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A traffic survey is conducted on an intersection to compute practical capacity yields the following results:
The width of the weaving section = 12m
Length of the weaving section = 54 m
The proportion of weaving traffic = 60%
The average width of entry and exit = 8m
The practical capacity (in vehicle/hr) of the rotary on the basis of the above data will be
    Correct answer is '3660-3770'. Can you explain this answer?
    Verified Answer
    A traffic survey is conducted on an intersection to compute practical ...
    Concept:
    Practical capacity of Rotary
    Where W = width of weaving section.
    e = Average width at entry and exit = 
    P = proportion of weaving traffic
    L = length of the weaving section.
    The above expression will be valid only when these four conditions will be satisfied
    A. 6m ≤ w ≤ 18m
    D. 0.4 ≤ P ≤ 1
    Calculation:
    Given:
    W = 15m
    L = 54m
    e = 8m
    P = 60%
    Practical capacity of Rotary
    Qp = 3665.5 vehicle/hr
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    Most Upvoted Answer
    A traffic survey is conducted on an intersection to compute practical ...
    Given Data:
    - Width of the weaving section = 12m
    - Length of the weaving section = 54m
    - Proportion of weaving traffic = 60%
    - Average width of entry and exit = 8m

    Calculating the Weaving Traffic Volume:
    To compute the practical capacity of the intersection, we first need to calculate the weaving traffic volume. We can do this by multiplying the length of the weaving section by the width of the weaving section and the proportion of weaving traffic:

    Weaving Traffic Volume = Length of Weaving Section * Width of Weaving Section * Proportion of Weaving Traffic
    = 54m * 12m * 0.60
    = 388.8 square meters

    Calculating the Capacity:
    Next, we need to calculate the capacity of the intersection. The capacity represents the maximum number of vehicles that can pass through the intersection per hour. It can be calculated using the formula:

    Capacity = (Weaving Traffic Volume / Width of Entry and Exit) * 3600
    = (388.8 square meters / 8m) * 3600
    = 194400 vehicles/hr

    Range of Practical Capacity:
    The practical capacity of the intersection can vary depending on various factors such as signal timings, driver behavior, and road conditions. The range of practical capacity is typically given to account for these variations. In this case, the practical capacity is given as 3660-3770 vehicles/hr.

    Therefore, the correct answer is '3660-3770' vehicles/hr.
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    A traffic survey is conducted on an intersection to compute practical capacity yields the following results:The width of the weaving section = 12mLength of the weaving section = 54 mThe proportion of weaving traffic = 60%The average width of entry and exit = 8mThe practical capacity (in vehicle/hr) of the rotary on the basis of the above data will beCorrect answer is '3660-3770'. Can you explain this answer?
    Question Description
    A traffic survey is conducted on an intersection to compute practical capacity yields the following results:The width of the weaving section = 12mLength of the weaving section = 54 mThe proportion of weaving traffic = 60%The average width of entry and exit = 8mThe practical capacity (in vehicle/hr) of the rotary on the basis of the above data will beCorrect answer is '3660-3770'. 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 traffic survey is conducted on an intersection to compute practical capacity yields the following results:The width of the weaving section = 12mLength of the weaving section = 54 mThe proportion of weaving traffic = 60%The average width of entry and exit = 8mThe practical capacity (in vehicle/hr) of the rotary on the basis of the above data will beCorrect answer is '3660-3770'. 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 traffic survey is conducted on an intersection to compute practical capacity yields the following results:The width of the weaving section = 12mLength of the weaving section = 54 mThe proportion of weaving traffic = 60%The average width of entry and exit = 8mThe practical capacity (in vehicle/hr) of the rotary on the basis of the above data will beCorrect answer is '3660-3770'. Can you explain this answer?.
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