--- In [email protected], "Tyler" <tfroatz4@...> wrote:
> Since my trigonometry is a little rusty, I went to double check my results
> with an online calculator and the results don't match. > Over a 4 foot module
> with a continuous slope;
> A 2% grade =1.68 inches of rise
> A 3% grade=2.52 inches of rise
> How would I calculate the amount of horizontal space necessary to transition
> to the maximum slope and how much height would be achieved over the course of
> the transition zone?
It's pretty simple if one works in rise times run, changing the percentage
to it's decimal equalent than keeping all the units in the common format(
inches with inches or feet with feet)so 2% is .02 times the run of 48
inches=.96 inch or a tiny bit less than an inch. and 3% is .03 times 48 =1.44"
or somewhat less than an inch and a half. A 16 foot run at 2% would be 16 feet
times 12 inches =192" times .02= 3.84" or about 3 7/8 inches. 16 feet run at
3% would be 5.76 inches or very close to 5 3/4" .
Transition vertical curves should be centered over the intersection of the
level segment and the grade segment.and extend far enough each direction to
keep the grade change under approximately 1% per car length. So for 80 foot
cars (15"in S scale)about 7.5 inches before the verticle curve point of
intersection(PIVC) start a 1% grade that extends for 15 inches to intersect
with the 2% grade. This is not a true verticle curve , it's actuall a compound
curve of a short segment of 1% then into the 2% ruling grade but the stiffness
of the roadbed will turn it into a verticle curve. For 3% it would start 15
inches before the PIVC and be 15 ihches at 1%, then another 15 inches at 2%, to
intersect the 3% grade at 15 inches beyond the PIVC.
In practice the roadbed and common sense will smooth all these out and not
cause a trainwreck :>) .....dave
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