Page:The Mathematical Principles of Natural Philosophy - 1729 - Volume 1.djvu/219

as a baſe, ſuppose the wheel GEF to move forward, revolving about its axis, and in the mean time by its point A deſcribing the cycloid ALI. which done, take GK to tEe perimeter GEFG of the wheel, in the ratio of the time in which the body, proceeding from A, deſribed the arc AP, to the time of a whole revolution in the ellipſis. Erect the perpendicular KL meeting the cycloid in L, then LP drawn parallel to KG will meet this ellipſis in P the required place of the body.

For about the centre O with the interval OA deſcribe the ſemi-circle AQB, and let LP, produced, if need be, meet the arc AQ in Q, and join SQ, OQ. Let OQ meet the arc EFG in F upon OQ let fall the perpendicular SR. The area APS is as the area AQS, that is, as the difference between the ſector OQA and the triangle OQS, or as the difference of the rectangle $\scriptstyle {\frac {1}{2}}OQ\times AQ$, and $\scriptstyle {\frac {1}{2}}OQ\times SR$, that is, becauſe $\scriptstyle {\frac {1}{2}}OQ$ is given, as the difference between the arc AQ and the right line SR; and therefore (becauſe the equality of the given ratios SR to the ſine of the arc AQ, OS to OA, OA to OG, to GF, and by diviſion. AQ - SR to GF - ſine of the arc AQ) as GK the difference between the arc GF and the ſine of the arc AQ. Q. E. D.

Scholium.

Plate 14, Figure 3

But ſince the deſcription of this curve is difficult, a ſolution by approximation will be preferable. Firſt let there be found a certain angle B which may be to an angle of 57,29578 degrees,