Answer:
The value is [tex] e = 1.92 [/tex]
Step-by-step explanation:
From the question we are told that
The formula for radius of curvature is [tex]R = \frac{1600}{15} +127e[/tex]
The design speed of the roadway is [tex]v = 40 \ km/hr[/tex]
The value of radius of curvature considered R = 350 m
From the equation given e is the superelevation rate
Now making the subject of the formula
[tex] e = \frac{R}{127} - \frac{1600}{1905} [/tex]
At R= 350
[tex] e = \frac{350 }{127} - \frac{1600}{1905} [/tex]
[tex] e = 1.92 [/tex]
Here, we are required to determine the maximum superelevation rate, e (rounded to 2 decimal places) recommended if the radius of curvature, R is 350 meters.
The answer is e = 1.92 (rounded to 2 decimal places).
First, from the information given, the minimum radius of curvature, R is related to the maximum superelevation rate, e by the formular;
R = (1600/15) + (127e)
Therefore, to evaluate the value of e for R = 350, we have;
350 - (1600/15) = 127e.
350 - (1600/15) = 127e.350 - 106.667 = 127e
Therefore, 243.333 = 127e.
Therefore, e = 1.9160meters
As a result, The maximum superelevation rate, e recommended if the radius of curvature, R is 350 meters is:
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