DelMonte, Guidubaldo, Le mechaniche

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          <chap id="N13354">
            <pb xlink:href="037/01/092.jpg"/>
            <p id="id.2.1.480.0.0" type="head">
              <s id="id.2.1.480.1.0">PROPOSITIONE IIII. </s>
            </p>
            <p id="id.2.1.481.0.0" type="main">
              <s id="id.2.1.481.1.0">Se la poſſanza mouerà il peſo appiccato nella leua, ſarà lo ſpatio
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              della poſſanza moſſa allo ſpatio del peſo moſſo, come la diſtan
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              za dal ſoſtegno alla poſſanza, alla diſtanza dall'iſteſſo ſoſtegno
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              fin allo appiccamento del peſo. </s>
            </p>
            <p id="id.2.1.482.0.0" type="main">
              <s id="id.2.1.482.1.0">
                <emph type="italics"/>
              Sia la leua AB, il cui ſoſtegno C, & ſia il peſo D attaccato al punto B, & ſia
                <lb/>
              la poſſanza in A mouente il peſo D con la leua AB. </s>
              <s id="id.2.1.482.2.0">Dico lo ſpatio della poſ­
                <lb/>
              ſanza in A allo ſpatio del peſo eſſere coſi come CA à CB. </s>
              <s id="id.2.1.482.3.0">Mouaſi la leua
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              AB, & affine che il peſo D ſi moua in sù, biſogna che B ſi moua in sù, & A in
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              giù. </s>
              <s id="id.2.1.482.4.0">& percioche C è punto immobile; però mentre A, & B ſi mouono, de­
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              ſcriueranno circonferenze di cerchi. </s>
              <s id="id.2.1.482.5.0">Mouaſi dunque AB in EF; ſaranno AEBF
                <lb/>
              circonferenze di cerchi, i me­
                <lb/>
              zi diametri de' quali ſono CA
                <lb/>
              CB. </s>
              <s id="id.2.1.482.6.0">compiſcaſi tutta la cir­
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              conferenza AGE, & tut­
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              ta la BHF, & ſiano KH
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              i punti doue AB, & EF ta­
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              gliano il cerchio BHF. </s>
              <s id="id.2.1.482.7.0">Hor
                <lb/>
              percioche l'angolo BCF è
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note137"/>
                <emph type="italics"/>
              eguale all'angolo HCK, ſa­
                <lb/>
              rà la circonferenza KH egua
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note138"/>
                <emph type="italics"/>
              le alla circonferenza BF, &
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              concioſia, che le circonferen­
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              ze AEKH ſiano ſotto l'i­
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              ſteſſo angolo ACE, & la
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              circonferenza AE à tutta
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              la circonferenza AGE ſia
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              come l'angolo ACE à quat
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              tro retti, & come l'iſteſſo an­
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              golo HCK à quattro retti,
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              coſi anche è la circonferenza
                <lb/>
              HK à tutta la circonferentia
                <lb/>
              HBK, ſarà la circonferentia
                <emph.end type="italics"/>
                <lb/>
                <figure id="id.037.01.092.1.jpg" xlink:href="037/01/092/1.jpg" number="86"/>
                <lb/>
                <emph type="italics"/>
              AE à tutta la circonferentia AGE, come la circonferentia KH à tutta la
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note139"/>
                <emph type="italics"/>
              KFH. </s>
              <s id="N13942">& permutando come la circonferentia AE alla circonferenza KH, cioè
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              BF, coſi tutta la circonferenza AGE à tutta la circonferenza BHF; ma tut­
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note140"/>
                <emph type="italics"/>
              ta la circonferenza AGE coſi ſi ha à tutta la BHF, come il diametro del cer­
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              chio AEG al diametro del cerchio BHF. </s>
              <s id="id.2.1.482.8.0">Come dunque la circonferenza AE
                <emph.end type="italics"/>
              </s>
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