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Elevation Grades
Elevation Grades
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Anonymous
Member since
April 2003
305,206 posts
Elevation Grades
Posted by
Anonymous
on Sunday, August 3, 2003 8:39 PM
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.
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Anonymous
Member since
April 2003
305,206 posts
Elevation Grades
Posted by
Anonymous
on Sunday, August 3, 2003 8:39 PM
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.
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Anonymous
Member since
April 2003
305,206 posts
Posted by
Anonymous
on Sunday, August 3, 2003 9:02 PM
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.
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Anonymous
Member since
April 2003
305,206 posts
Posted by
Anonymous
on Sunday, August 3, 2003 9:02 PM
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.
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Anonymous
Member since
April 2003
305,206 posts
Posted by
Anonymous
on Sunday, August 3, 2003 9:10 PM
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.
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Anonymous
Member since
April 2003
305,206 posts
Posted by
Anonymous
on Sunday, August 3, 2003 9:10 PM
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.
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Edit
tomnoy3
Member since
May 2002
223 posts
Posted by
tomnoy3
on Sunday, August 3, 2003 10:17 PM
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.
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tomnoy3
Member since
May 2002
223 posts
Posted by
tomnoy3
on Sunday, August 3, 2003 10:17 PM
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.
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Anonymous
Member since
April 2003
305,206 posts
Posted by
Anonymous
on Monday, August 4, 2003 10:02 AM
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!).
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Anonymous
Member since
April 2003
305,206 posts
Posted by
Anonymous
on Monday, August 4, 2003 10:02 AM
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!).
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Anonymous
Member since
April 2003
305,206 posts
Posted by
Anonymous
on Monday, August 4, 2003 2:10 PM
Here's how you figure it... 2% equals .02 so 12" X .02 = .24 So the answer is about 1/4 inch per foot.
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Anonymous
Member since
April 2003
305,206 posts
Posted by
Anonymous
on Monday, August 4, 2003 2:10 PM
Here's how you figure it... 2% equals .02 so 12" X .02 = .24 So the answer is about 1/4 inch per foot.
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Anonymous
Member since
April 2003
305,206 posts
Posted by
Anonymous
on Monday, August 4, 2003 6:08 PM
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.
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Anonymous
Member since
April 2003
305,206 posts
Posted by
Anonymous
on Monday, August 4, 2003 6:08 PM
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.
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Anonymous
Member since
April 2003
305,206 posts
Posted by
Anonymous
on Tuesday, August 5, 2003 8:58 AM
I wonder If my 40% grade will cause any problems.
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Anonymous
Member since
April 2003
305,206 posts
Posted by
Anonymous
on Tuesday, August 5, 2003 8:58 AM
I wonder If my 40% grade will cause any problems.
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Anonymous
Member since
April 2003
305,206 posts
Posted by
Anonymous
on Tuesday, August 5, 2003 2:32 PM
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.
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Anonymous
Member since
April 2003
305,206 posts
Posted by
Anonymous
on Tuesday, August 5, 2003 2:32 PM
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.
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Anonymous
Member since
April 2003
305,206 posts
Posted by
Anonymous
on Tuesday, August 5, 2003 3:17 PM
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.
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Anonymous
Member since
April 2003
305,206 posts
Posted by
Anonymous
on Tuesday, August 5, 2003 3:17 PM
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.
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