- Calculate the degree of curve in the present instance
Part c. Since the subtended central angle is directly proportional to its arc length, Z)/100 =
360/(2TrK); therefore,
D = 18,000/Trtf = 5729.58/tf degrees (24)Then, D = 5729.58/1200 = 4.7747° = 4°46.5'.- Repeat the calculations in steps 4, 5, and 6
INTERSECTION OF CIRCULAR CURVE
AND STRAIGHT LINE
In Fig. 14, MN represents a straight railroad spur that intersects the curved highway route
AB. Distances on the route are measured along the arc. Applying the recorded data, deter-
mine the station of the intersection point P.
Calculation Procedure:- Apply trigonometric relationships to determine three elements
in triangle ONP
Draw line OP. The problem resolves itself into the calculation of the central angle AOP,
and this may be readily found by solving the oblique triangle OTVP. Applying trigonomet-
ric relationships gives AV= T= 800 tan 54° = 1101.1 ft (335.62 m); AM= 1101.1 - 220 =
881.1 ft (268.56 m); AN=AM tan 28° = 468.5 ft (142.80 m); ON= 800 - 468.5 = 331.5 ft
(101.04 m); OP = 800 ft (243.84 m); ONP = 90° + 28° = 118°. - Establish the station of P
Solve triangle ONP to find the central angle; then calculate arc AP and establish the sta-
tion of P. By the law of sines, sin OPN= sin ONP(ON)IOP; therefore, OPN= 21°27.7';
FIGURE 14. Intersection of curve and straight line.