Gravesande, Willem Jacob 's, An essay on perspective

Table of contents

< >
[171.] Problem V.
Page: 173 (86)
[172.] Remarks.
Page: 177 (87)
[173.] CHAP. VIII. Of mechanically ſhortning the Operations of Perſpective. 1. WHEN the perſpective Plane is ſup-pos’d perpendicular or upright. Problem I.
Page: 177 (87)
[174.] Operation.
Page: 178 (88)
[175.] Method II.
Page: 178 (88)
[176.] Prob. II.
Page: 182 (89)
[177.] Operation.
Page: 182 (89)
[178.] Method II.
Page: 183 (90)
[179.] Method III.
Page: 183 (90)
[180.] The Demonſtration of the two laſt Ways.
Page: 184 (91)
[181.] II. When the Perſpective Plane is inclined. Prob. III.
Page: 185 (92)
[182.] Prob. IV.
Page: 185 (92)
[183.] Remarks.
Page: 185 (92)
[184.] III. When the Perſpective Plane is Parallel or Horizontal. Prob. V.
Page: 189 (93)
[185.] Operation.
Page: 189 (93)
[186.] Demonstration.
Page: 190 (94)
[187.] Remarks.
Page: 190 (94)
[188.] Prob. VI.
Page: 190 (94)
[189.] Demonstration.
Page: 191 (95)
[190.] Prob. VII.
Page: 192 (96)
[191.] CHAP. IX.
Page: 199 (97)
[192.] Prob. I. 122. To draw Vertical Dials.
Page: 200 (98)
[193.] Demonstration.
Page: 201 (99)
[194.] Remark.
Page: 201 (99)
[195.] Prob. II. 123. To draw inclining Dials.
Page: 201 (99)
[196.] The Uſe of the Camera Obscura in Deſigning. Advertisement.
Page: 206 (101)
[197.] The Uſe of the Camera Obscura in Deſigning. Definition.
Page: 208 (103)
[198.] Theorem I.
Page: 208 (103)
[199.] Theorem II.
Page: 209 (104)
[200.] The Deſcription of the Firſt Machine.
Page: 209 (104)
< >
page |< < (95) of 237 > >|
    <echo version="1.0RC">
      <text xml:lang="en" type="free">
        <div xml:id="echoid-div345" type="section" level="1" n="188">
          <p>
            <s xml:id="echoid-s2133" xml:space="preserve">
              <pb o="95" file="0165" n="191" rhead="on PERSPECTIVE."/>
            lar drawn from the Eye to the Geometrical
              <lb/>
            Plane, meets the ſaid Plane; </s>
            <s xml:id="echoid-s2134" xml:space="preserve">and T the Per-
              <lb/>
            ſpective of that Point, found by the aforegoing
              <lb/>
            Problem. </s>
            <s xml:id="echoid-s2135" xml:space="preserve">Now make F L and C R, equal to the
              <lb/>
            Length of the given Lines; </s>
            <s xml:id="echoid-s2136" xml:space="preserve">and about the
              <lb/>
            Points R and S, as Centers, with the Radius
              <lb/>
            M N or P Q, deſcribe two Arcs cutting the
              <lb/>
            Perpendiculars H D and G H, in the Points
              <lb/>
            X and S: </s>
            <s xml:id="echoid-s2137" xml:space="preserve">Then fix the Extremities of the
              <lb/>
            Threads, which before were faſten’d to the
              <lb/>
            Points F and E, to the Points L and S; </s>
            <s xml:id="echoid-s2138" xml:space="preserve">and alſo
              <lb/>
            the Extremities of thoſe two Threads, which
              <lb/>
            were before faſten’d to the Points C and D, in
              <lb/>
            the Points R and X: </s>
            <s xml:id="echoid-s2139" xml:space="preserve">Then moving the two
              <lb/>
            Rulers, until the Threads S P and X M, croſs
              <lb/>
            each other in the Point T, mark the Point O,
              <lb/>
            wherein the two other Threads croſs one another,
              <lb/>
            through which, and the Point B, draw the
              <lb/>
            indefinite Line B O V: </s>
            <s xml:id="echoid-s2140" xml:space="preserve">this being done, tranſ-
              <lb/>
            poſe the Lines of the Geometrical Plane, ſo that
              <lb/>
            the Point B, coincides with the Point O, and
              <lb/>
            the Line B O, with O V. </s>
            <s xml:id="echoid-s2141" xml:space="preserve">And by the precedent
              <lb/>
            Problem, find the Appearances of the Feet of
              <lb/>
            the Perpendiculars, and you will have the Re-
              <lb/>
            preſentations of their Extremities.</s>
            <s xml:id="echoid-s2142" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div347" type="section" level="1" n="189">
          <head xml:id="echoid-head203" xml:space="preserve">
            <emph style="sc">Demonstration</emph>
          .</head>
          <p>
            <s xml:id="echoid-s2143" xml:space="preserve">If a Plane be ſuppoſed to paſs through the
              <lb/>
            Extremities of the Perpendiculars, it will be
              <lb/>
            parallel to the Geometrical Plane, and conſe-
              <lb/>
            quently, likewiſe to the perſpective Plane, be-
              <lb/>
            cauſe all the Perpendiculars are ſuppoſed equal.
              <lb/>
            </s>
            <s xml:id="echoid-s2144" xml:space="preserve">But the Figure formed by the Extremities of
              <lb/>
            the Perpendiculars in the ſaid Plane, is ſimilar
              <lb/>
            and equal to that form’d by their Feet in the
              <lb/>
            Geometrical Plane: </s>
            <s xml:id="echoid-s2145" xml:space="preserve">and therefore, the </s>
          </p>
        </div>
      </text>
    </echo>
Impressum