Gravesande, Willem Jacob 's, An essay on perspective

Table of figures

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[61] page 88.Plate. 25.Fig. 60O G F f Z L R P D I T S M a Q E R H N A C B
[62] Plate 26Fig. 61O I F T N S Q S H E R M A
[63] Fig. 62C D S Q L C D R P H
[64] page 96.Plate. 27Fig. 63D E C F M H I G P A Q N
[65] Fig. 64X S D E T C R L F H I G P M B O V Q N
[66] page 98.Plate. 28Fig. 65L M F G D H C E I A B
[67] Fig. 66A B VII VIII IV V H C VI VI P V VII IV S VIII E O I III II I XII XIX IX F D
[68] page 100Plate. 29Fig. 675 6p 7 8 9 10 S V VI VII VIII IX X o XI ll l
[69] Fig. 68c P G e o Q
[70] Fig. 69P c G o e Q
[Figure 71]
[Figure 72]
[Figure 73]
[74] Page 120Plate. 30.Fig. 70.X I F B H D D P O M P R C C C C C E E Q
[75] Plate 31page 120Fig. 71D G C B A H F a I E
[76] Fig. 72P G C H A N B R Q M a F
[77] Fig. 73P G C H D N B I A R Q M a F
[78] Fig. 74G N B C H M a A
[79] Fig. 75D G B C A H F I E a
[80] page 120Plate. 32.Fig. 76.
[81] Fig. 77.R V T o
[82] Fig. 78.Z Z Y C M L I E A H D X G F B S Q P N 4 3 2
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        <div xml:id="echoid-div132" type="section" level="1" n="73">
          <p>
            <s xml:id="echoid-s941" xml:space="preserve">
              <pb o="37" file="0075" n="85" rhead="on PERSPECTIVE."/>
            Point A parallel to the Baſe Line, and made e-
              <lb/>
            qual to B C. </s>
            <s xml:id="echoid-s942" xml:space="preserve">Now if from the Extremities of
              <lb/>
            the ſaid Line A L, Perpendiculars are let fall,
              <lb/>
            meeting the Baſe Line in the Points P and M,
              <lb/>
            and from theſe Points, Lines are drawn to the
              <lb/>
            Point of Sight V; </s>
            <s xml:id="echoid-s943" xml:space="preserve">then a N will likewiſe be
              <note symbol="*" position="right" xlink:label="note-0075-01" xlink:href="note-0075-01a" xml:space="preserve">5, 16.</note>
            the Perſpective of A L; </s>
            <s xml:id="echoid-s944" xml:space="preserve">and ſince P M is equal
              <lb/>
            to D E, a N will be likewiſe equal to G H, and
              <lb/>
            conſequently a N will be likewiſe equal to a I,
              <lb/>
            which is equal to G H.</s>
            <s xml:id="echoid-s945" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div134" type="section" level="1" n="74">
          <head xml:id="echoid-head80" xml:space="preserve">
            <emph style="sc">Method</emph>
          II.</head>
          <p>
            <s xml:id="echoid-s946" xml:space="preserve">57. </s>
            <s xml:id="echoid-s947" xml:space="preserve">The ſame Things being given, as in the
              <lb/>
            precedent Method, about the Point A, as a
              <lb/>
              <note position="right" xlink:label="note-0075-02" xlink:href="note-0075-02a" xml:space="preserve">Fig. 23.</note>
            Center, and with the Radius B C, deſcribe the
              <lb/>
            Arc of a Circle L M, and draw the Line O L
              <lb/>
            from the Eye touching it; </s>
            <s xml:id="echoid-s948" xml:space="preserve">then about a, (which
              <lb/>
            is the Repreſentation of A) as a Center deſcribe
              <lb/>
            the Circular Arc G I touching the Line L O,
              <lb/>
            and cutting another Line drawn through a Per-
              <lb/>
            pendicular to the Baſe Line in the Point I: </s>
            <s xml:id="echoid-s949" xml:space="preserve">I
              <lb/>
            ſay the Point I is the Extremity of the Repre-
              <lb/>
            ſentation ſought.</s>
            <s xml:id="echoid-s950" xml:space="preserve"/>
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        <div xml:id="echoid-div136" type="section" level="1" n="75">
          <head xml:id="echoid-head81" xml:space="preserve">
            <emph style="sc">Demonstration</emph>
          .</head>
          <p>
            <s xml:id="echoid-s951" xml:space="preserve">To prove this, let fall the Perpendiculars A L
              <lb/>
            and a G upon the Line O L, which will meet
              <lb/>
            the ſaid Line in the Points wherein it touches
              <lb/>
            the circular Arcs M L and G I.</s>
            <s xml:id="echoid-s952" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s953" xml:space="preserve">Alſo aſſume D E in the Baſe Line equal to B C
              <lb/>
            or A L, and draw the Line D F; </s>
            <s xml:id="echoid-s954" xml:space="preserve">then through a,
              <lb/>
            draw a H parallel to the Baſe Line.</s>
            <s xml:id="echoid-s955" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s956" xml:space="preserve">Now let us conſider the Figure X, which re-
              <lb/>
              <note position="right" xlink:label="note-0075-03" xlink:href="note-0075-03a" xml:space="preserve">Fig. 24.</note>
            preſents a Plane paſſing through the Eye and
              <lb/>
            the Point A of the foregoing Figure, wherein
              <lb/>
            O f here, repreſents O F there; </s>
            <s xml:id="echoid-s957" xml:space="preserve">f e here, F </s>
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