The Quadrature of a Portion of the Epicycloid. By Mr. Caswel

By R, I understand the Radius of the Circle Generant; and by s, the Right Sine of the Arch BA, whose Versed Sine is VA. And, wherever in my whole Discourse of the Cycloid, or the Restored Trilinear (which is a Figure of Arches, and a Figure of Versed Sines) the Arch a is no Ingredient in the designation; such part or portion of them is capable of being Geometrically squared. But when I exclude a, I do therein exclude P (for that is an Arch also) and f = a + s, and e = a - s, because a is therein included. Mr. Caswell, (not being aware that I had squared these Figures) had done the same by a Method of his own, (which he shewed me lately) which I would have inserted here, but that he thought it not necessary; and instead thereof, hath given me the Quadrature of a Portion of the Epicycloid (which you will receive with this) and, I think, it is purely new. IV. The Quadrature of a Portion of the Epicycloid. By Mr. Caswel. Suppose DPV to be half of an exterior Epicycloid, VB its Axis, V the Vertex, VLB half of the generant Circle, E its Center; DB the Base, C its Center: Bisect the Arc of the Semicircle VB in L, and on the Center C through L draw a Circle cutting the Epicycloid in P: Then I say the Curvilinear Triangle VLP will be = BEq in \( \frac{CE}{CB} \); that is, the Square of the Semidiameter of the generant Circle will be to the Curvilinear Triangle VLP, as CB the Semidiameter of the Base, to CE: which CE in the exterior Epicycloid is the Sum of the Semidiameters of the Base and Generant, but in the Interior Epicycloid \( D p u \), 'tis the difference of the said Semidiameters. **COROLLARY.** In the Interior Epicycloid, if \( C E \) is \( \frac{1}{2} CB \), the Epicycloid then degenerating into a right Line, the Quadrature of the Triangle \( lp u \) will be in effect the same with the Quadrature of Hippocrates Chius. **COROL. II.** If the Semidiameter of the Base is supposed infinite, the Epicycloid then being the common Cycloid, the Area of the said Triangle will be equal to the Square of the Radius of the Generant, and so it falls in with that Theorem which Lalovera found, and calls Mirabile. Though I do not think the abovesaid Quadrature can easily be deduced from what has been yet published of the Epicycloid, I have not added the Demonstration; but think it enough to name a general Proposition from whence I deduced it, viz. The Segments of the Generant Circle are to the Correspondent Segments of the Epicycloid, as \( CB \) to \( 2CE + CB \). For Example, suppose \( FmG \) the Position of part of the Generant when the point \( F \) of the Exterior Epicycloid was designed, then the Segment \( FmGn \) is to the Segment \( DFng :: \) as \( CB \) to \( 2CE + CB \). And consequently the whole Epicycloid to the whole Generant in the same Proportion: Which is the only case demonstrated by Monsieur De la Hire. It follows also that in the Vulgar Cycloid, its Segments are triple of the Correspondent Sectors of the Generant, which was first shewn by Dr. Wallis. *A Demonstration hereof, with a General Proposition for all Curves of this kind, shall be given in the next Transaction.* V. A.