The one I used is called "1970 modified Davis Eq." Don't recall where I found it.
-Don (Random stuff, mostly about trains - what else? http://blerfblog.blogspot.com/)
Tim,
You are correct in that Don's version gives far lower values for journal resistance than the version I am using which is from "Management of Train Operation and Train Handling", Fourth Edition, published 1977 by the Air Brake Association.
Mac
PNWRMNMAt low speed the last item of the formula is not much, so look at the first three which resolve to 1.3 + .9 + .45 or 2.65 pounds per ton.
For whatever it is worth, relating to the example of a GP38-2 on level track...
I have a MILW tonnage ratings book that gives tonnages at arbitrary minimum continuous speeds (in other words, selected by the railroad to ensure locomotives don't stay below their actual rated minimum speeds.) In this book, the GP38-2 is rated 9880 tons at 14 MPH for level track, 8110 tons for 18 MPH. These tons are, to be clear, trailing tons.
This translates to the fact that on the MILW at least you could never assign a GP38-2 more than 9880 tons trailing. Using your 131.5 ton covered hoppers, that's 75 of them... or if you'd like, 74 and a cabin car.
This gives a 'real world, real railroad' parallel to the calculations that might be of interest. I'm sure others can find other ratings for other roads that might differ slightly.
-Will Davis
Railroad Locomotives - My Blog
Try it this way. At low speed the last item of the formula is not much, so look at the first three which resolve to 1.3 + .9 + .45 or 2.65 pounds per ton. At 61,0000# TE that is a tonnage rating of 23,000 tons or 175 cars of 131 tons each. As a practical matter, yes we can start that cut, may have to take slack, but can start it.
As to speed on .2% downgrade; the grade supplies 4 pounds per ton of accelerating force, without ANY locomotive, subject to only 2.65 pounds resistance at 10 MPH. As speed increases air resistance will increase, but on a several mile long tangent downgrade of .2% the train would accelerate to 60 MPH sooner or later since acceleration due to gravity is more than resistance due to all forces. Assuming a 100 square foot cross section at 60 MPH I came up with 1.37 pounds per ton of air resistance, which makes 60 MPH the approximate terminal velocity, assuming no locomotive power.
Note that if the resistance formula above is correct, then an ES44 with 140000? lb of continuous tractive effort can keep 90000+ tons moving on the level at 9 mph or so.
Put a train of 143-ton four-axle cars on a long 0.2% grade and it will accelerate downgrade to 60+ mph, if the formula is correct.
timz But if you're on level track-- then there are two separate questions: With how many cars can the engine keep moving at low speed, say 15 mph, or How many cars can the engine start from a standstill? On the level a GP38 might well be able to maintain 15 mph with 15000 tons or so. But can it start such a train? That we don't know.
But if you're on level track--
then there are two separate questions:
With how many cars can the engine keep moving at low speed, say 15 mph, or
How many cars can the engine start from a standstill?
On the level a GP38 might well be able to maintain 15 mph with 15000 tons or so. But can it start such a train? That we don't know.
Yeah, talking about level track is almost a theoretical exercise - kinda like having massless rope and frictionless pulleys.
BamaCSX83 All right, so by the Davis equation, my math formula set-up would be as follows: w=50,000lbs (200,000lb max weight by 4 axles) n=4 (4 axles on a covered hopper) V=10mph k=0.07 With that information, this should give me: R=0.6+20/50,000+.01(10)+0.07(10)`2/50,000(4) Which should mean that to move one covered hopper at 10mph R=0.70096lbs of resistance, or did i do my math wrong? And if I did my math right, then it only takes about 70.096lbs of tractive effort for a single locomotive to move a 100-ton covered hopper at 10mph... so something like a GP38-2 which makes 61,000 lb-f of TE could theoretically move 87 hoppers all by its lonesome. Again, if I did my math right.
All right, so by the Davis equation, my math formula set-up would be as follows:
w=50,000lbs (200,000lb max weight by 4 axles)
n=4 (4 axles on a covered hopper)
V=10mph
k=0.07
With that information, this should give me: R=0.6+20/50,000+.01(10)+0.07(10)`2/50,000(4)
Which should mean that to move one covered hopper at 10mph R=0.70096lbs of resistance, or did i do my math wrong?
And if I did my math right, then it only takes about 70.096lbs of tractive effort for a single locomotive to move a 100-ton covered hopper at 10mph... so something like a GP38-2 which makes 61,000 lb-f of TE could theoretically move 87 hoppers all by its lonesome. Again, if I did my math right.
Weight per axle should be in tons, so your answer is 1.5#/ton, or 150# per car.
61K of TE will get you 407 cars (really 406 plus the locomotive...)
But, RRs aren't flat. Even a 1/2% grade requires 10# ton additional grade resistance, bringing the total up to 11.5#/ton or 1150#/car.
61K of TE will get you 52 cars plus a locomotive.
1% grade drops it back to 27 car, 2% 13 cars.
Grades are killers. That's why no Saluda and no Tennessee Pass (and why the Raton route is a loser)
You've got a point there, I did forget the tare weight...
So my updated equation is as follows:
w=65,750lbs (263,000lb rail car by 4 axles)
n=4 axles
K=0.07
So with that, we have R=0.6+20/65,750+.01(10)+.07(10)*2/65,750(4)
Which means that R=0.70073lbs per ton
We have a 263000lb (or 131.5 ton) railcar, giving us an effort of 92.146lbs of effort
Again, with a GP38-2 making 61,000 lb-f of TE, it could theoretically move 66 covered hoppers all by itself (of course, again, if I did all my math right)
Bama,
I did not check it but your math looks correct. Your 200,000# per hopper is not correct since you forgot about the 63,000 typical tare weight. 200,000# is the net weight, 63,000# tare, 263,000# gross for a loaded "100 (net) ton hopper.
BaltACD As a practical matter - each carrier has developed Tonnage Ratings for their territories. When assigning power to a train, the tonnage of the train is know and those assigning power pick a group of engines whose combined tonnage rating just exceed the tonnage of the train on the ruling grade on the territory the train will traverse.
As a practical matter - each carrier has developed Tonnage Ratings for their territories. When assigning power to a train, the tonnage of the train is know and those assigning power pick a group of engines whose combined tonnage rating just exceed the tonnage of the train on the ruling grade on the territory the train will traverse.
Generally, they are developed using computer simulations - usually called train performance calculators - that use train, grade and curve resistance along with train consist and loco performance data to simulate a train running from A to B. They output includes run time and energy consumed, among other things. They are fed and validated with lots of road testing. (there were some pretty terrific articles in Trains in recent years about CSX and GE testing "heavy" AC units)
For simple calculation, you can add 20#/ton/%grade for grade resistance and 0.7#/ton/degree for curve resistance. Typically, though, ruling grade curves are "curve compensated" - that is the grade on the curve is flattened a bit to compensate for the added curve resistance such that the sum of grade + curve resistance is constant for the whole climb up the ruling grade. So, for purposes of figuring out whether your train will stall en route, you can usually ignore curve resistance.
Never too old to have a happy childhood!
BamaCSX83 For some reason I've been stuck on this. I've seen it with my own eyes and through video evidence, but still can't grasp it. How does one relate the tractive effort of a locomotive to the actual number of railcars the locomotive is capable of moving? First lets assume that you've got dead straight and level track, and the railcars that you're pulling are loaded 4750 cu ft covered hoppers (most I've seen are max weight of 200,000lbs). Lets say that the locomotive you're pulling with is a GP38-2 that's been recently (within 6 months) shopped and checked over. Any locomotive can be substituted for the sake of ease of finding TE or CTE figures, but I'm just wondering what the equation is (if there is such) that determines how many cars the locomotive can pull. Any help here?
For some reason I've been stuck on this. I've seen it with my own eyes and through video evidence, but still can't grasp it. How does one relate the tractive effort of a locomotive to the actual number of railcars the locomotive is capable of moving?
First lets assume that you've got dead straight and level track, and the railcars that you're pulling are loaded 4750 cu ft covered hoppers (most I've seen are max weight of 200,000lbs). Lets say that the locomotive you're pulling with is a GP38-2 that's been recently (within 6 months) shopped and checked over. Any locomotive can be substituted for the sake of ease of finding TE or CTE figures, but I'm just wondering what the equation is (if there is such) that determines how many cars the locomotive can pull.
Any help here?
The Davis equation, which calculates the rolling resistance of a train is what you need.
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