Question

A 1200-ft equal-tangent crest vertical curve is currently designed for 50 mi/h. A civil engineering student contends that 60 mi/h is safe in a van because of the higher driver’s-eye height. If all other design inputs are standard, what must the driver’s-eye height (in the van) be for the student’s claim to be valid?

Answer 1

Answer:

8.95ft

Explanation:

In order to develop this problem it is necessary to consider two concepts:

The first is the design of vertical curves through the general equation for the length of a curved vertical crest in terms of algebraic differences in grades. The second is the Design Controls for Crest vertical curves table (I attach a table at the end).

The aforementioned equation is given by:

$L = \frac{AS^2}{200(\sqrt{h_1}+\sqrt{h_2})^2}$

Where,

L = leght of vertical curve

S = Sight distance

A = Algebraic difference in grades

$h_1 =$Height of eye above roadway

$h_2 =$height of object above roadway surface

From the table we know that for design speed of 60 mi/h the S is 570 ft, while the value of the rate of vertival curve K, for design speed of 50mi/h is 84.

Then we can calculate the Algebraic difference in grades through:

$A= \frac{L}{K}$

$A = \frac{1200}{84}$

$A = 14.285$

Applying the equation to find $h_1$ we have:

$L = \frac{AS^2}{200(\sqrt{h_1}+\sqrt{h_2})^2}$

$1200 = \frac{14.32(570)^2}{200(\sqrt{h_1}+\sqrt{2})^2}$

Solving for$h_1$

$h_1 = 8.95ft$

Therefore the height of the driver's eye is 8.95ft