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Discorsi Proposition2/26-pr-10 |

PROBLEMA X, PROPOSITIO XXVI. | PROBLEM X, PROPOSITION XXVI |

Dato perpendiculo inter lineas parallelas horizontales, datoque spatio maiori eodem perpendiculo, sed minori quam duplum eiusdem, ex imo termino perpendiculi planum attollere inter easdem parallelas, super quo motu reflexo post descensum in perpendiculo conficiat mobile spatium dato aequale, et in tempore aequali tempori descensus in perpendiculo. | Given a vertical height joining two horizontal parallel lines; given also a distance greater than once and less than twice this vertical height, it is required to pass through the foot of the given perpendicular an inclined plane such that, after fall through the given vertical height, a body whose motion is deflected along the plane will traverse the assigned distance in a time equal to the time of vertical fall. |

{10} Inter parallelas horizontales AO, BC sit perpendiculum AB; FE vero maior sit quam BA, minor vero quam dupla eiusdem: oportet, ex B planum inter horizontales erigere, super quo mobile, post casum ex A in B, motu reflexo, in tempore aequali tempori descensus per AB, conficiat ascendendo spatium aequale ipsi EF. Ponatur ED aequalis AB; erit reliqua DF minor, cum tota EF minor sit quam dupla ad AB: sit DI aequalis DF, et ut EI ad ID, ita fiat DF ad aliam FX, atque ex B reflectatur recta BO aequalis EX: dico, planum per BO esse illud, super quo post casum AB mobile in tempore aequali tempori casus per AB pertransit ascendendo spatium aequale dato spatio EF. Ipsis ED, DF aequales ponantur BR, RS: {20} cum enim sit ut EI ad ID, ita DF ad FX, erit, componendo, ut ED ad DI, ita DX ad XF; hoc est, ut ED ad DF, ita DX ad XF, et EX {248} ad XD; hoc est, ut BO ad OR, ita RO ad OS. Quod si ponamus, tempus per AB esse AB, erit tempus per OB ipsa OB, et RO tempus per OS, et reliqua BR tempus per reliquum SB, descendendo ex O in B: sed tempus descensus per SB ex quiete in O est aequale tempori ascensus ex B in S post descensum AB: ergo BO est planum ex B elevatum, super quo post descensum per AB conficitur tempore BR, seu BA, spatium BS, aequale spatio dato EF: quod facere oportebat. | Let AB be the vertical distance separating two parallel horizontal lines AO and BC; also let FE be greater than once and less than twice BA. The problem is to pass a plane through B, extending to the upper horizontal line, and such that a body, after having fallen from A to B, will, if its motion be deflected along the inclined plane, traverse a distance equal to EF in a time equal to that of fall along AB. Lay off ED equal to AB; then the remainder DF will be less than AB since the entire length EF is less than twice this quantity; also lay off DI equal to DF, and choose the point X such that EI:ID = DF:FX; from B, draw the plane BO equal in length to EX. Then, I say, that the plane BO is the one along which, after fall through AB, a body will traverse the assigned distance FE in a time equal to the time of fall through AB. Lay off BR and RS equal to ED and DF respectively; then since EI:ID = DF:FX, we have, componendo, ED:DI = DX:XF = ED:DF = EX:XD = BO:OR = RO:OS. If we represent the time of fall along AB by the length AB, (Condition 2/03-th-03-cor) then OB will represent the time of descent along {248} OB, (Condition 2/02-th-02-cor2) and RO will stand for the time along OS, (Condition 2/11-th-11) while the remainder BR will represent the time required for a body starting from rest at O to traverse the remaining distance SB. (Condition 2/23-pr-09-schol4) But the time of descent along SB starting from rest at O is equal to the time of ascent from B to S after fall through AB. Hence BO is that plane, passing through B, along which a body, after fall through AB, will traverse the distance BS, equal to the assigned distance EF, in the time-interval BR or BA. Q. E. F. |

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Discorsi Proposition2/26-pr-10 |