Gravesande, Willem Jacob 's, An essay on perspective

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[41] page 64Plate. 18.d v
[42] Fig. 41E b a G F H C B A D
[43] Fig. 42G Q A M I S E H T B L Z F P a X Y N C D R O
[44] page 68Plate. 19Fig. 43B D E a G H I C F L
[45] Fig. 44O V X S H I T
[46] Fig. 45Q F V X S a H B C D E L M P T A
[47] page 72Plate. 20Fig. 46V I X a E M P A T
[48] Fig. 47V F X a Q G H D N A T R
[49] Page 34.Plate. 21Fig. 48F O D X S b a G N A E T B P C
[50] Fig. 49H F O D G X a M N L R Q
[51] Page 36Plate 22Fig. 50O R E G N S M
[52] Fig. 51I H T a X
[53] Fig. 52C D X I H G a F E L b T
[54] Fig. 53H I F T x d X L B C
[55] page 64.Plate 23.Fig. 54O M P Q t A X x Q R N
[56] Fig. 55G F b T L a
[57] Fig. 56I F a X b E T C P
[58] page 66.Plate. 24.Fig. 57E A Z C P B
[59] Fig. 58F O D I a b
[60] Fig. 59F E Z C A B
[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
[62] Plate 26Fig. 61O I F T N S Q S H E R M A
[63] Fig. 62C D S Q L C D R P H
[64] page 96.Plate. 27Fig. 63D E C F M H I G P A Q N
[65] Fig. 64X S D E T C R L F H I G P M B O V Q N
[66] page 98.Plate. 28Fig. 65L M F G D H C E I A B
[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
[68] page 100Plate. 29Fig. 675 6p 7 8 9 10 S V VI VII VIII IX X o XI ll l
[69] Fig. 68c P G e o Q
[70] Fig. 69P c G o e Q
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10248An ESSAY might be done) the Point on the Semicircle
h z m, as far as the Parallels (as 9 m) are uſeful:
For when theſe Parallels are uſeleſs, the Point
q will fall beyond the Point m:
But then the
Perſpective of the Torus is entirely drawn alrea-
dy, if thoſe Parallels were firſt begun to be drawn
near to 6 3 z, and the others continually going
from it.
In order to demonſtrate this Problem, the fol-
lowing Lemma is neceſſary.
Lemma.
65. If two Circles C D H E and D E F L cut
11Fig. 34. each other, thro’ whoſe Centers C and B the Line
C L paſſes, and D E joyns their Interſections;
then, if the Radius A C or A H be called a, and
B F or BL, b, and the Diſtance A B between the
two Centers c, I ſay A G is equal to {bb—aa/ec}—{1/2}C.
Demonstration.
Let us call A G, x, and G D or G E, y.
Then by the Property of the Circle, if y be conceiv’d
as an Ordinate of the Circle, C D H;
yy=aa—xx.
And if it be likewiſe conſider’d as an Ordinate of
the Circle F D L, yy=bb—cc—2cx—xx:
Whence
aa—xx=bb—cc—2cx—xx, and ſo 2cx=bb
—aa—cc;
and dividing each Side of this laſt Equa-
tion by 2c, we have a x={bb—aa/2c}{1/2} c.
Which was
to be Demonſtrated.

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