Discorsi Propositions 2/38-pr-16 Discorsi Proposition2/38-pr-16

 PROBLEMA XVI, PROPOSITIO XXXVIII. PROBLEM XVI, PROPOSITION XXXVIII Datis duobus planis horizontalibus a perpendiculo sectis, in perpendiculo punctum sublime reperire, ex quo cadentia mobilia, et in planis horizontalibus reflexa, conficiant, in temporibus aequalibus temporibus casuum in iisdem horizontalibus, in superiore nempe atque in inferiore, spatia quae inter se habeant quamcunque datam rationem minoris ad maiorem. Given two horizontal planes cut by a vertical line, it is required to and a point on the upper part of the vertical line from which bodies may fall to the horizontal planes and there, having their motion deflected into a horizontal direction, will, during an interval equal to the time of fall, traverse distances which bear to each other any assigned ratio of a smaller quantity to a larger. {20} Secta sint plana honzontalia CD, BE a perpendiculo ACB, sitque data ratio minoris ad maiorem, N ad FG: oportet, in perpendiculo AB punctum sublime reperire, ex quo mobile cadens, et in plano CD reflexum, tempore aequali tempori sui casus spatium conficiat, quod ad spatium confectum ab altero mobili, ex eodem puncto sublimi veniente, tempore aequali tempori sui casus, motu reflexo per BE planum, habeat rationem eamdem cum {266} data N ad FG. Ponatur GH aequalis ipsi N; et ut FH ad HG, ita fiat BC ad CL: dico, L esse punctum sublime quaesitum. Accepta enim CM dupla ad CL, ducatur LM, plano BE occurrens in O; erit BO dupla BL: et quia ut FH ad HG, ita BC ad CL, erit, componendo et convertendo, ut HG, hoc est N, ad GF, ita CL ad LB, hoc est CM ad BO. Cum autem CM dupla sit ad LC, patet, spatium CM esse illud quod a mobili veniente ex L post casum LC conficitur in plano CD, et eadem ratione BO esse illud quod conficitur post casum LB in tempore aequali tempori casus per LB, cum BO sit dupla ad BL. Ergo patet propositum. Let CD and BE be the horizontal planes cut by the vertical ACB, and let the ratio of the smaller quantity to the larger be that of N to FG. It is required to find in the upper part of the vertical line, AB, a point from which a body falling to the plane CD and there having its motion deflected along this plane, will traverse, during an interval equal to its time of fall a distance such that if another body, falling from this same point to the plane BE, there have its motion deflected along this plane and continued during an interval equal to its time of fall, will traverse a distance which bears to the former distance the {266} ratio of FG to N. Lay off GH equal to N, and select the point L so that FH:HG = BC:CL. Then, I say, L is the point sought. For, if we lay off CM equal to twice CL, and draw the line LM cutting the plane BE at O, then BO will be equal to twice BL. And since FH:HG = BC:CL, we have, componendo et convertendo, HG:GF = N:GF = CL:LB = CM:BO. (Condition 2/23-pr-09-schol1) It is clear that, since CM is double the distance LC, the space CM is that which a body falling from L through LC will traverse in the plane CD; and, for the same reason, since BO is twice the distance BL, it is dear that BO is the distance which a body, after fall through LB, will traverse during an interval equal to the time of its fall through LB. Q. E. F.

 Discorsi Propositions 2/38-pr-16 Discorsi Proposition2/38-pr-16