Page:Newton's Principia (1846).djvu/325

In the experiments of the 4th column there were equal motions lost in 535 oscillations made in the air, and 1$\scriptstyle {\frac {1}{5}}$ in water. The oscillations in the air were indeed a little swifter than those in the water. But if the oscillations in the water were accelerated in such a ratio that the motions of the pendulums might be equally swift in both mediums, there would be still the same number 1$\scriptstyle {\frac {1}{5}}$ of oscillations in the water, and by these the same quantity of motion would be lost as before; because the resistance it increased, and the square of the time diminished in the same duplicate ratio. The pendulums, therefore, being of equal velocities, there were equal motions lost in 535 oscillations in the air, and 1$\scriptstyle {\frac {1}{5}}$ in the water; and therefore the resistance of the pendulum in the water is to its resistance in the air as 535 to 1$\scriptstyle {\frac {1}{5}}$. This is the proportion of the whole resistances in the case of the 4th column.

Now let AV + CV² represent the difference of the arcs described in the descent and subsequent ascent by the globe moving in air with the greatest velocity V; and since the greatest velocity is in the case of the 4th column to the greatest velocity in the case of the 1st column as 1 to 8; and that difference of the arcs in the case of the 4th column to the difference in the case of the 1st column as $\scriptstyle {\frac {2}{535}}$ to $\scriptstyle {\frac {16}{85{\frac {1}{2}}}}$, or as 85½ to 4280; put in these cases 1 and 8 for the velocities, and 85½ and 4280 for the differences of the arcs, and A + C will be = 85½, and 8A + 64C = 4280 or A + 8C = 535; and then by reducing these equations, there will come out 7C = 449½ and C = 64$\scriptstyle {\frac {3}{14}}$ and A = 21$\scriptstyle {\frac {2}{7}}$; and therefore the resistance, which is as $\scriptstyle {\frac {7}{11}}$AV + ¾CV², will become as 13$\scriptstyle {\frac {6}{11}}$V + 48$\scriptstyle {\frac {9}{56}}$V². Therefore in the case of the 4th column, where the velocity was 1, the whole resistance is to its part proportional to the square of the velocity as 13$\scriptstyle {\frac {6}{11}}$ + 48$\scriptstyle {\frac {9}{56}}$ or 61$\scriptstyle {\frac {12}{17}}$ to 48$\scriptstyle {\frac {9}{56}}$; and therefore the resistance of the pendulum in water is to that part of the resistance in air, which is proportional to the square of the velocity, and which in swift motions is the only part that deserves consideration, as 61$\scriptstyle {\frac {12}{17}}$ to 48$\scriptstyle {\frac {9}{56}}$ and 535 to 1$\scriptstyle {\frac {1}{5}}$ conjunctly, that is, as 571 to 1. If the whole thread of the pendulum oscillating in the water had been immersed, its resistance would have been still greater; so that the resistance of the pendulum oscillating in the water, that is, that part which is proportional to the square of the velocity, and which only needs to be considered in swift bodies, is to the resistance of the same whole pendulum, oscillating in air with the same velocity, as about 850 to 1, that is as, the density of water to the density of air, nearly.

In this calculation we ought also to have taken in that part of the resistance of the pendulum in the water which was as the square of the velocity; but I found (which will perhaps seem strange) that the resistance in the water was augmented in more than a duplicate ratio of the velocity. In searching after the cause, I thought upon this, that the vessel was too