Vertical Curve Question

Please advise me regarding this quetion. When I calculated, my answer came 1520' based on CERM formula.

" A Vertical curve joins -1.2% grade to a + 0.8% grade. The PI of the vertical curve is at station 225+00 and elevation 167 feet above seal level. the centerline of the roadway must clear a pipe located at station 226+20 by 3.0 feet. The elevation of the top of the pipe is 167.65 feet above sea level. What is the minimum length of the vertical curve that can be used for 60 mph design speed ?

A. 280 feet B. 480 feet C.500 feet D. 300 feet

My answer 1520 ft doesn't match from above.

Please advise.

Here's what I was trying but got stuck, first I was going to use the SAG min. Vert. Curve equations based on headlight. The CERM has equations based on the algebraic differece and Stopping Sight Distance given speed. But I couldn't get a result from assuming S<L or L>S? Can someone help us out? Then I was going to compare this with the Lenght of Curve passing a point equation and see which equation governs? Does this seem right?

monty74 said:

Please advise me regarding this quetion. When I calculated, my answer came 1520' based on CERM formula.
" A Vertical curve joins -1.2% grade to a + 0.8% grade. The PI of the vertical curve is at station 225+00 and elevation 167 feet above seal level. the centerline of the roadway must clear a pipe located at station 226+20 by 3.0 feet. The elevation of the top of the pipe is 167.65 feet above sea level. What is the minimum length of the vertical curve that can be used for 60 mph design speed ?

A. 280 feet B. 480 feet C.500 feet D. 300 feet

My answer 1520 ft doesn't match from above.

Please advise.

Click to expand...

I was trying to make this problem using vertical curve data. See example 78.7 on page 78-12 CERM. But looks like not enough data, because our problem little different then given example (X is after PI). So i used table 78.2 CERM, found SSD=570 for V=60mph. Then using table 78.4 solve L. My answer is 422. I vote for answer B , where L=480

do you have solution, let me know if I am wright

I am not a transportation expert by any means, but I'll take a stab at this problem.

To check the MAXIMUM curve length, use the following:

x = 0.5*L + 120 ft

ELEVx = 167.65 + 3 = 170.65 ft

ELEVbvc = 167.65 + .012*(L/2) = 167.65 + .006*L

R = (.008 + .012)/L = .02/L

Using equation 78.47 in the CERM 11th ed to solve for L, the answer turns out to be 1518 ft. If the curve length is any longer, the roadway will be too flat to raise above the minimum elevation.

For the minimum curve length, use Figure 78.15 in the CERM. In the figure, K=136 for v=60. Since K = L/A, you can solve for L.

L = K*A = 136*2 = 272 ft. My choice is 280 ft.

So you used the known elevation at the top of pipe to solve for L?

After thinking about this problem I believe it is 1520'. This is the shortest length of vertical needed to make the clearance over the pipe. The length of 280' will cause not make the 3' clear from the top of pipe. This is not the maximum length. There is really no maximum constraint unless the problem states there's obstruction which needs to provide a certain vertical clearance above the road. Since the obstruction is below, there is a minimum to provide just enough cover. But then again, the answers do not match what I got. If we go by the design speed and K value, yes we will get the 280'. Maybe the tricky part in this question is to ignore the first part of pipe culvert clearance.

Sorry Guys ! working in NY is headache, long commute, longer working hrs!

Anyway, what about if we use formula based on CERM 78.17. As per CERM 78.17 " IF THE STATION AND ELEVATION OF THE POINT P, THE STATION AND ELEVATION OF THE BVC OR THE VERTEX AND THE GRADIENT VALUES G1 AND G2 ARE KNOWN, THE CURVE LENGTH CAN BE DETERMINED EXPLICITLY".

This problem is not same as per fig. 78.11 but it reverse ( obstruction is below ) Based on this formula I get 1520 ft answer.

TANYA : Based on your solution , I am not getting 422 ft as per CERM table 78.4. Which formula you used.

monty74 said:

Please advise me regarding this quetion. When I calculated, my answer came 1520' based on CERM formula.
" A Vertical curve joins -1.2% grade to a + 0.8% grade. The PI of the vertical curve is at station 225+00 and elevation 167 feet above seal level. the centerline of the roadway must clear a pipe located at station 226+20 by 3.0 feet. The elevation of the top of the pipe is 167.65 feet above sea level. What is the minimum length of the vertical curve that can be used for 60 mph design speed ?

A. 280 feet B. 480 feet C.500 feet D. 300 feet

My answer 1520 ft doesn't match from above.

Please advise.

Click to expand...

Using the Formula, L + 2h/L -2h = (v-G1h/v-G2h)^1/2. My answer is A. Can you guys solve the problem

monty74 said:

Please advise me regarding this quetion. When I calculated, my answer came 1520' based on CERM formula.
" A Vertical curve joins -1.2% grade to a + 0.8% grade. The PI of the vertical curve is at station 225+00 and elevation 167 feet above seal level. the centerline of the roadway must clear a pipe located at station 226+20 by 3.0 feet. The elevation of the top of the pipe is 167.65 feet above sea level. What is the minimum length of the vertical curve that can be used for 60 mph design speed ?

A. 280 feet B. 480 feet C.500 feet D. 300 feet

My answer 1520 ft doesn't match from above.

Please advise.

Click to expand...

I tried to solve the problem, my answer is A. Is that the correct answer

Samuel said:

monty74 said:

Please advise me regarding this quetion. When I calculated, my answer came 1520' based on CERM formula.
" A Vertical curve joins -1.2% grade to a + 0.8% grade. The PI of the vertical curve is at station 225+00 and elevation 167 feet above seal level. the centerline of the roadway must clear a pipe located at station 226+20 by 3.0 feet. The elevation of the top of the pipe is 167.65 feet above sea level. What is the minimum length of the vertical curve that can be used for 60 mph design speed ?

A. 280 feet B. 480 feet C.500 feet D. 300 feet

My answer 1520 ft doesn't match from above.

Please advise.

Click to expand...

Using the Formula, L + 2h/L -2h = (v-G1h/v-G2h)^1/2. My answer is A. Can you guys solve the problem

Click to expand...

Sorry?!? I don't know where you found that formula " L + 2h/L -2h = (v-G1h/v-G2h)^1/2" Is "h" the clearance height?

Thanks,

s_davis00 said:

Samuel said:

monty74 said:

Please advise me regarding this quetion. When I calculated, my answer came 1520' based on CERM formula.
" A Vertical curve joins -1.2% grade to a + 0.8% grade. The PI of the vertical curve is at station 225+00 and elevation 167 feet above seal level. the centerline of the roadway must clear a pipe located at station 226+20 by 3.0 feet. The elevation of the top of the pipe is 167.65 feet above sea level. What is the minimum length of the vertical curve that can be used for 60 mph design speed ?

A. 280 feet B. 480 feet C.500 feet D. 300 feet

My answer 1520 ft doesn't match from above.

Please advise.

Click to expand...

Using the Formula, L + 2h/L -2h = (v-G1h/v-G2h)^1/2. My answer is A. Can you guys solve the problem

Click to expand...

Sorry?!? I don't know where you found that formula " L + 2h/L -2h = (v-G1h/v-G2h)^1/2" Is "h" the clearance height?

Thanks,

Click to expand...

The answer is indeed 1520 feet. After calculating the length using the "pass-though" equation, I verified the length using the traditional parabolic equation and the elevation at the pipe location is calculated to be 170.65 feet, i.e. 3 feet above the pipe.

The 60 mph is irrelevant to the problem.

sac_engineer said:

s_davis00 said:

Samuel said:

monty74 said:

Please advise me regarding this quetion. When I calculated, my answer came 1520' based on CERM formula.
" A Vertical curve joins -1.2% grade to a + 0.8% grade. The PI of the vertical curve is at station 225+00 and elevation 167 feet above seal level. the centerline of the roadway must clear a pipe located at station 226+20 by 3.0 feet. The elevation of the top of the pipe is 167.65 feet above sea level. What is the minimum length of the vertical curve that can be used for 60 mph design speed ?

A. 280 feet B. 480 feet C.500 feet D. 300 feet

My answer 1520 ft doesn't match from above.

Please advise.

Click to expand...

Using the Formula, L + 2h/L -2h = (v-G1h/v-G2h)^1/2. My answer is A. Can you guys solve the problem

Click to expand...

Sorry?!? I don't know where you found that formula " L + 2h/L -2h = (v-G1h/v-G2h)^1/2" Is "h" the clearance height?

Thanks,

Click to expand...

The answer is indeed 1520 feet. After calculating the length using the "pass-though" equation, I verified the length using the traditional parabolic equation and the elevation at the pipe location is calculated to be 170.65 feet, i.e. 3 feet above the pipe.

The 60 mph is irrelevant to the problem.

Click to expand...

Upon further review of the problem, it is asking for the MINIMUM length of vertical curve. The "pass-through" equation is calculating the MAXIMUM curve length. Since we're only concerned with the minimum length, then 280 feet would be correct which would satisfy the 3' clearance for the pipe.

monty74 said:

Sorry Guys ! working in NY is headache, long commute, longer working hrs!
Anyway, what about if we use formula based on CERM 78.17. As per CERM 78.17 " IF THE STATION AND ELEVATION OF THE POINT P, THE STATION AND ELEVATION OF THE BVC OR THE VERTEX AND THE GRADIENT VALUES G1 AND G2 ARE KNOWN, THE CURVE LENGTH CAN BE DETERMINED EXPLICITLY".

This problem is not same as per fig. 78.11 but it reverse ( obstruction is below ) Based on this formula I get 1520 ft answer.

TANYA : Based on your solution , I am not getting 422 ft as per CERM table 78.4. Which formula you used.

Click to expand...

actualy, Iwas wrong, it L=271.3

Based on V=60, use tbl 78.2 solve SSD= 570

then, ignore LP elev, use formula L=A*SSD^2 / 400+(3.5*SSD) see table 78.4 page 78-14

the answer is A

Tanya,

If you use that equation, doesn't L have to greater than S? According to Greenbook, if S<L then you use that equation. What do you think?

maximus808 said:

Tanya,If you use that equation, doesn't L have to greater than S? According to Greenbook, if S<L then you use that equation. What do you think?

Click to expand...

You are right, i forgot to check. I spend so much time to think about this problem, that when I recalculated it, I didn't check. But you may use the other formula for S<L. It's 2 of them, isn't it. but I think this is the way to work out this problem. Hope I am right

I have checked with our sr.engineer within our company.

Correct Answer is 480 ft.

Following is procedure :

L=KA , K=136 ( EXH- 3-78) , L= 272 feet.

As per ITE we need to check clearnace criteria - page 75.

L= 4w-2z+4(w^2-wz)^1/2

Based on this eqation you can get 4.76stations.

Thanks for your support to resolve this problem.