Q2) A proposed highway has two tangents of bearings N 45°54'36" E and N 1°22'30" W. The highway design engineer, attempting to obtain the best fit for the simple circular curve to join these tangents, decides that the external ordinate is to be 43.00 ft. The PI is at station 65+43.21 Determine: (a) The central angle of the curve; (b) The radius of the curve; (c) The length of

Consider the same vertical curve from Problem 1 that has a +3.2% initial grade and -1.1% final grade. The point of vertical intersection is at station 98+20. The highest point of the curve is at station 100+79.35 to meet drainage design. What is the station of the point of vertical curvature (Also known as beginning of vertical curve). 92+88.95 92+90.81 92+88.05 (INCORRECT) 92+91.23 This is Question 1:  Consider a vertical curve with a +3/2% initial grade and -1.1% final grade. The point of vertical intersection is at station 98+20. The highest point of the curve is at station 100+79.35 to meet drainage design. What is the design speed, in mph? Correct answer here is 70 MPH

A simple circular curve is designed to connect a single lane highway at PI station 64+27.46 sta. with an interior angle (A) = 8° 24'. The terrain conditions require a minimum radius of 2864.79 ft. Compute the following parameters to complete the design process: 1) The station of PC and PT 2) Tangent length (T) 3) Degree of curvature (D) 4) Long Cord (LC) 5) External and middle ordinate (E & M)