2/16-th-13-cor
| | ||||
2/16-th-13-cor | | | | | |
| {20} COLLARIUM. | COROLLARY |
 | |
| Ex hac atque ex praecedenti constat, spatium quod conficitur in perpendiculo post casum ex sublimi, tempore eodem quo conficitur planum inclinatum, minus esse eo quod conficitur tempore eodem atque in inclinato non praecedente casu ex sublimi, maius tamen quam idem planum inclinatum. | From this and the preceding proposition, it is clear that the vertical distance covered by a freely falling body, after a preliminary fall, and during the time-interval required to traverse an inclined plane, is greater than the length of the inclined plane, but less than the distance traversed on the inclined plane during an equal time, without any preliminary fall. |
| Cum enim modo demonstratum sit, quod mobilium venientium ex termino sublimi A, tempus conversi per EC brevius sit tempore procedentis per EB, constat, spatium quod conficitur per EB tempore aequali tempori per EC, {30} minus esse toto spatio EB. Quod autem idem spatium perpendiculi maius sit quam EC, manifestum fit sumpta figura praecedentis propositionis, in qua partem perpendiculi BG confici demonstratum est tempore eodem cum BC post casum AB: hanc autem BG {235} maiorem esse quam BC, sic colligitur. Cum BE, FB aequales sint, BA vero minor BD, maiorem rationem habet FB ad BA quam EB ad BD, et, componendo, FA ad AB maiorem habet quam ED ad DB; est autem ut FA ad AB, ita GF ad FB (est enim AF media inter BA, AG), et, similiter, ut ED ad BD, ita est CE ad EB; ergo GB ad BF maiorem habet rationem quam CB ad BE: est igitur GB maior BC. | (Condition 2/16-th-13) For since we have just shown that bodies falling from an elevated point A will traverse the plane EC in Fig. 71 in a shorter time than the vertical EB, it is evident that the distance along EB which will be traversed during a time equal to that of descent along EC will be less than the whole of EB.But now in order to show that this vertical distance is greater than the length of the inclined plane EC, (Condition 2/15-pr-03) we reproduce Fig. 70 of the preceding theorem in which the vertical length BG is traversed in the same time as BC after a preliminary fall through AB. That BG is greater than BC is shown as follows: since BE and FB are equal {235} while BA is less than BD, it follows that FB will bear to BA a greater ratio than EB bears to BD; and, componendo, FA will bear to BA a greater ratio than ED to DB; but FA:AB = GF:FB (since AF is a mean proportional between BA and AG) and in like manner ED:BD = CE:EB. Hence GB bears to BF a greater ratio than CB bears to BE; therefore GB is greater than BC. |
| | ||||
2/16-th-13-cor | | | | | |