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
[tex]y=a x^{2}+b x+ ele p v c[/tex]
x = distance from pvc to point on curve
The highest point on vertical curve is at a distance "x" is given by
[tex]= -g1L/g2-g1[/tex]
so, elevation of highest point [tex]y=a x^{2}+b x+ele p v c[/tex]
here,
[tex]a=\frac{g_{2}-g_{1}}{2 L} \quad ; \quad b=g_{1} ; x=\frac{-g_{1} C}{g_{2}-g_{1}}[/tex]
substitute above values in "y" equation,
[tex]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 }[/tex]
[tex]\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 }[/tex]
[tex]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 }[/tex]
[tex]y=-\frac{1}{2} \frac{g_{1}^{r} L}{\left(g_{2}-g_{1}\right)}+\text { elepvc }[/tex]
[tex]\text { elepvc }-y=\frac{1}{2} \frac{g_{1}^{r} L}{\left(g_{2}-g_{1}\right)}[/tex]
Algebric sum of grades = A = [tex]g_{2}-g_{1}[/tex]
As per AASHTO,for V= 60 mph, K =151 based on Stopping sight distance
Minimum length of vertical curve
[tex]L=K A\\[/tex]
[tex]L=151 A[/tex]
[tex]A=\frac{L}{151}[/tex]
We know that,
[tex]\text { ele pvc }-y=\frac{1}{2} \frac{g_{1}^{\prime} L}{A}[/tex]
[tex]\frac{1}{2} \times \frac{(0.04)^{r} \times 4}{ L/{151}}[/tex]
solving above equation we get, elevation distance between highest point and pvc as 0.1208 ft
