S
scubasteve
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!!
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!!