Any suggestions for HO gradient?

Can anyone suggest the minimum length required to meet a rise of 100mm
or 4 inches in HO scale.
Any help is greatly appreciated.
Reply to
Scuzzie
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with what sort of motive power and what length/weight of trains?
Reply to
a_a_a
Thanks for the reply . . . nothing special in terms of power, just an average Hornby or similar and 4 - 5 passenger cars or 10 - 15 goods wagons
thanks
Reply to
Scuzzie
8 feet length on a 4 inch rise will give you about a 2 percent grade, which would be my recommendation. The formula you use to determine the percent of grade is: one quarter inch rise for every 1 foot of length equals about 1 percent grade.
Fred Ellis
Reply to
Fred Ellis
At a grade of 1%, that would be 100*100mm or 10,000mm [1,000cm, 10m] (or 100*4in or 400 inches [33-1/3 feet]). 1% is fairly slight. At 2%, it would be 50*100mm or 5,000mm [500cm, 5m] (or 50*4in or 200 inches [16-2/3 feet]). 2% is beginning to get 'steep' for many locos pulling trains of non-trivial size. Note: grades are effectively *worse* on curves: a grade on a curve 'behaves' (in terms on traction / power needed) as if it was steeper than the same grade on a straight section, so the grade on a spiral would have to be more gentle. A 36" radius circle has a circumfrence of 226.19448", so once around at 1% is a rise of 2.2619". This is *barely* enough to 'stack' trackage (unless you are running double stacks or Superliners (double deck passenger cars)). Separating 1/2 circles by 4' inproves things:
1st 1/2 circle (36" radius) @ 1%: 1.1309" rise 1st 4' straight @ 2%: 0.9600" rise (2.0909" cumulative) 2nd 1/2 circle (36" radius) @ 1%: 1.1309" rise (3.2218" cumulative) 2nd 4' straight @ 2%: 0.9600" rise (4.1818" cumulative)
Reply to
Robert Heller
Scuzzie skriver:
2% incline works fine.
2% = 20 mm incline per meter
100mm = 5m
Klaus
Reply to
Klaus D. Mikkelsen
That's about a 4% grade, Fred: 8ft = 96 inches. Etc. Too steep.
But you're right to recoemmend a 2% grade (or less).
Cheers,
wolf k.
Reply to
Wolf K
4% Grade (way to steep) takes 8ft to get 4in height. 3% Grade (most commercial pier sets in HO) take 12ft to get 4.5in height. 2% Grade (max generally recommended) takes 16ft for 4in height.
Len
: Can anyone suggest the minimum length required to meet a rise of 100mm : or 4 inches in HO scale. : : Any help is greatly appreciated.
Reply to
Len
I created a test layout with XTrkCAD that raises 4", using a mix of 2% (on straight sections) and 1% (on curved sections):
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This should get you started.
Reply to
Robert Heller
Don't forget to allow for the vertical transition curves at the beginning and end of the grade. Take a look at:
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Unless you're doing mountain, logging, or old time railroading, a 2% grade is aout as high as you should go to stay realistic. That would take 200 inches or a little over 16 feet, plus a little extra for transition curves.
Reply to
Larry Blanchard
Don't you mean 2 cm per meter?
Reply to
Larry Blanchard
Wolf, thanks for the correction. I thought I had made a miscalculation. I should of wrote "16 feet of length on a 4 inch rise will give you about a 2 percent grade."
Fred Ellis
Reply to
Fred Ellis
He probably doesn't but as 20mm = 2cm you are just as right.
Reply to
Erik Olsen
Larry Blanchard skriver:
Since 20 mm equals 2 cm I mean both....
Klaus
Reply to
Klaus D. Mikkelsen
I *knew* my eyes were going bad. I originally read that as 2 mm, not 20. My mistake.
Reply to
Larry Blanchard
Depends on the sort of railroad you want to model and how much space you have to model it.
Major prototype roads try very hard to hold their grades to less than 2% (that would be a two-foot rise for every hundered feet the rails cover horizontally) as it starts getting very expensive in terms of the number of locomotives required to pull a given load -and the fuel they must burn to do so as the grades get steeper and steeper. But in a modelling situation we rarely have enough space to climb to the heights we need unless we use 2% or more. (We are fortunate in the fact that most model railroads are quite short in terms of scale milage, and as a result they do not generally allow us to run prototype length trains anyway.) And the requirement that we *must* pull shorter trains also means that we can run them up grades that would be steeper than those that generally occur in real life. (If, OTOH, you are modeling a mining or logging line, it wasn't unusual to see them use 4% grades or higher, and even 7% or steeper wasn't unheard of.)
While the currently accepted ideal for model railroad grades hovers around 2%, you should know that many classic model railroads have been built using 4%+ grades, and that they were built that way on purpose. John Allen's famed Gorre & Daphetid railroad used 4% ruling grades simply because the heavy grades allowed him to pack a lot of mountain railroading into a relatively small basement with track running from 29" above the floor all the way up to 60" at the highest point: a climb of over 2 1/2 feet!
John favored short (six to fifteen car) trains on steep grades anyway, because they allowed him to do some interesting prototype moves such as double-heading steam locomotives or doubling -or even tripling- a hill (One locomotive breaking the train into sections and taking them up the grade one chunk at a time; then re-assembling the train at the top.) rather than just run his trains around in circles.
In short; you need to first define what sort of railroad you want to build and what you want to do with it. The answers to those questions will dictate what sort of grades you can use for your particular purposes.
~Pete
Reply to
Twibil
It would be of great help if you accepted the metric system. Then that 2%, which is what I also use as a maximum, translates into 2 cm rise per meter. Or 2 m per 100 m. It can be so simple..
Reply to
Wim van Bemmel
Correct. 20 mm = 2.0 cm = 2 hundredth of a meter.
?????? Absolute nonsens.
Reply to
Wim van Bemmel

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