<57r>

Mr Newton

Sir

The Letter herewith sent proclaimes the Authors great worth knowledge and Candour, and the method therein seemes unto me most admirable for Geometrick Curves wherein the Ordinates are expressed by an Æquation, but howit will performe in Mechanick Curves, wherein there is no such habitude expressed I humbly crave your sentiment: at first reading it was not immediately obvious how he came to find $y=\frac{2z{r}^{2}}{rr+zz}$ his designe is first out of these Data $AQ=r$ the Radius and ${\mathrm{Q}}_{1}\mathrm{N}=z$ to find the Square of the Chord $AD=\frac{4{r}^{4}}{rr+zz}$ for the finding whereof suppose the Chord of the Complement to the Semicircle to be likewise drawne, and then there is given the ratio of these Chords such as r to z and the Sum of their Squares $=4rr$: out of such data by an Analyticall processe $\square AD$ is found $=\frac{4{r}^{4}}{rr+zz}$

And then it holds
$\begin{array}{cccccccc}& \square AN& :& \square NQ& \colon\colon & \square AD& & {\mathrm{D}}_{1}{\mathrm{B}}_{1}\phantom{\rule{1em}{0ex}}\text{which he calls}\phantom{\rule{0.5em}{0ex}}{y}^{2}\\ \text{that is}& rr+zz& :& zz& \colon\colon & \frac{4{r}^{4}}{rr+zz}& :& \frac{4{r}^{4}zz}{\begin{array}{|c|}\hline 2\\ \hline\end{array}\begin{array}{c}rr\end{array}\begin{array}{c}+zz\end{array}}\phantom{\rule{1em}{0ex}}\text{the roote}\end{array}$
whereof is $\frac{2z{r}^{2}}{rr+zz}=y$ as he makes it, and changing $zz$ in the second tearme of the Proportion for $rr$, by such meanes ${\mathrm{A}}_{1}\mathrm{B}$ or x is found $=\frac{2{r}^{3}}{rr+zz}$ as he likewise makes it, excuse me for troubling you with this impertinence I remaine

John Collins

Whereas he sayes habita ergo recta ${}_{1}\mathrm{B}_{1}\mathrm{D}\left(=\frac{2z{r}^{2}}{rr+zz}\right)$
et recta ${}_{1}\mathrm{B}_{2}\mathrm{B}_______________\left(=\frac{4{r}^{3}z\beta }{\begin{array}{|c|}\hline 2\\ \hline\end{array}\begin{array}{c}rr\end{array}\begin{array}{c}+zz\end{array}}\right)$
habebitur valor rectanguli, ${}_{1}\mathrm{D}_{1}{\mathrm{B}}_{2}\mathrm{B}$ multiplicatis eorum Valoribus in

<57v>

Out of $r=QA$ and $z={\mathrm{Q}}_{1}N$ he finds $x=$${\mathrm{A}}_{1}B$ (which is supposed not to fall out of the line AQ though the figure represents otherwise) thus
$r:z\colon\colon x:\frac{zx}{r}={}_{1}\mathrm{B}_{1}\mathrm{D}$.
This squared viz $\frac{zzxx}{rr}=2rx-{x}^{2}$ whence by reduction $x=\frac{2{r}^{3}}{zz+rr}$ his Calculus of the areall ordinate ${}_{1}\mathrm{N}_{1}\mathrm{P}$ is faulty but I hope ere long to send you the Calculation true.

$\begin{array}{c}\begin{array}{c}2r-x\\ \phantom{1}x\phantom{-0}\end{array}\\ \begin{array}{c}2rx-xx\end{array}\end{array}$