Caverni, Raffaello, Storia del metodo sperimentale in Italia, 1891-1900

Table of figures

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    <archimedes>
      <text>
        <body>
          <chap>
            <p type="main">
              <s>
                <pb xlink:href="020/01/2693.jpg" pagenum="318"/>
              (PN+PQ+3OQ)/2=(QN+2PQ+3OQ)/2=(QN+6OQ+3OQ)/2=
                <lb/>
              (QN+9OQ)/2, onde avremo di qui 2QN=QN+9OQ, ossia QN=9Oq. </s>
              <s>
                <lb/>
              Rispetto a quell'altra parte della linea, abbiamo AQ=AP+PQ=
                <lb/>
              PN+PQ=QN+PQ+PQ=QN+2PQ=QN+6Oq. </s>
              <s>Sosti­
                <lb/>
              tuitovi il valore di QN, sarà AQ=9OQ+6OQ=15OQ, e perciò in ul­
                <lb/>
              timo AQ:QN=15OQ:9OQ=5:3, ciò che dimostrato, come si doveva,
                <lb/>
              ritorniamo a trascrivere il discorso del Torricelli. </s>
            </p>
            <p type="main">
              <s>“ Seghisi MA, sicchè la AH sia doppia di HM, e tirato il piano GHF
                <lb/>
              parallelo alla faccia DG, sarà in esso il centro del prisma. </s>
              <s>Segando poi MA
                <lb/>
                <emph type="italics"/>
              bifariam
                <emph.end type="italics"/>
              in I, e tirato il piano IL parallelo al DG, saranno segate per mezzo
                <lb/>
              quattro linee della piramide, per la XVII dell'XI, ed in esso sarà il centro
                <lb/>
              della piramide. </s>
              <s>Ora pongasi che il centro del prisma sia R, e della piramide
                <lb/>
              sia S, e tirisi la RS quale segherà per forza la AN, quale va da A al cen­
                <lb/>
              tro della faccia DG. ” </s>
            </p>
            <p type="main">
              <s>“ Poichè, se in NR è il centro di tutto, ed è anco in AN, però devono
                <lb/>
              concorrere, e sarà il concorso il centro di tutto. </s>
              <s>Sia dunque Q: sarà SQ alla
                <lb/>
              QR come il prisma alla piramide, cioè tripla. </s>
              <s>Immaginiamoci prolungato in
                <lb/>
              infinito il piano LI, sicchè seghi AN, v. </s>
              <s>g. </s>
              <s>in P. </s>
              <s>Sarà dunque PQ tripla di
                <lb/>
                <expan abbr="Oq.">Oque</expan>
              Ma essendo AP, PN, siccome sono AI, IM, uguali, ed essendo AO du­
                <lb/>
              pla di ON, siccome AH è dupla di HM; fatto il conto, sarà tutta la AQ, alla
                <lb/>
              QN, come 15 a 9, ovvero come 5 a 3 ” (ivi, fol. </s>
              <s>186). </s>
            </p>
            <p type="main">
              <s>“ PROPOSIZIONE XXVII. —
                <emph type="italics"/>
              Il centro dell'emisfero è nell'asse in sul
                <lb/>
              luogo, che sta come cinque a tre. </s>
              <s>”
                <emph.end type="italics"/>
              </s>
            </p>
            <p type="main">
              <s>“ Sia il quadrante BAC (fig. </s>
              <s>177), il cui asse AC. </s>
              <s>Immaginisi AD uguale
                <lb/>
              alla semiperiferia del circolo, e sia l'angolo BAD retto, e finiscasi il rettan­
                <lb/>
                <figure id="id.020.01.2693.1.jpg" xlink:href="020/01/2693/1.jpg" number="682"/>
              </s>
            </p>
            <p type="caption">
              <s>Figura 177.
                <lb/>
              golo BD, che sarà uguale al suo cerchio. </s>
              <s>Poi tirisi
                <lb/>
              la BC, e sopra il triangolo BAC facciasi il prisma
                <lb/>
              BGA, con l'altezza AD, è prodotta AH eguale ad
                <lb/>
              AC tirisi la HD, sicchè seghi la CG prodotta in E,
                <lb/>
              e facciasi la piramide FDGE. Tirisi, tra l'appli­
                <lb/>
              cata MN e per essa, un piano LO parallelo a
                <lb/>
              piano BD. ” </s>
            </p>
            <p type="main">
              <s>“ Che il rettangolo HAC, al rettangolo HNC,
                <lb/>
              sia come il rettangolo BD ad LO,
                <emph type="italics"/>
              ratio est
                <emph.end type="italics"/>
              perchè
                <lb/>
              il rettangolo HAC, al rettangolo HNC, ha ragion
                <lb/>
              composta di AH ad HN, cioè AD ad NO, e di AC
                <lb/>
              a CN, ovvero AB ad NL: però sarà il rettangolo
                <lb/>
              LO eguale al circolo MN. ” </s>
            </p>
            <p type="main">
              <s>“ Ora il cerchio AB, al cerchio MN, sta come
                <lb/>
              il quadrato AB al quadrato MN, cioè come il rettangolo HAC al rettangolo
                <lb/>
              HNC, ovvero come il rettangolo BD ad LO. </s>
              <s>Gli antecedenti sono uguali, ergo
                <lb/>
              ed i consequenti. </s>
              <s>” </s>
            </p>
          </chap>
        </body>
      </text>
    </archimedes>
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