Fabri, Honoré
,
Tractatus physicus de motu locali
,
1646
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ſito æquali tempore percurrerentur, quod falſum eſt; </
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<
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id
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">nam ſit AC ad A
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N vt AN ad AE; </
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<
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">ſitque BC ad BO vt BO ad BI; </
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<
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">certè tempus, quo
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percurritur BC eſt ad tempus, quo percurritur CI vt CB ad CO, &
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tempus quo percurritur BC eſt ad tempus quo percurritur CE vt BC ad
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CN; </
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<
s
id
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N1D81A
">ſed CN eſt minor quàm CO, vt conſtat ex Geometria, quod bre
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uiter in tironum
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gratiã
">gratiam</
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in terminis rationabilibus oſtendo, ſit planum
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inclinatum AE 9. ſitque AE id eſt 9. ad AD. 6. vt AD ad AC 4. ex
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centro C aſſumpta CH 3. ducatur arcus HB & ex A ad prædictum ar
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cum Tangens AB, tùm ex BC G indefinitè & ex E, EG perpendicularis
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in EA; </
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<
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id
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">haud dubiè triangula CGE, CAB ſunt proportionalia; </
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<
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">igitur vt
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CB;.ad CA. 4.ita CE 5. ad CG 6. 2/3; </
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<
s
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N1D836
">igitur tota BG eſt 9. 2/3; ſitque B
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G ad BF, vt BF ad DC, quod vt fiat BG 9. 2/3 in BC 3. productum erit
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29. igitur BF eſt Rad. </
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<
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">quad. </
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<
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">29.igitur eſt maior 5. ſed ſi eſſet maior 5. C
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M & CD eſſent æquales; </
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<
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">igitur CF eſt maior CD; </
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<
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">eſt enim BF ferè 3.
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1/2 paulò minùs: </
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<
s
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">vt autem reperiatur linea inclinata, quæ percurratur æ
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quali tempore cum BC ſuppoſito præuio motu per BC, aſſumatur CK
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æqualis CB id eſt 3.partium,
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fiatq́ue
">fiatque</
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vt AC ad AK, ita AK ad AN; </
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<
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">haud
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dubiè percurret CN æquali tempore, quo BC; </
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<
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">vt verò habeatur pun
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ctum in horizontali, ſit AF perpendicularis bifariam diuiſa in K, ſit K
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F diuiſa in 4. partes æquales, quibus addatur FP 1/4 KFEK V dupla FA,
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& producatur in X; </
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<
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id
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">ita vt EX ſit 1/4 EK: </
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<
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">dico quod præuio motu ex A in
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K, & deinde deflexo per KX conficietur KX æquali tempore cum AK; </
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ſi enim caderet mobile ex V primo tempore percurreret VL, id eſt 1/4 V
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K eo tempore, quo percurreret AK per Th.6. igitur ſecundo tempore
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æquali LK, id eſt 3/4 VK; </
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<
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id
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">igitur tertio tempore æquali KX 5/4 VK; nam eo
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dem modo ſe habet in k ſiue deſcendat ex V, ſiue ex A per Th.20. </
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<
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">Porrò vt habeatur in horizontali FS; </
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">ſit FR æqualis KF; </
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<
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qualis KR; </
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">ſit arcus TS ex k: </
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">Dico quod ks eſt linea quæſita; </
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<
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">nam ſi ſit
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vt BS ad BZ, ita BZ ad BK, kz erit æqualis KF, vel AK; </
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<
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id
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">ſed tempus
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quo percurritur AK eſt ad tempus quo percurritur Dk vt BK ad AK
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per Th.23.& ad tempus, quo percurritur BS, vt Bk ad BZ, & ad tem
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pus quo percurritur ks vt Bk ad kz; ergo Ak & ks percurruntur æ
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quali tempore, ſi kz ſit æqualis KF, quod ſic breuiter demonſtro, cùm
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figura apud Galileum deſideretur. </
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<
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">ſint AFFE æquales; </
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<
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">ducatur AE
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quæ transferatur iu FG, ſitque GI æqualis AG, ſic tota AG mihi repræ
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ſentat totam BS ſuperioris figuræ, vt conſtat; </
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<
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">ſit autem AG ad AH vt A
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H ad AI: </
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<
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">Dico GH eſſe æqualem AF; </
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">ſit enim quadratum HD mediæ
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proportionalis: </
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<
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">Dico eſſe æquale rectangulo IC, dùm AC ſit æqualis A
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G; </
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<
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id
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">igitur quadratum PR cuius latus eſt æquale FG, ſeu AE continet
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duo quadrata RDSN; </
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<
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id
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">ergo GH eſt æqualis VN; igitur GH quod erat
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demonſtrandum. </
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<
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">Quintò, hinc nunquam ks vel kx poteſt eſſe tripla Ak donec tan
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dem perueniatur ad perpendiculum kH; </
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<
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id
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">nam ſecundo tempore percur
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ritur kH triplum Ak, ſi primo percurritur Ak; </
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<
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">nunquam etiam ks vel
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vlla alia inclinata poteſt eſſe dupla tantùm Ak; </
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<
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