Page:Newton's Principia (1846).djvu/294

arc PQ is the fourth part of the decrement of the arc PR. Whence also if the area QSr be taken equal to the area PSQ, the decrement of the arc PQ will be equal to half the lineola Rr; and therefore the force of resistance and the centripetal force are to each other as the lineola ½Rr and TQ which they generate in the same time. Because the centripetal force with which the body is urged in P is reciprocally as SP², and (by Lem. X, Book I) the lineola TQ, which is generated by that force, is in a ratio compounded of the ratio of this force and the duplicate ratio of the time in which the arc PQ is described (for in this case I neglect the resistance, as being infinitely less than the centripetal force), it follows that TQ $\scriptstyle \times$ SP², that is (by the last Lemma), ½PQ² $\scriptstyle \times$ SP, will be in a duplicate ratio of the time, and therefore the time is as PQ $\scriptstyle \times$ $\scriptstyle {\sqrt {SP}}$; and the velocity of the body, with which the arc PQ is described in that time, as $\scriptstyle {\frac {PQ}{PQ\times {\sqrt {SP}}}}$ or $\scriptstyle {\frac {1}{\sqrt {SP}}}$, that is, in the subduplicate ratio of SP reciprocally. And, by a like reasoning, the velocity with which the arc QR is described, is in the subduplicate ratio of SQ reciprocally. Now those arcs PQ and QR are as the describing velocities to each other; that is, in the subduplicate ratio of SQ to SP, or as SQ to $\scriptstyle {\sqrt {SP\times SQ}}$; and, because of the equal angles SPQ, SQr, and the equal areas PSQ, QSr, the arc PQ is to the arc Qr as SQ to SP. Take the differences of the proportional consequents, and the arc PQ will be to the arc Rr as SQ to SP - $\scriptstyle {\sqrt {SP\times SQ}}$, or ½VQ. For the points P and Q coinciding, the ultimate ratio of SP - $\scriptstyle {\sqrt {SP\times SQ}}$ to ½VQ is the ratio of equality. Because the decrement of the arc PQ arising from the resistance, or its double Rr, is as the resistance and the square of the time conjunctly, the resistance will be as $\scriptstyle {\frac {Rr}{PQ^{2}\times SP}}$. But PQ was to Rr as SQ to ½VQ, and thence $\scriptstyle {\frac {Rr}{PQ^{2}\times SP}}$ becomes as $\scriptstyle {\frac {{\frac {1}{2}}VQ}{PQ\times SP\times SQ}}$, or as $\scriptstyle {\frac {{\frac {1}{2}}OS}{OP\times SP^{2}}}$. For the points P and Q coinciding, SP and SQ coincide also, and the angle PVQ becomes a right one; and, because of the similar triangles PVQ, PSO, PQ becomes to ½VQ as OP to ½OS. Therefore $\scriptstyle {\frac {OS}{OP\times SP^{2}}}$ is as the resistance, that is, in the ratio of the density of the medium in P and the duplicate ratio of the velocity conjunctly. Subduct the duplicate ratio of the velocity, namely, the ratio $\scriptstyle {\frac {1}{SP}}$, and there will remain the density of the medium in P, as $\scriptstyle {\frac {OS}{OP\times SP}}$. Let the spiral be given, and, because of the given ratio of OS to OP, the density of the medium in P will be as $\scriptstyle {\frac {1}{SP}}$. Therefore in a medium whose