Stelliola, Niccol� Antonio, De gli elementi mechanici, 1597

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    <archimedes>
      <text>
        <body>
          <chap id="N10C0F">
            <pb xlink:href="041/01/043.jpg" pagenum="42"/>
            <p id="N10C2D" type="head">
              <s id="N10C2F">
                <emph type="italics"/>
              Dimoſtratione.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="N10C35" type="main">
              <s id="N10C37">
                <emph type="italics"/>
              Sia la linea che rappreſenta il piano orizontale A B: la linea del pia
                <lb/>
              no inchinato A C: il circolo della rota D E FG: il toccamento D: e dal
                <lb/>
              ponto D, tiriſi perpendicolare all'orizonte B D F: è manifeſto che detta
                <lb/>
              linea, ſia la perpendicolare del ſoſtenimento: dico che'l centro del peſo
                <lb/>
              è fuori di detta linea. </s>
              <s id="N10C43">Si moſtra: perche del triangolo D B A: l'angolo,
                <lb/>
              E, che fa la perpendicolare con l'orizonte, è retto: reſta l'angolo B D A,
                <lb/>
              a cuto: e perciò la portione D G F, e maggiore del ſemicircolo; & in eſ
                <lb/>
              ſa ſarà il centro del circolo, che è anco centro di peſo. </s>
              <s id="N10C4B">è dunque il cen­
                <lb/>
              tro del peſo fuori della linea del
                <expan abbr="ſoſtenimẽto">ſoſtenimento</expan>
              . </s>
              <s id="N10C53">De ſcriuaſi alla D E, la por
                <lb/>
              tione di circolo D H F, ſimile a D E F; ſaranno dette portioni vgua­
                <lb/>
              li, e faranno equipondio. </s>
              <s id="N10C59">reſta dunque la figura lunare ſenza equi
                <lb/>
              pondio: & il momento della rota appoggiata ſarà meno che della ro
                <lb/>
              ta ſoſpeſa, ſecondo la ragione della figura lunare a tutto il circolo: cio è
                <lb/>
              ſecondo la ragione dell'ecceſſo delle portioni, al circolo tutto. </s>
              <s id="N10C61">Il che
                <lb/>
              ſi hauea da moſtrare.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="N10C67" type="head">
              <s id="N10C69">
                <emph type="italics"/>
              Appendice. </s>
              <s id="N10C6D">I.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="N10C71" type="main">
              <s id="N10C73">E l'iſteſſo che si è moſtrato nella rota c'ha grauezza;
                <lb/>
              si moſtra nelle rote al cui aſſe appoggi altro peſo. </s>
            </p>
            <p id="N10C77" type="main">
              <s id="N10C79">
                <emph type="italics"/>
              Percio che ſe in vece del peſo appoggiato all'aſſe, intendiamo darſi
                <lb/>
              l'iſteſſo peſo alle rote: eſſendo peſi vguali con loro centri nell'iſteſſe li­
                <lb/>
              nee, & la linea del ſoſtenimento l'iſteſſa, harranno li peſi l'iſteſſi
                <expan abbr="momẽti">momenti</expan>
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="N10C86" type="head">
              <s id="N10C88">
                <emph type="italics"/>
              Appendice, II.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="N10C8E" type="main">
              <s id="N10C90">Et è manifeſto che detta rota correrà verſo la parte
                <lb/>
              del piano inferiore. </s>
            </p>
            <p id="N10C94" type="main">
              <s id="N10C96">
                <emph type="italics"/>
              Percioche tirata dal centro I, la IG K perpendicolare del momento
                <lb/>
              tutto ſin che
                <expan abbr="s'incõtri">s'incontri</expan>
              col piano per oue camina: ſarà il ponto G della cir
                <emph.end type="italics"/>
              </s>
            </p>
          </chap>
        </body>
      </text>
    </archimedes>
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