336

NEW EXPERIMENTS

TRACT 34*

32. To find therefore the resistance of the air against theball in any case : it is first to be considered, that the resist-ance to a plane moving perpendicularly through a fluid atrest, is nearly equal to the weight or pressure of a columnof the fluid, whose altitude is the height through which thebody must fall, by the force of gravity,- to acquire thevelocity with which it moves through the fluid, the base ofthe column being equal to the plane. So that, if a denotethe area of the plane, v the velocity, n the specific gravityof the fluid, and h — 16-09 feet; the altitude due to thevelocity v being the whole resistance, or motive force m,

Mil v v a.nw

will be c x M x rr = -tt**.

4 h 4 h

Now, if d denote the diameter of the ball, and k — ’7854,then shall a — k d x be a great circle of the ball; and conse-quently m = = the motive force on the surface of a

circle equal to a great circle of the ball.

But the resistance on the hemispherical surface of theball is only one half nearly of that o‘n the flat circle ofthe same diameter ; therefore m = is the motive force

O fl

on the ball; and if w denote its weight, — will beequal to f the retarding force.

Since •§■ k d l is the magnitude of the sphere, if n denote itsdensity or specific gravity, its weight w will be = j-kd 3 n;consequently the retarding forced' or A

. knd 2 v* 3 3nw

tvill e — —jf£— x — TYd/Ti?

But by the laws of forces vv — 2 hf x — and

-- = * == ~ e x, where x is the space passed over,

putting e = and making the value negative, becausethe velocity v is decreasing. And the correct fluent of thisis loo-, v — log. v or log. — = ex, where v is the first or

o ° o u

greatest velocity of projection. Or if a be := 2-718281828&c, the number whose hyperbolic logarithm is 1, then is-A = A ex , and hence the velocity == = vA —e *. So that