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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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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.
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,
From headlight sight distance,
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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 bea)130 mb)30 mc)310 md)630mCorrect answer is option 'A'. Can you explain this answer?
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