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If a descending gradient of 1 in 25 meets an ascending gradient of 1 in 40, then the length ofvalley curve required for a headlight sight distance of 100 m will be

  • a)
    130 m 

  • b)
    30 m

  • c)
    310 m

  • d)
    630m

Correct answer is option 'A'. Can you explain this answer?
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If a descending gradient of 1 in 25 meets an ascending gradient of 1 i...
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If a descending gradient of 1 in 25 meets an ascending gradient of 1 i...
Given:
Descending gradient = 1 in 25
Ascending gradient = 1 in 40
Headlight sight distance = 100 m

To find: Length of valley curve required

Formula used:
L = 3.6aV²/2gd

where
L = Length of valley curve
a = Rate of change of centrifugal acceleration (m/s³)
V = Design speed (m/s)
g = Acceleration due to gravity (m/s²)
d = Rate of change of gradient (%)

Calculation:
1. Rate of change of centrifugal acceleration (a):
a = V²/r

where
V = Design speed (m/s)
r = Radius of curve (m)

To find the radius of curve, we need to find the degree of curvature (D):
D = 2arctan(1000G)

where
G = Rate of change of grade (% per 30 m)

G = -1/25 - 1/40 = -0.068

D = 2arctan(1000(-0.068)/30) = -8.23°

The radius of curve (r) can be found using the formula:
r = 5729.58/D (in meters)

r = 5729.58/-8.23 = -695.4 m (negative sign indicates a left-hand curve)

As radius cannot be negative, we take the absolute value:
r = 695.4 m

Now, we can find the rate of change of centrifugal acceleration:
a = V²/r

2. Design speed (V):
Design speed is usually taken as 80 km/h (22.22 m/s) for two-lane highways.

3. Acceleration due to gravity (g):
g = 9.81 m/s²

4. Rate of change of gradient (d):
d = G/100 = -0.068/100 = -0.00068

5. Length of valley curve (L):
L = 3.6aV²/2gd

L = 3.6(22.22)²/(2*9.81*-0.00068)

L = 130.7 m

Therefore, the length of valley curve required for a headlight sight distance of 100 m is 130 m (approx). Hence, option (a) is the correct answer.
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If a descending gradient of 1 in 25 meets an ascending gradient of 1 i...
Given SSD = 100 m
From headlight sight distance, 
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