2/36-th-22
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2/36-th-22 |
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| THEOREMA XXII, PROPOSITIO XXXVI. | THEOREM XXII, PROPOSITION XXXVI |
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| {30} Si in circulo ad horizontem erecto ab imo puncto elevetur planum non maiorem subtendens circumferentiam quadrante, a terminis cuius duo {262} alia plana ad quodlibet circunferentiae punctum inflectantur, descensus in planis ambobus inflexis breviori tempore absolvetur, quam in solo priori plano elevato, vel quam in altero tantum ex illis duobus, nempe in inferiori. | If from the lowest point of a vertical circle, a chord is drawn subtending an arc not greater than a quadrant, and if from the two ends of this chord two other chords be drawn to any point on the arc, the time of descent along the two latter chords will be shorter than along the first, and shorter also, by the same amount, than along the lower of these two latter chords. |
| Sit circuli ad horizontem erecti ab imo puncto C circunferentia CBD, non maior quadrante, in qua sit planum elevatum CD, et duo plana A terminis D, C inflexa ad quodlibet punctum B, in circunferentia sumptum: dico, tempus descensus per ambo plana DBC {10} brevius esse tempore descensus per solum DC, vel per unicum BC ex quiete in B. Ducta sit per D horizontalis MDA, cui CB extensa occurrat in A; sintque DN, MC ad MD, et BN ad BD perpendiculares, et circa triangulum rectangulum DBN semicirculus describatur DFBN, secans DC in F; et ipsarum CD, DF media sit proportionalis DO, ipsarum autem CA, AB media sit AV. Sit autem PS {20} tempus quo peragitur tota DC, vel BC (constat enim, tempore eodem peragi utramque), et quam rationem habet CD ad DO, hanc habeat tempus SP ad tempus PR: erit tempus PR id, in quo mobile ex D peragit DF; RS vero id, in quo reliquum FC. Cum vero PS sit quoque tempus quo mobile ex B peragit BC, si fiat ut BC ad CD, ita SP ad PT, erit PT tempus casus ex A in C, cum DC media sit inter AC, CB, ex ante demonstratis. Fiat tandem ut CA ad AV, ita TP ad PG: erit PG tempus quo mobile ex A venit in B, GT vero tempus residuum motus BC consequentis post motum ex A in B. Cum vero DN, circuli DFN diameter, ad horizontem sit erecta, temporibus aequalibus peragentur DF et DB lineae: quare si demonstratum {30} {263} fuerit, mobile citius permeare BC post casum DG, quam FC post peractam DF, habebimus intentum. At eadem temporis celeritate conficit mobile veniens ex D per DB ipsam BC, ac si venerit ex A per AB, cum ex utroque casu DG, AB aequalia accipiat velocitatis momenta; ergo demonstrandum erit, breviori tempore peragi BC post AB, quam FC post DF. Explicatum est autem, tempus quo peragitur BC post AB, esse GT; tempus vero ipsius FC post DF esse RS: ostendendum itaque est, RS maius esse quam GT. Quod sic ostenditur: quia ut SP ad PR, ita CD ad DO, per conversionem rationis et convertendo, ut RS ad SP, ita OC ad CD, ut autem SP ad PT, {10} ita DC ad CA; et quia est ut TP ad PG, ita CA ad AV, per conversionem rationis erit quoque ut PT ad TG, ita AC ad CV; ergo, ex aequali, ut RS ad GT, ita OC ad CV: est autem OC maior quam CV, ut mox demonstrabitur: ergo tempus RS maius est tempore GT: quod demonstrare oportebat. Cum vero CF maior sit CB, FD vero minor BA, habebit CD ad DF maiorem rationem quam CA ad AB; ut autem CD ad DF, ita quadratum CO ad quadratum OF, cum sint CD, DO, DF proportionales; ut vero CA ad AB, ita quadratum CV ad quadratum VB; ergo CO ad OF maiorem rationem habet quam CV ad VB: igitur, ex lemmate praedicto, CO maior est quam CV. Constat insuper, tempus per DC ad tempus per DBC esse ut DOC ad DO cum CV. | {262] Let CBD be an arc, not exceeding a quadrant, taken from a vertical circle whose lowest point is C; let CD be the chord {planum elevatum] subtending this arc, and let there be two other chords drawn from C and D to any point B on the arc. Then, I say, the time of descent along the two chords [plana] DB and BC is shorter than along DC alone, or along BC alone, starting from rest at B. Through the point D, draw the horizontal line MDA cutting CB extended at A: draw DN and MC at right angles to MD, and BN at right angles to BD; about the right-angled triangle DBN describe the semicircle DFBN, cutting DC at F. Choose the point O such that DO will be a mean proportional between CD and DF; in like manner select V so that AV is a mean proportional between CA and AB. Let the length PS represent the time of descent along the whole distance DC (Condition 2/06-th-06) or BC, both of which require the same time. Lay off PR such that CD:DO = timePS.timePR. (Condition 2/02-th-02-cor2) Then PR will represent the time in which a body, starting from D, will traverse the distance DF, (Condition 2/11-th-11) while RS will measure the time in which the remaining distance, FC, will be traversed. But since PS is also the time of descent, from rest at B, along BC, and if we choose T such that BC:CD = PS:PT (Condition 2/02-th-02-cor2) then PT will measure the time of descent from A to C, for we have already shown [Lemma] that DC is a mean proportional between AC and CB. Finally choose the point G such that CA:AV = PT:PG, (Condition 202C) then PG will be the time of descent from A to B, (Condition 2/11-th-11) while GT will be the residual time of descent along BC following descent from A to B. But, since the diameter, DN, of the circle DFN is a vertical line, (Condition 2/06-th-06) the chords DF and DB will be traversed in equal times; wherefore if one can prove that a body will traverse BC, after descent along DB, in a shorter time than it will FC after descent along DF he will have proved the theorem. But a body descending from D along DB will traverse BC in the same time as if it had come from A along AB, seeing that the body acquires the same {263} momentum in descending along DB as along AB. Hence it remains only to show that descent along BC after AB is quicker than along FC after DF. But we have already shown that GT represents the time along BC after AB; also that RS measures the time along FC after DF. Accordingly it must be shown that RS is greater than GT, which may be done as follows: Since SP:PR = CD:DO, it follows, invertendo et convertendo, that RS:SP = OC:CD; also we have SP:PI = DC:CA. And since TP:PG = CA:AV, it follows, invertendo, that PT:TG = AC:CV, therefore, ex aequali, RS:GT = OC:CV. But, as we shall presently show, OC is greater than CV; hence the time RS is greater than the time GT, which was to be shown. Now, since [Lemma] CF is greater than CB and FD smaller than BA, it follows that CD:DF>CA:AB. But CD:DF = CO:OF, seeing that CD:DO = DO:DF; and CA:AB = CV2xxx:VB2xxx. Therefore CO:OF>CV:VB, and, according to the preceding lemma, CO>CV. Besides this it is clear that the time of descent along DC is to the time along DBC as DOC is to the sum of DO and CV. |
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2/36-th-22 |
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