Page:Newton's Principia (1846).djvu/290

II of this Book) the moment KL of AK will be equal to $\scriptstyle {\frac {2APQ+2BA\times PQ}{Z}}$ or $\scriptstyle {\frac {2BPQ}{Z}}$, and the moment KLON of the area AbNK will be equal to $\scriptstyle {\frac {2BPQ\times LO}{Z}}$ or $\scriptstyle {\frac {BPQ\times BD^{3}}{2Z\times CK\times AB}}$.

Case 1. Now if the body ascends, and the gravity be as AB² + BD², BET being a circle, the line AC, which is proportional to the gravity, will be $\scriptstyle {\frac {AB^{2}+BD^{2}}{Z}}$, and DP² or AP² + 2BAP + AB² + BD² will be AK $\scriptstyle \times$ Z + AC $\scriptstyle \times$ Z or CK $\scriptstyle \times$ Z; and therefore the area DTV will be to the area DPQ as DT² or DB² to CK $\scriptstyle \times$ Z.

Case 2. If the body ascends, and the gravity be as AB² - BD², the line AC will be $\scriptstyle {\frac {AB^{2}-BD^{2}}{Z}}$, and DT² will be to DP² as DF² or DB² to BP² - BD² or AP² + 2BAP + AB² - BD², that is, to AK $\scriptstyle \times$ Z +

AC $\scriptstyle \times$ Z or CK $\scriptstyle \times$ Z. And therefore the area DTV will be to the area DPQ as DB² to CK $\scriptstyle \times$ Z.

Case 3. And by the same reasoning, if the body descends, and therefore the gravity is as BD² - AB², and the line AC becomes equal to $\scriptstyle {\frac {BD^{2}-AB^{2}}{Z}}$; the area DTV will be to the area DPQ, as DB² to CK $\scriptstyle \times$ Z: as above.

Since, therefore, these areas are always in this ratio, if for the area