Gravesande, Willem Jacob 's, An essay on perspective

Table of figures

< >
[41] page 64Plate. 18.d v
Page: 135
[42] Fig. 41E b a G F H C B A D
Page: 135
[43] Fig. 42G Q A M I S E H T B L Z F P a X Y N C D R O
Page: 135
[44] page 68Plate. 19Fig. 43B D E a G H I C F L
Page: 142
[45] Fig. 44O V X S H I T
Page: 142
[46] Fig. 45Q F V X S a H B C D E L M P T A
Page: 142
[47] page 72Plate. 20Fig. 46V I X a E M P A T
Page: 149
[48] Fig. 47V F X a Q G H D N A T R
Page: 149
[49] Page 34.Plate. 21Fig. 48F O D X S b a G N A E T B P C
Page: 154
[50] Fig. 49H F O D G X a M N L R Q
Page: 154
[51] Page 36Plate 22Fig. 50O R E G N S M
Page: 159
[52] Fig. 51I H T a X
Page: 159
[53] Fig. 52C D X I H G a F E L b T
Page: 159
[54] Fig. 53H I F T x d X L B C
Page: 159
[55] page 64.Plate 23.Fig. 54O M P Q t A X x Q R N
Page: 170
[56] Fig. 55G F b T L a
Page: 170
[57] Fig. 56I F a X b E T C P
Page: 170
[58] page 66.Plate. 24.Fig. 57E A Z C P B
Page: 175
[59] Fig. 58F O D I a b
Page: 175
[60] Fig. 59F E Z C A B
Page: 175
[61] page 88.Plate. 25.Fig. 60O G F f Z L R P D I T S M a Q E R H N A C B
Page: 180
[62] Plate 26Fig. 61O I F T N S Q S H E R M A
Page: 187
[63] Fig. 62C D S Q L C D R P H
Page: 187
[64] page 96.Plate. 27Fig. 63D E C F M H I G P A Q N
Page: 194
[65] Fig. 64X S D E T C R L F H I G P M B O V Q N
Page: 194
[66] page 98.Plate. 28Fig. 65L M F G D H C E I A B
Page: 197
[67] Fig. 66A B VII VIII IV V H C VI VI P V VII IV S VIII E O I III II I XII XIX IX F D
Page: 197
[68] page 100Plate. 29Fig. 675 6p 7 8 9 10 S V VI VII VIII IX X o XI ll l
Page: 204
[69] Fig. 68c P G e o Q
Page: 204
[70] Fig. 69P c G o e Q
Page: 204
< >
page |< < (20) of 237 > >|
4420An ESSAY
Now, it is evident, that to prove the 1127. pearance of A is in the Line C H, we need but
demonſtrate that O D is parallel to A E; which
may be done thus:
Becauſe the Triangles O G V, and A B F, are
ſimilar.
A F: A B: : O G: O V:
altern.
A F: O G: : A B: O V:
Divid. and altern. the firſt
Proportion.
AF—AB (=CF): O G—O V=HG: :AB: OV.
But becauſe the Triangles E C F, H G D are
ſimilar.
C F: H G : : E F : G D.
Now, by obſerving the two laſt Proportions of
the other two Triangles,
E F: G D: : A F: O G,
And the Angle A F E, being equal to the Angle
O G D, the Triangles A E F and O D G are
ſimilar; and therefore A E is parallel to O D:
Which was to be demonſtrated.
After the ſame manner we prove, that the
Appearance of the Point A is in the Line L I,
and conſequently is in the Interſection of this
Line and HC.
Remark.
Altho’ this Method appears more difficult than
the precedent one, as to the Geometrical Conſi-
deration thereof, yet the Operation is eaſier, if
the Points are not too far diſtant from the Baſe
Line: For Lines may well enough be drawn by
Gueſs, or Sight only, to touch Circles, and Cir-
cles to touch Lines.

Text layer

  • Dictionary

Text normalization

  • Original

Search


  • Exact
  • All forms
  • Fulltext index
  • Morphological index
Impressum