEASUEE.
PROBLEM I.
To find the superficial content of a hoard or plank.
Rule. Multiply the length by the breadth, ar.d the product will
be the area.
Uote. — When the plank is broader at one end than at the other, add both
ends together, and take half the sum for a mean breadth.
BY THE carpenters' RULE.
Set 12 on B, to the breadth in inches on A; then against the
length in feet, on B, will be found the superficies on A, in feet.
1. If a board be 12 feet 6 inches long, and 2 feet 3 inches
broad, how many feet are contained in it ?
12 . 6 12-5
2 . 3 2-25
25 . 0 625
3.1.6 250
28.1 . 6 Ans. ^^^
28-125 Ans,
BY THE carpenters' RULE.
As 12 on B : 27 on A : : 12-5 on B : 28-125 on A.
2. What is the value of a board whose length is 8 feet 6 inches,
and breadth 1 foot 3 inches, at 5c?. per foot? Ans. 45. 5c?.
3. What is the value of a board whose length is 1 2 feet 9 inches,
and breadth 1 foot 3 inches, at 5o?. per foot? Ans. Qs. l\d,
4. What is the value of a plank whose breadth at one end is 2 feet,
and at the other end 4 feet, at Qd. per foot, the length being 12 feet?
Ans. 185.
5. How many square feet in a board, whose breadth at one end
is 15 inches, and at the other 17 inches, the length being 6 feet?
Ans. 8.
6. How many square feet in a plank, whose length is 20 feet, and
mean breadth 3 feet 3 inches ? Ans, 65.
PROBLEM II.
To find the solid content of squared or four sided timber.
Rule. Take half the sum of the breadth and depth in the middle,
(that is, the quarter girt,) square this half sum, and multiply it by
the length for the solid content.*
* This rule, which is generally employed in practice, is far from bcina
correct, when the breadth and depth differ materially from each other, and
the Limber does not taper.
TIMBER MEASURE. 89
BY THE carpenters' RULE.
As 12 on D : length on C : : quarter girt on D : the solid content
onC.
1. If a piece of squared timber be 3 feet 9 inches broad, 2 feet 7
inches deep, and 20 feet long ; how many solid feet are contained
therein ?
3.9
2. 7
2)6 .4:
3 . 2 quarter girt.
3. 2
, 6
6.4
10 . 0 . 4 square of the quarter girt.
20 length of the piece.
200 . 6 . 8 solid content.
BY THE carpenters' RULE.
As 12 on D : 20 on C : : 38 on D : 200^ on C.
2. A squared piece of timber is 15 inches broad, 15 inches deep,
and 18 feet long; how many solid feet does it contain?
Ans. 28^ feet, wliich is the accurate content, as the breadth and
depth are equal.
3. What is the solid content of a piece of timber, whose breadth is
16 inches, depth 12 inches, and length 12 feet? Ans. 16 feet.
Rule II. Multiply the breadth in the middle by the depth in the
middle, and that product by the length, for the solidity.*
4. The length of a piece of timber is 18 feet 6 inches ; the
breadths at the greater and less end 1 foot 6 inches, and 1 foot
3 inches, and the thickness at the greater and less end 1 foot 3 inches,
and 1 foot ; what is the solid content ?
1-5 ^ 1-25
1-25 ■** 1
2)2-75 2)2-25
1*375 mean breadth. 1*125 mean depth.
1-125 mean depth.
1 375 mean breadth.
1-546875
18-5 length.
28-6171875 solid content.
* This rule is correct when the timber does not taper; but when the timber
takers considerably, and the breadth and depth are neany equal, the rule is
very erroneous. The measurer, therefore, ought to consider the shapo of tha
timber he is about to measure before he appUes either of the above rules.
90 MENSURATION OF TIMBER AND OF ARTIFICERS' WORK.,
BY THE SLIDING RULE.
B A B A
As 1 : 13i: : 16i : 223 the mean square.
CD CD
As 1 : 1 :: 223 : 14-9 quarter girt.
C D D C
As 184 : i2 : :14'9 : 28-6 the content.
Note. — When the piece to be measured tapers regularly from one end to th»
other, either take the mean breadth and depth in the middle, or take the
dimensions at both ends, and half their sum for the mean dimension. This,
however, though very easy in practice, is but a very imperfect approximation.
When the piece to be measured does not taper regularly, but is thick in
some i^arts and small in others, in this case take several dimensions ; add them
all together, and divide their sum by the number of dimensions so taken and
use the quotient as the mean dimension.
Rule III. Multiply the sum of the breadths of the two ends by
the sum of the depths, to which add the product of the breadth and
depth of each end ; one-sixth of this sum, multiplied by the length,
will give the exact solidity of any piece of squared timber tapering
regularly.*
5. How many feet in a tree, whose ends are rectangles, the length
and breadth of one being 14 and 12 inches, and the corresponding
dimensions of the other 6 and 4 inches ; also the length 30^ feet ?
14 12 12x14=168
6 4 6x 4= 24
20x16 = 320
612 square inches =-^ square feet.
Then i x y x 30| = 18^ feet, the solidity.
6. How many solid inches in a mahogany plank, the length and
breadth of one end being 91| and 55 inches, the length and breadth
of the other end 41 and 29^ inches, and the length of the plank
47i inches? Ans, 126340-59375 cubic inches.
PROBLEM III.
Given (he breadth of a rectangular plank in inches^ to find how much
in length will make afoot^ or any other required quantity.
Rule. Divide 144, or the area to be cut off, by the breadth in
inches, and the quotient will be the length in inches.
The Carpenters' Rule is furnished with a scale which answers the
• This rule is correct, being that given for finding the solidity of the pria-
moid — which see.
Let B and 6 be the breadths of the two ends, D and d the depths, and
L the length : | (B D + (B + 6) x (D + d) + 6 rf) x L =the true solidity, a^
iu the rule for the prismoid.
20 16
TDIBER MEASURE.
91
purpose of this rule. It is called a Table of Board Measure, and ia
in the following form : —
0
0
0
0
5
0
Si
6
Inches.
12
6
4
3
2
2
1
1
Feet.
1
2
3
4
6
6
7 1
8
Breadth.
If the breadth be 1 inch, the length standing against it is 12 feet ;
if the breadth be 2 inches, the length standing against it is 6 feet ; ir
the breadth be 5 inches, the length is 2 feet 5 inches, &c.
When the breadth goes beyond the limits of the table on the rule,
it must be shut, and then you are to look for the breadth in the line
of board measure, which runs along the rule from the table of board
measure, and over against it on the opposite side, in the scale of
inches, will be found the length required. For example, if the
breadth be 9 inches, you will find the length against it to be 16
inches; if the breadth be 11 inches, the length will be found to be a
little above 13 inches.
1. If a board be 6 inches broad, what length of it will make a
square foot? Ans. 2 feet.
2. If a board be 8 inches broad, whi^ length of it will make 4
square feet? Ans. 6 feet.
3. If a board be 16 inches broad, what length of it will make 7
square feet ? ■ Ans. 5^ feet.
When the board is broader at one end than at the other, proceed
according to the following
Rule. To the square of the product of the length, and narrow
end, add twice the continual product of tuese quantities, viz., the
length, the difference between the breadths of the ends, and the area
of the part required to be cut off; extract the square root of the sum ;
.from the result deduct the product of the length and narrow end, and
divide the remainder by the difference betw^een the breadths of the
ends.*
If it were required to cut off 60 square
inches from the smaller end of a board,
A D being 3 inches, C E 6 inches, and
A B 20 inches.
1
Here kx=
-(V{(ABxAD)2 +
> '
0
B
s
1
^r-—--^.^
B
"2 BG^
4BCxABx60}-ABxAD) = J
(V{(20x3)2 + 6x20x60}~20x3) =
14-64, the length required.
' PROBLEM IV.
To find how much in length will make a solid foot., or any other
required quantity., of squared timber ., of equal dimensions
from end to end.
Rule. Divide 1728, the solid inches in a foot or the solidity to be
cut off, by the area of the end in inches, and the quotient will be the
end in inches.
• See Appendix, Demonstration 111.
92 MENSURATION OF TIMBER AND OF ARTIFICERS' WORK.
1. If a piece of timber be 10 inches square, how much in length
will make a solid foot?
10 X 10=100 the area of the end ; then 1728^100 = 17*28. Ans,
2. If a piece of timber be 20 inches broad, and 10 inches deep, how
much of it will make a solid foot? Ans. 8^ inches.
3. If a piece of timber be 9 inches broad, and 6 inches deep, how
much of it will make 3 solid feet? Ans. 8 feet.
On some Carpenters' Rules, there is a table to answer the purpose
t'lf the last rule ; it is called a Table of Timber, and is in the following
form : —
0
0
0
0
0
0
11
3
9 Inches. '
144
36
16
9
5
4
2
2
1 Feet.
1
2
3
4
5
6
7
8
9 Side of square.
PROBLEM V.
To find the solidity of round or unsquared timber.
Rule I. Gird the piece of timber to be measured round the middlr
with a string, take one-fourth part of tlie girth, and square it, an
multiply this square by the length for the solidity.
BY THE SLIDING RULE.
As the length on C : 12 or 10 on D : : quarter girt, in 12ths or
lOths on D : content on C.
Note — When the tree is ^eiy irregnlar, divide it into several lengths and
find the solidity of each part separately; or add all the girths together, and
divide the sum by the number of them.
1. Let the length of a piece of round timber be 9 feet 6 inches, and
its mean quarter girt 42 inches ; what is its content ?
3.5 quarter girt. 3 . 6 quarter girt.
3.5 3.6
12-25
9-5 length.
10
1
6
9
116-375 content.
12
9
3
6 length.
110
6
3
1 .6
116 . 4 . 6 content.
BY THE SLIDING RULE.
As 9-5 on C : 30 on D : : 35 on D : 116| on C ;
Or 9-5 : 12 :: 42 : iie^.
Rule II. Multiply the area corresponding to the quarter girt in
inches, by the length of the piece in feet, and the product will be tho
solidity.
Note. — It may sometimes happen that the quarter girt exceeds the limits of
Ihe table : in this case, take half of it» aud four times the content thus found
will give the required coutenc.
TIMBER MEASURE.
A TABLE FOR MEASURING TIMBER.
93
Quarter
Girt.
Area.
Feet.
250
272
294
■317
340
364
390
417
444
472
•501
531
562
594
626
659
694
730
766
803
840
■878
918
959
Quarter .
Girt. ^'^®*-
Feft.
1-000
1-042
1-085
1-129
1-174
1-219
1-265
1-313
1-361
1-410
1-460
1-511
1-662
1-615
1-668
1-722
1-777
1-833
1-890
1-948
2-006
2-066
2-126
2-187
Quarter
Girt.
Inches.
18
18^
19
19i
20
20i
21
21J
22
22i
23
23i
24
25
25J
26
26A
27
27J
28
28J
29
29^
Fett.
2-250
2-376
2 506
2-640
2-777
2-917
3-062
3-209
3-362
3.516
3-673
3-835
4-000
4-168
4-340
4-516
4-694
4-876
5-062
6-252
6-444
6-640
5-840
6-044
2. If a piece of round timber be 10 feet long, and the quarter girt
I2i inches; required the solidity ? Ans. lO'^b.
To find the solid content by this table, look for the quarter girt
12 J, in the column marked Quarter Girt, and in adjoining column,
marked Area, will be found 1-085, which multiplied by the length,
10 feet, will give 10-85 feet for the solid content.
3. A piece of round timber is 20 feet long, and the quarter girt
14^; how many feet are contained therein? Ans. 28*2 feet.
4. How many solid feet are contained in a tree 40 feet long, its
quarter girt being 9 inches? Ans. 22*48.
5. How many solid feet in a tree 32 feet long, its quarter girt
being 8 inches. Ans. 14-208.
6. How many solid feet in a tree 8J feet long, its quarter girt being
7i inches. Ans. 3-316 feet.
^4 MENSURATION OF TIMBER AND OF ARTIFICERS' WORK.
7. Required the content of a tree, whose length is 40 feet, and
qiiai'ter girt 27i inches? Ans. 210*08 feet.
8. What is the content of a tree, whose length is 30 feet 6 inches,
and quarter girt 27^ inches? Ans. 160-186 feet.
9 Required the content of a piece of timber, whose length is 25 feet
9 inches, and quarter girt 12| inches? Ans. 29 071 feet.
10. What is the solid content of a piece of timber, whose length ia
12 feet, and quarter girt 13^ inches? Ans. 15*18 feet.
11. What is the solid content of a piece of timber, whose quarter
girt is 14f inches, and length 38 feet? Ans. 57*418 feet.
When the square of the quarter girt is multiplied by the length,
the product gives a result nearly one-fourth less than the true quantity
m the tree. This rule, however, is invariably practised by timber
merciiants, and is not likely to be abolished. W^hen the tree is in
the form of a cylinder, its content ought to be found by Prob. lY.
Sec. IV., which gives the content greater than that found by the last
rule, nearly in the proportion of 14 to 11. Notwithstanding that the
true content is not found by means of the square of the quarter girt,
yet some allowance ought to be made to the purchaser on account of
the waste in squaring the wood so as to be fit for use. If the cylindrical
tree be reckoned no more than what the inscribed square will amount
to, the last rule, which is said to give too little, gives too mucli.
When the tree is not perfectly circular, the quarter girt is always too
great, and therefore the content, on that account, will be too great.
Dr Button recommends the following rule, which will give the
Dontent extremely near the truth : —
Rule. Multiply the square of one-fifth of the girt, or circumfer-
ence, by twice the length, and the product will be the content.
BY THE SLIDING RULE.
As double the length on C : 12 or 10 on D : : ^ of the girt, in
I2ths or lOths on D : content on C.
12. Required the content of a tree, its length being 9 feet 6 inches^
and its mean girt 14 feet.
14-5 = 2-8
ft.
2-8 9.
2*8
ft
i = 2
in.
6
2
. in. p.
.9. 7= J- of the girt; then
2.9.7
2.9.7
7-84 19 .
19
0
•
6 :
6.7. 2
2.1. 2 . 3
1 . 7. 7.1
148*96 content.
7. 9 . 11 . 10. 1
19
C
As 19 : 10 :: 28
Or 19 : 12 :; 33-
148 . 9 . 8 . 11 . 7 content.
D D C
149 content by the Sliding Rule,
149 content without it.
TIMBER MEASURE. 96
Dr Gregory recommends the following rules given by Mr Andrews : —
Let L denote the length of the tree in feet and decimals, and G
the mean girt in inches.
Rule I. Making no allowance for bark.
L Gr^ T r ^
2oqt= cubic feet, customary; and -^—= cubic feet, true content.
Rule II. Allowing J for bark.
L G^ L r^
gTr^= cubic feet, customary; oufiO^^^^^^^ ^^^*' *"^® content.
Rule III. Allowing ^ for bark.
LQ.2 LG^
^g|^= cubic feet, customary; 22oT= cubic feet, true content.
Rule IV. Allowing y\ for bark.
2^^-=^ cubic feet, customary; ^y—= cubic feet, true content.
What is the solid content of a tree, whose circumference, or girt,
is 60 inches, and length 40 feet ?
By Rule I.
QOQ4 = o2J cubic feet, customary.
40x602 ^^„ ,. n .
— -.orvy = 79 f cubic feet, customary.
By Rule II.
40x60^ ^,7 OK 1. r . .
-^^Q— = 47*85 cubic feet, customary.
40x602 ^, ;. r ^ ,
gRfio ~ cubic feet, true content.
By Rule HI.
• ^^.^ =50-61 cubic feet, customary
40x60* nt ^A v.- c ^ .
• =64-54 cubic feet, true content
By Rule IV.
40x602 ^„ _ , • r .
■ =52-47 cubic feet, customary-
40 X 60^
2150
= 66-97 cubic feet, true content.
Then rlie two ends are very unequal, calculate its content by the
I given for finding the solidity of the frustum of a cone, and
tict the usual allowance from the result.
When it is required to find the accurate content of an irregular
body, not reducible to any figure of which we have already treated,
provide a cylindrical or prismatic vessel, capable of containing the
a
96 MENSURATION OF ARTIFICERS' WORK.
solid to be measured ; put the solid into the vessel, and pour in water
to cover it, marking the height to which the water reaches. Then
take out the solid, and observe how much the water has descended in
consequence of its removal ; calculate the capacity of the part of the
vessel thus left dry, and it will evidently be equal to the solidity of
the body whose content is required.
AETIFICERS' WOEK.
Artificers compute their works by several different measures : —
Glazing and masonry by the foot.
Plastering, painting, paving, &c., by the yard of 9 square feet.
Partitioning, roofing, tiling, flooring, &c., by the square of 100
square feet.
Brick work is computed, either by the yard of 9 square feet, or by
the perch, or square rood, containing 272^ square feet, or 30^ square
yards ; 272^ and 30;^ being the squares of 16^ feet and 5 J yards re-
spectively.
CAEPENTEES' AND JOINEES' WOEK
OF FLOORING.
To measure joists, multiply the breadth, depth, and length toge-
ther for the content.*
1. If a floor be 50 feet 4 inches long, and 22 feet 6 inches broad;
how many squares of flooring are in that room?
50-333 60 . 4
22-5 22 . 6
251665 1107 . 4
100666 25 . 2
100666
100)1132-4925
100)11-32 . 6
11-3249 squares.
11-325
Ans. 11 squares 32^ feet.
2. If a floor be 51 feet 6 inches long, and 40 feet 9 inches broad;
low many squares are contained in that floor? Ans. 20*986 squares.
3. If a floor be 36 feet 3 inches long, and 16 feet 6 inches broad,
how many squares are contained in that floor ?
Ans. 5 squares 98g feet.
4. If a floor be 86 feet 11 inches long, and 21 feet 2 inches broad ;
how many squares are contained in it? Ans. 18-3972.
5. In a naked floor the girder is 1 foot 2 inches deep, 1 foot broad,
and 22 feet long ; there are 9 bridgings, the scantling of each (viz.,
* Joists receive various names from their position ; such as girders, bindingf-
joists, ti-imming-joists, common-joists, ceiling-joists, &c. When girders and
joists of flooring are designed to bear considerable weight, they should be lot
irto the wall at each end about two-thirds of the thickness of the waU.
carpenters' and joiners* work 97
bieadth and depth) being 3 inclies, by 6 inches, and length 22 feet;
9 binding joists, the length of each being 10 feet, and scantlings 8
inches by 4 inches ; the ceiling joists are 25 in number, each 7 feet
long, and their scantlings 4 inches by 3 inches ; what is the solidity
of the whole? ^^^5. 85 feet.
6. What would the flooring of a house three storeys high come to,
at £5 per square ; the house measures 30 feet long, and 20 broad ;
there are seven fire-places,* two of which measure, each 6 feet by i
feet ; two others, eacli 6 feet by 5 feet 6 inches ; two, each of 5 feet
6 inches by 4 feet ; and the seventh 5 feet by 4 ; the well-hole for
the stairs is 10 feet by 8 ? Ans, £69, 2s.
OF partitioning.
Partitions are measured by squares of 100 feet, as flooring ; their
dimensions are taken by measuring from wall to wall, and from floor
to floor ; then multiply the length and height for the content in feet,
which bring to squares by dividing 100, as in flooring. When doors
and windows are not included by agreement, deductions must be
made for their amount, f
1. A partition measures 173 feet 10 inches in length, and 10 feet
7 inches in height ; required the number of squares in it ?
Ans. 18*3972 squares.
2. A partition between two rooms measures 80 feet in length, and
60 feet 6 inches in height ; how many squares in it?
Ans. 40f squares.
3. If a partition measure 10 feet 6 inches in length, and 10 feet 9
inches in height ; how many squares in it? Ans. 1 square 12^ feet.
4. What is the number of squares in a partition, whose length is
50 feet 6 inches, and height 12 feet 9 inches ?
Ans. 6 squares, 43 feet, 10| inches.
In roofing, the length of the rafters is equal to the length of a
string stretched from the ridge down the rafter till it meets the top of
the wall.
To find the content, multiply this length by the breadth and depth
of the rafters, and the result will be the content of one rafter ; and
that multiplied by the number of them will give the content of all the
rafters 4
1. If a house within the walls be 42 feet 6 inches long, and 20 feet
3 inches broad ; how many squares of roofing in that house ?
* Fire-places, <fec., are of course to be deducted.
t The best and strongest partitions are those made with framed work. The
king-posts are measured as roofing, the rest as flooring.
t Workmen generally take the flat and half the flat of any house, taken
within the walls, to be the measure of the roof of the same house. This,
however, is only when the roof is of a true pitch. The usual pitches are tho
common, or true pitches, in which the rafters are three-fourths of the breadth
of the building; the Gothic pitch is when the length of the principal rafters is
equal to the breadth of the building : the pediment pitch is when the perpen-
dicular height is two-ninths of the breadth.
"When the covering of the building is to be plain tiles or slates, the roof la
generally of a tnie or common pitch ; the Gothic pitch is used when the
covering is of pantiles ; the pediment pitch is used when the roof is to be
covered with lead.
98 MENSURATION OF ARTIFICERS' WORK.
ft. ft. in.
42-5 42 . 6
20-25 .20 . 3
2125 840
850 6| 10 . 1
8500 3i 10 . 7
860*625 flat. 860 . 8 flat.
430-3125 430 . 4
100)1290-9375 100)1291
12-91 squares. 12 . 91
2. What cost the roofing of a house at lis. per square ; the length
within the walls being 50 feet 9 inches, and the breadth 30 feet ; the
roof being of a true pitch? Ans. £12, lis. 2^d.
3. What number of squares are contained in a house, whose length
within the walls is 40 feet, and breadth 18 feet ; the roof being com-
mon pitch? Ans. 10 squares and 80 feet.
4. How many squares in the roof of a building, the length of the
Vouse being 60 feet, and the length of the rafter 14 feet 6 inches?
Ans. 17 squares and 40 feet.
6. How many squares in a building, whose length is 50 feet, and
Ihe length of the rafter 15 feet? Ans. 15 squares.
6*. How many squares in the roof of a building, whose length is
B7 feet, the length of the rafter being 13 feet ?
Ans. 9 squares and 62 feet.
7. How many squares in the roof of a building, whose length is 70
feet 6 inches, the length of the rafter being 14 feet 6 inches?
Ans. 20 squares and 44^ feet.
8. How many squares in the roof of a building, whose length if
50 feet, and the length of a string reaching across the ridge from
eave to eave being 30 feet ? Ans. 15 squares.
Note. — All the timbers employed in roofing are measured like those used in
flooring, except where there is a necessity of cutting out parallel pieces equal
to, or exceeding 2^ inches broad and two feet long. In this case the amount
of the pieces so cut out must be deducted from the content of the whole piece
found from its greatest scantlings. When the pieces cut out do not amount to
the above dimensions, they are considered as useless, and therefore no deduc-
tion is to be made for them.*
9. Let the tie-beam T B be 36 feet
long, 9 inches broad, and 1 foot 2 inches
thick ; the king-post K 11 feet 6 inches
high, 1 foot broad at the bottom, and
5 inches thick ; out of this post are
sawn two equal pieces from the sides,
each 7 feet long and 3 inches broad.
The braces B B are 7 feet 6 inches
* In the above figure K is called the king-post, and in measuring the piecei
cut out of it, the shortest length ia to be taken. T B is called the tie-beam,
which prevents the rafters R R from pressing out the wall. The braces B B
Serve to strengthen the rafters ; the struts S S servo for a similar purpose.
Besides strengthening tlie rafters, the braces and struts serve to bind the root
together. Wlsen head-room is required, the rafters are braced simply by B B.
carpenters' and JOTNEltS' WORK. 99
long, and 5 inches by 5 inches square ; the rafters R R are 19 feet
long, 5 inches broad, and 10 inches deep ; the struts S S are 3 feet
6 inches long, 4 inches broad, and 5 inches deep ; what is the mea-
surement for workmanship and also for materials?
ft. in. p
31 . 6 . 0 solidity of the tie-beam T B.
4.9.6 solidity of the king-post K.
2,7.3 solidity of the braces B B.
13.2.4 solidity of the rafters R R.
11 . 8 solidity of the struts S S.
53 . 0 . 9 solidity for workmanship.
1.5.6 solidity cut from the king-post.
51 . 7 . 3 solidity for materials.
OF WAINSCOTTINO.
Wainscotting is measured by the yard square, which is 9 square feet.
In taking the dimensions, the string is made to ply close over the
cornice, swelling panels, moulding, &c. The height of the room
from the floor to the ceiling being thus taken, is one dimension, and
the compass of the room taken all round the floor is the second
liimension.
Doors, windows, shutters, &c., where both their sides are planed,
are considered as work and half; therefore in measuring the room,
they need not be deducted ; but the superficial content of the whole
room found as if there were no door, window, &c., then the contents
of the doors and windows must be found, and half thereof added to
the content of the whole room.
When there are no shutters, the content of the windows must be
deducted ; chimneys, window-seats, check-boards, sopheta- boards,
linings, &c., must be measured by themselves.
Windows are sometimes valued at so much per window, and some-
times by the superficial foot. The dimensions of a window are taken
in feet and inches, from the under side of the sill to the upper side
of the top-rail ; and from the outside to outside of the jambs.
When the doors are pannelled on both sides, take double the mea-
sure for the workmanship.
For the surrounding architrave, girt round it and inside the jambs,
for one dimension, and add the length of the jambs to the length
of the cap-piece (taking the breadth of the opening for the length),
for the other dimension.
Weather- boarding is measured by the yard square, and some-
limes by the square.
Frame- doors are measured by the foot, or sometimes by the yard
square.
Staircases are measured by the foot superficial. The dimensions
are taken with a string passing over the riser and tread for one
dimension, and the length of the step for the other. By the length
of the step is meant the length of the front and the returns at the
two ends.
Per the balustrade, take the whole length of the upper part of tho
100 MENSURATION OF ARTIFICERS' WORK.
hand-rail, and girt it over its end till it meet the top of the newel-
post, for one dimension; and twice the length of the baluster upon
tlie landing, with the girt of the hand-rail, for the other dimension.
Modillion cornices, coves, &c., are generally measured by the foot
superficial.
Beads, stops, astragals, copings, fillets, boxings to windows, skirt-
ing-boards, and water-trmiks, are paid for by lineal measure.
Frontispieces are measured by the foot superficial, and the archi-
trave, frieze, and cornice, are measured separately.*
To find the contents of the foregoing work, multiply the two
corresponding dimensions together for the superficial content.
1. A room, or wainscot, being girt downwards over the mouldings,
measures 12 ft. 6 in. and 130 ft. 9 in. in compass ; how many yards
does that room contain ?
ft. in. ft.
130.9 130-75
12 . 6 12-5
1560
65,
.4.
. 6
6,
. 0.
.0
3,
. 0,
. 0
9)1634 .
. 4 .
. 6
65375
26150
13075
9)1634-375
181 yards, 6 feet.
181 . 5 A?is.
2. If the wainscot of a room be 15 ft. 6 in. high, and the compass
vf the room 142 ft. 6 in. ; how many yards are contained in it ?
Ans. 245t\ yards.
8. If the window shutters about a room be 60 ft. 6 in. broad, and
6 ft. 4 in. high; how many yards are contained therein, at w^ork and
a half? ^ Ans. 63f^ yards.
4. A rectangular room measures 129 feet 6 inches round, and is to
be wainscotted at 3s. 6d. per square yard ; after due allowance for
girt of cornice, &c., it is 16 feet 3 inches high ; the door is 7 feet by
3 feet 9 inches ; the window shutters, two pair, are 7 feet 3 inches
by 4 feet 6 inches; the cheek-boards round them come 15 inchea
below the shutters, and are 14 inches in breadth ; the lining-boards
* Baluster is a small column or pillar, used for balustrades.
Balustrade is a row of balusters, joined by a rail ; serving for a rest to tLe
arms, or as an inclosure to balconies, staircases, altars, &c.
Cornice is the third and uppermost part of the entablature of a column, or
the uppermost ornament of any wainscotting, (fee.
Bead is a round moulding carved Iiao beads in necklaces. There is also a
kind of plain bead, often set on the edge of each fascia of an architrave, on the
upper edge of skirting-boards, on the lining-board of a door-case, <fec
Architrave is that part of a column that bears immediately on the capitaL
It is supposed to represent the principal beam in timber buildings, iu which
it is sometimes called the master-piece or reason-piece. In chimneys it ib
called the mantel-piece. Architrave doors are those which have an architrave
on the jambs and over the doors. Architrave windows of timber are usually
raised out of the solid timber, and sometimes the mouldings are struck anC
laid on.
Astragal is a small round moulding, encompassing the top of the shaft of a
column, like a ring or bracelet. The shaft terminates at the top with au
astragal, and at bottom with a fillet, which in this place is called azia.
BKICKLAYERS' WORK. 101
round the doorway are 16 inclies broad: the door and window-
filiutters being worked on both sides, are reckoned as work and half,
and paid for accordingly ; the chimney 3 feet 9 inches by 3 feet, not
being enclosed, is to be deducted from the superficial content of the
room. The estimate of the charge is required. Ans. £43, is. 6^d,
5. The height of a room, taking in the cornice and mouldings, is
12 feet 6 inches, and the whole compass 83 feet 8 inches; the
three window-shutters are each 7 feet 8 inches by 3 feet 6 inches,
and the door 7 feet by 3 feet 6 inches ; the door and shutter, being
worked on both sides, are reckoned work and a half. Required the
estimate, at 6s. per square yard? Ans. £36, 12s. 2^d,
OE BEICKLAYEES' WOEK
OF TILING OR SLATING.
Tiling and slating are measured by the square of 100 feet. There
5s no material difference between the method employed for finding the
estimate of roofing and tiling ; bricklayers sometimes require double
measure for hips and valleys.
"When gutters are allowed double measure, the usual mode is, to
measure the length along the ridge tile, and add it to the contents of
the roof: this makes an allowance of one foot in breadth along the
hips or valleys. Double measure is usually allowed for the eaves,
60 much as the projector is over the plate, which is generally 18 or
20 inches.
AVhen sky-lights and chimney- shafts are not large, no allowance
is to be made for them ; but when they are large, their amount is to
be deducted.
1. There is a roof covered with tiles, whose depth on both sidea
(with the usual allowance at the eaves) is 30 feet 6 inches, and the
length 42 feet ; how many squares of tiling are contained therein ?
ft. in. ft.
30 . 6 30-5
42 42
1260 610
21 1220
100)1281 100)12,810
12 . 81 12 squares 81 feet.
2. There is a roof covered with tiles, whose depth on both sidea
(with the usual allowance at the eaves), is 40 feet 9 inches, and the
length 47 feet 6 inches ; required the number of squares contained
therein? Ans. 19 squares 35f feet.
3. What will the slating of a house cost at £1, 55. 6d. per square;
the length being 43 feet 10 inches, and the breadth 27 feet 5 inches,
on the flat; the eaves projecting 16 inches on each side — true pitch?
Ans. £24, 9^. b^d.
4. What is the content of a slated roof, the length being 45 feet
^ inches, and the whole giit 34 feet 3 inches ? Ans. 174*104 yards.
102 MENSURATION OF ARTIFICERS' WORK.
OF WALLING.
Brick- work is estimated at the rate of a brick and a half thick ; so
that if a wall be more or less tlian the standard thickness, it must
be reduced to it : thus, multiply the superficial content of the wall by
the number of half bricks in the thickness, and divide the product by 3.
The superficial content is found by multiplying the length by the
height. Bricklayers estimate their work by the rod of I6j feet, or
2725- square feet. Sometimes 18 feet are allowed to the rod, that is,
324 square feet; sometimes the work is measured by the rod of 21
feet long, and 3 feet high, that is, 63 square feet ; in this case, no
regard is paid to the thickness of the wall in measuring, but the
price is regulated according to the thickness.
When a piece of brick- work is to be measured, the first thing to be
done is to ascertain which of the above measures is to be employed ;
then, having multiplied the length and breadth together (the dimen-
sions being feet) the product is to be divided by the proper divisor,
namely, 272*25, 324, or 63, according to the measure of the rod,
and the quotient will be the measure in square rods of that measure.
To measure any arched way, arched window, or door, &c., the
height of the window or door from the crown or middle of the arch,
to tlie bottom or sill, is to be taken, and likewise from the bottom 01
sill to the spring of the arch, that is, where the arch begins to turn.
Then to the latter height add twice the former, and multiply the sum
by the breadth of the window, door, &c., and one-third of the pro-
duct will be the area sufficiently near the truth for practice.
1. If a wall be 72 feet 6 inches long, and 19 feet 3 inches high,
and 5 bricks and a half thick, how many rods of brick-work ara
contained therein, when reduced to the standard ?
Note. — ^Tlie standard means a wall a brick and a half thick ; therefore, to
reduce any wall to the standard, multiply the superficial content of it by the
n?imber of half bricks in its thickness, and divide by 3,
ft. in.
72 . 6
19 . 3
648
72
18,
, 1 .
6
9.
6.
0
1395.
.7.
6
11
3)15351 .
,10
. 6
272)5117(18 rods.
2397
68)221(3 quarters.
~T7 feet.
77crfc'.--That 68-06 is the fourth part of 272 26, and 68 is one-fourtU of 272.
BRICKLAYERS' WORK. 103
In reducing feet into rods, it is usual to divide by 272, rejecting the
decimal '25. By this method, the answer found above is about
i^ feet too much.
2. How many rods of standard brick-work are in a wall whose
length is 57 feet 3 inches, and height 24 feet 6 inches ; the wall being
2i bricks thick? Ans. 8*5866 rods.
3. The end wall of a house is 28 feet 10 inches long, and 55 feet
8 inches high to the eaves ; 20 feet high is 2^ bricks thick, another
2C feet high is 2 bricks thick, and the remaining 15 feet 8 inches ia
1^ half bricks thick, above which is a triangular gable 1 brick
thick, which rises 42 courses of bricks, of which every 4 courses
make a foot. What is the whole content in standard measure?
Ans, 253-62 yards.
OF CHIMNEYS.
When a chimney stands by itself, without any party-wall beinjoj
adjoined, take the girt in the middle for the length, and the height
of the storey for the breadth ; the thickness is to be the same as the
depth of the jambs ; if the chimney be built upright from the mantel-
piece to the ceiling, no deduction is to be made for the vacancy
between the floor (or hearth) and mantel- tree, on account of the
gatherings of the breast and wings, to make room for the hearth in
the next storey.
When the chimney-back forms a party-wall, and is measured by
itself, then the depth of the two jambs is to be measured, and the
length of the breast for a length, ^nd the height of the storey for the
breadth ; the thickness is the same as the depth of the jambs. That
part of the chimney which appears above the roof, called the chimney-
shaft, is measured by girding it round the middle for the length, and
the height is taken for the breadth.
In consideration of plastering and scaffolding, the thickness ia
generally reckoned half a brick more than it really is ; and in soma
places double measure is allowed on account of extra trouble.
1. Let the dimensions of a chimney, having a double funnel to-
wards the top, and a double shaft, be as follows, viz., in the parfour
the breast and two jambs measure 18 feet 9 inches, and the height
of the room is 12 feet 6 inches ; in the first floor, the breast and two
iambs girt 14 feet 6 inches, and the height 9 feet ; in the second
floor, the breast and the jambs girt 10 feet 3 inches, and the height
is 7 feet ; above the roof, the compass of the shaft is 13 feet 9 inches,
and its height 6 feet 6 inches; lastly, the length of the middle parti-
tion, which parts the funnel, is 12 feet, and its thickness 1 foot
3 inches ; how many rods of brick- work, standard measure, are
contained in the chiniuey, double measure being allowed, and thick-
104 MENSURATION OF ARTIFICERS' WORK.
ft. in. ft. in. p.
1st. 18 . 9 6th. 1.3.0
12 . 6 12
ii25 . 0 15 . 0 . 0 partition.
9.4.6 234 . 4 . 6 parlour.
■-— — - — - 130 . 6 . 0 first floor.
234 . 4. 6 71 . 9 . Osecondfloor.
89 . 4 . 6 shaft.
ft.
in.
2d.
14
9
.6
130
. 6
ft.
in.
ad.
10
7
.3
71
.9
ft.
in.
1th..
13
.9
6
. 6
541 . 0 . 0 sum
2
272)1082 .0.0 double.
68)266 (3 rods 3 quarters.
62 feet.
82 . 6
6 . 10 . 6
89. 4 76
Arts. 3 rods, 3 quarters, and 62 feet.
MASONS' WOEK.
To masonry belong all sorts of stone-work. The work is some-
times measured by the foot solid, sometimes by the foot in length,
and sometimes by the foot superficial. Masons, in taking dimensions,
girt all their mouldings, in the same manner as joiners.
Walls, columns, blocks of stone or marble, &c., are measured by
the solid foot, and pavements, slabs, chimney-pieces, &c., by the
square foot.
In estimating for the workmanship, square measure is generally
used, but for the materials, solid measure.
In the solid measure, the length, breadth, and thickness are
multiplied together.
In the superficial measure, there must be taken the length and
breadth of every part of the projection, which is seen without the
general upright face of the building.
1. If a wall be 82 feet 9 inches long, 20 feet 3 inches high, and 2
feet 3 inches thick ; how many solid feet are contained in that wall ?
PLASTERERS' WORK. 105
ft. in. ft.
82 . 9 82-75
20 . 3 20-25
1640 41375
3 = J 20.8.3 16550
6 = ^ 10 . 3 . 0 165500
^"^^ 5.0-Q 1675-6875
1675 .8.3 2-25
^ • ^ 83784375
3351 .4.6 33513750
3 = i 418.11.01 33513750
3770 . 3 . 6| 3770-296875 Ans.
2. If a wall be 120 feet 4 inches long, and 30 feet 8 inches liigh ;
how many superficial feet are contained therein ?
Ans. 36901 feet.
3. If a wall be 112 feet 3 inches long, and 16 feet 6 inches high;
now many superficial rods of 63 square feet are contained therein ?
Ans. 29 rods 25 feet.
4. What is the value cf a marble slab at 8s. per foot, the length
being 5 feet 7 inches, and breadth 1 foot 10 inches?
Ans. £4, Is. lOJd
PLASTEREES' WORK.
Plasterers' work is cf two kinds, viz., ceiling, which is plastering
apon laths ; and renderhig, which is plastering upon walls. These
are measured separately.
The content is sometimes estimated by the foot, sometimes by the
yard, and sometimes by the square of 100 feet. Enriched mould-
ings are calculated by the running foot or yard.
Deductions are made for chimneys, doors, windows, &c.
In plastering timber partitions, where several of the large braces
and otlier large timbers project from the plastering, a fifth is usually
deducted.
Whitening and colouring are measured in the same manner as
plastering. In timbered partitions, one-fourth or one-fifth of tha
wliole area is usually added, to compensate for the trouble of coloui--
ing the sides of the quarters and braces.
In arches, the girt round them is multiplied by the length for the
S'tiperficial content.
1. If a ceiling be 40 feet 3 inches lon^, and 16 feet 9 inches broad,
how many s(7uare yards contained therein?
106
MENSURATION OF ARTIFICERS' WORK.
640
6 = ^ 20.
3 = J 10.
3 = i 4
1
0,
0
6
9
0
9)674
. 2
.3
ft. in. ft.
40 . 3 40-25
16 . 9 16-75
20125
28175
24150
4025
9)674-1875
Ans, 74 yards 8 feet. Ans. 74.9097 yards.
2. The length of a room is 14 feet 5 inches, breadth 13 feet 2
mclies, and height 9 feet 3 inches to the under side of the cornice,*
which projects 5 inches from the wall, on the upper part next the
ceiling ; required the quantity of rendering and plastering, there
being no deduction but for one door, which is 7 feet by 4 ?
Ans. 53 yards 5 feet of rendering, 18 yards 5 feet of ceiling.
3. The circular vaulted roof of a church measures 105 feet 6 inches
in the arch, and 275 feet 5 inches in length ; what will the plastering
come to at Is. per yard? Ans. £161, Ss. 6|d
4. The length of a room is 18 feet 6 inches, the breadth 12 feet
3 inches, and height 10 feet 6 inches ; to how much amount the
ceiling and rendermg, the former at Sd. and the latter at 3c?. per
yard ; allowing for the door of 7 feet by 3 feet 8, and a fire-place of
6 feet square? Ans, £1, 135. 3d,
PLUMBEES' WORK
Plumbers' work is rated by the pound, or by the hundredweight
of 112 lbs. Sheet lead, used in roofing, guttering, &c., weighs from
6 to 12 pounds per square foot, according to the thickness ; and
leaden pipes vary in weight per yard, according to the diameters of
the bore.
The following table shews the weight of a square foot of sheet
lead, according to its thickness ; and the common weight of a yard
of leaden pipe according to the diameter of its bore.
Thickness
Pounds to a
Bore of
Pounds per
of lead.
square foot.
leaden pipes.
yard.
Inch.
1^
5-899
Of
10 ^
^
6-554
1
12
i
7-373
1
18 ^-
1
8-427
h
18
■^
9-831
If
21
i
11-797
2
24
• Cornices, festoons, &c., are put on after the room ia plastered and are not,
ti course, taken into account by the plasterer.
PAINTERS' WORK. 107
1. A piece of sheet lead measures 20 feet 6 inches in length, and 7
feet 9 inches in breadth ; what is its weiglit at 8;^ pounds to the square
foot?
ft. in. ft.
20 . 6 20-5
7 . 9 7-75
143 . 6 1025
15 . 4.6 1435
1435
158-875
158 . 10 . 6
1271-000
39*719
- cwt. qrs. lbs.
112)1310-719(11 . 2 . 22J, nearly.
112
190
112
28)78(2
56
22
2. What weight of lead ^ of an inch thick will cover a flat, 15
feet 6 inches long, and 10 feet 3 inches broad, the lead weighing
6 pounds to the square foot? Ans. 8 cwt., 2 qrs., Ij pounds.
3. What will be the expense of covering and guttering a roof with
lead, at 185. per cwt. ; the length of the roof being 43 feet, and the girt
over it 32 feet ; the guttering being 57 feet in length, and 2 feet in
breadth, allowing a square foot of lead to weigh 8| pounds ?
Ans. £104, 155. 3fJ.
4. What will be the expense of 130 yards of leaden pipe of an inch
and half bore, at 4d. per pound, admitting each yard to weigh 18
pounds? Ans, X39.
PAINTEES' WOEK.
Painters' work is computed in square yards. Every part ia
measured where the colour lies, and the measuring line is forced into
all the mouldings and corners. Double measure is allowed for carved
mouldings, &c.
Windows are done at so much a-piece ; sash-frames at a certain
price per dozen; sky-lights, window-bars, casements, &c., are charged
at a certain price per piece.
To measure balustrades, take the length of the hand-rail for one
dimension, and twice the height of the baluster upon the landing,
added to the girt of the hand-rail, for the other dimension.
No general rule can be given for measuring trellis-work; but,
108 MENSURATION OF ARTIFICERS' WORK.
however, double the area of one side is often taken for the nneasure of
both sides.
1. If a room be painted, whose height (being girt over the mould-
ing) is 16 feet 4 inches, and the compass of the room 120 feet 9
inches ; how many yards of painting in it ?
ft. in. ft.
120-75
16-3
120
16
. 9
. 4
1920
4=^ 40 .
G = i 8 .
3 = i 4 .
3
0
0
9)1972 .
3
36225
72450
12075
9)1968-225
Atis. 218-691 yards.
Ans. 219 yards 1 foot.
2. A gentleman had a room to be painted, its length being 24 feet
6 inches, breadth 16 feet 3 inches, and height 12 feet 9 inches ; also
the size of the door 7 feet by 3 feet 6 inches, and the size of the
window-shutters to each of the windows, there being two, is 7 feet
9 inches by 3 feet 6 inches ; but the breaks of the windows tltemselvea
are 8 feet 6 inches high, and 1 foot 3 inches deep ; wiiat will be the
expense of giving it three coats, at 2d. per yard each ; the size of
the fire-place to be deducted, being 5 feet by 5 feet 6 inches ?
Ans. £3, 35. lO^d.
3. The length of a room is 20 feet, iis breadth 14 feet 6 inches,
and height 10 feet 4 inches; how many yards of painting in it,
deducting a fire-place of 4 feet by 4 feet 4 inches, and tw^o window-
fehutters each 6 feet by 3 feet 2 inches ? Ans, 73t^V yards.
GLAZIEES' WOEK.
Glaziers take their dimensions either in feet, inches, and parts ; or
feet, tenths, and hundredths. They compute their work in square feet.
Windows are sometimes measured by taking the dimensions of one
pane, and multiplying its superficies by the number of panes. But
generally they take tire length and breadth of the whole frame for the
glazing. Circular windows are measured as if they were square,
taking for their dimensions their greatest length and breadth.
1. If a pane of glass be 3 feet 6 inches and 9 parts long, and 1
foot 3 inches and 3 parts broad ; how many feet of glass in thai
pane?
3.6.9 3-56
1.3.3 1-277
3.
6 .
9
2492
10.
8 .
3
2492
10 •
8 .
3
712
Ans. 4 .
6.
3.
11 .
3
Ans.
356
4-54612
VAULTED AND ARCHED HOOFS. 109
2. If there be 10 panes of glass, each 4 feet 8 inches 9 parts long,
and 1 foot 4 inches and 3 parts broad ; how many feet of glass are
contained in the 10 panes ? Ans. 64*0407.
3. There are 20 panes of glass, each 3 feet 6 inches 9 parts long,
and 1 foot 3 inches and 3 parts broad ; how many feet of glass are in
the 20 panes ? Ans. 90*9224 feet.
4. If a window be 7 feet 6 inches high, and 3 feet 4 inches broad ;
how many square feet of glass contained therein ? Ans. 25.
5. How many feet in an elliptical fan-light of 14 feet 6 inches in
length, and 4 feet 9 inches in breadth ? A7is. 68 feet 10 inches.
6. What will the glazing of a triangular sky-li^ht come to at 20d. ;
the base being 12 feet 6 inches, and the perpendicular height 6 feet
9 inches? Ans. £3, 105. S^d.
PAYEES' WOEK
Pavers' work is computed by the square yard ; and the content ia
found by multiplying tlie length by the breadth.
1. What will be paid for paving a foot-path, at 45. the yard, ihp
length being 40 feet 6 inches, and the breadth 7 feet 3 inches ?
ft. in. ft.
40 . 6 40-5
7 . 3 7-25
283 . 6 2025
10 . 1 . 6 810
Ans. 293 . 7 . 6
2835
Ans. 293-625 feet.
2. What will be the expense of paving a rectangular court-yard,
whose length is 62 feet 7 inches, and breadth 44 feet 5 inches ; and in
which there is a foot-path, whose whole length is 62 feet 7 inches,
and breadth 5 feet 6 inches, this at 3s. per yard, and the rest at
2s. 6d. per yard ? Ans. £39, 1 Is. S^d.
3. What is the expense of paving a court, at 35. 2d. per yard ; the
length being 27 feet 10 inches, and the breadth 14 feet 9 inches?
Ans. £7, 45. S^c?.
4. What will the paving of a walk round a circular bowling-green
come to at 25. Ad. per vard, the diameter of the bowling-green being
40 feet, and the breadth of the walk 5 feet? Ans. £9, 35. 3^d.
5. How many yards of paving in an elliptical walk 4 feet broad,
the longer diameter being 60 feet, and shorter 50 ?
Ans. 82-3797 yards.
VAULTED AND AECHED EOOFS.
Arched roofs are either domes, vaults, saloons, or groins.
Domes are formed of arches springing from a circular, or polygonal
base, and meeting in a point directly over the centre of that base.
no
MENSURATION OF ARTIFICERS' WORK.
Saloons are made by arches connecting the side walls of a building
to a flat roof or ceiling.
Groins are made by the intersection of vaulted roofs with each
other.
Vaulted roofs are sometimes circular, sometimes elliptical, and
sometimes Gothic.
Circular roofs are those of which the arch is a part of the circum-
ference of the circle.
Elliptical roofs are those of which the arch is a part of the circum-
ference of an ellipsis.
Gothic roofs are made by the meeting of two equal circular arches,
exactly above the span of the arch.
Groins are generally measured like a parallelopipedon, and the
content is found by multiplying the lerigtn and breadth of the base
by the height.
Sometimes one-tenth is deducted from the solidity thus found, and
the remainder is reckoned as the solidity of the vacuity.
PROBLEM I.
To find the solidity of a circular^ elliptical^ or Gothic vaulted
roof.
Rule. Find the area of one end, by one of the foregoing rules,
and multiply the area of the end by the length of the roof, or vault,
and the product will be the content.
Note. — When the arch is a segment of a circle, the area is found by Pi-ob.
XXVIII. Sec. II. When the arch is a segment of an ellipsis, multiply the
span by the height, and that product by '7854 for the area of the end. When
it is a Gothic arch, find the area of an isosceles triangle, whose base is equal to
the span of the arch, and its sides equal to the two chords of the circular seg-
ment of the arch ; then add the areas of the two segments to the area of the
triangle, and the sum will give the area of the end.
1. What is the content of a concavity of a
semi- circular vaulted roof, the span being 30
feet, and the length of the vault 150 feet?
30x30=900; then 900 x •7854 = 706-86,
hence
706-86-^2 = 353-43, the area of the end ;
Then 353*43 x 150 = 63014-5, the content.
2. What is the solid content of the vacuity
A 0 E B of a Gothic vault, whose span A B is
60 feet, the chord B 0, or A 0, of each arcli 60 feet ; the distance of
each arch from the middle of the chords as DE=12 feet, and tl\e
length of the vault 40 feet ?
In this example, the triangle ABO equilateral, and its area is \
DE3
AB2V3 = 900V3=1557. Again, f (B 0 x D E) + 2gQ=|(6ox
12*
12) + gQ^=494f=area of segment 0 E B, and 494|x2 = 988| the
areas of the two segments 0 E B and 0 H A; then (1557 + 988|) x 40
= 101832, the solidity required.
VAULTED AND ARCHED ROOFS. Ill
Let M N K L represent a perpendicular section of a vaulted rooi
((TOthic). The span A B is 60 feet, the thickness of the wall M A,
or B L, at the spring of the arch — 4 feet, the thickness OP at the
crown of the arch = 3, and the length of the roof =40 feet, the chord
A 0 or 0 B:=60 feet, and the versed sine D E=12 feet ; required the
«iolidity of the materials of tlie arch?
First, V(A03— AC2)=V(602— 302) = 51'96 = SO the height of
the vacuity of the arch, and S 0 + 0 P = 5 1 -96 + 3 = 54-96 = S P ;
again, AB + MA + B L=60 + 4 + 4 = 68 = M L, and MLxSP = the
area of the rectangle M N K L ; hence M L x S P x 40—101832 (tlie
solidity of the vacuity AO B by the last Problem), gives the solidity
of the materials ; that is,
68 X 54-96 X 40—101832 = 47659-2 feet, the solidity required.
Note.— When the arcli A O B is an elliptical segment, its area multiplied by
the length of the roof gives the soHdity of the vacuity, and M L multiplied by
S P, and the product by the length of the arch, gives the solidity of the cubic
figure whose end is M N K L ; and the difference of the two solidities is the
solidity of the mixed solid whose section isAMNKLBEOHA. The materials
of a bridge may be calculated after the same manner, by adding the solidities
of T, T, and of the battlements, to the solidity as found in this Problem.
3. Required the capacity of the vacuity of an elliptical vault, whose
gpan is 30 feet, and height 15 feet, the length of the vault being 90
feet. Ans. 31808-7 feet.
PROBLEM II.
2'ojind the concave or convex surface of a circular^ elliptical^ or
Gothic vaulted roof
Rule. Multiply the length of the arch by the length of the vault^
and the product will be the superficies.
Note. — To find the length of the arch, make a line ply close to it, quite across
.rom side to side.
1. What is the surface of a vaulted roof, the length of the arch
being 45 feet, and the length of the vault 140 feet?
140 X 45 = 6300 square feet.
2. Required the surface of a vaulted roof, the length of the arch
being 40 feet 6 inciies, and the length of the vault 100 feet ?
Ans. 4050 feet.
3. What is the surface of a vaulted roof, the length of the arch
being 40-5 feet, and the length of the vault 60 feet ? Ans. 2430 feet.
PROBLEM IIL
To find the solidity of a dome., having the height and the dimen-
sions of its base given.
Rule. Multiply the area of the base by the height, and two-thirda
of the product will give the solid content.*
* This rule is correct only in one case, namely, when the dome is half a
sphere, and in this case the height is equal to the radius of the circular base.
It is a well-known property that the solidity of a sphere is two-thirds of that
of a cylinder having the same base and height. But the solidity of a cylinder
is found by multiplying the area of its base by the height. Hence the reason
n
112 MENSURATION OF ARTIFICERS' WORK.
1. What is the solid content of a dome, in the form of a hemisphere,
tlie diameter of the circular base being 40 feet?
402 X -7854 = 1256 -64 = the area of base.
I (1256-64x20) = I (25132-8) = 16755'2, answer.
2. What is the solid content of an octagonal dome, each side of its
' ^se being 20 feet, and the height 2 L feet ?
Ans. 27039-1917 cubic feet.
3. Required the solidity of the stone-work of an elliptical dome,
the two diameters of its base being 40 and 30 feet, the height 17*32
feet, and the stone-work in every part 4 feet thick ?
Ans. 9479-086848 cubic feet.
PHOBLEM IV.
To find the superficial content of a dome., the height and dimensions
of its base being given,
Rule. Multiply the square of the diameter of the base by 1*5708,
and the product will be the superficial content.*
For an elliptical dome, multiply the two diameters of its base
together, and the product resulting by 1-5708 for the superficial
content, sufficiently correct for practical purposes.
1. The diameter of the base of a circular dome is 20 feet, and its
height 10 feet; required its concave superficies?
202x 1-6708 = 628*32 feet, the answer.
2. The two diameters of an elliptical dome are 40 and 30 feet, and
its height 17*32 feet; required the concave surface?
Ans. 1884*96 square feet.
3. What is the superficies of a hexagonal spherical dome, each side
of the base being 10 feet? Ans. 519-6152.
PROBLEM V.
To find the solid content of a saloon.
Rule. Multiply the area of a transverse section by the compass or
circumference of the solid part of the saloon, taken round the middle
part. Subtract this product from the whole vacuity of the room,
supposing the walls to ^o upright from the spring of the arch to the
flat ceiling, and the diff'erence will be the answer, as will appear
evident from the following example.
1 . What is the solid content of a saloon with a circular quadrantal
of the rule when ajjplied to this particular case. No general rule can be given
to answer every case, as some domes are circular, some elliptical, some poly-
gonal, &c. ; they are of various heights, and their sides of diflerent curvature.
>Yhen the height of the dome is equal to the radius of its base, (the curved
sides being circular or elliptical quadrants), or to half the mean proportional
between the two axes of its elliptical base, the above i-ule will answer pretty
vrcll ; but with any other dimensions it ought not to be used.
* This rule is correct only when the dome is circular, and its height oqual
to the radius of the base.— See Appendix. Demonstration 112.
VAULTED AND ARCHED ROOFS.
113
arch of 2 feet radiuvS, springing over a rectangular room of 20 feet
long and 16 feet wide?
23 X -7854 = 3-1416 = area of the quadrant
CDAF. 2 x2-^2 = 2 = areaofthe triangle CDF;
then 3-1416— 2 = 1*1416 = area of the segment
D AF. Now, 2x2 = 4= area of the rectangle
CDEF; then 4— 3-1416= -8584 = area of the
section DEFAD. V(22 + 22)= V8 =2-8284271.
2 X 16 + 2 X 20 = 72 = the compass within the
walls, i (2-8284271 — 2) = -4142136 = ES and
2-8284271 : -4142136 : : 2 : -2928932 = E ?/;
hence 72 = (-2928932 x 8) = 69-6568544 = the
circumference of the middle of the solid part of the saloon ; therefore
69-6568544 X •8584 = 59-79344381696 = the content of the solid part
of the saloon.
20 X 16 = 320 the area of the roof or floor, and 320 x 2 = 640 = the
»?olidity of the upper part of the room; then 640— 59*79344 =
i80-20656 feet, the solidity of the saloon.
2. If the height D E of the saloon be 3*2 feet, the chord J) F=4'5
feet, and its versed sine = 9 inches ; what is the solid content of the
lolid part, the mean compass being 60 feet? Ans. 138-26489 feet.
PROBLEM VI.
To find the superficies of a saloon.
Rule. Find its breadth by applying a string close to it across the
Burface ; find also its length by measuring along the middle of it,
quite round the room ; then multiply these two dimensions together
for the superficial content.
1. The girt across the face of the saloon is 5 feet, and its mean
compass 100 feet ; what is its superficial content ?
100 x 5 = 500, the answer.
2. The girt across the face of the saloon is 12 feet, and its mean
x>mpa8s 98 ; required its siuface? Ans, 1176 feet.
114 SPECIFIC GRAYITT.
SPECIFIC GRAVITY.
SECTION YIII.
1. The specific gi-avity of a body is the relation which the weight
of a given magnitude of that body has to the weight of au equal
magnitude of a body of another kind.
In this sense a body is said to be specifically heavier than another,
when under the same bulk it weighs more than that other. On the
contrary, a body is said to be specifically lighter than another, when
under the same bulk it weighs less than that other. Thus, if there
be two equal spheres, each one foot or one inch in diameter, the one
of lead and the other of wood, then since the leaden sphere is found
to be heavier than the wooden one, it is said to be specifically, or in
Bpecie, heavier, and the wooden sphere specifically lighter.
2. If two bodies be equal in bulk, their specific gravities are to
each other as their weights, or as their densities.
3. If two bodies be of the same specific gravity or density, thei\
nbsolute weights will be as their magnitudes or bulks.
4. If two bodies be of the same weight, the specific gravities will
be reciprocally as their bulks.
6. The specific gravities of all bodies are in a ratio compounded of
the direct ratio of their weights, and the reciprocal ratio of their
magnitude. Hence, again, the specific gravities are as the densities.
6. The absolute weights or gravities of bodies are in the compound
ratio of their specific gravities and magnitudes or bulks.
7. The magnitudes of bodies are directly as their weights, and
reciprocally as their specific gravities.
8. A body specifically heavier than a fluid, loses as much of its
weight, when immersed in it, as is equal to the weight of a quantity
of the fluid of the same bulk or magnitude ; if the body be of equal
density with the fluid, it loses all its weight, and requires no force
but the fluid to sustain it. If it be heavier, its weight in the fluid
will be only the difference between its own weight and the weight of
the same bulk of the fluid ; and therefore it will require a force equal
to this difl\3reuce to sustain it. But if the body immersed be liglitei-
than the fluid, it will require a force equal to the difterence between
its own weight and that of the same bulk of the fluid, to keep it from
rising in the fluid.
9. In comparing the weights of bodies, it is necessary to consider
SPECIFIC GRAVITY. 115
soiTie one as the standard with which all other bodies may be com-
pared. Rain water is generally taken as the standard, it being found
to be nearly alike in all places.
A cubic foot of rain water is found, by repeated experiments, to
weigh 62J pounds avoirdupois, or 1000 ounces, and a cubic foot
containing 1728 cubic inches, it follows that a cubic inch weighs
•03616898148 of a pound. Therefore if the specific gravity of any
body be multiplied by •03616898U8, the product will be the weight
of a cubic inch of that body in pounds avoirdupois ; and if ttds
weight be multiplied by 175, and the product be divided by 144, the
quotient will be the weight of a cubic inch in pounds troy, 144
pounds avoirdupois being exactly equal to 175 pounds troy.
10. Since the specific gravities of bodies are as their absolute
gravities under the same bulk, the specific gravity of a fluid will be
to the specific gravity of any body immersed in it, as the part of the
weight lost by the solid is to the whole weight. Hence the specific
gravities of different fluids are as the weights lost by the same solid
immersed in them.
PROBLEM I.
To find the specific gravity of a lady.
Case I. When the tody is heavier than water.
Weigh the body first in water, and afterwards in the open air; the
difference will give the weight lost in water ; then say, as the weight
lost in water is to the absolute weight of the body, so is the specific
gravity of water to the specific gravity of the body.
Case II. When the body is lighter than water.
Fix another body to it, so heavy as that both may sink in water
together as a compound mass. Weigh the compound mass and the
heavier body separately, both in the water and open air, and find
how much each loses in water, by taking its weight in water from its
weight in the open air. Then say, as the difference of these remainders
is to the weight of the lighter body in air, so is the specific gravity
of water to the specific gravity of the lighter body.
Case III. For a fluid of any kitid.
Weigh a body of known specific gravity both in the fluid and open
air, and find the loss of weight, by subtracting the weight in water
from the weight out of it. Then say, as the whole, or absolute
weight, is to the loss of weight, so is the specific gravity of the solid
to the specific gravity of the fluid.
The usual way of finding the specific gravities of bodies is the
following, viz. : —
On one arm of a balance suspend a globe of lead by a fine thread,
and to the otl>er arm of the balance fasten an equal weight sufficient
to balance it in the open air ; immerse the globe into the fluid, and
observe what weight balances it then, by which the lost weight is
ascertained, which is proportional to the specific gravity.
116 SPECIFIC GRAVITT.
Immerse the globe successively in all the fluids whose proportional
Bpecific gravity you require, observing the weight lost in each ; then
these weights lost in each will be the proportions of the fluids sought.
Examples. — Case I.
1. A piece of platina weighed 83*1886 pounds out of water, and in
water only 79 '5717 pounds; what is its specific gravity, that of water
being 1000?
83-] 886— 79-5717 = 3-61G9 pounds, which is the weight lost m
water ; then 3-6169 : 83*1886 : : 1000 : 23000 the specific gravity,
or the weight of a cubic foot of metal in ounces.
2. A piece of stone weighed 10 pounds in the open air, but in
water only 6 J pounds ; what is its specific gravity ? Arts, 3077.
Examples. — Case II.
3. If a piece of elm weigh 15 pounds in the open air, and that a
piece of copper, which weighs 18 pounds in open air, and 16 pounds
in water, is affixed to it, and that the compound weighs 6 pounds in
water ; required the specific gravity of the elm ?
Copper. Compound.
18 in air. 33
16 in water. 6
~2 loss. 27
2
As 25 : 16 : : lOOO : 600, the specific gravity of the elm.
4. A piece of cork weighs 20 pounds in open air, and a piece of
granite being affixed to it, which weighs 120 pounds in air, and
only 80 pounds in water, the compound mass weighs 16| pounds in
water ; required the specific gravity of the cork V Ans 240.
Examples. — Case III.
5. A piece of cast iron weighed 259*1 ounces in a fluid, and 298*1
ounces out of it ; required the specific gravity of the fluid, aflowing
the specific gravity of the cast-iron to be 7645 ?
298-1—259*1=39 loss of weight in the iron; then 298*1 : 39 : :
7645 : 1000, the specific gravity of the fluid ; shewing the fluid to be
water.*
6. A piece of lignum vitje weighed 42| ounces in a fluid, and 166|
ounces out of it ; what is the specific gravity of the fluid, that of the
lignum vitae being 1333 ?
Ans. 991 is the specific gravity of the fluid, which shews it to be
liquid turpentine or Burgundy wine.
* In this manner may the species of a fluid or solid be ascertained, by meana
of its specific gravity, and the annexed table. This table has been taken from
Gregoiy's work for practical men.
SPECIFIC GRAVITY.
117
TABLE OF SPECIFIC GRAVITIES.
Platina
Do. hammered
Cast zinc
Cast iron
Cast tin
Bar iron
Hard steel
Cast brass
Cast copper
Pure cast silve
Cast lead
Mercury
Pure cast gold
Amber
Brick .
Sulphur
Cast nichel
Cast cobalt
Paving stones
Common stone
Flint and spar
Green glass
White ghm
Pebble
Slate .
Pearl .
Alabaster
Marble
Chalk .
Limestone
Wax .
Tallow
Camphor
Bees wax
Honey
Bone of an ox
Ivory .
Air at the earth's surface
Liquid turpentine *^ 'Vw^^
Olive oil
Burgundy wine ^- r-^^i^
Distilled water .
Sea water
Milk .
Spec. grav.
19500
20S3Q
7190
7-207
7-291
7-788
7-816
8595
.^.
10474
11^352
13668
19258 ^
1078
2000
2033
7-807
imi
2416
2520
2594
2642
2892
2664
2672
2684
2730
2742
2784
3179
897
945
989
965
1456
1659
1822
14
>- 991-
915
991-
1-000
1028
1030
wt. cub. in.
oz.
11-285
11-777
4-161
4-165
4-219
4-507
4-523
4-858
5-085
6-061
6-569
7-872
11-145
wt. cub. ftw
125-00
127 06
4513
4520
151-00
157-50
162-12
166-50
lbs
167-00
171-38
174-00
193-68
IIS
Beer ..
Cork .
Poplar .
Larch .
Elm and West India fir
Mahogany-
Cedar .
Pitch pine
Pear tree
Walnut
Elder tree
Beech .
Cherry tree
Maple and Riga fir
Ash and Dantzic oak
Apple tree
Alder .
Oak, Canadian
Box, French
Log'W'ood
Oak, English
Oak, 60 years old
Ebony .
Lignum vitae
PROBLEM IL
2'he specific gravity of a hody^ and its weight, heing given, to
Jind it's solidity.
HuLE. Say, as the tabular specific gravity of the body is to its
weight, in ounces avoirdupois, so is 1 cubic foot to the content.
1. What is the solidity of a block of marble that weighs 10 tons,
its specific gravity being 2742?
First, 10 tons =200 hundreds =22400 pounds =358400 ounces;
then
2742 : 358400 : : 1
1
SPECIFIC GRAVITY.
Spec. grav.
wt. ctib. ft
lbs.
1-034
240 ,
15-00
383 .
2-394
544 .
34-00
556 .
34-75
660 .
35-00
696 .
37-25
660 .
41-25
661 .
41-31
671 .
41-94
695 .
43-44
696 .
43-50
715 .
44-68
750 .
46-87
760 .
47-50
793 .
49-56
800 .
60-00
872 .
64 50
912 .
67-00
913 .
67-06
970 .
61-87
H70 .
73-12
1-331 .
83-18
1-333 .
83-31
2742)358400(130xV7T-
2742
8420
8226
^2742^'^^^
2. How many cubic inches in an irregular block of marble which
weighs 112 pounds, allowing its specific gravity to be 2520?
Ans. 12281^^ cubic inches.
3. How many cubic inches of gunpowder are there in 1 pound
weight, its specific gravity being 1745? Ans. 15 J, nearly.
SPECIFIC GRAVITY. 119
4. How many cul)ic feet are there in a ton weight of dry oak, \i%
Bpecific gravity being 925 ? Ans. 38f ||-.
PROBLEM III.
The linear dimensions^ or magnitude of a hody^ heing given^ and
also its specific gravity^ tojind its weight.
Rule. One cubic foot is to the solidity of the body, as the tabular
gpocific gravity of the body is to the weight in avoirdupois ounces.
1. What is the weight of a piece of dry oak, in the form of a
parallelopipedon, whose length is 56 inches, breadth 18 inches, and
depth 12 ?
56 X 18 X 12= 12096 cubic inches, the solid content.
Then 1728 : 12096 :: 932 ; 6524 ounces = 407| pounds, the
weight required.
2. What is the weight of a block of dry oak, which measures
10 feet long, 3 feet broad, and 2J feet deep , its specific gravity
being 925 ? Ans. 4335^^ pounds.
3. What is the weight of a block of marble, whose length is
53 feet, and its breadth and thickness, each 12 feet ?
Ans. 694xVV tons.
PROBLEM ly.
Tojind the quantities of two ingredients in a given compound.
Rule. Take the difference of every pair of the three specific
gravities, viz., of the compound and each ingredient; and multiply
the difference of every two by the third.
Then as the greater product is to the whole weight of the com-
pound, so is each of the other products to the weights of the two
ingredients.*
1. A composition of 112 pounds being made of tin and copper,
whose specific gravity is found to be 8784 ; what is the quantity of
each ingredient, the specific gravity of tin being 7320, and of copper
9000?
9000 9000 8784
7320 8784 7320
1680 216 1464 diff.
8784 7320 9000
14757120 1581120 13176000 Then
ij.7K7i9n • 110 •• J 13176000 : 100 pounds copper.
i4/i>/izu . iiz .. -J 1581120 : 12 pounds tin.
2. Hiero, king of Sicily, furnished a goldsmith with a quantity of
gold, to make a crown. When it came home, he suspected that
the goldsmith had used a greater quantity of silver than was neces-
sary in the composition ; and applied to the famous mathematician,
Archimedes, a Syracusian, to discover the fraud, without defacing
the crown.
* For the reason of this rule, see Alligation Total in the Second Book of
/Lrithmetic, published by the Commissioners.
120 SPECIFIC GRAVITY.
To ascertain the quantity of gold and silver in the crown, ha
procured a mass of gold and another of silver, each exactly of the
same weight with the crown ; justly considering that if the crown
were of pure gold, it would be of equal bulk, and therefore displace
an equal quantity of water with the golden mass ; and if of silver, it
would be of equal bulk and displace an equal quantity of water with
the silver mass ; but if of a mixture of the two, it would displace an
intermediate quantity of water.
Now suppose that each of the three weighed 100 ounces ; and that
on immersing them severally in water, there were displaced 5 ounces
of water by the golden mass, 9 ounces by the silver mass, and
6 ounces by the crown ; what quantity of gold and silver did the
crown contain ?
. j 75 ounces of gold.
Ans. j 25 ounces of silver.
JVo/e.— Questions relating to specific gravities may be wrought by the rulets
of Alligation in Arithmetic, as well as by any Algebraic process that might be
employed.
PROBLEM V.
To find Jiow many inches a floating body will sink in a fluid.
Rule. Find, by Problem III. the weight of the floating body from
its solidity and specific gravity, and that will be the weight of the
fluid which it will displace.
Then say, as the specific gravity of the fluid is to 1728 cubio
inches, so is the weight of the body, in ounces, to the cubic inches
immersed. The depth will be found from the given dimensions.
1. Suppose a piece of dry oak, in the form of a parallelopipedon,
whose length is 56 inches, breadth 18, and depth 12, is to be floated
upon common smooth water, on its broadest side ; how many inches
will it sink, its specific gravity being 932?
By Problem III., the weight of the piece of oak is 6524 ounces,
which by the preliminary part of this section, is the weight of water
displaced.
Then 1000 : 1728 : : 6524 : 11273-472 cubicinches of oak immersed.
Therefore, 11273-472-t-(56x 18) = 11-184 inches, the depth it will
sink.
To find how far it will sink, allowing it to float on its narrower
side, 11273'472-f-(56x 12) = 16-776 inches.
2. How many inches will a cubic foot of dry oak sink in common
water, allowing the specific gravity of the oak to be 970?
Ans. 11-64.
PROBLEM VI.
To find what weight may he attached to a fioating body^ so that it
may he just covered with a given fluid.
Rule. Multiply the cubic feet in the body by the difference between
its specific gravity and that of the fluid, and the product will be the
weight in ounces avoirdupois, just sufficient to immerse it in the
fluid.
. 1. What weight must be attached to a piece of dry oak, 56 inches
SPECIFIC GRAVITY. 121
lon^, 18 inches broad, and 12 inches deep, to keep it from rising
above the surface of a fresh-water lake ; the specific gravity of the
water being 1000, and that of the oak 932 ?
Here 56 x 18 x 12= 12096 cubic inches.
Then 120964-1728 = 7 feet.
Then (1000—932) x 7 = 68 x 7 = 476 ounces = 29 pounds 12 ounces.
2. What weight fixed to a piece of dry oak, 9 inches long, 6 inches
broad, and 3 inches deep, will keep it from rising above the surface
of common water, the specific gravity of water being 1000, and that
of the oak 970 ? Ans. 2^^ ounces.
3. A sailor had half an anker of brandy, the specific gravity of the
liquor was 927, the cask was oak, and contained 216 cubic inches,
and its specific gravity was 932 ; to secure his prize from the custom-
house officers, he fixed just as much lead to the cask as would keep it
under water, and then threw it into the sea; what weight of lead
was necessary for his purpose ?
Ans. The cask of brandy contained 1371 cubic inches, the weight
of sea- water of an equal bulk was 81 7*20486 ounces, the cask weighed
116-5 ounces, the brandy 619*609375, both together weighed
736*19375 ounces. The difference between the specific gravity of
lead and sea-water is to this remainder, as the specific gravity of
lead to its weight in ounces, which will be found to be 89 '09495
ounces, or 6 pounds 9 ounces.
PROBLEM VII.
I'd find the solidity of a oody^ lighter than a fluids which will he
sufficient to prevent a body much heavier than the fiuid^ from
sinking.
Rule. Find the solidity of the body to be floated, from its weight
and specific ^avity, by Problem II. Find also the weight of an equaj
bulk of the fluid by Problem J II. Then say, as the difference between
the specific gravity of the fluid, and that of the body lighter than the
fluid, is to the difference between the weight of the body to be floated
and the weight of an equal bulk of the fluid, so is 1728 to the soli-
dity of the lighter body in cubic inches.
1. How many solid feet of yellow fir, whose specific gravity is 657,
will be sufficient to keep a brass cannon, weighing 56 cwt., afloat at
sea, the specific gravity of brass being 8396, and of sea- water 1030?
First, 56 cwt. = 100352 ounces, weight of the body to be floated.
Then, 8396 : 100352 : : 1728 : 20653*675 cubic inches in the cannon.
And, 1728 : 20653-675 : : 1030 : 12310*9289, the weight of sea-water
equal in bulk to that of the cannon. Hence, 1030—657 : 100352 —
12310-9289 ;; 1728 : 407868*5545 cubic inches = 236*036 feet, the
answer.
2. The specific gravity of lead is 11325, of cork 240, and of sea-
water 1030; now it is required to know how many cubic inclies of
cork will be sufficient to keep 49| pounds of lead afloat at sea?
Ans, 1570-84 cubic inches.
122 t<;nnage of ships.
TO FIND THE TONNAGE OF SHIPS.
1st.— VESSELS AGROUND.
By the Parliamentary Rule,
PROBLEM VIII.
For a ship or vessel, the length is to be measured on a straight
line along the rabbet of the keel, from a perpendicular, let fall from
the back of the main post, at the height of the wing-transom, to a
perpendicular at the height of the upper deck (but the middle deck
of three-decked ships), from the fore-part of the stern ; then from the
length between these perpendiculars subtract three-fifths of the
extreme breadth for the rake of the stern, and 2J inches for every
foot of the height of the wing-transom above the lower part of the
rabbet of the keel, for the rake abaft ; and the remainder will be the
length of the keel for tonnage.
The main breadth is to be taken from the outside of the outside
plank, in the broadest part of the ship, either above or below the
wales, deducting therefrom all that it exceeds the thickness of the
plank of the bottom, which shall be accounted the main breadljh ; so
that the moulding breadth, or the breadth of the frame, will then be
less than the main breadth, so found, by double the thickness of the
plank of the bottom.
Then multiply the length of the keel for tonnage, by the main
breadth, so taken, and the product by half the breadth, then divide
the whole by 94, and the quotient will give the tonnage.
In cutters and brigs, where the rake of the stern-post exceeds 2J
inches to every foot in height, the actual rake is generally subtracted
instead of the 2J inches to every foot, as before mentioned.
1. Let us suppose the length from the fore-part of the stern, at the
height of the upper deck, to the after-part of the stern-post, at the
height of the wing-transom, to be 155 feet 8 inches, the breadth
from out to outside 40 feet 6 inches, and the height of the wing-
transom 21 feet 10 inches, what is the tonnage?
ft.
40-6 breadth. ^
deduct 3
40*3
3
5)120-9
24-lf = 24-15
21*10 height of wing-transom.
2|- multiply.
12)54tV
4-55-t-24-]5 = 28-70
155-66-28-70 = 126-96 = length.
126-96 X 40-25 + 20-125 ,^^, ,,
^7 = 1094, the answer.
94
FLOATING BODIES. 123
2. Suppose the length of the keel to be 50*5 feet, breadth of the
midship -beam 20 feet ; required the tonnage? Ans. 107*4.
3. If the length of the keel be 100 feet, and the breadth of the
beam 30 feet ; what is the tonnage? Ans, 478.
2d.— VESSELS AFLOAT.
Drop a plumb-line over the stern of the ship, and measure the
distance between such line and the after-part of the stern-post, at
the load water-mark : in a paraM direction with the water, to a
perpendicular point immediately over the load water- mark, at the
fore-part of the main-stern, subtracting from such measurement the
above distance, the remainder will be the ship's extreme length ;
from which is to be deducted three inches for every foot of the load
draught of water for the rake abaft, and also three-fifths of the ship's
breadth for the rake forward, the remainder shall be esteemed tlie
just length of the keel to find the tonnage ; and the breadth shall be
taken from outside to outside of the plank, in the broadest part of
the ship, either above or below the main-wales, exclusive of all
manner of sheathing or doubling that may be wrought upon the
sides of the ship ; then multiply the length of the keel, taken as
before directed, by the breadth, as before taught, and that product
by half the said breadth, and dividing the product by 94, the quotient
IS the tonnage.
3d.— STEAM VESSELS.
The length shall be taken on a straight line, along the rabbet ol
the keel, from the back of the main- stern-post to a perpendicular
line from the fore-part of the main-stem under the bow-sprit ; from
which deducting the length of the engine-room, and subtracting
Ihree-fifths of the breadth, the remainder shall be esteemed the just
length of the keel to find the tonnage ; and the breadth shall be taken
from the outside of the outside plank in the broadest place of the ship
or vessel, be it either above or below the main-wales, exclusively of
all manner of doubling planks that may be wrought upon the sides
of the ship or vessel ; then multiply the length and breadth so fjund
together, and that product by half the same breadth, and dividing
by 94, the quotient will be the tonnage, according to which all such
vessels ^hall be measured.
Note. — Under certain penalties nothing but the fuel can be stowed in the
engine-room.
Some divide the last product by 100, to find the tonnage of king's
ehips, and by 95, to find that of merchants' ships.
FLOATING BODIES.
1. The buoyancy of casks, or the load which they will carry with-
out sinking, may be estimated by reckoning 10 pounds avoirdupois to
the ale gallon, or 8^ potmds to the wine gallon
124 PLOATINQ BODIES.
2. The buoyancy of pontoons may be estimated at about half a
lumdred weight, or 56 pounds for each cubic foot. Iherefore a
pontoon which contained 96 cubic feet, would sustain 48 hundred-
weight before it would sink.
N.B. — This is an approximation, In which the difference between -^1 and J,
viz , 1 of the whole weight, is allowed for that of the pontoou itself.
3. The principles of buoyancy are very ingeniously applied in the
self-acting flood-gate, which, m the case of common sluices to a
mill-dam, prevents inundation when a sudden flood occurs. By
means of the same principle, it is that a hollow ball attached to a
metallic lever of about a foot long, is made to rise with the liquid in
a water-cask, and thus to close the cock and stop the supply from the
pipe, just before the time when the water would otherwise run over
the top of the vessel.
The property of buoyancy has also been successfully employed in
raising ships which had sunk under water, and in pulling up old
piles in a river wlien the tide ebbs and floiws. A large barge is
brought over a pile as the water begins to rise; a strong chain
which has been previously fixed to the pile by a ring, &c., is made to
gird the barge, and is then firmly fastened ; then, as the tide rises,
the barge rises also, and by means of its buoyant force draws up the
pile with it.
In a case which actually occurred, a barge of 50 feet long, 12 feet
wide, 6 deep, and drawing 2 feet water, was employed. Then
50 X 12 X (6-2) X 4^50x12x16^ ^^2 x 7f = 1344 + 27f = 1371f
cwt. = 66 J tons, nearly, which is the measure of the force with which
the barge acted in pulling up the pfle>
WEIGHT AND DIMENSIONS OF BALLS AND SHELLS. 126
WEIGHT AND DIMENSIONS OF BALLS
AND SHELLS.
SECTION IX.
The foregoing problems furnish rules for finding the weight and
dimensions of balls and shells. But they may be found much easier
by means of the experimental weight of a ball of a given size, and
from the well-known geometrical property, that similar solids are aa
the cubes of their diameters.
PROBLEM I. H j'
To find the weight of an iron hall from its diameter. ^ i*
Rule. Nine times the cube of the diameter being divided by 64,
will express the required weight in pounds.*
1. The diameter of an iron shot is 5 inches ; required its weight?
6 X 5 X 5 = 125 = cube of the ball's diameter.
Then 125x 9-^64: = 17f} pounds, the answer.
2. The diameter of an iron shot being 3 inches; required ita
weight? Ans. 3-8 pounds.
3. The diameter of an iron shot is 5*54 inches; what is its weight?
Ans, 24 pounds.
PROBLEM II. ^d
' To find the weight of a leaden ball^ by having its diameter given, ^
Rule. Multiply the cube of its diameter by 2, and divide the pro-
duct by 9, and the quotient will give the weight in pounds.f
1. AVhat is the weight of a leaden ball of 5 inches diameter?
5x5 x 5= 125 cube of ball's diameter.
Then, 125 x 24-9 = 250-^9 = 27J pounds, answer.
2. What is the weight of a leaden ball, whose diameter is 6*6
inches? Ans. 63*888 pounds.
3. What is the weight of a leaden ball, whose diameter is 3*5
inches? Ans. 9*53 pounds.
4. What is the weight of a leaden ball, whose diameter is 6 inches?
Ans. 48 pounds.
* See Appeudix, DemonstJ^tioii 113. f Ibid. 111.
'126 WEIGHT AND DIMENSIONS OF BALLS AND SHELLS.
PROBLEM III.
Having the weight of an iron ball^ to determine its diameter.
Rule. Multiply the weight by 7J, then take the cube root of the
product for the diameter.*
1. What is the diameter of an iron ball, whose weight is 42 pounds ?
42x7|-=298|.
Then, ^298 = 6*685 inches, the answer.
2. Required the diameter of an iron ball, whose weight is 24
pounds? ^725. 5 '54 inches.
3. What is the diameter of an iron ball, whose weight is 3-8
pounds ? Ans. 3 inches.
PROBLEM IV.
Having the weight of a leaden ball, to determine its diameter.
Rule. Multiply the weight by 9, and divide the product by 2 ;
and the cube root of the quotient will express the diameter.f
1. What is the diameter of a leaden ball, whose weight is 64 pounds?
64x9 = 576.
Then, 576-^2 = 288.
Hence, >^288 = 6*6 inches, the answer.
2. Required the diameter of a leaden ball, whose weight is 27{
pounds? Ans. 6 inches.
3. What is the diameter of a leaden ball, whose weight is 63-888
pounds? Ans. 6*6 inches.
PROBLEM V. ^
Having given the external and internal diameter of an iron shell,
to find its weight.
Rule. Find the difference between the cubes of the two diameters,
iind multiply it by 9 ; divide the product by 64, and the quotient will
express the weight in pounds. J
1. What is the weight of an 18-inch iron bomb-shell, whose mean
thickness is Ij inches?
18— 2j= 15| = mternal diameter.
Then, 18^ = 5832 the cube of external diameter.
(15-5)^ = 3723 -875 the cube of internal diameter.
And, 6832-3723-875 = 2108-125 = difference of cubes.
Hence, 2108-125 x 9^-64 = 296*45 pounds, the answer.
2. What is the weight of a 9 -inch iron bomb-shell whose mean
thickness is 1^ inch? Ans. 72-14 pounds.
3 AVhat is the weight of an iron bomb- shell, whose external
diameter is 9-8 inches, and internal diameter 7 inches?
Ans. Si I pounds.
* This niTe is obvious from Problem I., being the converse thereof,
t Tliis rule is manifest from Problem III,, haing its converse,
i bee Appendix, Demonstration 1 S
WEIGHT AND DIMENSIONS OF BALLS AND SHELLS. 127
PROBLEM VI.
To find 7iow much powder will fill a shell of given dimensions.
Rule. Divide the cube of the internal diameter in inches, by 57*3,
and the quotient will express the answer.*
1. What quantity of powder will fill a shell, whose internal
diameter is 10 inches?
First, 10 X 10xlO=1000=cube of diameter.
57'3)1000(17-45 pounds, answer.
673
4270
4011
2590
2292
115, &c.
f^ Kote.— In some recent works, the cube of the diameter is divided by 59*32, I
y for the weight of powder in pounds. L
J 2. How many pounds of gunpowder are required to fill a hollow (
I shell, whose internal diameter is 13 inches? j
>^ Ans. 37 pounds, according to the note. ^
jf^ ^. Required the number of pounds of powder that will fill a shell,
whose internal diameter is 7 inches ?
Ans. 6 pounds, by the rule in the text.
PROBLEM VII.
To find how much powder will fill a rectangular box oj^ given
dimensions.
Rule. Multiply the length, breadth, and depth together in inches,
and the last result by '0322, and the last product will give the
weight in pounds.f
1. How many pounds of powder will fill a rectangular box, whose
length is 16 inches, breadth 12 inches, and depth 6 inches?
16x12x6=1 152 = content of the box.
Then, 1152 x •0322 = 37*0944, the answer.
2. How many pounds of powder will fill a rectangular box, whose
length is 10 inches, breadth 5 inches, and depth 2 inches?
Ans. 3-22 pounds.
3. How many pounds of powder will fill a rectangular box, whose
length is 5 inches, breadth 2 inches, and depth 10 inches?
Ans. 3-22 pounds.
PROBLEM YIII.
Having the length and diameter of a cylinder^ to determine how
many pounds of gunpowder will fill it.
Rule. Multiply the square of the diameter by the length, and
divide the product by 40, for the weight in pounds. J
* bee Appendix, Demonstration 116. f Ibid. 117. X Ibid. 11
1
128 PILING OF BALLS AND SHELLS.
1. The diameter of a hollow cylinder is 10 inches, and the length
14 inches ; how many pounds will it hold ?
10 X 10 = 100 = square of diameter.
Then, 100x14=1400.
Hence, 1400-^40 = 35 pounds, the answer.
2. The diameter of a hollow cylinder is 5 inches, and its length 40
inches ; how much powder will it hold ? Ans. 25 pounds.
3. The diameter of a hollow cylinder is 5 inches, and the length
12 inches ; how many pounds will it hold? Ans. 7*5 pounds.
PROBLEM IX.
To find wJiat portion of a cylinder will be occupied by a given
quantity of powder , the diameter of the cylinder being given.
Rule. Multiply the given weight of powder by 40, and divide
the product by the square of the diameter of the cylinder, and the
quotient will be the pounds required.*
1. The diameter of a hollow cylinder is 10 inches; how much of it
will hold 50 pounds of powder ?
50x40 = 2000.
Then, 2000-7-100=20 inches, the answer.
2. How much of a cylinder of 14 inches diameter will hold 10
pounds of powder? Ans. 2*05 inches.
3. How much of a cylinder, 12 inches in diameter, will hold 144
pounds of powder ? Ans. 40 inches.
PILING OF BALLS AND SHELLS.
Iron-shot and shells are usually piled in horizontal courses, eithel
in a pyramidical or in a wedge-like form ; the base being either an
equilateral triangle, a square, or a rectangle.
Those piles whose bases are triangles or squares, terminate in ono
ball at the top ; but piles whose bases are rectangles terminate in a
single row of balls.
In triangular and square piles, the number of horizontal rows or
courses, is always equal to the number of balls in one side of the
bottom row.
And in rectangular piles the number of rows is equal to the num-
ber of balls in the breadth of the bottom.
Also the number in the top row or edge, is one more than the dif-
ference between the length and breadth of the bottom row.
PROBLEM I. ^ • ^. .^^
To find the number of balls in a mtstoB^ttar pile.
Rule. Multiply the number in one side of the bottom row, by that
number increased by 1, and the result by the same number increased
by 2 ; then the one-sixth of the last product will give the number of
balls required.f
♦ See ADDendijL Demonstration 119. t Ibid. 120.
PILING OF BALLS AND SHELLS. 129
1. Required the number of shot in a complete triangular pile, one
f J whose sides contains 22 balls?
22 = the number in one side of base.
23 = the numbers 1.
66
44
"We
24= the number +2.
2024
1012
6)12144
2024 = the number of shot in the pile.
2. Required the number of sliot in a complete triangular pile, ono
Bide of whose base contains 15 balls ? Ans. 680 balls.
3. Required the number of balls in a triangular pite, each side of
the base containing 30 balls ? Ans, 4960.
PROBLEM II.
Tojind the number of halls in a square pile.
Rule. Multiply continually together the number in one side of the
bottom course, that number increased by 1, and double the same
number increased by 1 ; then one- sixth of the last product will be
the answer.* M^(^^^^ ♦^/^gt^^vM
1. How many balls are in a square pile of 30 rowsr" " *^
30 = number in one side. ^
31 = number in one side + 1.
930
61 = twice the number in one side + 1.
6)56730
9455 answer.
2. Required the number of shot in a complete square pile, one side
of whose base contains 19? Ans. 2470.
3. How many shot in a finished square pile, when a side of the
base contains 21 shot? Ans, 3311.
PROBLEM III.
Tojind the numter of shot in a finished rectangular pile.
Rule. Adi 1^ to three times the number of shot contained in the
length of the base, subtract the number of shot in the breadth of the
base, multiply the remainder by the said number increased by 1, and
this result again by the number in the breadth ; then one-sixth of
the last result will give the number of shot in the rectangular pile.f
• See Appendix, Demonstration 121. f Ibid. 122.
tC^c^
130 PILING OF BALLS AND SHELLS.
1. Required the number of shot in a finished rectangular pile, the
length of the base containing 59, and its breadth containing 20 balls?
69 = the number of shot in the length.
3
177; then 177 + 1 = 178, and 178-20 = 158.
158x21 = 3818, and 3318x20 = 66360. Hence 66860-5-6 =
11060, the answer.
2. How many balls are in a rectangular complete pile, the length
of the bottom course being 46, and its breadth 16 V Ans. 4960.
PROBLEM IV.
To determine the number of balls contained m a pile which is not
jinished^ ike highest course being complete^ and the number oj
balls in each side thereof being given.
Rule. Find the number of shot whicli would be contained in the
pile if it were complete. Find also the number in that complete pile,
each side of whose base contains one shot fewer than the correspond-
ing side of the uppermost course of the unfinished pile, and the
difference between these results will evidently give the number of
balls in the unfinished pile.
1. How many shot are there in an unfinished triangular pile, a
Bide of whose base contains 23, and a side of the uppermost course 7
shot?
23 = number of balls in the base.
24 = number of balls in the base + 1.
"652
25
6)13800
2300= number of the pile when complete.
6
7
42
8
6)336
66 number of balls in the imaginary pile.
Therefore, 2300 — 56 = 2244, the answer.
2. How many balls in an incomplete square pile, the side of the
base being 24, and of the top 8? Ans. 4760.
3. How many balls are there in the incomplete rectangular pile of
12 courses, the length and breadth of the base being 40 and 20?
Ans. 6146.
DETERMINING DISTANCES BY SOUND. 131
DETERMINING DISTANCES BY SOUND.
The velocity of sound, or the space through which it is propa-
gated in a given time, has been very differently estimated by phi-
losophers who have written on this subject. We shall, however, take
it to be 1142 feet in a second.
From repeated experiments it has been ascertained that sound
moves uniformly, or to speak more philosophically, that the pulses of
air which excite it move uniformly. The velocity of sound is the
same with that of the aerial waves, and does not vary much whether
it go with the wind or against it. By the wind, no doubt, a certain
quantity of air is carried from one place to another, and the sound
is somewhat accelerated while its waves move through that part of
the air, if their direction be the same as that of the wind. But as
the velocity of sound is vastly swifter than the wind, the accelera-
tion it will thereby receive is but inconsiderable, being at most but
^ of the whole velocity.
The chief effect perceptible from the wind is, that it increases and
iliminishes the space tnrough which sound is propagated. The
utmost distance at which sound has been heard is about 200
miles. It is said that the unassisted human voice has been heard
from Old to New Gibraltar, a distance of about 12 miles. Dr
Derham, placing cannon at different distances, and causing them to
be fired off*, observed the intervals between the flash and report, by
means of which he found the velocity of sound to be as above stated.
1. Having observed the flash of a cannon, I noticed by my watch
that 5 seconds elapsed previous to my hearing the report ; determine
iny distance from the gun.
1U2
5
5710 feet, the answer.
2. Bemg at sea I saw the flash of a cannon, and counted 8 seconds
between the flash and the report ; required the distance?
dns. 1^ mile.
132 GAUGINa
GAUGING.
SECTIOIsr X.
Gauging is the art of measuring the capacities of vessels, such as
casks, vats, &c.
The business of gauging is generally performed by means of two
instruments, namely, the gaugmg or sliding rule, and the gauging or
diagonal rod.
1. OF THE GAUGING RULE.— LEADBETTER'S.
By this instrument is computed the contents of casks, &c., after the
dimensions have been taken. It is a square rule, having various
logarithmic lines on its four faces, and jjhree sliding pieces capable of
being moved through grooves in whicTthey fit, in three of these
faces.
On the first face are delineated three lines, namely, two marked
A.B, on which multiplication and division are performed; and the
third marked M D, signifies malt depth, and serves to gauge malt.
The middle one B is on the slider, and is a kind of double line, being
marked at both edges of the slider, for applying it to both the lines
A and M D. These three lines are all of the same radius, or distance
from 1 to 10, each containing twice the length of the radius. A and
B are numbered and placed exactly alike, each commencing at 1,
which may be either 1, or 10, 100, &c., or -1, or -01, or -001, &c.
"Whatever the 1 at the beginning is estimated at, the middle division
10 will be 10 times as much, and the last division 100 times as
much. But 1 on the line MD is opposite 2220, or more exactly
2218*2 on the other lines, which number 2218*2 denotes the cubic
inch in an imperial malt bushel ; and its divisions numbered retrograde
to those of A and B. On these two lines are also several other marks
and letters ; thus on the line A are M B, or sometimes only B, for
malt bushel, at the number 2218*2, and A for ale, at 282, the cubic
inches in an old ale gallon ; and on the line B is W, for wine, at 231,
the cubic inches in an old wine gallon.
These marks are now usually omitted upon the rule, since the late
new act of parliament for uniformity of weights and measures, and
G for gallon is put at 277*274 the inches in an imperial gallon,*
whether of ale, wine, or spirits.
* Until 5 George IV., in which a uniform system of weights and measures
was established under the denomination of imperial weights and measures,
there were, amongst other sources of inconvenience, different measures, though
GAUGING. 133
On many sliding rules are also found s 2, for square inscribed, at
♦707, the side of a square inscribed in a circle whose diameter is 1 ;
ge^ for square equal, ^t '886, the side of a square which is equal to
the same circle ; and c, for circumference, at 3 '14 16, the circumference
of the same circle.
On the second face, or that opposite the first, are a slider and four
lines marked D, C, D, E, at one end, and root square, root cube at
the other end ; the lines C and D containing, respectively, the squares
and cubes of the opposite numbers on the lines D, D ; the radius of
D being double to that of A, B, C, and triple to that of E ; therefore
whatever the first 1 on D denotes, the first on C is its square, and
the first on E its cube ; that is, if D begin with 1, C and E will begin
with 1 ; but if D begin with 10, C will begin with 100, and E with
1000 ; and so on.
On the line 0 are marked o c at -0796, for the area of the circle
whose circumference is 1 ; and o d, at '7854, for the area of the circle
whose diameter is 1.
On the line D are marked G S, for gallon square, at 16*65, and
G R for gallon round at 18-789 ; also MS for malt square at 47*097,
and M R for malt round at 53*144.
These are the respective gauge-points for gallons and bushels.
The first 16*65 is the side of a square, which at an inch depth holds
a gallon; the second 18*789, the diameter of a circle, which at an
mch depth holds a gallon ; the third 47*097 the side of a square,
which at an inch depth holds a bushel; the fourth 53*144, the
diameter of a circle, which at an inch depth holds a bushel.
On the third face are three lines : one on a slider marked N ; and
two on the stock, marked S S and S L, for segment standing and
segment lying, which serve ullaging, standing and lying casks.
And on the fourth side, or opposite face, are a scale of inches, and
three other scales, marked spheroid, ©r 1st variety, 2d variety, 3d
variety ; the scale for the 4th or conic variety, being on the inside
of the slider in the third face. The use of these lines is, to find the
mean diameter of casks. On the inside of the two first sliders,
besides all those already described, are two other lines, being con-
tinued from one slider to the other.
The one of these is a scale of inches, from 2J to 36, and the other
is a scale of ale gallons, between the corresponding number 435 and
3*61, which form a table, to shew, in ale gallons, the contents of all
cylinders whose diameters are from 12j to 36 inches, their common
altitude being 1 inch.
of the same name, for ale and wine. A gallon of ale contained 282 cubic
inches, and a gallon of wine 231; a bushel of malt contained 2150 •42 cubic
inches.
To reduce old measure into new, say, as the number of cubic inches in the
imperial standard is to the number of cubic inches in the old standard, so is
the number of gallons or bushels, &c., old measure, to the number of gallons,
&c., imperial measure.
When great accuracy is not required, old wine gallons may be reduced to
imperial gallons by dividing by 1'2; and old ale gallons may be reduced to
imperial gallons by multiplying by 60, and dividing the product by 69 ; and
old, or Winchester bushels, maybe reduced to imperial bushels by multiplying
by 31, and dividing the product by 32.
x
134 GAUGINa.
verie's sliding rule.
Tills rule is in the form of a parallelopipedon, and is generany
made of box.
1. The line marked A, on the face of this rule, is called Gunter's
line, and is numbered 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. At 2218-192 is
fixed a brass pin, marked IM, B, signifying the cubic inches in an
imperial bushel; at 277*274 is fixed another brass pin, marked IM,
G, denoting the number of cubic inches in an imperial gallon.
2. The hue marked B is on the slide, and is divided exactly like
that marked A. There is another slide B, on the opposite side,
which is used along with this. The slide on the first face is called
the second radius^ and that on the opposite face, the first radius.
The two brass ends, when placed together, make a double radius,
numbered from the left hand towards the right. At 277*274, on the
second radius, is a fixed brass pin, marked IM, G, denoting the cubic
inches in an imperial gallon; at 314 is fixed another brass pin,
marked C, signifying the circumference of a circle whose diameter is
i; "These lines are used and read exactly as the lines A and B on
tlie Carpenter's Rule, which have been already described.
3. Tlie back of one slide or radius, marked B, has the dimensions
for imperial gallons, and bushels, green starch, dry starch, hard soap
hot, hard soap cold, green soft soap, white soft soap, flint glass, &c.,
&c., as in Table, page 136.
The back of the other slide or radius, marked B, contains the
gauge- points corresponding to these divisors, where S denotes squares,
and C circles.
4. The line M D on the rule, denoting malt depth, is a line of
numbers commencing at 2218*192, and is numbered from the left to
the right hand 2, 10, 9, 8, 7, 6, 6," 4, 3. This rule is used in malt-
gauging.
5. The two slides B, just described, are always used together,
either with the line A, M D, or the line D, which is on the opposite
face of the rule to that already described. The line D is numbered
from the left hand towards the right, 1, 2, 3, 31, to 32, which is at
the right-hand end ; it is then continued from the left-hand end of
the other edge of the rule, 32, 4, 5, 6, 7, 8, 9, 10. A.t 16*651 is a
brass pin G S, signifying a gauge square^ being the square gauge*'
point for imperial gallons. ' At 18*789 is fixed a brass pin, marked
G R, denoting gauge rounds or circular gauge-point for imperial
gallons. At 47*097, M S signifies malt square^ the square gauge-
point for malt bushels. At 53*144, ME, denotes malt rounds the
round or circular gauge-point for malt bushels. The line D on this
rule is of the same nature as the line marked D on the Carpenter's
Rule, which has been already described. The line A and the two
glides B, are used together, for performing multiplication, division,
simple proportion, &c. ; and the line D, and the same slides B, are
used together for extracting the square and cube roots.
6. The other two slides belonging to this rule are marked C, and
are divided in the same manner, and used together, like the slides B.
The back of the first slide or radius, marked C, is divided, next
tlic edge, into inches, and numbered from the left hand towards the
QAUGINO. 185
right 1, 2, 3, 4, 5, &c., and these inches are again subdivided into
ten equal parts. The second line is marked spheroid, and is numbered
from the left hand towards the right 1, 2, 3, 4, 5, 6, 7, 8. The third
line is marked second variety, and is numbered 1, 2, 3, 4, 5, 6. These
Mnes are used, vv^ith the scale of inches, for finding a mean diameter.
The back of the second slide or radius, marked C, has several
factors for reducing goods of one denomination to others of equiva-
lent values. Thus I X. to VI. 6. | signifies that to reduce strong
beer at 85. per barrel, to small beer at Is. M. you are to multiply by
6. I VI. to X. 17. I signifies that to reduce small beer at Is. 4^. per
barrel to strong beer at 8s. per barrel, you are to multiply by -17.
I C 4 ^ to X. 27. I signifies that 27 is the multiplier for reducing
cider at 4s. per barrel to another at 8s., &c.
7. The two slides C, just described, are always used together,
with the lines on the rule marked Seg. St., or SS, segments stand-
ing ; and Seg. L ?/ or S L, segments lying ; for uUaging casks. The
former of these lines is numbered 1, 2, 3, 4, 5, 6, 7, 8, which stands
at the right-hand end ; it then goes on from the left hand on the
other edge 8, 9, 10, &c., to 100. The latter is numbered in the same
manner 1, 2, 3, 4, which stands at the right-hand end ; it then
goes on from the left hand on the other edge, 4, 6, 6, 7, &c., to
100.
PROBLEM I.
To find the several multipliers^ divisors^ and gauge-pointSy
belonging to the several measures now used,
MULTIPWERS FOR SQUARES.
As 277*274 solid inches are contained in one imperial gallon, and
2218*192 solid inches in an imperial bushel ; then it is obvious that
if 1 be divided by 277*274, and 2218*192, respectively, the quotients
will be the multipliers for imperial gallons and bushels respectively.
Hence the method of finding the following multipliers is obvious : —
277*274)1 -OOOOOC-OOSGOeS multiplier for imperial gallons.
2218*192)l*00000(-0004508 multipher for imperial bushels.
Now it is manifest that if the solid inches contained in any vessel
be multiplied by the first of these multipliers, the product will be
the imperial gallons that vessel will contain ; and if multiplied by the
other, the product will be the imperial bushels.
MULTIPLIERS AND DIVISORS FOR CIRCLES.
It has been shewn that when the diameter of a circle is 1, the area
of that circle is -785398, &c., or -7854, nearly ; then by dividing the
solid capacity of any figure by '7854, the quotient will be the proper
iivisor for the square of the diameter of a circular figure. Then to
reduce the area at one inch deep into gallons, divide '7854, or
•785398, &c., by 277*274, and 2218*192, and the quotients will give
the multipliers for imperial gallons and bushels respectively ; and
•7S54 divided into 277*274 and 2218*192, will give the divisors for
the imperial gallons and bushels.
277'274)'785398(. 002832 multiplier for imperial gallons.
2218'192)'785398(.00354 multiplier for imperial bushels.
•785398)277 •274(3530362 divisor for imperial gallons.
•785398)2218*192(2824-2897 divisor for imperial bushels.
136
GAUGING.
The gauge-points are found bv extracting the square root of tho
divisors.
GAUGE- POINTS FOR SQUARES.
V 277-274 = 16-651 imperial gallons.
V 2218-192 = 47-097 imperial bushels.
GAUGE-POINTS FOR CIRCLES.
V 353-0362 = 18-789 imperial gallons.
V 2824-2897=53-144 imperial bushels.
In this manner the numbers in the following Table were calculated
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GAUGING. 137
Kote. — ^It very often happens in the practice of prauc^ng-, that when the on.3
given number is set to the gauge-point on the sUding rule, the other given
number will fall ofif the rule ; hence in many cases it will be necessary to find
a second, or new gauge-point. The second gauge-points are the square roots
of ten times the divisors in the above table. Thus, for squares, the new gauge-
point for imperial gallons is 52-65, for bushels 14.8*93 ; and for circles, the new
gauge-point for gallons is 59*42, for malt bushels 168 05.
PROBLEM II.
Tojind the area^ in imperial gallons^ of any rectilineal plane
figure.
Rule. By the rules given in Mensuration of Superficies, find the
area of the figure in inches, which bein^ divided by 277*274, or mul-
tiplied by '0036065, w^ill give the area in gallons.*
1. Suppose a back or cooler in the form of a parallelogram to be
100 inches in length, and 40 in breadth ; required the area in impe-
rial gallons ?
100x40 = 4000 the area in inches, which divided by 277*274, the
quotient 14 '426 = the number of imperial gallons; or if we multiply
4000 by "0036065, the product 14.426 is the number of imperial gal-
lons as before.
BY THE SLIDING RULE.
On A. On B, On A. On B.
As 277*274 : 40 : : 100 : 14-4, nearly.
2. If the side of a square be 40 inches, what is the area in imperial
gallons? ^ ^??5. 5*77 gallons.
3. If the side of a rhombus be 40 inches, and its perpendicular
Dreadth 37 inches ; required its area in wine gallons? Ans. 5*41.
4. What is the area of a square cooler, in imperial gallons, the side
being 144 inches? Ans. 74*785.
5. Allowing the side of a hexagon to be 64 inches, and the per-
pendicular from the centre to the middle of one of the sides 55*42
inches ; required its area in imperial gallons and malt bushels ?
A no (38*38 imperial gallons.
^^^' \ 4*8 malt bushels.
PROBLEM III.
The diameter of a circular vessel being given in inches^ to find
its area in imperial gallons^ Sfc.
Rule. Multiply the square of the diameter by '002832 ; or divide
the square of the diameter by 353*036, the product or quotient will
give the area in imperial gallons.
When it is required to find the area in any other denomination than
imperial gallons, use the proper multiplier or divisor for the required
denomination, as given in the Table, page 136.
* The areas of plane figures, in gauging, are expressed in gallons, or bushels.
For there will be as many solid inches in any vessel of one inch deep, as there
are superficial inches in its base. What is called in gauging a surface or area,
is in reality a solid of one inch deep, which multiplied by the height will givo
the whole content in gallons or bushels.
138 GAUGING
1. The diameter of a circular vessel is 32 '6 inches ; required the
area in imperial £:allons ?
(32-6y=1062'76. Then,
1062-76 X •002832 = 3-01 gallons.
Or, 1062-76-r-353-036 = 3-01.
BY THE SLIDING RULE.
As 18 '78 is the circular gauge-point for imperial gallons, say,
On D. On B. On D. On B.
As 18-78: l::32-6 :3
2. If the diameter of a circular vessel be 10 inches, what is the
area in imperial gallons ? Ans, -283.
3. Suppose the diameter of a circular vessel is 30 inches, what is
its area in imperial gallons ? Ans. 2-548.
4. What is the area in imperial gallons of a round vessel, whose
diameter is 24 inches? Ans, 1*631.
PROBLEM IV.
Given the transverse and conjugate diameter of an elliptical vessel^
to find its area m imperial measure.
Rule. Multiply the product of the two diameters by '002832 ; or
divide the product of the two diameters by 353 '036; the product or
quotient will give the imperial gallons required.
When any other denomination is required, the proper multiplier or
divisor in the table is to be employed.
1. Suppose the longer diameter of an elliptical vessel is 10, and
Vat shorter diameter 6, required the area in ale and wine gallons ?
Here, 10x6 = 60.
Then, 60 x -003832 = -17 of a gallon.
2. The transverse or longer diameter of an elliptical vessel is 20,
and the conjugate or shorter diameter 10 inches ; what is the area in
imperial measure ? Ans, '666 of a gallon.
On A. On B. On A. On B.
As 353 : 20 : : 10 : -566 of a gallon.
8. Suppose the transverse diameter of an elliptical vessel is 70
inches, conjugate 50 inches; required its area in imperial gallons
and malt bushels? ^^_ (9*914 gallons.
^^^' "11-24 malt bushels.
Note. — As vessels are seldom or never made truly elliptical, being generally
ovals, the area found by the above rule is not correct, except the vessel be a
truly mathematical ellipsis ; when the vessel is of an oval form, the area is
best found by the method of equidistant ordinates.
Let A B C D be the oval vessel whose area is required, and let
AB and CD be the transverse and conjugate diameters, at right
angles to each other, the former behig 102-8 inches. Divide this
transverse (102-8) by some even number which will leave a small
remainder, the quotient will be the distance of the ordinates ; which
distance may be laid off on both sides of the conjugate diameter
GAUGING.
139
a number of times equal to half the
even number by which the transverse
was divided ; then with chalk and a
parellel ruler, draw the ordinates
through the points 1, 2, 3, 4, &c.
Then, by Problem XXL, Sec. III.,
the area may be found, which being
multiplied or divided by the proper
tabular number?, will give the area
in gallons, &c. Or,
1st, Add together the first and last ordinates.
2d, Add together the even ordinates, that is, the 2, 4, 6, 8, 10,
&c., and multiply the sum by 4.
3d, Add together the odd ordinates, except the first and last ; that
IS, add the ordinates, 3, 5, 7, 9, &c., and multiply the sum by 2.
4th, Multiply the sum of the extreme ordinates by their distance
from the curve.
5th, Add the three first found sums together, and multiply the
Bum by the common distance of the ordinates, and to the product
add the fourth found sum, and divide the total by 3, and the quotient
resulting by 277*274, or 2218*192, for the area in imperial gallons,
or malt bushels, respectively.
First, 102*8-h10 = 10 the distance of the ordinates asunder, and
the remainder 2 8 is double the distance of the extreme ordinatea
from the curve; that is, 1*4 = A 1, or B 11.
Now let us suppose the lengths of the ordinates to be 20, 40*2,
57, 66*6, 73, 75, 73, 66*6, 67, 40*2, 20, respectively beginning at 1,
and proceeding to 11.
1st,
{.l:
= 20
20
40 inches, sum of the firat and last
1-4
fl:
2A, -! 6 =
dd,
56
2 = 40-2
66*6
750
8 = 66*6
10=40*2
288*6x4 = 1154'4
3 = 57
5 = 73
7 = 73
9 = 57
"260 >< 2=620
HO GAUGING.
Then, 40+ 11 54 4 + 520 = 171 4 -4 sum of first three sums.
10
17144
56
"3)17200
5733-3; then,
5733-3-+277-274 = 20-64 gallons.
5733-3-T-2218-192 = 2-58 malt bushels.
When the vessel is not circular, or elliptical, it is best to measure
the equidistant ordinates, which though ever so unequal, will, by
proceeding as above, serve to find the area of the base. Whenever
the vessel is an irregular curved figure, the area should be invariably
found by the method of equidistant ordinates, as the true result
cannot be found by any other method.
4. What is the area, in imperial measure, of an ellipse, whose .
transverse axis is 24, and conjugate 18? Ans, 1-2234 gallons.
PROBLEM V.
To find the content ofaprism^ in imperial gallons.
Rule, Find the area of the base, by Problem II. in Gauging,
which being multiplied by the depth within, will give the content in
gallons.
Or, find the solid content by mensuration, and divide that content
by 277-274 for imperial gallons.
A vessel, whose base is a right-angled parallelogram, is 49-3
inches in length, the breadth 36*5 inches, and the depth 42*6 inches ;
required its content in imperial gallons ?
Here, 49-3 x 36-5 x 42-6 = 76656-57.
Then, 76656-57-t-277-274 = 276'465 gallons.
And 76656-57-r-2218-192 = 34-558 malt bushels.
BY THE SLIDING RULE.
OnB. OnD. OnB.
49-3: 49-3:: 36-5: 42-42.
OnD. OnB. OnD.
16-65> . .o-e . . 42-42 • |27-6 gallons.
46-371 . *^ *> . . *^ ^^ . -(34.5 malt bushels.
2. Each side of the square base of a vessel is 20 inches, and its
depth 10 inches, what is the content in old ale gallons ?
Ans. 14-28 gallons.
3. The side of a vessel in the form of a rhombus is 20 inches,
breadth 15 inches, and depth 10 inches ; required the content in old
ale gallons? Ans. 10*638 gallons.
4. What is the content, in old wine gallons, of a vessel in the form
of a rhomboid, whose longest side is 20 inches, breadth from side to
Bide 8 inches, and depth 10 inches? Ans. 6-88 wine gallons.
GAUGING. - 14|
PROBLEM VI.
To find the content of any vessel, whose ends are squares or
rectangles, of any dimensions.
Rule. Multiply the sum of the lengths of the two ends, by the
8um of their breadths, to which add the areas of the two ends ; this
sum, multiplied by one-sixth of the depth, will give the solidity in
cubic inches ; then divide by 277-274, or 2218'192 for the content in
imperial gallons, or malt bushels.
1. Suppose the top and bottom of a vessel are parallelograms, the
length of the top is 40 inches, and its breadth 30 inches ; the length
of the bottom is 30 inches, and its breadth 20 ; and the depth 60
inches ; required the contents in imperial gallons ?
40 + 30 = 70 sum of the lengths.
30 + 20 = 50 sum of the breadths.
3500 product.
40 X 30 = 1200 area of the greater base.
80 X 20= 600 area of the lesser base.
5300
10 one-sixth of the depth.
53000 solidity in cubic inches.
Then, 53000-277-274= 191 '146.
BY THE SLIDING RULE.
Find a mean proportional (V(40 x 30) = 34*64,) between the
length and breadth at the top, and a mean proportional ( V(^0 x 20)
= 24-49,) between the length and breadth at the bottom; the sum
of these is 59*13, twice a mean proportional between the length and
breadth in the middle. Then,
On D. On B. On D. On B.
:-64 :
(34-6
:: ^24-4
(59-1
16*65 : V : i •< 24-49 : }- sum 191*146 imperial gallons.
•13 : )
2. Suppose the top and bottom of a vessel are parallelograms, the
length of the top is 100 inches, and its breadth 70 inches ; the length
of the bottom 80, and its breadth 56, and the depth 42 inches ; what
is its content in imperial gallons? Ans, 862*59 imperial gallons.
THE GAUGING OR DIAGONAL ROD.
The diagonal rod is a square rule, having four faces, and is gene-
rally 4 feet long. It folds together by joints. This instrument is
employed both for gauging and measuring casks, and computing
their contents; and that from one dimension only, namely, the
diagonal of the cask, or the length from the middle of the bung-hole
to the meeting of the cask with the stave opposite the bung ; being
142 GAUGING.
the longest line that can be drawn from the middle of the bung-hole
to any part within the cask.
On one face of the rule is a scale of inches for measuring this
diagonal; to which are placed the areas, in ale gallons, of circles to
the corresponding diameters, in like manner as the lines on the under
sides of the three slides in the Sliding Rule.
On the opposite face, there are two scales of ale and wine gallons,
expressing the contents of casks having the correspondent diagonals.
All the other lines on the instrument are similar to those on the
Sliding Rule, and are used in the same manner.
Example. The diagonal, or distance between the middle of 7he
bung-hole to the most distant part of the cask, as found by the dia-
gonal rod, is 34*4 inches ; what is the content in gallons?
To 34*4 inches correspond, on the rod, 90| ale gallons, or 111 wme
gallons, 92 J imperial gallons, the content required ?
Note. — ^The contents shewn by the rod answer to the most commou form of
casks, and fall in between the 2d and 3d varieties following.
OF CASKS AS DIVIDED INTO VARIETIES.
Casks are usually divided into four varieties, which are easily
distinguished by the curvature of their sides.
1. The middle frustum of a spheroid belongs to the first variety.
2. The middle frustum of a parabolic spindle belongs to the second
rariety.
3. The two equal frustums of a paraboloid belong to the third
Variety.
4. And the two equal frustums of a cone belong to the fourth
variety.
If the content of any of these be found in inches by their proper
rules, and this divided by 277*274, or 2218*2, the quotient will be
the content in imperial gallons, or bushels, respectively.
PROBLEM VII.
To find the content of a vesselin the form of the frustum of a cone.
Rule. To three times the product of the two diameters add the
square of their difference; multiply the sum by one-third of the
depth, and divide the product by 353-0362 for imperial gallons, and
by 2824-289 for malt bushels.
1. What is the content of a cone's frustum, whose greater dia-
meter is 20 inches, least diameter 15 inches, and depth Ai inches ?
20x15x3 =900
20— 15 = 5 and 62= 25
"925x7=6475. Then,
353'0362)6475(18'34 imperial gallons.
294-12)6475(22-01 wine gallons.
2. The greater diameter of a conical frustum is 38 inches, the less
diameter 20*2, and depth 21 inches ; what is the content in old ale
gallons? Ans. 51*07 gallons.
GAUGINa. 143
PROBLEM VIII.
To find the content of the frustum of a square pyramid,
RuLF To three times the product of the top and bottom sides, add
the square of their difference, multiply their sum by one-third of the
depth, and divide the product by 282 and 231, for old ale and wine
gallons, respectively ; and by 277*274, for imperial gallons.
1. Suppose the greater base is 20 inches, the less base 15 inches,
and depth 21 inches ; required the content in old wine measure ?
20x15x3 = 900
20—15 = 5
Then, 5x5= 25
925 X 7-=-231 = 27-8 gallons.
Note. — ^The content of the frustum of a pyramid is found just like that of a
cone, with the exception of the tabular divisor, or multiplier, the cone require
ing the circular factor, and the pyramid the square one.
PROBLEM IX.
To find the content of a globe.
Rule. Multiply the diameter of the globe by its cir^jumference,
and the resulting product by one-sixth of the diameter; then the
last product multiplied or divided by the circular factor, will give the
content in gallons.
1. Let the diameter be 34 inches, what is its content?
34 X 34 X 34 X •5236 = 20579-5744.
Then, 20579-5744-t-282 = 72-9772 old ale gallons.
And, 20579-5744-f-231 = 89-08 old wine gallons.
Rule. II. Or cube the diameter of the globe, which multiply by
•001888 (I of -002832) for the content in imperial gallons.
343 = 39304; then 39304 x •001888 = 74-2 imperial gallons.
2. What is the coirtent of a globe in old ale and wine measme, the
diameter being 20 inches ?
. J 14-848 old ale gallons.
'^^^' I 18-128 old wine gallons.
3. Required the content of a globular vessel, whose diameter is 100
inches? An's. 1888^ imperial gallons.
PROBLEM X.
To find the content of the segment of a sphere^ as the rising crowr
of a copper stilly ^c.
Rule. Measure the diameter, or chord of the segment, and the
altitude just in the middle. Multiply the square of half the diameter
by 3 ; to the product add the square of the altitude ; multiply this
8um by the altitude, and the product again by •001856, or -002266,
for old ale or wine measure, respectively, and by -001888 forimperiaj
gallons.
144
GAUGING.
1. The diameter of the crown of a copper still is 27*6, its depth
9'2 ; required its content?
Here 27-6-7-2= 13-8.
Then 13-8 x 13-8 x 3 = 571*32
9-2x9-2= 84-64
655-96 sum.
9-2 depth.
6034-832 X -001888 = 13-39 imperial gallons.
PROBLEM XL
To gauge a copper having either a concave or convex bottom; or
what is called a falling bottom, or rising crown.
Rule. If the side of the vessel be straight, with a falling bottom,
find the content of the segment Q y D, by Prob. X. ; find also the
content of the upper part ABDC,
by Prob. VIL ; the sum of both ^^
will give the content of the
'3opper.
When the copper has a ris-
ing crown, find the content of
ABCD, by Prob. VIL; from
which deduct the content of the
segment C a; D, and the remainder will be the content of the vessel
A B D a; C.
PROBLEM XII. •
To gauge a vessel whose side is curved from top to bottom.
Take the diameters at equal distances of 2, 3, 4, or 5 inches,
according as the case may require ; if the side of the vessel be con-
siderably curved, the number of diameters that will be required will
be considerable ; the less the curvature of the side, the less the num-
ber of diameters that will be required.
To gauge the vessel, or cop-
per, ABDC, fasten a piece of
pack-thread at A and B, as
AFB; then with some con-
venient instrument find the
distance aC of the deepest
part of the copper, which let
us suppose to be 47 inches.
By means of the same in-
strument measure the distance oF from the top of the crnwn to F
the middle of A B ; which let us suppose to be 42 inches, this de-
ducted from a C, 47, wiU leave 5 ( = o G) the height of the crown.
To find the diameter CD, of the bottom of the crown.
Measure the top diameter AB, which suppose to be 99 inches:
then hold a thread, so that a plummet attached to the end thereof
may hang just over C, and measure Aa=B F, each of which let U3
GAUGING.
Hff
admit to be 17*5 inches; add these together, and deduct their sura
(85) from 99, and the remainder (64) will evidently be equal to CD,
the diameter at the bottom of the crown. Measure the diameter
mon^ which touches the top of the crown, which suppose is 65
inches.
Now, as this copper is not considerably curved, the diameters may
be taken in the middle of every 6 inches of the depth, whicfi suppose
to be as in the second column of the following table; to each diameter
find the area in imperial gallons, by Prob. III., which write in the
third column ; find also the content of every 6 inches, corresponding
to these diameters, which write in the fourth column of the table ;
lastly, find the content of the crown by Prob. X., and subtract it
from the content of A B D G C, the remainder will give the capacity
of the copper.
Or thus, CD being 64 inches, the area answering to it is 11'6022 ;
this multiplied by half the altitude of the crown, viz., by 2 -5, gives
.^9*0055 gallons, the content of the crown. The content of the part
mnDC is 58'9222 gallons, from which the content of the crown
being deducted, the remainder (29*9167 gallons) is the quantity of
Vquor which covers the crown.
Parts of
the depth.
Diameters.
Areas.
Content of every
6 inches.
6
95-3
25-7257
154-3542
6
90-1
22-9948
137-9688
6
85-
20-4653
122-7918
6
80-
18-1284
108-7704
6
75-2
16-0183
96-1098
6
70-5
14-0786
84-4716
6
(jQ-
12-3387
74-0322
The sum
778-4988
To cover crown
29-9167
The whole content
808'4155
PKOBLEM yill.
To find the content of any close cash.
Whatever be the torm of the cask, the following dimensions must
be taken ; that is,
The bung diameter,
The head diameter.
The length of the cask.
On account of the difficulty in ascertaining the figure of the cask.
It is not, in many cases, easy to find the exact contents of casks.
> within.
146
GAUGING.
In taking the dimensions of a cask, it is essential that the bung*
hole be in the middle of the cask, and also that the bung-stave, and
the stave opposite to it, are both regular and even within.
It is likewise essential that the heads of casks are equal and truly
circular ; and if so, the distance between the inside of the chimb to
the outside of the opposite stave will be the head diameter within the
cask, nearly.
From the variety in the forms of casks, no general rule could be
given to answer every form ; two casks may have equal head dia-
meters, equal bung diameters, and equal lengths, and yet their con-
tents may be very unequal.
PROBLEM XIV.
To find the content of a cask of the first variety.
Rule. To the square of the head diameter add double the square
of the bung diameter, and multiply the sum by the length of the cask.
Then multiply the last product by -OOOOf , or divide by 1059 '1, the
product or quotient will be the content in imperial gallons.
1. What is the content of a spheroidal
whose length is 40 inches, bung
diameter 32 inches, and head diameter
24 inches ?
24x24= 576
32x32=1024
2
2624x40=104960
•00091
944640
34987
11662
99*1289 imperial gallons.
BY THE GAUGING RULE.
Set 40 on C, to the G R 18-79 on D, against
24 on D, stands 64-99 on C,
32 on D, stands 116-2 on C,
+ 116 2
3)297-39
99 13 gallons.
GAUGTNa.
m
2. What is the content of a spheroidal cask, whose length is 20
inches, bung diameter 16 inches, and head diameter 12 inches?
4 (12-36 old ale gallons.
'^ ^* \U'86d old wine gallons.
Tojind the content of a cask hy the mean diameter.
Rule. Multiply the difference of the head and bung diameters by
'68 for the first variety ; by -62 for the second variety ; by -55 for
the third ; and by "5 for the fourth, when the difference between the
head and bung diameters is less than 6 inches ; but when the difference
between these exceeds 6 inches, multiply that difference by '7 for the
first variety ; by '64 for the second ; by '57 for the third ; and by '52
for the fourth. Add this product to the head diameter, and the sum
will be a mean diameter. Square this mean diameter, and multiply
the square by the length of the cask; this product multiplied, or
divided, by the proper multiplier or divisor^ will give the content.
By resuming the last example but one, we have
Bung diameter 32 29-6 mean diameter.
Head diameter 24 29*6
876-16 square.
40 length.
6-6
24
859-5)35046-40
97-6 gallons.
Mean diameter 29*6
In the same manner the content for the second variety will be
94-46 ale gallons; for the third variety 90*87 ale gallons; and for
vhe fourth variety 83*34 gallons.
PROBLEM XV.
Tojind the content of a cask of the second variety.
Rule. To the square of the head diameter add double the square-
of the bung diameter, and from the sum
deduct two- fifths of the square of the dif-
ference of the diameters* multiply the
remainder by the length, and the product
again by -00091 for the content in imperial
gallons.
1 . What is the content of a cask, whose
length is 40 inches, bung diameter 32
inches, and head diameter 24 inches ?
32—24 = 8; then 8^ = 64, and f of 64 = 25*6
242 = 576, and 32^ = 1024, then 1024x2 = 2048
2048 + 576 = 2624, and 2624-256 = 2598*4
40
103936 X •00091 = 98*1617 gallons.
103936
148
GAUGINQ.
PROBLEM XVI.
To find the content of a cask of the third variety.
Rule. To the square of the bung diameter add the square of the
head diameter ; multiply the sum by the length, and the last product
Vy '001416 for the answer in imperial gallons.
Let us resume the last example : thus
322=
242=
= 1024
= 576
1600x40= 64000
•001416
90*624 imperial gallons.
PROBLEM XVII.
To find the content oj a cask of the fourth variety.
Rule. Add the square of the difference of the diameters to 3 timej'
the square of their sum ; multiply the sum \
by the length, and the last product by ^^^^^■^-^ I
•000236 for the content in gallons.
Resuming still the last example, 32 + 24 =
56, and 562x3 = 3136x3 = 9408, and 82=
64, then 9408 + 64 = 9472 ; then 9472 x 40 =
378880, and 378880 x •000236 = 89*41668
imperial gallons.
PROBLEM XVIII.
To find the content of any cask by Dr Hutton''s general rule.
Rule. Add into one sum, 39 times the square of the bung dia-
meter, 25 times the square of the head diameter, and 26 times the
product of the two diameters ; then multiply the sum by the length,
and the product again by -000311 for the content in gallons.
1. What is the content of a cask, whose length is 40 inches, and
the bung and head diameters 32 and 24 ?
322=1024 242=576 32x24 = 768
39 25 26
39936
14400
19968
14400
19968
74304 X 40 = 2972160
•000311-
93*4579 gallona.
QAUGINCr. 149
ULLAGING.
PROBLEM XIX.
To ullage a lying cask.
This is the finding what quantity of liquor is contained in a cask
when partly empty.
To ullage a lying cask, the wet and dry inches must be known, as
also the content of the cask and bung diameter.
Rule. Take the wet inches, and divide them by the bung dia-
meter ; find tlie quotient in the column of
versed sines, in the Table at the end of the
practical part of this book, and take out
its corresponding segment; multiply this
segment by the whole content of the cask,
and the product arising by IJ for the
ullage required, nearly.
1. Find the ullage for 8 wet inches, the
bung diameter being 32 inches, and the
content 92 ale gallons ?
32)8('25, whose tabular segment is -ISSS^G.
Then, •153546x92=14-126232.
And, 14-126232 X 1^=17-65779 gallons.
PROBLEM XX.
To ullage a standing cask.
Rule. Add together the square of the diameter at the surface of
the liquor, the square of the diameter of the nearest end, and the
square of double the diameter taken in the middle between the other
two ; multiply the sum by the length between the surface and nearest
end, and the product arising by -000472 for the gallons in the less
part of the cask whether empty or filled.
1. What is the ullage for 10 wet inches, the three diameters being
24, 27, and 29 inches?
242= 576 43330
292= 841 -000472
(2x27)^=^2916 ^^
4333 303310
10 173320
43330 20-45176 gallons.
PROBLEM XXL
To find the content of an ungula^ or hoof, of the frustum of a cone.
Rule. For the less hoof, multiply the product of the less diameter
and height, by the product of the greater diameter multiplied by a
150
GAUGING.
mean proportional between both diameters, less the square of the less
diameter, and this last divided by three times the circular factor
multiplied by the difference of the diameters, gives the content of the
less hoof.
1. CD = 30, AB = 40, C(Z=:20; required the ^ ZZ — --v^b
content of the less hoof ?
40x30 =1200, and V 1200 = 34-6 mean.
80 X 20 = 600, 1st product.
40 X 34-6 = 1384, 2d product.
30x30 = 900
**? ., 484 remainder.
484x600 = 290400
40->^30=10, then 359 x 3x 10 = 10770Vaiiii':r;i7rO7?<'i^"^
290400-^10770 = 26-96 gallons. \^ ^J "^—^
Rule. For the greater hoof multiply the product of the greater
diameter and the height of the frustum, by the square of the greater
diameter made less by the product of the less diameter multiplied
by a mean proportional between those diameters ; this remainder,
divided by three times the circular divisor multiplied by the differ-
ence of the diameters, gives the content of the greater hoof.
Resuming the last example, vre have
^ 40x40 = 1000
20x40 = 800, 1st product.
40x30=1200, and V 1200 = 34-6
34-6 X 30 = 1038, 2d product. / *- ]
40-30=10. lA/JndLJ^J^
Then 1600-1038 = 562 -/-"7?^^
800
859 X 3 X 10= 10770)449600, last product.
41*74 old ale gallons.
PROBLEM XXII.
To gauge a still.
Fill the still with water, and draw it off in another vessel of some
regular form, whose content is easily computed. This is by far the
most accurate method that can be employed.
Or gauge the shoulder by itself, and gauge the body by taking a
greater number of diameters at near and equal distances throughout,
first covering the bottom, if there be any cavity, with water, the
uantity of which is known.
LAND-SURVEYING. IBl
LAND-SUEYEYING.
/^.-sp.-.. .^'^ ,/ 71
SECTION XI.
Land-surveying is that art which enables us to give a true plan or
representation of any field or parcel of land, and to determine the
superficial content thereof.
In measuring land, the area or superficial content is always ex-
pressed in acres, or in acres, roods, and perches ; each acre containing
4 roods, and each rood 40 perches.
Land is measured with a chain, called Gunter's chain, of 4 poles or
22 yards in length, which consists of 100 equal links, each link
being ^^V of a yard long, or ,% of a foot, or 7-92 inches. 10 square
chains, o: 10 chains in length and 1 in breadth, make an acre ; or
4840 square yards, 160 square poles, or 100,000 square links make
an acre. The length of lines measured with a chain are generally
set down in links as integers ; every chain being 100 links in length.
Therefore, after the content is found, it will be in square links, and
as 100,000 square links make an acre, it will be necessary to cut of!
five of the figures on the right-hand for decimals, and the rest will
be acres. The decimals are reduced to r6ods by multiplying by 4,
and cutting off" five figures as before for decimals, which decimal part
is reduced to perches by multiplying by 4^, and cutting ofi*five figures
from the product. As an example : —
Suppose the length of a rectangular piece of ground to be 792
links, and its breadth 385 ; required the number of acres, roods, and
perches it contains ?
792 3-04920
385 4
3960 -19680
6336 40
2376 .
■ 7-87200
304920
Ans. 3 acres, 0 roods, 7 perches.
The statute perch is 5 J yards, but the Irish plantation perch is 7
yards ; hence the length of a plantation link is 10*08 inches.
PROBLEM I.
To measure a line or distance on the ground, two persons are
employed ; the foremost, for the sake of distinction, is called
the leader, and the hindermost, the follower,..
Ten small arrows or rods, to stick in the ground at the end of each
152 LAND-SURVEYING.
chain, are provided; also some station -staves, or long poles with
coloured flags, to set up in the direction of the line to be measured,
if there do not appear some marks naturally in that direction.
The leader takes the 10 aiTOvs^s in one hand, and one end of the
chain by the ring;, in the other ; the follower stands at the beginning
of the line, holding the ring at the end of the chain in his hand,
while the leader drags forward the chain by the other end of it, till it
is stretched straight, and the leader directed by the follower, by
moving his hand, to the right or left, till the follower see him exactly
in a line with the mark or direction to be measured to ; then both of
them holding the chain level and stretched, the leader sticks an arrow
upright in the ground, as a mark for the follower to come to, and
advances another chain forward, being directed in his position by
the follower standing at the arrow, as before, as also by himself, now
and at every succeeding chain's length, by moving himself from side
to side, till the follower and back-mark be in a direct line. Having
then stretched the chain, and stuck down an arrow, as before, the
follower takes up the arrow, and thus they proceed till the 10 arrows
are employed, or in the hands of the follower, and the leader, without
an arrow, is arrived at the end of the eleventh chain-length. The
follower then sends or brings the 10 arrows to the leader, who puts
one of them down at the end of his chain, and advances with his
chain, as before. And thus the arrows are changed from one to the
other at every 10 chains' length, till the whole line is finished, if il
exceed 10 chains; and the number of changes shews how many
times 10 chains the line contains, to which the follower adds the
arrows he holds in his hand, and the number of links of another
chain over to the mark or end of the line. Thus, if the whole line
measure 36 chains 45 links, or 3645 links, the arrows have been
changed three times, the follower will have Arrows in his hand, the ^
leader 4, and it will be 45 links from the la^ arrow, to be taken up
by the follower, to the end of the line.
In works on Surveying, it is usual to describe the various instru-
ments used in the art. The pupil, however, will best learn the use
of these instruments when actually engaged in the practice. The
chief instruments employed are the chain, the plane table, tlie theo-
dolite, the cross, the circumferentor, the offset staff, the perambulator,
used in measuring roads, and other great distances.
Levels, with telescopic or other sights, are used to find the levels
between two or more places, or how much one place is higher or
lower than the other.
Besides all these, various scales are used in protracting and
measuring on paper ; such as plane scales, line of chords, protractor,
compasses, reducing scales, parallel and perpendicular rulers, &c.
THE FIELD-BOOZ.
In surveying with the plane table, a field-book is not required, as
everything is drawn on the table immediately when it is measured.
But when the theodolite, or any other instrument, is us^, some sort
of a field-book is used in order to register all that is done relative to
the survey in hand. This book every one contrives and rules as he
r
LAND-SURVEYING.
153
thinks fit. It is, however, usuall}' divided into three columns. The
middle column contains the different distances on tlie chain-line,
angles, bearings, &c., and the columns on the right and left are
for the offsets on the right and left, which are set against their cor-
responding distances in the middle column ; as also for such remarks
as may occur, and may be proper to note in drawing the plan ; sucn
as houses, ponds, castles, churches, rivers, trees, &c. &c.
But in smaller surveys, an excellent way of setting down the work
is, to draw by the eye, on a piece of paper, a figure resembling that
which i? to be measured ; and then write the dimensions, as they are
found, against the corresponding parts of the figure. This method
may be practised even in larger surveys, and is far superior to any
other at present practised. A specimen of this plan will be seen
further on.
FORM OF THE FIELD-BOOK.
Offsets and remarks on the
left.
stations,
Bearings, and
Distances.
Offsfets and remarks on the
right.
D 1
104° 25'
00
Cross a hedge 24
67^
Brown's bam.
a brook 30
120
u ' . /'^
734
954
Tree.
w.- "-1^ ./ .. .i'
y 736
67 stile.
82
62"' 25'
00
.L
House comer €1
40
67
Foot path 15
84
95
44
467
14 Spring.
J \, 1 '
976
ns
'^ W .,-:,. ., 1
54*= ir
62
20 Pond.
124
Clayton's hedge 24
630
767
767
30 Stile.
' <. '^'
305
760 -
I5i LAND-SURVEYING.
In this form of a field-book D 1 is the first station, where the
angle or bearing is 104° 25'. On the left, at 67 links in the distance
or principal line, is an offset of 24 ; and at 120 an offset of 30 to a
brook on the right ; at 67 Brown's barn is situated ; at 954 ia an
offset of 20 to' a tree, and at 736 an offset to a stile.
And so on for the other stations.
A line is drawn under the work, at the end of every station, tO
prevent confusion.
PROBLEM II.
To take angles and bearings.
Let it be required to take the bearings of the two
objects B, C, from the station A.
In this problem it is required to measure the angle ^
at A, formed by two lines, passing from the station
A, through two objects B and -G.
1. By measurement with the chain., ^c.
Measure with the chain any distance along the two lines A B, AC,
as A 6, Ac; then measure the distance be; and this being done^
transfer the three sides of the triangle Abe to paper, on which
measm-e the angle c A &, as in Problem XV., Practical Geometry.
2. With the magnetic needle and compass.
Turn the instrument, or compass, so that the north end of tho
needle may point to the flower-de-luce. Then direct the sights to a
mark at B, noting the degrees cut by the needle. Next direct the
sights to another mark at C, noting the degrees cut by the needle as
before. Then their sum or difference, as the case may be, will give
the number of degrees in the angle GAB.
3. With the theodolite, ^c.
Direct the fixed sights along the line A B, by turning the instru-
iiient about till you see the mark B through these sights, and in that
position screw the instrument fast. Then turn the moveable index
about till, through its sights, you see the other mark G. Then the
degrees cut by the index, on the graduated limb or ring of the instru-
ment, shew the number of degrees in the angle GAB.
4. With the plane table.
Having covered the table with paper, and fixed it on its standi
plant it at the station A, and fix a fine pin, or a point of the compass,
in a proper point of the paper, to represent the station A. Glose by
the side of this pin, lay the fiducial edge of the index, and turn it
about, still touching the pin, till one object B can be seen through
the sights ; then by the fiducial edge of the index draw a line. By
a similar process draw another line in the direction of the object C.
And it is done.
LAND-SURVEYING.
155
PROBLEM IlL
To measure the offsets.
Let Abed efg be a crooked hedge, river, or brojok, &c., and
A Gr a base line.
Begin at the point A, and measure towards G; and when yon
come opposite any of the corners bcd^ &c., which is ascertained by
means of the cross- staff, measure the offsets B &, C c, D J, &c., with
the chain, and register the dimension, as in the annexed field-book.
FIELD-BOOK.
91
57
98
70
84
62
785 = AG.
634
510
340
220
45
D A go North.
,
Offsets
Left.
Base line A G, or
□ Station.
Offsets
Right.
2'o lay down the plan.
Draw the line AG of an indefinite length ; then by a diagonal
scale, set off A B equal to 45 links ; at B erect the perpendicular B b
equal to 62 links taken from the same scale. Next set oft' A C equal
to 220 links, or 2 chains 20 links, and at C erect the perpendicular
C c, equal to 84 links ; in the same way set off A D equal to 340
links, or 3 chains 40 links, and at D erect the perpendicular D d
equal to 70 links. Proceed in a similar manner with the remaining
offsets, and straight lines joining the points A, &, c, d^ e, &c., will
complete the figure.
To find the content.
Some authors direct to add up all the perpendiculars B&, Cc, &c.^
and divide their sura by the number of them, tlien multiply the quo-
tient by the length AG. This method, however, should never be
used, except when the offsets B &, Cc, &c., are equally distant from
each other.
When the offsets are not equally distant from each other, which
indeed is generally the case, this method is erroneous ; therefore the
following method ought to jje employed.
159
LAND-SURVEYINa.
Find the content of the space A B & as a triangle, by Problem V.,
Section II. Find the contents of the figures BCcb^ C D c? c, &c.,
as trapezoids, by Problem XIII., Section II., the sum of all these
separate results will be the content of the figure A G gfe dch A.
The actual calcwlation is as follows : —
CALCULATION.
AB= 45
b6= 62
AC = 220
AB= 45
AD = 840
A 0 = 220
AE = 510
AD = 340
AF=634
AE=510
AG = 785
AF = 634
90
BC=175
CI>=120
DE = 170
EF = 124
GF=151
270
2790
Bb=; 63
Cc= 84
Cc= 84
Dc^= 70
Dd= 70
Ee= 98
Ee= 98
?/= 57
F/= 57
Gg= 91
Pr
Sum 146
BC=175
S»m 154
CD=120
Sum 168
DE = 170
Sum 155
EF=124
Sum 148
FG=151
od. 25550
18480
28560
19220
22348
These respective products are evidently double the true contents of
the respective figures A B &, B C c 6, C D df c, &c., that is,
2790 = double area of A B 6.
25550 = double area of B C c &.
18480 = double area ofCJ>de.
28560= double area of D E e d.
19220 = double area of E F/e.
22348 = double area of F G gf.
2)116948= double area of the whole in square linkb.
68474 = area in square links.
y<(>
•58474= area in acres =0a., 2r., 13 -5584?. ^r^
2. Required the plan and ccaitent of part of a field, from the fol-
lowing field-book : —
A
t>
AC 45
62 C^
A 6/ 220
84 di
Ae 340
70 ek
A/ 510
88 fl
Acr 634
AB785
57 am
91 Bn
Ans, Oa., 2r., 12P.
h-
LAND-SURVEYINO. 157
PROBLEM IV.
To measure ajield of a triangular form, — 1. By the chain.
Set up marks at the three corners A, B, C, and c
measure with the chain, the distance A D, D being
the point at which a perpendicular demitted from
C, would meet the line A B ; measure also the
distance D B ; hence you have the measure of A B.
Next measure the perpendicular D C ; then from
the two dimensions A B and D C, the contei^t may A,/ ^ ^F
be found by Problem IV., Section II. ^
Let AD = 794, AB=1321, DC = 826 links.
1321 X 826-7-2 = 545573 links.
Then 545573-^-100000 = 5-45573 acres.
45573 X 4 = 1 -82292 roods.
82292x40 = 32-91680 perches.
Hence the answer is 5a., 1r., 33p., nearly.
2. What is the area of a triangular field, whose base is 12*25
chains, and height 8-5 chains? Ans. 5A., Dr., 33p.
2. By taking one or more of the angles.
Measure two sides A B, AC, and the angle A, included between
them; then half the continual product of the two sides, and the
natural sine of the contained angle will give the area.*
Or, measure the two angles A and B, and the adjacent side A B,
) from which the figure may be planned, and the perpendicular C D
found, which perpendicular being multiplied by half the base A B,
will give the area. Or by measuring the three sides of the triangle,
its area may be found by Problem V., Sectiou II.
PROBLEM V.
To survey a four -sided field. — 1. By the chain.
Measure the diagonal A C, and, as be-
fore directed, measure the perpendiculars
D E and B F ; then the area of each of
the triangles A B C, A D C n>ay be found,
as in the last problem, and both are^s
being added together, wjU give the ccHi-
tent of the four-sided Sgure A B C D.
1. Let AC = 592, DE=210, BF=30e links.
592 X 210= 124320 double area of A D C.
592x 306 = 181152 double area of ABC.
2)305472 double area of A B C D.
1-52736 = area of A B C D.
4
2-10944
40
4-37760 Hence lA., 2r., 4r., th« answer.
' See Appendix, Demonstratioa 11.
158
LAND-SURVEYING.
2. By talcing one or more of the angles.
Measure the diagonal AC, also the sides AD and AB. Next
measure the angles DAG and BAG; then the area of each of the
triangles ABC and ADC may be found by case 2, last problem.
2. Required the plan and content of a field by the following field-
book : —
FIELD- BOOK.
1360 = AB.
1190
600
n D go East.
625
342
Offsets
Left.
Station D?
or base line.
Offsets
Right.
\J"
Ans. 6a., 2r., 12p.
3. How many acres are there in a four-sided field, whose diagonal
is 4*75 chains, and the two perpendiculars falling on it, from its
opposite angles, 2-25 and 3*6 chains, respectively?
Ans. lA., IR., 22-3P.
PROBLEM VL
m.
To survey afield of many sides by the chain only.
Let ABCDEFG be the field
whose content is required. Set up
marks at the corners of the field, if
there be none there naturally. Con-
sider how the field may be best
divided into trapeziums and triangles ;
measure them separately, as in the two
last problems ; and the sum of all the
separate results will give the area of
the whole field.
In this way of measurmg with the
chain, the field should be divided into
trapeziums and triangles, by drawing diagonals from corner to corner,
so that all the perpendiculars may be within the figure.
The last figure is divided into two trapeziums A B C G, G D E F,
and the triangle G C D. In the first trapezium measure the diagonal
A C, and the two perpendiculars G m and B n. In the triangle G C D,
measure the base G C, and the perpendicular D q. Finally, measure
the diagonal F D, and the two perpendiculars G o and Ep. Having
drawn & rough figure resembling the field, set all these measures
against the corresponding parts of the figure. Or set them down
thus : —
LAND-SURVEYING.
139
A n 41o>
A C 550)
180 w B
^SS^SO^D
F^ 288^20 0 0
130 + 180
550-^2 = 275, 275 X
310 =
440x230-^2 =
120 + 80 = 200,
5204-2 = 260, 260 X
200 =
CALCULATION.
310,
85250=ABCG.
50600 = CGD.
52000=DEFG.
ABCDEFG.
1-87850 =
4
3 51400
40
20-56000
lA., 3r., 20'56p., answer.
Other methods will naturally present themselves to an ingenious
practitioner who has read the preceding part of this work, or who
lias been previously acquainted with the principles of Mathematics.
Every surveyor ought to be well acquainted with Plane Geometry at
least. This, with a knowledge of Trigonometry, would be sufficient
for the purpose of most surveyors.
The content of the last figure may be found by measuring the sides
AB, BC, CD, DE, E F, FG, GA; and the diagonals A 0, CG,
G D, D F, by which the figure is divided into triangles, the content
of each of which may be found by Problem V., Section II.
2. Required the plan and content of a field of an irregular form
from the following
FIELD-BOOK.
268
900 = EG
550
D E, go S.W.
280
1100 = HE
790
350
n H, go East.
410
140
1180=CH
710
350
D C, go S.W.
280
200
900 = AC
430
300
a A, go S.E.
450
Oflfsets
Left.
1-
Stations, D» or
Base Lines.
Offsets
Right.
Ans. IOA..
IB., 24'64P.
L
160
lAND- SURVEYING.
PROBLEM VII.
To survey a field with the theodolite^ ^c.
1. From one point or station.
When all the angles can be seen from one point, as suppose C.
Having placed the instrument at C, turn it
about till, through the fixed sights, the mark B
may be seen. Fixing the instrument in this posi-
tion, turn the moveable index about, till the mark
A is seen through the sights, and note the degrees
on the instrument. In the same manner, turn the
index successively to the angles E and D, taking
care to note the degrees cut off at each ; by which you have all the
angles, viz., B C A, B C E, B C D. Now, having obtained the angles,
measure the lines C B, C A, C E, CD; entering the respective mea-
sures against the corresponding part of a rough figure, drawn to
resemble the figure.
2. By going round the field.
Set up marks a!, B, C, D, &c. Place the instrument at the point
A, and turn it about till the fixed index
be in the direction AB, and then screw
it fast : turn the moveable index in the
direction A F, and the degrees cut off
will be the angle A; next measure AB,
and planting the instrument at B, mea-
sure, as before, the angle B ; measure
the line B C, and the angle C : and so
proceed round the figure, always mea-
suring the side as you go along, as also
the angles.
The 32d Proposition of the 1st Book of Euclid affords an easy
method of proving the work : thus, add all the internal angles. A,
B, C, &c., of the figure together, and their sum must be equal to
twice as many right angles as the figure has sides, wanting four
right angles. But when the figure has a re-enterant angle as F,
measure the external angle, which is less than two right angles,
and deduct it from four right angles, or 360 degrees, the remainder
vsqll give the internal angle (if such it may be called), which ia
greater than 180 degrees.
When the field is surveyed from one station, as in the first case
shewn above, the content of the figure is found as in'the second case
of Prob. IV., since we have two sides and the angle included between
them in each triangle of the figure.
PROBLEM VIII.
To survey a field with crooked hedges.
Measure the lengths and positions of lines running as near tho
sides of the field as you can ; and, in proceeding along these lines,
measure the offsets to the different corners, as before taught, and
LAND-SURVEYIxS'G.
101
join the ends of the offsets ; these connecting lines will represent the
required fi^nire. When the plane table is used, the plan will be truly
represented on the paper which covers it. But wlien the survey is
made with the theodolite, or other instrument, the different measures
are to be noted in the field-book, from which the sides and angles are
laid down on a map, after returning from the field.
In surveying the piece
ABCDEFGHIKLM,set
up marks at 5 E Fa;. Begin
at the station 5, and
measure the lines sE, E^?,
px^ xs^ as also their posi-
tions, or the angles E 5 a:,
^EF, Ejoar, and pxs;
and in going along the
four-sided figure s Ep x^
measure the offsets at «, &,
d^ g^ k, Z, m^ as before
taught. By means of the
figure sEpx, and of the offsets, the ground is easily planned.
When the principal lines are taken within the figure, as in tha
Kbove case, the contents of the ex-
terior portions sCBA, ODE, &c.,
must be added to the area of the
quadralateral 5a:FE. But when
the principal lines are taken outside
the figure, the portions included
oetween them and the boundaries of
the field. are to be deducted from the
content of the quadralaieral, and the
remainder will give the true content
of the field.
When there are obstructions within the
jvater, hills, &c., measure the lengths and positions of the four- sided
figure atccl^ taking care to measure the offsets from the different
corners as you go along.
PROBLEM IX.
To survey any puce of land by two stations.
Choose two stations, from which all the corners of the ground can
be seen, if possible ; measure the distance between the stations ; at
each station take the angles formed by every object, from the station
line, or distance. Then, the station line and these different angles
being laid down from a regular scale, and the external points of in-
tersection connected, the connecting lines will give the boundary.
The two stations may be taken within the bounds, in one of the
«;ides, or without the bounds of the ground to be surveyed.
Let m and n be two stations, from which all the marks A,* B, C
figure, such as wood,
iG2
LAND-SURVEYING.
&c., can be seen, plant the in-
strument at m, and by it mea-
sure the angles kmn^ ^mn^
C m 72, &c. Next measure m w,
and planting the instrument at
71, measure the angles knm^
BwTTi, Cwm, &c. These ob-
servations being planned, the
lines joining tlie points of ex-
ternal intersection will give a
true map of the ground. The
method of finding the content
will be shewn further on.
The principal objects on the ground may be delineated on the map,
by measuring the angles at each station, which every object makes
with the station line m n. When all the objects to be surveyed cannot
be seen from two stations, then three or four may be used, or as
many as may be found necessary ; taking care to measure the dis-
tance from one station to another ; placing the instrument at every
station, and observing the angles formed by all the visible objects
with the respective station line ; then the intersection of the lines
forming these respective angles will give the positions of all the re-
markable objects thus observed.
In this manner may very extensive surveys be taken; and the
positions of hills, rivers, coasts, &c., ascertained.
PROBLEM X.
To survey a large estate.
The following method of surveying a large estate was first given
by Emerson, in his '' Surveying," page 47. It has been followed by
Hutton and Keith.
When the estate is very large, and contains a great number of
fields, it cannot be accurately surveyed and planned by measuring
each field separately, and then adding all the separate results together;
nor by taking all the angles, and measuring the boundaries that
enclose it. For in these cases the small errors will be so multiplied as
to render it very much distorted.
1. Walk over the estate two or three times, in order to get a per-
Tect idea of its figure. And to help your memory, make a rough
draft of it on paper, inserting the names of the different fields within
it, and noting down the principal objects.
2. Choose two or more elevated places in the estate for your
stations, from which you can see all the principal parts of it ; and
let these stations be as far distant from each otlier as possible, as the
fewer stations you have to command the whole, the more exact the
work will be.
In selecting the stations, care should be taken that the lines which
connect them may run along the boundaries of the estate, or some of
the hedges, to which ofi^sets may be taken when necessary.
8. Take such angles, between the stations, as you think necossary.
LAND-SURVEYING. 1 63
and measure the distance from station to station, always m a right
line ; these thinj]js must be done till you get as many lines and angles
as are sufficient for determining all the station points. In measuring
any of these station distances, mark accurately where these lines
meet with any hedges, ditclies, roads, lanes, paths, rivulets, &c.,
and where any remarkable object is placed, by measuring its distance
from the station line ; and where a perpendicular from it cuts that
line ; and always mind, in any of these observations, that you be in
a right line, which you may easily know by taking a back-sight and
fore-sight, along the station line. In going along any main station
line, take offsets to the ends of ail hedges, and to any pond, house,
mill, bridge, &c., omitting nothing that is remarkable. All these
things must be noted down; for these are the data by which the
places of such objects are to be determined on the plnn.
Be careful to set up marks at the intersections of all hedges with
the station line, that you may know where to measure from when you
come to survey the particular fields that are crossed by this line.
These fields must be measured as soon as you have completed your
station-line, whilst they are fresh in your memory. In this manner
ail the station lines must be measured, and the situations of all
adjacent objects determined. It will be proper to lay down the work
on paper every night, that you may see how you go on.
4. With respect to the internal parts of the estate, they must be
determined by new station lines ; for, after the main stations are
determined, and everything adjoining to them, then the estate must
be subdivided into two or three parts by new station lines : taking
the inner stations at proper places, where you can have tlie best view.
Measure these station lines as you did the first, and all their intersec-
tions with hedges, ditches, roads, &c., also take offsets to the bends
of hedges, and to such objects as appear near these lines. Then
proceed to survey the adjoining fields, by taking the angles which
the sides make with the station line at the intersections, and measur-
ing the distances to each corner from these intersections ; for every
station line will be a basis to all future operations, the situation of
every object being entirely dependent on them ; and therefore the^-
should be taken of as great length as possible ; and it is best io<
them to run along some of the hedges or boundaries of one or raorj
fields, or to pass through some of their angles.
All things being determined for these stations, you must take more
inner stations, and continue to divide and subdivide, till at last you
come to single fields ; repeating the same work for the inner stations
as for the outer ones, till the whole is finished. The oftener you close
vour work, and the fewer lines you make use of, the less you will be
liable to error.
5. An estate may be so situated that the whole cannot be surveyed
together, because one part of the estate may not be seen from another.
In this case you may divide it into three or four parts, and survey
these parts separately, as if they were lands, belonging to diff"erent
persons, and at last join them together.
6. As it is necessary to protract or lay down the work as you
proceed in it, you must have a scale of due length to do it by» Tc
164
LAND-SURVEYING.
^et such a scale, measure the wliole length of the estate in chams ;
then consider how many inches long the map is to be : and from t hese
you will know how many chains you must have in an inch ; then
make your scale accordingly, or choose one already made.
7. The trees in every hedge-row may be placed in their proper
pituation, which is soon done by the plane table ; but may be done
by the eye without an instrument ; and being thus taken by guess in
a rough draft, they wili be exact enough, being only to look at ;
except it be such as are at any remarkable places, as at the ends of
hedges, at stiles, gates, &c., and these must be measured or taken
with the plane table, or some other instrument. But all this need
not be^ done till the draft is finished. And observe, in all hedges,
what side the gutter or ditch is on, and to whom the fence belongs.
PROBLEM XT.
To survey a town or city.
To survey a town or city, it wdll be proper to hax^e an instrument
for taking angles, such as a theodolite or plane table ; the latter is a
very convenient instrument, because the minute parts may be drawn
upon it on the spot. A chain of 50 feet long, divided into 60 links,
will be more convenient than the common surveying chain, and an
offset staff of 10 feet long will be very useful. Begin at the meeting
of two or more of the principal streets, through which you can have
the longest prospects, to get the longest station lines. There having
fixed the instruments, draw lines of direction along these streets,
using two men as marks, or poles set in wooden pedestals, or perhaps
some remarkable places in the houses at the farther ends, as windows,
doors, corners, &c. Measure these lines with the chain, taking
offsets with the staff, at allcomers of streets, bendings, or windings,
and to all remarkable objects, as churches, markets, halls, colleges,
eminent buildings, &c. Then remove the instrument to another
station, along one of these lines, and there repeat the same process
as before. And so continue until the w-hole is finished.
Thus, fix the instrament at A, and draw lines in the directions of
all the streets meeting there ; then measure A C, noting the street
at X. At the second station C, draw the directions of all the streets
meeting there ; measure from C to D, noting the place of the street
K, as you pass by it. At
the third station D, take
the direction of all the
streets meeting there, and
measure DS, noting the
cross street at T. Proceed
in like manner through
all the principal streets;
after which proceed to the
smaller and intermediate
Ftreets ; and last of all to
the lanes, alleys, courts,
yards, and every other place which i| may be thought proper to
represent in the plan.
LAND-SURVEYING.
IfiS
PROBLEM XII.
To compute the content of any survfi.y,
1. In small and separate pieces, the method generally employed is,
to compute their contents from the measm^es of the lines taken in
purveying them, without drawing any correct map of them : rulet
for this purpose have been given in the preceding part of the work.
But in large pieces, and whole estates, consisting of a great number
of fields, the usual method is, to make an unfinished but correct plan
of tlie whole, and from this plan, the boundaries of which include the
whole estate, comjjute the contents quite independent of the measure,
of the lines and angles that were taken in surveying. Divide the
plan of the survey into triangles and trapeziums, by drawing new lines
through it : measure all the bases and perpendiculars of all these new
figures, by means of the scale from which the plan was drawn, and
from these dimensions compute the contents, whether triangles or
trapeziums, by the proper rules for finding the areas of such figures.
The chief difficulty in computing consists in finding the contents of
land bounded by curved or very irregular lines, or in reducing ' such
crooked sides or boundaries to straight lines, that shall enclose an
equal area with those crooked sides, and so obtain the area of the
curved figure by means of the right-lined one, which in general wilJ
be a trapezium.
The reduction of crooked sidee to straight ones is easily performed,
thus : —
Apply a hors€-hair or silk thread across th^ crooked sides in such
a manner, that the small parts cut off from the crooked figure by it,
may be equal to those taken in. A iittle practice will enable you to
exclude exactly as much as you include ; then, with a pencil draw
a line along the thread, or horse-hair. Do the same by the other
^ides of the figure, and you will thus have the figure reduced to a
straight-sided figure equal to the curved one : the content of which,
1 eing computed as before directed, will be the content of the curved
figure proposed.
The best way of using the thread or horse-hair is, to string a small
slender bow with it, either of whalebone or wire, which will keep it
stretched.
If it were required to find the contents of the following crooked-
sided figure ; draw the four dotted straight lines A B, B C, CD, and
D A, excluding as much
from the survey as is _^ y" \ B ,
taken in by the straight
lines; by which the
crooked figure is re-
duced to a right-lined
one, both equal in area.
Then draw the diagonal
B D, which being mea-
sured by a proper scale,
and multiplied by half
the sum. of the perpea-
266 LAND-SURVEYINO.
diculars let fall from A and C upon BD (measured on the same
scale), will odve the area required.
Many other methods might have been given for computing the
contents of a survey, but they are omitted, the above being, perhaps,
tho most expeditioijs.
}^
MISCELLANEOUS PROBLEMS. 167
MISCELLANEOUS PEOBLEMS.
1. The three sides of a triangle are 12, 20, and 28; what is the
area ? ^ Ans. 60 V 3.
2. Find the difference between the area of a triangle whose sides
are 3, 4, and 5 feet, and the area of an equilateral triangle having
an equal perimeter. Ans. '928 of a square foot.
3. There is a segment of a sphere, the diameter of whose base is
24 inches, and its altitude 10 inches ; required its solidity ?
Ans. 2785-552 inches.
4. There is a bushel in the form of a cylinder, whose depth is 8»
inches, and breadth 18^ inches ; required to determine the breadth
of another cylindrical vessel of the same capacity as the former,
whose depth is only 7j inches? Ans. 19-107 inches.
5. A ladder, 40 feet long, may be so planted that it shall reach a
window 33 feet from the ground on one side of the street ; and by
only turning it over, without moving the foot out of its place, it will
do the same by a window 21 feet high on the other side ; what ia
the breadth of the street? j.'Lns. 56 feet 7| inches.
6. In turning a one-horse chaise within a ring of a certain
diameter, it was observed that the outer wheel made two turns while
the inner made but one ; the wheels were both 4 feet high ; and
supposing them fixed at the statutable distance of 5 feet asunder on
the axle-tree, what was the circumference of the track described by
the outer wheel ? Ans. 63 feet, nearly.
7. A cable which is 3 feet long, and 9 inches in compass, weigiis;
22 pounds ; what will a fathom of that cable weigh which measures
a foot about? Ans. 78f pounds.
8. How many solid cubes, a side of which equals 4 inches, may
be cut out of a large cube, whose side is 8 inches? Ans. 8.
9. Determine the areas of an equilateral triangle, a square, a
hexagon, the perimeter of each being 40 feet.
Ans. 76-980035- 100- -115-47.
10. A person wants a cylindrical vessel 3 feet deep, that shall
contain twice as much as another cylindrical vessel whose diameter
is 3 J feet, and altitude 5 feet ; find the diameter of the required
vessel. Ans. 6*39 feet.
11. Three persons having bought a conical sugar-loaf, wish to
divide it into three equal parts by sections parallel to the base ; it is
required to find the altitude of each person's share, the altitude of
the loaf being 20 inches.
Ans. Altitude of the upper part= 13-867 of the middle part=s
8*604, of the lower part = 2-528 inches.
1C8 MISCELL.ANEOUS PROBLEMS.
12. There is a frustum of a pyramid, whose bases are regular
octagons; each side of the greater base is 21 inches, and each side
of the less base 9 inches, and its perpendicular length 1 5 feet ; how
many solid feet are contained in it? Aiis. 119-2 feet.
13. Requiring to find the height of a May-pole, I procured a staff
5 feet in length, and placing it in the sunshine, perpendicular to the
horizon, I found its shadow to be 4*1 feet. Next I measured the
shadow of the May-pole, whicli I found to be 65 feet ; from this data
the heiglit of the pole is required. Ans. 79*26 feet.
14. Given two sides of an obtuse-angled triangle, whicli are 20
and 40 poles ; required the third side, that the triangle may contain
just an acre of land? Ans. 58-876 or 23-099.
15. A circular fish-pond is to be made in a garden, that shall take
np just half an acre; what must be the length of the cjiord that
strikes the circle ? Ans. 27f yards.
16. A gentleman has a garden 100 feet long, and 80 feet broad •,
now a gravel walk is to be made of an equal width all round it ;
what must the breadth of the walk be, to take up just half the ground?
Ans. 12-9846 feet.
1 7. A silver cup, in form of a frustum of a cone, whose top
diameter is 3 inches, its bottom diameter 4, and its altitude 6 inches,
being filled with liquor, a person drank out of it till he could see the
middle of the bottom ; it is required to find how much he drank?
Ans. -152127 ale gallons.
18. I have a right cone, which cost me £5, 135. 7(1.^ at 10^. u
cubic foot, the diameter of its base being to its altitude as 5 to 8 ;
and would have its convex surface divided in the same ratio, by a
plane parallel to the base ; the upper part to be the greater ; required
the slant height of each part ?
A (3-9506486, the slant height of the upper part.
^ns. -11.0854612, the slant height of the under part.
19. How many acres of the earth's surface may be seen from the
top of a steeple whose height is 400 feet, the earth being supposed to
be a perfect sphere, whose circumference is 25000 miles?
Ans. 12120981-338267112 acres.
20. Two boys meeting at a farm-house, had a tankard of milk set
down to them ; the one being very thirsty drank till he could see the
centre of the bottom of the tankard ; the other drank the rest. Now,
if we suppose that the milk cost 4id, and the tankard measured
.4 inches diameter at the top and bottom, and 6 inches in depth ; it
is required to know what each boy had to pay, proportionable to th-^!
quantity of milk he drank.
. (14-1 802815 farthings, for the first,
^^^- \ 3-8197185 farthings, for the second.
21. If the linear side of a certain cube be increased one inch, the
surface of the cube will be increased 246 square inches : determine
the side of the cube. Ans, 20 inches.
22*. If from a piece of tin, in the form of a sector of a circle, whose
radius is 30 inches, and the length of its arc 36 inches, be cut
another sector whose radius is 20 inches ; and if then the remaining
frustum be rolled up so' as to form the frustum of a cone; it is
MISCELLANEOUS PROBLEMS. 109
required to find its content, supposing one- eighth of an inch to be
allowed off its slant height for the bottom, and the same allowance
of the circumference, of both top and bottom, for what the sides fold
over each other, in order to their being soldered together.
Ans. 685-3263 cubic inches.
23. Three men bought a grinding-stone of 40 inches diameter,
which cost 20.9., of which sum the first man paid 9^., the second 6s. ^
and the third 55. ; how much of the stone must each man grind down,
proportionably to the money he paid ?
Ans. The first man must grind down 5*167603 inches of the
radius ; the second 4*832397 inches, and the third 10 inches.
24. There is a frustum of a cone, whose solid content is 20 feet,
and its length 12 feet ; the greater diameter is to the less as 5 to 2 ;
what are the diameters ?
.^,^ (2-02012 feet.
^"*- t -80804 feet.
25. A farmer borrowed of his neighbour part of a hay-rick, which
measured 6 feet in length, breadth, and thickness ; at the next Lay-
time he paid back two equal cubical pieces, each side of which was
I feet. Has the debt been discharged?
Ans. No ; 88 cubic feet are due.
P6. There is a bow-' in form of the segment of an oblong spheroid,
whose axes are to each other in the proportion of 3 to 4, the depth of
the bovjl one-fourth of the whole transverse axis, and the diameter
of its top 20 inches ; it is required to determine what number of
glasses a company of 10 persons would have in the contents of it,
when filled, using a conical glass, whose depth is 2 inches, and the
diameter of its top an inch and a half.
Ans. 114*0444976 glasses each.
27. If a cubical foot of br^ss were to be drawn into wire, of -^ of
an inch in diameter; it is required to determine the length of the
8aid wire, allowing no loss in the metal. Ans. 55f miles.
28. How many shot are there in an unfinished oblong pile, the
length and breadth of whose base being 48 and 30, and the length
and breadth of the highest course being 24 and 6 ? Ans. 17356.
29. How many shot are there in an unfinished oblong pile of
12 courses ; length and breadth of the top contain 40 and 10 shot
respectively ? Ans. 8606 shot.
30. Of what diameter must the bore of a cannon be cast for a ball
of 24 pounds weight, so that the diameter of the bore may be -j^ of
an inch more than that of the ball? Ans. 5*757098 inches.
31. What is the content of a tree, whose length is 17^ feet, and
which girts in five different places as follows, viz., in the first place
9-43 feet, in the second 7 '92, in the third 6*15, in the fourth 4*74,
and the fiflh 3-16? Ans. 42*5195.
32. What three numbers will express the proportions subsisting
between the solidity of a sphere, that of the circumscribing cylinder,
and circumscribing equilateral cone ? Ans. 4, 6, 9.
■" 33. Given the side of an equilateral triangle 10, it is required to
find the radii of its circumscribing circle. Ans. 5-7786.
34. Given the perpendicular of a plane triangle 300, the sum of
170 MISCELLANEOUS PROBLEMS.
the two sides 1150, and the difference of the segment of the base 495 ;
required the base and the sides? Ans. 945, 375, and 780.
35. A side wall of a house is 30 feet high, and the opposite one 40,
the roof forms a right angle at the top, the lengths of the rafters are
10 feet and 12 ; the end of the shorter is placed on the higher wall,
and vice versa; required the length of the upright which supports
the ridge of the roof, and the breadth of the house ?
Ans. 41-803, length of upright, and 12 feet the breadth of thj
house.
AREAS OF THE SEGMENTS OF A CIRCLE.
171
A TABLE
OF THE AREAS OF THE SEGMENTS OF A CIRCLE,
IVhose diameter is 1, and supposed to he divided into 1000 equal parts.
Height.
Area Seg.
Height.
Area Seg. H
eight.
Area Seg.
Height.
Area Seg.
•001
•000042
•044
•012142
087
•033307
•130
•059999
•002
•000119
•045
•012554 •
088
•033872
•131
•060672
•003
•000219
•046
•012971
089
•034441
•132
•061348
•004
•000337
•047
•013392
090
•035011
•133
•062026
•005
•000470
•048
•013818
091
•035585
•134
•062707
•006
•000618
•049
•014247
092
•036162
•135
•063389
•007
•000779
•050
•014681
093
•036741
•136
•064074
•008
•000951
•051
•015119
094
•037323
•137
•064760
•009
•001135
•052
•015561
095
•037909
•138
•065449
•010
•001329
•053
•016007
096
•038496
•139
•066140
•Oil
•001533
•054
•016457
097
•039087
•140
•066833
•012
•001746
•055
•016911
098
•039680
•141
•067528
•013
•001968
•056
•017369
099
•040276
•142
•068225
•014
•002199
•057
•017831
100
•040875
•148
•068924
•015
•00243S
•058
•018296
101
•041476
•144
•069625
•016
•002685
•059
•018766
102
•042080
•145
•070328
•017
•002940
•060
•019239
103
•042687
•146
•071033
•018
•003202
•061
•019716
104
•043296
•147
•071741
•019
•003471
•062
•020196
105
•043908
•148
•072450
•020
•003748
•063
•020681
•106
•044522
•149
•073161
•021
•004031
•064
•021168
107
•045139
•150
•073874
•022
•004322
•065
•021659
108
•045759
•151
•074589
•023
•004618
•066
022154
109
•046381
•152
•075306
•024
•004921
•067
•022652
110
•047005
•153
•076026
•025
•005230
•068
•023154
111
•047632
•154
•076747
•026
•005546
•069
•023659
•112
•048262
•155
•077469
•027
•005867
•070
•024168
•113
•048894
•156
•078194
•028
•006194
•071
•024680
114
•049528
•157
•078921
•029
•006527
•072
•025195
115
•050165
•158
•079649
•030
•006865
•073
•025714
116
•050804
•159
•080380
•031
•007209
•074
•026236
117
•051446
•160
•081112
•032
•007558
•075
•026761
•118
•052090
•161
•081846
•033
•007913
•076
•027289
119
•052736
•162
•082582
•034
•008273
•077
•027821
•120
•053385
•163
•083320
•035
•008438
•078
•028356
121
•054036
•164
•084059
•036
•009008
•079
•028894
•122
•054689
•165
•084801
•037
•009383
•080
•029435
123
•055345
•166
•085544
•038
•009763
•081
•029979
124
•056003
•167
•086289
•039
•010148
•082
•030526
125
•056663
•168
•087036
•040
•010537
•083
•031076
126
•057326
•169
•087785
•041
•010931
•084
•031629
•127
•057991
•170
•088535
•042
•011330
•085
•032180
128
•058658
•171
•089287
•043
•011734
•086
•032745
•129
•059327
•172
•090041
172
AREAS OF THE SEGMENTS OF A CIRCLE.
Height. 2
\.rea Seg.
Height.
Area Seg.
Height.
Area Seg. H
eight.
Area Seg.
•173
•090797
•224
•131438
•275
•175542
•326
•222277
I'M
•091554
•225
•132272
•276
•176435
327
•223215
175
•092313
•226
•133108
•277
•177330
•328
•2^24154
176
•093074
•227
•133945
•278
•178225
•329
•225093
177
•093836
•228
•134784
•279
•179122
330
•226033
178
•094601
•229
•135624
•280
•1C'0019
331
•226974
179
•095366
•230
•136465
•281
•180918
332
•227915
180
•096134
•231
•137307
•282
•18:81'/ .
333
•228858
•181
•096903
•232
•138150
•283
•182718
334
•229801
182
•097674
•233
•138995
•284
•183619
335
•230745
183
•098447
•234
•139841
•285
'184521
336
•231689
184
•099221
•235
•140688
•286
•185425
337
•232634
185
•099997
•236
•141537
•287
•186329
338
•233580
186
•100774
•237
•142387
•288
•187C34
339
•23:^526
187
•101553
•238
•143238
•289
•188140
340
•235473
188
•102334
•239
•144091
•290
•189047
341
236421
189
•103116
•240
•144944
•291
•189955
342
•237369
190
•103900
•241
•145799
•292
•190864
343
•238318
191
•104685
•242
•146655
•293
•191775
344
•239268
192
•105472
•243
•147512
•294
•192684
345
•240218
193
•106261
•244
•148371
•295
•193596
346
•241169
194
•107051
•245
•149230
•296
•194509
347
•242121
195
•107842
•246
•150091
•297
•195422
348
•243074
•196
•108636
•247
•150953
•298
•196337
349
•244026
•197
•109430
•248
•151816
•299
•197252
350
•244980
•198
•110226
•249
•152680
•300
•198168
351
•245934
•199
•111024
•250
•153546
•301
•199085
352
•246889
•200
•111823
•251
•154412
•302
•200003
353
•247845
•201
•11262J
•252
•155280
•303
•200922
354
•248801
•202
•113426
•253
•156149
•304
•201841
355
•249757
•203
•114230
•254
•157019
•305
•20-2761
356
•250715
•204
•115035
•255
•157890
•306
•203683 -
357
•251673
•205
•115842
•256
•158762
•307
•204605
358
•252631
•206
•116650
•257
•159636
•308
•205527
359
•253590
•207
•117460
•258
•160510
•309
"206451
1)60
•254550
•208
•118271
•259
•161386
•310
207376
361
•255510
•209
•119083
•260
•162^263
•311
•208301
362
•256471
•210
•119897
•261
•163140
•312
•209227
363
•257433
•211
•120712
•262
•164019>
•3i3
;210154
364
•258395
•212
•121529
•263
•164899
•314
/211082
365
•259357
•213
•12-2347
•264
•165780
•315
•212011
366
•260320
•2U
•123167
265
•166663
•316
•212940
367
•261284
•215
•123988
•266
•167546
•317
•213871
368
•262248
•216
•124810
•267
•168430
•318
•214802
369
•263213
•217
•125634
•268
•169315
•319
•215733
370
•264178
•218
•126459
•2q^
•170202
•320
•216666
371
•265144
•219
•127285
•270
•171089
•321
•217599
372
•266111
•220
•128113
•271
•171978
•322
•218533
373
•267"7S
•221
•128942
•272
•172867
•323
•219468
374
•268045
•222
•129773
•273
•173758
•324
•220404
375
•261013
•2-23
•130605
•274
•174649
•325
•221340
•376
•269982
AREAS OF THE SEGMENTS OF A CIRCLE.
173
Height.
Area Seg.
Height.
Area Seg.
Height.
Area Seg.
Height.
Area Seg.
•377
•270951
•408
•301220
•439
•331850
•470
•362717
•378
•271920
•409
•302203
•440
•332843
•471
•363715
•879
•272890
•410
•303187
•441
•333836
•472
•364713
•380
•273861
•411
•304171
•442
•334829
•473
•365712
•381
•274832
•412
•305155
•443
•335822
•474
•366710
•382
•275803
•413
•306140
•444
•336816
•475
•367709
•383
•276775
•414
•307125
•445
•337810
•476
•368708
•384
•277748
•415
•308110
•446
•338804
•477
•369707
•385
•278721
•416
•309095
•447
•339798
•478
•370706
•386
•279694
•417
•310081
•448
•340793
•479
•371705
•387
•280668
•418
•311068
•449
•341787
•480
•372764
•388
•281642
•419
•312054
•450
•342782
•481
•373703
•389
•282617
•420
•313041
•451
•343777
•482
•374702
•390
•283592
•421
•314029
•452
•344772
•483
•375702
•391
•284568
•422
•315016
•453
•345768
•484
•376702
•392
•285544
•423
•316004
•454
•346764
•485
•377701
•393
•286521
•424
•316992
•455
•347759
•486
•378701
•394
•287498
•425
•317981
•456
•348755
•487
•379700
•395
•288476
•426
•318970
•457
•349752
•488
•380700
•396
•289453
•427
•319959
•458
•350748
•489
•381699
•397
290432
•428
•320948
•459
•351745
•490
•382699
•398
•291411
•429
•321938
•460
•352742
•491
•383699
•399
•292390
•430
•322928
•461
•353739
•492
•384699
•400
•293369
•431
•323918
•462
•354736
•493
•385699
•401
•294349
•432
•324909
•463
•355732
•494
•386699
•402
•295330
•433
•325900
•464
•356730
•495
•387699
•403
•296311
•434
•326892
•465
•357727
•496
•388699
•404
•297292
•435
•327882
•466
•358725
•497
•389699
•405
•298273
•436
•328874
•467
•359723
•498
•390699
•406
•299255
•437
•329866
•468
•360721
•499
•391699
•407
•300238
•438
•330858
•469
•361719
•500
392699
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Index, cloth lettered, - - - -\- -^-26
^"^-i?- — —
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THE UNIVERSITY OF CALIFORNIA LIBRARY
HISTORICAL AND CLASSIGASi GEOGRAPHY,
Pocket Atlas op Historical Geography, 16 Maps, Imp. 16mo,
The Crown Atlas of Historical Geography, 16 Maps, with
Lettei^press, by Wm. F. Collier, LL.D., Imperial 16mo, cloth.
Pocket Atlas of Classical Geography, 15 Maps, Imp. 16mo, cl..
The Crown Atlas op Classical Geography, 15 Maps, with
Letterpress, by Leon. Schmitz, LL.D., Imperial 16mo, cloth^
v^ w. r^ SCRIPTURE GEOGRAPHY.
The Atlas op Scripture Geography, 16 Maps, with Questio^g^
on each Map, Stiff Cover, - /^-l
London, Edinburgh, and Herriot Hill Works, Glasgow.
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