Crest Vertical Curve: Suppose you are designing a crest vertical curve for a new

Question

Crest Vertical Curve: Suppose you are designing a crest vertical curve for a new highway, where the speed limit is 65 mi/h. The curve is 1100 ft long and has its PVC at station 125+00. The PVC elevation is 1000 ft. The initial grade leading to the curve is +2.5% and the final grade is -2.0%. (a) Determine the elevation and stationing of the high point, PVI, and PVT. (b) Does this curve satisfy stopping sight distance requirements?

a) PVI 130+55 1013.75, PVT 136+00 1002.75, xhi 131+11, 1007.64 b) Yes

Explanation / Answer

Given data:

PVC at station 125+00=sta PVC=12500. and PVC elevation (elev.PVC)= 1000 ft,

Curve length L=1100 ft, initail grade leading=+2.5%, final grade=-2.0%

a) Absolute value difference in grade A= |G2-G1|=|2.5+2|=4.5%

FOR PVI:

elevation PVI=elev PVC+(G1*L/2)=1000+(2.5%*1100/2)=1013.75

stationing of highpoint PVI= sta PVC+(L/2)= 12500+(1100/2)=130+55

FOR PVT:

Yfinal= (AL/200)= (4.5*1100)/200=24.75

elevation PVT=elev PVC+(G1*L)-Y final=1000+(2.5%*1100)-24.75=1002.75 ft

stationing of highpoint PVI= sta PVI+(L/2)= 13055+(1100/2)=136+00

HIGHPOINTS:

a=-A/2L=-(4.5/100)/(2*1100)=-2.045*(10^-5)

b=G1=0.025

xhigh=(-b/2a)= 611

staHigh=sta PVC+xhigh=12500+611=131+11

elev High= elev PVC+(G1*dist High)-Yhigh= 1007.64 ft

b) stopping sight distance SSD= (V12/2g(a/g)+/-(A))+V1*tr

where V1= speed=65mi/h, a=3.4m/s2, t=2.5s and A= 4.5

from above calculations SSD <L;

Hence it is safe.

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