Question

Determine the minimum radius of a horizontal curve required for a highway if the design speed is 70 mph and the superelevation rate is 0.08.

Answer 1

Answer:

1819.4 ft

Explanation:

Let us assume  the coefficient of side friction is 0.10

The radius of curve is given by the formula:

$R=\frac{V^2}{g(f_s+\frac{e}{100} )}$

Where R is the minimum radius of a horizontal curve, V is the speed, f is coefficient of side friction and e/100 = super elevation rate, g =acceleration due to gravity

Given that:

V = 70 mph = (70 * 1.467) ft/s = 102.69 ft/s, g =32.2 ft/s², e/100 = 0.08, f= 0.1

$R=\frac{V^2}{g(f_s+\frac{e}{100} )}=\frac{102.69^2}{32.2(0.1+0.08)}=1819.4\ ft$

the minimum radius of a horizontal curve is 1819.4 ft