Casati, Paolo, Fabrica, et uso del compasso di proportione, dove insegna à gli artefici il modo di fare in esso le necessarie divisioni, e con varij problemi ...

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              <pb o="158" file="0172" n="174" rhead="CAPOV."/>
              <figure xlink:label="fig-0172-01" xlink:href="fig-0172-01a" number="48">
                <image file="0172-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0172-01"/>
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            uendo la proportione delle baſi, per
              <lb/>
            la 11 del lib. </s>
            <s xml:id="echoid-s2932" xml:space="preserve">12. </s>
            <s xml:id="echoid-s2933" xml:space="preserve">Dunque il cilindro
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            CG d’argento è vguate alla sfera S
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            d’argento. </s>
            <s xml:id="echoid-s2934" xml:space="preserve">Or volendoſi vn cilindro
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            quadrato, che fia vguale al cilindo
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            CG, e per confeguenza alla sfera da-
              <lb/>
            ta S, tra il diametro della baſe CF, e
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            l’altezza FG ſitroui la ſeconda delle
              <lb/>
            quattro continuatamente proportio-
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            nali, per la Queſt. </s>
            <s xml:id="echoid-s2935" xml:space="preserve">1. </s>
            <s xml:id="echoid-s2936" xml:space="preserve">del Cap. </s>
            <s xml:id="echoid-s2937" xml:space="preserve">4. </s>
            <s xml:id="echoid-s2938" xml:space="preserve">col
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            mezzo della linea cubica, e ſia CO,
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            diametro della baſe del cilindro, à cui
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            eſſendo vguale l’altezza OL, ſarà il ci-
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            lindro CL quadrato vguale al cilindro
              <lb/>
            CG, cioè alla sfera; </s>
            <s xml:id="echoid-s2939" xml:space="preserve">eſſendo che le ba-
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            ſi, e l’altezze di queſti due cilindri ſo-
              <lb/>
            no reciproche, come s’è dimoſtrato
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            nella Queſt. </s>
            <s xml:id="echoid-s2940" xml:space="preserve">6. </s>
            <s xml:id="echoid-s2941" xml:space="preserve">del Cap. </s>
            <s xml:id="echoid-s2942" xml:space="preserve">4. </s>
            <s xml:id="echoid-s2943" xml:space="preserve">perche per
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            la coſtruttione il circolo del diametro
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            CF al circolo del diametro CO è co-
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            mela prima alla terza proportionale,
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            tra le quali la linea CO è la ſeconda.
              <lb/>
            </s>
            <s xml:id="echoid-s2944" xml:space="preserve">Or cſſendo come la prima alla terza, così la ſeconda alla
              <lb/>
            quarta, cioè CO, ouero OL vguale altezza, all’ altezza FG,
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            ſi rende manifeſto, che ſi reciprocano le baſi, e l’altezze. </s>
            <s xml:id="echoid-s2945" xml:space="preserve">Tra-
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            portato dunque CO nella linea metallica all’interuallo AA
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            dell’argento, prendaſi l’interuallo OO dell’oro, e ſia la linea
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            IM diametro della baſe, & </s>
            <s xml:id="echoid-s2946" xml:space="preserve">MK altezza vguale: </s>
            <s xml:id="echoid-s2947" xml:space="preserve">onde il cilin-
              <lb/>
            dro d’oro IK eſſendo ſimile al cilindro CL d’argento, & </s>
            <s xml:id="echoid-s2948" xml:space="preserve">eſ-
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            ſendo per la coſtruttione dello ſtromento nella </s>
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