AOP = 180° - (118° + 21°27.7') = 40°32.3' arc AP = 2ir (800)(40°32.3')/360° = 566.0 ft
(172.52 m); station of P = (22 + 00) + (5 + 66) = 27 + 66.
REALIGNMENT OF CIRCULAR CURVE BY
DISPLACEMENT OF FORWARD TANGENT
In Fig. 15, the horizontal circular curve AB has a radius of 720 ft (219.5 m) and an inter-
section angle of 126°. The curve is to be realigned by rotating the forward tangent
through an angle of 22° to the new position VB while maintaining the PT at B. Compute
the radius, and locate the PC of the new curve.
Calculation Procedure:- Find the tangent distance
of the new curve
Solve triangle B V V to find the tangent
distance of the new curve and the loca-
tion of V. Thus, A' = 126° - 22° =
104°; VB = 720 tan 63° = 1413.1 ft
(430.71 m). By the law of sines, VB =
14131 sin 126°/sin 104° = 1178.2 ft
(359.12 m); VV= 1413.1 sin 22°/sin
104° = 545.6 ft (166.30m). - Compute the radius R'
By Eq. 19, R' = 1178.2 cot 52° = 920.5
ft (280.57 m). - Determine the station
ofA
Thus, AV= VB = 1413.1 ft (403.71 m);
FIGURE 15. Displacement of forward tan- A' V = VB = 1178.2 ft (359.12 m); A 'A
gent. = A' V + V V-AV=310.7 ft (94.7Om);
station of new PC = (34 + 41) - (3 +
10.7) = 31+30.3. - Verify the foregoing results
Draw the long chords AB and A'B. Then apply the computed value of R' to solve triangle
BA'A and thereby fmdA'A. By Eq. 20, A'B = 2R' sin^1 X 2 A' = 1450.7 ft (442.17 m); AA'B =
(^1) M' = 52°; A'AB = 180° - (^1) M = 117°; ABA (^9) = 180° - (52° + 117°) = 11°. By the law of
sines, A'A = 1450.7 sin 1 l°/sin 117° = 310.7 ft (94.70 m). This is acceptable.
CHARACTERISTICS OFA
COMPOUND CURVE
The tangents to a horizontal curve intersect at an angle of 68°22'. To fit the curve to the
terrain, it is necessary to use a compound curve having tangent lengths of 955 ft (291.1 m)