Re: HP-35S Vertical Curve: Elevation at Peak and at End Point Message #21 Posted by Eddie W. Shore on 23 May 2013, 3:56 p.m., in response to message #20 by Thomas Klemm
Quote:
Given your solution I get:
G = (1,030.469 - 1,000) / 870.545 = 0.035
H = (1,020.520 - 1,030.469) / (1,368 - 870.545) = -0.02
So it seems a factor 2 is missing somewhere.
As I'm not an engineer I wonder why the length L is divided into two parts similar to G:H (or rather |G|:|H|) to determine the peak. Why would somebody want to do that? What's the real world application of it?
Kind regards
Thomas
I am looking into this. I got the formulas from the Fundamental of Engineering Handbook reference book, if it helps.
You determination for G and H are straight forward.
Update: It seems to be a factor of 2 at peak point. Vertical curves, to my understanding, are symmetrical in a sense of tangent lines: one created from the beginning of the curve and on at the end of the curve. If a line is drawn at the intersection of the tangent line perpendicular to the base, that line bisects the length of the base.
The grade is determined from where the tangent lines meet. In a perfectly symmetrical vertical curves the tangent lines meet at the curve's peak elevation. For this case:
G = (y0 - I)/(L/2) and H = (I - y0)/(L/2) where y0 is the peak elevation.
Some more information is provided here:
http://www.wsdot.wa.gov/publications/manuals/fulltext/M22-23/Vertical.pdf
http://www.iowadot.gov/design/dmanual/02b-01.pdf
I am not an engineer either, I just have an interest in the subject. Hope this helps, Eddie
Edited: 23 May 2013, 10:09 p.m.
|