weight of one of the same length supported at each end.
If the end of a beam, instead of being only supported, be _fixed_,
its strength will be in the proportion of 3 to 2.
From the foregoing results it will be seen that the strength of a
rectangular beam varies, as the breadth multiplied by the depth
squared, divided by the length, (b × d^2)/l and if the breaking
weight of any material, 1 inch square, and 1 foot long, be found, it
will represent a _constant multiplier_ for the above equation.
Thus the breaking weight of a beam of Riga fir, 1 inch square, and
1 foot long (_vide following_ TABLE), is ·164 of a ton; and to find
the breaking weight of a beam of any other dimensions, the rule is
simply
W = (b d^2)/l × ·164.
_Example._—What will be the breaking weight of a beam of Riga fir, 8
inches broad, 12 inches deep, and 20 feet long?
(8 × 12^2)/20 = 57·6 57·6 × ·164 = 9·44 tons, breaking weight.
_Table of constants_, for beams of different materials, being the
breaking weights of such beams, 1 inch square, and 1 foot long.
Riga fir ·164 of a ton. English oak ·248 of a ton.
Red pine ·199 ” Canadian do. ·261 ”
Pitch pine ·242 ” Dantzic do. ·219 ”
Beech ·231 ” Teak ·366 ”
Elm ·150 ” Cast iron, mean. 1·000 ”
Ash ·301 ” Wrought do. 1·083 ”
From the foregoing rules
Length = (b d^2)/W × constant.
Breadth = (l W)/d^2 × constant.
Depth = √((l W)/b) × constant.
_The practical weight_ that a beam will carry _with safety_,
permanently, should only be taken at one-fourth of the above
computations.
ADHESION OF NAILS, AND SCREWS.
_The percussive force_ required to drive the common sixpenny nail
(73 to the pound) to the depth of an inch and a half into deal, with
a weight of six pounds and a quarter, is four blows, or strokes,
falling freely the space of one foot; and _the steady pressure_
to produce the same effect is four hundred pounds. A _sixpenny
nail_ driven into dry elm to the depth of one inch across the
grain requires a force of 327 pounds to extract it; and the same
nail, driven into the same wood endways, or longitudinally, can be
extracted with a force of 257 pounds.
To extract a sixpenny nail from a depth of one inch out of dry oak
requires 507 pounds, and out of dry beech 667 pounds. A sixpenny nail
driven two inches into dry oak would require a steady force of more
than half a ton to extract it.
_A common screw_ of one-fifth of an inch diameter has an adhesive
force of about three times that of a sixpenny nail.
TRIGONOMETRY.
_Plane trigonometry_ treats of the relations, and calculations of the
sides, and angles of plane triangles.
_The measure of an angle_ is an arc of any circle contained between
the two lines which form that angle, the angular point being the
centre; and it is estimated by the number of degrees contained in
that arc. Hence a right angle being measured by a quadrant, or
quarter of a circle, is an angle of 90 degrees. The sum of the three
angles of every triangle is equal to 180 degrees, or two right
angles; therefore, in a right-angled triangle, taking one of the
acute angles from 90 degrees, leaves the other acute angle; and the
sum of the two angles in any triangle, taken from 180 degrees, leaves
the third angle; or one angle being taken from 180 degrees leaves the
sum of the other two angles.
_Definitions._
_The sine of an arc_ is the line drawn from one extremity of the arc
perpendicular to the diameter of the circle which passes through the
other extremity.
_The supplement of an arc_ is the difference, in degrees, between the
arc, and a semicircle, or 180 degrees.
_The complement of an arc_ is the difference, in degrees, between the
arc, and a quadrant, or 90 degrees.
_The tangent of an arc_ is a line touching the circle in one
extremity of that arc, continued from thence to meet a line drawn
from the centre through the other extremity; which last line is
called _the secant_ of the same arc.
The _cosine_, _cotangent_, and _cosecant_ of an arc are the sine,
tangent, and secant of the complement of that arc, _the co_ being
only a contraction of the word complement.
The sine, tangent, or secant of an angle is the sine, tangent, or
secant of the arc by which the angle is measured, or of the degrees,
&c., in the same arc, or angle. _Vide also Definitions_, PRACTICAL
GEOMETRY.
_There are two Methods of resolving triangles_, or the cases of
trigonometry—viz., _Construction_, and _Computation_.
_1st method._—The triangle is constructed by making the sides from a
scale of equal parts, and laying down the angles from the protractor.
Then, by measuring the unknown parts by the same scale, the solution
will be obtained.
_2nd method._—Having stated the terms of the proposition, resolve
it like any other proportion, in which a fourth term is to be found
from three given terms, by multiplying the second and third terms
together, and dividing the product by the first.
_Note._—Every triangle has six parts—viz., three sides, and three
angles; and, in every case in trigonometry, there must be given three
of these parts to find the other three. Also of the three parts that
are given, one of them at least must be a side; because, with the
same angles, the sides may be greater, or less, in any proportion.
_Computation._
_Case 1._—_When a side and its opposite angle are two of the given
parts._
The sides of any triangle having the same proportion to each other,
as the sines of their opposite angles; then—
As any one side, is to the sine of its opposite angle; so is any
other side, to the sine of its opposite angle.
_To find an angle_, begin the proportion with a side, opposite to a
given angle; and, _to find a side_, begin with an angle opposite to a
given side.
_Case 2._—_When the three sides of a triangle are given, to find the
angles._
Let fall a perpendicular from the greatest angle, on the opposite
side, or base, dividing it into two segments; and the whole triangle
into two right-angled triangles: then the proportion will be—
As the base or sum of the segments, is to the sum of the other two
sides; so is the difference of those sides, to the difference of the
segment of the bases: then add half the difference of the segments
to the half sum, or the half base, for the greater segment; and
subtract the same for the less segment. Hence, in each of the two
right-angled triangles, there will be known two sides, and the right
angle opposite to one of them, consequently the other angle will be
found by the method in _Case 1_.
USEFUL THEOREMS, AND COROLLARIES.
1. When one line meets another, the angles, which it makes on the
same side of the other, are together equal to two right angles.
2. All the angles, which can be made at any point (by any number
of lines), on the same side of a right line, are, when taken all
together, equal to two right angles: and, as all the angles that
can be made, on the other side of the line, are also equal to two
right angles; therefore all the angles that can be made quite round
a point, by any number of lines, are equal to four right angles.
Hence also the whole circumference of a circle, being the sum of all
the angles that can be made about the centre, is the measure of four
right angles.
3. When two lines intersect each other, the opposite angles are equal.
4. When one side of a triangle is produced, or extended, the outward
angle is equal to the sum of the two inward opposite angles.
5. In any triangle, the sum of all the three angles is equal to two
right angles (180°). Hence, if one angle of a triangle be a right
angle, the sum of the other two angles will be equal to a right angle
(90°).
6. In any quadrilateral, the sum of all the four inward angles is
equal to four right angles.
7. In any right-angled triangle, the square of the hypothenuse (or
side opposite to the right angle) is equal to the sum of the squares
of the other two sides. Therefore, _to find the hypothenuse_, add
together the squares of the other two sides, and extract the square
root of that sum: and _to find one of the other sides_, subtract from
the square of the hypothenuse the square of the other given side, and
extract the square root of the remainder for the side required.
Or hypothenuse = √(base^2 + perpendicular^2)
Base = √((hypoth. + perpend.) × (hypoth. - perpend.))
Perpendicular = √((hypoth. + base) × (hypoth. - base.))
TRIGONOMETRY, WITHOUT LOGARITHMS.[49]
“In all the more elaborate, and refined operations of trigonometry,
it is not only desirable, but necessary to employ some of the larger
logarithmic tables, both to save time, and to ensure the requisite
accuracy in the results. But in the more ordinary operations, as
in those of common surveying, ascertaining inaccessible heights,
and distances, reconnoitring, &c., where it is not very usual to
measure a distance nearer than within about its thousandth part, or
to ascertain an angle nearer than within two or three minutes, it
is quite a useless labour to aim at greater accuracy in a numerical
result. Why compute the length of a line to the fourth, or fifth
place of decimals, when it must depend upon another line, whose
accuracy cannot be ensured beyond the unit’s place? Or, why compute
an angle to seconds, when the instrument employed does not ensure
the angles in the data beyond the nearest minute? In the following
Table are brought together the _natural sines, and cosines_, to every
degree in the quadrant, and this table will be found sufficiently
extensive, and correct for the various practical purposes above
alluded to. The requisite proportions must, it is true, be worked
by multiplication, and division, instead of by logarithms. Yet this
by no means involves such a disadvantage as might seem, at first
sight. For when the measured lines are expressed by three, or at
most, four figures, the multiplications, and divisions are performed
nearly as quick, and in some cases quicker, than by logarithms.
Then as to accuracy, even in cases where the computer will have to
take proportional parts for the minutes of a degree, the result may
usually, if not always, be relied upon to within about a minute.”
TRIGONOMETRIC RATIOS.
_Natural sines, and cosines to every degree in the quadrant, radius
being 1·000000._
+----+--------+--------+----++----+--------+--------+----+
|Deg.| Sines. |Cosines.| ||Deg.| Sines. |Cosines.| |
+----+--------+--------+----++----+--------+--------+----+
| 0 | ·00000 |1·00000 | 90 || | | | |
| 1 | ·01745 | ·99985 | 89 || 26 | ·43837 | ·89879 | 64 |
| 2 | ·03490 | ·99939 | 88 || 27 | ·45399 | ·89101 | 63 |
| 3 | ·05234 | ·99863 | 87 || 28 | ·46947 | ·88295 | 62 |
| 4 | ·06976 | ·99756 | 86 || 29 | ·48481 | ·87462 | 61 |
| 5 | ·08716 | ·99619 | 85 || 30 | ·50000 | ·86603 | 60 |
+----+----------------------------+--------+--------+----+
| 6 | ·10453 | ·99452 | 84 || 31 | ·51504 | ·85717 | 59 |
| 7 | ·12187 | ·99255 | 83 || 32 | ·52992 | ·84805 | 58 |
| 8 | ·13917 | ·99027 | 82 || 33 | ·54464 | ·83867 | 57 |
| 9 | ·15643 | ·98769 | 81 || 34 | ·55919 | ·82904 | 56 |
| 10 | ·17365 | ·98481 | 80 || 35 | ·57358 | ·81915 | 55 |
+----+--------+--------+----++----+--------+--------+----+
| 11 | ·19081 | ·98163 | 79 || 36 | ·58778 | ·80902 | 54 |
| 12 | ·20791 | ·97815 | 78 || 37 | ·60181 | ·79863 | 53 |
| 13 | ·22495 | ·97437 | 77 || 38 | ·61566 | ·78801 | 52 |
| 14 | ·24192 | ·97030 | 76 || 39 | ·62932 | ·77715 | 51 |
| 15 | ·25882 | ·96593 | 75 || 40 | ·64279 | ·76604 | 50 |
+----+--------+--------+----++----+--------+--------+----+
| 16 | ·27564 | ·96126 | 74 || 41 | ·65606 | ·75471 | 49 |
| 17 | ·29237 | ·95630 | 73 || 42 | ·66913 | ·74314 | 48 |
| 18 | ·30902 | ·95106 | 72 || 43 | ·68200 | ·73135 | 47 |
| 19 | ·32557 | ·94552 | 71 || 44 | ·69466 | ·71934 | 46 |
| 20 | ·34202 | ·93969 | 70 || 45 | ·70711 | ·70711 | 45 |
+----+--------+--------+----++----+--------+--------+----+
| 21 | ·35837 | ·93358 | 69 || | | | |
| 22 | ·37461 | ·92718 | 68 || | | | |
| 23 | ·39073 | ·92050 | 67 || | | | |
| 24 | ·40674 | ·91355 | 66 || | | | |
| 25 | ·42262 | ·90631 | 65 || | | | |
+----+--------+--------+----++----+--------+--------+----+
| |Cosines.| Sines. |Deg.|| |Cosines.| Sines. |Deg.|
+----+--------+--------+----++----+--------+--------+----+
“The preceding table is so arranged that for angles not exceeding
45 degrees, the sine, and cosine for any number of degrees will be
found opposite to the proposed number in the left hand column, and in
the column under the appropriate word. When the number of degrees in
the arc, or angle, exceeds 45 degrees, that number must be found in
the right hand column, and opposite to it in the column indicated by
the appropriate word at the bottom of the table. Thus, the sine, and
cosine of 36 degrees are ·58778 and ·80902 respectively, the radius
of the table being unity, or 1. The taking of proportional parts
for minutes can only be done correctly in those parts of the table
where the differences between the successive sines, &c., run pretty
uniformly. Suppose we want the natural sine of 20° 16′. The sine of
21 degrees is ·35837, that of 20 degrees is ·34202; their difference
is ·1635. This divided by 60 gives 27·25 for the proportional part
due to 1 minute, and that again multiplied by 16 gives 436 for the
proportional part for 16 minutes. Hence the sum of ·34202 and 436,
or ·34638, is very nearly the sine of 20° 16′. But the operation may
often be contracted by recollecting that 10 minutes are ⅙, 15 minutes
are ¼, 40 minutes are ⅔ of a degree, and so on. Observe, also, that
for cosines the results of the operations for proportional parts
are to be _deducted_ from the value of the required trigonometrical
quantity in the preceding degree.”
APPLICATION OF TRIGONOMETRY, WITHOUT LOGARITHMS, to the determination
of Heights, and Distances.
_Example 1._—Having measured a distance of 200 feet in a direct
horizontal line from the bottom of a steeple, the angle of elevation
of its top, taken at that distance, was found to be 47° 30′, from
hence it is required to find the height of the steeple?
By deducting 47° 30′ from 90°, the angle opposite the given side will
be found (42° 30′).
Then by _Case 1_. TRIGONOMETRY:—
As sine ∠ 42° 30′ : 200 :: sine ∠ 47° 30′:
Or ·67556 : 200 :: ·73723 : 208·2, &c., height required.
_By construction_—
The triangle is constructed by making the side from a scale of equal
parts, and laying down the angles from the protractor. Then by
measuring the unknown parts by the same scale, the solution will be
obtained.
_Example 2._—Being on the side of a river, and requiring the distance
to a house on the other side, 200 yards were measured in a straight
line by the side of the river, and at each end of this base line the
angles with the house were 68° 2′, and 73° 15′—required the distance
from each end of the base line to the house?
The sum of the given angles (68° 2′ + 73° 15′) subtracted from 180°
will give the third angle (38° 43′).
Then by _Case 1_. TRIGONOMETRY:—
As sine ∠ 38° 43′ : 200 :: sine ∠ 68° 2′
·62544 : 200 :: ·92739 : 296·5 first distance required.
As sine ∠ 38° 43′ : 200 :: sine ∠ 73° 15′
·62544 : 200 :: ·95753 : 306·1 second distance required.
Similarly to the preceding examples, HEIGHTS, AND DISTANCES may
be rapidly (and for military purposes, sufficiently accurately)
computed in the field, by means of the foregoing trigonometrical
table, if proper attention is paid to the principles by which the
unknown angles of triangles may be ascertained: a base line, and
requisite angle, or angles, having been given.
It will, however, be necessary to use advantageously the methods in
Cases 1, 2 (_vide_ Trigonometry), and also the properties in the
subsequent theorems, and corollaries.[50]
TABLE,
_Showing the reduction in feet, and decimals upon 100 feet, for the
following angles of elevation, and depression._
+--------+----------+--------+----------+--------+----------+
| Angle. |Reduction.| Angle. |Reduction.| Angle. |Reduction.|
+--------+----------+--------+----------+--------+----------+
| ° ′ | | ° ′ | | ° ′ | |
| 3 0 | ·14 | 9 0 | 1·22 | 15 0 | 3·40 |
| | | 9 30 | 1·38 | 15 30 | 3·64 |
| 4 0 | ·25 | 10 0 | 1·52 | 16 0 | 3·88 |
| | | 10 30 | 1·68 | 16 30 | 4·12 |
| 5 0 | ·38 | 11 0 | 1·84 | 17 0 | 4·37 |
| | | 11 30 | 2·01 | 17 30 | 4·63 |
| 6 0 | ·55 | 12 0 | 2·19 | 18 0 | 4·90 |
| 6 30 | ·65 | 12 30 | 2·37 | 18 30 | 5·17 |
| 7 0 | ·76 | 13 0 | 2·56 | 19 0 | 5·44 |
| 7 30 | ·86 | 13 30 | 2·77 | 19 30 | 5·74 |
| 8 0 | .98 | 14 0 | 2·97 | 20 0 | 6·08 |
| 8 30 | 1·10 | 14 30 | 3·18 | 20 30 | 6·33 |
+--------+----------+--------+----------+--------+----------+
The reduction for 100 feet (from the above table) multiplied by the
number of times 100 feet measured, will give the quantity to be
subtracted from the measured length of an inclination, to reduce it
to a horizontal position.
[Illustration: _Surveying, and Reconnoitring._
_Fig. 1._ _Fig. 2._ _Fig. 3._
_Fig. 4._ _Fig. 5._ _Fig. 6._
_J. W. Lowry, sc._]
TABLE,
_showing the rate of inclination of inclined planes, for the
following angles of elevation_.
+--------+--------+--------+--------+--------+--------+
| Angle. | One in | Angle. | One in | Angle. | One in |
+--------+--------+--------+--------+--------+--------+
| ° ′ | | ° ′ | | ° ′ | |
| 0 15 | 228 | 3 30 | 17 | 7 0 | 8 |
| 0 30 | 114 | 3 45 | 16 | 7 30 | 7¼ |
| 0 45 | 76 | 4 0 | 15 | 8 0 | 7 |
| 1 0 | 56 | 4 15 | 14 | 9 0 | 6½ |
| 1 15 | 46 | 4 30 | 13 | 10 0 | 6 |
| 1 30 | 38 | 4 45 | 12 | 11 0 | 5¾ |
| 1 45 | 32 | 5 0 | 11½ | 12 0 | 5½ |
| 2 0 | 28 | 5 15 | 11 | 13 0 | 5 |
| 2 15 | 26 | 5 30 | 10½ | 14 0 | 4½ |
| 2 30 | 23 | 5 45 | 10 | 15 0 | 4 |
| 2 45 | 21 | 6 0 | 9½ | 16 0 | 3¾ |
| 3 0 | 19 | 6 30 | 9 | 17 0 | 3½ |
| 3 15 | 18 | 6 45 | 8½ | 18 0 | 3¼ |
+--------+--------+--------+--------+--------+--------+
SURVEYING, AND RECONNOITRING.
HEIGHTS, AND DISTANCES.
The accurate determination of heights, and distances of objects being
required in various military operations, especially for the position
of batteries, the following methods for their attainment will be
found useful when the requisite instruments are at hand; by frequent
practice, the eye should, however, be enabled to determine, _nearly_,
either the height of, or distance from any object.
HEIGHTS.
1.—BY MEANS OF A “POCKET SEXTANT,”
_to ascertain the height of an object_.
When the sextant is used for taking the height of objects, it is to
be held vertically, and the quicksilvered part of the horizon glass
will be on the left hand of the observer, or on the left part of
the transparent glass. Altitudes are measured in the same manner as
horizontal angles, for if we conceive the horizontal triangle A B
C (_vide Plate 2, Fig. 2_), to be raised on its base A C with the
angle C next to the observer, then the perpendicular A B becomes
the height of the object B; and supposing the object to stand on a
horizontal plane, then the ground and the object form the right angle
at A; therefore, _if the object is accessible_, the sextant need only
be set at any of the angles mentioned for distances (_vide Art._
DISTANCES), and walking backward on the line A C until the top of
the object is brought down to the height of the observer’s eye from
the ground, then the distance from where the observer stands to the
object will be in the same proportion to its height as the base was
to the distance. Then add the height of the eye from the ground, and
the height of the object will be ascertained. If the object is not
accessible, the angle must be taken, and calculated by trigonometry.
2.—BY MEANS OF A PORTABLE BAROMETER, AND THERMOMETER,
_to ascertain the height of an object_.
Observe the altitude (B) of the mercurial column in inches, tenths,
and hundredths, at the bottom of the hill, or other object, the
height of which is required.
Observe, also, the altitude (b) of the mercurial column at the top
of the object. Observe the temperature on Fahrenheit’s thermometer
at the times of the two barometrical observations, and take the mean
between them. Then 55000 × (B - b)/(B × b) = the height of the hill
in feet, for the temperature of 55 degrees on Fahrenheit. Add 1/440
of this result for every degree which the mean temperature exceeds 55
degrees, and subtract as much for every degree below 55 degrees. This
will be a good approximation when the height of the hill is below
2000 feet.
3.—BY MEANS OF THE RECONNOITRING PROTRACTOR,[51]
_to measure the height of an inaccessible object_.
[_Plate_, SURVEYING, AND RECONNOITRING, _Fig. 1_.]
Place yourself at a convenient distance from the object whose height
is required, taking care to have a good base line to the second
station. Hold the protractor vertically, with a steady hand, the tube
side uppermost, and bring the top of the object in a line with the
centre of the tube. Allow the arm (or index) to vibrate freely, and,
when steady, note the angular height of the object (shown by the edge
of the index on the marginal scale of degrees). By the aid of points
taken through the tube, or by pickets, then pace, or measure a base
in a direct line from the object; and, when arrived at the second
station, again note the angular height of the object.
_Construction_—
Set off the angles, and draw the respective lines, which, by their
intersection, will determine the height of the perpendicular, to
which the height of the protractor above the ground must be added
for the altitude of the object. By using the scale of the measured
base line, the height required will be ascertained, or it may be
calculated by “TRIGONOMETRY, WITHOUT LOGARITHMS.”—_Page 303._
_To measure the height of an accessible object._
[_Plate_, SURVEYING, AND RECONNOITRING, _Fig. 2_.]
At an appropriate distance from the object, take its angular height
and measure the distance to its base.
_Construction_—
Draw a line representing this distance, at one end of which
draw another line at the angle found, and at the other erect a
perpendicular; the intersection of these lines will determine the
altitude of the object.
_To measure the vertical height of a hill, or mountain._
[_Fig. 3, Plate_, SURVEYING, AND RECONNOITRING.]
From a station a short distance from the hill, take, and note down
its angular height; then select a rear position for a base line,
using the tube of the protractor to insure a straight direction;
proceed to the requisite distance on the base, and again note the
altitude of the hill.
_Construction_—
The intersection of lines drawn from each end of the base line, at
the angles found, will determine the altitude; the perpendicular
height of which, added to that of the protractor above the ground,
will give the altitude required.
_To measure the altitude of a tower, &c., on a height._
[_Fig. 4, Plate_, SURVEYING, AND RECONNOITRING.]
From the first station, near the base, take the altitude of the hill,
and also that of the tower above it, and note down these angles;
proceed to another station in a straight line with the former one,
measuring its length, and again observe the angular height of the
hill, and also that of the top of the tower.
Similarly to the previously described mode, ascertain, first, the
height of the hill; second, the height of the hill, and tower; deduct
the first calculation from the second, which will leave the height of
the tower.
In all the foregoing cases the heights may be correctly ascertained
by trigonometrical calculations (_vide_ TRIGONOMETRY, WITHOUT
LOGARITHMS, _page 303_).
4.—BY THE SHADOW OF THE OBJECT,
_to ascertain the height_.
Set up vertically a staff of known length, and measure the length of
its shadow upon a horizontal, or other plane; measure also the length
of the shadow of the object of which the altitude is required. Then,
by the property of similar triangles,
As the length of the shadow of the staff
is to the altitude of the staff,
so is the length of the shadow of the object
to the altitude of the object.
5.—WHEN THERE IS NO SHADOW,
_to ascertain the height_.
Place a staff (equal in length to the height of the observer’s eye)
vertically at such a distance from the foot of the required altitude,
that the observer, having laid himself upon his back, with his feet
against the bottom of the stick, may see the top of the staff, and
object in the same line. Then, by similar triangles, the height may
be readily ascertained.
6.—BY MEANS OF THE TANGENT SCALE OF A GUN,
_to ascertain the height of an object, the distance being known_.
Lay the gun for the top of the object the height of which is
required, then raise the tangent scale until the top of it, and the
notch on the muzzle are in line with the bottom of the object: then,
by similar triangles,
As the length of the gun
is to the length of the raised part of the tangent scale,
so is the distance from the gun to the object,
to the height required.
[Illustration: _Plate 2._
HEIGHTS.
_Fig. 1._ _Fig. 2._
DISTANCES.
_Fig. 3._ _Fig. 4._ _Fig. 5._
PRACTICAL GEOMETRY.
_Fig. ½._ _Fig. 21._ _Fig. 22._]
7.—BY MEANS OF TWO PICKETS,
_to ascertain the height of an object_.
[_Vide 2nd Plate_, HEIGHTS, AND DISTANCES, _Fig. 1_.]
Let two pickets C D (4 feet), E F (6 feet), be placed with their
bases in the line C A passing through A the height required, and
move them nearer to, or farther from each other, until the summit
B of the object is seen in the same line as D, and F, the tops of the
rods. Then, by the principles of similar triangles,
As D H (= C E) : F H :: D G (= C A) : B G.
To which add A G = C D for the whole height A B.
Thus, supposing C E to be 6 feet, F H 2 feet, and C A 150 feet, the
proportion will be,
As 6 : 2 :: 150 : 50 feet.
Then 50 + C D will be the altitude required.
DISTANCES.
1.—BY MEANS OF THE SEXTANT,[52]
_to find the distance from an object, whose height is known_.
Let A B represent the height of the object; C your station; and C B
the distance to be found.
[Illustration: (right angle triangle A B C)]
Take the angle B C A with the sextant,[52] and note it in minutes;
then A B, in feet × 573 ÷ B C A, in minutes = A C in fathoms. Or A B
in feet × 573 ÷ B C A, in minutes × 2 = A C in yards.
573 is a constant multiple.
This method requires no table of sines, &c., the number of minutes in
the angle being used instead of the sine.
2.—BY MEANS OF A POCKET SEXTANT,
_to measure inaccessible distances_.
When used for taking the distance of objects, the sextant is to be
held horizontally, and the quicksilvered part of the glass will be
uppermost, or above the transparent part.
To ascertain the distance A B (_vide Plate 2, Fig. 2_), obtain, by
observation, the direction A C perpendicular to A B, which is thus
performed:—Set the instrument at 90°, and place yourself at the
point A, with your right towards the point B; then look through the
sextant, and direct a picket to be placed in the line A C at 100
yards, or feet, from you, so that the point B will appear right above
it. Then set the sextant at 45°, and walk along the line towards
C until you bring the points A, and B to coincide; the base and
perpendicular will then be of equal length, and A C being known,
or measured, the distance A B will also be ascertained. But if you
cannot walk far enough to find angle C 45°, find it equal to 63° 26′,
and then A C = ½ A B; at 71° 34′ = ⅓ A B; at 75° 58′ = ¼ A B; at 78°
41′ = ⅕ A B; at 80° 32′ = ⅙ A B; at 82° 52′ = ⅛ A B; and at 84° 17′
the distance will be ⅒ A B.
Should the object be far distant, it will be necessary to take a long
base, and the side A B must be calculated, therefore, by trigonometry.
3.—BY MEANS OF THE PRISMATIC COMPASS,
_to measure inaccessible distances_.
Having fixed the instrument to the stand, place it over the
station-point, spreading the legs so as to give sufficient firmness,
and observing that the card is level enough to allow it to play
freely; raise the prism by means of the slide, until the divisions of
the compass-card are distinctly seen; then look through the slit, and
turn the box round until the thread bisects the object whose distance
is required; allow the card to settle, and the division on it, which
coincides with the thread of the vane, will be the azimuth, or
bearing of the object, reckoned from the north, or south point of the
needle, when the card is divided into twice 180 degrees. The angular
distance between any two objects will, of course, be the difference
of their bearings; thus, suppose one to bear 15° N.E., and the other
165° S.E., the angular distance between them will be 150°.
In military sketching, the compass is often supported merely by the
hands, using the little spring to check the vibrations of the card.
In windy weather, the mean of these vibrations must be taken for the
bearing sought.
The directions for surveying, &c., &c., by means of “The
Reconnoitring Protractor,” apply similarly to the “Prismatic Compass.”
4.—BY MEANS OF “THE RECONNOITRING PROTRACTOR,”
_to ascertain the distance from inacessible objects_.
[_Plate_, SURVEYING, AND RECONNOITRING, _Fig. 6_.]
Select a good position for a base line; fix the protractor on the
tripod at the first station, placing the instrument in a direct line
between the first station and the point selected for the second
station. Direct the index consecutively at the objects, the relative
distances of which are to be ascertained, and note correctly their
respective angles. When the object is above the horizontal line, the
sliding-sight must be sufficiently raised to take its bearing; and,
should the object be below the level of the protractor, its angle may
be taken by observation through the upper holes of the near sight; or
the feet of the tripod may be adjusted, by raising, or sinking them
in the ground, so that the index may be correctly directed to the
object. Then proceed to the second station, measuring, or carefully
pacing the base line, at the end of which fix the protractor in
a straight line between the two stations; direct the index at
the objects previously noted at the first station, taking their
respective angles as before.
_Construction_—
Draw the base of the length required, according to the scale; from
each end of which set off the angles found, and draw the lines
required; the intersection of these will determine the position of
the several objects, and their relative distances may be ascertained
by measurement on the scale of the base line; or they may be
calculated trigonometrically.
5.—BY MEANS OF TWO PICKETS,
_to ascertain the distance from an object_.
Take two pickets of unequal lengths, drive the shortest into the
ground, say close to the edge of a river; measure some paces back
from it, and drive in the other, till you find, by looking over
the tops of both, that your sight cuts the opposite bank. Pull up
the first picket, measure the same distance from the second in any
direction the most horizontal, and drive it as deep in the ground as
before. Then, if you look over them again, and observe where the line
of sight falls, or terminates, you will have the distance required.
This method is only applicable to short distances.
6.—_To ascertain the distance of the object A from B._
[_Vide Plate 2, Fig. 3._]
Place a picket at B, and another at C at a few yards’ distance,
making A B C a right angle, or B C perpendicular to A B.[53] Divide B
C into 4, 5, or any number of equal parts, make another similar angle
at C in a direction from the object, and walk along the line C D
until you bring yourself in a line with the object A, and any of the
divisions (say O) of the line B C. Then (having measured C D)
as C O : C D :: B O : B A.
Or, as 10 : 53 :: 30 : 159 yards.
7.—_To find the distance between two objects_, C, _and_ D.
[_Vide Plate 2, Fig. 4._]
From any point A, taken in the line C D, erect the perpendicular A
E, in which set off from A to E 40 yards, set off from E to G, in
the prolongation of A E, 10 yards, at G raise the perpendicular G F,
and produce it towards I, plant pickets at E, and G, then move with
another picket on G F, till F is in a line with E, and D; and on the
prolongation of the perpendicular F G place another picket at I in
the line with E, and C: measure F I (54 yards), then—
as G E : A E :: F I : C D;
Or, as 10 : 40 :: 54 : 216 yards.
8.—_To find the inaccessible length_, A, B, _of the front of a
fortification_.
[_Plate 2, Fig. 5._]
Plant a picket at C, from whence both points may be seen; find the
lengths C A, C B (by the method in No. 5); make C E one-fourth,
or any part of C B, and make C D bear the same proportion to C A:
measure D E; then
as C D : D E :: C A : A B.
Nearly in the same manner the distance from B to A may be
ascertained, when the point B is accessible; for having measured the
line C B, and made the angle C E D equal to C B A, the proportion
will be as C E : D E :: C B : B A.
9.—BY MEANS OF THE TANGENT SCALE OF A GUN,
_to ascertain the distance, the height of the object at the required
distance being known_.
Lay the gun by the line of metal for the top of the object; then
raise the tangent scale till the top of it and the notch on the
muzzle are in line with the foot of the object, and note what length
of scale is required.
Then,—by similar triangles—
As the length of the raised part of the tangent scale
is to the length of the gun;
so is the height of the distant object
to the distance required.
Thus, supposing the height of the object to be 9 feet, the length of
that part of the tangent scale which is raised, 3 inches, and of the
gun 6 feet, the proportion will be—
As 3 : 72 :: 108 : 2592 inches, or 216 feet.
10.—BY MEANS OF THE PEAK OF A CAP,
_to measure the breadth of a river_.
Place yourself at the edge of one bank, and lower the peak of your
cap till you find the edge of it cut the other bank, then steady your
head by placing your hand under your chin, and turn round gently to
some level spot of ground on your side of the river, and observe
where your eyes, and the edge of the peak again meet the ground;
measure the distance, which will be _nearly_ the breadth of the river.
11.—BY THE REPORT OF FIRE-ARMS, TO ASCERTAIN THE DISTANCE OF ANY
OBJECT, _vide_ SOUND, _page 316_.
_To estimate distances, in the field._
Good eyesight recognises masses of troops at 1700 yards; beyond this
distance the glitter of arms may be observed. At 1300 yards infantry
may be distinguished from cavalry, and the movement of troops may
be seen; the horses of cavalry are not, however quite distinct, but
that the men are on horseback is clear. A single individual detached
from the rest of the corps may be seen at 1000 yards, but his head
does not appear as a round ball until he has approached up to 700
yards; at which distance white cross-belts, and white trousers may be
seen. At 500 yards the face may be observed as a light coloured spot;
the head, body, arms, and their movements, as well as the uniform,
and the firelocks (when bright barrels) can be made out. At between
200 and 250 yards all parts of the body are clearly visible, the
details of the uniform are tolerably clear, and the officers may be
distinguished from the men.
_Vide_ “UNITED SERVICE MAGAZINE.”—No. CCCXXXI.
BY MEANS OF THE RECONNOITRING PROTRACTOR,
_to traverse roads_.
[_Plate_, SURVEYING, AND RECONNOITRING, _Fig. 5_.]
Fix the protractor on the tripod at the first station, placing it so
that the side tube may be in a direct line with the intended second
station. From each end of the tube observe the objects in sight (or
place pickets) in order to secure a straight line in pacing, or
measuring, from the first to the second station. Mark the distance
between the stations, and place the protractor, by means of the
tube, in a direct line with the first station. Then select the third
station, and direct the arm or index correctly to it (using the upper
holes of the near sight for a declivity, or raising the sliding-sight
for an ascent); note the angle thus found, and notice the objects in
front, and rear (if any, if not, place pickets) for points to enable
you to pace towards, and work with accuracy at the third station.
Select station 4, place the tube in line with the third, and second
stations; note the bearing of No. 4, and pace the distance to it.
Proceed thus from station to station, entering the angles, and
distances in your note-book, as well as the offsets (which must also
be carefully measured) from the lines taken, until the survey is
completed.
_Construction_—
The day’s work will be easily plotted on paper, by setting off the
angles found, and drawing lines for the measured distances, according
to scale.
SOUND.
The movement communicated to the particles of air by the vibrations
of a sonorous body is the cause of the sensation of sound; and it is
because the particles are driven from the point of vibration in every
direction, as from a centre, that the sound is perceived at once,
everywhere within the surface of a sphere of a certain extent.
The velocity of sound; or the space through which it is propagated
in a given time, has been differently estimated by authors who have
written on this subject. Roberval states it to be at the rate of
560 feet in a second; Gassendus at 1473; Mersenne at 1474; Duhamel
at 1338; Newton at 960; Derham, in whose measure Flamsteed and
Halley acquiesce, at 1142. By accounts in the Memoirs of the Royal
Academy of Sciences, at Paris, 1738, where cannon were fired at
various distances, under many varieties of weather, wind, and other
circumstances, and where the measures of the different places had
been settled with the utmost exactness, it was found that sound was
propagated, on a medium, at the rate of 1038 French feet in a second
of time, which is equivalent to 1107 _English feet_, the French foot
being in proportion to the English as 15 to 16.
From various experiments made with great care by Dr. O. Gregory, it
has been found that sound flies through the air uniformly at the
rate of about 1100 feet per second, when the air is quiescent, and
at a medium temperature. At the temperature of freezing, or a little
below, the velocity is about 1120. The approximate velocities under
different temperatures may be found by adding to 1100 _half a foot_
for every degree on Fahrenheit’s thermometer above the freezing
point. The mean velocity may be taken at 370 yards per second, or
a mile in 4-7/9 second. Hence, multiplying any time employed by
sound in moving by 370, will give the corresponding space in yards,
or dividing any space in yards by 370 will give the time which
sound will occupy in passing uniformly over that space. If the wind
blow briskly, as at the rate of 20 to 60 feet per second, in the
direction in which the sound moves, the velocity of the sound will
be proportionally augmented; if the direction of the wind is opposed
to that of the sound, the difference of their velocities must be
employed. The velocity of sound is not affected by its intensity, the
smallest sound moving as rapidly as the loudest.
_To ascertain the distance of any object by the report of fire-arms._
(_Vide 11. Page 314._)
Multiply the number of seconds which elapse between the time of
seeing the flash, and hearing the report by 1100, and the product
will be the distance in feet, with sufficient accuracy for ordinary
purposes. If greater accuracy be required, this rule must be
modified, on account of the velocity, and direction of the wind, and
state of the thermometer.
_Sound will be louder_ in proportion to the condensation of the air.
Water is one of the greatest conductors of sound; it can be heard on
water nearly twice as far as upon land.
GRAVITY.
Gravity is downward pressure, or weight, being the natural tendency
of all bodies towards the centre of the earth. (_Vide Gravity_,
MOTION, FORCES. _Page 320._)
_Absolute gravity_ denotes the whole force with which a body tends
downwards, as when the body is in empty space.
_Specific gravity_ denotes the relative or comparative gravity of any
body, in respect to that of another body of equal bulk, or magnitude.
_Centre of gravity_ is that point in a body, or system of bodies, on
which, if rested, or suspended, the whole would remain in a state of
equilibrium about that point.
_The centre of gravity_ of a circle, regular polygon, prism,
cylinder, or sphere, is in its centre.
_The centre of gravity_ of a triangle is found by bisecting any two
of its sides, and drawing lines from the points of bisection to the
opposite angles; the intersection of these lines will be the centre
of gravity.
_Force of gravity, or gravitation_, is an accelerated velocity, which
bodies acquire in falling freely from a state of rest.
1. The space through which a body will fall in feet, in any given
time equals the product of the square of the time multiplied by
16·0833.
_Example._—Required the space a falling body will pass through in
five seconds?
16·0833 × 25 = 412·0825 feet.
2. The velocity in feet, which a body in descending freely will
acquire in a given time, equals the product of the time in seconds
multiplied by 32·1666.
_Example._—What is the velocity acquired at the end of seven seconds?
32·1666 × 7 = 225·1662 feet.
3. The velocity in feet per second that a body will acquire, in
falling through a given space, equals the square root of the
product of the time multiplied by 64·3333.
_Example._—The space through which a body has fallen is 201 feet;
required its velocity at the end of the fall?
√(64·3333 × 201) = √(12931) = 1137 feet.
SPECIFIC GRAVITIES OF SEVERAL SOLID, AND FLUID BODIES.
Air,[54] in a mean state 1·232
Brass, cast 8000
Brick 2000
Coal[54] 1250
Copper 9000
Cork 240
Clay 2160
Earth, common 1984
Flint 2570
Gold, standard 18888
Gun metal 8784
Gunpowder—solid 1745
” loose 868
Granite 3000
Iron, cast 7425
Lead 11325
Pitch 1150
Sand[54] 1520
Silver, standard 10535
Steel 7850
Stone, common 2520
Tin 7320
Water, rain 1000
[54] sea 1030
Wood—alder 800
ash, the trunk 845
beech 852
elm, and larch 540
fir, Riga, & maple 750
pine, pitch & red 660
oak 950
walnut 671
These numbers represent the weight of a cubic foot (or 1728 cubic
inches) of each of the bodies in ounces (avoirdupois).
_To find the magnitude of any body from its weight._
As the tabular specific gravity of the body
is to its weight in avoirdupois ounces;
so is one cubic foot (or 1728 cubic inches)
to its content in feet, or inches, respectively.
_To find the weight of a body, from its magnitude._
As one cubic foot (1728 cubic inches)
is to the content of the body;
so is its tabular specific gravity
to the weight of the body.
_To find the specific gravity of a body._
1.—_When the body is heavier than water._
Weigh it both in water, and out of water, and take the difference:
Then,—As the weight lost in water
is to the whole or absolute weight;
so is the specific gravity of water
to the specific gravity of the body.
2.—_When the body is lighter than water_, so that it will not
sink, annex to it another body heavier than water, so that the mass
compounded of the two may sink together. Weigh the denser body, and
the compound mass separately, both in water, and out of it; then find
how much each loses in water, by subtracting its weight in water from
its weight in air; and subtract the less of these remainders from the
greater.
Then,—As the last remainder
is to the weight of the light body in air;
so is the specific gravity of water
to the specific gravity of the body.
3.—_For a fluid of any sort._
Take a piece of a body of known specific gravity, weigh it both
in, and out of the fluid, finding the loss of weight by taking the
difference of the two:
Then,—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.
_To find the quantities of two ingredients in a given compound._
Take the three differences of every pair of the three specific
gravities, namely, the specific gravities of the compound, and each
ingredient, and multiply each specific gravity by the difference of
the other two:
Then,—As the greatest product
is to the whole weight of the compound;
so is each of the other two products
to the weights of the two ingredients.
_To find the diameter of any small sphere, or globule, whose specific
gravity is given_ (_or can be found in the Table_) _and weight known._
Divide its weight in grains by the number expressing its specific
gravity; extract the cube root of this quotient, and multiply it by
1·9612 for the diameter.
WEIGHT OF A CUBIC FOOT OF THE FOLLOWING MATERIALS,
_in pounds_.
Ash 49
Beech 43
Birch 49
Box 60
Cork 15
Elm 36
Fir 30
Mahogany, Spanish 50
Pine, red 41
Teak 41
Walnut 41
Coke 46
Clay 125
Earth, loose 95
Gravel 120
Granite 166
Brick, common 98
Chalk 145
Coal, Newcastle 78
Antimony 418
Brass, cast 525
Copper 538
Gold, pure 1203
Iron, cast, variable 444
Lead 717
Silver, standard 644
Tin 455
By means of the foregoing table, the weight of any quantity of the
materials specified (in cubic feet) may readily be found.
MOTION, FORCES, &c.
_Body_ is the mass or quantity of matter in any material substance,
and it is always proportional to its weight, or gravity, whatever its
figure may be.
_Density_ is the proportional weight, or quantity of matter in any
body.
_Velocity, or celerity_, is an affection of motion by which a body
passes over a certain space in a certain time.
_Momentum, or quantity of motion_, is the power, or force, in moving
bodies.
_Force_ is a power exerted on a body to move it, or to stop it. If
the force act constantly, it is a _permanent force_, like pressure,
or the force or gravity; but if it act instantaneously, or for an
imperceptibly short time, it is called _impulse_, or _percussion_,
like the smart blow of a hammer.
_A motive, or moving force_, is the power of an agent to produce
motion.
_Accelerative, or retardative force_, is that which affects the
velocity only, or it is that by which the velocity is accelerated, or
retarded.
The change, or alteration of motion by any external force, is always
proportional to that force, and in the direction of the right line in
which it acts.
_If a body be projected_ in free space, either parallel to the
horizon, or in an oblique direction, by the force of gunpowder, or
any other impulse: it will, by this motion, in conjunction with the
action of gravity, describe the curve line of a parabola.
_A parabola_ is the section formed by cutting a cone, with a plane,
parallel to the side of the cone.
_Gravity_ (_vide page 316_) is a force of such a nature that all
bodies, whether light or heavy, fall perpendicularly through equal
spaces in the same time, abstracting the resistance of the air; as
lead, and a feather, which, in an exhausted receiver, fall from
the top to the bottom in the same time. The velocities acquired by
descending, are in the exact proportion of the times of descent, and
the spaces descended are proportional to the squares of the times,
and, therefore, to the squares of the velocities. Hence, then, it
follows that the weights, or gravities of bodies near the surface
of the earth are proportional to the quantities of matter contained
in them; and that the spaces, times, and velocities generated
by gravity, have the relations contained in the three general
proportions before laid down.
A body in the latitude of London falls nearly 16-1/12 feet in the
first second of time, and consequently, at the end of that time, it
has acquired a velocity double, or of 32⅙ feet.
The times being as the velocities, and the spaces as the squares of
either; therefore,
if the times be as the Nos.
1, 2, 3, 4, 5, 6, 7, 8, 9, 10;
the velocities will also be as
1, 2, 3, 4, 5, 6, 7, 8, 9, 10;
and the spaces as their squares
1, 4, 9, 16, 25, 36, 49, 64, 81, 100;
and the spaces for each time,
1, 3, 5, 7, 9, 11, 13, 15, 17, 19.
Namely, as the series of the odd numbers, which are the differences
of the squares denoting the whole spaces. So that if the first series
of natural numbers be seconds of time,
namely: the times in seconds 1 2 3 4 &c.
the velocities in feet will be 32⅙ 64⅓ 96½ 128⅔, &c.
the spaces in the whole times 16-1/12 64⅓ 144¾ 257⅓, &c.
and the space for each second 16-1/12 48¼ 80-5/12 112-7/12, &c.
of which spaces the common difference is 32⅙ feet, the natural and
obvious measure of the force of gravity.
Thus, a body falling from a state of rest acquires a velocity to pass
through 9 spaces in the fifth second of time; 7 in the fourth; 5 in
the third; 3 in the second; and 1 in the first. Thus it is 9 + 7 + 5
+ 3 + 1 = 25, which shows that the whole spaces passed through in 5
seconds equal the square of 5.
_The momentum_, or force, of a body falling through the atmosphere is
the mass or weight, multiplied by the square root of the height it
has fallen through, multiplied by 8·021.
Suppose a weight of 10 tons to be raised 9 feet, and to drop thence
suddenly on a bridge; the momentum is 10 × (3 × 8·021) = 240·63 tons.
That is, a weight of 10 tons, so falling, would exert as great a
strain to break down the bridge, as the pressure of 240·63 tons of
dead weight.
Thus, a one-ounce ball falling from a height of 400 feet, would
strike the earth with a momentum of
oz. feet. oz. lb.
1 × (20 × 8·021) = 160·42 = 10·026.
By experiments to ascertain the effect of Carnot’s vertical fire, it
was found that 4-oz. balls only penetrated ½0 of an inch into deal
board, and from 2 to 3 inches into meadow ground.
_Amplitude_ signifies the range of a projectile, or the right line
upon the ground, subtending the curvilinear path in which it moves.
_The time of flight_ of different shot, and shells is equal to the
time a heavy body takes to descend freely from the highest point
described by the curve of the projectile.
_To find the time of descent_:
Divide the given height, or altitude, by 16-1/12, and the square root
of the quotient will be the time required. Thus, if the altitude is
1200 feet, and the time of descent is required,
1200 ÷ 16-1/12 = 74·61, the square root of which is 8·637, the
time required.
When a body is projected vertically _downwards_ with a given
velocity, the space described is equal to the time multiplied by the
velocity, together with the product of 16-1/12 by the square of the
time; but, if the body is projected _upwards_, the latter product
must be subtracted from the former.
PRACTICAL GEOMETRY.
DEFINITIONS.[55]
_A line is perpendicular to another_ when it inclines not more on the
one side than on the other, the angles on both sides being equal.
_Parallel lines_ are those which have no inclination to each other,
being everywhere equi-distant, however far produced, or extended.
_An angle_ is the inclination, or opening of two lines, which meet in
a point called the _vertex, or angular point_: and the two lines are
called the _legs, or sides_ of the angle.
_The measure of an angle_ is estimated by the number of degrees
contained in the arc between its two legs.
_A rectilinear angle_ has its legs or sides, _right_, or straight
lines.
_A curvilinear angle_ has its legs _curves_.
_A right angle_ is formed by one line perpendicular to another; the
measure of which is an arc of 90°.
_An acute angle_ is less than a right angle, or than 90°.
_An obtuse angle_ is greater than a right angle.
_An oblique angle_ may be either acute, or obtuse.
_The circumference, or periphery of a circle_ is the curved line
which bounds it, being everywhere equally distant from the _centre_.
The circumference is supposed to be divided into 360 degrees (marked
thus °); each degree into 60 minutes, each minute (′) into 60 seconds
(″).
_An arc_ is any part of the circumference of a circle.
_A chord, or subtense_, is a right line joining the extremities of an
arc.
_The radius of a circle_ is a right line drawn from the centre to the
circumference.
_The diameter of a circle_ is a right line drawn through the centre,
and terminated by the circumference.
_A semicircle_ (180°) is that part of a circle which is contained
between the diameter, and half the circumference.
_A quadrant_ is the fourth part of a circle, being contained between
two radii, and an arc of 90°.
_A segment_ is that part of a circle which is cut off by a chord.
_A sector_ is that part of a circle contained between two radii, and
an arc.
_A secant_ is a line which cuts a circle, lying partly within, and
partly without it.
_A tangent_ is a line which touches a circle, or curve, without
cutting it.
_The point of contact_ is where a tangent touches an arc.
_Triangles_ are figures having three sides, and three angles.
_An equilateral triangle_ has its three sides equal.
_An isosceles triangle_ has only two equal sides.
_A scalene triangle_ has all its sides unequal.
_A rectangular, or right-angled triangle_ has one of its angles a
right one, or 90°; and the square of the side opposite the right
angle is equal to the sum of the squares of the sides containing that
angle; hence a triangle, having its sides proportional to the numbers
3, 4, 5, will be right-angled.
_The hypothenuse_ is the side opposite the right angle in a
rectangular triangle.
_An obtuse-angled triangle_ has one of its angles obtuse.
_An acute-angled triangle_ has all its angles acute.
_The three angles of any triangle_, taken together, are equal to two
right angles, or 180°.
_The difference of the squares of two sides of a triangle_ is equal
to the product of their sum and difference.
_The sides of a triangle are proportional_ to the sines of their
opposite angles.
_Quadrangles, or quadrilaterals_, are plane figures bounded by four
right lines.
_A square_ is a quadrilateral having all its sides equal, and all
its angles right angles. The _diagonal of a square_ is equal to
the square root of twice the square of its sides: and _the side of
the square_ is equal to the square root of half the square of its
diagonal.
_The diagonal_ is a right line drawn across a quadrilateral figure,
from one angle to another. The sum of the squares of the two
diagonals of every parallelogram is equal to the sum of the squares
of the four sides.
_A parallelogram_ is a quadrilateral, whose opposite sides are
parallel.
_A rectangle_ is a parallelogram having four right angles.
_A rhomboid_ is an oblique-angled parallelogram.
_A rhombus, or lozenge_, is a quadrilateral, whose sides are all
equal but its angles oblique.
_A trapezium_ is a quadrilateral, which has none of its sides
parallel to each other.
_A trapezoid_ is a quadrilateral, which has only two of its sides
parallel.
_Polygons_ are plane figures bounded by more than four sides.
_A regular polygon_ has all its sides, and angles equal.
_The perimeter_ of a figure is the sum of all its sides.
_To bisect_—is to divide into two equal parts.
_To trisect_—is to divide into three equal parts.
_To inscribe_—is to draw one figure within another, so that all the
angles of the inner figure touch either the angles, sides, or planes
of the external figure.
_To circumscribe_—is to draw a figure round another, so that either
the angles, sides, or planes of the circumscribing figure touch all
the angles of the figure within it.
LINES, ANGLES, AND FIGURES.
_To divide a given right line into two equal parts._
From the extremities of the line as centres, and with any opening
in the compasses, greater than half the given line, as a radius,
describe arcs intersecting each other above, and below the given
line. A line being drawn through these intersections will divide the
given line into two equal parts.
_An arc of a circle_ is bisected in the same manner.
_To bisect an angle._
From the angular point, measure equal distances on the two lines
(forming the angle), and from these points, with the same distance
as radius, describe arcs intersecting each other. A line drawn from
their intersections to the angular point will bisect the angle.
_To erect a perpendicular._
From the point A set off any length 4 times to C; from A as a centre
with 3 of those parts describe an arc at B, and from C with 5 of them
cut the arc at B. Draw A B, which will be the perpendicular required.
Any equimultiples of these numbers, 3, 4, 5, may be used for erecting
a perpendicular. _Plate 2_, HEIGHTS AND DISTANCES, and PRACTICAL
GEOMETRY, _Fig. ½_.
_To erect a perpendicular._
Set off on each side of the point A, any two equal distances, A D, A
E. From D and E as centres, and with any radius greater than half D
E, describe two arcs intersecting each other in F. Through A, and F
draw the line A F, and it will be the perpendicular required.
_Fig. 1.—Plate_, PRACTICAL GEOMETRY.
_To let fall a perpendicular._
From D as a centre, and with any radius, describe an arc intersecting
the given line. From the points of intersection C, and E, with any
radius greater than half, describe two arcs, cutting each other at
F. Through D, and F draw a line, and D F will be the perpendicular
required. _Fig. 2._
_To draw a line parallel to a given line._
From any point D in the given line with the radius D C, describe the
arc C E, and from C with the same radius describe the arc D F. Take
E C, and set it off from D to F. Through C, and F draw C F for the
parallel required. _Fig. 3._
_To divide an angle into two equal parts._
From B as a centre with any radius describe an arc A C. From A, and C
with any radius describe arcs intersecting each other in D. Then draw
B D, and it will bisect the angle. _Fig. 4._
[Illustration: PRACTICAL GEOMETRY.
_Fig. 1-9._]
[Illustration:_Fig. 10-14._]
_To divide a right angle into three equal parts._
From B as a centre with any radius describe the arc A C. From A with
the radius A B cut the arc A C in D, and with the same radius from C
cut it in E. Then through the intersections D, and E draw the lines B
D, B E, and they will trisect, or divide the angle into three equal
parts. _Fig. 5._
_To find the centre of a circle._
Draw any chord A B, and bisect it by the perpendicular C D. Divide
C D into two equal parts, and the point of bisection O will be the
centre required. _Fig. 6._
_To describe an equilateral triangle._
From the points A, B, as centres, and with A B as radius, describe
arcs intersecting each other in C. Draw C A, C B, and the figure A B
C will be the triangle required. _Fig. 7._
_To describe a square._
From the point B, draw B C perpendicular, and equal to A B. On A,
and C, with the radius A B, describe arcs cutting each other in D.
Draw the lines D A, D C, and the figure A B C D will be the square
required. _Fig. 8._
_To inscribe a square in a circle._
Draw the diameters A B, C D perpendicular to each other. Then
draw the lines A D, A C, B D, B C; and A B C D will be the square
required. _Fig. 9._
_To inscribe an octagon in a circle._
Bisect any two arcs A C, B C of the square A B C D in G, and E.
Through the points G, and E, and the centre O draw lines, which
produce to F, and H. Join A F, F D, D H, &c. and they will form the
octagon required. _Fig. 9._
_On a line to describe all the several polygons, from the hexagon to
the dodecagon._
Bisect A B by the perpendicular C D. From A as a centre, and with A B
as a radius, describe the arc B E, which divide into six equal parts;
and from E as a centre describe the arcs 5 F, 4 G, 3 H, &c. Then from
the intersection E as a centre, and with E A as a radius, describe
the circle A I D B, which will contain A B six times. From F in like
manner as a centre, and with F A as radius, describe the circle A
K L B, which will contain A B seven times; and so on for the other
polygons. _Fig. 10._
_To inscribe in a circle an equilateral triangle._
From any point D in the circumference as a centre, and with the
radius D O of the given circle, describe an arc A O B cutting the
circumference in A, and B. Through D, and O draw D C. Then, join
A B, A C, B C; and the figure A B C will be the triangle required.
_Fig. 11._
_To inscribe a hexagon in a circle._
Bisect the arcs A C, B C in E, and F, and join A D, D B, B F, &c.,
which will form the hexagon. Or carry the radius six times round the
circumference, and the hexagon will be obtained. _Fig. 11._
_To inscribe a dodecagon in a circle._
Bisect the arc A D of the hexagon in G, and A G being carried twelve
times round the circumference, will form the dodecagon. _Fig. 11._
_To inscribe a pentagon, hexagon, or decagon, in a circle._
Draw the diameter A B, and make the radius D C perpendicular to A B.
Bisect D B in E. From E as a centre, and with E C as radius, describe
an arc cutting A D in F. Join C F, which will be the side of the
pentagon, C D that of the hexagon, and D F that of the decagon. _Fig.
12._
_To find the angles at the centre, and circumference of a regular
polygon._
Divide 360 by the number of the sides of the given polygon, and
the quotient will be the angle at the centre; and this angle being
subtracted from 180, the difference will be the angle, at the
circumference, required.
_Table, showing the angles at the centre, and circumference._
Names. No. of Angles Angles at
sides. at centre. circumference.
Trigon 3 120° 60°
Tetragon 4 90° 90°
Pentagon 5 72° 108°
Hexagon 6 60° 120°
Heptagon 7 51° 25-5/7′ 128° 34-2/7′
Octagon 8 45° 135°
Nonagon 9 40° 140°
Decagon 10 36° 144°
_To inscribe any regular polygon in a circle._
From the centre C draw the radii C A, C B, making an angle equal
to that at the centre of the proposed polygon, as contained in the
preceding table. Then the distance A B will be one side of the
polygon, which, being carried round the circumference the proper
number of times, will complete the polygon required. _Fig. 13._
[Illustration: _Fig. 15-20._]
_To circumscribe a circle about a triangle._
Bisect any two of the given sides, A B, B C by the perpendiculars E
F, D F. From the intersection F as a centre, and with the distance of
any of the angles, as a radius, describe the circle required. _Fig.
14._
_To circumscribe a circle about a square._
Draw the two diagonals A C, B D intersecting each other in O. From O
as a centre, and with O A, or O B, as a radius, describe the required
circle. _Fig. 15._
_To circumscribe a square about a circle._
Draw the two diameters A B, C D perpendicular to each other, through
the points A, C, B, D, draw the tangents E F, E G, G H, F H, and E G
H F will be the square required. _Fig. 16._
_To reduce a map, or plan, from one scale to another._
Divide the given figure A C by cross lines, forming as many squares
as may be thought necessary. Draw a line E F, on which set off as
many parts from the scale M, as A B contains parts of the scale
N. Draw E H, and F G perpendicular to E F, and each equal to the
proportional parts contained in A D, or B C. Join H G, and divide
the figure E G into the same number of squares as the original A
C. Describe in every square what is contained in the corresponding
square of the given figure; and E F G H will be the reduced plan
required. The same operation will serve either to reduce, or enlarge
any map, plan, drawing, or painting. _Fig. 17._
MENSURATION OF PLANES, AND SOLIDS.
_Mensuration is of three kinds_, viz., lineal, superficial, and solid.
_Lineal measure_ has reference to length only.
_Superficial measure_ (_or the surface_) includes length, and breadth.
_Solid measure_ (_or the content_) comprehends length, breadth, and
thickness.
MENSURATION OF PLANES.
_The area_ of any plane figure is the superficial measure contained
within its extremes, or bounds. This area is estimated by the number
of small squares that may be contained in it, the side of these
measuring squares being an inch, a foot, or any other fixed quantity,
and hence the area is said to be so many square inches, square feet,
&c. _Vide Table, Square measure._ _Page 275._
_To find the area of a parallelogram, whether a square, rectangle,
&c._
Multiply the length by the breadth, or perpendicular height, for the
area required.
_Example._—Required the area of a rectangle, whose length is 9 feet,
and breadth 4 feet.
9 × 4 = 36 feet. The required area, or surface.
_To find the area of a triangle, its base, and perpendicular height
being given._
Multiply the base by the perpendicular height, and half the product
will be the area.
_Example._—Required the number of square yards contained in a
triangle, whose base is 20 yards, and perpendicular height 14 yards.
(20 × 14) / 2 = 140 square yards. Area required.
_To find the area of a triangle, whose three sides are given._
From half the sum of the three sides, subtract each side severally;
multiply the half sum, and the three remainders together, and the
square root of the product will be the area required.
_Example._—Required the area of a triangle, whose sides are 50, 40,
and 30 feet.
(50 + 40 + 30) / 2 = 60, half the sum of the three sides.
60 - 30 = 30 First difference.
60 - 40 = 20 Second difference.
60 - 50 = 10 Third difference.
30 × 20 × 10 × 60 = 360000.
Square root of 360000 = 600. Area required.
_Two sides of a right-angled triangle being given, to find the third
side._
1. When the two sides forming the right angle are given, to find the
hypothenuse, or side opposite the right angle.
Take the square root of the sum of the two sides squared for the side
required.
_Example._—Required the length of the interior slope of a rampart,
whose perpendicular height is 17 feet, and the base of the slope 20
feet.
17 × 17 = 289
20 × 20 = 400
-----
The square root of 689 = 26·24. The length required.
2. When the hypothenuse, and one of the perpendicular sides are given.
From the square of the hypothenuse, subtract the square of the given
side, and the square root of the remainder will be the side required.
_Example._—The hypothenuse being 5 yards, and the base 4 yards,
required the other side.
5 × 5 = 25
4 × 4 = 16
------------
The square root of 9 = 3 yards. The side required.
_To find the area of a trapezium, A B C D._
Draw the diagonal A C, upon which let fall from its opposite angles
B, and D, the perpendiculars B F, D E. Find by measurement the
diagonal A C, and the perpendiculars B F, D E, then multiply the sum
of the perpendiculars by the diagonal, and half the product will be
the area of the trapezium. _Fig. 18._
_Example._—Required the area of the trapezium, whose diagonal A C is
100 feet, and perpendiculars B F 30 feet, and D E 40 feet.
((30 + 40) × 100) / 2 = 3500 square feet. Area required.
Or, divide the trapezium into two triangles by a diagonal, then find
the areas of these triangles, and add them together.
_To find the area of a trapezoid, A B C D._
Multiply the sum of the parallel sides A B, D C by the perpendicular
distance E C, and half the product will be the area. _Fig. 19._
_Example._—Required the area of the trapezoid A B C D, of which
the parallel sides A B, D C are 120 feet, and 90 feet, and the
perpendicular distance E C 40 feet.
((120 + 90) × 40) / 2 = 4200 square feet. Area required.
_To find the area of an irregular figure, or polygon._
Draw diagonals dividing the figure into trapeziums, and triangles;
then, having found the area of each, add them together, and the sum
will be the area required.
_To find the area of a figure, having a part bounded by a curve._
Draw a right line joining the extremities of the curve, then find
the area of the trapezium. On the right line let fall as many
perpendiculars as the several windings of the curve may require. Find
their lengths, and divide their sum by the number of perpendiculars,
and the quotient will be the mean breadth; which being multiplied by
the length of the right line, will give the area of the curved part.
This area being added to that of the trapezium will give the area of
the required figure.
_To measure long irregular figures._
Measure the breadth at both ends, and at several places _at equal
distances_. Add together all these intermediate breadths, and half
the two extremes, which sum multiply by the length, and divide by the
number of parts for the area. If the perpendiculars, or breadths, _be
not at equal distances_, compute all the parts separately, as so many
trapezoids, and add them all together for the whole area.
_Example._—The breadths of an irregular figure at five equi-distant
places being 8, 2, 7, 9, 4, and the whole length 40, required the
area.
8 + 4 = 12 12 ÷ 2 = 6
6 + 2 + 7 + 9 = 24
(24 × 40) / 4 = 240. Area required.
_To find the number of square acres in any of the preceding
figures._[56]
Divide the superficial content in feet by 43560, and the quotient
will be the number required.
_To bring square chains to acres._
Of square chains strike off two decimal places to the right, and the
rest of the figures will be acres.
_To bring square links to acres._
Of square links cut off five of the figures on the right hand, for
decimals, and the rest will be acres; then multiply these decimals by
4, for roods, cutting off five figures as before; and the decimals
of these again by 40, for perches, when five figures are again to be
struck off.
_To find the area of a regular polygon._
Multiply the _perimeter_ (or sum of the sides) of the polygon by the
perpendicular drawn from its centre on one of its sides, and take
half the product for the area.
Or, multiply the area of one of the triangles by the number of sides
of the polygon, and the product will be the area of it.
_Example._—Required, the area of a regular hexagon, whose side is 40
feet, and the perpendicular 34·64 feet.
40 × 6 = 240 the perimeter.
(240 × 34·64) / 2 = 4156·8 square feet. Area required.
_To find the diameter, and circumference of any circle, the one from
the other._
Use either of the following proportions:
as 7 is to 22 } { so is the diameter
or as 1 is to 3·1416 } { to the circumference.
as 22 is to 7 } { so is the circumference
or, as 3·1416 is to 1 } { to the diameter.
or, instead of dividing the diameter by 3·1416, multiply it by ·3183,
for the circumference.
_Example 1._—Required, the circumference of a circle, whose diameter
is 20 feet.
As 7 : 22 :: 20 : 62·857 feet. Circumference required.
_Example 2._—Required, the diameter of a circle, whose circumference
is 36 inches.
As 22 : 7 :: 36 : 11·45 inches. Diameter required.
_To find the diameter of a circle, the area being given._
Divide the area by ·7854, and the square root of the quotient will be
the diameter required.
_Example._—Required, the diameter of a circle, whose area is 176·715
square feet.
176·715 ÷ ·7854 = 225.
Square root of 225 = 15 feet. Diameter required.
_To find the area of a circle._
1. Multiply half the circumference by half the diameter, or multiply
the whole circumference by the whole diameter, and take ¼ of the
product.
2. Or, square the diameter, and multiply that square by ·7854 for the
area.
3. Or, square the circumference, and multiply that square by 0·7958.
_Example 1._—Required the area of a circle, whose circumference is
55·548 inches, and its diameter 18 inches.
55·548 / 2 = 27·774 half circumference.
18 / 2 = 9 half diameter.
27·774 × 9 = 249·966, square inches. Area required.
_Example 2._—Required the area of a circle whose diameter is 12 feet.
12 × 12 = 144, square of the diameter.
·7854 × 144 = 113·0976 square feet. Area required.
_Example 3._—Required the area of a circle, whose circumference is 22
feet.
22 × 22 = 484.
484 × ·07958 = 38·51672 square feet. Area required.
_To find the area of a circular ring_,
or space included between the circumferences of two circles, the one
within the other.
1. Subtract the square of the less diameter from the square of the
greater, and multiply their difference by ·7854.
2. Or, find the area of each circle separately, and subtract one from
the other, for the area required.
3. Or, multiply the sum of the diameters by the difference of the
same, and that product by ·7854 for the area.
_Example._—Required the area of a ring, the diameters of whose
bounding circles are 10, and 20.
_By Rule 3._
20 + 10 = 30, sum of diameters.
20 - 10 = 10, difference of diameters.
30 × 10 × ·7854 = 235·62. The area.
_To find the length of any arc of a circle._
1. As 360° is to the number of degrees in the arc, so is the
circumference to the length of the arc.
2. Or, multiply the degrees in the given arc by the radius of the
circle, and the product by ·01745 for the length of the arc.
_Example._—_Rule 2._—Required the length of an arc of 30°, the radius
being 9 feet.
30 × 9 × ·01745 = 4·7115. Length of arc.
_To find the area of the sector of a circle._
Multiply the radius by the arc, and half the product will be the
area.
_Example._—Required the area of the sector, whose radius is 30
inches, and the length of the arc 36·6 inches.
(36·6 × 30) / 2 = 549 square inches. Area required.
_To find the area of the segment of a circle._
Find the area of the sector, by the preceding rule. Then find the
area of the triangle formed by the chord of the segments, and the
radii of the sector. Then, if the segment be less than a semicircle,
subtract the area of the triangle from it; or, if the segment be
greater than a semicircle, add the area of the triangle to it; for
the area of the segment.
_Example._—Required the area of a segment less than a semicircle, the
radius being 20 inches, the chord 22·42 inches, the length of the arc
24·43 inches, and the perpendicular 16·56 inches.
(24·43 × 20) / 2 = 244·3 square inches. Area of the sector.
(22·42 × 16·56) / 2 = 185·6376 square inches. Area of the triangle.
244·3 - 185·6376 = 58·6624 square inches. Area required.
_To find the area of a semicircle._
1. Multiply ¼ of the circumference by the radius, and the product
will be the area.
2. Or, multiply the square of the diameter by ·7854, and half the
product will be the area.
_Example._—_Rule 2._—Required the area of a semicircle, the diameter
being 50 inches.
(50 × 50 × ·7854) / 2 = 981·75 square inches. Area required.
_To find the area of an ellipsis, or oval._
Multiply the longest diameter, or axis, by the shortest, then
multiply the product by ·7854 for the area.
_Example._—Required the area of the ellipse, whose diameters are 25
inches, and 18 inches.
25 × 18 × ·7854 = 353·43 square inches. Area required.
_To find the area of a parabola, or its segment._
Multiply the base by the perpendicular height, and take two-thirds of
the product for the area.
_Example._—Required the area of a parabola, whose base is 20 feet,
and height 12 feet.
20 × 12 = 240
⅔ of 240 = 160 square feet. Area required.
MENSURATION OF SOLIDS.
_A solid_ is a body containing length, breadth, and thickness.
_Solids are measured_ by cubes, whose sides are each an inch, a foot,
a yard, &c., and the solidity, capacity, or content of any figure is
computed by the number of such cubes as are contained in it.—_Vide
Cubic measure, page 276._
_A cube_ is a solid contained by six equal square sides.
_A pyramid_ is a solid whose sides are all triangles meeting together
in a point, the base being any plane figure whatever. It is called a
triangular pyramid when its base is a triangle; a square pyramid when
its base is a square, &c.
_The segment of a pyramid, cone, or any other solid_ is a part of D
E F G cut off from the top by a plane D E F, parallel to the base
A B C.—_Vide Fig. 21, Plate 2_, HEIGHTS, DISTANCES, and PRACTICAL
GEOMETRY.
_A frustrum, or trunk_, is a part A B C D E F, that remains at the
bottom after the segment is cut off.
_A cone_ is a round pyramid, of which the base is a circle.
_The axis of a solid_ is a line from the vertex (or point) to the
centre of the base, or through the centres of the two ends. When the
axis is perpendicular to the base, it is a right prism, pyramid, or
cone; otherwise it is oblique.
_A sphere_ is a solid contained under one convex surface, and is
described by the revolution of a semicircle about its diameter, which
remains fixed.
_The centre of the sphere_ is such a point within the solid as is
everywhere equally distant from the convex surface, or circumference
of it.
_The diameter_ (_or axis_) _of a sphere_ is a straight line, which
passes through the centre, and is terminated by the convex surface.
_A segment of a sphere_ is a part cut off by a plane, the section of
which is always a circle, called the _base of the segment_.
_A sector of a sphere_ is that which is composed of a segment (less
than an hemisphere) and of a cone.
_A prism_ is a solid, the sides of which are parallelograms, having
its ends equal, and similar plane figures.
_Prisms are named_ according to the number of angles in the base.
_A cylinder_ is a solid, the two ends of which are circular; and
it is described, or formed, by the revolution of a right-angled
parallelogram about one of its sides, which remains fixed.
_To find the superficies of a prism, or cylinder._
Multiply the perimeter of one end of the prism by the length, or
height of the solid, and the product will be the surface of all its
sides. To which add also the area of the two ends of the prism, when
required.
Or, compute the areas of all the sides, and ends separately, and add
them all together.
_Example._—Required the surface of a cube, whose sides are each 5
inches.
5 + 5 + 5 + 5 = 20, perimeter of one end.
20 × 5 = 100, surface of sides.
5 × 5 = 25, area of one end.
100 + 25 + 25 = 150 square inches. Surface of cube.
_To find the surface of a pyramid, or cone._
Multiply the perimeter of the base by the slant height, or length of
the side, and half the product will be the surface of the sides; to
which add the area of the base when required.
_Example._—Required the upright surface of a triangular pyramid, the
slant height being 20 feet, and each side of the base 3 feet.
3 + 3 + 3 = 9, perimeter of base.
(9 × 20) / 2 = 90 feet. Surface required.
_To find the surface of the frustrum of a pyramid, or cone._
Add together the perimeters of the two ends, multiply their sum by
the slant height, and take half the product.
_Example._—How many square feet are in the surface of the frustrum
of a square pyramid, whose slant height is 10 feet, each side of the
base 3 feet, and each side of the less end 2 feet.
3 + 3 + 3 + 3 = 12, perimeter of base.
2 + 2 + 2 + 2 = 8, perimeter of less end.
((12 + 8) × 10) / 2 = 100 feet. Surface required.
_To find the solid content of a prism or cylinder._
Find the area of the base, or end, and multiply it by the length of
the prism, or cylinder. Fora cube, multiply its side twice by itself;
and for a parallelopipedon, multiply the length, breadth, and depth
together for the content.
_Example._—Required the solid content of a cube, whose side is 24
inches.
24 × 24 × 24 = 13824 square inches. Content required.
_To find the content of the solid part of a hollow cylinder._
From the content of the whole cylinder considered as a solid,
subtract the content of the hollow part, also considered as a solid,
and the difference will be the solidity required.
_Example._—Required the content of the solid part of the hollow
cylinder whose exterior diameter is 12 inches, the interior diameter
8 inches, and height 20 inches.
12 × 12 × ·7854 = 113·0976, area of base of cylinder.
113·0976 × 20 = 2261·952, solidity of whole cylinder.
8 × 8 × ·7854 = 50·2656, area of base of hollow cylinder.
50·2656 × 20 = 1005·312, content of hollow part.
2261·952 - 1005·312 = 1256·64 cubic inches. Solidity required.
_To find the solidity of the frustrum of a cylinder._
Multiply the area of the base by half the greatest, and the least
lengths, and the product will be the solidity.
_Example._—Required the solidity of a frustrum, whose diameter is 24
inches, the greatest length 36 inches, and the least length 20 inches.
24 × 24 = 576. Square of the diameter.
576 × ·7854 = 452·3904. Area of the base.
452·3904 × (36 + 20) / 2 = 12666·9312 Cubic inches.
Solidity required.
_To find the content of a pyramid, or cone._
Find the area of the base, and multiply that area by the
perpendicular height, and take ⅓ of the product.
_Example._—Required the solidity of a square pyramid, each side of
its base being 30, and its perpendicular height 25.
30 × 30 = 900, area of base.
(900 × 25) / 3 = 7500, solidity required.
_To find the solidity of the frustrum of a cone, or pyramid._
Add into one sum the areas of the two ends, and the mean proportional
between them: take ⅓ of that sum for the mean area, which multiply by
the perpendicular height, or length of the frustrum.
_Note._—_To find a mean proportional._
As one of the sides of the base is to the homologous, or
corresponding side of the other end, so is the area of the base to
the mean proportional required.
_Example._—Required the number of solid feet in a piece of timber,
whose bases are squares, each side of the greater end being 15
inches, and each side of the less end 6 inches; also the length of
the perpendicular altitude 24 feet.
15 × 15 = 225, area of the base.
6 × 6 = 36, area of the top.
As 15: 6 :: 225: 90, mean proportional.
24 feet = 288 inches.
((225 + 36 + 90) × 288) / 3 = 33696 cubic inches
= 19½ cubic feet.
_To find the surface of a sphere, or any segment._
Multiply the circumference of the sphere by its diameter, which will
give the whole surface.
Or, square the diameter, and multiply by 3·1416.
Or, square the circumference, and multiply by ·3183;
or divide by 3·1416.
_Note._—_For the surface of the segment, or frustrum_, multiply the
whole circumference of the sphere by the height of the part required.
_Example._—Required the superficies of a globe whose diameter is 24
inches.
24 × 24 × 3·1416 = 1809·5616 square inches.
_To find the solidity of a sphere, or globe._
1. Multiply the surface by the diameter, and take ⅙ of the product.
Or, multiply the square of the diameter by the circumference, and
take ⅙ of the product.
2. Cube the diameter, and multiply by ·5236.
3. Cube the circumference, and multiply by ·01688.
_Example._—Required the content of a sphere, whose axis is 12.
12 × 12 × 12 × ·5236 = 904·7808. Content required.
_To find the solidity of an hemisphere._
Find the solidity of the sphere, and half the content will be that of
the hemisphere.
_Note 1._—Any sphere, or globe _twice_ the diameter of another
contains _four times_ the superficies, or area of the other, and
_eight times_ the solid content. Hence the superficies of spheres are
as the squares, and the solidity as the cubes of their diameters.
_Note 2._—The cube of the diameter of a sphere in inches, multiplied
by ·00188, will give the number of _imperial gallons it will contain_.
_To find the solid content of a spherical segment._
1. From three times the diameter of the sphere, take double the
height of the segment; then multiply the remainder by the square of
the height, and this product by ·5236.
2. Or, to three times the square of the radius of the segment’s base
add the square of its height; then multiply the sum by the height,
and the product by ·5236.
_Example._—Required the content of a spherical segment 2 feet in
height, cut from a sphere of 8 feet diameter.
(3 × 8) - (2 × 2) = 20
20 × 2^2 × ·5236 = 41·888 cubic feet. Content required.
_To find the diameter of a sphere, its solidity being given._
Divide the solidity by ·5236, and take the cube root of the quotient.
_Example._—The solidity of a sphere being 113·0976 solid inches, what
will be its diameter?
113·0976 / ·5236 = 216, the cube root of which is 6 inches,
the diameter required.
_To find the weight of an iron shot, its diameter being given._
Take ⅛ of the cube of the diameter, and ⅛ of that eighth, and the sum
of these two quotients will be the weight in pounds.
Or, as 64 is to 9 lb. so is the diameter cubed to its weight.
_Example._—Required the weight of an iron shot whose diameter is 3·5
inches?
3·5 cubed = 42·875, cube of diameter.
42·875 / 8 = 5·359 5·359 / 8 = ·669
5·359 + ·669 = 6·028 pounds. Weight required.
_To find the weight of a leaden ball, its diameter being given._
Take ⅓ of the cube of the diameter, and from it subtract ⅓ of this
third, and the remainder will be the weight, nearly.
Or, take 3/14 of the cube of the diameter.
_Example._—What is the weight of a leaden ball whose diameter is 3·3
inches?
3·3 cubed = 35·937, cube of diameter.
35·937 / 3 = 11·979 11·979 / 3 = 3·993
11·979 - 3·993 = 7·986 pounds. Weight required.
_To find the diameter of an iron shot, its weight being given._
Multiply the cube root of the shot’s weight by 1·923 for the
diameter.
Pr. Cube root. Diameter.
42 3·4760, &c. { Multiplied } 6·684, &c.
32 3·1748 { by 1·923, } 6·103 ”
24 2·8844 { diameter } 5·545 ”
18 2·6207 { of a } 5·038 ”
12 2·2894 { 1 lb. } 4·401 ”
9 2·0800 { shot. } 3·999 ”
6 1·8171 { } 3·494 ”
3 1·4422 { } 2·772 ”
_To find the diameter of a leaden ball, its weight being given._
To 4 times the weight add half the weight, and 3/100 of half the
weight; and the cube root of this sum will be the diameter in inches,
nearly.
_Example._—What is the diameter of a leaden ball, whose weight is 8
pounds?
8 x 4 = 32 8 / 2 = 4 3 / 100 of 4 = ·12.
32 + 4 + ·12 = 36·12, of which the cube root is 3·3 inches,
nearly. Diameter required.
_To find the weight of an iron shell, its interior and exterior
diameter being given._
Take 9/64 of the difference of the cubes of the external and internal
diameters, for the weight of the shell in pounds.
_Example._—What is the weight of a shell whose exterior diameter is
12·85 inches, and interior diameter 8·75 inches?
12·85 cubed = 2121·8241, 8·75 cubed = 669·9218.
2121·8241 - 669·9218 = 1451·9022.
9/64 of 1451·9022 = 204·1737 pounds. Weight required.
_To find the quantity of powder a shell will contain._
Divide the cube of the interior diameter in inches by 57·3, and the
quotient will be the weight in pounds.
Or, multiply the cube of the diameter by 11, and divide by 21 for the
quantity in half ounces.
_Example._—How much powder will fill a shell, whose internal diameter
is 7 inches?
7 cubed = 343.
343 / 57·3 = 6 pounds nearly. Powder required.
_To find the side of a cubical box to contain a given quantity of
powder._[57]
Multiply the weight in pounds by 30, and the cube root of the product
will be the side of the box in inches.
_Example._—Required the side of a cubical box to hold 50 pounds of
powder?
50 × 30 = 1500, the cube root of which is 11·44, which will be
the side of the box in inches.
_To find the quantity of powder to fill the chamber of a mortar, or
howitzer._
Multiply the content of the chamber in inches by 55, and divide the
product by 1728, and the quotient will be the quantity of powder in
pounds.
_Note._—The chamber of a mortar, or howitzer, is formed of a hollow
frustrum of a right cone, and of a hollow hemisphere.
_Example._—Required the quantity of powder to fill the chamber of a
13-inch mortar in which the diameter A B is 9·5 inches, the diameter
C E 6·5 inches, and the length D G 21·5 inches. _Vide Fig. 22. Plate
2._ HEIGHTS AND DISTANCES, and PRACTICAL GEOMETRY.
The content of the chamber must be found by finding the content of
the hollow frustrum of the cone, and that of the hemisphere (_vide
preceding rules_): which in this example will be 999·9741875.
Then (999·9741875 × 55) / 1728 = 31 pounds, nearly.
_To find the quantity of powder to fill a rectangular box._
Divide the content (viz., length × breadth × depth) of the box in
inches by 30 for the pounds of powder.
_Example._—How much powder will fill a box, the length being 15
inches, the breadth 12, and the depth 10 inches.
15 × 12 × 10 / 30 = 1800 / 30 = 60 pounds. Number required.
_To find the quantity of powder to fill a cylinder._
Multiply the square of the diameter by the length, then divide by
38·2 for the pounds of powder.
_Example._—How much powder will the cylinder contain, whose diameter
is 10 inches, and length 20 inches?
(10 × 10 × 20) / 38·2 = 52⅓ pounds, nearly.
_To find the size of a shell, to contain a given weight of powder._
Multiply the pounds of powder by 57·3, and the cube root of the
product will be the diameter in inches.
_Example._—Required the diameter of a shell to contain 6 lb. of
powder?
6 × 57·3 = 343·8, the cube root of which is 7, the diameter
required, in inches.
_To find what length of a cylinder (or bore of a gun) will be filled
by a given weight of powder._
Multiply the weight in pounds by 38·2, and divide the product by the
square of the diameter in inches, for the length.
_Example._—What length of a cylinder 8 inches in diameter will be
filled with 20 lb. of powder?
(20 × 38·2) / (8 × 8) = 11-15/16 inches.
_To find the content, and weight of a piece of ordnance._
Divide the length of the gun into as many sections as may be found
necessary. Find the content of each (_by preceding rules_) and from
their sum subtract the content of a cylinder, whose length is equal
to that of the bore, and its diameter equal to that of the calibre of
the piece; multiply the difference (if it be a brass gun) by 5·0833,
(if an iron gun) by 4·2968, and the product will be the weight in
ounces.
_Note._—A cubic inch of gun metal weighs 5·0833 ounces.
Ditto of _cast_ iron 4·2968 ounces.
_To find the content of a cask._
Multiply half the sum of the areas of the two interior circles, viz.
at the head, and bung, by the interior length, for the content.
Or, to the area of the head add twice the area at the bung, multiply
that sum by the length, and take one-third of the product.
_Example._—Required the content of a cask, its greatest interior
diameter being 24 inches, its least interior diameter 20 inches, and
the interior length 30 inches.
24 × 24 × ·7854 = 452·3904, area of large circle.
20 × 20 × ·7854 = 314·1600, area of small circle.
(452·3904 + 314·1600) / 2 = 383·2752, half sum.
Then 383·2752 × 30 = 11498·256, the content; which being divided by
277¼ (the number of cubic inches in a gallon) will give the number of
gallons contained in the cask.
Thus 11498·256 / 277·25 = 41·4725, &c. Number of gallons required.
_Note._-The content of any vessel in cubic feet, multiplied by 6·232
(or if in inches by ·003607) will give the number of _imperial
gallons it will contain_.
EPITOME OF MENSURATION.
OF THE CIRCLE, CYLINDER, SPHERE, ETC.
1. The circle contains a greater area than any other plane figure,
bounded by an equal perimeter, or outline.
2. The areas of circles are to each other as the squares of their
diameters; any circle twice the diameter of another contains four
times the area of the other.
3. The diameter of a circle being 1, its circumference equals 3·1416.
4. The diameter of a circle is equal to ·31831 of its circumference.
5. The square of the diameter of a circle being 1, its area equals
·7854.
6. The square root of the area of a circle, multiplied by 1·12837,
equals its diameter.
7. The diameter of a circle, multiplied by ·8862, or the
circumference multiplied by ·2821, equals the side of a square of
equal area.
8. The sum of the squares of half the chord, and versed sine,
divided by the versed sine, the quotient equals the diameter of the
corresponding circle.
9. The chord of the whole arc of a circle taken from eight times the
chord of half the arc, one-third of the remainder equals the length
of the arc.
10. Or, the number of degrees contained in the arc of a circle,
multiplied by the diameter of the circle, and by ·008727, the product
equals the length of the arc in equal terms of unity.
11. The length of the arc of the sector of a circle multiplied by its
radius, half the product is the area.
12. The area of the segment of a circle equals the area of the
sector, minus the area of a triangle whose vertex is the centre; and
base equals the chord of the segment.
13. The sum of the diameters of two concentric circles multiplied by
their difference, and by ·7854, equals the area of the ring, or space
contained between them.
14. The sum of the thickness, and internal diameter of a cylindric
ring multiplied by the square of its thickness, and by 2·4674, equals
its solidity.
15. The circumference of a cylinder multiplied by its length, or
height, equals its convex surface.
16. The area of the end of a cylinder multiplied by its length,
equals its solid content.
17. The area of the internal diameter of a cylinder multiplied by its
depth, equals its cubical capacity.
18. The square of the diameter of a cylinder multiplied by its
length, and divided by any other required length, the square root
of the quotient equals the diameter of the other cylinder of equal
solidity, or capacity.
19. The square of the diameter of a sphere multiplied by 3·1416
equals its convex surface.
20. The cube of the diameter of a sphere multiplied by ·5236, equals
its solid content.
21. The height of any spherical segment, or zone, multiplied by the
diameter of the sphere, of which it is a part, and by 3·1416, equals
the area, or convex surface of the segment;
22. Or, the height of the segment multiplied by the circumference of
the sphere of which it is a part, equals the area.
23. The solidity of any spherical segment is equal to three times the
square of the radius of its base, plus the square of its height, and
multiplied by its height, and by ·5236.
24. The solidity of a spherical zone equals the sum of the squares of
the radii of its two ends, and one-third the square of its height,
multiplied by the height, and by 1·5708.
25. The solidity of the middle zone of a sphere equals the sum of
the square of either end, and two-thirds the square of the height,
multiplied by the height, and by ·7854.
26. The capacity of a cylinder 1 foot in diameter, and 1 foot in
length, equals 4·895 imperial gallons.
27. The capacity of a cylinder 1 inch in diameter, and 1 foot in
length, equals ·034 of an imperial gallon.
28. The capacity of a cylinder 1 inch in diameter, and 1 inch in
length, equals ·002832 of an imperial gallon.
29. The capacity of a sphere 1 foot in diameter, equals 3·263
imperial gallons.
30. The capacity of a sphere 1 inch in diameter, equals ·001888 of an
imperial gallon.
31. Hence the capacity of any other cylinder in imperial gallons is
obtained by multiplying the square of its diameter by its length; or
the capacity of any other sphere by the cube of its diameter, and by
the number of imperial gallons contained as above in the unity of its
measurement.
OF THE SQUARE, RECTANGLE, CUBE, ETC.
1. The side of a square equals the square root of its area.
2. The area of a square equals the square of one of its sides.
3. The diagonal of a square equals the square root of twice the
square of its side.
4. The side of a square is equal to the square root of half the
square of its diagonal.
5. The side of a square, equal to the diagonal of a given square,
contains double the area of the given square.
6. The area of a rectangle equals its length multiplied by its
breadth.
7. The length of a rectangle equals the area divided by the breadth;
or the breadth equals the area divided by the length.
8. The side, or end of a rectangle, equals the square root of the sum
of the diagonal, and opposite side to that required, multiplied by
their difference.
9. The diagonal in a rectangle equals the square root of the sum of
the squares of the base, and perpendicular.
10. The solidity of a cube equals the area of one of its sides
multiplied by the length of one of its edges.
11. The edge of a cube equals the cube root of its solidity.
12. The capacity of a 12-inch cube equals 6·232 gallons.
_Surfaces, and solidities of the regular bodies, when the linear edge
is 1._
+-------------+--------------+------------+-----------+
|No. of Sides.| Names. | Surfaces. | Solids. |
+-------------+--------------+------------+-----------+
| 4 | Tetrahedron | 1·7320508 | 0·1178513 |
| 6 | Hexahedron | 6· | 1· |
| 8 | Octahedron | 3·4641016 | 0·4714045 |
| 12 | Dodecahedron | 20·6457788 | 7·6631189 |
| 20 | Icosahedron | 8·6602540 | 2·1816950 |
+-------------+--------------+------------+-----------+
The tabular surface multiplied by the square of the linear edge, the
product equals the surface required:
Or, the tabular solidity, multiplied by the cube of the linear edge,
the product is the solidity required.
OF TRIANGLES, POLYGONS, ETC.
1. The complement of an angle is its defect from a right angle.
2. The supplement of an angle is its defect from two right angles.
3. The sine, tangent, and secant of an angle, are the cosine,
cotangent and cosecant of the complement of that angle.
4. The hypothenuse of a right-angled triangle being made radii, its
sides become the sines of the opposite angles, or the cosines of the
adjacent angles.
5. The three angles of every triangle are equal to two right angles;
hence the oblique angles of a right-angled triangle are each other’s
complements.
6. The sum of the squares of the two given sides of a right-angled
triangle is equal to the square of the hypothenuse.
7. The difference between the square of the hypothenuse, and given
side of a right-angled triangle is equal to the square of the
required side.
8. The area of a triangle equals half the product of the base
multiplied by the perpendicular height;
9. Or, the area of a triangle equals half the product of the two
sides, and the natural sine of the contained angle.
10. The side of any regular polygon multiplied by its apothem, or
perpendicular, and by the number of its sides, half the product is
the area.
_Table of the areas of regular polygons whose sides are unity._
+----------+------+-----------+-----------+------------+------------+
| Name of |No. of| Apothem, | Area, when| Interior | Central |
| polygon. |sides.| or perpen-|side is one| angle. | angle. |
| | | dicular. | or unity. | | |
+----------+------+-----------+-----------+------------+------------+
| | | | | ° ′ | ° ′ |
| Triangle | 3 | 0·2886751 | 0·4330127 | 60 0 | 120 0 |
| Square | 4 | 0·5 | 1· | 90 0 | 90 0 |
| Pentagon | 5 | 0·6881910 | 1·7204774 | 108 0 | 72 0 |
| Hexagon | 6 | 0·8660254 | 2·5980762 | 120 0 | 60 0 |
| Heptagon | 7 | 1·0382607 | 3·6339124 | 128 34-2/7 | 51 25-5/7 |
| Octagon | 8 | 1·2071068 | 4·8284271 | 135 0 | 45 0 |
| Nonagon | 9 | 1·3737387 | 6·1818242 | 140 0 | 40 0 |
| Decagon | 10 | 1·5388418 | 7·6942088 | 144 0 | 36 0 |
| Undecagon| 11 | 1·7028436 | 9·3656399 | 147 16-4/11| 32 43-7/11|
| Dodecagon| 12 | 1·8660254 |11·1961524 | 150 0 | 30 0 |
+----------+------+-----------+-----------+------------+------------+
The tabular area of the corresponding polygon multiplied by the
square of the side of the given polygon, equals the area of the given
polygon.
OF ELLIPSES, CONES, FRUSTRUMS, ETC.
1. The square root of half the sum of the squares of the two
diameters of an ellipse multiplied by 3·1416 equals its circumference.
2. The product of the two axes of an ellipse multiplied by ·7854
equals its area.
3. The curve surface of a cone is equal to half the product of the
circumference of its base multiplied by its slant side, to which, if
the area of the base be added, the sum is the whole surface.
4. The solidity of a cone equals one-third of the product of its base
multiplied by its altitude, or height.
5. The squares of the diameters of the two ends of the frustrum
of a cone added to the product of the two diameters, and that sum
multiplied by its height, and by ·2618, equals its solidity.
THE END.
LONDON:
PRINTED BY W. CLOWES AND SONS, STAMFORD STREET
AND CHARING CROSS.
FOOTNOTES:
[1] _Vide page_ vi.
[2] The Articles omitted consist chiefly of directions, &c., or are
not generally required.
[3] NOTE. _In the “Exercise and Movements.”_
Commander’s Words are printed in SMALL CAPITALS.
Executive Small print.
Directions, &c. _Italics_.
[4] _Note._—Vide “Motion,” “Forces,” &c., Velocity, Gravity, and
Amplitude.
[5] Vide “Tables,” “Excentric Shot, Experiments.”
[6] _When a shot is jammed in a gun, and cannot be rammed home to the
cartridge_, destroy the charge by pouring water down the vent, and
muzzle until the ingredients are dissolved, and cleared out of the
bore; then introduce a small quantity of powder through the vent, and
blow out the shot.
[7] The recoil of guns on sleighs varies from four to five feet when
on rough ground or in deep snow; to twenty or thirty yards when on
glare ice. In the latter case it is of course necessary to send
the ammunition sleighs further to the rear; but the recoil may be
considerably lessened by placing a small chain round each of the
runners.
Ice of eight inches thick will bear with safety a weight of 1115 lb.
(or nearly half a ton) on the square foot.
[8] Old pattern.
[9] Further information relative to mixing the composition, and
filling combustibles, &c., &c., may be obtained from the “Aide
Mémoire,” under the head, “Pyrotechny, Military.”
[10] This will be discontinued when Shrapnell Diaphragm shells are
generally introduced into the Service.
[11] _Vide_ Practice Tables for Ranges, Elevations, &c.
[12] The composition for French cannon tubes is two parts of
fulminate of mercury and two of mealed powder, mixed together: then
formed into a paste with distilled water, slightly impregnated with
gum arabic.
[13] Extracted from “Instructions and Regulations for Field Battery
Exercise and Movements” for the Royal Regiment of Artillery: the
Sections, &c., being similarly numbered.
Commander’s Words are printed in SMALL CAPITALS.
Executive Common type.
Directions, &c. _Italics._
[14] The Sections, of which merely the heads are given, consist
chiefly of details too long for the limited size of the Manual, and
they are therefore necessarily omitted.
[15] When Guns are in action, and “CEASE FIRING” is given, all Guns
then loaded are to be fired off, and on no account is a Gun to be
limbered up, or to move whilst loaded.
[16] The Commanding officer’s Word of command is always to be
repeated by the officers.
[17] From “Field Battery Exercise.”
[18] From “Field Battery Exercise,” &c.
[19] From “Field Battery Exercise,” &c.
[20] In the transport of horses to Turkey (July, 1854,) in the
Himalaya and Simla steamers, the distance between the upright posts
was 2 feet 1 inch in the clear per horse, and the length 9 feet.
[21] “For the guidance of the Farriers of the Royal Artillery.
Suggested by Charles Percival, Veterinary Surgeon; and approved of
by the Right Honourable the Master-General, and Honourable Board of
Ordnance.”
[22] In administering draughts to horses, the greatest possible
care and attention are required; should the horse cough, or make an
attempt to do so, his head must be instantly lowered, otherwise a
portion of the drink will be apt to find its way into the trachea or
windpipe, which will produce most distressing symptoms, and often be
followed by death. In lowering the head, a can or vessel of any kind
should be held under the mouth to catch the drink as it escapes.
[23] From “INSTRUCTIONS FOR THE SERVICE OF HEAVY ORDNANCE.”—_Article
15._
[24] _Words of command_—SMALL CAPITALS.
[25] From “Instructions, and Regulations for the Service, and
Management of Heavy Ordnance, for the Royal Regiment of Artillery.”
Fourth edition. The Parts, and Articles are numbered in conformity
thereto.
[26] _Words of command_—SMALL CAPITALS.
[27] Vide PART 12, “ARTILLERIST’S MANUAL,” etc., The Mechanical
powers. The Lever.
[28] By the ballistic experiment, conducted in May, 1837, it was
found that, with a heavy 6-pounder gun, a charge of 1½ lb. gave a
velocity of 1740 feet, and a charge of 2 lb. a velocity of 1892 feet
per second. The shot employed were of a high gauge, windage only ·078
inch, and the powder was of the strongest quality; the weight of the
pendulum fired into was 58 cwt. 3 qrs. 16 lb. A light 6-pounder, two
feet shorter than the heavy 6-pounder, with similar charges, gave
velocities of about 190 feet less.
[29] Extracted from PART 2, and APPENDIX of General Sir Howard
Douglas’ highly valued work, entitled “A TREATISE ON NAVAL GUNNERY.”
3rd edition.
[30] _On wads for Heavy Ordnance._
The presence of a compressible body, between the powder and the ball,
is necessary for the preservation of the gun. The results of the
experiments at Fere, in 1844; at Ruelle in 1844, and 1846; and at
Gavres in 1848; with cast iron 30, and 24-pounders, proved that all
the pieces, fired without a thin piece of cork interposed between the
powder and the ball, burst before 500 discharges were made; whilst
those, with which this precaution was taken, sustained 1800 and 2000
discharges without any damage, except an enlargement of the vent.
_Vide_ United Service Magazine, September, 1855.
[31] _Vide_ “TREATISE ON NAVAL GUNNERY.” 3rd Edition. By General Sir
H. Douglas.
[32] _In Extreme training of a gun to the Right_: Nos. 3, 5, 7, 11,
13, are placed outside; Nos. 8, 6, inside the tackle. No. 13 keeps
the end of the fall coiled up.
_In Extreme training to the Left_: Nos 4, 6, 8, 2, are placed
outside; Nos. 13, 7, 5, inside the tackle. No. 2 keeps the end of the
fall coiled up.
[33] _In running out the right guns_, Nos. 3, 5, 5, 7, man the left
tackle; Nos. 4, 6, 6, 2, man the right tackle.
_In running out the left guns_, Nos. 3, 5, 7, 5, man the left tackle;
Nos. 4, 6, 6, 2, man the right tackle.
[34] _Note._—When the direction of the gun is to be altered, the word
“Traverse” is to be given, if the gun is in, and “Point,” when the
gun is out.
[35] _Vide_ Sir Howard Douglas’s highly-valued publication, entitled
“A TREATISE ON NAVAL GUNNERY.” Fourth edition.
[36] _Vide_—“United Service Magazine,” No. CCCVIII.
[37] _Vide_ FIELD FORTIFICATION, pages 246, 247.
[38] _Vide_ Preface.
[39] For a square, the length of the perpendicular is ⅛th the
exterior side; for a pentagon ⅐th; for the hexagon, and other
polygons, ⅙th.
[40] _Vide Tables of Weights, and Measures._
[41] _Vide Tables_ of Weights, and Measures.
[42] In reducing fractions to a common denominator, and in
multiplication of fractions, the work may be considerably diminished
by cancelling any figures, which are in all the multiples; or by
dividing a figure in each of them by any figure which can divide all
without any remainder.
[43] See Note, page 268.
[44] _To multiply decimals by 1, with any number of ciphers, as
10, 100_, &c.—This is done by only removing the decimal point so
many places farther to the right hand, as there are ciphers in the
multiplier, and subjoining ciphers, if need be.
[45] The best way of doubling the root, to form the new divisor, is
by adding the last figure always to the last divisor, as appears in
the following example.
After the figures belonging to the given number are all exhausted,
the operation may be continued into decimals, by adding any number of
periods of ciphers, two in each period.
[46] _This rule is only applicable to the very best-made new cordage.
The circumference squared should be divided by 6 instead of 5 for the
description of rope generally employed._
[47] When the board is tapering, add the breadths at the two ends
together, and take half the sum for the mean breadth. _Or else_, take
the mean breadth in the middle.
[48] _To strengthen a beam, &c. which is required to support a great
weight over a cavity, or ditch._—Place a prop, or short skid, under
the centre of the beam, and pass a strong rope, or chain, over the
beam lengthways, and under the skid, hauling it very tight, and
making fast.
[49] In Lieut.-Colonel B. Jackson’s scientific “Treatise on
Military Surveying, &c., &c., &c.,” _Portable trigonometry without
logarithms_, is thus introduced—
“The following useful application of Trigonometry, by means of the
natural sines, tangents, &c., is taken from an early number of that
valuable periodical, ‘The Mechanics’ Magazine,’ and will be found
particularly suited to the purposes of the military surveyor.”
[50] For further information on Surveying, and Reconnoitring,
reference should be made to the highly-valued publication, entitled
“A TREATISE ON MILITARY SURVEYING, INCLUDING SKETCHING IN THE
FIELD, PLAN DRAWING, LEVELLING, MILITARY RECONNOISSANCE, &c.,” by
Lieut.-Colonel Basil Jackson, containing a full account of every
surveying instrument, and the right adaptation of them.
[51] 1. The Reconnoitring protractor is not intended to supply the
place of the Theodolite, or other expensive instruments, when very
great accuracy is required in surveying, or in trigonometrical
observations; but, in the hands of officers accustomed to the use of
it, bearings may be rapidly taken, heights and distances ascertained,
roads traversed, &c., &c., with sufficient accuracy for a military
survey, or reconnoissance.
The protractor has a tripod, on which it is to be steadily fixed
for taking angles, &c.; but the instrument can nevertheless be
used without the tripod; and mounted officers may, after a little
practice, make a reconnoissance with the protractor alone, especially
if they are able to measure, or calculate the distance of base lines,
by the length of the paces of their horses.
2. A survey, &c., may be very rapidly taken in the field, by laying
drawing-paper on the face of the protractor, under the marginal
scale, fixing it firmly by means of drawing-pins in the sides, and
using, at the first station, the edge of the index as a ruler to set
off on the paper, at once, by observation through the sights, the
angles of the objects whose distance is required; drawing a base
line parallel to the tube side of the instrument, and also lines at
the angles found. At the second station, the paper must be moved a
few inches, for a base line to be drawn; at the termination of which
(the second station) the index is to be directed to the objects,
as before, and lines are to be produced until they intersect those
drawn at the first station: thus the position of the objects will be
obtained; and, by using the scale on the index for the length drawn
for the measured base line, as well as for the lines directed to the
objects, their respective distances will be ascertained.
3. The reconnoitring protractor, and all other instruments for
surveying, &c., &c. can be readily obtained from Messrs. Elliott, 56,
Strand, London.
[52] Or Reconnoitring protractor.
[53] To erect a perpendicular, _vide_ “Practical Geometry.”
[54]
3 inch cube full of air floats 1 lb. in water.
3 inch cube of water weighs 1 lb. in air.
1 cubic foot of water weighs 64 lb. in air.
1 ditto coal ditto 80 - 64 = 16 in water.
1 ditto sand ditto 95 - 64 = 31 in water.
A suit of clothes and a pair of boots, which weigh 7 lb. in air, when
well saturated with water, only weigh in water 1 lb.
[55] _Vide also Definitions_—TRIGONOMETRY, page 301.
[56] _Gunter’s chain_ is in length 4 poles = 22 yards = 66 feet, and
is divided into 100 links. Each link is therefore 22/100 of a yard,
or 66/100 of a foot, or 7·92 inches. _Land is estimated_ in acres,
roods, and perches. _An acre_ contains 10 square chains, or as much
as 10 chains in length and 1 chain in breadth; or in yards it is 220
× 22 = 4840; or in poles it is 40 × 4 = 160 square poles; or in links
it is 1000 X 100 = 100,000 square links. An acre is divided into 4
parts called roods, and a rood into 40 parts called perches, which
are square poles, or the square of a pole of 5½ yards long, or the
square of a quarter of a chain, or of 25 links, which is 625 links.
Thus the divisions of land measure are—
625 square links = 1 pole, or perch.
40 perches = 1 rood.
4 roods = 1 acre.
The length of lines, measured with a chain, should be set down in
links as integers, instead of in chains, and decimals. Therefore,
after the content is found, it will be in square links.
[57] 57·3 is the number of pounds of powder contained in a cubic
foot, when shaken; and 55 pounds when not shaken. According to the
first case, one pound of powder will occupy 30 cubic inches; and
according to the second case one pound will occupy 31·4182 cubic
inches.
TRANSCRIBER’S NOTE
Footnote [37] is referenced six times from page 237;
footnote [52] is referenced twice from page 311;
footnote [54] is referenced four times from pages 317 and page 318.
Obvious typographical errors and punctuation errors have been
corrected after careful comparison with other occurrences within
the text and consultation of external sources.
Some hyphens in words have been silently removed, some added,
when a predominant preference was found in the original book.
Some { bracketing in some tables has been adjusted or removed for
readability.
Except for those changes noted below, all misspellings in the text,
and inconsistent or archaic usage, have been retained.
Pg viii: page number ‘8’ replaced by ‘48’.
Pg xvi: Added new section ‘CONGREVE ROCKETS.’ to the ToC.
Pg xviii: ‘Embrasures’ replaced by ‘Embrazures’.
Pg xxi: ‘312, 313, 314’ replaced by ‘312–4’.
Pg 61: ‘thirty ronnds of’ replaced by ‘thirty rounds of’.
Pg 69: in the table header ‘Fore ... Diameter’ replaced by
‘Fore ... Hind’.
Pg 72: in second column of the table ‘3¾’ replaced by ‘2¾’.
Pg 80: ‘4⅗ inch Mortar’ replaced by ‘4⅖ inch Mortar’.
Pg 85: in the table ‘5½ in c’ replaced by ‘5½ inch’.
Pg 91: in LEVERS section ‘lb. oz.’ replaced by ‘ft. in.’
Pg 171: in the table ‘1’ replaced by ‘10’.
Pg 171: in the table ‘1209’ replaced by ‘1200’.
Pg 176: ‘to facilite the’ replaced by ‘to facilitate the’.
Pg 187: ‘assist 2 at’ replaced by ‘assists 2 at’.
Pg 195: ‘at an elevavation’ replaced by ‘at an elevation’.
Pg 220: the Remarks column has been moved under the table to
conserve table space.
Pg 222: ‘a longe range’ replaced by ‘a long range’.
Pg 226: ‘they ricoched and’ replaced by ‘they ricocheted and’.
Pg 226: the italic markup on the small table has been removed.
Pg 227: ‘the same is in’ replaced by ‘the same as in’.
Pg 234: ‘left betweeen the’ replaced by ‘left between the’.
Pg 235: ‘placed at tho top’ replaced by ‘placed at the top’.
Pg 238: ‘embrasures should be’ replaced by ‘embrazures should be’.
Pg 291: ‘(5 × 1)’ replaced by ‘(5 + 1)’.
Pg 318: the footnote in the original book ‘See note, p. 317’ was
redundant and has been removed.
Pg 320: ‘is is a’ replaced by ‘it is a’.
Footnote [7]: ‘to five eet’ replaced by ‘to five feet’.