Benedetti, Giovanni Battista de
,
Io. Baptistae Benedicti ... Diversarvm specvlationvm mathematicarum, et physicarum liber : quarum seriem sequens pagina indicabit ; [annotated and critiqued by Guidobaldo Del Monte]
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IO. BAPT. BENED.
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à centro circuli, ipſum triangulum circunſcribentis, terminatur, & à baſi, vt in tertio
<
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propoſito decimæſeptimæ quartidecimi Eucli. probatur, ex quo ſequitur proportio
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nem huiuſmodi perpendicularis ad axem Tetraedri, hoc eſt ad
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ſeſquioctauam
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eſſe in potentia, ex penultima primi Eucli. </
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xml:space
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tertia pars ſit ipſius
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vt
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etiam ex .2. propoſito, ſeu corollario decimæſeptimæ .14. lib. diſcurrere licet, cum ex
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dicto corollario
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ſit ſexta pars ipſius
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. </
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quarta pars erit ipſius
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vn
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de
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var
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ſeſquitertia erit ipſi
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in longitudine,
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quadratum ipſius
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ad qua-
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dratum ipſius
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erit vt .9. ad .16: </
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">& ita duplum quadrati ipſius
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hoc eſt quadra-
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tum ipſius
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ad quadratum ipſius
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erit, vt .18. ad .16. hoc eſt ſeſquioctauum, er-
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go
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æqualis erit dictæ perpendiculari, ex .9. quinti.</
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<
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erit partium .1539838576570176.</
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<
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<
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xml:space
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">Pro Octaedro deinde, accipies productum diametri in ſemidiametrum, quod
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productum, æquale erit quadrato diuidenti per æqualia Octaedron, hocigitur pro-
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ductum, multiplicando per .100000. ſemidiametrum ſphæræ, tibi dabit columnam
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quadrilateram cuius tertia pars, erit partium .666666666666666. cuius duplum
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erit ipſum Octaedron partium .1333333333333.</
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<
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xml:space
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">Pro Icoſaedro autem, oportet prius quantitatem perpendicularis inuenire, quæ
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perpendicularis, per æqualia diuidit baſim ipſius Icoſaedri, quæ vt radix quadrata
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trium quartarum quadrati lateris ipſius baſis, erit partium .91055. talium, qualium
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dictum latus erit partium .105142. cuius medietas eſt .52571. quæ medietas ſi mul-
<
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tiplicata fuerit cum dicta perpendiculari, dabit totam baſim ſuperficialem, hoc eſt
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ſuperficiem vnius trianguli æquilateris partium ſuperficialium .4786852405. quo
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facto, accipe quadratum duarum tertiarum ipſius, hic ſupra dictæ perpendicularis,
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deme ex quadrato ſemidiametri ſphæræ, hoc eſt, ex quadrato
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.100000
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radix poſtea quadrata reſidui, erit partium .79468. & hæc erit perpendicularis à cen
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tro ſphærę ad vnam baſim ipſius Icoſaedri, quam volueris, quam perpendicularem
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ſi multiplicaueris cum quantitate ſuperficiali, hic ſuperius reperta, vnius baſis, con-
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ſequeris columnam trilateram partium .380401586920540. cuius tertia pars, erit
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partium .126800528973513. pro vna ex .20. </
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nentibus. </
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parte ipſius perpendicularis, hanc poſtea pyramidem multiplicando per .20. habebis
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totam corpulentiam ipſius Icoſaedri partium .2536010579470260.</
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<
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">Pro Duodecaedro demum, accipe ſinum gra .36. qui
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ſunt pro dimidio quin
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tæ partis totius gyri circularis,
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ſinus, erit partium .58778. cuius quadratum
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ſi
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pſeris ex quadrato
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.100000. ſemidiametri circuli
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tago</
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num æquilaterum, & æquiangulum, </
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<
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xml:space
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">tunc radix reſidui, erit perpendicularis du-
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cta à centro dicti circuli ad medium vnius lateris ipſius pentagoni, quæ perp endicu
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laris, erit partium .80902. talium qualium medietas lateris dicti fuerit .58778.
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Nunc verò dicendo ſi .58778. dat .80902. quid nobis dabit .35684? </
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<
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">medietas lateris
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ipſius Duodecaedri, vnde da bit .49116. pro perpendiculari, à centro ipſius penta-
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goni, ad latus ipſius Duodecaedri, quæ multiplicata cum me dietate ſupradicta ip-
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ſius lat eris, hoc eſt cum .35684. producet vnum ex quinque triangulis componenti-
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bus vn um pentagonum, ſeu vnam baſim ipſius Duodecaedri, quod quidem triangu
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lum, erit partium .1752655344. ſu perficialium, quas ſi per quinque multiplicaueris
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habeb is vnam baſim pentagonam dicti corporis partium .8763276720. </
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poſtea eſt, ſi ad .80901. conuenit ſemidiameter circularis partium .100000. quid
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ueniet partibus .49116. dabit .60711. pro tali ſemidiametro circulari, cuius quadra- </
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