Chapter XXXIV: Part XII: Mathematics (2)

THE MECHANICAL POWERS.

_Power is compounded of the weight, or expansive force of a moving body multiplied into its velocity._

The power of a body, which weighs 40 lb., and moves with the velocity of 50 feet in a second, is the same as that of another body which weighs 80 lb., and moves with the velocity of 25 feet in a second: for the products of the respective weights, and velocities are the same.

40 × 50 = 2000; and 80 × 25 = 2000.

_Power cannot be increased by mechanical means._

Power is applied to mechanical purposes—

1. By the lever; 4. By the inclined plane;
2. By the wheel, and axle; 5. By the wedge;
3. By the pulley; 6. By the screw;

which are the simple elements of all machines.

The whole theory of these elements consists simply in causing the weight, which is to be raised, to pass through a greater or a less space than the power which raises it; for, as power is compounded of the weight, or mass of a moving body, multiplied into its velocity, a weight passing through a certain space may be made to raise, through a less space, a weight heavier than itself.

THE LEVER.

_The lever_ is the most simple of all machines, being only a straight bar of iron, wood, &c., supported on, and moveable round a prop, called the _fulcrum_.

_Case 1._—When the fulcrum of the lever is between the power, and the weight.

_Rule._—Divide the weight to be raised by the power to be applied; the quotient will give the difference of leverage necessary to support the weight in equilibrio. Hence, a small addition either of leverage, or weight, will cause the power to preponderate.

_Example 1._—A ball weighing 3 tons is to be raised by 4 men, who can exert a force of 12 cwt.; required the proportionate length of lever?

3 tons = 60 cwt.; and 60/12 = 5

In this example, the proportionate lengths of the lever to maintain the weight in equilibrio, are as 5 to 1. If, therefore, an additional pound be added to the power, the power side of the lever will preponderate, and the weight will be raised. But, although the ball is raised by a force of only one-fifth of its weight, no power is gained, for the weight passes through only one-fifth of the space. The products, therefore, arising from the multiplication of the respective weights, and velocities are the same.

_Example 2._—A weight of 1 ton is to be raised with a lever 8 feet in length, by a man who can exert, for a short time, a force of rather more than 4 cwt.; required at what part of the lever the fulcrum must be placed?

(20 cwt.) / (4 cwt.) = 5; that is, the weight is to the power
as 5 to 1,
therefore, 8 / (5 + 1) = 1 foot and a third, from the weight.

_Example 3._—A weight of 40 lb. is placed 1 foot from the fulcrum of a lever; required the power to raise the same, when the length of the lever on the other side of the fulcrum is 5 feet?

(40 × 1) / 5 = 8 lb. _Ans._

_Case 2._—_When the fulcrum is at one extremity of the lever, and the power at the other._

_Rule._—As the distance between the power, and the fulcrum is to the distance between the weight, and the fulcrum, so is the effect to the power.

_Example 1._—Required the power necessary to raise 120 lb., when the weight is placed 6 feet from the power, and 2 feet from the fulcrum?

As 8 : 2 :: 120 : 30 lb. _Ans._

_Example 2._—A beam, 20 feet in length, and supported at both ends, bears a weight of 2 tons at the distance of eight feet from one end; required the weight on each support?

(40 cwt. × 8 feet) / (20 feet) = 16 cwt. on the support that is
furthest from the weight; and (40 × 12) / (20 feet) = 24 cwt.
on the support nearest to the weight.

_Case 3._—_When the weight to be raised is at one end of the lever, the fulcrum at the other, and the power is applied between them._

_Rule._—As the distance between the power, and the fulcrum, is to the length of the lever, so is the weight, to the power.

_Example._—The length of the lever being 8 feet, and the weight at its extremity 60 lb., required the power to be applied 6 feet from the fulcrum to raise it?

As 6 : 8 :: 60 : 80 lb. _Ans._

_Velocity is gained at the expense of power by the lever, and wheel, and axle._

Note 1.—When two men are carrying a load on a pole between them,
the strongest man should have the weight placed nearer to him than
the other man.

Note 2.—_To carry guns, &c._—If the burden can be carried by four
men; after having made it fast to the middle of a large lever, fix
the extremities of this lever on two shorter levers, and place a
man at each of the points, C, D, E, F. _Vide plate, Mechanics, Fig.
1._ In _Fig. 2_, the weight is equally divided between eight men,
and in _Fig. 3_, between sixteen men.

THE WHEEL, AND AXLE.

The advantage gained is in proportion as the circumference of the wheel exceeds that of the axle; therefore, the larger the wheel, and the smaller the axle, the stronger is the power of this machine, but then the weight will rise proportionally slower. A winch may be used instead of a wheel, for in turning the winch the hand will describe a circle, and there is no difference in the result, whether an entire wheel be turned, or a single spoke which the winch as a lever represents.

_Rule._—As the radius of the wheel is to the radius of the axle, so is the effect, to the power.

_Example._—A weight of 50 lb. is exerted on the periphery of a wheel, whose radius is 10 feet; required the weight raised at the extremity of a cord wound round the axle, the radius being 20 inches.

(50 lb. × 10 feet × 12 inches) / (20 inches) = 300 lb. _Ans._

THE PULLEY.

_The pulley_ consists of a grooved wheel, called _a sheave_, moveable on an axis, or gudgeon, and enclosed in a frame, or case, called a _block_. By passing a cord over the pulley, a man will be enabled _to draw up_ a weight equal to that which his own body supplies in pulling downwards.

_By combining a number of pulleys_, as many assistants are obtained as there are wheels: thus, two pulleys will have double the power of one, because half the weight is sustained by the frame to which one end of the cord is attached; but then it requires _double the time_ to do the work. As the _friction of the pulley_ is very great, particular attention must be paid that all the turns or kinks of a rope be taken out, before it is made use of, and it should enter easily into the grooves of the sheaves.

_Rule._—Divide the weight to be raised by twice the number of pulleys in the lower block; the quotient will give the power necessary to raise the weight.

_Example._—What power is required to raise 600 lb., when the lower block contains six pulleys?

600 / (6 × 2) = 50 lb. _Ans._

TACKLES.

_Tackles_ are indispensable in the service of the artillery.

_The fall_ is the rope of which the tackle is composed; that end of it which is fixed to the block is called _the standing part_, or _end_; the other, which is pulled, or hauled on by the men, is called _the running part_, or _end_; and the parts which pass from one block to the other are called the _returns of the fall_.

_In all operations with tackles, the following directions should be attended to_:—

1st. Make fastenings stronger than appears _actually_ necessary.

2nd. Examine the straps, and hooks of the blocks carefully.

3rd. Consider whether the cordage is new, half-worn, or almost worn out.

4th. Attend to the seizings of the clinches, the sheet bends, the _proper_ stoppering of the fall, the belaying the fall with two half hitches.

5th. Be very distrustful of selvages applied on smooth worn rope.

6th. Do not allow ropes to be struck, or trampled on, when the weight is suspended.

7th. The men should _stand as safe_ as the proper performance of the various duties will permit.

8th. In pulling a rope, the men ought to place themselves in a right line, and haul together. The most advantageous position for pulling is down a slope, or in a descending position.

COMBINATION OF PULLEYS.

_A leading block_ is a fixed pulley, which alters the direction of the power, but does not increase it: Power = Weight. On account of friction the power must exceed the weight a little, in order to raise it.

_Vide plate, Mechanics, Fig. 1._

_A whip_ is one moveable pulley, which increases the power without altering the direction.

Power = ½ weight (or 2 to 1).—_Vide Fig. 2._

_A whip upon whip_ will afford the same purchase as a tackle having a single and double block, and with much less friction.

_A gun tackle_ consists of two single blocks with fall fixed to the one, then rove through the other, and then through the first. Power = ½ weight (or 2 to 1): or Power = ⅓ weight (or 3 to 1). _Vide Fig. 3, and 4._

Two double blocks are generally used for very heavy guns.

_A luff tackle, or half watch tackle_, consists of one double and one single block: the fall is fixed to the single, then rove through first sheave of the double, then through sheave of single, and lastly through second sheave of double block. Power = ⅓ weight (3 to 1): or Power = ¼ weight (4 to 1). _Vide Fig. 5, and 6._

_A runner tackle_ is the same as a luff tackle, applied to the end of a large rope, called a runner, which is rove through a single block attached to a fixed point, or to a body that is to be moved, or raised; the standing end of the runner being secured to another point.

Power is either 6 to 1, or 7 to 1, or 8 to 1.

_A gyn tackle_ consists of one triple and one double block: the fall is fixed to the double, then rove through first sheave of triple, then through first sheave of double, then through second sheave of triple, then through second sheave of double, and lastly through third sheave of triple block.

Power = ⅕ weight (5 to 1): or Power = ⅙ weight (6 to 1). _Vide Fig. 7._

If the moveable block of a tackle be strapped with a tail, it is called a _tail_, or _jigger block_: and the tackle a _tail_, or _jigger tackle_: a block with a hook strapped to it, and attached to a selvage, answers the same purpose.

_Two double blocks_, with fall fixed to one of them, and then rove through the sheaves of both blocks, will either give Power = ¼ weight (4 to 1): or Power = ⅕ weight (5 to 1). _Fig. 8._

_Two triple blocks_, with fall fixed to one of them, then rove through sheaves of both blocks, will either give power = ⅙ weight (6 to 1): or Power = ⅐ weight (7 to 1). _Fig. 9._

Fig. 1 2 3 4 5 6 7 8 9]

Fig. 1 2 3]

_In the system of pulleys_ (_vide plate, Mechanics_) the Power is shown at the hooks of the moveable blocks, which are to be applied to the bodies, or weights, requiring to be moved or raised. The strain is also shown at the fixed blocks.

_In Fig. 3_, there are _three parts of the rope engaged_ in supporting the weight—_viz._, _the parts marked 1, 1, 1_. Each of them, hence, sustains _one-third_ of it, and the fall of the rope to which the power is to be attached requires the Power = 1, if weight = 3. The same principle of calculation is applicable to all systems of pulleys having one fixed block, any number of moveable wheels, and a single rope over all the wheels. Hence, in such a system of pulleys, gravity being applied, there will be an equilibrium, when the weight is as many times the power as there are portions of the rope employed in sustaining the weight. _For example_, in a system consisting of six moveable sheaves, the same rope going over them all, there will be 12 portions of the rope engaged; and to produce an equilibrium the power must be equivalent to 1/12 the weight, no allowance being made for friction.

From the foregoing observations, and by referring to the plate, it will be seen that _each tackle has two applications_, differing in power one from the other; _for example_, if the double block of a luff tackle is fixed to a weight to be moved, and the single block to a picket, or other fastening, _Fig. 6_, then, if _one man_ haul on the fall, the power of _four men_ will be applied to the weight (4 to 1), and the power of _three men_ to the picket; but if the double block be fixed to the picket, _Fig. 5_, and the single, block to the weight, then the force of only _three men_ will be applied to the weight (3 to 1), and a power of _four men_ to the picket, or fastening.

When the moveable block of one tackle is fixed to the fall of another tackle, their respective powers are to be multiplied into each other for the power of the combination: thus, if one luff tackle is fixed to the fall of another luff tackle (the double blocks of both tackles being moveable), the power will be 4 × 4 = 16 (16 to 1): in this, the men haul through 16 feet to move the weight one foot; therefore if the combination be increased until the men haul through 100 feet to move the weight one foot, then the power would be 100 to 1.

The foregoing powers are, however, only true in theory, and are, therefore, called _theoretical powers_: for owing to the great friction of the pulleys, the stiffness of the ropes, &c., the actual _practical powers_ are far less; so much so, that with a combination giving a power of 48 to 1, a 24-pr. (2½ tons weight) suspended, can scarcely overhaul the fall, the friction being so very great.

THE INCLINED PLANE.

_The inclined plane_ forms simply a gradual and sloping instead of a sudden and perpendicular ascent, by which heavy bodies may be raised to certain heights. The power necessary for raising a weight depends on the difference between the length of the plane and the height to be ascended. If the height be one-third of the length, then one pound will lift three pounds. The force with which a rolling body descends on an inclined plane is to the force of its absolute gravity, as the height of the plane is to its length.

_Parbuckling a gun_ on skids unites the advantage of one moveable pulley with that of the inclined plane.

_Rule._—As the length of the plane is to its height, so is the weight to the power.

_Example._—Required the power necessary to raise 540 lb. up an inclined plane, five feet long, and two feet high.

As 5 : 2 :: 540 : 216 lb. _Ans._

THE WEDGE.

_The wedge_ may be considered as two equally inclined planes joined together at their bases. It has a great advantage over all the other powers, arising from the force of percussion, or blow, with which the back is struck; which is a force incomparably greater than any dead weight, or pressure, such as is employed in other machines. The largest masses of timber may by this means be riven, and vessels of war, weighing many thousand tons, are lifted from their supports by the power of a few men, exerted by blows of mallets on wedges inserted for that purpose.

_The power of the wedge_ increases in proportion as its angle is acute. In tools intended for cutting wood the angle is commonly about 30°; for iron from 50° to 60°; and for brass from 80° to 90°.

_Case 1._—_When two bodies are forced from one another, by means of a wedge, in a direction parallel to its back._

_Rule._—As the length of the wedge is to half its back, or head, so is the resistance, to the power.

_Example._—The breadth of the back, or head of the wedge, being three inches, and the length of either of its inclined sides 10 inches, required the power necessary to separate two substances, with a force of 150 lb.

As 10 : 1½ :: 150 : 22½ lb. _Ans._

_Case 2._—_When only one of the bodies is moveable._

_Rule._—As the length of the wedge, is to its back, or head, so is the resistance, to the power.

_Example._—The breadth, length, and force, the same as in the last example.

As 10 : 3 :: 150 : 45 lb. _Ans._

THE SCREW.

_The screw_ is a spiral thread or groove cut round a cylinder, and everywhere making the same angle with the length of it. The force of a power applied to turn a screw round is to the force with which it presses upward, or downward, setting aside the friction, as the distance between two threads is to the circumference where the power is applied; or the advantage gained is as much as the circumference of a circle described by the handle of the winch exceeds the interval, or distance, between the spirals of the screw. Hence the force of any machine turned by a screw can readily be computed; for instance, in a press driven by a screw, whose threads are each a quarter of an inch asunder, and with a handle, to turn the screw, four feet long; then, if the natural force of a man, by which he can lift, pull, or draw, be 150 lb., and it is required to determine with what force the screw will press when the man turns the handle with his whole force; the diameter of the handle (_power_) being 4 feet, or 48 inches, its circumference is 48 × 3·1416, or 150⅘ nearly; and the distance of the threads being one-fourth of an inch, therefore the power is to the pressure as 1 to (150⅘ × 4) = 603⅕, but the power is equal to 150 lb., therefore as 1 : 603⅕ :: 150 : 90480, and consequently the pressure is equal to a weight of 90480 lb. independent of friction.

COMPOUND MACHINES.

Though each of the mechanical powers is capable of overcoming the greatest possible resistance in theory, yet in practice, if used singly for producing very great effects, they would frequently be so unwieldy and unmanageable as to render it impossible to apply them. For this reason it is generally found more advantageous to combine them together, by which means the power is more easily applied, and many other advantages are obtained. In all the mechanical powers, and their combinations, and in all machines, simple as well as compound, _what is gained in power is lost in time or velocity_; and _vice versâ_, or in other words, the product of the power, and the space through which it moves, is equal to the product of the weight, and the space through which it moves in the same plane. Suppose that a man, by means of a fixed pulley, raises a beam to the top of a house in two minutes, it is clear that he will be able to raise six beams in twelve minutes; but by means of a tackle with three lower pulleys, he will raise the six beams at once with the same ease as he before raised one, but then he will be six times as long about it, that is, twelve minutes; thus the work is performed in the same time whether the mechanical power is used, or not. But the convenience gained by the power is very great; for if the six beams are joined in one, they may be raised by the tackle, though it would be impossible to move them by the unassisted strength of one man. No real gain of force is obtained by mechanical contrivances; on the contrary, from friction and other causes, force is always lost; but by machines a more convenient direction can be given to the moving power, and so modify its energy as to obtain effects which it could not otherwise produce.

FRICTION.

_Friction_ arises from the irregularities of the surfaces which move upon one another. The surfaces of bodies of the same nature are moved with more facility over each other than those of a dissimilar nature. In proportion as the surfaces which are to be moved upon one another are rough, a greater force is requisite to produce motion. The same surfaces when under a greater pressure, are subject to still further friction. A double pressure doubles the amount of friction, a treble pressure trebles, and so on in nearly the same proportion. When surfaces are moving along each other in the direction of their grains, the friction is greater than when the direction of the grains is at right angles. Friction is little influenced by the velocity with which bodies move upon one another. Friction may be diminished in various ways, as will appear by the result of the following experiment with a block of square stone weighing 1080 lb.:—

1. In order to drag this stone along the floor of a quarry lb.
roughly chiselled, it required a force equal to 758

2. Over a floor of planks, ditto 652

3. Placed on a platform of wood, and dragged over a floor of
planks 606

4. After soaping the two surfaces of wood, which slide over
each other 182

5. Placed on rollers of three inches diameter, and moved along
the floor of the quarry 34

6. To drag it on these rollers over a wooden floor 28

7. Mounted on a wooden platform, and the same rollers placed
between the platform, and a plank floor 22

One of the most remarkable instances of the application of rollers is the transport of the rock which now serves as the pedestal of the equestrian statue of Peter the Great at St. Petersburg. This rock is a single block of granite weighing 1217 tons. A railway was formed, consisting of two lines of timber, furnished with hard metal grooves; similar, and corresponding metal grooves were fixed to the under side of the sledge, or frame, on which the stone was laid, and between these grooves were placed spheres of hard brass, about six inches in diameter. On these spheres the frame with its enormous load was easily moved by sixty men working at capstans with triple purchase blocks.

UNGUENTS.

Mr. G. Rennie found, from a mean of experiments, with different unguents, on axles in motion, and under different pressures, that, with the unguent tallow under a pressure of from 1 to 5 cwt., the friction did not exceed 1/39th of the whole pressure; when soft soap was applied, it became 1/34th; and with the softer unguents applied, such as oil, hogs’ lard, &c., the ratio of the friction to the pressure increased; but with the harder unguents, as soft soap, tallow, and anti-attrition composition, the friction considerably diminished; consequently, to render an unguent of proper efficiency, the nature of the unguent must be measured by the pressure, or weight, tending to force the surfaces together.

TRANSVERSE STRENGTH OF MATERIALS.

When a beam, of any material, is loaded, the surface in contact with the load is _compressed_, and the opposite surface _extended_; and there is a line between these, which is neither compressed, nor extended, called _the neutral line_.

_If the depth of a beam be doubled_, the breadth, and length between supports remaining the same, _its strength will be increased four times_.

_If its breadth be doubled_, the other dimensions being as above, _its strength will be doubled_.

By increasing the distance between the supports of any beam, its strength is decreased in the same ratio; twice the distance between the supports will weaken the beam one-half; half the distance between the supports will enable it to bear twice the load.[48]

The same beam will bear twice the load, if, instead of being concentrated in the middle, it be equally distributed over the whole length of the beam.

If the load on a beam be placed near to one of the supports, instead of in the middle, its effect will decrease in the ratio of its proximity to the support.

Let S s represent the beam, W the load or weight in the middle, w the weight near s; then the load which the beam will carry at the point where w is placed will be found by the following proportion:—

As S w × w s : S W × W s :: W : w.

A beam, fixed at one end, and loaded at the other, will bear half the 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.

_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.

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_.]