A / R - 1 doth also vanish, and thereby the Aggregate becomes = VR / R - 1. That is (as will appear by dividing VR by R - 1;) V + V ⁄ R + V ⁄ RR + V ⁄ R 3 + &c. = VR / R - 1[14]; (supposing the Progression to begin at V = 1.) That is (dividing all by R , that so the Progression may begin at V ⁄ R = 1 ⁄ m :) V / R - 1 = V ⁄ R + V ⁄ RR + V ⁄ R 3 + &c. , That is, in our present Case (because of V = 1, & R = m :) 1 ⁄ m + 1 ⁄ mm + 1 ⁄ m 3 &c. = 1 / m - 1. That is, (putting n = m - 1) 1 ⁄ n of what it would have been if there had been no Resistance.
This infinite Progression is fitly expressed by an Ordinate in the Exterior Hyperbola, parallel to one of the Asymptotes; and the several Members of that, by the several Members of this, cut in continual Proportion. As is there demonstrated at Prop. 15. For let SH , ( vid. Fig. 4. Tab. 5.) be an Hyperbola between the Asymptotes AB , AF : And let the Ordinate DH (in the Exterior Hyperbola, parallel to AF ,) represent the impressed Force undiminished; or the Line to be described in such time, by a Celerity answerable to such undiminished Force. And let BS (a like Ordinate) be 1 ⁄ m thereof; which therefore, being less than DH (as being equal to a Part of it) will be farther than it from AF . In AB (which I put = 1) let Bd be such a Part thereof, as is BS of DH . Now because (as is, well known) all the inscribed Parallelograms, in the Exterior Hyperbola, AS , AH , &c. are equal; and therefore their sides reciprocal: Therefore as