DelMonte, Guidubaldo, Le mechaniche

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    <archimedes>
      <text id="id.0.0.0.0.3">
        <body id="id.2.0.0.0.0">
          <chap id="N106DF">
            <p id="id.2.1.395.0.0" type="main">
              <s id="id.2.1.395.10.0">
                <pb pagenum="22" xlink:href="037/01/059.jpg"/>
                <emph type="italics"/>
              ſarà la circonferenza EA eguale alla circonferenza FB. </s>
              <s id="id.2.1.395.11.0">Hor percioche l'ango­
                <lb/>
              lo miſto CEA è eguale al miſto CFB, & HFB è maggiore di CFB, &
                <emph.end type="italics"/>
                <arrow.to.target n="note121"/>
                <lb/>
                <emph type="italics"/>
              l'angolo HEA è minore di CEA; ſarà l'angolo HFB maggiore dell'angolo
                <lb/>
              HEA. </s>
              <s id="id.2.1.395.12.0">Da quali ſe ſaranno leuati via gli angoli HFG HEK eguali, ſarà l'an
                <lb/>
              golo GFB maggiore dell'angolo KEA. </s>
              <s id="id.2.1.395.13.0">Adunque la diſceſa del peſo poſto in
                <lb/>
              E ſarà meno obliqua della ſalita del peſo poſto in F. </s>
              <s id="id.2.1.395.14.0">&
                <expan abbr="quãtunque">quantunque</expan>
              il peſo poſto in E
                <lb/>
              deſcendendo, & il peſo poſto in F ſalendo ſi mouino per eguali circonferenze, nondi
                <lb/>
              meno percioche il peſo poſto in E da queſto luogo diſcende piu dirittamente di quel
                <lb/>
              che il peſo F
                <expan abbr="aſcẽde">aſcende</expan>
              :pero la naturale poſſanza del peſo poſto in E ſupererà la
                <expan abbr="reſiſtẽza">reſiſten
                  <lb/>
                za</expan>
              della violentia del peſo F. </s>
              <s id="id.2.1.395.15.0">Onde grauezza maggiore hauerà il peſo posto in E,
                <lb/>
              che il peſo poſto in F. </s>
              <s id="id.2.1.395.16.0">Adunque il peſo poſto in E ſi mouerà in giù & il peſo poſto
                <lb/>
              in F in sù, fin che la bilancia EF ritorni in AB, che biſognaua moſtrare.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.397.0.0" type="margin">
              <s id="id.2.1.397.1.0">
                <margin.target id="note119"/>
                <emph type="italics"/>
              Per la
                <emph.end type="italics"/>
              4.
                <emph type="italics"/>
              del primo.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.398.0.0" type="margin">
              <s id="id.2.1.398.1.0">
                <margin.target id="note120"/>
                <emph type="italics"/>
              Per la
                <emph.end type="italics"/>
              28.
                <emph type="italics"/>
              del terzo.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.399.0.0" type="margin">
              <s id="id.2.1.399.1.0">
                <margin.target id="note121"/>
                <emph type="italics"/>
              Per la
                <emph.end type="italics"/>
              29.
                <emph type="italics"/>
              del primo.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.400.0.0" type="main">
              <s id="id.2.1.400.1.0">
                <emph type="italics"/>
              La ragione di queſto effetto poſta da Ariſtotele qui ſi puote vedere manifeſta. </s>
              <s id="id.2.1.400.2.0">Percio­
                <emph.end type="italics"/>
                <arrow.to.target n="note122"/>
                <lb/>
                <emph type="italics"/>
              che ſia il punto N doue le linee CS EF ſi tagliano inſieme. </s>
              <s id="id.2.1.400.3.0">& percioche HE
                <lb/>
              è eguale ad HF; ſarà NE maggiore di NF. </s>
              <s id="id.2.1.400.4.0">adunque la linea CS, che no­
                <lb/>
              ma perpendicolo, diuiderà la bilancia EF in parti diſuguali. </s>
              <s id="id.2.1.400.5.0">concioſia dunque, che
                <lb/>
              la parte della bilancia NE ſia maggiore della NF, & quel che è di più biſo­
                <lb/>
              gni, che ſia portato in giù, la bilancia EF dalla parte di E ſi mouerà in giu finche
                <lb/>
              ritorni in AB.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.401.0.0" type="margin">
              <s id="id.2.1.401.1.0">
                <margin.target id="note122"/>
                <emph type="italics"/>
              Ragione de Aristotele.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.402.0.0" type="main">
              <s id="id.2.1.402.1.0">
                <emph type="italics"/>
              Oltre à cio da quelle coſe, che
                <lb/>
              fin hora ſono ſtate dette,
                <lb/>
              ſi puote affermare, la bilan
                <lb/>
              cia EF da quel ſito mo­
                <lb/>
              uerſi piu velocemente in
                <lb/>
              AB; d'onde la linea EF
                <lb/>
              allungata a dirittura per­
                <lb/>
              uenga nel centro del mon­
                <lb/>
              do. </s>
              <s id="id.2.1.402.2.0">come ſia EFS vna
                <lb/>
              linea diritta. </s>
              <s id="id.2.1.402.3.0">& percioche
                <lb/>
              CD CK ſono tra loro
                <lb/>
              eguali. </s>
              <s id="id.2.1.402.4.0">ſe dunque col cen­
                <lb/>
              tro C, & con lo ſpatio
                <lb/>
              CD ſi deſcriuerà il cerchio
                <lb/>
              DHM, ſaranno i punti
                <lb/>
              DH nella circonferenza
                <lb/>
              del cerchio. </s>
              <s id="id.2.1.402.5.0">Ma perche la
                <lb/>
              CH è à piombo di EF,
                <lb/>
              toccherà la EHS il cer­
                <lb/>
              chio DHM nel punto
                <lb/>
              H. </s>
              <s id="id.2.1.402.6.0">il peſo dunque poſto in
                <lb/>
              H, (ſi come di ſopra hab
                <lb/>
              biamo prouato) ſarà piu
                <emph.end type="italics"/>
                <lb/>
                <figure id="id.037.01.060.1.jpg" xlink:href="037/01/060/1.jpg" number="42"/>
              </s>
            </p>
          </chap>
        </body>
      </text>
    </archimedes>
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