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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          <p>
            <s xml:id="echoid-s331" xml:space="preserve">
              <pb o="8" file="0026" n="26" rhead="An ESSAY"/>
            Sides a d and d c of the Figure a b c d. </s>
            <s xml:id="echoid-s332" xml:space="preserve">The
              <lb/>
            ſame may be demonſtrated of the other Sides;
              <lb/>
            </s>
            <s xml:id="echoid-s333" xml:space="preserve">and therefore the Figures are ſimilar.</s>
            <s xml:id="echoid-s334" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s335" xml:space="preserve">Now to prove the other Part of the Theorem:
              <lb/>
            </s>
            <s xml:id="echoid-s336" xml:space="preserve">If a perpendicular be ſuppoſed to be let fall from
              <lb/>
            the Eye upon the Plane of the Figure, and con-
              <lb/>
            tinued as is neceſſary; </s>
            <s xml:id="echoid-s337" xml:space="preserve">it is evident, that O D,
              <lb/>
            will be to O d, as this Perpendicular, which
              <lb/>
            meaſures the Diſtance from the Eye to the Plane
              <lb/>
            of the Figure, is to the Diſtance of the Eye
              <lb/>
            from the Perſpective Plane, which is meaſur’d
              <lb/>
            by the Part of the perpendicular, contain’d be-
              <lb/>
            tween the Eye and the perſpective Plane. </s>
            <s xml:id="echoid-s338" xml:space="preserve">Now
              <lb/>
            this before was manifeſt; </s>
            <s xml:id="echoid-s339" xml:space="preserve">viz. </s>
            <s xml:id="echoid-s340" xml:space="preserve">that
              <lb/>
            O D : </s>
            <s xml:id="echoid-s341" xml:space="preserve">O d : </s>
            <s xml:id="echoid-s342" xml:space="preserve">: </s>
            <s xml:id="echoid-s343" xml:space="preserve">A D : </s>
            <s xml:id="echoid-s344" xml:space="preserve">a d :</s>
            <s xml:id="echoid-s345" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s346" xml:space="preserve">Whence there is the ſame Proportion between
              <lb/>
            A d one of the Sides of the Figure, and A D its
              <lb/>
            Appearance, as the Theorem expreſſes. </s>
            <s xml:id="echoid-s347" xml:space="preserve">The
              <lb/>
            ſame may be demonſtrated of the other Sides of
              <lb/>
            the Figure. </s>
            <s xml:id="echoid-s348" xml:space="preserve">Which was to be demonſtrated.</s>
            <s xml:id="echoid-s349" xml:space="preserve"/>
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        <div xml:id="echoid-div36" type="section" level="1" n="15">
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            <emph style="sc">Corollary</emph>
          I.</head>
          <p style="it">
            <s xml:id="echoid-s350" xml:space="preserve">10. </s>
            <s xml:id="echoid-s351" xml:space="preserve">If from a Point in the Geometrical Plane, three
              <lb/>
            right Lines proceed, which are equal between them-
              <lb/>
            ſelves, and parallel to the perſpective Plane; </s>
            <s xml:id="echoid-s352" xml:space="preserve">the firſt
              <lb/>
            of which is in the Geometrical Plane, the ſecond ele-
              <lb/>
            vated Perpendicular to the firſt, and the third in-
              <lb/>
            clined to it; </s>
            <s xml:id="echoid-s353" xml:space="preserve">the Appearances of theſe three right
              <lb/>
            Lines are equal.</s>
            <s xml:id="echoid-s354" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s355" xml:space="preserve">This will appear clear enough in conſidering
              <lb/>
            the Lines as a Figure parallel to the perſpective
              <lb/>
            Plane; </s>
            <s xml:id="echoid-s356" xml:space="preserve">and ſo conſequently they will have the
              <lb/>
            ſame Proportion as their Appearances.</s>
            <s xml:id="echoid-s357" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s358" xml:space="preserve">Note, The firſt of the aforeſaid Lines is always
              <lb/>
            parallel to the baſe Line; </s>
            <s xml:id="echoid-s359" xml:space="preserve">and the ſecond, when
              <lb/>
            the perſpective Plane is perpendicular or </s>
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