Page:Newton's Principia (1846).djvu/320

 60, 120 inches was described, the difference of the arcs described in the descent and subsequent ascent will be $$\scriptstyle \frac{1}{656}$$, $$\scriptstyle \frac{1}{242}$$, $$\scriptstyle \frac{1}{69}$$, $$\scriptstyle \frac{4}{71}$$, $$\scriptstyle \frac{8}{37}$$, $$\scriptstyle \frac{24}{29}$$ parts of an inch, respectively. But these differences in the greater oscillations are in the duplicate ratio of the arcs described nearly, but in lesser oscillations something greater than in that ratio; and therefore (by Cor. 2, Prop. XXXI of this Book) the resistance of the globe, when it moves very swift, is in the duplicate ratio of the velocity, nearly; and when it moves slowly, somewhat greater than in that ratio.

Now let V represent the greatest velocity in any oscillation, and let A, B, and C be given quantities, and let us suppose the difference of the arcs to be AV + $$\scriptstyle BV^{\frac{3}{2}}$$ + CV². Since the greatest velocities are in the cycloid as ½ the arcs described in oscillating, and in the circle as ½ the chords of those arcs; and therefore in equal arcs are greater in the cycloid than in the circle in the ratio of ½ the arcs to their chords; but the times in the circle are greater than in the cycloid, in a reciprocal ratio of the velocity; it is plain that the differences of the arcs (which are as the resistance and the square of the time conjunctly) are nearly the same in both curves: for in the cycloid those differences must be on the one hand augmented, with the resistance, in about the duplicate ratio of the arc to the chord, because of the velocity augmented in the simple ratio of the same; and on the other hand diminished, with the square of the time, in the same duplicate ratio. Therefore to reduce these observations to the cycloid, we must take the same differences of the arcs as were observed in the circle, and suppose the greatest velocities analogous to the half, or the whole arcs, that is, to the numbers ½, 1, 2, 4, 8, 16. Therefore in the 2d, 4th, and 6th cases, put 1, 4, and 16 for V; and the difference of the arcs in the 2d case will become $$\scriptstyle \frac{\frac{1}{2}}{121}$$ = A + B + C; in the 4th case $$\scriptstyle \frac{2}{35\frac{1}{2}}$$ = 4A + 8B + 16C; in the 6th $$\scriptstyle \frac{8}{9\frac{2}{3}}$$ = 16A + 64B + 256C. These equations reduced give A = 0,0000916, B = 0,0010847, and C = 0,0029558. Therefore the difference of the arcs is as 0,0000916V + $$\scriptstyle 0,0010847V^{\frac{3}{2}}$$ + 0,0029558V²: and therefore since (by Cor. Prop. XXX, applied to this case) the resistance of the globe in the middle of the arc described in oscillating, where the velocity is V, is to its weight as $$\scriptstyle \frac{7}{11}$$AV + $$\scriptstyle \frac{7}{10}BV^{\frac{3}{2}}$$ + ¾CV² to the length of the pendulum, if for A, B, and C you put the numbers found, the resistance of the globe will be to its weight as 0,0000583V + $$\scriptstyle 0,0007593V^{\frac{3}{2}}$$ + 0,0022169V² to the length of the pendulum between the centre of suspension and the ruler, that is, to 121 inches. Therefore since V in the second case represents 1, in the 4th case 4, and in the 6th case 16, the resistance will be to the weight of the globe, in the 2d case, as 0,0030345 to 121; in the 4th, as 0,041748 to 121; in the 6th, as 0,61705 to 121.