DelMonte, Guidubaldo, Le mechaniche

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    <archimedes>
      <text id="id.0.0.0.0.3">
        <body id="id.2.0.0.0.0">
          <chap id="N181D6">
            <pb xlink:href="037/01/240.jpg"/>
            <p id="id.2.1.1281.0.0" type="main">
              <s id="id.2.1.1281.1.0">
                <emph type="italics"/>
              Siano li cunei ABC DEF, & l'angolo ABC ſia minore dell'angolo DEF, &
                <lb/>
              AB BC DE EF ſiano tra loro eguali. </s>
              <s id="id.2.1.1281.2.0">ſiano dapoi quattro peſi eguali GH IL
                <lb/>
              NO QR rettangoli; & ſiano LM KH nella medeſima linea retta. </s>
              <s id="id.2.1.1281.3.0">ſimilmente
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note309"/>
                <emph type="italics"/>
              RS PO in linea retta; ſaranno GK IM egualmente diſtanti, & NP QS an
                <lb/>
              co egualmente diſtanti. </s>
              <s id="id.2.1.1281.4.0">ſia IBG la parte del cuneo fra i peſi GH IL; & la par
                <lb/>
              te del cuneo QEN fra i peſi NOQR; & ſiano IB BG QE EN tra loro
                <lb/>
              eguali. </s>
              <s id="id.2.1.1281.5.0">Dico che i peſi GH IL più ageuolmente ſaranno dalla poſſanza iſteſſa co'l
                <lb/>
              cuneo ABC moſsi, che i peſi NO QR dal cuneo DEF.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.1282.0.0" type="margin">
              <s id="id.2.1.1282.1.0">
                <margin.target id="note309"/>
                <emph type="italics"/>
              Per la
                <emph.end type="italics"/>
              28.
                <emph type="italics"/>
              del primo.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.1283.0.0" type="main">
              <s id="id.2.1.1283.1.0">
                <emph type="italics"/>
              Diuidanſi AC DF in due parti eguali in TV, & congiunganſi TBVE, ſaranno
                <lb/>
              gli angoli poſti al T, & V retti. </s>
              <s id="id.2.1.1283.2.0">congiungaſi IG, laquale tagli BT in X. </s>
              <s id="id.2.1.1283.3.0">Hor
                <emph.end type="italics"/>
                <lb/>
                <figure id="id.037.01.240.1.jpg" xlink:href="037/01/240/1.jpg" number="216"/>
                <lb/>
                <arrow.to.target n="note310"/>
                <emph type="italics"/>
              percioche IB è eguale à BG, & BA eguale à BC: ſarà IA eguale ad eſſa
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note311"/>
                <emph type="italics"/>
              GC. </s>
              <s id="id.2.1.1283.4.0">Per laqual coſa BI ad IA è coſi, come BG à GC; dunque IG è egualmen
                <lb/>
              te diſtante ad eſſa AC: & perciò gli angoli ad X ſono retti; ma gli angoli XGK
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note312"/>
                <emph type="italics"/>
              XIM ſono retti, peroche GM è rettangolo. </s>
              <s id="id.2.1.1283.5.0">Per laqual coſa TB è egualmente di­
                <lb/>
              ſtante da GKIM. </s>
              <s id="id.2.1.1283.6.0">dunque l'angolo TBC è eguale all'angolo BGK, & TBA è
                <lb/>
              eguale ad eſſo BIM. </s>
              <s id="N18630">
                <expan abbr="ſimilmẽte">ſimilmente</expan>
              moſtreremo che l'angolo VEF è eguale ad ENP,
                <lb/>
              & VED eguale ad EQS. </s>
              <s id="N18637">& per eſſere l'angolo ABC minore dell'angolo DEF;
                <lb/>
              ſarà anco l'angolo TBC minore di VEN. </s>
              <s id="id.2.1.1283.7.0">Per laqual coſa BGK ſarà anche mi­
                <lb/>
              nore di ENP. </s>
              <s id="id.2.1.1283.8.0">con ſimile modo BIM è minore di EQS. </s>
              <s id="id.2.1.1283.9.0">Hor percioche il cuneo
                <lb/>
              ABC moue con due leue AB BC, che hanno i ſoſtegni ſuoi in B, & i peſi in
                <lb/>
              GI. </s>
              <s id="N1864A">ſimilmente il cuneo DEF moue con due altre leue DE EF, i cui ſoſtegni ſo­
                <lb/>
              no in E; & i peſi in NQ: per la precedente i peſi GH IL ſi moueranno più ageuol
                <lb/>
              mente con le leue AB BC, che i peſi NO QR con le leue DE EF. </s>
              <s id="id.2.1.1283.10.0">i peſi dunque
                <lb/>
              GH IL, ſi moueranno più ageuolmente co'l cuneo ABC, che i peſi NO QR co'l
                <emph.end type="italics"/>
              </s>
            </p>
          </chap>
        </body>
      </text>
    </archimedes>
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