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2.5% is usually the normally accepted maximum grade for trains in any scale. However, more and less has been used. Several factors will determine what you can get away with on your railroad.

If you have this grade on a curve, you've just increased the grade and the tighter the curve the higher the grade. It is called 'effective grade' and it is a very important consideration when designing a railroad with grades.

While your locomotive may pull a string of ten cars straight up 2.5% grade, it may stall a third of the way through an 18" radius curve while the grade continues to climb.

It also will depend on what equipment you run. Do your locomotives use tractions tires? Are you pulling short excursion or short local freight trains? Do your steam locomotives have enough weight to provide better adhesion? Diesels too?

I have many different grades and effective grades on my railroad. and weeks went by before I was satisfied that what I had planned would actually work in practice. It does work, and it works well.

Here is a calculator for you to use to work out your grades:

Grade Calculator

 

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Most layouts don't have anywhere near the space available for proper grades. You can create a grade of just about any slope, but your trains probably won't climb it. I made that mistake on my very first layout, 50 years ago. Nice mountain track and trestle... no train could climb it, even without cars attached.

I would say flatter is ALWAYS better. Use the shallowest grade you can get away with, and nothing exceeding 3%. Use that with care, and have a short transition area (easement) of shallower slope at each end.
 
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I concur. And the longer the train is, may require 2 or 3 locomotives on the head end. I’ve reviewed theories and advice but never put any of it into 1st hand practice, so I’m expecting to run 2 locos minimum on each train. For context I plan a 2% oval helix having 24” minimum radius, with HO train lengths being around 11 actual feet. 1 loco might do it, but I expect 2. If 3 are needed, I’ve prepared for that already but not via what I point out below.

Which brings me to Helper Service. That’s an aspect some folks like to model. Where Helper Locos couple to the rear and push, uncouple at the top of the grade and coast back down. I think this is still done out in Cajon Pass? Another example is PRR’s Horseshoe curve.
 

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I concur. And the longer the train is, may require 2 or 3 locomotives on the head end. I’ve reviewed theories and advice but never put any of it into 1st hand practice, so I’m expecting to run 2 locos minimum on each train. For context I plan a 2% oval helix having 24” minimum radius, with HO train lengths being around 11 actual feet. 1 loco might do it, but I expect 2. If 3 are needed, I’ve prepared for that already but not via what I point out below.

Which brings me to Helper Service. That’s an aspect some folks like to model. Where Helper Locos couple to the rear and push, uncouple at the top of the grade and coast back down. I think this is still done out in Cajon Pass? Another example is PRR’s Horseshoe curve.
Norfolk Southern still uses helpers at Horseshoe Curve, and has a maximum grade of only 1.85%. They are based in Altoona and Cresson.
 
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I don’t want to detract/hijack the OP’s thread.

Suffice to say I’m duplicating Roy Smith’s Eastern(?) helix in HO scale. Maybe it’s his Western one… So his functional one in N is my proof of concept.
 

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On one of my N-scale layouts, I have a 2-1/4% grade on a 15" radius curve (approximately the equivalent of the same % on a 28" radius, HO-scale curve). A typical, 4-axle, quality N-scale loco (with no traction tires) can only pull about seven quality (properly weighted & MTL truck-equipped) freight cars up it without coming to wheel-spinning halt.

This is just on my particular layout with my particular equipment parameters, other folks will undoubtedly have different results. Possibly not a whole lot different, however?
 

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what is the recommended grade to raise the level of the track on HO scale?
Whatever works for your track plan, but please understand that it may not work for your trains. Grades require much more energy to ascend trailing dead weight tonnage than steel-on-steel rolling on level tracks. In fact, it takes prototype locomotives about 3.2 times the original horsepower to ascend even a 0.5% increase in gradient if the engineer is required to maintain track speed. Our models don't quite scale that way, but it's close enough that if you exceed about 3% you'll soon see why.

Just a bit of history: back in the 1830's, the Cumberland & Wheeling, part of what became of the Baltimore & Ohio, decided that their locomotives would not make money, or keep to schedule with typical tonnages, unless they graded their tracks no more than 2.2%. Later, when other roads began to do surveying and wanted government help, the feds said they'd have to conform to the grade conventions imposed on itself by the B&O Railroad of a 'standard grade', that being a grade not requiring a helper engine. Eventually, that standard applied on all US tracks and in Canada as well. Steeper sections became 'helper' districts where crews and engines were serviced and kept for getting consists up grades. Some modelers like to emulate this practice, and in time you may decide to do the same thing...use a helper engine.

What we usually advise to newcomers is to mock it up temporarily and test your locomotives. You may find that they do rather well with a grade near 3.5% pulling what you want them to pull, but you may find they can't do it even at 2.8%. Use empirical data for your own locomotives. Test.
 

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what is the recommended grade to raise the level of the track on HO scale?
Grades are calculated as length divided by length in other words they are “unit less”. A 2% grade, for example, is the same for all scales including the real railroads. The physics are essentially the same. I’m presently building a grade that will not exceed 2 %. I calculate the incremental rise from one riser to the next using the equation of a straight line y = mx + b, where b is the height of the previous riser, x is the track distance between risers, m is the slope or grade in % divided by 100, y is the height of the target or next riser.
 

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The simple way to understand it is the European logic. 1 in 100 is 1%. 2 in 100 is 2%.

So to maintain a maximum grade of 2%, and to obtain 4 inches in clearance, you need 200 linear inches of climb, with 2 inch height at 100 inches along the climb, 1” at 50 inch linear distance, 0.5” at 25 inches of linear distance.
And again for the down grade side.

8 feet is 96 inches so you can fudge it really by using 8 ft instead of 100”. It comes out to something around 2.12% grade or there abouts.
 

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What the above calculation does not include are the vertical easements at the top and bottom of the grades. My layout has a number of grades from 1% to 2.5%. Most of the grades have some amount of curve at the top where the track transitions back to level. Engines with "high rail" flanges work a little better than the engines with scale flanges. The engines with the scale flanges would just ride up over the rail and derail without adequate easements. I found with testing, the vertical easements on my layout had to be 1.5 times the length of the longest engine to assure derailment free operation.
If the longest engine is 10" (do not count the tender on steam engines) then the total easement length is 30". The grade between the vertical easements is closely approximated by subtracting half of the total easement length, 15" in this example. To have 2% on a 4" rise the length would be 215". Making the total grade with easements 200", subtracting 15" the maximum grade would be 2.22%.
My only point is it is important to allow for some amount of easement and try a few engines to see how little will work reliably.
 

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All good information to the OP's question back in February. Another drive-by poster, hopefully it helps someone else.
Agreed. Several new comers to the hobby around here lately. This hobby is in need of new and/or younger folks. I’m certain it’ll be educational for them.
 

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The simple way to understand it is the European logic. 1 in 100 is 1%. 2 in 100 is 2%.

So to maintain a maximum grade of 2%, and to obtain 4 inches in clearance, you need 200 linear inches of climb, with 2 inch height at 100 inches along the climb, 1” at 50 inch linear distance, 0.5” at 25 inches of linear distance.
And again for the down grade side.

8 feet is 96 inches so you can fudge it really by using 8 ft instead of 100”. It comes out to something around 2.12% grade or there abouts.
For a 2% slope, with riser distance increments of 100, using y = mx+ b, you have:
2 = .02(100) + 0
4 = .02(100) + 2
6 = .02(100) + 4
This of course does not include an initial vertical easement.
 

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And to go a bit finer, not that anybody is going to return to this moribund thread again, the two vertical cures at either end of the grade necessarily shorten the distance between the two curves that permits a reasonably constant rate of elevation change, the 'delta'. So, if you are building an overpass, and need X" of clearance to the nether surfaces of whatever is supporting the upper level trackage, you will have to factor the 'easements' at either end, and the constant elevation change between them to achieve the desired clearance. This will mean less distance to get to 3.5" of clearance for the typical HO scale tunnel portal, and that will require a grade somewhat steeper than the 2.2% you may have hoped was going to suffice.
 
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