Gravesande, Willem Jacob 's, An essay on perspective

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        <div xml:id="echoid-div170" type="section" level="1" n="91">
          <pb o="49" file="0093" n="106" rhead="on PERSPECTIVE."/>
          <p>
            <s xml:id="echoid-s1212" xml:space="preserve">The Demonſtration of the
              <emph style="sc">Problem</emph>
            .</s>
            <s xml:id="echoid-s1213" xml:space="preserve"/>
          </p>
          <p style="it">
            <s xml:id="echoid-s1214" xml:space="preserve">66. </s>
            <s xml:id="echoid-s1215" xml:space="preserve">The Torus of the Column muſt be conceiv’d as
              <lb/>
            made up of an Infinite Number of Circular Planes,
              <lb/>
            lying one upon another. </s>
            <s xml:id="echoid-s1216" xml:space="preserve">And it is evident that
              <lb/>
            the Reaſon why each of thoſe Circles cannot be wholly
              <lb/>
            ſeen, is becauſe that which is immediately under it
              <lb/>
            hides a Part thereof; </s>
            <s xml:id="echoid-s1217" xml:space="preserve">from whence it follows, that
              <lb/>
            if the Plane of one oſ theſe Circles be every way
              <lb/>
            continu’d, and the Circle immediately under it, be
              <lb/>
            thrown into Perſpective upon it, (which
              <note symbol="*" position="right" xlink:label="note-0093-01" xlink:href="note-0093-01a" xml:space="preserve">8.</note>
            is alſo a Circle) the two Points of Interſection of this
              <lb/>
            Repreſentation, and the Circle in the Plane, will deter-
              <lb/>
            mine the viſible Part of the ſaid Repreſentation; </s>
            <s xml:id="echoid-s1218" xml:space="preserve">and
              <lb/>
            conſequently if the Repreſentation of theſe two Points
              <lb/>
            of Interſection be found upon the Perſpective Plane, we
              <lb/>
            ſhall have two Points of the Perſpective of the Torus of
              <lb/>
            the propoſed Column. </s>
            <s xml:id="echoid-s1219" xml:space="preserve">This is what I have done in the
              <lb/>
            Solution of the Problem, as we ſhall now Analytically
              <lb/>
            demonſtrate.</s>
            <s xml:id="echoid-s1220" xml:space="preserve"/>
          </p>
          <p style="it">
            <s xml:id="echoid-s1221" xml:space="preserve">Let O be the Eye, A M a part of the Torus of
              <lb/>
              <note position="right" xlink:label="note-0093-02" xlink:href="note-0093-02a" xml:space="preserve">Fig. 35.</note>
            the Column, A P a Perpendicular to the Baſe paſſing
              <lb/>
            through the Center of the Column, and A B a
              <lb/>
            Parallel to the Baſe, drawn thro’ the Center B of
              <lb/>
            the Semicircle Concavity of the Torus. </s>
            <s xml:id="echoid-s1222" xml:space="preserve">Let M P be
              <lb/>
            a Semidiameter of one of the Circles ſpoken of in the
              <lb/>
            the beginning of this Demonſtration. </s>
            <s xml:id="echoid-s1223" xml:space="preserve">Then if the
              <lb/>
            Line m p be drawn parallel and infinitely near C M P
              <lb/>
            and the Lines m O and p O are drawn cutting M P
              <lb/>
            in D and T, it is evident that D T, which is in the
              <lb/>
            Plane of the Circle paſſing thro’ M P, will be the
              <lb/>
            Semidiameter of the Perſpective of the Circle imme-
              <lb/>
            diately underneath.</s>
            <s xml:id="echoid-s1224" xml:space="preserve"/>
          </p>
          <p style="it">
            <s xml:id="echoid-s1225" xml:space="preserve">Now let fall the Perpendicular O S from the Eye
              <lb/>
            to the Line A B, and continue the Lines M P and
              <lb/>
            m p, till they meet the ſaid Perpendicular in the Points
              <lb/>
            Q and q. </s>
            <s xml:id="echoid-s1226" xml:space="preserve">Moreover, continue the Line M P to </s>
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