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XVI.—On the Path of a Rotating Spherical Projectile. II.

Published online by Cambridge University Press:  06 July 2012

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Extract

The first instalment of this paper was devoted in great part to the general subject involved in its title, but many of the illustrations were derived from the special case of the flight of a golf-ball. Since it was read I have endeavoured, alike by observation and by experiment, to improve my numerical data for this interesting application, particularly as regards the important question of the coefficient of resistance of the air. As will be seen, I now find a value intermediate to those derived (by taking average estimates of the mass and diameter of a golf-ball) from the results of Robins and of Bashforth. This has been obtained indirectly by means of a considerable improvement in the apparatus by which I had attempted to measure the initial speed of a golf-ball. I have, still, little doubt that the speed may, occasionally, amount to the 300, or perhaps even the 350, foot-seconds which I assumed provisionally in my former paper:—but even the first of these is a somewhat extravagant estimate; and I am now of opinion that, even with very good driving, an initial speed of about 240 is not often an underestimate, at least in careful play. From this, and the fact that six seconds at least are required for a long carry (say 180 yards), I reckon the “terminal velocity” at about 108, giving v2/360 as the resistance-acceleration.

Type
Research Article
Information
Earth and Environmental Science Transactions of The Royal Society of Edinburgh , Volume 39 , Issue 2 , 1900 , pp. 491 - 506
Copyright
Copyright © Royal Society of Edinburgh 1900

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References

page 493 note * Cambridge and Dublin Mathematical Journal, ix. 145 (1854).

page 493 note † Messenger of Mathematics, vii. 14 (1878).

page 493 note * If l be the length (in feet) of the supporting straps, d the (small) horizontal deflection of the bob, its vertical rise is obviously d 2/2l, so that its utmost potential energy is

(M + m)gd 2/2l

where M is its mass and m that of the ball. But, if V was the horizontal speed of the ball, that of bob and ball was mV/(M+m). Equating the corresponding kinetic energy to the potential energy into which it is transformed, we find at once (M+m)gd 2/2l=m 2V 2/2(M+m) leading to the very simple expression

With the numerical values given in the text we easily find that this is equivalent to

where V is, of course, in foot-seconds, but the deflection is now (for convenience) expressed in inches, and called D. Hence the numerical result in the text.