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This might be a question for some of you engineering types who are out here...

I was reading about a railroad grade that was listed as something like 1.6% "compensated." I'm guessing that the term compensated means compensated for curves as a curved grade is a harder pull than the same grade that is tangent. So....... is there a relatively simple way to figure out this compensation? Let's say that someone had a helix with a 1.5% grade and a radius of 36"; what would the "compensated" grade be?

I'm curious because all of my grades are on curves.

Thanks!

dlm

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I was reading about a railroad grade that was listed as something like 1.6% "compensated." ....... is there a relatively simple way to figure out this compensation?

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Excellent question. I await a knowledgeable answer.

Also a definition of what "compensated grade" really means.

-- Happy Holidays Roger T.

of the Great Eastern Railway

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The compensation is indeed for the curves of the railroad. These values have been obtained empirically - they went out and did tests on what was happening in the curve and built charts and formulas later. On the model railroad, the difference between a curve and a straight isn't that much and, as such, compensation on a grade won't make much difference. If you do find the trains slowing down at a particular part of the grade, you will want to decrease the grade slightly at that point and maybe increase the rest of the grade if you have a height point that you need to match or lengthen the grade by the few inches necessary otherwise.

-- Bob May Losing weight is easy! If you ever want to lose weight, eat and drink less. Works evevery time it is tried!

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Jim Bernier

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Often 0.04% per degree of curvature [see note below] is used nowadays.

In track design, civil engineers will often try to design around a maximum compensated grade for a particular stretch of track. For example, if the controlling grade on a tangent section of track is 1.00% then the maximum actual grade used on a section of track with a 2.0 degree curve would be

0.92%.

The "compensated grade" in both cases is the same - 1.00% on the tangent and

0.92% + 2 x 0.04% = 1.00% on the curve -- and both grades will have about the same effect of the pulling capacity of a locomotive. Naturally, this applies to grades and curves long enough to affect the entire train; the compensation will be less for shorter curves.

Model curves are so much sharper and the physics affecting our model locos and cars is so much different that I don't think you could apply the 0.04% per degree of curvature rule with any sort of predictable or meaningful results. I don't think I've ever seen it discussed in an NMRA Data Sheet.

[Note] "Degree of curvature" is a way of measuring the sharpness/radius of a curve, not the total angle that the curve goes through. 1.0° curve (R=5,730 ft), HO scale =789 in. 5.0° curve (R=1,146 ft), HO scale =158 in. 10.0° curve (R=574 ft), HO scale =79 in. 20.0° curve (R=288 ft), HO scale =40 in. 30.0° curve (R=193 ft), HO scale =27 in. 40.0° curve (R=146 ft), HO scale =20 in. 50.0° curve (R=118 ft), HO scale =16 in.
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It's pretty technical. Requires a PDF viewer.

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=>The compensation is indeed for the curves of the railroad. =>These values have been obtained empirically - they went out and did tests on =>what was happening in the curve and built charts and formulas later. =>On the model railroad, the difference between a curve and a straight isn't =>that much and, as such, compensation on a grade won't make much difference. =>If you do find the trains slowing down at a particular part of the grade, =>you will want to decrease the grade slightly at that point and maybe =>increase the rest of the grade if you have a height point that you need to =>match or lengthen the grade by the few inches necessary otherwise.

1.6% "compensated" means that the effect of the grade + the effect of the curve = 1.6% on straight track. Bob is right AFAIK about how knowledge of this was gained in the early days of railroading.

But he's wrong about the effect of curves in model railroading. It's about the same as on the real thing - a curve will increase drag, and hence limit maximum trainlength. That's why if possible I limit curving grades to 2.5% in my designs - 2% on tight curves (

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Since the 1930s, railroads have been simulating train performance with "davis Formulas", which take into account resistance from grade, curve, and flange.

IIRC, each degree of curvature is equal to .04% grade. So, if you have a 1.6% grade and a 10 degree curve, the compensated grade is 2%

I doubt that the 1:1 frictional coefficients scale down to 1:87, personally I wouldn't sweat compensation on a scale model.

regards, Jerry

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Cheats Method of Model Grade Compensation 101 :-)

If you set your grades using a marked spirit level, you're nearly there. If you don't, find/borrow/buy/steal a 2' spirit level. If not already marked, mark graduations along the whole length every 1/4" or so. Mark the first one on one end "0" - just make sure it's right at the end of the level.

Well, just use the level exactly the same way on the cord across the arc of the curved roadbed. Because the distance travelled around the arc is greater than the chord described by the level, you travel further for the same rise in height - ie, less grade. IE, compensation. Now how much compensation? I can only quote that great Italian mathematician, Bugga Difino. Perhaps someone else could tell us. But I never knew, just that it works!

Steve Newcastle (stinking hot 100 deg F) NSW Aust

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Mark,

Any idea what the geometry might be here? I always considered myself to be fair at geometry, but I dont see any relationship to the curves in degrees and the related raduii... : (

dlm

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mathematician, Bugga Difino. Perhaps

Thanks for the explanation, but there is one thing that I think you are confused about... you are quoting an Italian mathematician here, but I think you have him confused with that great southern cipherer, "Bubba Dunno!" Other than that, I really appreciated the explanation! : )

dlm

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OUCH!!! If Mark Mathu's numbers are correct, then that means that my 2% grades are actually much more like 3.2% (30 x .04 = 1.2% + original 2.0%) since I have 30" radius curves.

dlm

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Good ole NMRA data sheet D3b provides the definition that "Degree of Curvature is the number of degrees in the angle formed at the center of curvature by radii drawn to the ends of a chord 100 scale feet long."

The data sheet offers the formula that the radius of the curve in feet (real or scale) is equal to 50 divided by SINE ( 1/2 of the degrees of curvature ) As I picture it, you have a right triangle with the side opposite the angle at the center of the curve being 1/2 of the 100 ft cord, the side adjacent being the perpendicular bisector of the 100 ft cord, and the hypotenuse being the radius of the curve. GQ

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But as I qualified my original statement: "Model curves are so much sharper and the physics affecting our model locos and cars is so much different [than the prototype] that I don't think you could apply the 0.04% per degree of curvature rule with any sort of predictable or meaningful results."

- Mark

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Dan; You confused radius with degrees of curvature. I'm too lazy to figure it out, but a 30" curve scaled up is:

30" x 87 / 12 = 217.5 foot radius. I believe that is something like a 25 degree curve, so your numbers are close, for the wrong reason, but in real life, the compensated grade is horrendous!

We need to remember that by real life standards, even a 30" radius is sharp. regards, Jerry

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R = 50 / sine(D/2)

where R = prototype radius in feet D = curvature in degrees sine = sine function

To convert to the radius of your favorite scale, divide the prototype radius by your scale factor (48, 64, 87.1, 160, etc.) and don't forget to change to inches.

For example, a prototype 26.5 degree curve (which is terribly sharp, as far as prototype railroads are concerned) would correspond to:

D = 26.5 degrees R = 50 / sine(26.5/2) = 218 feet radius

In HO scale, this would be (218 feet) x (12 inches / foot) / 87.1 = 30.0" radius curve.

- Mark

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"Mark Mathu" R = 50 / sine(D/2)

What?

All this mathematics. My head hurts.

-- Happy Holidays Roger T.

of the Great Eastern Railway

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I just estimated... I think one of Mark's numbers was fr a scale 27" curve; mine are 30" so I figured that using that number would be close enough...

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Oops, sorry. I'm a civil engineer by day - I guess it came out in that last message.

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Oh no, I don't join in that one! Rule No 1 for visitors to a US-mostly web site. Do not become involved in any discussion or arguement or flame war with a Sep - er, American, on domestic politics. You elect 'em (or vote agin 'em), you support/pan 'em. We have enough probs with our own little gold diggers - um, sorry, politicians.

BTW, you have 50 states, could you fit another one in and call it Difino? You could not only have the towns of Bugga (see above), but Farque, Stuft and others that somehow or other spring to mind....

Steve Newcastle NSW(*) Oz (* - Having only six states and two territories makes remembering their names much easier)

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