Jordanus de Nemore
,
[Liber de ratione ponderis]
,
1565
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">Quaestio decima.
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<
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id
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">Si per diuersarum obliquitatum uias duo pondera descen
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dant, fiantque declinationum, et ponderum vna proportio, eo
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dem ordine sumpta vna erit utriusque uirtus in descendendo.
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<
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id
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">Sit linea a, b, c, aequedistans orizonti, et super
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eam orthogonaliter erecta sit b, d, á qua descen
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dant hinc, inde lineae d, a, d, c, sitque d, c, maioris
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obliquitatis proportione igitur declinationum dico
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non angulorum, sed linearum usque ad aequedistan
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tem resecationem, in qua aequaliter sumunt de dire
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cto. </
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">Sit ergo e, pondus super d, c, et h, super d, a, et
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sit e, ad b, sicut d, c, ad a, d. </
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">Dico ea pondera esse vni
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us uirtutis in hoc situ, sit enim d, k, linea vnius ob
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liquitatis, cum d, c, et pondus super eam. </
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le est e, quae sit 6. </
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">Si igitur possibile est, descendat e,
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in l, et trahat h, in m, sitque 6, n, aequale h, m, quod
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etiam aequale est e, l, et transeat per 6. et h, perpen
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dicularis, super d, b. </
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">Sitque 6, h, y, et ab 1, sit l, t, sunt
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et tunc super 6, h, y, n, z, m, x, et super l, t, erit e, r,
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quia igitur proportio n, z, ad n, 6, sicut ad d, 6, d, y,
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propter similitudinem triangulorum, et ideo sicut
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d, b, ad d, k, et quia similiter m, x, ad m, h, sicut d,
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b, ad d, a. </
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">Erit propter aequalem proportionalitatem per
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turbata m, x, ad n, z, sicut d, k, ad d, a, et hoc est
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sicut 6, ad h, sed quia r, e, non sufficit attollere 6, in
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n, nec sufficiet attollere m, in m, sic ergo manebunt.
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<
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id
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">Quum sit responsa libre vnius ponderis,
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et grossiciei per totum: et ipsa in pondere
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data super inaequalia diuidatur, atque ex
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parte breuiore dependeat aequabiliter pon
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dus datum, erunt et portiones, et regulae,
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quae sunt a centro examinis similiter datae.
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<
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id
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">Sit responsa a, b, c, data in pondere, et aequalis in grossicie, et dependeat</
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