3. In order to the Computation hereof, I first premise this Lemma, (as the most rational that doth occur for my first footing,) That (supposing other things equal) the Resistance is proportional to the Celerity. For in a double Celerity, there is to be removed (in the same time) twice as much Air, (which is a double Impediment) in a treble, thrice as much; and so in other Proportions.
4. Suppose we then the Force impressed (and consequently the Celerity, if there were no Resistance) as 1; the Resistance as r. (which must be less than the Force, or else the Force would not prevail over the Impediment, to create a Motion.) And therefore the effective Force at a first Moment, is to be reputed as 1 - r: That is, so much as the Force impressed, is more than the Impediment or Resistance.
5. Be it as 1 - r to 1; so one to m. (which m is therefore greater than 1.)
6. And therefore the effective Force (and consequently the Celerity) as to a first Moment, is to be 1⁄m of what it would be, had there been no Resistance.
7. This 1⁄m is also the remaining Force after such first Moment; and this remaining Force is (for the same Reason) to be proportionally abated as to a second Moment; that is, we are to take 1⁄m thereof, that is 1⁄mm of the impressed Force. And for a third Moment (at equal distance of time) 1⁄mmm; for a fourth 1⁄m4; and so onward infinitely.
8. Because the length dispatched (in equal times) is proportional to the Celerities; the Lines of Motion (answering to those equal Times) are to be as 1⁄m, 1⁄m2, 1⁄m3, 1⁄m4, &c. of what they would have been, in the same Times, had there been no Resistance.
9. This therefore is a Geometrical Progression; and (because of m greater than 1) continually decreasing.
10. This decreasing Progression infinitely continued (determining in the same Point of Rest, where the Motion is supposed to expire) is yet of a finite Magnitude; and equal to 1 / m - 1, of what it would have been in so much Time, if there had been no Resistance. As is demonstrated in my Algebra, Chap. 95. Prop. 8. For (as I have elsewhere demonstrated) the Sum or Aggregate of a Geometrical Progression is VR - A / R - 1 (supposing V the greatest Term, A the least, and R the common Multiplier.) That is VR / R - 1 - A / R - 1. Now in the present Case, (supposing the Progression infinitely continued) the least Term A, becomes infinitely small, or = 0. And consequently 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⁄R3 + &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⁄R3 + &c., That is, in our present Case (because of V = 1, & R = m:) 1⁄m + 1⁄mm + 1⁄m3 &c. = 1 / m - 1. That is, (putting n = m - 1) 1⁄n of what it would have been if there had been no Resistance.
11. 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 Ad = 1 - 1⁄m (supposing Bd to be taken, from B towards A,) to AB = 1, or as m - 1 to m: so is BS = 1⁄mDH, to dh, which is therefore equal to 1 / m - 1 of DH; that is (as will appear by dividing 1, by m - 1,) to 1⁄m + 1⁄mm + 1⁄m3, &c. of DH.[15]
Or if Bd be taken beyond B; then as Ad = 1 + 1⁄m to AB = 1, or as m + 1 to m, so is 1⁄m DH to dh, which is therefore equal to 1 / m + 1DH; that is (as will appear by like dividing of 1 by m + 1;) = to 1⁄m - 1⁄mm + 1⁄m3 - &c. of DH.