# Page:Newton's Principia (1846).djvu/270

For let the force of gravity be expounded by the given line AC; the force of resistance by the indefinite line AK; the absolute force in the descent of the body by the difference KC: the velocity of the body by a line AP, which shall be a mean proportional between AK and AC, and therefore in a subduplicate ratio of the resistance; the increment of the resistance made in a given particle of time by the lineola KL, and the contemporaneous increment of the velocity by the lineola PQ; and with the centre C, and rectangular asymptotes CA, CH, describe any hyperbola BNS meeting the erected perpendiculars AB, KN, LO in B, N and O. Because AK is as AP², the moment KL of the one will be as the moment 2APQ of the other, that is, as AP $\scriptstyle \times$ KC; for the increment PQ of the velocity is (by Law II) proportional to the generating force KC. Let the ratio of KL be compounded with the ratio KN, and the rectangle KL $\scriptstyle \times$ KN will become as AP $\scriptstyle \times$ KC $\scriptstyle \times$ KN; that is (because the rectangle KC $\scriptstyle \times$ KN is given), as AP. But the ultimate ratio of the hyperbolic area KNOL to the rectangle KL $\scriptstyle \times$ KN becomes, when the points K and L coincide, the ratio of equality. Therefore that hyperbolic evanescent area is as AP. Therefore the whole hyperbolic area ABOL is composed of particles KNOL which are always proportional to the velocity AP; and therefore is itself proportional to the space described with that velocity. Let that area be now divided into equal parts