Gravesande, Willem Jacob 's, An essay on perspective

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            <s xml:id="echoid-s1823" xml:space="preserve">
              <pb o="78" file="0140" n="162" rhead="An ESSAY"/>
            cauſe the Triangles THX and a F X are ſimilar,
              <lb/>
            TH — a F: </s>
            <s xml:id="echoid-s1824" xml:space="preserve">a F : </s>
            <s xml:id="echoid-s1825" xml:space="preserve">: </s>
            <s xml:id="echoid-s1826" xml:space="preserve">Ta: </s>
            <s xml:id="echoid-s1827" xml:space="preserve">a X.</s>
            <s xml:id="echoid-s1828" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1829" xml:space="preserve">And becauſe the Triangles T I x and ax L, are
              <lb/>
            alſo ſimilar, we have
              <lb/>
            TI + a L : </s>
            <s xml:id="echoid-s1830" xml:space="preserve">: </s>
            <s xml:id="echoid-s1831" xml:space="preserve">a L: </s>
            <s xml:id="echoid-s1832" xml:space="preserve">Ta : </s>
            <s xml:id="echoid-s1833" xml:space="preserve">ax.</s>
            <s xml:id="echoid-s1834" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1835" xml:space="preserve">Now let PM NR be the perſpective Plane,
              <lb/>
              <note position="left" xlink:label="note-0140-01" xlink:href="note-0140-01a" xml:space="preserve">Fig. 54.</note>
            O the Eye, A Q the Perpendicular, whoſe
              <lb/>
            Perſpective is requir’d, and O t a perpendicular
              <lb/>
            let fall from the Eye upon the perſpective Plane,
              <lb/>
            and ſo t will be the ſame, as the Point T in the
              <lb/>
            aforegoing Figure, Now if the Lines O Q be
              <lb/>
            drawn, it is manifeſt that A x, or A X, is the
              <lb/>
            Perſpective of A Q, according as this Line is
              <lb/>
            above or below the perſpective Plane in reſpect
              <lb/>
            to the Eye. </s>
            <s xml:id="echoid-s1836" xml:space="preserve">Then becauſe the Triangles O t x
              <lb/>
            and Q A x are ſimilar, we have
              <lb/>
            O t — A Q: </s>
            <s xml:id="echoid-s1837" xml:space="preserve">A Q :</s>
            <s xml:id="echoid-s1838" xml:space="preserve">: t A: </s>
            <s xml:id="echoid-s1839" xml:space="preserve">Ax.</s>
            <s xml:id="echoid-s1840" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1841" xml:space="preserve">And ſince the Triangles O t X and X A Q are
              <lb/>
            ſimilar,
              <lb/>
            O t + A Q: </s>
            <s xml:id="echoid-s1842" xml:space="preserve">A Q :</s>
            <s xml:id="echoid-s1843" xml:space="preserve">: t A: </s>
            <s xml:id="echoid-s1844" xml:space="preserve">A X.</s>
            <s xml:id="echoid-s1845" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1846" xml:space="preserve">Now Ot is equal to TH or TI of the afore-
              <lb/>
            going Figure, and AQ to a F or a L of the
              <lb/>
            ſame Figure; </s>
            <s xml:id="echoid-s1847" xml:space="preserve">as likewiſe At, Ta: </s>
            <s xml:id="echoid-s1848" xml:space="preserve">Therefore
              <lb/>
            if theſe two laſt Proportions be compared with
              <lb/>
            the two precedent ones, we ſhall find A x = a X,
              <lb/>
            and A X = a x; </s>
            <s xml:id="echoid-s1849" xml:space="preserve">which was to be demon-
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            ſtrated.</s>
            <s xml:id="echoid-s1850" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div285" type="section" level="1" n="151">
          <head xml:id="echoid-head157" xml:space="preserve">
            <emph style="sc">Remarks</emph>
          .</head>
          <p>
            <s xml:id="echoid-s1851" xml:space="preserve">96. </s>
            <s xml:id="echoid-s1852" xml:space="preserve">When the two Circles interſect each other,
              <lb/>
            or fall within one another, and ſo this Way be-
              <lb/>
            comes uſeleſs; </s>
            <s xml:id="echoid-s1853" xml:space="preserve">a Line muſt be drawn at Pleaſure,
              <lb/>
            through the Point T, equal to the Diſtance of
              <lb/>
            the Eye from the perſpective Plane; </s>
            <s xml:id="echoid-s1854" xml:space="preserve">and then a
              <lb/>
            parallel equal to the given Perpendicular muſt be
              <lb/>
            drawn to the ſaid Line through the Point a, ei-
              <lb/>
            ther towards L or F, according as the </s>
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