Gravesande, Willem Jacob 's
,
An essay on perspective
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Point R, wherein it is cut by the Line m R per-
<
lb
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pendicular to m p. </
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<
s
xml:id
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echoid-s1227
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xml:space
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preserve
">Aſſume A S = e, OQ = x,
<
lb
/>
and M P = y. </
s
>
<
s
xml:id
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echoid-s1228
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xml:space
="
preserve
">Then in the ſimilar Triangles O q m,
<
lb
/>
and m R D, we have,
<
lb
/>
O q (x): </
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>
<
s
xml:id
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echoid-s1229
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xml:space
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">q m (e + y):</
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<
s
xml:id
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xml:space
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">: m R (d x) R D: </
s
>
<
s
xml:id
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echoid-s1231
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xml:space
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">({edx + ydx/x})
<
lb
/>
The ſimilar Triangles O p q and p T D, give,
<
lb
/>
oq (x): </
s
>
<
s
xml:id
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echoid-s1232
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xml:space
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">q p (e):</
s
>
<
s
xml:id
="
echoid-s1233
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xml:space
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preserve
">: p P (d x): </
s
>
<
s
xml:id
="
echoid-s1234
"
xml:space
="
preserve
">P T ({edx/x})
<
lb
/>
P R is equal to y + dy, and if P T ({edx/x}) be added
<
lb
/>
to it, and then from the Aggregate be taken R D
<
lb
/>
({edx + ydx/x}) we ſhall have T D = y + dy - {ydx/x}</
s
>
</
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<
s
xml:id
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xml:space
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">Now to find the Points of Interſection of the two
<
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/>
Circles, whoſe Radii are T D and P M, and Centers
<
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diſtant from each other, by the Space T P, the Square
<
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/>
of T D leſs the Square of P M muſt be divided
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xlink:label
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note-0094-01a
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xml:space
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">65.</
note
>
Twice P T, and then half of P T muſt be taken there-
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from, which may be here neglected, becauſe it is infinite-
<
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/>
ly ſmall in compariſon of the reſt; </
s
>
<
s
xml:id
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xml:space
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preserve
">and we ſhall have
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{xydy/dex} - {yy/e} for the Part of the Line PM included be-
<
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tween P and the Point wherein this Line is cut by a
<
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Line joyning the two Points of Interſection of the two
<
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Circles.</
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xml:space
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<
s
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xml:space
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">Now before what I have here demonſtrated be ap-
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ply’d to the Problem, we muſt obſerve, that if from
<
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the Point M, a Line be drawn thro’ the Center B,
<
lb
/>
the Triangles M P C and m R M will be ſimilar;
<
lb
/>
</
s
>
<
s
xml:id
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echoid-s1239
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xml:space
="
preserve
">for the Angle m M P is the Exterior Angle of the
<
lb
/>
Triangle m R M, and the Angle m M C is a right
<
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/>
one. </
s
>
<
s
xml:id
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echoid-s1240
"
xml:space
="
preserve
">And conſequently
<
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/>
m R (dx): </
s
>
<
s
xml:id
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xml:space
="
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">R M (d y):</
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>
<
s
xml:id
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xml:space
="
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">: M P (y): </
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>
<
s
xml:id
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xml:space
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">P C ({ydx/dx}).</
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<
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xml:space
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</
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s
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xml:space
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preserve
">Now if S A = s a in the 32d and 33d Figures
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xlink:href
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xml:space
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">Fig. 32,
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33.</
note
>
be repreſented by e in this Computation; </
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xml:space
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">as </
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