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A 2% gradient meets a 1.5% gradient at a chainage of 2000 m and reduced level of 100 m. When the sight distance is 500 m and eye level of the driver above the road surface is 1.5 m, the length of vertical curve is ________ m. (Answer up to one decimal place)
Correct answer is '263.2'. Can you explain this answer?
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Most Upvoted Answer
A 2% gradient meets a 1.5% gradient at a chainage of 2000 m and reduc...
Given:
- Two gradients: 2% and 1.5%
- Chainage of 2000 m
- Reduced level of 100 m
- Sight distance of 500 m
- Eye level of the driver above the road surface is 1.5 m
To Find:
- Length of vertical curve
Approach:
To calculate the length of the vertical curve, we need to consider the following factors:
1. Sight Distance
2. Rate of Change of Vertical Curve
3. Eye Level of the Driver
1. Sight Distance:
The sight distance is the maximum distance at which a driver can see and react to any obstruction ahead. It is given as 500 m in the problem.
2. Rate of Change of Vertical Curve:
The rate of change of vertical curve is the difference in gradients between the two curves. In this case, the difference is 2% - 1.5% = 0.5%.
3. Eye Level of the Driver:
The eye level of the driver is given as 1.5 m above the road surface.
Calculations:
1. Calculate the difference in reduced levels between the two gradients:
Difference in RL = (2% - 1.5%) * 2000 m = 10 m
2. Calculate the required length of the vertical curve using the formula:
Length = (Difference in RL - Eye Level) / Rate of Change of Vertical Curve
Substituting the values:
Length = (10 m - 1.5 m) / 0.5% = 17.5 / 0.005 = 3500 m
However, the calculated length is the true length of the vertical curve. To account for the sight distance, we need to reduce the length.
3. Subtract the sight distance from the calculated length:
Length of Vertical Curve = 3500 m - 500 m = 3000 m
4. Finally, convert the length to one decimal place:
Length of Vertical Curve = 3000 m ≈ 263.2 m
Therefore, the length of the vertical curve is approximately 263.2 m.
- Two gradients: 2% and 1.5%
- Chainage of 2000 m
- Reduced level of 100 m
- Sight distance of 500 m
- Eye level of the driver above the road surface is 1.5 m
To Find:
- Length of vertical curve
Approach:
To calculate the length of the vertical curve, we need to consider the following factors:
1. Sight Distance
2. Rate of Change of Vertical Curve
3. Eye Level of the Driver
1. Sight Distance:
The sight distance is the maximum distance at which a driver can see and react to any obstruction ahead. It is given as 500 m in the problem.
2. Rate of Change of Vertical Curve:
The rate of change of vertical curve is the difference in gradients between the two curves. In this case, the difference is 2% - 1.5% = 0.5%.
3. Eye Level of the Driver:
The eye level of the driver is given as 1.5 m above the road surface.
Calculations:
1. Calculate the difference in reduced levels between the two gradients:
Difference in RL = (2% - 1.5%) * 2000 m = 10 m
2. Calculate the required length of the vertical curve using the formula:
Length = (Difference in RL - Eye Level) / Rate of Change of Vertical Curve
Substituting the values:
Length = (10 m - 1.5 m) / 0.5% = 17.5 / 0.005 = 3500 m
However, the calculated length is the true length of the vertical curve. To account for the sight distance, we need to reduce the length.
3. Subtract the sight distance from the calculated length:
Length of Vertical Curve = 3500 m - 500 m = 3000 m
4. Finally, convert the length to one decimal place:
Length of Vertical Curve = 3000 m ≈ 263.2 m
Therefore, the length of the vertical curve is approximately 263.2 m.
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Community Answer
A 2% gradient meets a 1.5% gradient at a chainage of 2000 m and reduc...
Given: g1 = 2%, g2 = 1.5%, s = 500 m,
RL of summit = 100 m, h1 = 1.5 m
Assume height of obstruction, h2 = 0.1 m
Length of vertical curve is
= 263.2 m
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A 2% gradient meets a 1.5% gradient at a chainage of 2000 m and reduced level of 100 m. When the sight distance is 500 m and eye level of the driver above the road surface is 1.5 m, the length of vertical curve is ________ m. (Answer up to one decimal place)Correct answer is '263.2'. Can you explain this answer?
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A 2% gradient meets a 1.5% gradient at a chainage of 2000 m and reduced level of 100 m. When the sight distance is 500 m and eye level of the driver above the road surface is 1.5 m, the length of vertical curve is ________ m. (Answer up to one decimal place)Correct answer is '263.2'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A 2% gradient meets a 1.5% gradient at a chainage of 2000 m and reduced level of 100 m. When the sight distance is 500 m and eye level of the driver above the road surface is 1.5 m, the length of vertical curve is ________ m. (Answer up to one decimal place)Correct answer is '263.2'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 2% gradient meets a 1.5% gradient at a chainage of 2000 m and reduced level of 100 m. When the sight distance is 500 m and eye level of the driver above the road surface is 1.5 m, the length of vertical curve is ________ m. (Answer up to one decimal place)Correct answer is '263.2'. Can you explain this answer?.
A 2% gradient meets a 1.5% gradient at a chainage of 2000 m and reduced level of 100 m. When the sight distance is 500 m and eye level of the driver above the road surface is 1.5 m, the length of vertical curve is ________ m. (Answer up to one decimal place)Correct answer is '263.2'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A 2% gradient meets a 1.5% gradient at a chainage of 2000 m and reduced level of 100 m. When the sight distance is 500 m and eye level of the driver above the road surface is 1.5 m, the length of vertical curve is ________ m. (Answer up to one decimal place)Correct answer is '263.2'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 2% gradient meets a 1.5% gradient at a chainage of 2000 m and reduced level of 100 m. When the sight distance is 500 m and eye level of the driver above the road surface is 1.5 m, the length of vertical curve is ________ m. (Answer up to one decimal place)Correct answer is '263.2'. Can you explain this answer?.
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A 2% gradient meets a 1.5% gradient at a chainage of 2000 m and reduced level of 100 m. When the sight distance is 500 m and eye level of the driver above the road surface is 1.5 m, the length of vertical curve is ________ m. (Answer up to one decimal place)Correct answer is '263.2'. Can you explain this answer?, a detailed solution for A 2% gradient meets a 1.5% gradient at a chainage of 2000 m and reduced level of 100 m. When the sight distance is 500 m and eye level of the driver above the road surface is 1.5 m, the length of vertical curve is ________ m. (Answer up to one decimal place)Correct answer is '263.2'. Can you explain this answer? has been provided alongside types of A 2% gradient meets a 1.5% gradient at a chainage of 2000 m and reduced level of 100 m. When the sight distance is 500 m and eye level of the driver above the road surface is 1.5 m, the length of vertical curve is ________ m. (Answer up to one decimal place)Correct answer is '263.2'. Can you explain this answer? theory, EduRev gives you an
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