Gravesande, Willem Jacob 's, An essay on perspective

Table of contents

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[11.] Corollary I.
[12.] Corollary II.
[13.] Corollary III.
[14.] Theorem II.
[15.] Corollary I.
[16.] Corollary II.
[17.] Theorem III.
[18.] Theorem IV.
[19.] Corollary I.
[20.] Corollary II.
[21.] Corollary III.
[22.] Corollary IV.
[23.] Theorem V.
[24.] Theorem VI.
[25.] Corollary.
[26.] CHAP. III.
[27.] Problem I.
[28.] Operation.
[29.] Demonstration.
[30.] Remarks.
[31.] Method II.
[32.] Operation.
[33.] Demonstration.
[34.] Remarks.
[35.] Method III.
[36.] Operation.
[37.] Demonstration.
[38.] Remark.
[39.] Method. IV.
[40.] Operation, Without Compaſſes.
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        <div xml:id="echoid-div71" type="section" level="1" n="38">
          <pb o="21" file="0045" n="48" rhead="on PERSPECTIVE."/>
        </div>
        <div xml:id="echoid-div72" type="section" level="1" n="39">
          <head xml:id="echoid-head41" xml:space="preserve">
            <emph style="sc">Method</emph>
          . IV.</head>
          <p>
            <s xml:id="echoid-s617" xml:space="preserve">31. </s>
            <s xml:id="echoid-s618" xml:space="preserve">Draw the Line F O G through the Eye O,
              <lb/>
            parallel to the Baſe Line, then aſſume F O in
              <lb/>
              <note position="right" xlink:label="note-0045-01" xlink:href="note-0045-01a" xml:space="preserve">Fig. 10.</note>
            this Line, equal to the Height of the Eye, and
              <lb/>
            O G equal to the Length of the principal Ray.
              <lb/>
            </s>
            <s xml:id="echoid-s619" xml:space="preserve">A is the given Point.</s>
            <s xml:id="echoid-s620" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div74" type="section" level="1" n="40">
          <head xml:id="echoid-head42" xml:space="preserve">
            <emph style="sc">Operation</emph>
          ,
            <lb/>
          Without Compaſſes.</head>
          <p>
            <s xml:id="echoid-s621" xml:space="preserve">From the given Point A, draw the Lines AO,
              <lb/>
            A F, to the Points O and F, and from the
              <lb/>
            Point E, wherein A F cuts the Baſe Line, draw
              <lb/>
            the Line E G to the Point G; </s>
            <s xml:id="echoid-s622" xml:space="preserve">then the Point a,
              <lb/>
            the Interſection of A O, and E G, is the Repre-
              <lb/>
            ſentation ſought.</s>
            <s xml:id="echoid-s623" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div75" type="section" level="1" n="41">
          <head xml:id="echoid-head43" xml:space="preserve">
            <emph style="sc">Demonstration</emph>
          .</head>
          <p>
            <s xml:id="echoid-s624" xml:space="preserve">Let fall the perpendicular G M from the Point
              <lb/>
            G, upon the Baſe Line, and through the Eye O,
              <lb/>
            draw the Line O D to the Point D, the Inter-
              <lb/>
            ſection of the Horizontal Line, and the Line G E.</s>
            <s xml:id="echoid-s625" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s626" xml:space="preserve">Then becauſe the Triangles G D L, G E M are
              <lb/>
            fimilar,
              <lb/>
            G D: </s>
            <s xml:id="echoid-s627" xml:space="preserve">G E:</s>
            <s xml:id="echoid-s628" xml:space="preserve">: G L: </s>
            <s xml:id="echoid-s629" xml:space="preserve">G M.
              <lb/>
            </s>
            <s xml:id="echoid-s630" xml:space="preserve">But G O is equal to G L, and G F to L M. </s>
            <s xml:id="echoid-s631" xml:space="preserve">
              <lb/>
            whence
              <lb/>
            G D: </s>
            <s xml:id="echoid-s632" xml:space="preserve">G E:</s>
            <s xml:id="echoid-s633" xml:space="preserve">: G O: </s>
            <s xml:id="echoid-s634" xml:space="preserve">G F.</s>
            <s xml:id="echoid-s635" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s636" xml:space="preserve">And conſequenely the Triangles G O D: </s>
            <s xml:id="echoid-s637" xml:space="preserve">and
              <lb/>
            G F E, are ſimilar, and the Lines O D, and
              <lb/>
            A E F, are parallel between themſelves; </s>
            <s xml:id="echoid-s638" xml:space="preserve">and
              <lb/>
            therefore the Appearance of A E, is a
              <note symbol="*" position="right" xlink:label="note-0045-02" xlink:href="note-0045-02a" xml:space="preserve">13</note>
            of the Line E D G. </s>
            <s xml:id="echoid-s639" xml:space="preserve">It has alſo been
              <note symbol="*" position="right" xlink:label="note-0045-03" xlink:href="note-0045-03a" xml:space="preserve">27.</note>
            that the Repreſentation of the Point A, is in the
              <lb/>
            Line A O; </s>
            <s xml:id="echoid-s640" xml:space="preserve">therefore i@ is in a the </s>
          </p>
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