Question

A crest vertical curve is designed for 60 MPH. The initial grade is +4% and the final grade is negative. What is the elevation difference between the PVC and the high point on the curve?

Answer 1

The elevation difference between the PVC and the high point on the curve 0.1208 ft

Explanation:

g1 = 4.0%

    = 0.04

The elevation of curve at a distance “x”from pvc is given by

  $y=a x^{2}+b x+ ele p v c$

x = distance from pvc to point on curve

The highest point on vertical curve is at a distance "x" is given by

                                                                  $= -g1L/g2-g1$

so, elevation of highest point $y=a x^{2}+b x+ele p v c$

here,

$a=\frac{g_{2}-g_{1}}{2 L} \quad ; \quad b=g_{1} ; x=\frac{-g_{1} C}{g_{2}-g_{1}}$

substitute above values in "y" equation,

$y=\left(\frac{g_{2}-g_{1}}{2 C}\right) \cdot\left(\frac{-g_{1} L}{g_{2}-g_{1}}\right)^{2}+g_{1}\left(\frac{-g_{1} L}{g_{2}-g_{1}}\right)+\text { elepvc }$

$\frac{\left(g_{2}-g_{1}\right)}{2 L} \cdot \frac{g_{1}^{\prime} L^{\prime}}{\left(g_{2}-g_{1}\right)}-\frac{g_{1}^{\prime} L}{\left(g_{2}-g_{1}\right)}+\text { ele pvc }$

$y=\frac{g_{1}^{\prime} L}{2\left(g_{2}-g_{1}\right)}-\frac{g_{1}^{\prime} L_{1}}{\left(g_{2}-g_{1}\right)}+\text { ele pvc }$

$y=-\frac{1}{2} \frac{g_{1}^{r} L}{\left(g_{2}-g_{1}\right)}+\text { elepvc }$

$\text { elepvc }-y=\frac{1}{2} \frac{g_{1}^{r} L}{\left(g_{2}-g_{1}\right)}$

Algebric sum of grades = A = $g_{2}-g_{1}$

As per AASHTO,for V= 60 mph, K =151 based on Stopping sight distance

Minimum length of vertical curve

$L=K A$

$L=151 A$

$A=\frac{L}{151}$

We know that,

$\text { ele pvc }-y=\frac{1}{2} \frac{g_{1}^{\prime} L}{A}$

$\frac{1}{2} \times \frac{(0.04)^{r} \times 4}{ L/{151}}$

solving above equation we get, elevation distance between highest point and pvc as 0.1208 ft