Chapter XXXV: Part XII: Mathematics (3)

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.

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._

_Fig. 1-9._]

_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._

_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.