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.
Table,
showing the rate of inclination of inclined planes, for the following angles of elevation.
Six survey sketchesSurveying, and Reconnoitring.J. W. Lowry, sc.Fig. 1.Fig. 2.Fig. 3.Fig. 4.Fig. 5.Fig. 6.
Surveying, and Reconnoitring.J. W. Lowry, sc.Fig. 1.Fig. 2.Fig. 3.Fig. 4.Fig. 5.Fig. 6.
J. W. Lowry, sc.
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 (videPlate 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 theline 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 - bB × b= the height of the hill in feet, for the temperature of 55 degrees on Fahrenheit. Add1440of 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 sideuppermost, 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; proceedto 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 (videTrigonometry, 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 staffis to the altitude of the staff,so is the length of the shadow of the objectto 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 gunis 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.
Plate 2.Geometric figuresHeights.Fig. 1.Fig. 2.Distances.Fig. 3.Fig. 4.Fig. 5.Practical Geometry.Fig. ½.Fig. 21.Fig. 22.
Plate 2.
Heights.Fig. 1.Fig. 2.Distances.Fig. 3.Fig. 4.Fig. 5.Practical Geometry.Fig. ½.Fig. 21.Fig. 22.
Heights.Fig. 1.Fig. 2.
Distances.Fig. 3.Fig. 4.Fig. 5.
Practical Geometry.Fig. ½.Fig. 21.Fig. 22.
7.—BY MEANS OF TWO PICKETS,
to ascertain the height of an object.
[Vide 2nd Plate,Heights, and Distances,Fig. 1.]
Let two pickets C D (4 feet), E F (6 feet), be placed with their bases in the line C A passing through A the height required, and movethem 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.
Right triangle A B C
Take the angle B C A with the sextant,[52]and note it in minutes; then A B, in feet × 573 ÷ B C A, in minutes = A C in fathoms. Or A B in feet × 573 ÷ B C A, in minutes × 2 = A C in yards.
573 is a constant multiple.
This method requires no table of sines, &c., the number of minutes in the angle being used instead of the sine.
2.—BY MEANS OF A POCKET SEXTANT,
to measure inaccessible distances.
When used for taking the distance of objects, the sextant is to be held horizontally, and the quicksilvered part of the glass will be uppermost, or above the transparent part.
To ascertain the distance A B (vide Plate 2,Fig. 2), obtain, by observation, the direction A C perpendicular to A B, which is thus performed:—Set the instrument at 90°, and place yourself at the point A, with your right towards the point B; then look through the sextant, and direct a picket to be placed in the line A C at 100 yards, or feet, from you, so that the point B will appear right above it. Then set the sextant at 45°, and walk along the line towards C until you bring the points A, and B to coincide; the base and perpendicularwill 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,andD.
[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 scaleis to the length of the gun;so is the height of the distant objectto 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 benearlythe breadth of the river.
11.—BY THE REPORT OF FIRE-ARMS, TO ASCERTAIN THE DISTANCEOF ANY OBJECT,videSound,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 yardsinfantry 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.
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 havewritten 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 1107English 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 1100half a footfor 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.
(Vide11. 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 louderin 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 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 gravitydenotes the whole force with which a body tends downwards, as when the body is in empty space.
Specific gravitydenotes the relative or comparative gravity of any body, in respect to that of another body of equal bulk, or magnitude.
Centre of gravityis 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 gravityof a circle, regular polygon, prism, cylinder, or sphere, is in its centre.
The centre of gravityof 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.
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 bodyis 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 gravityto 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 wateris to the whole or absolute weight;so is the specific gravity of waterto 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 remainderis to the weight of the light body in air;so is the specific gravity of waterto 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 weightis to the loss of weight;so is the specific gravity of the solidto 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 productis to the whole weight of the compound;so is each of the other two productsto 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.
By means of the foregoing table, the weight of any quantity of the materials specified (in cubic feet) may readily be found.
Bodyis 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.
Densityis 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.
Forceis a power exerted on a body to move it, or to stop it. If the force act constantly,it is apermanent force, like pressure, or the force or gravity; but if it act instantaneously, or for an imperceptibly short time, it is calledimpulse, orpercussion, 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 projectedin 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 parabolais the section formed by cutting a cone, with a plane, parallel to the side of the cone.
Gravity(videpage 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,
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,
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
By experiments to ascertain the effect of Carnot’s vertical fire, it was found that 4-oz. balls only penetrated120of an inch into deal board, and from 2 to 3 inches into meadow ground.
Amplitudesignifies the range of a projectile, or the right line upon the ground, subtending the curvilinear path in which it moves.
The time of flightof 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 16112, 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 ÷ 16112= 74·61, the square root of which is 8·637, the time required.
When a body is projected verticallydownwardswith 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 projectedupwards, the latter product must be subtracted from the former.
DEFINITIONS.[55]
A line is perpendicular to anotherwhen it inclines not more on the one side than on the other, the angles on both sides being equal.
Parallel linesare those which have no inclination to each other, being everywhere equi-distant, however far produced, or extended.
An angleis the inclination, or opening of two lines, which meet in a point called thevertex, or angular point: and the two lines are called thelegs, or sidesof the angle.
The measure of an angleis estimated by the number of degrees contained in the arc between its two legs.
A rectilinear anglehas its legs or sides,right, or straight lines.
A curvilinear anglehas its legscurves.
A right angleis formed by one line perpendicular to another; the measure of which is an arc of 90°.
An acute angleis less than a right angle, or than 90°.
An obtuse angleis greater than a right angle.
An oblique anglemay be either acute, or obtuse.
The circumference, or periphery of a circleis the curved line which bounds it, being everywhere equally distant from thecentre. The circumference is supposed to be divided into 360 degrees (marked thus °); each degree into 60 minutes, each minute (′) into 60 seconds (″).
An arcis 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 circleis a right line drawn from the centre to the circumference.
The diameter of a circleis 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 quadrantis the fourth part of a circle, being contained between two radii, and an arc of 90°.
A segmentis that part of a circle which is cut off by a chord.
A sectoris that part of a circle contained between two radii, and an arc.
A secantis a line which cuts a circle, lying partly within, and partly without it.
A tangentis a line which touches a circle, or curve, without cutting it.
The point of contactis where a tangent touches an arc.
Trianglesare figures having three sides, and three angles.
An equilateral trianglehas its three sides equal.
An isosceles trianglehas only two equal sides.
A scalene trianglehas all its sides unequal.
A rectangular, or right-angled trianglehas 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 hypothenuseis the side opposite the right angle in a rectangular triangle.
An obtuse-angled trianglehas one of its angles obtuse.
An acute-angled trianglehas 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 triangleis equal to the product of their sum and difference.
The sides of a triangle are proportionalto the sines of their opposite angles.
Quadrangles, or quadrilaterals, are plane figures bounded by four right lines.
A squareis a quadrilateral having all its sides equal, and all its angles right angles. Thediagonal of a squareis equal to the square root of twice the square of its sides: andthe side of the squareis equal to the square root of half the square of its diagonal.
The diagonalis 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 parallelogramis a quadrilateral, whose opposite sides are parallel.
A rectangleis a parallelogram having four right angles.
A rhomboidis an oblique-angled parallelogram.
A rhombus, or lozenge, is a quadrilateral, whose sides are all equal but its angles oblique.
A trapeziumis a quadrilateral, which has none of its sides parallel to each other.
A trapezoidis a quadrilateral, which has only two of its sides parallel.
Polygonsare plane figures bounded by more than four sides.
A regular polygonhas all its sides, and angles equal.
The perimeterof 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 circleis 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 pointAset off any length 4 times toC; fromAas a centre with 3 of those parts describe an arc atB, and fromCwith 5 of them cut the arc atB. DrawA 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, andPractical Geometry,Fig. ½.
To erect a perpendicular.
Set off on each side of the pointA, any two equal distances,A D,A E. FromDandEas centres, and with any radius greater than halfD E, describe two arcs intersecting each other inF. ThroughA, andFdraw the lineA F, and it will be the perpendicular required.
Fig. 1.—Plate,Practical Geometry.
To let fall a perpendicular.
FromDas a centre, and with any radius, describe an arc intersecting the given line. From the points of intersectionC, andE, with any radius greater than half, describe two arcs, cutting each other atF.ThroughD, andFdraw a line, andD Fwill be the perpendicular required.Fig. 2.
To draw a line parallel to a given line.
From any pointDin the given line with the radiusD C, describe the arcC E, and fromCwith the same radius describe the arcD F. TakeE C, and set it off fromDtoF. ThroughC, andFdrawC Ffor the parallel required.Fig. 3.
To divide an angle into two equal parts.
FromBas a centre with any radius describe an arcA C. FromA, andCwith any radius describe arcs intersecting each other inD. Then drawB D, and it will bisect the angle.Fig. 4.