Page:Newton's Principia (1846).djvu/116

 (by Cor. 1, Prop. I.), and therefore, by construction, as the perpendiculars AP, BQ directly; that is, as the perpendiculars let fall from the point D on the tangents. Whence it is easy to infer that the points S, D, T, are in one right line. And by the like argument the points S, E, V are also in one right line; and therefore the centre S is in the point where the right lines TD, VE meet. Q.E.D.

For the versed sine in a given time is as the force (by Cor. 4, Prop. 1); and augmenting the time in any ratio, because the arc will be augmented in the same ratio, the versed sine will be augmented in the duplicate of that ratio (by Cor. 2 and 3, Lem. XI.), and therefore is as the force and the square of the time. Subduct on both sides the duplicate ratio of the time, and the force will be as the versed sine directly, and the square of the time inversely. Q.E.D.

And the same thing may also be easily demonstrated by Corol. 4, Lem. X.

. 1. If a body P revolving about the centre S describes a curve line APQ, which a right line ZPR touches in any point P; and from any other point Q of the curve, QR is drawn parallel to the distance SP, meeting the tangent in R; and QT is drawn perpendicular to the distance SP; the centripetal force will be reciprocally as the solid $$\scriptstyle \frac{SP^{2}\times QT^{2}}{QR}$$, if the solid be taken of that magnitude which it ultimately acquires when the points P and Q coincide. For QR is equal to the versed sine of double the arc QP, whose middle is P: and double the triangle SQP, or $$\scriptstyle SP\times QT$$ is proportional to the time in which that double arc is described; and therefore may be used for the exponent of the time.

. 2. By a like reasoning, the centripetal force is reciprocally as the solid $$\scriptstyle \frac{SY^{2}\times QP^{2}}{QR}$$; if SY is a perpendicular from the centre of force on PR the tangent of the orbit. For the rectangles $$\scriptstyle SY\times QP$$ and $$\scriptstyle SP\times QT$$ are equal.