Fabri, Honoré, Tractatus physicus de motu locali, 1646

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              ſito æquali tempore percurrerentur, quod falſum eſt; </s>
              <s id="N1D806">nam ſit AC ad A
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              N vt AN ad AE; </s>
              <s id="N1D80C">ſitque BC ad BO vt BO ad BI; </s>
              <s id="N1D810">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; </s>
              <s id="N1D81A">ſed CN eſt minor quàm CO, vt conſtat ex Geometria, quod bre­
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              uiter in tironum
                <expan abbr="gratiã">gratiam</expan>
              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; </s>
              <s id="N1D82C">haud dubiè triangula CGE, CAB ſunt proportionalia; </s>
              <s id="N1D830">igitur vt
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              CB;.ad CA. 4.ita CE 5. ad CG 6. 2/3; </s>
              <s id="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. </s>
              <s id="N1D83E">quad. </s>
              <s id="N1D841">29.igitur eſt maior 5. ſed ſi eſſet maior 5. C
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              M & CD eſſent æquales; </s>
              <s id="N1D847">igitur CF eſt maior CD; </s>
              <s id="N1D84B">eſt enim BF ferè 3.
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              1/2 paulò minùs: </s>
              <s id="N1D851">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,
                <expan abbr="fiatq́ue">fiatque</expan>
              vt AC ad AK, ita AK ad AN; </s>
              <s id="N1D85D">haud
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              dubiè percurret CN æquali tempore, quo BC; </s>
              <s id="N1D863">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; </s>
              <s id="N1D86D">ita vt EX ſit 1/4 EK: </s>
              <s id="N1D871">dico quod præuio motu ex A in
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              K, & deinde deflexo per KX conficietur KX æquali tempore cum AK; </s>
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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; </s>
              <s id="N1D880">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. </s>
            </p>
            <p id="N1D886" type="main">
              <s id="N1D888">Porrò vt habeatur in horizontali FS; </s>
              <s id="N1D88C">ſit FR æqualis KF; </s>
              <s id="N1D890">ſit FT æ­
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              qualis KR; </s>
              <s id="N1D896">ſit arcus TS ex k: </s>
              <s id="N1D89A">Dico quod ks eſt linea quæſita; </s>
              <s id="N1D89E">nam ſi ſit
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              vt BS ad BZ, ita BZ ad BK, kz erit æqualis KF, vel AK; </s>
              <s id="N1D8A4">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. </s>
              <s id="N1D8B2">ſint AFFE æquales; </s>
              <s id="N1D8B6">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; </s>
              <s id="N1D8BE">ſit autem AG ad AH vt A
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              H ad AI: </s>
              <s id="N1D8C4">Dico GH eſſe æqualem AF; </s>
              <s id="N1D8C8">ſit enim quadratum HD mediæ
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              proportionalis: </s>
              <s id="N1D8CE">Dico eſſe æquale rectangulo IC, dùm AC ſit æqualis A
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              G; </s>
              <s id="N1D8D4">igitur quadratum PR cuius latus eſt æquale FG, ſeu AE continet
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              duo quadrata RDSN; </s>
              <s id="N1D8DA">ergo GH eſt æqualis VN; igitur GH quod erat
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              demonſtrandum. </s>
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            <p id="N1D8E0" type="main">
              <s id="N1D8E2">Quintò, hinc nunquam ks vel kx poteſt eſſe tripla Ak donec tan­
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              dem perueniatur ad perpendiculum kH; </s>
              <s id="N1D8E8">nam ſecundo tempore percur­
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              ritur kH triplum Ak, ſi primo percurritur Ak; </s>
              <s id="N1D8EE">nunquam etiam ks vel
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              vlla alia inclinata poteſt eſſe dupla tantùm Ak; </s>
              <s id="N1D8F4">ſed ſemper eſt maior, do-</s>
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