How To Figure Length Of Inclines

[jacon12]

I think a 1.5% grade is 1/4th inch grade for every 16 inches of track, so to climb just 3 inches in incline would would take 12 feet of track, is that correct? Is there a chart somewhere on the web that has the figures for 1% thru... say.. 3% or do you simply manually figure as I did and hope you get it right?

[jwmurrayjr]

The rise in inches divided by the length of the grade will give you the %.

2" / 100" = .02 or a 2% grade.

The grade is the rise in units (inches in this case) per 100 units (inches in this case).

[jacon12]

John Armstrong's book " Track Planning For Realistic Operation" so I think it is going to be a big help also.

[jacon12]

Thanks for the formula Jim!

[CBQ_Guy]

Since a 1 percent grade=1 inch of rise in elevation in 100 inches...AND

Since 100 inches is only 4 inches more than 8 feet...

Just appoximate that a 1 percent grade is roughly 1 inch of rise in 8 feet.

(If you want to get fancy, you can remember the extra 4 inches, but 1 inch rise in 8 feet is pretty close enough.)

[Anonymous]

do these formulas apply to all gauges, or just HO?

[preceng]

All scales

[RedLeader]

If you use metric system is much more simpler.

1% = For every meter you should raise 1cm
1.5% = For evey meter you should raise 1.5cm
... and so on.. easy...

So if your track length is 6 meters and you need a 1.5% grade, you must raise your track (1.5 x 6) 9cm.

It works backwards also... if raise 12cm in a length of 6meters, then your grade is (12/6) 2%

[Anonymous]

thank you R.L. i was browsing thru some train books from the U.K. Apparently, the approach over there in dealing with the general lack of space for run-up's, is to utilize a fair number of flying bridges, where one grade crosses above the other, but the lower track is lowered, thereby having the upper track only having to go up by half the normal incline. Wait a second, i just confused myself. OK: let's say you need a grade of 3% to cross over a track on the ground below. Well, they just lower the bottom track by 1.5%,Now the top track incline can be 1.5% as well. An easier grade both ways. And it looks beautiful in action, very flowing. Of course, i prefer the challenge of the 3% since i dig industrial switchers, and sharp curves. Here's a related question: 2 questions actually: does putting an easement on the outer rail of a curved incline help or hurt? And second, who has put easements on an N gauge layout? Did you see any performance or derailment differences?

[Anonymous]

thank you R.L. i was browsing thru some train books from the U.K. Apparently, the approach over there in dealing with the general lack of space for run-up's, is to utilize a fair number of flying bridges, where one grade crosses above the other, but the lower track is lowered, thereby having the upper track only having to go up by half the normal incline. Wait a second, i just confused myself. OK: let's say you need a grade of 3% to cross over a track on the ground below. Well, they just lower the bottom track by 1.5%,Now the top track incline can be 1.5% as well. An easier grade both ways. And it looks beautiful in action, very flowing. Of course, i prefer the challenge of the 3% since i dig industrial switchers, and sharp curves. Here's a related question: 2 questions actually: does putting an easement on the outer rail of a curved incline help or hurt? And second, who has put easements on an N gauge layout? Did you see any performance or derailment differences?

[RedLeader]

I actually use that trick on my layout! esaments help everytime. The straighter, the better.

[Anonymous]

take the hieght you want: say 6" is how high you want to go.

Divide that by the run length (the length your dedicating to getting up that hieght: say 144"

times that number by 100 to get the percent and you'll have your grade: 4.1%

so ( hieght / run ) x 100 = grade.

using this simple formula, you can determine the run you need at a set hieght and grade as well.

say you want a 2% grade and a hieght of 5"

you'd have: 5/run x 100 = 2 which is 250" needed, or 20.83 ft