Page:Newton's Principia (1846).djvu/275

Therefore since, in the same time, the action of gravity generates, in a falling body, the velocity $\scriptstyle {\frac {2NI}{t}}$, the resistance will be to the gravity as $\scriptstyle {\frac {GH}{T}}-{\frac {HI}{t}}+{\frac {2MI\times NI}{t\times HI}}$ or as $\scriptstyle {\frac {t\times GH}{T}}-HI+{\frac {2MI\times NI}{HI}}$ to 2NI.

Now for the abscissas CB, CD, CE, put -o, o, 2o. For the ordinate CH put P; and for MI put any series Qo + Ro² + So³ +, &c. And all the terms of the series after the first, that is, Ro² + So³ +, &c., will be NI; and the ordinates DI, EK, and BG will be P - Qo - Ro² - So³ -, &c., P - 2Qo - 4Ro² - 8So³ -, &c., and P + Qo - Ro² + So³ -, &c., respectively. And by squaring the differences of the ordinates BG - CH and CH - DI, and to the squares thence produced adding the squares of BC and CD themselves, you will have oo + QQoo - 2QRo³ +, &c., and oo + QQoo + 2QRo³ +, &c., the squares of the arcs GH, HI; whose roots $\scriptstyle o{\sqrt {1+QQ}}-{\frac {QRoo}{\sqrt {1+QQ}}}$, and $\scriptstyle o{\sqrt {1+QQ}}+{\frac {QRoo}{\sqrt {1+QQ}}}$ are the arcs GH and HI. Moreover, if from the ordinate CH there be subducted half the sum of the ordinates BG and DI, and from the ordinate DI there be subducted half the sum of the ordinates CH and EK, there will remain Roo and Roo + 3So³, the versed sines of the arcs GI and HK. And these are proportional to the lineolae LH and NI, and therefore in the duplicate ratio of the infinitely small times T and t: and thence the ratio $\scriptstyle {\frac {t}{T}}$ is $\scriptstyle {\sqrt {\frac {R+3So}{R}}}$ or $\scriptstyle {\frac {R+{\frac {3}{2}}So}{R}}$; and $\scriptstyle {\frac {t\times GH}{T}}-HI+{\frac {2MI\times NI}{HI}}$, by substituting the values of $\scriptstyle {\frac {t}{T}}$, GH, HI, MI and NI just found, becomes $\scriptstyle {\frac {3Soo}{2R}}{\sqrt {1+QQ}}$. And since 2NI is 2Roo, the resistance will be now to the gravity as $\scriptstyle {\frac {3Soo}{2R}}{\sqrt {1+QQ}}$, that is, as $\scriptstyle 3S{\sqrt {1+qq}}$ to 4RR.

And the velocity will be such, that a body going off therewith from any place H, in the direction of the tangent HN, would describe, in vacuo, a parabola, whose diameter is HC, and its latus rectum $\scriptstyle {\frac {HN^{2}}{NI}}$ or $\scriptstyle {\frac {1+QQ}{R}}$.

And the resistance is as the density of the medium and the square of the velocity conjunctly; and therefore the density of the medium is as the resistance directly, and the square of the velocity inversely; that is, as