DelMonte, Guidubaldo, Le mechaniche

Table of figures

< >
[Figure 211]
[Figure 212]
[Figure 213]
[Figure 214]
[Figure 215]
[Figure 216]
[Figure 217]
[Figure 218]
[Figure 219]
[Figure 220]
[Figure 221]
[Figure 222]
[Figure 223]
[Figure 224]
[Figure 225]
[Figure 226]
[Figure 227]
[Figure 228]
[Figure 229]
[Figure 230]
[Figure 231]
[Figure 232]
[Figure 233]
[Figure 234]
[Figure 235]
[Figure 236]
[Figure 237]
[Figure 238]
< >
page |< < of 270 > >|
    <archimedes>
      <text id="id.0.0.0.0.3">
        <body id="id.2.0.0.0.0">
          <chap id="N18810">
            <pb xlink:href="037/01/262.jpg"/>
            <p id="id.2.1.1363.0.0" type="head">
              <s id="id.2.1.1363.1.0">Altramente. </s>
            </p>
            <p id="id.2.1.1364.0.0" type="main">
              <s id="id.2.1.1364.1.0">
                <emph type="italics"/>
              Sia data la vite AB, che habbia due helici eguali CDEFG; ſia dapoi vn'altro ci­
                <lb/>
              lindro
                <foreign lang="grc">α β</foreign>
              eguale ad eſſo AB, nel quale prendaſi OP eguale à CG; & diuidaſi
                <lb/>
              OP in tre parti eguali OR RT TP; & deſcriuanſi tre helici OQ RS TV P;
                <lb/>
              ſarà ciaſcuna delle OR RT TP minore di CE, & di EG; percioche la terza
                <emph.end type="italics"/>
                <lb/>
                <figure id="id.037.01.262.1.jpg" xlink:href="037/01/262/1.jpg" number="235"/>
                <lb/>
                <emph type="italics"/>
              parte è minore della metà. </s>
              <s id="id.2.1.1364.2.0">dico, che il peſo medeſimo ſi mouerà più facilmente ſo­
                <lb/>
              prale helici OQRS TVP, che ſopra CDEFG. </s>
              <s id="id.2.1.1364.3.0">facciaſi HIL triangolo di an
                <lb/>
              goli retti, in modo che HI ſia eguale à CG, & IL ſia eguale al doppio del Peri­
                <lb/>
              metro del cilindro AB, & per LI ſi intenda vn piano egualmente diſtante dall'o­
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note317"/>
                <emph type="italics"/>
              rizonte; ſarà HL eguale à CDEFG, & HLI ſarà l'angolo della inclinatione.
                <lb/>
              </s>
              <s id="id.2.1.1364.4.0">facciaſi ſimilmente il triangolo X
                <foreign lang="grc">Υ</foreign>
              Z di angoli retti, in modo che XZ ſia eguale
                <emph.end type="italics"/>
              </s>
            </p>
          </chap>
        </body>
      </text>
    </archimedes>