2/36-th-22-schol1
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2/36-th-22-schol1 |
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| {20} SCHOLIUM. | SCHOLIUM |
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| Ex his quae demonstrata sunt, colligi posse videtur, lationem omnium velocissimam ex termino ad terminum non per brevissimam lineam, nempe per rectam, sed per circuli portionem, fieri. | From the preceding it is possible to infer that the path of quickest descent [lationem omnium velocissimam] from one point to another is not the shortest path, namely, a straight line, but the arc of a circle. |
| In quadrante enim BAEC, cuius latus BC sit ad horizontem erectum, divisus sit arcus AC in quotcunque partes aequales, AD, DE, EF, FG, GC, et ductae sint rectae ex C ad puncta A, D, {30} E, F, G, et iunctae sint rectae quoque AD, DE, EF, FG, GC: manifestum est, lationem {264} per duas ADC citius absolvi quam per unam AC, vel DC ex quiete in D.Sed ex quiete in A citius absolvitur DC quam duae ADC: sed per duas DEC ex quiete in A verisimile est, citius absolvi descensum quam per solam CD: ergo descensus per tres ADEC absolvitur citius quam per duas ADC.Verum similiter, praecedente descensu per ADE, citius fit latio per duas EFC quam per solam EC; ergo per quatuor ADEFC citius fit motus quam per tres ADEC.Ac tandem per duas FGC, post praecedentem descensum per ADEF, citius absolvitur latio quam per solam FC; ergo per quinque ADEFGC breviori adhuc tempore fit descensus quam per quatuor ADEFC.Quo igitur per inscriptos polygonos magis ad {10} circumferentiam accedimus, eo citius absolvitur motus inter duos terminos signatos A, C. | In the quadrant BAEC, having the side BC vertical, divide the arc AC into any number of equal parts, AD, DE, EF, FG, GC, and from C draw straight lines to the points A, D, E, F, G, draw also the straight lines AD, DE, EF, FG, GC. (Condition 2/36-th-22) Evidently descent along the path ADC is quicker {264} than along AC alone or along DC from rest at D.But a body, starting from rest at A, will traverse DC more quickly than the path ADC; while, if it starts from rest at A, it will traverse the path DEC in a shorter time than DC alone. Hence descent along the three chords, ADEC, will take less time than along the two chords ADC. Similarly, following descent along ADE, the time required to traverse EFC is less than that needed for EC alone. Therefore descent is more rapid along the four chords ADEFC than along the three ADEC.And finally a body, after descent along ADEF, will traverse the two chords, FGC, more quickly than FC alone. Therefore, along the five chords, ADEFGC, descent will be more rapid than along the four, ADEFC.Consequently the nearer the inscribed polygon approaches a circle the shorter is the time required for descent from A to C. |
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