Stelliola, Niccol� Antonio, De gli elementi mechanici, 1597
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            <pb xlink:href="041/01/043.jpg" pagenum="42"/>
            <p id="N10C2D" type="head">
              <s id="N10C2F">
                <emph type="italics"/>
              Dimoſtratione.
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              </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
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              no inchinato A C: il circolo della rota D E FG: il toccamento D: e dal
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              ponto D, tiriſi perpendicolare all'orizonte B D F: è manifeſto che detta
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              linea, ſia la perpendicolare del ſoſtenimento: dico che'l centro del peſo
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              è fuori di detta linea. </s>
              <s id="N10C43">Si moſtra: perche del triangolo D B A: l'angolo,
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              E, che fa la perpendicolare con l'orizonte, è retto: reſta l'angolo B D A,
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              a cuto: e perciò la portione D G F, e maggiore del ſemicircolo; & in eſ
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              ſa ſarà il centro del circolo, che è anco centro di peſo. </s>
              <s id="N10C4B">è dunque il cen­
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              tro del peſo fuori della linea del
                <expan abbr="ſoſtenimẽto">ſoſtenimento</expan>
              . </s>
              <s id="N10C53">De ſcriuaſi alla D E, la por
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              tione di circolo D H F, ſimile a D E F; ſaranno dette portioni vgua­
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              li, e faranno equipondio. </s>
              <s id="N10C59">reſta dunque la figura lunare ſenza equi
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              pondio: & il momento della rota appoggiata ſarà meno che della ro
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              ta ſoſpeſa, ſecondo la ragione della figura lunare a tutto il circolo: cio è
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              ſecondo la ragione dell'ecceſſo delle portioni, al circolo tutto. </s>
              <s id="N10C61">Il che
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              ſi hauea da moſtrare.
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              </s>
            </p>
            <p id="N10C67" type="head">
              <s id="N10C69">
                <emph type="italics"/>
              Appendice. </s>
              <s id="N10C6D">I.
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              </s>
            </p>
            <p id="N10C71" type="main">
              <s id="N10C73">E l'iſteſſo che si è moſtrato nella rota c'ha grauezza;
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              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
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              l'iſteſſo peſo alle rote: eſſendo peſi vguali con loro centri nell'iſteſſe li­
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              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.
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              </s>
            </p>
            <p id="N10C8E" type="main">
              <s id="N10C90">Et è manifeſto che detta rota correrà verſo la parte
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              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
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              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>
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