TRACT 37.

OF GUNNERY.

235

ter d and specific gravity s. And if we further suppose theball to be cast iron, the specific gravity, or weight of onecubic inch of which, is ^GSSo'b, it becomes 290 '6d, for thatcoefficient: also 69259sd =zl8600d = ---, and — = 231-5.

1 md 27 m

Hence the foregoing fluent becomes 290*6 x hyp. log.

v* - 231-5U+ 18600

or

669d x com. log.

v* — 231-5V + ISoOOflE

18600 d ’ —. 186 'OlW ’

changing the hyperbolic for the common logs. And this is ageneral expression for the altitude in feet, ascended by anyiron ball, whose diameter is d inches, discharged with anyvelocity v feet. So that, substituting any values of d and v,the particular heights will be given, to which the balls willascend.

39. Exam. 1 . Suppose the ball be that belonging to thelast table of resistances, its weight being 18 oz, or 1-flb, andits diameter 2 inches, when discharged with the velocity2000 feet, being nearly the greatest velocity for any ironball. The calculation being made with these values of f/andv, the height ascended is found to be 2653 feet, or onlyabout half a mile; though found to be almost 12 miles with-out the air’s resistance. And thus the height may be foundfor any other diameter and velocity.

40. Exam. 2. Again, for the 24lb ball, with the samevelocity 2000, its diameter being nearly 5'6 = d. Here 669d

-3/46, and- - - = the log. of which is

1-54354; theref. T54354X 3746= 5782 —x the height, beinga little more than a mile.

We may now examine what will be the height ascended,considering the resistance always as the square of the velocity.

problem v.

41. To determine the Height ascended by a Ball, projected asin the two foregoing Problems; supposing the Resistance ofthe Air to be as the Square of the Pelocity.

Here it will be proper to commence with selecting someexperimented resistance corresponding to a medium kind of