Gravesande, Willem Jacob 's, An essay on perspective

Table of Notes

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          <p>
            <s xml:id="echoid-s2390" xml:space="preserve">
              <pb o="109" file="0185" n="214" rhead="on PERSPECTIVE."/>
            ved backwards or forwards, or elſe the Looking-
              <lb/>
            glaſs raiſed or lower’d, until the Rays proceed-
              <lb/>
            ing from the Statue may be reflected by the Mir-
              <lb/>
            rour upon the Convex Glaſs. </s>
            <s xml:id="echoid-s2391" xml:space="preserve">When theſe Alte-
              <lb/>
            rations of the Box, or Mirrour, are not ſufficient to
              <lb/>
            throw the Rays upon the Convex Glafs, the whole
              <lb/>
            Machine muſt be removed backwards or for-
              <lb/>
            wards.</s>
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        <div xml:id="echoid-div373" type="section" level="1" n="203">
          <head xml:id="echoid-head222" xml:space="preserve">
            <emph style="sc">Demonstration</emph>
          .</head>
          <head xml:id="echoid-head223" style="it" xml:space="preserve">Concerning the before-mention’d Inclination of the
            <lb/>
          Mirrours.</head>
          <p>
            <s xml:id="echoid-s2393" xml:space="preserve">19. </s>
            <s xml:id="echoid-s2394" xml:space="preserve">In order to demonſtrate, that the Mirrour
              <lb/>
            L hath been conveniently inclin’d, we need on-
              <lb/>
            ly prove, that the reflected Rays fall upon the
              <lb/>
            Table A under the ſame Angle, as the direct
              <lb/>
            Rays do upon a Plane, having the ſame Situation
              <lb/>
            as one would give to the Picture.</s>
            <s xml:id="echoid-s2395" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s2396" xml:space="preserve">Now let A B be a Ray falling from a Point of
              <lb/>
              <note position="right" xlink:label="note-0185-01" xlink:href="note-0185-01a" xml:space="preserve">Fig. 71.</note>
            ſome Object upon the Mirrour G H, and from
              <lb/>
            thence is reflected in the Point a upon the Table
              <lb/>
            of the Machine: </s>
            <s xml:id="echoid-s2397" xml:space="preserve">We are to demonſtrate, that if
              <lb/>
            the Line D I be drawn, making an Angle with
              <lb/>
            FE equal to the Inclination of the Picture; </s>
            <s xml:id="echoid-s2398" xml:space="preserve">that
              <lb/>
            is, if the Angle DIE be the double of the
              <note symbol="*" position="right" xlink:label="note-0185-02" xlink:href="note-0185-02a" xml:space="preserve">15, 16.</note>
            D F I; </s>
            <s xml:id="echoid-s2399" xml:space="preserve">I ſay, we are to demonſtrate, that the
              <lb/>
            Angle B a f is equal to the Angle BCD.</s>
            <s xml:id="echoid-s2400" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s2401" xml:space="preserve">The Angle DIE, by Conſtruction, is the double
              <lb/>
            of the Angle DFI; </s>
            <s xml:id="echoid-s2402" xml:space="preserve">and conſequently this laſt Angle
              <lb/>
            is equal to the Angle I D F; </s>
            <s xml:id="echoid-s2403" xml:space="preserve">and ſince the Angle
              <lb/>
            of Incidence C B D is equal to the Angle of Re-
              <lb/>
            flection a B F, the Triangle BCD is ſimilar to
              <lb/>
            the Triangle F a B: </s>
            <s xml:id="echoid-s2404" xml:space="preserve">Whence it follows, that the
              <lb/>
            Angle Ba F is equal to the Angle BCD. </s>
            <s xml:id="echoid-s2405" xml:space="preserve">Which
              <lb/>
            was to be demonſtrated.</s>
            <s xml:id="echoid-s2406" xml:space="preserve"/>
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Impressum