1. ADVANCING TOWARDS THE INTERSECTION OF TANGENTS. Given the angle of divergence, N, the initial P. C. at G, the distance GH, and the radii R, r, to find the central angles A and B.

GK=CG X cos. N = = R cos. N.
GL=GH × sin. N.

GK-GL= L K or EF, C F being drawn parallel to L E.
Cos. BDF÷DC = (r+ EF) ÷ (R+ r).

Angle GCK=90° - N; angle DCF 90°.

- B.

Angle AGCK - DCF (90° - N) (90° — B) = B

=

LK or EFGK-GL= 1,310 — 354 = 956.

2. RECEDING FROM THE INTERSECTION OF TANGENTS.

Given the angle of divergence, N, the initial P. C. at G, the distance G H, and the radii R, r, to find the central angles A and B.

LC or EF = KC × cos. N, the line C F being drawn parallel to LE.

Cos. B
Angle A manifestly

TO SHIFT A P. R. C. SO THAT THE TERMINAL TANGENT SHALL MERGE IN A GIVEN TANGENT PARALLEL THERETO.

Given the reversed curve E F G, ending in tangent GV: to find the angle of retreat, A, on the first branch, and the angle C of the second branch, ending in tangent HT, parallel to GV.

D, perpendicular to the terminal

In the figure, draw L K parallel to G V, and passing through centre of first branch.

.. (R+r) × cos. C― (R+r) × cos. B = D.
.. (R+r) X cos. C = (R+r) × cos. B + D.
... Cos. C= [(R+ r) × cos. B + D] ÷ (R+ r).
(90° C) (90° — B)=B-C.

A: =

Example.

R=1,433, r=819, B=34° 20', D=94.

Cos. C [(R+ r) cos. B + D] ÷ (R+r).

XLV.

TO PASS A CURVE THROUGH A FIXED POINT, THE ANGLE OF INTERSECTION BEING GIVEN.

Given the intersection angle, A, of two tangents, to find the radius, R, of a curve which shall pass through a point, C; the position of said point, with reference to the tangents or the point of intersection, being known.

1. By what data soever point C is located, they may be commuted by simple processes to the form shown in the figure; namely, the ordinate BC and the distance IC to apex. Call the angle BIC a, and complete the triangle I CO.