Track fiddlerHi Mark I Revisited the thread to check out that Compensated Grade formula because math has always interested me. The guys at my MR Club were talking about that formula one day and I asked them if anyone had it and nobody did. Actually after looking it seems to be pretty simple but I wanted to check with you if I'm thinking correctly here. CG = G + 28 / Radius So on my layout my steepest grade is 2% and my smallest radius is 18. CG = (2 + 28) ÷ 18 CG = 30 ÷ 18 = 1.667 CG = 1.667 + 2 (original grade) Adjusted grade = 3.667 Is that correct Mark? Thanks TF
I Revisited the thread to check out that Compensated Grade formula because math has always interested me.
The guys at my MR Club were talking about that formula one day and I asked them if anyone had it and nobody did. Actually after looking it seems to be pretty simple but I wanted to check with you if I'm thinking correctly here.
CG = G + 28 / Radius
So on my layout my steepest grade is 2% and my smallest radius is 18.
CG = (2 + 28) ÷ 18
CG = 30 ÷ 18 = 1.667
CG = 1.667 + 2 (original grade)
Adjusted grade = 3.667
Is that correct Mark?
Thanks
TF
It's 2 + 28/18 = 3.56%
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A lot of people are saying the second term of the equation is 32/radius, not 28/radius.
The original formula was G + 32/R, but the LDSIG's additional experiments refined the second term to the 28/R I quoted.
Using 32/R will give you a slightly higher grade, which just means you're being a bit conservative (pardon the word) in your calculations.
Mark P.
Website: http://www.thecbandqinwyoming.comVideos: https://www.youtube.com/user/mabrunton
doctorwayne gregc you wouldn't consider that case with a continuous 50' curve as 2 sections of 32" radius curve and therefore 4% = 2 + 2 * (32/32)? what if you considered the continuous curve as 4 adjacent 32" curves, would it be 6% = 2 + 4 * (32/32)? No, because it's still only one curve, while the first two curves on my layout are separated by straight track and curve in opposite directions, and likewise with the third curve, which is separated from the second by a short length of straight track and curves in the opposite direction. The final two curves are separated from the third curve by a length of straight track over a bridge, again curving in the opposite direction of the previous one, and then segueing directly into the final curve, with no straight track, again in the opposite direction of the previous curve, a true "S" bend. The direction of each curve counters that of the previous one, and all but the last two are separated by various lengths of straight track. A locomotive running light would be in only one curve at a time, the radius of that curve only affecting the compensation of the grade in that area. Where it goes through the two final curves, which are not separated by straight track, I can't say if the transition from a left curve directly into a right curve both add to the compensated grade or if the second one cancels out the previous one...by the time that happens, the loco will already be in the final curve, and onto level track again.My original tests never even took curves into consideration, as I was unaware, at the time, of the formula to determine the effect of curves on grades, although I guessed that it must have had some effect on performance. In some instances, the loco on a curve can pull better than on straight track, especially if the engine is on the curve, but the trailing train, at the same time, is mostly on straight track...less friction on the straight, more on the curve. Wayne
gregc you wouldn't consider that case with a continuous 50' curve as 2 sections of 32" radius curve and therefore 4% = 2 + 2 * (32/32)? what if you considered the continuous curve as 4 adjacent 32" curves, would it be 6% = 2 + 4 * (32/32)?
No, because it's still only one curve, while the first two curves on my layout are separated by straight track and curve in opposite directions, and likewise with the third curve, which is separated from the second by a short length of straight track and curves in the opposite direction.
The final two curves are separated from the third curve by a length of straight track over a bridge, again curving in the opposite direction of the previous one, and then segueing directly into the final curve, with no straight track, again in the opposite direction of the previous curve, a true "S" bend.
The direction of each curve counters that of the previous one, and all but the last two are separated by various lengths of straight track. A locomotive running light would be in only one curve at a time, the radius of that curve only affecting the compensation of the grade in that area. Where it goes through the two final curves, which are not separated by straight track, I can't say if the transition from a left curve directly into a right curve both add to the compensated grade or if the second one cancels out the previous one...by the time that happens, the loco will already be in the final curve, and onto level track again.My original tests never even took curves into consideration, as I was unaware, at the time, of the formula to determine the effect of curves on grades, although I guessed that it must have had some effect on performance.
In some instances, the loco on a curve can pull better than on straight track, especially if the engine is on the curve, but the trailing train, at the same time, is mostly on straight track...less friction on the straight, more on the curve.
Wayne
i don't believe you answered the question and that you still believe you sum the effective grade for each curved section on your grade rather than recognizing that the effective grades for each curve only affects the portion of the train occupying the curve.
doctorwayneThat means that if I run a train as long as that entire grade, the effect of the compensation for the curves shows it to be the equivalent of 6.82%.
are you suggesting that the grade depends on the length of the train going up the grade?
i beleive a 20 car (4 oz/car) train going up a 6.8% grade requires a loco weighing 27.3 oz. on the other hand, the loco would only need to weigh 12 oz for a 2.8% grade
greg - Philadelphia & Reading / Reading
Good morning
Thanks Mark
Interesting that the equation was revised as you say.
I see what I did. I had that dyslexia thing going on. I did the left side of the equation first instead of the fraction on the right that gave me the wrong answer.
I'll re-do it the right way here.
2+28/18
2+1.55555555555555555555
2+1.56=3.56
Gotcha
Now I have a thought here. Scales are different. Although grades remain exactly the same from Z scale to prototypical, radius does not. Although radius is the same proportionally to size between scales, An 18" radius in N scale is 33.1 in HO. .54375 smaller or 1.839 larger depending on which way you look at it.
So my question, Is the equation the same for all scales? Somehow I don't see how it can be but I suppose it could. Do you use one of the numbers I have listed above to convert the equation between scales...???
Don't Know
What the OP needs is tables, since he hasn't given his helix diameter, either physical or track-centerline. The table is simplified because he gives 18" as his desired rise; it could be further simplified if some arbitrary maximum grade is specified.
I'm sure these tables exist, perhaps in NMRA data, but I don't know a link.
Table needs two radii: track centerline, and outside physical radius based on NMRA clearance outside the track curve. Both are important for him to know, and he won't want a Mars Orbiter sort of solution if he mistakes one for the other!
Then the effective grade per turn is easily assigned, and he just adds turns to get to his 18" to keep in the ballpark.
We can discuss compensation at the top and bottom, and the effect of the 'last turn' being a partial one if the entrance and exit aren't in the same horizontal direction, but those are relative niceties at this stage of his thinking.
TF: see Mark Pruitt's example for part of the answer to your scale question. (NB, "compensation" of grade is technically the reduction of grade in a curve to keep the ruling grade minimized, not correction for the resistance factors together...)
joe323How much of an incline could l get away with if I want a train to climb 18 inches between levels. Thinking of building a helix.
The first reply was pretty much all that was needed.
Track fiddlerA 2% grade that is ideally the steepest you want is approximately 1/4 inch per foot. Therefore you would need 72 feet of run to climb to 18 inches of rise.
Joe, I would suggest you buy some Kato Unitrack, three loops worth, in the diameter you are considering. Mock up your spiral and run some trains on it.
An experiment like this will answer all your questions.
My layout plan requires a 5% incline with a 22 inch radius curve on the top. To verify this would work, I built the layout section from cardboard, laid the track, and tested it with real trains to be sure it was OK.
-Kevin
Living the dream.
I found a helix design calculator on line. This might answer a lot of questions without having to understand all the mathematical blather spouted above:
https://www.modelbuildings.org/helix-design-calculator/
maxmanwithout having to understand all the mathematical blather
If I were better at mathimatical blather, I'd be able to retire now comforably. My daughter did, and is an engineer at Lockheed Martin making more than me and she in only a couple years into her job there.
Rio Grande. The Action Road - Focus 1977-1983
riogrande5761If I were better at mathimatical blather, I'd be able to retire now comforably. My daughter did, and is an engineer at Lockheed Martin making more than me and she in only a couple years into her job there.
Back in 1970 I started work at GE for $9,600/year. A McDonalds associate this year would make about $17K/year plus all the burgers/fries you can eat. Both of my kids earn about 10X what I started at.
I was okay with the mathematical blather until they invented the "new math". Went downhill from there.
riogrande5761 My daughter did, and is an engineer at Lockheed Martin making more than me and she in only a couple years into her job there.
I am pretty sure two of my daughters have already exceeded my total lifetime income, and the youngest is on her way.
gregci don't believe you answered the question and that you still believe you sum the effective grade for each curved section on your grade rather than recognizing that the effective grades for each curve only affects the portion of the train occupying the curve.
I don't see the need to beat this topic to death, as it often happens, but you're not reading my replies correctly, either.
If the train is as long, or longer than the length of the grade, once the entire train is within that grade, all curves will have an effect. As the train leaves the grade, it also leaves the effect of the curves that are no longer occupied - seems pretty simple to me.
gregc..are you suggesting that the grade depends on the length of the train going up the grade?
Not at all: the grade is whatever it actually is. The effect of the grade and curves on the train will obviously vary depending on the train's length and location within the grade.The 71 car train used for the initial test was simply to see how many locos were needed to get to the top - at that time, I wasn't really concerned about the effects of the curves, because at that time, I never even thought about it.
gregci beleive a 20 car (4 oz/car) train going up a 6.8% grade requires a loco weighing 27.3 oz. on the other hand, the loco would only need to weigh 12 oz for a 2.8% grade
Ah, but you're forgetting that a 20 car train would not come anywhere near the length needed to occupy the entire grade at one time. About the only 4oz. cars I have are empty hoppers. The shortest ones, 30'-ers, loaded, weigh just under 8 oz.Before I did these tests, I did have diesel locomotives modified with two motors and ballasted to just over 33oz. each. One easily handled a 24lb. train of loaded hoppers up a similar, but much shorter grade.Around that time, I was back-dating my layout's era to the late '30s, and was selling-off most of my diesels. If I had stayed with such diesels, this wouldn't even be a discussion. My layout doesn't have enough room to run such long trains on a regular basis: the 71 car train with four locos occupied three or four of the layout's towns at that time, depending on which part of the layout they were located.For normal operations, train lengths range from 1, to perhaps 20 cars, and locos are allocated depending on their measured drawbar pull: each has a specific rating and I have enough to assign whatever's needed to get a train from here to there.
However, out of curiosity, I decided to run the train which you suggested. The locomotive was a Bachmann Consolidation, weighing-in at 1lb, 7oz. (engine & tender), followed by 20 empty hoppers and a caboose.
The train made it up the grade without issue.
I don't see much need to continue this discussion...my locos do what they're required to do, and I've tested all of them enough to know what's needed to move any particular train from point A to point B or beyond.
joe323 How much of an incline could l get away with if I want a train to climb 18 inches between levels. Thinking of building a helix.
How much of an incline could l get away with if I want a train to climb 18 inches between levels.
Thinking of building a helix.
doctorwayneI don't see the need to beat this topic to death, as it often happens, but you're not reading my replies correctly, either.
there's a need to correct a incorrect information
doctorwayneI have a 45' long grade on my layout that's at a constant 2.8% grade. However, there are several curves within that incline. The first two each have a radius of 34" and as horseshoe-type grades, extend somewhat beyond a half circle each. The third curve is the entry into an "S-curve, and relatively short, with a 40" radius. It's followed by two somewhat longer reverse curves, both at 48" radii. That means that if I run a train as long as that entire grade, the effect of the compensation for the curves shows it to be the equivalent of 6.82%.
The first two each have a radius of 34" and as horseshoe-type grades, extend somewhat beyond a half circle each. The third curve is the entry into an "S-curve, and relatively short, with a 40" radius. It's followed by two somewhat longer reverse curves, both at 48" radii.
That means that if I run a train as long as that entire grade, the effect of the compensation for the curves shows it to be the equivalent of 6.82%.
doctorwayne gregc you wouldn't consider that case with a continuous 50' curve as 2 sections of 32" radius curve and therefore 4% = 2 + 2 * (32/32)? what if you considered the continuous curve as 4 adjacent 32" curves, would it be 6% = 2 + 4 * (32/32)? No, because it's still only one curve, while the first two curves on my layout are separated by straight track and curve in opposite directions, and likewise with the third curve, which is separated from the second by a short length of straight track and curves in the opposite direction.
it doesn't matter how many separated curves there are. it simply matters how much of the grade occupied by the train is curved.
if the entire grade were a 32" radius (e.g. helix), then the effective grade is 3.8% = 2.8 + 32/32".
how can it possible be greater?
if half the train were on one or more sections of 32" curved track, the effective grade would be 3.3% = 2.8 + (32/32") /2.
if a quarter of the train were on 32" curved track, the effective grade is 3.05% = 2.8 + (32/32") /4
but you seem to think that each curved section adds 32/R% to the grade
that somehow, despite that fact that if the effective grade is 3.3% (see above) because it is entirely curved, that having one or more straight sections breaking up the curves increases the effective grade rather than descreases it
doctorwayneHowever, out of curiosity, I decided to run the train which you suggested. The locomotive was a Bachmann Consolidation, weighing-in at 1lb, 7oz. (engine & tender), followed by 20 empty hoppers and a caboose. The train made it up the grade without issue.
assuming the train (cars + loco) weighs 103 oz (20*4 + 23), a 6.8% grade adds a force of 7 oz.
tractive effort from a loco is 20-30% its weight, or 4.6 - 6.9 oz, which is insufficient to pull that train up a 6.8% grade, especially when ignoring wheel resistance of ~2%.
however, it would be might be enough on a continuous 3.3% grade + ~2% wheel resistance of 5.5 oz = 103 * (3.3 + 2) / 100
yes, of course the entire train isn't on a curve (see above)
Track fiddlerNow I have a thought here. Scales are different. Although grades remain exactly the same from Z scale to prototypical, radius does not. Although radius is the same proportionally to size between scales, An 18" radius in N scale is 33.1 in HO. .54375 smaller or 1.839 larger depending on which way you look at it. So my question, Is the equation the same for all scales? Somehow I don't see how it can be but I suppose it could. Do you use one of the numbers I have listed above to convert the equation between scales...??? TF
The formula was derived for HO scale. I think the constant in the second term of the equation (28 or 32 / Radius) would change depending on scale. For example, in N scale, where the rolling stock is roughly 54% of HO scale, the constant would probably be close to either 15.2 or 17.4. This is because the main effects creating the drag in the curve would remain proportional in all scales. an 18" radius in N is the same as a 33" radius in HO scale. Shorter cars means less angle between couplers, in turn meaning the pulling force of the loco is more in line with the direction of travel of the car, so less perceived curve "grade."
Not sure I explained that well - does it make sense?
Track fiddlerSo my question, Is the equation the same for all scales?
In Effective Slope caused by curves cuyama stated that for N scale, the effective grade is 17.5/R
Thanks Mark. Your explanation made complete sense to me. I see you had the constant number for N scale as 17.4 in place of 32 or the revised 28 for HO at 15.2. I see Greg pointed out Cuyama's prior reference to the constant number as 17.5
I like to understand where numbers come from so I fumbled around at the drawing board and figured out where the N scale constant number came from.
HO/N = 87/160 = .54375
32 x HO/N = N Constant
32 x .54375 = 17.4
Looks like Marks number, I'm sure Byron probably rounded his up.
Thanks Mark, Greg and Byron as well for the info. I find this mathematical stuff quite interesting. My radiuses and grades on my layout are completed. The numbers are not helping me plan or change anything at this point but it's interesting to know what they are.
maxmanBack in 1970 I started work at GE for $9,600/year. A McDonalds associate this year would make about $17K/year plus all the burgers/fries you can eat. Both of my kids earn about 10X what I started at. I was okay with the mathematical blather until they invented the "new math". Went downhill from there.
Because it's an empirically derived (i.e., generated from experimental data) formula, the constants will vary a bit depending on specific conditions of the tests performed.
Second-order effects also will play a part, even though they're not explicitly stated. Such effects might include material from which the wheels on the cars are made, surface finish of said wheels, smoothness of the radii in the curves (in other words, how much the radius varies around the curves), swing limits of the couplers, etc.
Geez. Reading through these posts can confuse folks more.
Work with similar units or you'll get in trouble.
I.E. inches/inches or feet/feet.
2% is 2/100 regardless of units. So you can rise 2" in 100" or 2ft in 100ft.