Just do question 2
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- The data given for a vertical sag curve on a roadway plan and profile sheet are as follows: PVI station @ 32+11.61, PVI elevation =54.18 ft, back tangent gradient g1 =- 4.00 percent, forward tangent gradient g2 = + 7.00 percent, and length of curve L= 600ft. Determine the curve elevations at half-station intervals along the curve, and compute the station and elevation of the lowest point.An old equal-tangent crest vertical curve was designed with a PVI at station 117+45 and elevation of 960 ft., and a PVC at station 110 + 85 and elevation of 927 ft. If the highest point on the curve is at station 117 + 65, what is the maximum safe design speed for this curve? (explore and report your answer in 5 mph increment) [hint: Use design tables. But this time, find the speed based on K]A parabolic vertical curve of length 280m connects a tangent grade = + 4% to a grade of -6%. What is the safe design speed based on stopping sight distance?
- When my first grade is (g1) and the second grade is (g2), what type or vertical curve is created? "Summit curve" or "valley curve"? A. g1=10% ; g2=4% B. g1=-2% ; g2=6% C. g1=-5% ; g2=-2% D. g1=2% ; g2=6%A highway intersection of 40 degrees is to be connected by a 6-degree simple curve and an 80-m long spiral on each end of the simple curve. Use arc basis. Determine the total length of the curve. Asap.The degree of the curve of the central curve of a spiral easement curve is 50 and its central angle is 380. . If the external distance of the spiral curve is 14.60 m, compute the length of throw.
- A blind horizontal curve has a length equivalent to 54 meters. If the design sight distance is equivalent to 67 meters, and the moddle ordinate is equivalent to 8.2 meters, calculate the minimum radius of the curve so it can be classified SAFE.A vertical parabolic sag curve has tangent grades of -1.428% and +0.707%. If the grade changes uniformly at 0.16% per 20 m, find the length of the curveA vertical sag curve has tangent grades of -3.6% and +4.8% meeting at point S whose elevation is 446.5 m. If the length of the curve is 420 m, find the elevation of the PC.
- A 500-m long equal tangent vertical curve has a point of vertical curvature at station 2 + 600 and elevation 450 m. The initial grade is -1% and final grade is +1% b) determine the highest and the lowest points on this vertical curve by using the K-value.1. A parabolic curve has an ascending grade of 4% which meets a descending grade of 3% at sta. 1+ 180 whose elevation is at 320 m. Compute the elevation of the PVT(point of vertical tangency) if the length of the curve is 280 mA parabolic curve has a descending grade of -0.8% and an ascending grade of +0.4% that intersect at station 10 + 020 with an elevation of 240.600 m. If the maximum allowable rate of change of grade for this curve is 0.15% per 20 m stations, determine the following: A. The elevation at station 10 + 000 *Draw the figure