TRACT 37.

OP GUNNERY.

281

body naturally descends by gravity in the first second of time,orbalf the velocity generated ordestroyed in 1 second. Then,

by the principles of constant forces, f

138. Now, for an example, I have found, by a mediumof several experiments, that a cast-iron ball, of 1'96 inches

diameter, discharged into the end of a block of elm wood,

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or in the direction of the fibres, with a velocity of 1600 feet

per second of time, penetrated 20 inches deep into its sub-stance. It is proposed then to determine the time of thepenetration; also the strength or resisting force of the wood,as compared to the force of gravity, supposing that force tobe a constant quantity.

Here we have v = 1600 feet, and s — 20 inches = -J. feet:therefore t == — = - lt * ; - = the time, or part of a se-

v S x 1600 480’ 5 1

corn), during the whole penetration. Also f = — =

1600’

:— -= 21000. That is, the resisting force of the wood,

4 x 16 x 9 &

in this instance, is to the force of gravity, as 24000 to 1.

139. Exam. 2. In another instance the same ball, with1200 feet velocity, penetrated 15 inches. — Here then f —

1200 ’

= 18000, the strength of the wood. And t=~

s . V

— , the time, the same as before.

C4f 64. j5

1800 480’

140. Exam. 3. In a third instance, the 3-pound ball, of2 - 78 inches diameter, with 1500 feet velocity, penetrated 30

inches deep.—Here then

/'= = 14062, the strength of the wood. And

the time in this case.

141. Exam. 4. In a 4th case, the last ball, with 1060velocity, penetrated 16 inches deep.—Hence

1060’

13200, the strength in this case.—And, the time in this case.

64. ',i

1060 398’