# Page:Newton's Principia (1846).djvu/322

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316
[Book II.
the mathematical principles

only $\scriptstyle \frac{1}{8}$ part of the motion was lost. I leave the calculation to such as are disposed to make it.

 First descent 2 4 8 16 32 64 Last ascent 1½ 3 6 12 24 48 Numb. of oscill. 374 272 162½ 83⅓ 41⅔ 22⅔

I afterward suspended a leaden globe of 2 inches in diameter, weighing 26¼ ounces troy by the same thread, so that between the centre of the globe and the point of suspension there was an interval of 10½ feet, and I counted the oscillations in which a given part of the motion was lost. The first of the following tables exhibits the number of oscillations in which $\scriptstyle \frac{1}{8}$ part of the whole motion was lost; the second the number of oscillations in which there was lost part of the same.

 First descent 1 2 4 8 16 32 64 Last ascent $\scriptstyle \frac{7}{8}$ $\scriptstyle \frac{7}{4}$ 3½ 7 14 28 56 Numb, of oscill. 226 228 193 140 90½ 53 30 First descent 1 2 4 8 16 32 64 Last ascent ¾ 1½ 3 6 12 24 48 Numb. of oscill. 510 518 420 318 204 121 70

Selecting in the first table the 3d, 5th, and 7th observations, and expressing the greatest velocities in these observations particularly by the numbers 1, 4, 16 respectively, and generally by the quantity V as above, there will come out in the 3d observation $\scriptstyle \frac{\frac{1}{2}}{193}$ = A + B + C, in the 5th observation $\scriptstyle \frac{2}{90\frac{1}{2}}$ = 4A + 8B + 16C, in the 7th observation $\scriptstyle \frac{8}{30}$ = 16A + 64B + 256C. These equations reduced give A = 0,001414, B = 0,000297, C = 0,000879. And thence the resistance of the globe moving with the velocity V will be to its weight 26¼ ounces in the same ratio as 0,0009V + $\scriptstyle 0,000208V^{\frac{3}{2}}$ + 0,000659V² to 121 inches, the length of the pendulum. And if we regard that part only of the resistance which is in the duplicate ratio of the velocity, it will be to the weight of the globe as 0,000659V² to 121 inches. But this part of the resistance in the first experiment was to the weight of the wooden globe of 57$\scriptstyle \frac{7}{22}$ ounces as 0,002217V² to 121; and thence the resistance of the wooden globe is to the resistance of the leaden one (their velocities being equal) as 57$\scriptstyle \frac{7}{22}$ into 0,002217 to 26¼ into 0,000659, that is, as 7⅓ to 1. The diameters of the two globes were 6$\scriptstyle \frac{7}{8}$ and 2 inches, and the squares of these are to each other as 47¼ and 4, or 11$\scriptstyle \frac{13}{16}$ and 1, nearly. Therefore the resistances of these equally swift globes were in less than a duplicate ratio of the diameters. But we have not yet considered the resistance of the thread, which was certainly very considerable, and ought to be subducted from the resistance of the pendulums here found. I could not determine this accurately, but I found it