so, I've been studying, and I can usually get the vertical curve through a point type problems.... but this one is kicking my butt... and I think I'm just doing something dumb... so please help... its a mission now as to why I can't solve it using the quad formula.
so:
L=2x+150
g1= 0.02
g2= -0.06
Elev@ station 14+25 ((75*-.02)+270.00)=268.50'
Elev@clear pipe is = 266.50' (elevation of curve at point)
y=2.00'
y=ax^2 a=(g2-g1)/2L
2= ((-.08)/(4x+300)) *x^2
8x+600=(-.08)x^2
.08x^2+8x+600=0
time for quad formula
(-b +/- sqrt(b^2-4ac))/2a
(-8 +/- sqrt(64-192))/ 0.16
can't have a neg square root....
I know there are other ways to solve it, and i have, but, i don't see what I'm doing wrong yet on this one, with this style.
thanks!
vertical curve question help
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shaunb
- Posts: 58
- Joined: Tue May 08, 2012 12:40 pm
vertical curve question help
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DerekHouston
- Posts: 6
- Joined: Tue Oct 26, 2010 10:49 am
In your equation, your y-value of 2 should be negative, not positive. The methodology for vertical offsets for a vertical curve and for your equation is:
y is negative when the curve is a crest curve.
y is positive when the curve is a sag curve.
Check out page 3 of this pdf, this can explain it better.
http://www.pdhcenter.com/courses/l121/l121content.pdf
y is negative when the curve is a crest curve.
y is positive when the curve is a sag curve.
Check out page 3 of this pdf, this can explain it better.
http://www.pdhcenter.com/courses/l121/l121content.pdf
Derek Houston, LSIT