Gravesande, Willem Jacob 's - An essay on perspective

Gravesande, Willem Jacob 's, An essay on perspective

Table of contents

[51.] Remarks.

[52.] Corollary.

[53.] Problem II.

[54.] Remark.

[55.] Problem III.

[56.] Method. II.

[57.] Problem IV.

[58.] Example I.

[59.] Example II.

[60.] Remarks.

[61.] Example III. 48. To throw a circle into Perſpective.

[62.] Remarks.

[63.] Prob. V. 50. To find the Repreſentation of a Point, elevated above the Geometrical Planc.

[64.] Operation.

[65.] Demonstration.

[66.] Prob. VI. 52. To throm a Pyramid, or Cone, into Perſpective.

[67.] 53. To determine the viſible Part of the Baſe of a Cone.

[68.] Operation.

[69.] Demonstration.

[70.] Remarks.

[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.

8739on PERSPECTIVE. the Station Point, and A the Point wherein the
Perpendicular meets the Geometrical Plane.

Operation,
Without Compaſſes.

Having firſt found the Perſpective a of 1131. Point A, draw the Line A S cutting the Baſe
Line in E, through which Point E draw the
Line Ea; then from the Point B draw a Line
B a to the Point a, cutting the Horizontal Line
in F. Again through F draw a Line to the
Point L, cutting E a in I; and a I is the Repre-
ſentation ſought.

Demonstration.

To prove this, let G N be a Perpendicular to
the Baſe Line drawn from the Point G, wherein
the ſaid Baſe Line is cut by the Line B F; alſo
let G D be equal to the Perpendicular whoſe Ap-
pearance is ſought, and a H parallel to the Baſe
Line.

It is plain that the Perſpective of E A is
E a: But E A paſſes through the Station Point;
and conſequently its Repreſentation is 2241. dicular to the Baſe Line; therefore we are 3356. to prove, that a I is equal to a H.

Now the Triangles B G C and B F M are ſimi-
lar; and ſo

B C : B M : : B G: B F.

But B M by Conſtruction is the double of B C;
whence B F is alſo the double of B G, which,
conſequently, is equal to G F.

Becauſe the Triangles F G N and F B L are
ſimilar, therefore

F G : F B : : G N : B L.

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