DelMonte, Guidubaldo, Le mechaniche

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      <text id="id.0.0.0.0.3">
        <body id="id.2.0.0.0.0">
          <chap id="N106DF">
            <pb xlink:href="037/01/076.jpg"/>
            <figure id="id.037.01.076.1.jpg" xlink:href="037/01/076/1.jpg" number="64"/>
            <p id="id.2.1.244.0.0" type="main">
              <s id="id.2.1.244.1.0">
                <emph type="italics"/>
              Sia
                <expan abbr="finalmẽte">finalmente</expan>
              la
                <expan abbr="bilãcia">bilancia</expan>
              AB, & da i
                <expan abbr="pũti">punti</expan>
              AB ſiano
                <expan abbr="pẽdenti">pendenti</expan>
              i peſi EF, & ſia il centro
                <lb/>
              della bilancia C fra i peſi, & diuidaſi la AB in D, talche AD verſo DB
                <lb/>
              ſia come il peſo F al peſo E. </s>
              <s id="id.2.1.244.2.0">Dico che i peſi EF peſano tanto in AB, quan
                <lb/>
              to ſe ambidue foſſero pendenti dal punto D. </s>
              <s id="id.2.1.244.3.0">facciaſi CG eguale à CD; & co­
                <lb/>
              me DC à CA, coſi facciaſi il peſo E ad vn'altro peſo H, ilquale ſia attac
                <lb/>
              cato in D. </s>
              <s id="N12CFD">& come GC verſo CB, coſi facciaſi il peſo F ad vn'altro che
                <lb/>
              ſia K, & attachiſi K in G. </s>
              <s id="id.2.1.244.4.0">Hor percioche, come il BC è verſo il CG, cioè
                <lb/>
              verſo il CD, coſi il peſo K ad F; ſarà il K maggiore del peſo F. </s>
              <s id="id.2.1.244.5.0">Per laqual
                <lb/>
              coſa diuidaſi il peſo K in L & in MN, & facciaſi la parte L eguale ad F,
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note73"/>
                <emph type="italics"/>
              ſarà come BC à CD, coſi tutto LMN ad L; & diuidendo, come BD
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note74"/>
                <emph type="italics"/>
              verſo DC, coſi la parte MN alla parte L. </s>
              <s id="N12D1D">come dunque BD à DC, coſi
                <lb/>
              la parte MN ad F. </s>
              <s id="N12D21">& come AD à DB, coſi F ad E. </s>
              <s id="id.2.1.244.6.0">Per laqual coſa
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note75"/>
                <emph type="italics"/>
              per la egual proportione, come AD verſo DC, coſi MN ad E. </s>
              <s id="N12D2F">& eſſendo AD
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note76"/>
                <emph type="italics"/>
              maggiore di CD; ſarà anco la parte MN maggiore del peſo E. </s>
              <s id="id.2.1.244.7.0">Diuidaſi dun
                <lb/>
              que MN in due parti MN, & ſia M eguale ad E. </s>
              <s id="id.2.1.244.8.0">ſarà come AD à
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note77"/>
                <emph type="italics"/>
              DC, coſi NM ad M; & diuidendo, come AC verſo CD, coſi N ad M:
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note78"/>
                <emph type="italics"/>
              & conuertendo, come DC verſo CA, coſi M ad N. </s>
              <s id="id.2.1.244.9.0">& come DC à
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note79"/>
                <emph type="italics"/>
              CA, coſi è E ad H; ſarà dunque M ad N come E ad H; & permutan
                <lb/>
              do come M ad E, coſi N ad H. </s>
              <s id="id.2.1.244.10.0">Ma per eſſere ME tra loro eguali, ſaran­
                <lb/>
              no anche NH tra ſe eguali. </s>
              <s id="id.2.1.244.11.0">& percioche coſi è AC verſo CD, come H
                <lb/>
              ad E: i peſi HE peſeranno egualmente. </s>
              <s id="id.2.1.244.12.0">ſimilmente percioche, come è GC à CB,
                <lb/>
              coſi F verſo K, i peſi etiandio KF peſeranno egualmente. </s>
              <s id="id.2.1.244.13.0">Adunque i peſi
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note80"/>
                <emph type="italics"/>
              EK HF nella bilancia AB, il cui centro ſia C peſeranno egualmente. </s>
              <s id="id.2.1.244.14.0">& con
                <lb/>
              cioſia che GC ſia eguale à CD, & il peſo H ſia pur eguale ad N, i peſi NH
                <emph.end type="italics"/>
              </s>
            </p>
          </chap>
        </body>
      </text>
    </archimedes>
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