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shot in one experiment, and v', c', b', the same quantities in another, then

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v: v' =√3c:√3 c', when b is constant,

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1. Find the initial velocity of a shell weighing 48 lbs., the charge being 3 lbs.

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2. The weight of a ball is 32 lbs. ; what must be the charge of powder necessary to give it a velocity of 1500

feet?

c =

b

2 v

=

(1600

32,1500 2

32 225 = X =9.375 lbs. 1600 3 256

3. The velocity of a ball, with a charge of 10 lbs. of powder, is 1200 feet; what would be its velocity with a charge of 12 lbs. ?

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1응=

12 1200

=1200√ TO

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=120 x 10.954 = 1314.

EXERCISES.

120

1. What is the velocity of a shell weighing 36 lbs., when discharged with 4 lbs. of powder?

Ans. 923.7. 2. With what velocity will a 48 lb. ball be impelled by a charge of 21 lbs. ?

3. The weight of a shell is 100 lbs.; powder is necessary to project it with a feet?

Ans. 632-46. what charge of velocity of 1000 Ans. 13.02 lbs.

4. A ball is discharged with a velocity of 900 feet by a charge of 2 lbs. of powder; required its weight.

Ans. 18-96 lbs. 5. The velocity of a ball of 24 lbs. weight is 800 feet; what would be the velocity of a ball of 18 lbs., impelled with the same charge? Ans. 924.

221. PROBLEM II.-Given the range for one charge, to find the range for another charge, and conversely.

The ranges are proportional to the charges; that is, one charge is to another charge, as the range corresponding to the former to that corresponding to the latter.'

Or,

c: c'=r:r',

and

•, also c' = C.

r

EXAMPLE.—If a shell range 4000 feet when discharged with 9 lbs. of powder, what will be the charge necessary to project it 3000 feet?

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3000
4000

=63 lbs.

It was found (208) that r is proportional to v2, or r÷ v2;

b

2

and since (219) c = (1600), therefore e is proportional

to v2, when b is given, or cv2; but r÷v2; therefore r÷c, or r: r' = c:c.

It could be similarly shown that, when c is given or con

stant, r

1 1

1

or r:r' =

ō or r:rb':b, and bb"

r

r=r.. also l' = b; so that the range is inversely b"

as the weight of the ball, all other circumstances being the

same.

EXERCISES.

1. If a shell range 2500 when projected with a charge of 5 lbs., what will be its range when the charge is 8 lbs. ?

Ans. 4000.

2. If a charge of 6 lbs. is sufficient to impel a ball over a range of 3600 feet, what charge will be required that the range may be 4500 feet? Ans. 7.5 lbs.

ADDITIONAL EXERCISES.

1. At what elevation, on the parabolic theory, must a 13-inch mortar be pointed, in order that a shell of 196 lbs. may range 6745 feet on a descending plane, whose inclination is 8° 15', the charge being 411 lbs.? Ans. 32° 46'.

2. On the parabolic theory, with what impetus, velocity, and charge, must a 13-inch shell, weighing 196 lbs., be fired, at an elevation of 32° 12′, to strike an object at the distance of 3250 feet?

Ans. Impetus 1802, velocity 340, charge 3 lbs. nearly. 3. Find the velocities with which the shells given in the following table will be discharged, with the respective charges given there :

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222. Some important problems in practical gunnery can be solved by means of the following table, calculated by Mr Robins, in which the actual and potential ranges for the same elevation are given in terms of the terminal velocity.

223. The actual range is the range in a resisting medium; the potential range is the range in a non-resisting medium or vacuum; and the potential random is the greatest range in a vacuum.

224. The terminal velocity of a projectile is that velocity which it has in a resisting medium, when the resistance against it is equal to its weight.

225. The resistance to a plane surface, moving with a moderate velocity in a resisting medium, is nearly equal to the weight of a column of the fluid, having the surface for its base, and a height equal to that due to the velocity in a vacuum. The resistance on a hemisphere, or on the ante

rior surface of a ball, is only half that on a surface equal to the area of one of its great circles; and hence the resistance to a ball, moving with a small velocity in the atmosphere, is nearly half the weight of a column of air, having a great circle of the ball for its base, and a height equal to that due to the velocity; for the resistance to a sphere is equal to only half the resistance to the end of a cylinder of the same diameter. When the velocity is not considerable, the resistance is about instead of of the above column, as appears by computing the example in article 288, but for great velocities it is considerably greater.

226. There are several formulas for determining the terminal velocity of a ball. One of these is given by Dr Hutton, and is as follows:-Let the resistance in avoirdupois pounds, d = the diameter of the ball in inches, and the velocity in feet; then

r=(·000007565 v2

·00175 v) d2, or r=0000044 d2v2; the former value referring to considerable, and the latter to smaller, velocities.

227. In order to find the terminal velocity, for which r=w, the weight of the ball,

•578

u = •5236 d

× 7.25·137134 d3 ;

16

and when rw, the terminal velocity' will be found from the equation, 137134 d3·0000044 d2v2,

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The height due to this velocity is h' =

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and for a shell, the weight of which is of that of a ball of equal diameter,

=

4x31167

d = √ 24934 d = 158 √↓ d.

228. Mr Robins found that the resistance to a 12-pound ball, moving with a velocity of about 25.5 feet in a second, was ounce avoirdupois. Now, for velocities less than 1100 feet per second, the resistance is nearly proportional to the

squares of the velocities, and it is also as the squares of the diameter; hence, if c is the constant to be determined,

r= cd22, or lb. = c × 4·452 × 25.52.

It will be found from this equation that c is = '000002427; and the value of 'would be found as above to be 238√d, and h' 883 d. In the following table, Robins has taken this quantity to be 900 d, and denotes it by F, that is, F- = 900 d. This appears to be the origin of this quantity F, which has not before been accounted for. Robins had likely found, by other experiments, that 900 would generally afford more correct results than 883.

229. This quantity, namely, the height due to the terminal velocity in a vacuum, may be called the terminal height.

230. PROBLEM III.—To find the terminal height.

The terminal height is found by multiplying the diameter of the ball by 900.

When the ball has a different specific gravity from iron, find the height for iron; then the specific gravity of iron is to the specific gravity of the ball, as the height for an iron ball is to the required height.

For iron,

F900 d.

For a ball of other material, whose specific gravity is s, 7.25: s = 900 d: F,

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s = 11:35, and F =1409 d.

231. The following table gives the weight of an iron ball when its diameter is known, and conversely. The weight is in avoirdupois pounds, and the diameter in inches:

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