, the resistance will be to the gravity as to or as to . Now for the abscissas CB, CD, CE, put -o, o, 2o. For the ordinate CH put P; and for MI put any series Qo + Ro2 + So3 +, &c. And all the terms of the series after the first, that is, Ro2 + So3 +, &c., will be NI; and the ordinates DI, EK, and BG will be P - Qo - Ro2 - So3 -, &c., P - 2Qo - 4Ro2 - 8So3 -, &c., and P + Qo - Ro2 + So3 -, &c., respectively. And by squaring the differences of the ordinates BG - CH and CH - DI, and to the squares thence produced adding the squares of BC and CD themselves, you will have oo + QQoo - 2QRo3 +, &c., and oo + QQoo + 2QRo3 +, &c., the squares of the arcs GH, HI; whose roots , and are the arcs GH and HI. Moreover, if from the ordinate CH there be subducted half the sum of the ordinates BG and DI, and from the ordinate DI there be subducted half the sum of the ordinates CH and EK, there will remain Roo and Roo + 3So3, the versed sines of the arcs GI and HK. And these are proportional to the lineolæ LH and NI, and therefore in the duplicate ratio of the infinitely small times T and t: and thence the ratio is or ; and , by substituting the values of , GH, HI, MI and NI just found, becomes . And since 2NI is 2Roo, the resistance will be now to the gravity as to 2Roo, that is, as to 4RR. And the velocity will be such, that a body going off therewith from any place H, in the direction of the tangent HN, would describe, in vacuo, a parabola, whose diameter is HC, and its latus rectum or . And the resistance is as the density of the medium and the square of the velocity conjunctly; and therefore the density of the medium is as the resistance directly, and the square of the velocity inversely; that is, as[Pg 270] directly and inversely; that is, as Q.E.I. COR. 1. If the tangent HN be produced both ways, so as to meet any ordinate AF in will be equal to , and therefore in what has gone before may be put for . By this means the resistance will be to the gravity as 3S × HT to 4RR × AC; the velocity will be as , and the density of the medium will be as . COR. 2. And hence, if the curve line PFHQ be defined by the relation between the base or abscissa AC and the ordinate CH, as is usual, and the value of the ordinate be resolved into a converging series, the Problem will be expeditiously solved by the first terms of the series; as in the following examples. EXAMPLE 1. Let the line PFHQ be a semi-circle described upon the diameter PQ, to find the density of the medium that shall make a projectile move in that line. Bisect the diameter PQ in A; and call AQ, n; AC, a; CH, e; and CD, o; then DI2 or AQ2 - AD2 = nn - aa - 2ao - oo, or ee - 2ao - oo; and the root being extracted by our method, will give , &c. Here put nn for ee + aa, and DI will become , &c.