Gravesande, Willem Jacob 's, An essay on perspective

Table of contents

< >
[71.] Problem VII. 55. To find the Perſpective of a Line, perpendicular to the Geometrical Plane.
[72.] Operation.
[73.] Demonstration.
[74.] Method II.
[75.] Demonstration.
[76.] Method III.
[77.] Operation, Without Compaſſes.
[78.] Demonstration.
[79.] Scholium.
[80.] Corollary.
[81.] Problem VIII.
[82.] To do this another Way.
[83.] Demonstration.
[84.] Problem IX.
[85.] Problem X.
[86.] Demonstration.
[87.] EG: EN:: GY: NM.
[88.] Definition.
[89.] Problem XI.
[90.] Lemma.
[91.] Demonstration.
[92.] Remarks.
[93.] Problem IX.
[94.] Operation.
[95.] Demonstration.
[96.] Problem X.
[97.] Operation.
[98.] Demonstration.
[99.] Remarks.
[100.] Method II. 70. By the accidental Point of inclin’d Lines.
< >
page |< < (61) of 237 > >|
    <echo version="1.0RC">
      <text xml:lang="en" type="free">
        <div xml:id="echoid-div215" type="section" level="1" n="113">
          <p>
            <s xml:id="echoid-s1446" xml:space="preserve">
              <pb o="61" file="0113" n="130" rhead="on PERSPECTIVE."/>
            continued, it would cut the Horizontal Line in
              <lb/>
            the Point of Sight.</s>
            <s xml:id="echoid-s1447" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1448" xml:space="preserve">This is done in aſſuming C H equal to {1/3}, or
              <lb/>
            {1/4} Part, &</s>
            <s xml:id="echoid-s1449" xml:space="preserve">c. </s>
            <s xml:id="echoid-s1450" xml:space="preserve">of the Diſtance from the Point C,
              <lb/>
            to the Foot of the vertical Line; </s>
            <s xml:id="echoid-s1451" xml:space="preserve">and in raiſing
              <lb/>
            the Perpendicular H E, in the Point H, equal
              <lb/>
            to {1/3} or {1/4} Part, &</s>
            <s xml:id="echoid-s1452" xml:space="preserve">c. </s>
            <s xml:id="echoid-s1453" xml:space="preserve">of the Height of the Eye.
              <lb/>
            </s>
            <s xml:id="echoid-s1454" xml:space="preserve">Now A is a given Point, whoſe Appearance is
              <lb/>
            ſought.</s>
            <s xml:id="echoid-s1455" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div217" type="section" level="1" n="114">
          <head xml:id="echoid-head120" xml:space="preserve">
            <emph style="sc">Operation</emph>
          .</head>
          <p>
            <s xml:id="echoid-s1456" xml:space="preserve">Draw a Parallel A B, through the Point A,
              <lb/>
            to the Baſe Line, meeting the Line C D in the
              <lb/>
            Point B, and let a ſecond Eye be ſuppoſed at
              <lb/>
            the ſame Height and Diſtance as the firſt; </s>
            <s xml:id="echoid-s1457" xml:space="preserve">then
              <lb/>
            find F G the Repreſentation of A B for
              <note symbol="*" position="right" xlink:label="note-0113-01" xlink:href="note-0113-01a" xml:space="preserve">42.</note>
            ſecond Eye, which continue until it meets the
              <lb/>
            Line C E in b, and in this Continuation aſſume
              <lb/>
            b a equal to F G; </s>
            <s xml:id="echoid-s1458" xml:space="preserve">then a will be the Perſpective
              <lb/>
            ſought.</s>
            <s xml:id="echoid-s1459" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div219" type="section" level="1" n="115">
          <head xml:id="echoid-head121" xml:space="preserve">
            <emph style="sc">Demonstration</emph>
          .</head>
          <p>
            <s xml:id="echoid-s1460" xml:space="preserve">Becauſe the Height and Diſtance of the ſecond
              <lb/>
            Eye, is equal to the Height and Diſtance of the
              <lb/>
            firſt; </s>
            <s xml:id="echoid-s1461" xml:space="preserve">the ſaid two Eyes are both in one parallel
              <lb/>
            Line A B; </s>
            <s xml:id="echoid-s1462" xml:space="preserve">and conſequently , the
              <note symbol="*" position="right" xlink:label="note-0113-02" xlink:href="note-0113-02a" xml:space="preserve">18.</note>
            of A B muſt be a Part of F G continued, and
              <lb/>
              <note symbol="*" position="right" xlink:label="note-0113-03" xlink:href="note-0113-03a" xml:space="preserve">12.</note>
            alſo equal to F G: </s>
            <s xml:id="echoid-s1463" xml:space="preserve">And therefore becauſe the Perſpective of B is in the Line C E, a b is the
              <lb/>
              <note symbol="*" position="right" xlink:label="note-0113-04" xlink:href="note-0113-04a" xml:space="preserve">16.</note>
            Perſpective of A B; </s>
            <s xml:id="echoid-s1464" xml:space="preserve">and a , that of A. </s>
            <s xml:id="echoid-s1465" xml:space="preserve">Which was to be demonſtrated.</s>
            <s xml:id="echoid-s1466" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1467" xml:space="preserve">78. </s>
            <s xml:id="echoid-s1468" xml:space="preserve">Note, as to Lines perpendicular, and inclined
              <lb/>
            to the Geometrical Plane, ſee n. </s>
            <s xml:id="echoid-s1469" xml:space="preserve">76. </s>
            <s xml:id="echoid-s1470" xml:space="preserve">This is
              <lb/>
            ſcarcely uſeful, unleſs for the Decorations of a
              <lb/>
            Theatre.</s>
            <s xml:id="echoid-s1471" xml:space="preserve"/>
          </p>
        </div>
      </text>
    </echo>