[TEX: CI = \frac{\overline{AC^2}}{CD'}] 11. In order to avoid all calculation, we may proceed thus: If I wish to arrange the instrument so that C I represents a given quantity (§ 8), I take (Fig. 7) the length Ci = CI/n, where n is any entire number whatever. [Illustration: Fig. 7.] In other terms, Ci is the reduction to the scale of CI. I describe the circumference C b i a, and arrange the instrument as seen in the figure, and measure the length C b. It is visible that C i 1 C b C d ----- = --- = ----- = ------; then C B = n.C b (5) C I n C B C D CD = n.C d; (6) and, consequently, the position of the needles which are found at A and B are determined. 12. The question treated in § 10, then, is simply solved. In fact, on describing the circumference C b i a with any radius whatever, I shall have C B n = -----; (7) c b and, consequently, C I = n.C i (8) 13. As may be seen, the instrument composed of three firmly united rulers is the simplest of all and easy to use. Any one can construct it for himself with a piece of cardboard, and give the angle 2 [alpha] the value that he thinks most suitable for each application. The greater 2 [alpha] is, the shorter is the distance at which we should put the needles for a given point of meeting. 14 The jointed instrument may be constructed as shown in Figs 8, 9, and 10. The three pieces, A. B, and C, united by a pivot, O, in which there is a small hole, are of brass or other metal. Rulers may be easily procured of any length whatever. The instrument is Y-shaped. In the particular case in which [alpha] = 180° it becomes T-shaped, and serves to draw parallel lines. [Illustration: Fig. 8, Fig. 9, Fig. 10] 15. The instrument may be used likewise, as we have seen, to draw arcs of circles of the diameter C I or of the radius A O = r, whose center o falls outside the paper. The pencil will be rested on C. We may operate as follows (Fig. 2): Being given the direction of the radii A O and B O, or, what amounts to the same thing, the tangents to the curve at the given points, A and B to be united, we draw the line A D and raise at its center the perpendicular D C, which, prolonged, passes necessarily through the center. It is necessary to calculate the length C D. We shall have ___ ___ ___ 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}}.]