Given,
Length of vertical curve, L = 1500 ft = 15 stations
G1 = -3%
G2 = +6%
PVC station = (3+70)
PVC elevation = 1500 ft
a) Station of PVC = Station of PVI - L/2
Station of PVI = Station of PVC + L/2 = (3+70) + (7+50) = (11+20)
Station of PVT = Station of PVI + L/2 = (11+20) + (7+50) = (18+80)
b) Elevation of PVC = Elevation of PVI - G1 * L/2
Elevation of PVI = Elevation of PVC + G1 * L/2 = 1500 + (-0.03) * 750 = 1477.5 ft
Elevation of PVT = Elevation of PVI + G2 * L/2 = 1477.5 + (0.06) * 750 = 1522.5 ft
c) Equation of the curve:
a = (G2-G1)/2L = (0.06 - (-0.03))/(2*1500) = -0.00003
b = G1 = -0.03
c = elevation of PVC = 1500 ft
Lowest point of curve:
X = (-b/2a) = (-(-0.03)/2*(0.00003)) = 500 ft = (5+00) stations
Station of lowest point = station of PVC + X = (3+70) + (5+00) = (8+70)
Elevation of lowest point = (0.00003 * 500 * 500) + (-0.03 * 500) + 1500 = 1492.5 ft
Question 3 (10 marks) A 1,500ft long sag vertical curve (equal tangent) has a PC at...
A 200m equal tangent sag vertical curve has the PVC at station 3 + 700.000 and elevation 321m. The initial grade is -3.5% and the final grade is 0.5%. Determine the elevation and stationing of the PVI, PVT, and lowest point on the curve.
a) a 200 m vertical crest curve is designed to connect a +4.5% tangent with a -2% tangent. What should the design speed be to provide ample stopping sight distance? SSD t Pra) [10 marks) b) A 300 m sag parabolic vertical curve has a PVC at station 2+600.000 and elevation 320.000 m. the initial grade is -4.0% and the final grade is +1.0%. Determine the stationing and elevation of PVI, PVT and the lowest point on the curve. Also...
A 500-ft-long equal-tangent crest vertical curve connects tangents that intersect at station 340 + 00 and elevation 1322 ft. The initial grade is +4.0% and the final grade is -2.5%. Determine the elevation and stationing of the high point, PVC, and PVT. Solve this problem using parabolic equation and offset method.
Help please 2 (36 Pts) A 4250 ft equal tangent parabolic vertical curve has a back tangent grade (gl) of + 3.6 % and a forward tangent grade (g2) of-1.8 %. If the elevation of the PVI is 4264.82 ft. at station 78+ 25.00; what is: (a) the elevation and stationing of the PVC and PVT; (b) the elevation of the curve at station 83 + 63.00; and (c) the elevation and stationing of the highest and lowest points on...
13. A 400-ft long equal tangent sag vertical curve has its PVC at station 100+00 and o b elevation 500 ft, the initial grade g,--4% and the final grade g2 elevation of the lowest point of the curve. +2.5%. Determine the
7. A sag vertical curve connects a-4.75% grade and a +3.75% grade. The PVI is at station 45+90.00 at an elevation of 611.00 ft. The design speed is 50 mi/h. Determine the minimum length of the vertical curve using the AASHTO method ("K" factors). Round the minimum length up to the next 10 ft multiple for design of the vertical curve and associated calculations. Using this length, determine: The station and elevation of the PVC The station and elevation of...
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 gi4.00 percent, forward tangent gradient g, 7.00 percent, and length of curve L- 600 ft. Determine the curve elevations at half-station intervals along the curve, and compute the sta elevation of the lowest point. tion and い600 ft PVT 35+11.61 Elev. 75.18 35 34 PVC 29 11.61 Elev 66.18 Low point 31...
Please provide answer with full detailed working and any relevant diagrams. Thanks a lot Question 8 (20 marks) A 400-m equal-tangent crest vertical curve has the PVC at station 2+500 and elevation at 300 m. The initial grade is 2% and the final grade is -1%. First plot the vertical road geometry (all parameters associated with the calculation of this question must be labelled), and then determine the stationing and elevation of the PVI, PVT, and the highest point on...
(10 pts) A sag vertical curve is designed with the PVC at station 10900 and elevation 950.00 ft, the PVI at station 110 +77 and elevation 947.34 ft, and the low point at station 110 + 30 a. b. Determine the sight distance provided by the geometry. Determine the design speed of the curve
A 1080 ft long tangent crest vertical curve connects tangents ( G1= 4% , G2 = -2.8%). The two stations intersect at ghost station 640+00 where elevation is 1325.000 ft. Determine the stationing and elevations of the following points. 1. PVC 2. PVT 3. High point 4. 640+20