Login
or
Register
Subscriber & Member Login
Login, or register today to interact in our online community, comment on articles, receive our newsletter, manage your account online and more!
Login
Register
Home
»
Model Railroader
»
Forums
»
Layouts and layout building
»
How To Figure Length Of Inclines
How To Figure Length Of Inclines
2320 views
11 replies
Order Ascending
Order Descending
jacon12
Member since
November 2002
From: US
4,641 posts
How To Figure Length Of Inclines
Posted by
jacon12
on Saturday, September 25, 2004 7:52 PM
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?
HO Scale DCC Modeler of 1950, give or take 30 years.
Reply
jwmurrayjr
Member since
February 2002
From: US
517 posts
Posted by
jwmurrayjr
on Saturday, September 25, 2004 8:37 PM
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).
Jim Murray
The San Juan Southern RR
Reply
jacon12
Member since
November 2002
From: US
4,641 posts
Thanks Jim. I found my copy of
Posted by
jacon12
on Saturday, September 25, 2004 8:42 PM
John Armstrong's book " Track Planning For Realistic Operation" so I think it is going to be a big help also.
HO Scale DCC Modeler of 1950, give or take 30 years.
Reply
jacon12
Member since
November 2002
From: US
4,641 posts
Posted by
jacon12
on Saturday, September 25, 2004 8:43 PM
Thanks for the formula Jim!
HO Scale DCC Modeler of 1950, give or take 30 years.
Reply
CBQ_Guy
Member since
September 2003
From: North Central Illinois
1,458 posts
Posted by
CBQ_Guy
on Sunday, September 26, 2004 5:44 PM
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.)
"Paul [Kossart] - The CB&Q Guy" [In Illinois] ~ Modeling the CB&Q and its fictional 'Illiniwek River-Subdivision-Branch Line' in the 1960's. ~
Reply
Anonymous
Member since
April 2003
305,205 posts
Posted by
Anonymous
on Sunday, September 26, 2004 7:57 PM
do these formulas apply to all gauges, or just HO?
Reply
Edit
preceng
Member since
August 2003
From: Pittsburgh, PA
208 posts
Posted by
preceng
on Sunday, September 26, 2004 9:11 PM
All scales
Allan B.
Reply
RedLeader
Member since
January 2001
From: Barranquilla, Colombia
327 posts
Posted by
RedLeader
on Monday, September 27, 2004 4:59 PM
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%
Reply
Anonymous
Member since
April 2003
305,205 posts
Posted by
Anonymous
on Monday, September 27, 2004 9:36 PM
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?
Reply
Edit
Anonymous
Member since
April 2003
305,205 posts
Posted by
Anonymous
on Monday, September 27, 2004 9:43 PM
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?
Reply
Edit
RedLeader
Member since
January 2001
From: Barranquilla, Colombia
327 posts
Posted by
RedLeader
on Tuesday, September 28, 2004 10:08 AM
I actually use that trick on my layout! esaments help everytime. The straighter, the better.
Reply
Anonymous
Member since
April 2003
305,205 posts
Posted by
Anonymous
on Tuesday, September 28, 2004 11:28 AM
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
Reply
Edit
Subscriber & Member Login
Login, or register today to interact in our online community, comment on articles, receive our newsletter, manage your account online and more!
Login
Register
Users Online
There are no community member online
Search the Community
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT
Model Railroader Newsletter
See all
Sign up for our FREE e-newsletter
and get model railroad news in your inbox!
Sign up