In preparation for the December CST III exam I watched on YouTube Vertical Curves - Part 1 problem:
grade in = - 4.00%
grade out = + 3.80%
PVI station 52 + 00
PVI elevation = 1261.50'
Railroad crossing station 53 + 50
Railroad crossing elevation = 1271.20'
Design an equal tangent parabolic curve to meet a railroad crossing. I need to solve for the length (L) of the vertical curve. I know the elevation (Y) of the railroad crossing therefore Yrailroad = Ypvc + g1X + (r/2)X^2. The professor fills in the numbers therefore
1271.20 = [1261.50 + 4.00 (L/2)] + [ - 4.00 (L/2 + 1.5)] + [(3.80 + 4.00 / L) (L/2 + 1.5)^2]. The professor simplifies the equation to a quadratic equation. That's where I'm lost. He gets 0.975L^2 - 9.85L + 8.775 = 0. Then L = [ - b +/- SQRT(b^2 - 4ac] / 2a. The answer is 9.1152 stations. Can anyone explain how to simplify the equation to a quadratic equation?
Here's a screen shot of the original problem and equation. Your parentheses in the last term do not quite match the equation on the slide.
The trick is to multiply both sides of the equation by the L in the denominator of the last term on the right-hand side of the equation. When you do that and collect terms, you'll have what the professor has.
But, note: He made a change in red whereby he eliminated the 2 in the denominator of that last term. His solution includes the 2. I have no idea whether it belongs there or not; he seemed to indicate that it does not.
No doubt, many readers of this post will know whether the 2 belongs or not.
Thanks for your help!
For those of you interested in investigating this further, Jerry Mahun has a section on Vertical Curve Fitting in his Open Access Library of survey education material at jhttp://errymahun.com/&apos ;">errymahun.com. The curve fitting section is at http://jerrymahun.com/library/Curves/Vertical/B6.htm .