Gravesande, Willem Jacob 's, An essay on perspective

Table of Notes

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            Point R, wherein it is cut by the Line m R per-
              <lb/>
            pendicular to m p. </s>
            <s xml:id="echoid-s1227" xml:space="preserve">Aſſume A S = e, OQ = x,
              <lb/>
            and M P = y. </s>
            <s xml:id="echoid-s1228" xml:space="preserve">Then in the ſimilar Triangles O q m,
              <lb/>
            and m R D, we have,
              <lb/>
            O q (x): </s>
            <s xml:id="echoid-s1229" xml:space="preserve">q m (e + y):</s>
            <s xml:id="echoid-s1230" xml:space="preserve">: m R (d x) R D: </s>
            <s xml:id="echoid-s1231" xml:space="preserve">({edx + ydx/x})
              <lb/>
            The ſimilar Triangles O p q and p T D, give,
              <lb/>
            oq (x): </s>
            <s xml:id="echoid-s1232" xml:space="preserve">q p (e):</s>
            <s xml:id="echoid-s1233" xml:space="preserve">: p P (d x): </s>
            <s xml:id="echoid-s1234" xml:space="preserve">P T ({edx/x})
              <lb/>
            P R is equal to y + dy, and if P T ({edx/x}) be added
              <lb/>
            to it, and then from the Aggregate be taken R D
              <lb/>
            ({edx + ydx/x}) we ſhall have T D = y + dy - {ydx/x}</s>
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            <s xml:id="echoid-s1235" xml:space="preserve">Now to find the Points of Interſection of the two
              <lb/>
            Circles, whoſe Radii are T D and P M, and Centers
              <lb/>
            diſtant from each other, by the Space T P, the Square
              <lb/>
            of T D leſs the Square of P M muſt be divided
              <note symbol="*" position="left" xlink:label="note-0094-01" xlink:href="note-0094-01a" xml:space="preserve">65.</note>
            Twice P T, and then half of P T muſt be taken there-
              <lb/>
            from, which may be here neglected, becauſe it is infinite-
              <lb/>
            ly ſmall in compariſon of the reſt; </s>
            <s xml:id="echoid-s1236" xml:space="preserve">and we ſhall have
              <lb/>
            {xydy/dex} - {yy/e} for the Part of the Line PM included be-
              <lb/>
            tween P and the Point wherein this Line is cut by a
              <lb/>
            Line joyning the two Points of Interſection of the two
              <lb/>
            Circles.</s>
            <s xml:id="echoid-s1237" xml:space="preserve"/>
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            <s xml:id="echoid-s1238" xml:space="preserve">Now before what I have here demonſtrated be ap-
              <lb/>
            ply’d to the Problem, we muſt obſerve, that if from
              <lb/>
            the Point M, a Line be drawn thro’ the Center B,
              <lb/>
            the Triangles M P C and m R M will be ſimilar;
              <lb/>
            </s>
            <s xml:id="echoid-s1239" xml:space="preserve">for the Angle m M P is the Exterior Angle of the
              <lb/>
            Triangle m R M, and the Angle m M C is a right
              <lb/>
            one. </s>
            <s xml:id="echoid-s1240" xml:space="preserve">And conſequently
              <lb/>
            m R (dx): </s>
            <s xml:id="echoid-s1241" xml:space="preserve">R M (d y):</s>
            <s xml:id="echoid-s1242" xml:space="preserve">: M P (y): </s>
            <s xml:id="echoid-s1243" xml:space="preserve">P C ({ydx/dx}).</s>
            <s xml:id="echoid-s1244" xml:space="preserve"/>
          </p>
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            <s xml:id="echoid-s1245" xml:space="preserve">Now if S A = s a in the 32d and 33d Figures
              <lb/>
              <note position="left" xlink:label="note-0094-02" xlink:href="note-0094-02a" xml:space="preserve">Fig. 32,
                <lb/>
              33.</note>
            be repreſented by e in this Computation; </s>
            <s xml:id="echoid-s1246" xml:space="preserve">as </s>
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Impressum