Supposing what is above demonstrated, I say, that if the tangents of the angles of the sector of a circle, and of an hyperbola, be taken proportional to the velocities, the radius being of a fit magnitude, all the time of the ascent to the highest place will be as the sector of the circle, and all the time of descending from the highest place as the sector of the hyperbola. To the right line AC, which expresses the force of gravity, let AD be drawn perpendicular and equal. From the centre D with the semi-diameter AD describe as well the quadrant AtE of a circle, as the rectangular hyperbola AVZ, whose axis is AK, principal vertex A, and asymptote DC. Let Dp, DP be drawn; and the circular sector AtD will be as all the time of the ascent to the highest place; and the hyperbolic sector ATD as all the time of descent from the highest place; if so be that the tangents Ap, AP of those sectors be as the velocities. CASE 1. Draw Dvq cutting off the moments or least particles tDv and qDp, described in the same time, of the sector ADt and of the triangle ADp. Since those particles (because of the common angle D) are in a duplicate ratio of the sides, the particle tDv will be as , that is[Pg 266] (because tD is given), as . But is , that is, , or ; and qDp is . Therefore tDv, the particle of the sector, is as ; that is, as the least decrement pq of the velocity directly, and the force Ck which diminishes the velocity, inversely; and therefore as the particle of time answering to the decrement of the velocity. And, by composition, the sum of all the particles tDv in the sector ADt will be as the sum of the particles of time answering to each of the lost particles pq of the decreasing velocity Ap, till that velocity, being diminished into nothing, vanishes; that is, the whole sector ADt is as the whole time of ascent to the highest place. Q.E.D. CASE 2. Draw DQV cutting off the least particles TDV and PDQ of the sector DAV, and of the triangle DAQ; and these particles will be to each other as DT2 to DP2, that is (if TX and AP are parallel), as DX2 to DA2 or TX2 to AP2; and, by division, as DX2 - TX2 to DA2 - AP2. But, from the nature of the hyperbola, DX2 - TX2 is AD2; and, by the supposition, AP2 is AD × AK. Therefore the particles are to each other as AD2 to AD2 - AD × AK; that is, as AD to AD - AK or AC to CK: and therefore the particle TDV of the sector is ; and therefore (because AC and AD are given) as ; that is, as the increment of the velocity directly, and as the force generating the increment inversely; and therefore as the particle of the time answering to the increment. And, by composition, the sum of the particles of time, in which all the particles PQ of the velocity AP are generated, will be as the sum of the particles of the sector ATD; that is, the whole time will be as the whole sector. Q.E.D.