Gravesande, Willem Jacob 's, An essay on perspective

Table of contents

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[71.] Problem VII. 55. To find the Perſpective of a Line, perpendicular to the Geometrical Plane.
[72.] Operation.
[73.] Demonstration.
[74.] Method II.
[75.] Demonstration.
[76.] Method III.
[77.] Operation, Without Compaſſes.
[78.] Demonstration.
[79.] Scholium.
[80.] Corollary.
[81.] Problem VIII.
[82.] To do this another Way.
[83.] Demonstration.
[84.] Problem IX.
[85.] Problem X.
[86.] Demonstration.
[87.] EG: EN:: GY: NM.
[88.] Definition.
[89.] Problem XI.
[90.] Lemma.
[91.] Demonstration.
[92.] Remarks.
[93.] Problem IX.
[94.] Operation.
[95.] Demonstration.
[96.] Problem X.
[97.] Operation.
[98.] Demonstration.
[99.] Remarks.
[100.] Method II. 70. By the accidental Point of inclin’d Lines.
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            the Circular Baſe of the Cone will be deter-
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            min’d.</s>
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        <div xml:id="echoid-div126" type="section" level="1" n="69">
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            <emph style="sc">Demonstration</emph>
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            <s xml:id="echoid-s871" xml:space="preserve">To prove this, draw the Lines B C and L F, cut-
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            ting the Line A S in the Points N and M; </s>
            <s xml:id="echoid-s872" xml:space="preserve">and make
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            the Line G n equal to A N, and draw the Line
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            n D m. </s>
            <s xml:id="echoid-s873" xml:space="preserve">It is now manifeſt, that if the Cone be
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            continued out above its Vertex, (that is, if the oppo-
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            ſite Cone be form’d) it will cut the Horizontal Plane
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            in a Circle equal to B E C, whoſe Seat will be BEC:
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            </s>
            <s xml:id="echoid-s874" xml:space="preserve">So that the Point S, in reſpect of B E C, is in the
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            ſame Situation as the Eye hath, with reſpect to the
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            Circle form’d in the Horizontal Plane, by the Conti-
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            nuation of the Cone. </s>
            <s xml:id="echoid-s875" xml:space="preserve">Whence it follows, that B C
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            is the Seat of the viſible Portion of that Circle. </s>
            <s xml:id="echoid-s876" xml:space="preserve">For,
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            by Conſtruction, B and C are the Points of Contact
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            of the Tangents to the Circle B E C, which paſs
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            thro’ the Point S; </s>
            <s xml:id="echoid-s877" xml:space="preserve">becauſe the Angle ABS, which
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            is in a Semicircle, is a right one.</s>
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            <s xml:id="echoid-s879" xml:space="preserve">Now, if a Plane be conceiv’d, as paſſing thro’ ſome
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            Points in the Horizontal Plane, whoſe Seats are
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            B and C, and which cuts the two oppoſite Cones
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            thro’ their Vertex; </s>
            <s xml:id="echoid-s880" xml:space="preserve">it is evident, that this Plane
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            continued, will cut the Geometrical Plane in a Line
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            parallel to B N C; </s>
            <s xml:id="echoid-s881" xml:space="preserve">and that this Line upon the
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            ſaid Plane, will determine the viſible Part of the
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            Cone’s Baſe. </s>
            <s xml:id="echoid-s882" xml:space="preserve">So, ſince G n was made equal to
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            A N, we have only to prove, that P m is equal to
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            A M: </s>
            <s xml:id="echoid-s883" xml:space="preserve">For, it follows from thence, that L M F is
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            the Common Section of the Geometrical Plane, and
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            the Plane which we have here imagin’d.</s>
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            <s xml:id="echoid-s885" xml:space="preserve">The Triangles D Q P and G H D are ſimilar, whence
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            D G: </s>
            <s xml:id="echoid-s886" xml:space="preserve">D P:</s>
            <s xml:id="echoid-s887" xml:space="preserve">: G H: </s>
            <s xml:id="echoid-s888" xml:space="preserve">P Q.</s>
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