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">
            <pb xlink:href="037/01/074.jpg"/>
            <p id="id.2.1.221.0.0" type="head">
              <s id="id.2.1.221.1.0">PROPOSITIONE V. </s>
            </p>
            <p id="id.2.1.222.0.0" type="main">
              <s id="id.2.1.222.1.0">Due peſi attaccati nella bilancia, ſe la bilancia ſarà tra loro in modo
                <lb/>
              diuiſa, chele parti riſpondano ſcambieuolmente à peſi; peſeranno
                <lb/>
              tanto ne'punti doue ſono attaccati, quanto ſel'uno & l'altro foſſe
                <lb/>
              pendente dal punto della diuiſione. </s>
            </p>
            <figure id="id.037.01.074.1.jpg" xlink:href="037/01/074/1.jpg" number="62"/>
            <p id="id.2.1.224.0.0" type="main">
              <s id="id.2.1.224.1.0">
                <emph type="italics"/>
              Sia la bilancia AB, il cui centro ſia C, & ſiano due peſi EF pendenti da' punti
                <lb/>
              BG: & diuidaſi BG in H, ſi fattamente, che BH ad HG habbia la pro­
                <lb/>
              portione isteſſa, che hà il peſo E al peſo F. </s>
              <s id="id.2.1.224.2.0">Dico i peſi EF peſare tanto in BG,
                <lb/>
              quanto ſe amendue pendeſſero dal punto H. </s>
              <s id="id.2.1.224.3.0">facciaſi AC eguale à CH. </s>
              <s id="N12A6C">& ſi
                <lb/>
              come AC à CG, coſi facciaſi il peſo E al peſo L. </s>
              <s id="N12A70">ſimilmente come AC à
                <lb/>
              CB, coſi facciaſi il peſo F al peſo M. </s>
              <s id="N12A74">& ſiano attaccati i peſi LM al punto
                <lb/>
              A. </s>
              <s id="id.2.1.224.4.0">Hor percioche AC è eguale à CH, ſarà BC verſo CH come il peſo
                <lb/>
              M al peſo F. </s>
              <s id="N12A7D">& percioche piu grande è BC di CH; ſarà anche il peſo M
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note58"/>
                <emph type="italics"/>
              maggiore di F. </s>
              <s id="id.2.1.224.5.0">Diuidaſi dunque il peſo M in due parti QR, & ſia la parte di
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note59"/>
                <emph type="italics"/>
              Q eguale ad F; ſarà BC à CH, come RQ à Q: & diuidendo, come BH
                <lb/>
              ad HC, coſi R à Q. </s>
              <s id="N12A96">Dapoi conuertendo, come CH ad HB, coſi Q ad
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note60"/>
                <emph type="italics"/>
              R. </s>
              <s id="id.2.1.224.6.0">Oltre à ciò perche CH è eguale à CA, ſarà HC verſo CG come il peſo
                <lb/>
              E al peſo L: maè piu grande HC di CG, però ſarà anche il peſo E maggio­
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note61"/>
                <emph type="italics"/>
              re del peſo L. </s>
              <s id="id.2.1.224.7.0">Onde diuidaſi il peſo E in due parti NO, ſi fattamente, che la
                <lb/>
              parte di O ſia eguale ad L, ſarà HC à CG come tutto lo NO ad O; &
                <lb/>
              diuidendo, come HG à GC, coſi N ad O. </s>
              <s id="N12AB6">& conuertendo, come CG à
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note62"/>
                <emph type="italics"/>
              GH, coſi O ad N. </s>
              <s id="N12AC1">& di nuouo componendo, come CH ad HG, coſi ON
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note63"/>
                <emph type="italics"/>
              ad N. </s>
              <s id="id.2.1.224.8.0">& come GH ad HB, coſi è F ad ON. </s>
              <s id="id.2.1.224.9.0">Per la qual coſa per la pro
                <lb/>
              portione vguale come CH ad HB, coſi F ad N. </s>
              <s id="id.2.1.224.10.0">Ma come CH ad HB
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note64"/>
                <emph type="italics"/>
              coſi è Q ad R: ſarà dunque Q ad R come F ad N. </s>
              <s id="N12AE0">& permutando co­
                <lb/>
              me Q ad F; coſi R ad N. </s>
              <s id="id.2.1.224.11.0">ma la parte di Q è egual ad eſſo F. </s>
              <s id="N12AE7">per la qual
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note65"/>
                <emph type="italics"/>
              coſa la parte di R ancora ſarà eguale ad N. </s>
              <s id="id.2.1.224.12.0">eſſendo dunque il peſo L eguale
                <emph.end type="italics"/>
              </s>
            </p>
          </chap>
        </body>
      </text>
    </archimedes>
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