On 17 Mar 97 at 20:25, n0kfb@mediaone.net wrote:
> Would it be OK with you if I posted your "Physics of Trains" explainations on my web page?
OK, with proper credit of course.
Enclosed is an updated copy correcting a few mistakes and adding some material.
> So what your saying is:
> Horsepower/(loco weight/powered axles)
Not certain of what you mean by that?
Horsepower is pull times speed. You can have pull with _no_ HP. Example: A 15,000 ton coal train sitting on a 1% grade and being held there by the engine brakes alone has 300,000 lb pull on that first coupler. But there is no horsepower being produced. The factor of adhesion for a steel wheel on a steel rail is between 20% and 30% of the weight on the wheel. So to prevent that coal train from pulling (sliding) those locos back down the grade the locos need to have at least 300,000 lbs of adhesion. This means they must weigh at least 1,000,000 lbs because 30% of 1 million is 300,000 lbs. The maximum weight per wheel, to prevent crushing the rail, is 35,000 lbs (70,000 lbs per axle). So we need a loco with at least 29.6 wheels, each of which is weighted to 35,000 lbs. 30 wheels is 15 axles or four 4 axle units minimum. It could also be 3 six axle units which would be 18 axles. Or it could be two 6 axle units and one 4 axle unit for a total of 16 axles. The point is that you have to have that one million pounds on the wheels and you are limited to no more than 35,000 lbs per wheel.
Adhesion as described above and Tractive Effort are closely related and can be though of as the same thing in many cases in that the amount of adhesion limits the maximum TE that can be used. TE is usually quoted at a specific speed. TE is the pull at the drawbar.
HP is the amount of pull (TE) times the speed. So while our coal train is just sitting there on the grade there is no HP required. But try to move it at 1 mph up that hill and HP is required. The required HP is the TE needed (300,000 lbs) times the speed (1 mph or 1.47 ft per second) divided by the definition of a HP (550 lb-ft per second). So the HP required is 801 HP! Yes just 800 hp will move this coal train up the hill. Amazing isn't it? But will one 800 hp unit do it? No! Because that one 800 hp unit must have at least 1 million pounds on its drivers to prevent it from sliding back down the hill. You MUST have the adhesion required. This means each wheel of an 800 hp 6 axle unit would have to have 84,000 lbs on it. Oh my the busted rails that would leave behind! As I said above, the minimum number of axles we need to spread out the required weight is 15 axles. Now it doesn't matter whether we have one 800 hp 4 axle unit and three 4 axle slugs, or whether we have four 200 hp units. It is all the same to the coal train.
One mph is kind of slow. It would take us 24 hours just to get up Parkman hill. So we want to go up the hill about 15 mph. 15 mph is a good compromise between taking forever or extreme high power costs. To go up this hill at 15 mph (22 ft per second) requires 12,000 HP. That comes from (300,000 lbs) x (22 ft per sec) divided by (550 lb-ftper sec).
Gee now that sounds familiar doesn't it. That is exactly the HP of four 3,000 hp units! So we can use four 3000 HP GP40s or three 4000 HP SD70MACs. Either way you satify _BOTH_ the adhesion needed and the HP needed. Will two 6000 HP units work. No. While you have the required 12,000 hp you can only have 70,000 lbs per axle weight, which times the 12 axles is only 840,000 lbs total and 30% of 840,000 lbs is only 252,000 lbs, not enough to hold or move the coal train. Remember you need at least 300,000 lbs of adhesion (traction).
All the above figures are based on a 30% adhesion factor. Sand will increase this factor somewhat. Modern locos such as SD70MACs and Dash 9-44CW claim adhesion factors of 33 to 43% !!!! Personally I don't believe the higher figures. In any case that would be the absolute best on good, dry, sanded rail. The other night I was running a freight up hill at 12 mph with a Dash 9-44CW on the point. I hadpreviously calculated that we should have gone up the hill at 16 mph, so why are we only doing 12? The rail was slightly frosty. I punched up the loco monitor screen on the computer, it showed that this supposedly 4400 hp unit was only putting out 2650 HP!!! It had derated to prevent slipping in spite of the sanders being on. So the adhesion factor of this loco at that time was not the touted 36%. But if they could get the adhesion up to 36% then _theoretically_ the two 6000 HP units could do the job for our coal train above.
But then you run into other problems such as traction motor overheating. Can we pour 1,000 hp into each TM continuously at 15 mph without frying them? Low gearing helps here but that lowers the top speed because the TMs will fly apart at high rpms. Here is where AC locos shine. Their TM rotors are much more solid than DC motors so they can be geared lower and still have a high top end.
Note that HP is TE times speed. If you have a fixed maximum HP, such as a loco has, then as speed increases the TE _MUST_ come down. The product of the two must remain a constant and is directly related to the HP rating of the loco. In the 12,000 hp coal train above why can't we go faster than 15 mph on the 1% grade? Because 15 mph times the required 300,000 lb of drawbar pull divided by 550 lb-ft per second (the definition of a HP) equals the HP. If the train went faster the product of the speed times the pull would be higher and thus the required HP would be higher, but we are limited to 12,000 hp on this consist. So it trudges along at 15 mph. Similarly if we throttle down a notch or two, reduce the HP, then the speed is going to drop because there is less HP available. The drawbar pull remains the same at 300,000 lbs and the lower HP means a lower pull x speed figure so the speed _must_ drop until the product is proportional to the new lower HP. This why I say HP is speed.
Ok what happens when we encounter a steeper grade with this train? The drawbar pull needed to hold or move a train is 20 lbs per ton per grade percent. That is where the 300,000 lb figure came from for the 1% grade of the example above. If this train were to roll onto a 1.5% grade what happens? The drawbar pull needed now is 450,000 lbs. We still have only 12,000 hp available however so the speed will drop. It will drop to 10 mph, the point where the product of the new 450,000 pull and the speed divided by 550 equals the available HP. But we are in real serious trouble here folks. Our locos can only deliver 300,000 lbs of pull because of their weight on drivers and the 30% adhesion factor. So our locos are going to slip and stall on the hill. And you had better set the train airbrakes as you stall or the train will drag you back down the hill.
So how do we proceed?
Well we can't increase the weight on drivers by adding weight to our existing locos because they are already weighted to the max for the rail. The only other way to increase traction, weight on drivers, is to increase the number of drivers, add more units. We will need to add two more units to get the weight up to 1.5 million lbs so the 30% factor gives us 450,000 lbs of adhesion. We do _NOT_ need the added HP of the units however, they could be just slugs. Without additional HP the train will go up the hill at 10 mph. At this point I would like to point out that if we simply add two 4 axle slugs, which get the power for their traction motors from the original 4 locos, that we can do the same thing by switching from GP models to SD models. An SD40 is simply a GP40 with two more axles (with traction motors) and more weight. In other words we've added "a half a slug" to each of the 4 units. In this manner we once again have the required number of drivers and the required weight on drivers by using just four SD40s. These are the same weight and number of drivers as 4 four axle units and two 4 axle slugs. If we want to go up the hill at 15 mph instead of 10 mph however, we must add the HP. Slugs or converting to SDs will not do. If the additional units added are 3000 hp like the rest then we will again go up the hill at 15 mph. HP is speed.
In either case we will not go up the hill very far, probably not at all! Why not? The figures say we will. The one word is "KAPOW"! You are going to break in two. Remember we are now trying to put 450,000 lbs of pull into a drawbar rated at 350,000 lbs, it is going to break. So what can we do? Well we could double the hill. Take half the train up to the top and leave it there then come back with the engines and get the second half. When you get both halves to the top, recouple them, make an air test, and proceed. By taking half the train up the hill at a time the required drawbar pull is only half that of the entire train or only 225,000 lbs. Well under the 350,000 lb strength of our drawbars. This method also requires no more slugs, SDs, or other units. The original 4 GP40s have enough traction to haul half the train up at a time. Unfortunately doubling the hill requires a lot of time. The line is blocked while it is being done and this train and others are delayed for the duration. Alternatively we could put the added two units on the rear of the train to PUSH. we would need another engineer (a helper engineer) or a distributed power set-up (radio controlled slaves). The physics are the same, the coal train and grade could care less where you apply the power just so you have the right amount to move it. But the drawbars do care where you put all that power. If you try to put it all thru one drawbar, the first one, it is going to say "screw you, I ain't gonna take this abuse" and it will break to prove its point. (You never knew drawbars were so animated did you.)
Now for the purists, I know I have left out a few things.
That 1 million pounds of locos does not go up the hill for free. It takes just as much HP to move each of those pounds up the hill as each of the train's pounds. So you should add their weight to the train when calculating speeds & HP required. In fact this is one reason for 4 axle high HP lococs. The more the locos weigh the more of their HP is required to just to move the loco upgrade. Lets look at the difference between 6 axle SD40s and 4 axle GP40s. Suppose we want to run a 5250 ton priority manifest train up a 1% grade at 30 mph. This requires 8400 HP. TE is not a consideration because even three 250,000 lb GP40s will have 225,000 lbs of adhesion. Our train only requires 105,000 lbs of adhesion on this grade. These three GP40s weigh a total of 750,000 lbs or 375 tons. These locos require 600 HP just to move themselves up the grade at 30 mph. So our 5250 tons of train requires 8400 HP and the 375 tons of locos require 600 Hp to go up this grade at 30 mph, total 9,000 HP. What happens if we use SD40s instead of GP40s. Same HP at 3000 each but the SDs weigh much more. Ours are ballasted for lots of TE needed on coal & grain trains. Our SD40s weigh 420,000 lbs each. Three of them weigh 630 tons! To move these SDs up the grade at 30 mph requires 1008 HP. This means we only have 7992 HP left for the train. That means we can only haul 4995 tons at 30 mph instead of the original 5250 that the GP40s hauled. While this may not seem like much, it _is_ 5% and a 5% efficiency improvement is a big deal. It is in high speed freight service where the new 6000 HP single unit locos shine. True they are all 6 axle units, needed for the weight to get that HP to the rail, but you are trading 8 axles of two 3000 HP GP40s for the 6 axles of the new units and you have much less HP wasting weight in the single unit. Remember that TE decreases as speed increases so as long as they keep the HP per ton ratio of the trains high enough to maintain high speeds then the TE will be low enough that these high HP single units won't slip. But try to use them in low speed drag service and they will slip as noted in the coal train discussion above. This is because at low speeds the TE available exceeds the adhesion.
Back to the coal train and what I left out.
As the grade gets steeper less and less of the loco weight is felt as pressing directly down on the rails so effective weight on drivers decreases slightly. (Do the geometry yourself if you want). I neglected rolling resistance of the train. At low speeds such as these, on straight rail, there is little rolling resistance to a loaded coal train. The resistance is mostly due to the grade. However there is some rolling resistance that will increase the total pull & HP required somewhat.
I neglected acceleration. The figures given are for steady state running. To accelerate requires more pull than steady speed.
The force of acceleration is mass times acceleration.
Force = mass x acceleration.
A coal train is a _very_ big mass! So even small accellerations need a lot of force. If you keep the acceleration low by notching out one notch at a time and allowing speed to increase slowly you can minimize the force of acceleration. If you are reckless and try to accelerate quickly then that force adds to the drawbar pull account of the grade alone and you can break the train in two that way also.
The main point of all this is to hopefully dispell the myth that high HP means lots of pull. It does not. Higher HP means higher pull at higher speeds but the total maximum pull is strictly related to weight on drivers. No HP required. None! Therefore a switch engine which only operates at low speeds does not need, nor can it use, high HP. It needs to be heavy. (but not too heavy that it breaks or turns over light industrial or yard rails). Life is a compromise.
Whew!
AK
Posted (with permission) to the World Wide Web by n0kfb@skypoint.com
V.101 03-29-97