Gravesande, Willem Jacob 's, An essay on perspective

Table of contents

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[121.] Problem I.
[122.] Problem II.
[123.] Operation.
[124.] Demonstration.
[125.] Remark.
[126.] Problem III.
[127.] Method II.
[128.] Operation.
[129.] Demonstration.
[130.] Method III.
[131.] Operation.
[132.] Demonstration.
[133.] Remark.
[134.] Problem IV.
[135.] Problem V.
[136.] Operation.
[137.] Demonstration.
[138.] Problem VI.
[139.] Method II.
[140.] Operation.
[141.] Demonstration.
[142.] Method III.
[143.] CHAP. VI.
[144.] Prob. I.
[145.] Prob. II.
[146.] Demonstration.
[147.] Corollary.
[148.] Method II.
[149.] Operation,
[150.] Demonstration.
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          <p>
            <s xml:id="echoid-s1651" xml:space="preserve">
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            this Appearance paſſes thro’ that of the Point
              <lb/>
            L, it will be a Part of C X. </s>
            <s xml:id="echoid-s1652" xml:space="preserve">But becauſe that
              <lb/>
            Line, drawn from the Point L, paſſes thro’ the
              <lb/>
            propos’d Point; </s>
            <s xml:id="echoid-s1653" xml:space="preserve">the Repreſentation of the ſaid
              <lb/>
            Point is alſo in C X; </s>
            <s xml:id="echoid-s1654" xml:space="preserve">and ſo in X, the common
              <lb/>
            Interſection of C X, and T X.</s>
            <s xml:id="echoid-s1655" xml:space="preserve"/>
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        <div xml:id="echoid-div238" type="section" level="1" n="125">
          <head xml:id="echoid-head131" xml:space="preserve">
            <emph style="sc">Remark</emph>
          .</head>
          <p>
            <s xml:id="echoid-s1656" xml:space="preserve">83. </s>
            <s xml:id="echoid-s1657" xml:space="preserve">If the Point T ſhould be at too great a
              <lb/>
            Diſtance; </s>
            <s xml:id="echoid-s1658" xml:space="preserve">or if T B X, or C X, ſhould too ob-
              <lb/>
            liquely cut each other; </s>
            <s xml:id="echoid-s1659" xml:space="preserve">the perſpective Plane
              <lb/>
            muſt then be ſuppos’d to be reduc’d to a
              <note symbol="*" position="left" xlink:label="note-0122-01" xlink:href="note-0122-01a" xml:space="preserve">81.</note>
            pendicular, or upright one; </s>
            <s xml:id="echoid-s1660" xml:space="preserve">and the Repreſen-
              <lb/>
            tation of a Point, above the Geometrical Plane,
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            (whoſe Seat is L, and Height M E) muſt be
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            found .</s>
            <s xml:id="echoid-s1661" xml:space="preserve"/>
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          <note symbol="*" position="left" xml:space="preserve">50.</note>
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        <div xml:id="echoid-div240" type="section" level="1" n="126">
          <head xml:id="echoid-head132" xml:space="preserve">
            <emph style="sc">Problem</emph>
          III.</head>
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            <s xml:id="echoid-s1662" xml:space="preserve">84. </s>
            <s xml:id="echoid-s1663" xml:space="preserve">To find the Repreſentation of a Line, perpendi-
              <lb/>
            cular to the Geometrical Plane`.</s>
            <s xml:id="echoid-s1664" xml:space="preserve"/>
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          <p>
            <s xml:id="echoid-s1665" xml:space="preserve">The Appearance of the Extremity of the Per-
              <lb/>
              <note position="left" xlink:label="note-0122-03" xlink:href="note-0122-03a" xml:space="preserve">Fig. 45.</note>
            pendicular muſt be found , in conſidering
              <note symbol="*" position="left" xlink:label="note-0122-04" xlink:href="note-0122-04a" xml:space="preserve">82.</note>
            ſaid Extremity as a Point above the Geometri-
              <lb/>
            cal Plane, by the Height of the propos’d Perpendi-
              <lb/>
            cular: </s>
            <s xml:id="echoid-s1666" xml:space="preserve">Then if a Line be drawn from the Point
              <lb/>
            D, to the Point of Sight; </s>
            <s xml:id="echoid-s1667" xml:space="preserve">its Interſection
              <note symbol="*" position="left" xlink:label="note-0122-05" xlink:href="note-0122-05a" xml:space="preserve">16.</note>
            T X, will give the Appearance a of the Seat of
              <lb/>
            the Perpendicular propos’d.</s>
            <s xml:id="echoid-s1668" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1669" xml:space="preserve">Note, When there is a Neceſſity of having re-
              <lb/>
            courſe to the Remarks of the foregoing Problem,
              <lb/>
            in order to find the Point X; </s>
            <s xml:id="echoid-s1670" xml:space="preserve">then the Point a
              <lb/>
            may be found, in drawing A S and D V, and
              <lb/>
            afterwards joining the Points B and X by a
              <lb/>
            Line. </s>
            <s xml:id="echoid-s1671" xml:space="preserve">And when B X and D V cut each </s>
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