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02/26/2012 09:50:40 AM · #1 |
i need help. this is for a takehome test. i know there are few people here with this background
if anyone knows anything about vertical curves. i have a problem and i am just stumped.
i have to tangent sections, the west section is -3% and the east section is +2% the first curve is a sag curve, the second is a crest. I need to find the grade between the two pvis. the design speed if 50mph. the distance between the pvis is 1275'
extra info in case you forgot or don't know it:
K=L/A
K=rate of curvature of the parabola
L is the length of the vertical curve
A is the algebraic difference between the tangent grades
at 50mph (from AASHTO) K=96 for the sag and 84 for the crest curve.
so i think i am close but the math doesn't work. im trying to relate the two equations to each other since:
K=L/A and A=abs(G2-G1) so for the sag 96=L/(3+X) and the crest 84=L/(2+X) X has to be the same in both equations.
so if i set them equal to each other i get (L-288)/96 = (L-168)/84, i get an L= -672 which corresponds to a grade of -10% and an A value of -7 for the sag and -8 for the crest. A can't be negative.
so where did i screw this up?
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02/26/2012 10:03:59 AM · #2 |
while typing it, i think i cracked it. i think i had my equation set wrong
it should have been A = X-3; X+2
(288-L)/96 = (168+L)/84 this gives a x of .67, which works.
i have another problem if anyone wants to help confirm. it much simpler. has to do finding the max acceleration of a car :) |
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02/26/2012 11:48:34 AM · #3 |
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02/26/2012 03:29:49 PM · #4 |
curious to see how others attempt this one:
the driver of a vehicle on a level road determined that she could increase her speed from rest to 50mph in 35 seconds and from rest to 65 mph in 95 seconds. assume that the acceleration varies linearly with speed (with a non-zero intercept). determine the maximum acceleration in each case. |
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