Answer:
The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.
Explanation:
Length of curve is given as
[tex]L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft[/tex]
[tex]G_2[/tex] is given as
[tex]G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%[/tex]
The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as
[tex]A=\frac{L}{K}\\A=\frac{460}{115}\\A=4[/tex]
A is given as
[tex]-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%[/tex]
With initial grade, the elevation of PVC is
[tex]E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\[/tex]
The station is given as
[tex]St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\[/tex]
Low point is given as
[tex]x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft[/tex]
The station of low point is given as
[tex]St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\[/tex]
The elevation is given as
[tex]E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft[/tex]
So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.
