DelMonte, Guidubaldo, Le mechaniche

Table of figures

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[Figure 11]
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[Figure 12]
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[Figure 13]
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[Figure 18]
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[Figure 19]
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[Figure 22]
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[Figure 23]
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[Figure 27]
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[Figure 29]
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[Figure 30]
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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.2.1.75.0.0" type="main">
              <s id="id.2.1.75.1.0">
                <pb xlink:href="037/01/020.jpg"/>
              che in Italia communemente vi ſi confaccia, ne alcuno di
                <expan abbr="qneſti">queſti</expan>
              ſarebbe inteſo
                <lb/>
              per tutto. </s>
              <s id="id.2.1.75.2.0">Onde io ho ſcritto coſi la Trutina, ſperando, che ſi habbia à fare termi
                <lb/>
              ne, & parola generale à tutte le nationi d'Italia. </s>
            </p>
            <p id="id.2.1.76.0.0" type="main">
              <s id="id.2.1.76.1.0">Perpendicolo vuol dire quella linea, che ſporge in fuori dal centro della Bilancia al
                <lb/>
              mezo di detta Bilancia, ilqual Perpendicolo è ſolamente nelle Bilancie, lequali han
                <lb/>
              no il centro di fuori della Bilancia, o ſia di ſotto, ò ſia di ſopra. </s>
              <s id="id.2.1.76.2.0">Ma quando il cen­
                <lb/>
              tro della Bilancia è nel mezo di eſſa, all'hora non vi è queſto Perpendicolo per eſ
                <lb/>
              ſere il centro della Bilancia, & il mezo di eſſa vn'iſteſſo punto. </s>
              <s id="id.2.1.76.3.0">Et queſto Perpen­
                <lb/>
              dicolo è coſa imaginata dall' Autore ſolamente, & non da altri, per ageuolare al­
                <lb/>
              cune dimoſtrationi della Bilancia, che di nouo ha inueſtigate: & non è la linguet­
                <lb/>
              ta, ne meno la linea della direttione, ò dirittura che ſi habbia à dire. </s>
            </p>
            <p id="id.2.1.77.0.0" type="head">
              <s id="id.2.1.77.1.0">LEMMA. </s>
            </p>
            <p id="id.2.1.78.0.0" type="main">
              <s id="id.2.1.78.1.0">Sia la linea AB à piombo dell'orizonte, & col diametro AB ſi deſcri­
                <lb/>
              ua il cerchio AEBD, il cui centro ſia C. </s>
              <s id="id.2.1.78.2.0">Dico il punto B eſſere
                <lb/>
              l'infimo luogo della circonferenza del cerchio AEBD, & il pun­
                <lb/>
              to A il piu alto, & quali ſi voglian punti, come DE, i quali ſiano
                <lb/>
              però egualmente diſtanti da A eſſere egualmente poſti di ſotto, &
                <lb/>
              quelli che ſtanno piu da preſſo ad eſſo A, eſſere più alti di quelli, che
                <lb/>
              ſono più da lunge. </s>
            </p>
            <p id="id.2.1.79.0.0" type="main">
              <s id="id.2.1.79.1.0">
                <arrow.to.target n="note1"/>
                <emph type="italics"/>
              Allunghiſi la linea AB fin al centro del mondo,
                <lb/>
              che ſia F. </s>
              <s id="id.2.1.79.2.0">Dapoi ſia preſo nella circonferenza
                <lb/>
              del cerchio qual ſi voglia punto, come G, & ſi
                <lb/>
              congiungano le linee FG FD FE. </s>
              <s id="id.2.1.79.3.0">Hor per­
                <lb/>
              cioche BF è la minima linea di tutte quelle,
                <lb/>
              che dal punto F ſono tirate alla circonferenza
                <lb/>
              AEBD, ſarà la BF minore della FG. </s>
              <s id="id.2.1.79.4.0">Per
                <lb/>
              laqual coſa il punto B ſarà piu da preſſo al pun­
                <lb/>
              to F, che il G. </s>
              <s id="id.2.1.79.5.0">Et per cotesta ragione ſi dimo­
                <lb/>
              strerà, che il punto B ſta più da preſſo al centro
                <lb/>
              del mondo di qual ſi voglia altro punto della cir­
                <lb/>
              conferenza del cerchio AEBD. </s>
              <s id="id.2.1.79.6.0">Sarà dunque
                <lb/>
              il punto B l'infimo luogo della circonferenza del
                <lb/>
              cerchio AEBD. </s>
              <s id="id.2.1.79.7.0">Dapoi perche AF tirata
                <lb/>
              per lo centro è maggiore di GF, ſarà il punto A
                <lb/>
              più alto non ſolamente di G, ma etiandio di qual
                <lb/>
              ſi voglia altro punto della circonferenza del cer­
                <lb/>
              chio AEBD. </s>
              <s id="id.2.1.79.8.0">Oltre à ciò perche DF, & FE
                <lb/>
              ſono eguali, i punti DE ſaranno egualmente di
                <lb/>
              stanti dal centro del mondo. </s>
              <s id="id.2.1.79.9.0">Et eſſendo DF
                <lb/>
              maggiore di FG, ſarà il punto D, che è più da
                <lb/>
              preſſo al punto A, più alto del punto G, lequali
                <lb/>
              coſe tutte erano da moſtrarſi.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.80.0.0" type="margin">
              <s id="id.2.1.80.1.0">
                <margin.target id="note1"/>
                <emph type="italics"/>
              Per la ottaua del terzo.
                <emph.end type="italics"/>
              </s>
            </p>
            <figure id="id.037.01.020.1.jpg" xlink:href="037/01/020/1.jpg" number="3"/>
          </chap>
        </body>
      </text>
    </archimedes>
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