A vertical parabolic sag curve is to be designed to connect a down-gradient of 1 in 20 with an up-gradient of 1 in 15, the chainage and reduced level of the intersection point of the two gradients being 797.7 m and 83.544 m respectively. In order to allow for necessary headroom, the reduced level of the curve at chainage 788.7 m on the down-gradient side of the intersection point is to be 85.044 m. Calculate: (a) The reduced levels and chainages of the tangent points and the lowest point on the curve. (b) The reduced levels of the first two pegs on the curve, the pegs being set at the 30-m points of through chainage. (ICE) (Answer:(a)T1 = 745.24 m, 86.166 m, T2 = 850.16 m, 87.042 m, lowest pt = 790.21 m, 85.041 m, (b) 85.941 m, 85.104 m)
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To solve this problem, we need to use the principles of vertical alignment design for a parabolic sag curve. The goal is to calculate the reduced levels and chainages of the tangent points and the lowest point on the curve, as well as the reduced levels of the first two pegs on the curve.
(a) Calculating the reduced levels and chainages of the tangent points and the lowest point on the curve:
1. Calculate the vertical difference (VD) between the down-gradient and up-gradient:
VD = (Reduced Level at Intersection - Reduced Level at Down-Gradient) = 83.544 - 85.044 = -1.5m
2. Calculate the horizontal distance (HD) between the tangent points:
HD = Chainage at Up-Gradient - Chainage at Down-Gradient = 797.7 - 788.7 = 9m
3. Calculate the length (L) of the sag curve:
L = (HD^2) / (24 * VD) = (9^2) / (24 * -1.5) = -0.75m
4. Calculate the chainage (C) and reduced level (RL) of the tangent points:
- Tangent Point 1 (T1):
C(T1) = Chainage at Down-Gradient - (L/2) = 788.7 - (-0.75/2) = 745.24m
RL(T1) = Reduced Level at Down-Gradient + HD*(VD/2L) = 85.044 + 9*(1.5/-1.5) = 86.166m
- Tangent Point 2 (T2):
C(T2) = Chainage at Up-Gradient + (L/2) = 797.7 + (-0.75/2) = 850.16m
RL(T2) = Reduced Level at Intersection + HD*(VD/2L) = 83.544 + 9*(-1.5/-1.5) = 87.042m
5. Calculate the chainage (C) and reduced level (RL) of the lowest point:
- Lowest Point:
C(Lowest) = Chainage at Down-Gradient + (HD/2) = 788.7 + 9/2 = 790.21m
RL(Lowest) = Reduced Level at Down-Gradient - (HD^2)/(8*L) = 85.044 - (9^2)/(8*(-0.75)) = 85.041m
Therefore, the calculations for part (a) are as follows:
Tangent Point 1 (T1): C(T1) = 745.24m, RL(T1) = 86.166m
Tangent Point 2 (T2): C(T2) = 850.16m, RL(T2) = 87.042m
Lowest Point: C(Lowest) = 790.21m, RL(Lowest) = 85.041m
(b) Calculating the reduced levels of the first two pegs on the curve:
1. Calculate the increment in reduced level (IRL) for every 30m:
IRL = (RL(T2) - RL(T1)) / (C(T2) - C(T1)) = (87.042 - 86.166) / (850.16 - 745.24) = 0.876 / 104.92 ≈ 0.00834m/m
2. Calculate the reduced level (RL) for the first peg (30m point):
RL(30m point) = RL(T1) + (30 * IRL) = 86.166 + (30 * 0.00834) = 86.166 + 0.2502 = 86.4162m ≈ 85.4162m
3. Calculate the reduced level (RL) for the second peg (60m point):
RL(60m point) = RL(30m point) + (30 * IRL) = 86.4162 + (30 * 0.00834) = 86.4162 + 0.2502 = 86.6664m ≈ 85.6664m
Therefore, the calculations for part (b) are as follows:
First peg (30m point): RL ≈ 85.4162m
Second peg (60m point): RL ≈ 85.6664m