Elevation Grades

[Anonymous]

I'm sketching out my layout to begin creating the benchwork. I am pretty much a novice at this and I need to know what the mathematical formula is for figuring out how many inches I can elevate the roadbed at any given point of the grade. For example I know that if I have a 2% grade and I want it to eleveate from 0 inches to 8 inches that I need 144 inches of track to do so. How do I figure out how many inches of elevation per foot or inches I can raise the roadbed.

Thanks in advance for any help you can give me.

[Anonymous]

I'm sketching out my layout to begin creating the benchwork. I am pretty much a novice at this and I need to know what the mathematical formula is for figuring out how many inches I can elevate the roadbed at any given point of the grade. For example I know that if I have a 2% grade and I want it to eleveate from 0 inches to 8 inches that I need 144 inches of track to do so. How do I figure out how many inches of elevation per foot or inches I can raise the roadbed.

Thanks in advance for any help you can give me.

[Anonymous]

No you would need 400 inches of track to elevate 8" at a 2% grade. Look at it thisway, 1" elevation in 100" of track is 1%, 2" in 100" is 2%, and so on. Hope this helps.

[Anonymous]

No you would need 400 inches of track to elevate 8" at a 2% grade. Look at it thisway, 1" elevation in 100" of track is 1%, 2" in 100" is 2%, and so on. Hope this helps.

[Anonymous]

Thanks for pointing out my mistake, but I still need to know how many inches of elevation per foot or inches I can raise the roadbed.

[Anonymous]

Thanks for pointing out my mistake, but I still need to know how many inches of elevation per foot or inches I can raise the roadbed.

[tomnoy3]

Instead of using inches and feet, why dont you go with the metric system. A 2% grade will be a 2% grade no matter what you use to measure. And the units based on ten really help with percents. If your still wanting to stay with customary, its roughly 1/4 inch rise for every foot of track.

[tomnoy3]

Instead of using inches and feet, why dont you go with the metric system. A 2% grade will be a 2% grade no matter what you use to measure. And the units based on ten really help with percents. If your still wanting to stay with customary, its roughly 1/4 inch rise for every foot of track.

[Anonymous]

Metric or not, just use the same units for horizontal and vertical distances. You've got three numbers to play with, the vertical distance to want your tracks to go (rise), the horizontal distance over which they will be rising (run), and the grade. Grade (in percent) = rise * 100 / run. So for the 2% grade, 8" rise example, 2 = 8 * 100 / run, and a little algebra tells you that a run of 400" is needed. For the 8" rise, 144" run example, grade = 8 * 100 / 144, or about 5.5% (too much!).

[Anonymous]

Metric or not, just use the same units for horizontal and vertical distances. You've got three numbers to play with, the vertical distance to want your tracks to go (rise), the horizontal distance over which they will be rising (run), and the grade. Grade (in percent) = rise * 100 / run. So for the 2% grade, 8" rise example, 2 = 8 * 100 / run, and a little algebra tells you that a run of 400" is needed. For the 8" rise, 144" run example, grade = 8 * 100 / 144, or about 5.5% (too much!).

[Anonymous]

Here's how you figure it... 2% equals .02 so 12" X .02 = .24 So the answer is about 1/4 inch per foot.

[Anonymous]

Here's how you figure it... 2% equals .02 so 12" X .02 = .24 So the answer is about 1/4 inch per foot.

[Anonymous]

I used two percent grades on my layout. I spaced the joists on 16" centers. However, on curves, the dimensions become longer or shorter, so you have to measure the total length of the curve from support to support.

Since 2% is 2" in 100" the first fomula for straight track was 2" / 100" * 16" = .32" or about 5/16".

A curve creating 19" between supports is 2 / 100 * 19 = .38" or about 3/8".

Hope this helps you on the curves!

Happy modeling.

[Anonymous]

I used two percent grades on my layout. I spaced the joists on 16" centers. However, on curves, the dimensions become longer or shorter, so you have to measure the total length of the curve from support to support.

Since 2% is 2" in 100" the first fomula for straight track was 2" / 100" * 16" = .32" or about 5/16".

A curve creating 19" between supports is 2 / 100 * 19 = .38" or about 3/8".

Hope this helps you on the curves!

Happy modeling.

[Anonymous]

I wonder If my 40% grade will cause any problems.

[Anonymous]

I wonder If my 40% grade will cause any problems.

[Anonymous]

Just mount a big magnet under your locomotive to give it magna-traction. I have one that will do loop de loops, so 40% should be no problem.

[Anonymous]

Just mount a big magnet under your locomotive to give it magna-traction. I have one that will do loop de loops, so 40% should be no problem.

[Anonymous]

2% is 2% no matter what the distance. If you want a 2% grade per foot, the elevation is 2% of 1 foot per foot. That's 0.24 inches per foot of track.

per inch, it is 2% of an inch, or .02 inch elevation per inch of track.

For curves, you need to know the radius and PI (3.14) to figure the arc length for the same thing. You might want to reduce grade to 1% on curves if running long trains due to added friction.

[Anonymous]

2% is 2% no matter what the distance. If you want a 2% grade per foot, the elevation is 2% of 1 foot per foot. That's 0.24 inches per foot of track.

per inch, it is 2% of an inch, or .02 inch elevation per inch of track.

For curves, you need to know the radius and PI (3.14) to figure the arc length for the same thing. You might want to reduce grade to 1% on curves if running long trains due to added friction.