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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          <p>
            <s xml:id="echoid-s1980" xml:space="preserve">
              <pb o="103" file="0117" n="119" rhead="Linea Geometrica"/>
            56. </s>
            <s xml:id="echoid-s1981" xml:space="preserve">9. </s>
            <s xml:id="echoid-s1982" xml:space="preserve">12. </s>
            <s xml:id="echoid-s1983" xml:space="preserve">Ecosì preſo l’interuallo 56. </s>
            <s xml:id="echoid-s1984" xml:space="preserve">56, deuo trouar’il la-
              <lb/>
            to del quadrato noncuplo, e perciò l’applico al 4. </s>
            <s xml:id="echoid-s1985" xml:space="preserve">4, il cui
              <lb/>
            noncuplo è 36, el’interuallo 36. </s>
            <s xml:id="echoid-s1986" xml:space="preserve">36 ſarà il lato del quadrato
              <lb/>
            noncuplo del primo. </s>
            <s xml:id="echoid-s1987" xml:space="preserve">E perche à queſto ſi deue trouar’il duo-
              <lb/>
            decuplo, applico queſto ſecondo interuallo al 5. </s>
            <s xml:id="echoid-s1988" xml:space="preserve">5, e piglio il
              <lb/>
            duodecuplo, che ſarà all’interuallo 60. </s>
            <s xml:id="echoid-s1989" xml:space="preserve">60, e con queſto ope-
              <lb/>
            rando nelle linee Aritmetiche, come s’è detto, trouo la ra-
              <lb/>
            dice quadrata del numero dato 604812 eſſere 777, e quaſi
              <lb/>
            778, poiche nella linea deſcritta ſi può leuare ſette volte
              <lb/>
            l’interuallo 100. </s>
            <s xml:id="echoid-s1990" xml:space="preserve">100, & </s>
            <s xml:id="echoid-s1991" xml:space="preserve">il reſtante è quaſi 78.</s>
            <s xml:id="echoid-s1992" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1993" xml:space="preserve">Mà cercando la Radice Quadrata d’vn Rotto, prendi nel-
              <lb/>
            le linee Geometriche li due interualli corriſpondenti al Nu-
              <lb/>
            meratore, & </s>
            <s xml:id="echoid-s1994" xml:space="preserve">al Denominatore: </s>
            <s xml:id="echoid-s1995" xml:space="preserve">dipoi traportali nelle linee
              <lb/>
            Aritmetiche, aprendo lo ſtromento in modo, che capiſca,
              <lb/>
            l’interuallo del numero, che vuoi ritenere; </s>
            <s xml:id="echoid-s1996" xml:space="preserve">poiche l’altro in-
              <lb/>
            teruallo nelle ſteſſe linee darà il numero cercato.</s>
            <s xml:id="echoid-s1997" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1998" xml:space="preserve">Sia il Rotto {4/9}, di cui ſi cerca la Radice Quadrata: </s>
            <s xml:id="echoid-s1999" xml:space="preserve">prendo
              <lb/>
            nelle linee Geometriche 4.</s>
            <s xml:id="echoid-s2000" xml:space="preserve">4, con vn Compaſſo, e con vn’al-
              <lb/>
            tro 9. </s>
            <s xml:id="echoid-s2001" xml:space="preserve">9. </s>
            <s xml:id="echoid-s2002" xml:space="preserve">Dipoi volendo ritener il Numeratore 4; </s>
            <s xml:id="echoid-s2003" xml:space="preserve">apro lo
              <lb/>
            ſtromento in modo, che l’interuallo del primo Compaſſo ſi
              <lb/>
            addatti alli punti 4.</s>
            <s xml:id="echoid-s2004" xml:space="preserve">4, nelle linee Aritmetiche; </s>
            <s xml:id="echoid-s2005" xml:space="preserve">poiche l’altro
              <lb/>
            Compaſſo ſi addattarà alli punti 6. </s>
            <s xml:id="echoid-s2006" xml:space="preserve">6: </s>
            <s xml:id="echoid-s2007" xml:space="preserve">onde dirò che la radi-
              <lb/>
            ce cercata è {4/6}, cioè {2/3}. </s>
            <s xml:id="echoid-s2008" xml:space="preserve">Ouero addattando il ſecondo Com-
              <lb/>
            paſſo, che corriſponde al Denominatore, alli punti 9. </s>
            <s xml:id="echoid-s2009" xml:space="preserve">9, tro-
              <lb/>
            uo che l’altro corriſponde alli 6. </s>
            <s xml:id="echoid-s2010" xml:space="preserve">6: </s>
            <s xml:id="echoid-s2011" xml:space="preserve">onde dirò, che la Radice
              <lb/>
            cercata è {6/9}. </s>
            <s xml:id="echoid-s2012" xml:space="preserve">E perche il 4, & </s>
            <s xml:id="echoid-s2013" xml:space="preserve">il 9 ſono interualli troppo pic-
              <lb/>
            coli, in lor vece ſi prendano li moltiplici, cioè 40, e 90, ò
              <lb/>
            qualſiuoglia altro. </s>
            <s xml:id="echoid-s2014" xml:space="preserve">II che molto più ſerue, quando il Rotto
              <lb/>
            dato non hà la Radice preciſa, poiche ſi trouarebbe la Radi-
              <lb/>
            ce più vicina alla vera. </s>
            <s xml:id="echoid-s2015" xml:space="preserve">Così cercando la Radice di {4/10} ſi </s>
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