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computing per cent of grade

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  • Member since
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  • 64 posts
computing per cent of grade
Posted by jimsabol on Thursday, June 19, 2008 1:50 PM
What is an algorithm for figuring--for example--the grade needed to rise 35" in 952" or 5" in 132"?  (Talking O scale here) Thanks. NisquallyJim here.
  • Member since
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  • From: SE Minnesota
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Posted by jrbernier on Thursday, June 19, 2008 2:05 PM

  There is nothing  'scale specific' about calculating a grade.  It is simply the rise(unit) in 100 units of length.  For example, a 2% grade rises 2" in a run of 100".  Your example would go into the equation like this:

(35*100)/952 = 3.67%

(5*100)/132 = 3.78%

 

Jim

Modeling BNSF  and Milwaukee Road in SW Wisconsin

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  • From: Northern CA Bay Area
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Posted by cuyama on Thursday, June 19, 2008 2:06 PM

 jimsabol wrote:
What is an algorithm for figuring--for example--the grade needed to rise 35" in 952" or 5" in 132"?  (Talking O scale here) Thanks. NisquallyJim here.

Grade is easier than you might think and works the same for all scales. It's just rise divided by run (use the same units for each).

35" in 952" is nearly 3.7% (35/952). 5" in 132" is nearly 3.8% (5/132). Fairly steep for mainline running and you'll probably want some of that distance at each end for a vertical transition, which makes the remaining grade a bit steeper.

[Edit: Just like Jim B. said while I was typing!]

Byron
Model RR Blog

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Posted by jimsabol on Thursday, June 19, 2008 2:17 PM
Thanks, JR.  (I realize the computation is not scale-specific.  It's just that the nmbers are so large that N scale guys think it's a misprint.)  Why didn't you sit behind me in algebra?  Nisqually Jim here.
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Posted by jimsabol on Thursday, June 19, 2008 2:25 PM
Thanks to you, too, Byron.  I realize that 3.67% is a steep grade.  I was just playing with made-up numbers to see how the algebra works.  Thanks to you and JR, I get it.  Nisqually Jim here.

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