# Vertical curve problem #113 in the '08 sample exam book.

4 replies to this topic

### #1 Guest_scubasteve_*

Guest_scubasteve_*
• Guests

Posted 30 December 2007 - 12:35 AM

Hi All,

I'm not a transportation guy and never took it in school, so this question is probably laughable to you folks. I can follow the solution (vaguely) part of the way, but get stuck. I've also referenced topic 17 of page 79-12 of Lindeburg's CE Ref Man. 10th Ed., talking about vertical curves, and still can't figure this problem out, although I follow the manual's examples ok. Anyway, if you don't have it handy, here's the question:

The tangent vertical alignment of a section of proposed highway is shown in figure below. The vert. clearance (ft) between the bridge structure at Sta 73+00 and the vert curve is nost nearly..

The figure is (from what I can tell) your basic vertical curve problem, with PVC on left, PVI in middle below curve at vertex, and PVT at the exti on right. Entry grade g1=-3% downhill and ext grade g2 = 2% uphill. L=1500 ft or 15 sta. PVI coord's are 76+00 and 334.56' elev. Point on bridge structure is at Sta 73+00 and 356.94 elev.

I can find the vert. dist. from PVI to the curve at Sta 73+00 (under point on bridge) using geometry, but I get bogged down after that. There is a similar sample problem in the manual that I follow ok but, best I can tell, is totally different from the solution in the practice exam. Here is the sample exam solution:

Compute tangent elevation at Sta 73+00 = 334.56+(3)(3)=343.56' (figured this out)

Compute tangent offset, e, at the PVI station = (L/8)(g2-g1)=(15/8)(2-(-3))=9.375' (where does the 8 come from, and why do we need this offset?)

Compute tangent offset, y, at Sta 73+00 = (4e/L^2)x^2 = ((4)9.375(4.5)^2)/(15)(15)=3.38' (where does this equation come from, and what is the 4 and the x?)

It's basic math after that, and the answer is 19'. I try it with the alternative method in the ref. manual, and come up with 42'.

Any light you can shed on this problem is greatly appreciated!!

Veteran

• 2,815 posts
• Gender:Male
• Location:California
• Discipline:Military Engineer

Posted 30 December 2007 - 07:02 PM

It would help if you could scan the problem and post here to make sure we're all on the same page (so the speak).

Are you sure the bridge elevation is 356.94 and not 365.94? A quick run through the simple vertical curve calculations gives me an answer of 10 ft (which would be 19 ft if the bridge elevation was 365.94). Verify, and I'll be happy to explain.

A bit of advice if you're preparing for the exam: Don't worry about where the formulas come from (that's what college was for!) - just understand how they're applied and what the variables mean. Sure, it's great to understand what's actually going on behind the equation, but isn't it more important to pass the exam? Regarding your question about the tangent offset (M or sometimes called E), it's probably a straightforward result of geometry and calculus, but so long as you understand the formula, does it matter?

### #3 Guest_traffic_*

Guest_traffic_*
• Guests

Posted 03 January 2008 - 10:26 PM

I'm not sure I'm following the problem as desrcibed but the formulas I used also gave me 10.95

R= g2-(-g1)/2L L in sta.'s

Then E at any point = (Epvc +(G1*X) + R*(X^2))

Scan in the prob and I'll take a look, traffic

### #4 Guest_eric08_*

Guest_eric08_*
• Guests

Posted 06 January 2008 - 05:31 PM

QUOTE (IlPadrino @ Dec 30 2007, 01:02 PM) <{POST_SNAPBACK}>
It would help if you could scan the problem and post here to make sure we're all on the same page (so the speak).

Are you sure the bridge elevation is 356.94 and not 365.94? A quick run through the simple vertical curve calculations gives me an answer of 10 ft (which would be 19 ft if the bridge elevation was 365.94). Verify, and I'll be happy to explain.

A bit of advice if you're preparing for the exam: Don't worry about where the formulas come from (that's what college was for!) - just understand how they're applied and what the variables mean. Sure, it's great to understand what's actually going on behind the equation, but isn't it more important to pass the exam? Regarding your question about the tangent offset (M or sometimes called E), it's probably a straightforward result of geometry and calculus, but so long as you understand the formula, does it matter?

I agree I've worked it and come up w/ 10' as well

### #5 Guest_scubasteve_*

Guest_scubasteve_*
• Guests

Posted 11 January 2008 - 12:51 AM

QUOTE (eric08 @ Jan 6 2008, 10:31 AM) <{POST_SNAPBACK}>
I agree I've worked it and come up w/ 10' as well

thanks everyone - i will check that I have the elevation # correct, and scan in the problem and solution as well. thank you for the advice not to get to bogged down in understanding everything insdie and out. Ive started realizing that as I've been studying and practicing the problems - I'll never get through it if I try to theorize on every one of them!

#### 0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users

=