**Si elevationes duorum planorum duplam habuerint rationem eius quam habeant eorumdem planorum longitudines, lationes ex quiete in ipsis, temporibus aequalibus absolventur.** | **If the heights of two inclined planes are to each other in the same ratio as the squares of their lengths, bodies starting from rest will traverse these planes in equal times. ** |

Sint plana in aequalia et inaequaliter inclinata AE, AB, quorum elevationes sint FA, DA; et quam rationem habet AE ad AB, eamdem duplicatam habeat FA ad DA: dico, tempora lationum super planis AE, AB ex quiete in A esse {10} aequalia. Ductae sint parallelae horizontales ad lineam elevationum EF et DB quae secet AE in G: et quia ratio FA ad AD dupla est rationis EA ad AB, et ut FA ad AD, ita EA ad AG, ergo ratio EA ad AG dupla est rationis EA ad AB; ergo ab media est inter EA, AG. Et quia tempus descensus per AB ad tempus per AG est ut AB ad AG, tempus autem descensus per AG ad tempus per AE est ut AG ad mediam inter AG, AE, quae est AB, ergo, ex aequali, tempus per AB ad tempus per AE est ut AB ad se ipsam; sunt {20} igitur tempora aequalia: quod erat demonstrandum. | Take two planes of different lengths and different inclinations, AE and AB, whose heights are AF and AD: let AF be to AD as the square of AE, is to the square of AB; then, I say, that a body, starting from rest at A, will traverse the planes AE and AB in equal times. From the vertical line, draw the horizontal parallel lines EF and DB, the latter cutting AE at G.Since FA:DA = EA^2:BA^2, and since FA:DA = EA:GA, it follows that EA: GA = EA^2: BA^2. Hence BA is a mean proportional between EA and GA. (Condition 2/03-th-03-cor) Now since the time of descent along AB bears to the time along AG the same ratio which AB bears to AG (Condition 2/02-th-02-cor2) and since also the time of descent along AG is to the time along AE as AG is to a mean proportional between AG and AE, that is, to AB, it follows, ex aequali, that the time along AB is to the time along AE as AB is to itself. Therefore the times are equal. Q. E. D. |