2/29-th-18
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| THEOREMA XVIII, PROPOSITIO XXIX. | THEOREM XVIII, PROPOSITION XXIX |
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| Dato quolibet spatio horizontali, ex cuius termino erectum sit perpendiculum, in quo sumatur pars aequalis dimidio spatii in horizontali dato, mobile ex tali altitudine descendens et in horizontali conversum {250} conficiet horizontale spatium una cum perpendiculo breviori tempore, quam quodcunque aliud spatium perpendiculi cum eodem spatio horizontali. | Given a limited horizontal line, at one end of which is erected a limited vertical line whose length is equal to one-half the given horizontal line; then a body, falling through this given height and having its motion deflected into a horizontal direction, will traverse the given horizontal distance and vertical line in less time than it will any other vertical distance plus the given horizontal distance. |
| Sit planum horizontale, in quo datum sit quodlibet spatium BC, et ex termino B sit perpendiculum, in quo BA sit dimidium ipsius BC: dico, tempus, quo mobile ex A demissum conficiet ambo spatia AB, BC, esse temporum omnium brevissimum, quibus idem spatium BC cum parte perpendiculi, sive maiori sive minori parte {10} AB, conficeretur. Sit sumpta maior, ut in prima figura, vel minor, ut in secunda, EB: ostendendum est, tempus quo conficiuntur spatia EB, BC, longius esse tempore quo conficiuntur AB, BC. Intelligatur, tempus per AB esse ut AB; erit quoque tempus motus in horizontali BC, cum BC dupla sit ad AB, et per ambo spatia ABC tempus erit dupla BA. Sit BO media inter EB, BA; erit BO tempus casus per EB: sit praeterea horizontale spatium BD duplum ipsius BE; constat, tempus ipsius post casum EB esse idem BO. Fiat ut DB ad BC, seu ut EB ad BA, {20} ita OB ad BN, et cum motus in horizontali sit aequabilis, sitque OB tempus per BD post casum ex E: erit NB tempus per BC post casum ex eadem altitudine E. Ex quo constat, OB cum BN esse tempus per EBC; cumque dupla BA sit tempus per ABC, ostendendum relinquitur, OB cum BN maiora esse quam dupla BA. Cum autem OB media sit inter EB, BA, ratio EB ad BA dupla est rationis OB ad BA; et cum EB ad BA sit ut OB ad BN, erit quoque ratio OB ad BN dupla rationis OB ad BA: verum ipsa ratio OB ad BN componitur ex rationibus OB ad BA et AB ad BN: ergo ratio AB ad BN est eadem cum ratione OB ad BA. Sunt igitur BO, BA, BN tres continue proportionales, et OB cum BN maiores {30} quam dupla BA: ex quo patet propositum. | {250} Let BC be the given distance in a horizontal plane; at the end B erect a perpendicular, on which lay off BA equal to half BC. Then, I say, that the time required for a body, starting from rest at A, to traverse the two distances, AB and BC, is the least of all possible times in which this same distance BC together with a vertical portion, whether greater or less than AB, can be traversed. Lay off EB greater than AB, as in the first figure, and less than AB, as in the second. It must be shown that the time required to traverse the distance EB plus BC is greater than that required for AB plus EC. Let us agree that the length AB shall represent the time along AB, (Condition 2/23-pr-09-schol1) then the time occupied in traversing the horizontal portion BC will also be AB, seeing that BC = 2AB; consequently the time required for both AB and BC will be twice AB. Choose the point O such that EB:BO = BO:BA, (Condition 2/02-th-02-cor2) then BO will represent the time of fall through EB. Again lay off the horizontal distance BD equal to twice BE; (Condition 2/23-pr-09-schol1) whence it is clear that BO represents the time along BD after fall through EB. Select a point N such that DB:BC = EB:BA = OB:BN. (Condition 1/01-th-01) Now since the horizontal motion is uniform and since OB is the time occupied in traversing BD, after fall from E, it follows that NB will be the time along BC after fall through the same height EB. Hence it is clear that OB plus BN represents the time of traversing EB plus BC; and, since twice BA is the time along AB plus BC, it remains to be shown that OB+BN>2BA. But since EB:BO = BO:BA, it follows that EB:BA = OB2xxx:BA2xxx. Moreover since EB:BA = OB:BN it follows that OB:BN = OB2xxx:BA2xxx. But OB:BN = (OB:BA)(BA:BN), and therefore AB:BN = OB:BA, that is, BA is a mean proportional between BO and BN.Consequently OB+BN>2BA. Q. E. D. |
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2/29-th-18 |
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