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vertical curve layout easy Q

vertical curve layout easy Q

(OP)

I am trying to build a ramp to a an elevation of 13' and want to use 2 vertical curves with a 20 % grade that are 20 ft long.  How could I find the elevation at each point?

RE: vertical curve layout easy Q

13/0.20=65' long tangent (VPI to VPI). But the 20' long curves at each end add 10' from VPI to VPC. So say starting point is beginning of sag curve at ground level, 20' is end of curve, 65' is beginning of crest curve, 85' is VPT at +13'.  EL at 20' is 10 * .20=2'; EL at 65' is 55' * 0.20 = 11'.

RE: vertical curve layout easy Q

You can also use an Old Surveyor's trick.  Just draw the vertical curve on graph paper.  Use a horizontal scale of , say, 1"=10'.  But, here's the trick, use a very exaggerated vertical scale of say 1"= 1'  or even 1" = 0.5'.  Then, just scale the elevations you are interested in.

good luck

RE: vertical curve layout easy Q

From a little program...

Sta= 0.00   Elev= 0.00
Sta= 2.00   Elev= 0.02
Sta= 4.00   Elev= 0.08
Sta= 6.00   Elev= 0.18
Sta= 8.00   Elev= 0.32
Sta= 10.00    Elev= 0.50
Sta= 12.00    Elev= 0.72
Sta= 14.00    Elev= 0.98
Sta= 16.00    Elev= 1.28
Sta= 18.00    Elev= 1.62
Sta= 20.00    Elev= 2.00
<20% slope betwen 20' & 65'>
Sta= 65.00    Elev= 11.00
Sta= 67.00    Elev= 11.38
Sta= 69.00    Elev= 11.72
Sta= 71.00    Elev= 12.02
Sta= 73.00    Elev= 12.28
Sta= 75.00    Elev= 12.50
Sta= 77.00    Elev= 12.68
Sta= 79.00    Elev= 12.82
Sta= 81.00    Elev= 12.92
Sta= 83.00    Elev= 12.98
Sta= 85.00    Elev= 13.00

RE: vertical curve layout easy Q

(OP)
thanks for you help

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