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If any iwo of the above quantities are given to find the angle elevation we must substitute, instead of sin. (A+B), its value : viz.

sin. (A + B)=sin. A cos. B+ sin. B cos. A

u hence sin. A, or cos. A may be obtained. Din all teie cases, we shall find sin. (A+B) cos. A equal to a known quantity, which let be denoted by C, then

cos. A sin. A cos. B+sin. B cos. ?A=C
Let sin. A=r, then cos A=(1-x2); and we have

с
*V(1-7)+tang. B(1-7)=

A quadratic equation from which two values of r will always be determined ; and whence we learn that there are always two angles of elevation, which equally answer the conditions of the Problem.

191. To tind the greatest height of the projectile above the point of projection, we must observe that the body will continue to ascend till the velocity of descent from gravity, is equal to the uniform velocity of ascent from projection ; that is, calling the time at which the height i; greatest t', we sl all have (41) 2gt'=the velocity of descent from gravity; and v sin. A will denote the uniform velocity of ascent estimated in a direction perpendicular to he horizon, whence we have

2gt'=v sin. A, or

v sin. A is

2

bözt the descent in the tinie i' is gt'e, in v hic , substituting the above value of l', we have

sin. A

gt?

48

and the ascent in the same time from projection, is t'v sin. A, or substiuting for t' as abi ve, we have

ve sin. :A t'u sin. Az

2 g

and consequently, Je difference of these will be the greatest height of the projectile Above the point A ; calling therefore the greatest height h, we have

А
ha
4 g

(Equation IV.

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or

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4lig=sin. A

2

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3

3

By article, 192, we have 81

whence, the angle of elevation and the velocity being given, the greatest height will be immediately determined; and if r or i be given, The value of v2 may first be determined from the proper equation, and then the value of h from equation (IV.)

192. All the preceding equations are rendered much more simple if suppose ihe plane AB to become horizontal For then the angle B=0, and consequently sin. B=0, and cos. B=i. Afier this reduction, Equation I becomes &'=sin. A

stan. A

į siu. 24

= sin. 24

ve which formulas involve all the conditions of a projectile while the plane is horizontal, and passes through the point of projection.

When the plane is not borizontal, it is obvious ihat in the furmula in which sin (A + B) occurs, B must be accounied posiiive when the plane descends, and negative when A ascends.

193. 'If the velocity of the projectile in any point of the curve be required, or the relocity atter any time (','it is obvious that this is compounded of the constant horizontal velocity of projection v cos. A, and the difference of the vertical velocities of ascent from projection =v sin. A, wid that of descent from gravity =2gt', that is, calling the required velocity V, we shall have V= v cos. A+ (v sin. A-23't) (Equation V.)

Examples on a Horizontal Plane, 1. Let the angle in which a body is projected by 45°, and the time of its flight 12"; what is the horizontal range?

=tan. A

4hg

1615x12 151' x122 whence is

=2318'4 feet. tan. 45°

1 2. In what time will a shell range 3250 feet, in an elevation of 32".

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3. If the elevation of the piece de 30o and the horizontal range 2000, what is the greatest height to which the ball will ascend ?

By article (192)

4hg.-sin.?A

ve sin.'A whence h=

Now by the same article = sin. 2 A, or v2=

2gr vs

sin. 2 Asin. A cos. A and substituting this value, we obtain

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h=

or

-tan. A=28807 feet.

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Example on an Inclined Plane. How far will a shot range on a plane which descends 8° 15', the projectile velocity being 440 feet per second, and the elevation of the piece 32° 30' ?

By Equation III., (article 190) we have sin. (A + B) cos. A V2

sin. (A + B) cos. A

8
whencer=
(440)? , sin. 40° 45' x cos. 32° 30'

cos. 8° 15'
Log. sin. 40° 45'

= 9814753 cos. 32° 30'

9.926029 2 log. 440...

5:286906 2 log. cos. 8° 15' = 19.989832

25.027688 log. 1077 = 1.206376

{sin.

• A}

g cos.? B

cos.B

16ra

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log. range

= 3.83 1480 whence the range is 6784 feet nearly. The learner will observe, that in the preceding reasonings, no notice has been takta of the effects of the resistance of the air on the motion of projectiles. Now, as this is very considerable, especially when they are discharged with great velocity, the theory requires to be modelled and corrected by ex. perimental investigations, before it can be applied in practice. There are indeed some cases, such as in the throwing of shells, when the velocity does not exceed 400 feet per second, in which the results by the theory do not differ much from the truth. But when the velocity is great, the resistance of the air occasions a diminution of motion so prodigions, as to revder the theory, withont the aid of data derived from experiment, of very little use. Thus, a musket ball, discharged with the ordinary allotment of powder, issues from the piece with the velocity of 1670 feet per second. At the elevation of 45°, it should therefore range 16 miles, whereas it does not range above balf a mile. Tbus, alsn, a 24 Ib. ball, discharged with 16 lbs of powder, which should range about 16 miles, does not range 3 miles.

Again, the path of a projectile, when the velocity is great, is not parabolical: but is much less incurvaled in the weending tbau in the descending brmat:

the greatest range, therefore, is not made with an elevation of_45° as in racun, but with an elevation lower, as the first velocity is greater. Thus it is found, that although the larger shells with small velocities range farthest wisen projected at an elevation of about 15°, yet the smaller shells with great velocities range farthest at an elevation vot much above 30'.

These instances sufficiently show, that such rules as are deduced from the theory alone, without the aid of experiment, are upfit for directing practice.

PRACTICAL GUNNERY. Of Projections made on the Horizontal Plane. 194. Prob. 1. To find the velocity of any shot or shell.

Rule. Divide double of the weight of the charge of powder by the weight of the shot, both in lbs. Then if the square root of the quotient be multiplied by 1000, the product will be the velocity, or the number of feet the shot passes over in a second.

Or say, As the square root of the weight of the shot is to the square root of double the weight of powder, so is 1600 feet to the first velocity of the shot.

Ex. With what velocity will a 13-inch shell, weighing 196 lbs. be discharged by 9 lbs. of powder ?

Nox 1600=485 feet per second, nearly. Ans.

Or thus, v 196 : V18 :: 1600 : 485 feet, nearly. 195. Prob. 2. To find the terminal velocity of a ball or shell ; that is, the greatest velocity it can acquire by descending through the air, by its own weight.

Rule. For talls, the terminal velocity is found by multiplying the square root of the diameter of the ball in inches by 175-5; and for shells, by multiplying the square root of the diameter of the shell in inches by 147.3.

Er. What is the terminal velocity of a 2 lbs. iron ball, its diameter being 2:45 inches ?

175'5 V2:45=175.5 x 15.6524=275 feet. Ans 196. Prob. 3. To find the height from which a body must fall, in vacuo, in order to acquire a given velocity.

Rule. Since the spaces descended by falling bodies are as the square of the velocities, and a fall of 1617 feet or 193 inches produces a velocity of 32 feet or 386 inches, therefore 386': the square of the given velocity in inches :: 193 : height required in inches. Or, omitting the fractions, if the square of the given velocity in feet be divided by 64, the quotient will be the height required in feet, nearly.

Er. From what height must a body descend, in order to acquire the velocity of 1670 feet per second ?

1670'+

+64=43576 feet. Ans. 197. Prob. 4. To find the greatest range of a ball or shell ; and the elevation of the piece to produce that range.

Rule. Divide the given initial velocity by the termina, velocity of the ball or shell, and find the quotient in the first column of the fol. lowing table ; against which, in the second column, will be found the elevation to give the greatest range; and the corresponding nunibet in

the third column multiplied by the height producing the terminal velocity will give the greatest range, nearly.

Elevations giving the greatest range.

Initial velo.
Range Initial velo.

Range divided by Elevation. divided by divided by | Elevation. divided by term. velo. altitude term. velo.

altitude.

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the range

Ex. What is the greatest range of a 24 lb. iron ball, when discharged with a velocity of 1640 feet, and the elevation to produce that range, ihe diameter of the ball being 5.6 inches?

175.5 15:6=415= the terminal velocity.
4159 +69=2691=the height producing that velocity.
1640+415=3.95, which nearly corresponds to 34° 15'=th

elevation required. The range in the third column against 34°15' is 2.9094, and 2·9094 x 2691=7829 feet=the greatest range, nearly. 198. Prob. 5. The range of any one elevation being given, to find

of any other elevation, and the converse. Rule. As the sine of twice the first elevation is to the sine of twice the second, so is the range at the former to that at the latter, the velocity being the same in both cases.

199. Prob. 6. The range for one charge being given, to find the range for another charge, or the charge for another range.

Řule. The ranges at the same elevation are nearly proportional to the charges.

Er. If, at an elevation of 45°, with a charge of 9 lbs. of powder, a shell range 4000 feet, what charge, at the same elevation, will be required to throw it 3000 feet?

4000 : 3000 ::9:64 lbs. Ans. 200. Prob.7. The range and elevation being given, to find the time of the flight.

Rule. As radius is to the tangent of elevation so is the range in feet, to the square of 4 times the number of seconds taken up in the fight, pearly.

When the elevation is 45°, then of the square root of the range in feet will be the number of seconds required, nearly. Ex. In what time will a shell range 3000 feet at an elevation of 35°?

Rad : Tan 35 :: 3000 : 2100:6

$121006=11:45 seconds nearly. Ans.

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