TRACT 37;

OF GUNNEKf.

891

„ • / ,. Qgmnacd* %

Hence, by forces, vv = Zgfx = —-— X —;the fluent of which is v 1 = 4g m ^ gc - x hyp. log. of x.

But, when v = o, then x — a ; therefore, by correction,t> 2 _ ^am^acd ^ jjyp_ ] 0 g # JL is the correct fluent; conseq.

v = •/ x hyp. log. ~) istheveloc. of the ball at c; or

v — x hyp. log. is the velocity with which

the ball issues from the muzzle at E; where h denotes thelength of the cylinder filled with powder; a being the lengthto the hinder part of the ball; which will be more than hwhen tiie ball does not touch the powder.

But, the content of the ball being ■§ -cd l , its weight is w ~\cd? ei therefore = v = ^±2^21^ _

17S3v'^’. Consequently the rule is

= 1783 •/(-£ X hyp. log. =s 2706</(~ e X com. log. •£).

When the ball is of cast iron, e = 7400, and the rule isv = 31 - 45y(^ X Jog. for the velocity of the iron ball.

Or, when the ball is of lead, then e = 11325, andv = 25 , 42V'Cj x log. ~) for the velocity of the leaden ball;in which two theorems, a, b, d, h, may be taken in any mea-sures, either feet or inches, &c.

161. Exam. For an example, it has been found that themedium velocity of the gun n° 2, with a charge of 4 ouncesof powder, has been about 1180 feet. Now, the ball being1‘96 inches in diameter, and the charge of powder with itscontaining bag occupying 3'45 inches of the cylinder,but thepowder alone only 2-54; the numeral values of the lettersin the theorem v = 3\‘45V(^~ X log. *), will be a = 3‘45>b = 38'43, d — 1-96, h = 2'54 ; and if we suppose w= 1000,as assumed by Mr. Robins; these substituted in that theorem,give v = 1159, being about 21 feet too little.

162. Such then is the solution of the problem in its most