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Vertical Curve Data fit parabola 1

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stanfea

Structural
Sep 30, 2003
29
I have some info on a vertical curve. Everything is given, however I do not have reference to the exact nature of the data. The elevations of the start and end of the curve is given as well as the PVI at the middle. The curve is a crest curve symmetric also. I have the following.

2% slope into and out of curve
V.C.L. = 410' (vertical curve length)
SSD = 369 (stopping sight distance ??)
E=20.5 (no idea?)
K=102 (rate of change = VCL/(sum of grades))?

I know what the VCL and the K are. What I am trying to accomplish is determining a parabolic curve that would fit to this data. I also do not know the high point of the curve (at least I can't figure it out). If someone could give me some constructive direction on fitting a parabola it would be greatly appreciated. I have fit parabolas before when I knew the high point of the curve.

Thanks
Stan
 
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The curve is a parabola. Because it is symmetrical you can easily calculate the high point in one of two ways.

1. Subtract the "E" distance from the elevation of the VPI.

or

2. The high point lies exactly halfway between the VPI and the line joining the BVC and EVC.

You should be able to find this, and much more information in any good land surveying book.

Good luck
 
So the "E" distance is the distance from the intersection of the two tangents to the high point of the parabola? I am fairly annoyed that the units were only put on the first number on the drawing. If the E had been 20.5" I might have figured it out a bit easier. I, as you can tell do not do survey math, and am a bit unfamiliar with the various terms.

Thanks for your Help, much appreciated
Stan
 
Use the second method. The "E" value doesn't sound like it could be correct for this very long, flat curve.

good luck
 
Thanks
By sketching the curve in ACAD with a quadratic spline (parabola) I figured the height of the curve above the Level end points to be about 37" or so. The difference in elevation between the end points and the center intersection of the tnagents is 49.2". So I don't think the E value is 1) correct or 2) Is not what we think it is.
These values are on a bridge drawing and represent the camber of the deck, the bridge being 400' long, and the curve being 410' long (5' past each end). I really need to know what these values represent, I am from Canada and was hoping these were standard AASHTO variables. They are not the same on my side of the border the best that I can tell...Still struggling.

Stan
 
Stan,

Please do get hold of a good surveying text or perhaps a Highway Design Manual. There you will find illustrations of these values which will be much clearer than the word descriptions. This web page doesn't allow us to send sketches to each other so it is often difficult to convet geometric data. I believe parabolas are calculated the same way in the US and in Canada.

I don't know what "E" is. Don't think it is standard terminology.

You may be able to find some design manuals on the internet. Have you tried searching ?

good luck

Russ

 
E is calculated as the difference in % of Grade divided by 8, times the length of the vertical curve in stations. Station is 100'.

E = ((-2.0 - 2.0)/8 ) X (410/100) = 2.05'

Looks like a misplaced decimal on the plans.

I wouldn't depend on Autocad's curve routines to give a good answer.
 
Thanks guys for the help.

I was able to calculate the station elevations and determine the camber at midspan after going thru my survey books. I did not use the 'E' in my calcs but got the same number for camber, so looks good. By my calcs the camber is 2.05' which as noted above matches the 'actual' E value. I love working with other peoples numbers...

Thanks
Guys
 
New development.

In AutoCAD there is a system variable called splinetype. If the variable is set as '5' you get a perfect parabola, if it is set as '6', you get a cubic spline (not a parabola). Somehow my variable got reset. Now in AutoCAD I get exactly the same answer.

Stan
 
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