___ ___ ___
CD (2r - CD) = AD².CD² - 2r.CD + AD² = o.
[TEX: CD (2r - CD) = \overline{AD^2}.\overline{CD^2} - 2r.CD +
\overline{AD^2} = o.]
_________
/ ___
CD = r ± \ / r² - AD² .
\/
[TEX: CD = r ± \sqrt{r^2 - \overline{AD^2}}.]
It is evident that the lower sign alone suits our case, for d < r;
consequently,
_________
/ ___
CD = r - \ / r² - AD² . (9)
\/
[TEX: CD = r - \sqrt{r^2 - \overline{AD^2}}.]
Having obtained C, we put the instrument in the direction A B C. Then
each point of C F describes a circumference of the same center o.
16. If the distance of the points A and B were too great, then it
would be easy to determine a series of points belonging to the arc of
circumference sought (Fig. 4).
Being given C, the direction C I, and C I = R, on C I I lay off C E = d,
draw A E B perpendicularly, and calculate C A or A E. I shall have
___
d = (R - d) = AE²;
[TEX: d = (R - d) = \overline{AE^2};]
or, as absolute value,
__________
/
A E = \ / d (R - d) . (10)
\/
[TEX: AE = \sqrt{d (R-d)}]
The instrument being arranged according to A C B, I prolong C B and take
B C' = B C, when C' will be one of the points sought. It will be readily
understood how, by repeating the above operations, but by varying the
value of d, we obtain the other intermediate points, and how we may
continue the operation to the right of C' with the process pointed out.
17. If the three rulers were three arcs of a large circle of a sphere,
the instrument might serve for drawing the meridians on such sphere.
18. If we imagine, instead of three axes placed in one plane and
converging at one point, a system of four axes also converging in one
point, but situated in any manner whatever in space, and if we rest
three of them against three fixed points, we shall be able to solve in
space problems analogous to those that have just been solved in a plane.
If we had, for example, to draw a spherical vault whose center was
inaccessible, we might adopt the same method.--_Le Génie Civil_.
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