Fabri, Honoré
,
Tractatus physicus de motu locali
,
1646
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<
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nec tandem perueniat ad horizontalem KY, quæ eſt dupla AK, quia in
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horizontali non acceleratur motus; </
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<
s
id
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">igitur cum impetu acquiſito in deſ
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cenſu AK, conficiet motu æquabili KY duplum AK per Th.42.l.3. poſito
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quòd non deſtruatur; atque ex his ſatis facilè intelligentur, quæcumque
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habes apud Galileum in dialog.3.à propoſitione 3.ad 23. </
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<
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<
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id
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">Sextò non probat Galileus, ſed tantùm ſupponit mobile ad
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eãdem
">eandem</
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<
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alti-tudinẽ
">alti
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tudinem</
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aſcendere poſſe motu reflexo ex qua deſcendit, quod examinabi
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mus lib.
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ſequẽti
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, hinc non laborabimus in
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examinãdis
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prop. 24.25.26.27. </
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<
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">Septimò, cognito tempore, quo percurrit mobile perpendiculum EC
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quod ſit diameter circuli; </
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<
s
id
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">ſciri poteſt quo tempore percurrat duas chor
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das ſimul EGGC; </
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<
s
id
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N1D931
">ſit enim Tangens EF, ſitque vt FG ad FD, ita FD ad
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FC; </
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<
s
id
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N1D937
">cum EG & EC deſcendat æquali tempore per Th.27. cum in G ſit
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idem motus, ſiue ex E, ſiue ex F deſcendat per Th.20. certè ſi deſcendit
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per EG dato tempore, quod ſit vt EG, deſcendit per GC tempore, quod
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eſt vt GD; igitur tempus, quo deſcendit per EC eſt ad tempus, quo deſ
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cendit per EGC, vt EG ad EGD. </
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</
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<
s
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">Obſeruabis autem GF eſſe ad EF vt EF ad FC; </
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<
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">igitur FD eſt media
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inter FC GF, & eſt æqualis FE, igitur anguli FDE.FED æquales; </
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<
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id
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">ſed FD
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E eſt æqualis duobus DCE.DEC, & FEG, eſt æqualis DCE; igitur duo G
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DE DEC ſunt æquales. </
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<
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id
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">Octauò, ſi accipiantur æquales horizontalis, & perpendicularis, v.g.
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BA AC, ducaturque BC: </
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<
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id
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">Dico nullum duci poſſe planum inclinatum à
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puncto B ad perpendiculum AEM, quod breuiori tempore percurratur,
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quàm BC, nec intra angulum vt BR, nec extra vt BM; </
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<
s
id
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">ſit enim vt BC ad
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BI ita BI ad BH, eſt autem BI æqualis BA, igitur ſi BA, ſit 4.BC eſt v.g.
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32. & BH radix q.8.igitur HI eſt ferè I paulò plùs; igitur cum BH percur
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ratur æquali tempore cum AC, eſt tempus, quo percurritur BH ad tem
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pus quo percurritur HC vt BH ad HI. </
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<
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<
s
id
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">Sit autem BR dupla AR, ſitque perpendicularis AK in BR; </
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<
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id
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">certè KR
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eſt ſubquadrupla BR; </
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<
s
id
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">igitur percurritur BL æqualis KR eo tempore quo
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percurritur AR; </
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<
s
id
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">igitur BL ſit ad BV vt BV ad BR; </
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<
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">igitur temporibus æ
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qualibus percurruntur BL LR; </
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<
s
id
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">igitur ſi tempus quo percurritur BL ſit vt
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BH, tempus quo percurretur LR erit etiam vt BH; </
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>
<
s
id
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">igitur totum tempus
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quo percurritur tota BR erit vt tota BE, ſed tempus quo percurritur tota
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BC eſt tantum vt BI quę eſt minor BC; </
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<
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id
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">igitur BC breuiori tempore per
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curritur quàm BR; ſit
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etiã
">etiam</
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vt BP ad BX ita BX ad BM, ſi BO eſt 4. OP 2.
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certè BP eſt rad.q. </
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<
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">12.id eſt ferè 3.1/2 paulò minùs, BM verò eſt dupla BA
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vel BO; </
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<
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">igitur eſt 8. ducatur ergo 8. in 4. 1/3 productum erit 28. cuius radix
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eſt ferè 5.1/3 paulò minùs; </
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<
s
id
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">igitur BX eſt 5.1/3 paulò minùs; </
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<
s
id
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N1D9BC
">cum autem BH
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ſit 2.q.8.eſt ferè 2.5/6, paulò minùs; </
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>
<
s
id
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">igitur ſit vt BP 3.1/2 ad BX 5.1/3, ita BH
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2.5/6 ad aliam; </
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<
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">certè erit 144. id eſt 4.(26/63), licèt minùs acceptum ſit; </
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<
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">igitur
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126.eſt maior BI, quæ eſt tantùm 4; igitur BE breuiori tempore percur
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ritur, quàm BM. </
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<
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id
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">Nonò, per duas chordas quadrantis deſcendit breuiori tempore mo
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bile, quàm per alteram tantùm inferiorem ſcilicet ſit enim tantùm </
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