An Extract of a Letter from the Reverend Dr. John Wallis to Richard Waller, Esq; Secretary to the Royal Society, Concerning the Spaces in the Cycloid, which are Perfectly Quadrable

III. An Extract of a Letter from the Reverend Dr. John Wallis to Richard Waller, Esq; Secretary to the Royal Society, concerning the Spaces in the Cycloid, which are perfectly Quadrable. Oxford, August 22. 1695. SIR, I find it is thought by most, that there is no other part of the Semicycloid Figure (adjacent to the Curve) that is capable of being Geometrically Squared, but these two, viz. 1. The Segment $ABV$, Figur. 1. taking $AV = \frac{1}{2} Aa$, (which was first observed by Sir Christopher Wren, and after him by Hugenius and others) and it is $\frac{3}{4} sR = \frac{3}{4} R^2 \sqrt{3}$. 2. The Trilinear $AdD$ (taking $dD$, in the Parallel $dDC$, passing through the Center $C$,) which is $= R^2$. But it is otherwise (as I have shewed in my Treatise De Cycloide, and that De Motu; the Figures of which latter I retain here, so far as they concern this occasion;) there being other portions of it, equally capable of Quadrature. In order to which, I there shew (De Motu, Cap. 5. Prop. 20. A. pag. 802, 803, 804.) that not only the Cycloid is Triple to the Circle Generant, (which was known before) but that the respective parts of that are Triple to those of this: which is the Foundation on which I build my whole Process concerning the Cycloid in both Treatises, (and which is not pretended, that I know of, to have been observed, or known, by any body before me:) That is $b\beta aA$ Figur. 1, triple to the Sector $BaA$ (taking $b\beta$ parallel to $Ba$) wherever, in the Curve $Ar$, we take the point $b$. I then shew, that the Cycloid is a Figure compounded of these two; the Semicircle $AD\alpha$, and the Trilinear $AD\alpha\tau bA$, lying between the two Curves $AD\alpha$ and $Ad\tau$, (and therefore, to Square any part of these, is the same as to Square the respective part of the Cycloid. I shew further (Ibidem, pag. 804.) that this Trilinear is but a distorted Figure, (by reason of the Semicircle thrust in between it and its Axis) which being restored to its due Position (by taking out the Semicircle into a different Figure, as Figur. 2. and thrusting the Lines $bB$ home to the Axis, so as that $BV$ be the same point) is the same with $Ar\alpha$, Figur. 3. (the Parallelograms $b\beta\alpha B$ being set upright, which in the Cycloid stand sloping; and the Circular Archs $b\beta$, Figur. 1, becoming straight-lines in Figur. 3. and the Lines $bB$ being, in both, equal to the respective Archs $BA$, everywhere;) which therefore I call Trilineum Restitutum (the Trilinear restored to its due Position, which Figure I do not find that any before me has considered:) So that to Square any part of this, is the same as to Square the respective part of the Cycloid, (or of the Trilinear in the Cycloid:) That which in the Cycloid lies between two Archs of the Circle Generant in different Positions, answering to that which, in the restored Figure, lies between the respective straight-lines. And therefore $AdDA = \tau d\delta\tau$, Figur. 1, = $AdDA = \tau d\delta\tau$, Figur. 3, = $R^2$. And $AbkDA$, $\tau b k \delta \tau$, Figur. 1, = $AbkDA$, $\tau b k \delta \tau$, Figur. 3, = $s R$. And $bkd$, Figur. 1, = $bkd$, Figur. 3, = $R^2 - s R$, Ibidem, Cap. 17. B. pag. 756. Where, if $b$ be taken above $dkDC$ (passing through the Center $C$,) these Figures are within the Cycloid, and within the restored Figure; but without them, if $b$ be taken below that Line, and adjacent to the Curve $Ab\tau$, in both cases. By 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