Simple Curve Problems

Problem The angle of intersection of a circular curve is 45° 30' and its radius is 198.17 m. PC is at Sta. 0 + 700. Compute the right angle offset from Sta. 0 + 736.58 on the curve to tangent through PC.

Problem The angle of intersection of a circular curve is 36° 30'. Compute the radius if the external distance is 12.02 m. Cos(1/2)I=R/(R+E)

Length of curve from PC to A: s=736.58−700s=736.58−700 s=36.58 ms=36.58 m

cos18∘15′=R/R+12.02cos18∘15′=RR+ 12.02 Rcos18∘15′ +12.02cos18∘15′=RRcos18∘15′ +12.02cos18∘15′=R Angle subtended by arc s from the center of the curve: sθ=2πR360∘sθ=2πR360∘

R−Rcos18∘15′=12.02cos18∘15′R−Rco s18∘15′=12.02cos18∘15′ R(1−cos18∘15′)=12.02cos18∘15′R(1− cos18∘15′)=12.02cos18∘15′

36.58θ=2π(198.17)360∘36.58θ=2π(19 8.17)360∘

R=12.02cos18∘15′1−cos18∘15′

θ=10.58∘θ=10.58∘

R=12.02cos18∘15′/1−cos18∘15′ R=226.94 m

Length of offset x: cosθ=R−xRcosθ=R−xR x=R−Rcosθ=198.17−198.17cos10.58 ∘x=R−Rcosθ=198.17−198.17cos10.5 8∘ x=3.37 mx=3.37 m

Given the following elements of a circular curve: middle ordinate = 2 m; length of long chord = 70 m. Find its degree of curve, use arc basis. Apply Pythagorean theorem to find the radius: (R−2)2+352=R2(R−2)2+352=R2

(R2−4R+4)+1225=R2(R2−4R+4)+12 25=R2 4R=12294R=1229 R=307.25 mR=307.25 m

Degree of curve (arc basis): 20D=2πR360∘20D=2πR360∘ 20D=2π(307.25)360∘20D=2π(307.25) 360∘ D=3.7∘