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In a BG railway track, the specified ruling gradient is 1 in 250. The horizontal curve of 3° on a gradient of 1 in 250 will have the permissible compensated gradient of
- a)1 in 257
- b)1 in 357
- c)1 in 457
- d)1 in 512
Correct answer is option 'B'. Can you explain this answer?
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In a BG railway track, the specified ruling gradient is 1 in 250. The ...
Grade compensation = 0.04 x 3 = 0.12%
Permissible gradient =
Permissible gradient =
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In a BG railway track, the specified ruling gradient is 1 in 250. The ...
Degrees is to be laid on this gradient. Determine the super elevation required for the curve.
We can use the following formula to calculate the super elevation required for a given curve:
e = (v²/127R) - G
Where e is the super elevation, v is the speed in meters per second, R is the radius of the curve in meters, and G is the ruling gradient.
First, let's convert the 3 degree curve to its radius in meters:
R = 1000/(2π/360 × 3)
R = 573.6 meters
Next, let's calculate the speed required for the curve. Assuming a maximum speed of 80 km/h (22.2 m/s), we can use the following formula:
v² = gR(1 + G/e)
Where g is the acceleration due to gravity (9.81 m/s²).
Plugging in the values, we get:
v² = 9.81 × 573.6 × (1 + 0.004)
v² = 3,566.61
v = 59.7 m/s
Finally, we can calculate the super elevation:
e = (59.7²/127 × 573.6) - 0.004
e = 0.16 meters (rounded to two decimal places)
Therefore, the super elevation required for the curve is 0.16 meters.
We can use the following formula to calculate the super elevation required for a given curve:
e = (v²/127R) - G
Where e is the super elevation, v is the speed in meters per second, R is the radius of the curve in meters, and G is the ruling gradient.
First, let's convert the 3 degree curve to its radius in meters:
R = 1000/(2π/360 × 3)
R = 573.6 meters
Next, let's calculate the speed required for the curve. Assuming a maximum speed of 80 km/h (22.2 m/s), we can use the following formula:
v² = gR(1 + G/e)
Where g is the acceleration due to gravity (9.81 m/s²).
Plugging in the values, we get:
v² = 9.81 × 573.6 × (1 + 0.004)
v² = 3,566.61
v = 59.7 m/s
Finally, we can calculate the super elevation:
e = (59.7²/127 × 573.6) - 0.004
e = 0.16 meters (rounded to two decimal places)
Therefore, the super elevation required for the curve is 0.16 meters.
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In a BG railway track, the specified ruling gradient is 1 in 250. The horizontal curve of 3° on a gradient of 1 in 250 will have the permissible compensated gradient ofa)1 in 257b)1 in 357c)1 in 457d)1 in 512Correct answer is option 'B'. Can you explain this answer?
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In a BG railway track, the specified ruling gradient is 1 in 250. The horizontal curve of 3° on a gradient of 1 in 250 will have the permissible compensated gradient ofa)1 in 257b)1 in 357c)1 in 457d)1 in 512Correct answer is option 'B'. Can you explain this answer?, a detailed solution for In a BG railway track, the specified ruling gradient is 1 in 250. The horizontal curve of 3° on a gradient of 1 in 250 will have the permissible compensated gradient ofa)1 in 257b)1 in 357c)1 in 457d)1 in 512Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of In a BG railway track, the specified ruling gradient is 1 in 250. The horizontal curve of 3° on a gradient of 1 in 250 will have the permissible compensated gradient ofa)1 in 257b)1 in 357c)1 in 457d)1 in 512Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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