Homework 7 Solutions (revised 2 Nov 06) / Due: Fri. 27 October 2006
CE361 HW7 Solutions - 3 -
GEOMETRIC DESIGN FOR SAFETY
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In the uphill direction, the road will begin as a sag vertical curve with an initial grade of 1.5 percent. After 571 feet, the sag curve ends at G2 = 4.0 percent. After 320 feet as a tangent section, a crest vertical curve with L = 680 feet will be used to complete the climb, ending with a terminal gradient of 1.1 percent.
1. Vertical curve elevations.
A. (10 points) Draw the side view of the connector road’s climb up the hillside. Label the G’s, VPTs, and VPCs. The drawing does not have to be to scale.
B. (10 points) Calculate the elevation of the sag curve’s VPT.
Use (7.3) with datum set so that a = 0 = YVPC1. b = 0.015 = G1.
c = (G2-G1)/2L = (0.040-0.015)/(2*571) = 0.000022.
YVPT(sag) = a + bx + cx2 = 0 + (0.015*571) + (0.000022*5712) = 8.565 + 7.14 = 15.70 ft.
C. (15 points) Calculate the elevation of the crest curve’s VPC and VPT.
Increase in elevation over tangent section: DY = 0.04 * 320 = 12.80 ft.
Use (7.3) with a = 15.70 + 12.80 = 28.50 ft = YVPC(crest). b = 0.040 = G1.
c = (G2-G1)/2L = (0.011-0.040)/(2*680) = -000021.
YVPT(crest) = a + bx + cx2 = 28.50 + (0.040*680) + (-000021*6802) = 28.50 + 27.20 + (-9.86) = 45.84 ft.
2. Horizontal alignment. As the connector road climbs the hill, it follows a circular path that subtends a 90-degree angle from River Road to SR361.
A. (10 points) What is the length of the connector road described in Problem 1, as measured along its centerline (not horizontally)? Describe your method clearly.
You could attempt to integrate the parabolic function in (7.3) or do a piecewise linear approximation using a spreadsheet. Using Dx = 10 feet to calculate the DY over that Dx, then computing DL = sqrt(Dx2 + DY2), the length of each linear piece is determined. Summing the linear approximations over the sag VC = 571.23 ft. Summing the linear approximations over the crest VC = 680.24 ft. The length of the tangent section along its centerline = sqrt(3202 + 12.802) = 320.26 ft. Total length = 571.23 + 320.26 + 680.24 = 1571.73 ft. The piecewise linear approximations give a slight underestimate, but the estimated length of the parabola is not a lot different from DL = sqrt(Dx2 + DY2) = sqrt[(571+320+680)2 + 45.842] = 1571.67 ft, which is the straight line distance between the sag curve’s VPC and the crest curve’s VPT. Perhaps I should have tried a smaller Dx in the piecewise linear approximation.
B. (10 points) What is the radius of the connector road? Use L = 1571 ft.
(7.9) with D = 90 degrees: D = (100D)/L = (100*90)/1571 = 5.7288.
(7.8) R = 5729.6/D = 5729.6/5.7288 = 1000.13 ft.
3. SSD on Vertical Curves.
A. (10 points) Can a speed of 55 mph be safely maintained on the sag VC described in Problem 1?
From FTE Table 7.4, S = 495 ft for 55 mph. |A| = |G2 – G1| = |4.0-1.5| = 2.5 percent.
Equations (7.18) with standard h = 2 ft and b = 1 degree:
Lmin1 = 287.85 ft if S<L; 495 < 287.85? No.
Lmin2 = 990 - 662.71 = 138.78 ft if S>L; 495 138.78? Yes.
Lmin = 138.78 ft, which is exceeded by the sag VC with L = 571 ft.
55 mph can be maintained.
B. (10 points) Can a speed of 55 mph be safely maintained on the crest VC described in Problem 1?
From FTE Table 7.4, S = 495 ft for 55 mph. |A| = |G2 – G1| = |1.1-4.0| = 2.9 percent.
Equations (7.17) with standard h1 = 3.5 ft and h2 = 2 ft:
Lmin1 = 329.23 ft if S<L; 495 < 329.23? No.
Lmin2 = 990 – 744.14 = 245.76ft if S>L; 495 > 245.76? Yes.
Lmin2 is exceeded by the crest VC with L = 680 ft. 55 mph can be maintained.
4. SSD on Horizontal Curves. (15 points) Two lanes of standard width. How close to the roadway can any sight obstruction be?
FTE Figure 7.17 and equation 7.19: Rv = R – (12ft/2) = 1000.13 – 6 = 994.13 ft. D = 90 degrees. S = 495 ft for 55 mph.
Ms = 994.13 (1 – cos 14.26) = 994.13 (1-0.9692) = 30.65 ft.