Simple-Curve-Lecture PDF

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1. The angle of intersection of a simple curve is 8°24’. The station of PI is 64 + 27.46 and that terrain conditions require the minimum radius permitted by the specifications is 2864.79 ft (arc definition). Calculate the stationing of PC and PT, and the external and middle ordinate distances of the curve.

2. The angle of intersection of a curve will be 8°24’. The station of PI is at 6 + 427.464. The terrain conditions require that the minimum radius of the curve be 900 m. Calculate the station of PC and PT, also E, M, LC and other elements of the simple curve.

3. 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.

4. The angle of intersection of a circular curve is 36°30’. Compute the radius if the external distance is 12.02 m.

5. The radius of a simple curve is equal to the length of the long chord. What is the angle of intersection?

6. Given the following elements of a circular curve: middle ordinate = 2 m, length of long chord = 70 m. Find the degree of the curve using arc basis.

8. A simple horizontal curve of radius 125 m connects two tangents that intersect at an angle of 105°40’. Compute the parts of the curve, including T, C, LC, E and M. Also, what is the degree of the curve?

9. From example no. 8, if the station of the PI is 22 + 050, what are the stations of the PC and PT?

10.The radius of a simple curve is twice its tangent distance. What is the angle of intersection?

11.The deflection angles between two intermediate points A and B of a highway curve are 6°30’ and 12°30’. The chord distance between the two points is 30 m. If the length of the long chord is 160 m long, calculate: a. The angle of intersection of the simple curve. b. Compute the tangent distance of the curve. c. Compute the external distance of the curve....