Page:Newton's Principia (1846).djvu/305

 the areas YmtZ, XhmY, proportional to them, will be also equal; and the lines SX, SY, SZ, that is, AH, EM, QT continually proportional, as they ought to be. And if the lines SA, SE, SQ, obtain any other order in the series of continued proportionals, the lines AH, EM, QT, because of the proportional hyperbolic areas, will obtain the same order in another series of quantities continually proportional.


 * Let the density of any fluid be proportional to the compression, and its parts be attracted downwards by a gravitation reciprocally proportional to the squares of the distances from the centre: I say, that if the distances be taken in harmonic progression, the densities of the fluid at those distances will be in a geometrical progression.

Let S denote the centre, and SA, SB, SC, SD, SE, the distances in geometrical progression. Erect the perpendiculars AH, BI, CK, &c., which shall be as the densities of the fluid in the places A, B, C, D, E, &c., and the specific gravities thereof in those places will be as $$\scriptstyle \frac{AH}{SA^2}$$, $$\scriptstyle \frac{BI}{SB^2}$$, $$\scriptstyle \frac{CK}{SC^2}$$, &c. Suppose these gravities to be uniformly continued, the first from A to B, the second from B to C, the third from C to D, &c. And these drawn into the altitudes AB, BC, CD, DE, &c., or, which is the same thing, into the distances SA, SB, SC, &c., proportional to those altitudes, will give $$\scriptstyle \frac{AH}{SA}$$, $$\scriptstyle \frac{BI}{SB}$$, $$\scriptstyle \frac{CK}{SC}$$, &c., the exponents of the pressures. Therefore since the densities are as the sums of those pressures, the differences AH - BI, BI - CK, &c., of the densities will be as the differences of those sums $$\scriptstyle \frac{AH}{SA}$$, $$\scriptstyle \frac{BI}{SB}$$, $$\scriptstyle \frac{CK}{SC}$$, &c. With the centre S, and the asymptotes SA, Sx, describe any hyperbola, cutting the perpendiculars AH, BI, CK, &c., in a, b, c, &c., and the perpendiculars Ht, In, Kw, let fall upon the asymptote Sx, in h, i, k; and the differences of the densities tu, uw, &c., will be as $$\scriptstyle \frac{AH}{SA}$$, $$\scriptstyle \frac{BI}{SB}$$, &c. And the rectangles tu $$\scriptstyle \times$$ th, uw $$\scriptstyle \times$$ ui, &c., or tp, uq, &c., as $$\scriptstyle \frac{AH\times th}{SA}$$, $$\scriptstyle \frac{BI\times ui}{SB}$$, &c., that is, as Aa, Bb, &c. For, by the nature of the hyperbola, SA is to AH or St as th to Ac, and therefore $$\scriptstyle \frac{AH\times th}{SA}$$ is equal to Aa. And, by a like