Designing a helix for a new layout. For the life of me I can not find a table that lists the effective slope induced by various radii. In other words, the smaller the radii of a helix, the more friction on the pulling force, thus a 2% slope at a 22" radius actually pulls like a 3.5% slope - or something. I remember using this table before. I believe it is from a J Armstrong book. Any help is welcomed
Preceng,
Just a Suggestion,,,,Check out ''HO scale Grades'' Thread in General Discussion,on these Forums,,you may find your answers..
Cheers,
Frank
I believe the right term is "grade compensation." It refers to the practice of either widening curves or lessening the grades in curves (or both) so as to permit the combination of curve and grade from slowing pusher or drag freight operations
In the original TPFRO on page 51, there is a table that shows the reduction in tractive effort in terms of cars pulled up % of grade. If a loco can pull 100 cars on the flat, it can pull 36 cars up a 2.5% grade. It sits next to an up-and-over grade separation chart, making it easy to figure some things out in planning a layout. Be happy to look up a particular grade other than 2.5%.
That's not quite what you're looking for, but it may be what you are remembering.
There's really no set answer here, as it also depends very much on what equipment you're using in terms of what will go up without stalling -- and your train handling skills
You have you choice of whether to lessen the grade (add more spirals to the helix) or make the radius wider. a 6" wider curve difference will probably make more difference than 1/2% in grade, if you want a rough seat of the pants answer. Maybe someone with engineering sklillz has a formula?
Mike Lehman
Urbana, IL
This link will be helpful to you.
http://model-railroad-hobbyist.com/node/833
John Allen did the initial calculations, and John Armstrong later confirmed these calculations.
Rich
Alton Junction
mlehmanI believe the right term is "grade compensation."
compensation for curvature
greg - Philadelphia & Reading / Reading
greg,
Thanks.
To be specific (although not exact as done by pro engineers), quoting from wikipedia:
To compensate for this, the gradient should be a little less steep the sharper the curve is; the necessary grade reduction is assumed to be given by a simple formula such as 0.04 per cent per "degree of curve", the latter being a measure of curve sharpness used in the United States. On a 10-degree curve (radius 573.7 feet) the grade would thus need to be 0.4% less than the grade on straight track.
Obviously, not a direct translation from degrees of curvature to inches of radius, but I think there's a formula for that around somewhere, also.
Check this link that I provided in my previous post. I believe this is what the OP is looking for.
mlehmanObviously, not a direct translation from degrees of curvature to inches of radius, but I think there's a formula for that around somewhere, also.
i thought this was an interesting question. thanks for posting it. (bear in mind, a 10 degree curve, 574' full-scale is 6.6" HO).
here's a table indicating HO curve radius, the degree of curvature, and effective grade due to the curvature using the factor of 0.04 from the wikipedia article. I'd guess that the value of 0.04 would be on the low side for the small radii we use on model railroads.
radius degCurve gradient 18.0 3.66 0.15 20.0 3.29 0.13 22.0 2.99 0.12 24.0 2.74 0.11 26.0 2.53 0.10 28.0 2.35 0.09 30.0 2.20 0.09 32.0 2.06 0.08 34.0 1.94 0.08 36.0 1.83 0.07 38.0 1.73 0.07 40.0 1.65 0.07 42.0 1.57 0.06 44.0 1.50 0.06 46.0 1.43 0.06 48.0 1.37 0.05 50.0 1.32 0.05here's an article describing degree of curvature. I properly calculated the values for a 2 degree curve when I scaled our model radii (33") up by 87.
i looked at the other link, but think it just calculates the grade of the helix based on the radius, clearance above the rails and change in height (railhead to railhead) between each layer. (the grade of a helix is the change in height / circumference)
What is your scale?
The most common rule-of-thumb is that proposed by John Allen. For HO, this is expressed by the formula 32/R, "R" being the radius in inches. The result is the additional grade percentage to be added to the nominal grade. For 22" radius, the result would be 1.45%, added to your 2% nominal grade would be 3.45%.
Empirical testing is underway on this calculation and it seems to be relatively accurate. The table posted earlier doesn't seem to relate to observed curve effects on grades.
But in the first place, a 22" radius at 2% nominal grade is not sufficient for an HO helix. 22" radius at 2% gives only 2 3/4" elevation gain railhead-to-railhead, not nearly enough when you consider that you need to allow for subroadbed and the height of the track, etc. Experienced HO modelers have found 26" to 28" radius helixes more workable.
If you are in N scale, the equivalent calculation is roughly 17.5/R. In that case, a 22" radius would calculate to .79% added to your nominal 2% grade for a total of 2.79% and you would have enough clearance for N scale equipment.
Layout Design GalleryLayout Design Special Interest Group
Rich,
Yep, That's It. THANK YOU so much, particularly for being succinct.
Allan B.
(Should have known to back check my JF Videos - sorry Joe)
preceng Rich, Yep, That's It. THANK YOU so much, particularly for being succinct. Allan B. (Should have known to back check my JF Videos - sorry Joe)
Allan,
Mr Fugate forgives you.
cuyamaThe most common rule-of-thumb is that proposed by John Allen. For HO, this is expressed by the formula 32/R, "R" being the radius in inches. The result is the additional grade percentage to be added to the nominal grade. For 22" radius, the result would be 1.45%, added to your 2% nominal grade would be 3.45%. Empirical testing is underway on this calculation and it seems to be relatively accurate. The table posted earlier doesn't seem to relate to observed curve effects on grades.
Thanks Byron.
As I said, I'm interested in this topic wanted to see what Byron's recommendation means for different helix configurations.
here are 3 tables for board thicknesses of 1, 3/4 and 1/2 in. Clearance above the rails is assumed 2.75" suitable for HO. The actual grade is the rise / 100. Compensation is 32/Radius and the total is the sum of the grade and compensation.
If these calculations are correct, a target compensated grade of 2% would require a radius of 46, 44 or 42" for board thicknesses of 1, 3/4 and 1/2".
clearance board radius grade(%) compensation total 2.75 1.00 18 3.3 1.78 5.09 % 2.75 1.00 20 3.0 1.60 4.58 % 2.75 1.00 22 2.7 1.45 4.17 % 2.75 1.00 24 2.5 1.33 3.82 % 2.75 1.00 26 2.3 1.23 3.53 % 2.75 1.00 28 2.1 1.14 3.27 % 2.75 1.00 30 2.0 1.07 3.06 % 2.75 1.00 32 1.9 1.00 2.87 % 2.75 1.00 34 1.8 0.94 2.70 % 2.75 1.00 36 1.7 0.89 2.55 % 2.75 1.00 38 1.6 0.84 2.41 % 2.75 1.00 40 1.5 0.80 2.29 % 2.75 1.00 42 1.4 0.76 2.18 % 2.75 1.00 44 1.4 0.73 2.08 % 2.75 1.00 46 1.3 0.70 1.99 % 2.75 1.00 48 1.2 0.67 1.91 % 2.75 1.00 50 1.2 0.64 1.83 % clearance board radius grade(%) compensation total 2.75 0.75 18 3.1 1.78 4.87 % 2.75 0.75 20 2.8 1.60 4.39 % 2.75 0.75 22 2.5 1.45 3.99 % 2.75 0.75 24 2.3 1.33 3.65 % 2.75 0.75 26 2.1 1.23 3.37 % 2.75 0.75 28 2.0 1.14 3.13 % 2.75 0.75 30 1.9 1.07 2.92 % 2.75 0.75 32 1.7 1.00 2.74 % 2.75 0.75 34 1.6 0.94 2.58 % 2.75 0.75 36 1.5 0.89 2.44 % 2.75 0.75 38 1.5 0.84 2.31 % 2.75 0.75 40 1.4 0.80 2.19 % 2.75 0.75 42 1.3 0.76 2.09 % 2.75 0.75 44 1.3 0.73 1.99 % 2.75 0.75 46 1.2 0.70 1.91 % 2.75 0.75 48 1.2 0.67 1.83 % 2.75 0.75 50 1.1 0.64 1.75 % clearance board radius grade(%) compensation total 2.75 0.50 18 2.9 1.78 4.65 % 2.75 0.50 20 2.6 1.60 4.19 % 2.75 0.50 22 2.4 1.45 3.81 % 2.75 0.50 24 2.2 1.33 3.49 % 2.75 0.50 26 2.0 1.23 3.22 % 2.75 0.50 28 1.8 1.14 2.99 % 2.75 0.50 30 1.7 1.07 2.79 % 2.75 0.50 32 1.6 1.00 2.62 % 2.75 0.50 34 1.5 0.94 2.46 % 2.75 0.50 36 1.4 0.89 2.33 % 2.75 0.50 38 1.4 0.84 2.20 % 2.75 0.50 40 1.3 0.80 2.09 % 2.75 0.50 42 1.2 0.76 1.99 % 2.75 0.50 44 1.2 0.73 1.90 % 2.75 0.50 46 1.1 0.70 1.82 % 2.75 0.50 48 1.1 0.67 1.74 % 2.75 0.50 50 1.0 0.64 1.67 %
A 2% grade using a 22" radius would give you an effective grade of just a tad under 3.5%
i agree. But i don't see how the grade of a 22" helix can be 2%
A 22" radius has a circumference of 138". A 2% grade rises 2.76" in that distance. That rise is just enough for the recommended clearance above the track but doesn't account for the thickness of the track and roadbed.
even with a minimal roadbed thickness of 0.5", the grade for a 22" radius curve needs to be 2.4% to have a clearance of 2.75". Add the effective grade due to a curve (32/rad, 1.45 = 32/22) and the effective grade is 3.8%.
gregceven with a minimal roadbed thickness of 0.5", the grade for a 22" radius curve needs to be 2.4% to have a clearance of 2.75". Add the effective grade due to a curve (32/rad, 1.45 = 32/22) and the effective grade is 3.8%.
And even that is probably too tight in most situations in HO, since hand room is needed for emergencies and occasional maintenance. For standard-gauge steam- or transition-era equipment, the NMRA S-7 minimum clearance recommended over the railhead is just over 3”. [I personally wouldn’t recommend a turn-to-turn helix clearance that tight to my layout design clients, but it's theoretically possible.]
Assuming a minimum subroadbed of ½” and a track height of 1/8” above, that’s a railhead-to-railhead climb of 3 21/32” in 138” for a nominal grade of 2.65%. Add to that the effective grade caused by friction of 1.45% (32/22) and that’s 4.1%. Pretty steep on its own, and certainly raises the concern of stringlining.
Modern-era equipment demands more clearance and would make the grade even worse.
Byron
...and, if you ever want to run a wrecker crane, or a GG1 or E33 and other electrics with pantograph (even compacted and locked), or double stacks...............good luck.
According to this article, superelevation has a lot to do with track wear and friction. A little superelevation in your helix might negate some of the friction on your track.
that article is for prototype railroads.
of course you know that flanges on railcars are actually a safety feature. The slight camber of the wheel, that the outer edge is slightly smaller in diameter than the innermost edge next to the flange causes the wheel to stay centered on the track. The wheel with its edge closest to the flange covers more distance and forces itself more toward the center of the track.
as you can imagine this doesn't work on a curve. Super elevation, raising the outer rail, tilts the railcar creating a force that pulls the car torward the center of the curve.
ideally, when a train travels at the proper speed for a curve with a specific amount of superelevation, the forces balance to keep the wheels centered, avoiding the flanges rubbing against the rails.
model railroad typically have such tighter curves that we rely on the flanges to keep the wheels on the track.
gregc ... The wheel with its edge closest to the flange covers more distance and forces itself more toward the center of the track. as you can imagine this doesn't work on a curve...
... The wheel with its edge closest to the flange covers more distance and forces itself more toward the center of the track.
as you can imagine this doesn't work on a curve...
Greg, you misunderstand. When two wheels fixed to a common axle negotiate a curve, one or both tread surfaces will experience scrubbing due to the disparity in distance that they cover as they move. With truncated cones, this can be obviated by having the inner tire move radially outward, thus rolling a slightly lesser distance on the slightly smaller diameter, while the outer tire lengthens its travel by having the flange and the thicker part of the cone get closer to the outer rail. So, there is centering neither encouraged nor desired on curves. Instead, in order to prevent dynamic rocking and 'hunting' of the axles, there is greater stablity and better tracking, and less mechanical friction and wear, because the respective truncated cones of each tire surface on either end of the axle can now roll as they do on tangent rails.
The car tilting imparted by super-elevation moves the center of gravity of the car, with the bulk of its weight at the frame and up through the superstructure, including stacked goods as cargo, back toward the middle of the rails and helps to counteract centrifugal force. All the flanges and conical shapes of the wheels accomplish is to help the trucks to be stable as they run at speed down the rails, whether tangential or curved. The super-elevation does not help the car's wheels to stay centered on the tread because they can't BE centered on curves for the very reasons given.
SouthPennsuperelevation in your helix
In the tight curves of a model multi-turn helix, that would increase the tendency to stringline -- derailing across the center of the curve. Not a good idea.
The physics of the model and of real life are completely different.
cuyama SouthPenn superelevation in your helix In the tight curves of a model multi-turn helix, that would increase the tendency to stringline -- derailing across the center of the curve. Not a good idea. The physics of the model and of real life are completely different.
SouthPenn superelevation in your helix
Although I have seen no empirical evidence of this that I can point to, I agree with Byron on this. Superelevation on our models is great for realism, especially in photography, but when on grades and with substantial trailing tonnage, or even on level curves with strong resistance in the train behind the locomotive, you get increased forces pulling the couplers inward toward the center of the circle the more you super-elevate.
Great to look at, but just another variable that can work against you in hidden, hard to reach trackage. My helix tracks were level across them. Worked very well.
selectorThe super-elevation does not help the car's wheels to stay centered on the tread because they can't BE centered on curves for the very reasons given
can't the wheels, when offset from the centerline of the rail, where the outer and inner wheel travel a different distance because of their slightly conic shape move across the curved rails without any undo friction?
and when you consider both the centripedal and the force pulling the car toward the center of the curve due to the uneven superelevated rails balance one another at one speed for curves of sufficient radius?