How to make sense of power/strength metrics?

[ATLANTIC CENTRAL]

wrench567

Thank you Sheldon for the great read. Worth every letter.

    Pete.

 

You are most welcome, happy to share. Like any good engineering student, I paid attention when they said "you will not remember everything, you just need to remember what and where to look it up"

Sheldon

[wrench567]

Thank you Sheldon for the great read. Worth every letter.

    Pete.

[wrench567]

     It's not how many cars. It's tonnage that can be pulled. Twenty cars of ping pong balls could equal one 100 ton coal hopper.

  When some steam locomotives were upgraded they actually lost tractive effort. Most times the TE was gagged at speed. When the PRR fitted poppet valves to a K4s pacific it showed better economy in steam usage but a greater loss of adhesion. But when some were fitted with solid pilots instead of the slat pilot, the TE was greater because of the added weight. Even though it was not on the drivers.

     Pete.

[dehusman]

The locomotive manufacturers don't really decide how many cars an engine can pull, the railroads do that.  Most railroads don't use tractive effort for day to day calculation of how much an engine can pull.  They use formulas to decide how much tonnage a particular class of engine can pull on a particular territory.  Sometimes its segments of a subdivision, sometimes entire regions.  

They might decide that a GP7 is worth 1000 tons and an SD40 is worth 2200 tons.  If they have a 7000 ton train they would assign 7 GP7's or 4 SD40's.

Commonly up until the 1970's or 1980's the railroad would have tonnage charts for each engine class on each portion of the railroad or would have some form of tonnage adjustment factor for each engine class and territory.  

In the 1980's until even today, a popular method was horsepower per trailing ton (hptt).  The trairoad would set hptt ratings for different types of trains on different territories.  For example a bulk train on a territory would have a rating of .5 hptt, a manifest train 1 hptt, an intermodal train 2 hptt and a premium Z train 3 hptt.  If a manifest rain was 8500 tons it would need 8500 hp, if a premium Z train was 3000 tons it would need 9000 hp.  3 SD40's could haul 18,000 tons of bulk, 9000 tons of manifest, or 4500 tons of intermodal.

Real general rule of thumb, you need about the ruling grade of hptt to make it up the hill.  If the max grade on a route is 2%, then you need at least 2 hptt.

Later on some roads went to equivalent powered axles (EPA)  The tractive effort of an SD40 was the baseline, with 6 powered axles and then other engines were set relaive to that.  Engines with way more tractive effort got higher EPA ratings, for example a C44AC might be considered 10 powered axles.  Then trains had a tons per powered axle (TPA) rating.  If a C44AC is rated at 10 powered axles, and a 5000 ton train was rated at 250 tpa, then it needed 20 powered axles, which would mean 2 C44AC's (2@ 10 EPA = 20) or 4 SD40's (4 @ 6 EPA= 24). 

[gregc]

BradenD

If so how do manufacturers like GE know how many cars their designs will pull?

the maximum tractive effort depends on the adhesive weight of a locomotive, the weight on the driving wheels of a steam locomotive.    exceeding that force results in wheel slip.

Newtons Laws translate force to acceleration and speed.   but the tractive effort needs to overcome train resistance which includes both bearing friction and aerodynamic forces (see chart).   it also needs to over the force of gravity due to grade.   the net force results in positive acceleration when it exceeds the sum of the resistance force

horsepower is force / speed.  doubling the speed reduces the force by half.  excessive resitance forces can be overcome by slowing down.

205

 

 

[BradenD]

Thanks, Sheldon. Very interesting read.

[SeeYou190]

Locomotives with bigger numbers can pull more train cars.

My answer was easier to read.

-Kevin

[ATLANTIC CENTRAL]

Railroad Facts and Figures
Copyright AA Krug

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Tractive Effort vs Horsepower

Many people confuse Horsepower (Hp) and Tractive Effort (TE). With this essay I hope to clear that confusion. I use some terms and formulas in this essay that may not be familiar. I suggest that you read the Definitions at the end of this document.

Assume a train that weighs 15,000 tons is stopped on a 1% grade. For every ton of train weight on a 1% grade a force of 20 pounds is acting to roll the train down the hill. A 15,000 ton train produces a force of 300,000 lbs. (15,000 tons x 20 lbs per ton = 300,000 lbs).

To prevent this train from rolling back down the hill we must apply an equal force in the opposite direction to the coupler of the first car. Imagine yourself holding onto the coupler of the first car and trying to hold the train from rolling back down the hill. It should be obvious that try as you might you will not be able to hold the train. No matter how hard you grasp the coupler you cannot hold the train because your shoes will simply slide across the ties and ballast. The adhesion of your shoes to the ties and ballast is not equal to 300,000 lbs. Thus not even if you were Superman could you develope 300,000 lbs of pull (traction) to hold the train.

The level of force required to break the adhesion of your shoes to the ties and thus slide your shoes is dependent upon two things.
1. The coefficient of friction between your shoes and the ties.
2. The weight on your shoes.
Increase either one of those parameters and your adhesion to the ties increases enabling you to pull harder before sliding.

For the purpose of this article we will say the factor of adhesion of a steel wheel on a steel rail is about 30%. (Actually it is more like 20%-25% but we will get to that later). A factor of adhesion of 30% means that a wheel will stick to the rail so that a pull equal to 30% of the weight on that wheel is required to break that adhesion and slide the wheel. Locomotives have a lot of weight on their wheels. A typical 4 axle GP40 might weigh 280,000 lbs. Four of them weigh 1,120,000 lbs. Thirty percent of 1.12 million pounds equals 336,000 lbs. In other words, it requires a force greater than 336,000 lbs to slide four GP40s. Therefore these locomotives, with their brakes set, will hold the train on the hill since the weight of that train on the grade is only producing 300,000 lbs of pull.

Tractive Effort is the amount of Pull.
You can have pull with no HP. Notice that in the example above we are applying a force, a pull, a Tractive Effort, of 300,000 lbs to that coupler of the first car but we don't need any diesel engines. All we need is at least 1 million pounds of weight on the wheels and a factor of adhesion of 30%. We have Tractive Effort (TE) without any Horsepower (HP).

If I don't need any HP why did I use 4 locomotives. Why not just use one very heavy loco? Because the maximum weight on any single wheel that steel rails can withstand is 35,000 lbs. If we put more weight than that on any wheel it will crush the rail head and create excessive wear or outright failure of the rail. The two wheels on opposite ends of one axle can each carry 35,000 lbs of weight. So the maximum weight on an axle assembly is 70,000 lbs. Since we need a total locomotive weight of at least 1 million pounds that means we need a minimum of 14.3 axles. Three 4 axle locos would only be 12 axles so we must use four which gives us 16 axles.

From this discussion we can see that total locomotive weight determines the maximum amount of tractive effort a locomotive can produce. The number of wheels or the number of traction motors has nothing to do with it. We only add additional wheels to spread the required weight out along the rails to avoid damaging the track. I said that 3 GP40s will not work in this case because they only have 12 axles and each can be weighted no more than 70,000 lbs for total locoweight of 840,000 lbs. At 30% adhesion that only produces 252,000 lbs of TE and we need at least 300,000 lbs. If we take those same 3 GP40s and add two more axles to each unit but DO NOT increase the weight of each unit, then we still only have 840,000 lbs of total weight and 30% of that is still only 252,000 lbs. Adding axles and wheels to locomotives does not increase tractive effort.

HP is the Tractive Effort (pull) times the Speed.
Burn that statement into your brain. It is crucial to understanding this essay.
While our coal train is just sitting on the grade, held there by the locomotives, there is 300,000 lbs of "pull" on the first car's drawbar. Because the train is not moving 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 for the grade (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).

(300,000 lbs) x (1.47 ft per sec) / (550 lb-ft per sec) = 801 HP

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 4 or 6 axle 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 weight to get the adhesion required. That means each wheel of an 800 hp 6 axle unit would have to have 84,000 lbs on it. (That is 168,000 lbs. per axle). Oh my the crushed and broken rails that would leave behind! Not to mention the overloading of bridges and the track structure itself. As I said above, the minimum number of axles we need to spread out the required weight is 14.3 axles. It doesn't matter whether we have four 200 Hp 4 axle units, or whether we have one 800 Hp 4 axle unit and three 4 axle engineless slugs . 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. We want to go up the hill about 15 mph. 15 mph is a good compromise between taking forever and extreme high power costs. To go up this hill at 15 mph (22 ft per second) requires:

(300,000 lbs) x (22 ft per sec) / (550 lb-ft per sec) = 12,000 HP

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. SD70MACs have 6 axles and weigh 420,000 lbs. Again 70,000 lbs per axle. So three of them give us 18 axles and 1.26 million pounds on the wheels, more than enough to produce our required 300,000 lbs of traction. SD70MACs are rated at 4000 Hp so three of them are 12,000 Hp. Four GP40s or three SD70s, either way you satisfy both the adhesion needed and the HP needed.

Horsepower Alone.
Will two 6000 HP units work? No. While you have the required 12,000 hp you can only have 70,000 lbs weight per axle, which times the 12 axles is only 840,000 lbs total. With 30% adhesion and only 840,000 lbs of weight the two 6000 HP units have only 252,000 lbs of adhesion. Not enough to hold or move the coal train. The weight of the train on the grade will slide these two locomotives backwards down the hill. Remember you need at least 300,000 lbs of adhesion (traction).

High HP locomotives on 6 axles create other problems such as traction motor (TM) overheating. Can we pour 1,000 hp into each TM continuously at 10-15 mph without frying them? Low gearing helps. But low gearing lowers the top speed because the TMs will fly apart at high rpms. Here is where AC locos have an advantage over their DC counterparts. AC TM rotors are much more solid than DC TM armatures so they can be geared lower and still have a high top end.

Varying Adhesion.
All of the above figures are based on a 30% adhesion factor. IE, the wheels grip the rail with a force equal to 30% of the loco weight. Locomotives of the GP40/SD40 era and their Dash 2 offspring are generally considered to have an adhesion factor of about 25% not 30%. Sand will increase this factor to about 30%. Thus to achieve the 30% adhesion needed for the examples used in this essay so far, these locos may need to be on sanded rail.

Modern locos such as SD70MACs and C44s claim adhesion factors of 36 to 43%! They do this by using sophisticated anti-wheelslip circuits. These circuits allow the wheels to spin slightly faster than the rail speed warrants. It is called creep. Strangely enough, a creeping wheel has a higher factor of adhesion than a stationary or rolling wheel. Thus in theory two 6,000 HP SD90s weighing 420,000 lbs each and achieving an adhesion factor of 36% will produce a TE of 302,400 lbs and should pull the train up the hill at 15 mph.

However, in my experience you cannot count on that 36% adhesion factor in all types of weather and rail conditions. On wet or frosty rail these units slip and you stall. And when you stall you had better set the train airbrakes in a hurry or the train will slide these units back down the hill. On the other hand I have had C44s, SD90s and SD70MACs absolutely astound me with what they are pulling. At times they attain greater than 40% adhesion on dry, sanded, rail. It is that "at times" that concerns me. You cannot count on them to do that reliably time after time.

An Actual Experience with High HP Locos and Adhesion.
One night I was running a freight up hill at 7 mph with a Dash 9-44CW on the point. I had previously calculated that we should have gone up the hill at 11 mph, so why were we only doing 7 mph? 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 2930 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 36-43% but instead only 22%. The railroad had paid for a 4400 HP locomotive with 36% adhesion but was only getting a 2930 HP locomotive with 22% adhesion. The common SD40-2 would have done as good or better in this situation than the hi-tech wonder. This was not a one time occurrance. I have seen similar performances on many occasions.

Horsepower is Speed.
Up to now we have assumed that a locomotive has enough power to slip its wheels. That is true only at low speeds. Note that Hp is TE times speed. If the speed remains the same and the TE (pull) increases then the Hp requirement increases. If the TE remains the same and the speed increases then the Hp requirement increases. 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. On 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 total locomotive 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, reducing the HP, then the speed is going to drop because there is now less than 12,000 HP available. The drawbar pull account of the grade 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 like to say Hp is speed.

Steeper Grades.
We are proceeding at 15 mph up the 1% grade with our 15,000 ton train powered by four 4 axle GP40s. What happens when we encounter a steeper grade with this train? The drawbar pull needed to hold or move a train on a grade 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.

(15,000 tons) x (20 lbs per ton) x (1.5 grade) = 450,000 lbs

We still have only 12,000 hp available. Since HP equals pull times speed, if the pull goes up then the speed must come down. The train speed will drop to 10 mph, the point where the product of the new 450,000 pounds pull and the speed divided by 550 (HP definition) equals the available HP. But we are in real serious trouble here folks. Our four 4 axle locos can only deliver 336,000 lbs of pull because of their 1.12 million lbs of weight on drivers and the 30% adhesion factor. So our locos are going to slip and stall on the hill. Although we have enough Hp to pull this train up the 1.5% grade at 10 mph we do not have enough traction to do so. And, again, you had better set the train air brakes as you stall or the train will drag you back down the hill.

Slugs, and SDs.
So how do we proceed? Well we can't increase the weight on drivers by adding weight to our existing four 4 axle 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 (8 more axles) to get the weight up to at least 1.5 million lbs so the 30% adhesion factor gives us 450,000 lbs of adhesion. We do not need the added HP of those two units however, they could be just engineless 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 through electrical cables from the original 4 locos, that we can do the same thing by switching from GP models to SD models. From 4 axle units to 6 axle units. An SD40 is simply a GP40 with two more axles (with traction motors) and 50% more weight. In other words we've added "half a slug" to each of the 4 units. In this manner we once again have the required weight on drivers by using just four SD40s. These are the same weight and number of drivers as 4 four axle units plus 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.

Helpers.
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. We are now trying to put 450,000 lbs of pull into a coupler rated at 390,000 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 coupler pull is only half that of the entire train or only 225,000 lbs. Well under the 390,000 lb strength of our couplers. 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 couplers do care where you put all that power. If you try to put it all thru one coupler, 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 couplers were so animated did you.)

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Some things left out.

Now for you purists, I know I have left out a few things. (If you know enough about this to know that I left things out then you sure don't need to be reading this document).

·         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 the weight of the locomotives. They don't go up hill for free.

·         I neglected the efficiency of the locomotive's mechanical and electrical transmission.

·         I neglected rolling resistance of the train. At low speeds such as these, on straight rail, rolling resistance for a loaded coal train is only 10-15% relative to the grade resistance. However that will increase the total pull and HP required.

·         I neglected acceleration. The figures given are for steady state running. To accelerate requires more pull than steady speed.

The Weight of the Locomotive.

First we'll look at the loco weight. Four 280,000 lB GP-40-2s weigh 1,120,000 lbs. 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 traction, speeds, & Hp required. In fact this is one reason for 4 axle high HP locos. The more the locos weigh the more of their Hp is required to just to move the loco upgrade.

Heavy Haul vs High Speed.
You may have noticed that most railroads tend to use 6 axle power on heavy trains such as coal and grain while they use 4 axle power on their high speed lighter weight intermodal trains.
1. If a particular railroad has a good mix of high speed and heavy haul trains its locomotive roster will be a mixed bag of 4 axle and 6 axle power.
2. If a railroad, such as BN, has a preponderance of coal and grain trains and/or operates its trains in mountainous territory where grades are steep and speeds are low its roster will be dominated by heavy 6 axle power.
3. If a railroad, such as ATSF, has a majority of lighter weight high speed intermodal trains and a lot of relatively flat territory its roster will reflect that with lots of high Hp 4 axle power and/or lighter weight 6 axle power.

Lets look at the difference between 6 axle SD40s and 4 axle GP40s. Suppose we want to run a 5200 ton priority manifest train up a 1% grade at 30 mph. This requires 8320 HP. TE is not a consideration because even three 280,000 lb GP40s will have 252,000 lbs of adhesion. Our train only requires 104,000 lbs of adhesion on this grade. Three GP40s weigh a total of 840,000 lbs or 420 tons. These locos require 672 Hp just to move themselves up the grade at 30 mph. So our 5200 tons of train requires 8320 Hp and the 420 tons of locos require 672 Hp to go up this grade at 30 mph, total 8,992 Hp. The three GP40s produce 9,000 Hp. What happens if we use three 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 5200 that the GP40s hauled. While this may not seem like much difference it is over 4% and a 4% efficiency improvement is a big deal when you are burning 1.6 billion gallons of fuel per year. Trains that run at high speeds don't need heavy locos with lots of Tractive Effort. What they need are light weight high horsepower locos.

Here is a table showing the theoretical Tractive Effort a 3000 Hp loco produces at various speeds.

Speed	Tractive Effort
60 mph	18,707 lbs
40 mph	28,060 lbs
30 mph	37,415 lbs
25 mph	44,898 lbs
15.0 mph	75,000 lbs
13.4 mph	84,000 lbs
8.9 mph	126,000 lbs

Using a 30% adhesion factor, a locomotive that weighs 280,000 lbs has 84,000 lbs of adhesion. From the above chart we can see that such a loco can operate as slow as 13.4 mph without slipping. Suppose our railroad has a large proportion of service sensitive intermodal trains and we want to operate those trains no slower than 30 mph on our worst grades. At 30 mph a 3,000 Hp loco is only capable of producing 37,415 lbs of Tractive Effort. Therefore as long as we put enough Hp on our trains to maintain 30 mph on our worst grades we do not need to make them weigh 280,000 lbs. In fact they only need to weigh about 125,000 lbs because 30% of 125,000 is 37500 lbs of adhesion and we only need 37,415 lbs. We could save a lot of fuel by using these light weight locos instead of the 280,000 lbs heavy weights. If we have four such locos on a train we are saving 310 tons of wasted weight and that translates into saved fuel.

A loco that cannot use full throttle below 30 mph without slipping would be rather restricted in its service. So railroads tend to compromise. If instead of 280,000 lb locos or 125,000 lb locos we use a loco weighing 250,000 lbs we still get some of the fuel savings and the loco becomes much more versatile since it can now operate as low as 15 mph in full throttle without slipping.

It is in high speed freight service where locos with high Hp to weight ratios shine. This why the ATSF had 3800 Hp 4 axle GP60s and 4,000 Hp 4 axle B40-8s. Since BN had a preponderance of heavy coal & grain trains and even its freights had to contend with steep grades it had 3,800 Hp heavy SD60s and 4,000 Hp heavy SD70MACs instead of the GP60s and B-40-8s that the ATSF had.

Lets look at this Hp to loco weight ratio from another angle. The following chart shows the maximum Tractive Effort of various loco models and the speed at which that maximum TE is achieved. All units are 3,000 Hp and we assume a 30 % adhesion factor.

Model	Weight	Max TE	Speed	Total tons on 1% grade	Trailing tons
Light GP40	250,000	75,000	15.0	3750	3625
Heavy GP40	280,000	84,000	13.5	4200	4060
Light SD40	380,000	114,000	9.8	5700	5540
Heavy SD40	420,000	126,000	8.9	6300	6090

A first glance at the table looks as if the heavy SD40 is the best loco. It can pull the most trailing tonnage up the 1% grade. But we have to ask ourselves "What is the job?". If the job is to haul as much tonnage up the grade as is possible, then indeed the heavy SD40 is the loco we want. But if the job is to haul as much trailing tonnage up the grade at 15 mph then the SD40 is not the best choice. The following chart shows how much tonnage each of these locos can haul up the 1% grade at 15 mph. Since all the locos are 3,000 Hp they all produce the same 75,000 lbs of TE at 15 mph. But the weight of the loco uses up some of that TE. What is left over can pull the freight that is paying the bills.

Model	Weight	TE	Speed	Total tons on 1% grade	Trailing tons
Light GP40	250,000	75,000	15	3750	3625
Heavy GP40	280,000	75,000	15	3750	3610
Light SD40	380,000	75,000	15	3750	3590
Heavy SD40	420,000	75,000	15	3750	3540

The light weight GP40s can haul 85 more tons of paying freight per unit up the grade at 15 mph than the heavy SD40 can.

Those Superpower units.
If you haven't been paying attention you might think that the new 6000 HP single unit locos are destined for heavy haul service. True they are all heavy 6 axle units. But that is because the weight is needed to put that 6,000 HP to the rail without slipping. A 6,000 Hp unit that weighs 420,000 lbs and can attain a 43% adhesion factor has an adhesion of 180,600 lbs. The 6,000 Hp diesel engine can deliver that 180,600 lbs of Tractive Effort at a speed of 13 mph. Below that speed you cannot use full throttle on these locos because they will slip. That was for an astounding adhesion factor of 43%. What if they cannot maintain that extreme level of adhesion? What if they "only" get 36%? 36% of 420,000 lbs is 151,200 lbs of TE. The 6000 hp diesel can deliver that TE at 15 mph so the loco cannot operate below 15 mph in full throttle without slipping. At an adhesion factor of 30% the lowest full throttle speed is 18 mph. If the rail is wet or frosty can these modern marvels maintain even a 30% adhesion factor? My experience with 4400 Hp units is a definite no. The C44s often have trouble maintaining 22% adhesion with bad rail conditions. If a 6,000 Hp unit gets down to 22% adhesion it can only operate at full throttle above 24 mph! Thus if you want these behemoths to reliably move your trains over the hills in all kinds of weather you had better dispatch them with trains light enough that they can maintain 24 mph or greater on your steepest hills. That means they are only useful for trains such as intermodals which get a high Hp to tonnage ratio. When it is frosty they won't work on heavy freights or coal or grain trains which routinely pull up the hills at 10-12mph.

The railroad I work for uses 12,000 Hp on their coal trains through here and we go up the hills at about 12-13 mph. Note that you can replace the 12,000 Hp of 3 SD70MACs, or the 12,000 Hp of 4 SD40-2s, with the 12,000 Hp of just two SD90s. You have the same Hp so you should go up the hills at the same 12-13 mph. But it will be awfully iffy. That is because the minimum speed these 6,000 Hp units can operate at full throttle is 13 mph even with an adhesion factor of 43%. If anything causes the train speed to fall below 13 mph even momentarily, you will never regain the lost speed. The train might be temporarily slowed for various reasons. Perhaps the SD90s temporarily lost that 43% adhesion factor and slipped or reduced Hp to prevent slipping. Perhaps a wind came up and increased train restance. At 12 mph the 6,000 Hp locos cannot operate in full throttle even if they regain that 43% factor of adhesion. They will slip. Operating at reduced throttle the locos are not producing the 12,000 Hp this train needs to travel up the hill at 13 mph. So the train will never accelerate back up to 13 mph where it could again operate at full throttle. Four SD40s or 3 SD70MACs would have no difficulty re-accellerating the train back up to 13 mph. That is because they are not operating at the limit of their adhesion as the SD90s are. The 4 SD40s have 12,000 Hp just like the two SD90s but the SD40s have a total weight of 1,680,000 lbs and even at a 30% factor of adhesion can operate in full throttle down to 9mph! The 3 SD70MACs weigh 1,260,000 lbs and with only a 30% factor adhesion they can operate at full throttle down to 11.9 mph. If they achieve a 36% factor of adhesion they can operate at full throttle down to 9.9 mph. So either the SD40s or the SD70s have enough reserve adhesion they can operate at full throttle after being temporarily slowed. That allows them to accellerate the train back up to the 13 mph.

Thus on an equal total Hp basis these high Hp units are not equal to their lower Hp cousins when used in heavy haul service. And heaven help you (more like helpers help you) if the factor of adhesion on these brutes ever falls below 36% because you won't have enough adhesion to pull that 15,000 ton train up that 1% grade, period. You had better hope that it does not rain, frost, or snow.

Keep the high Hp units in high speed freight service where they do the most good. You are trading 8 axles of weight on two 3000 HP GP40s or 12 axles of weight on two SD40s for the 6 axles of the new units and you have 25-50% less Hp-wasting weight with the two high Hp units. 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. The slower the train goes up a hill the closer these high powered 6,000 Hp wonders perform like the good old 3,000 Hp SD40.

The Efficiency of the Locomotive.

Next we'll look at the efficiency of the locomotives transmission. Their transmission consists of the generator, traction motors, and gearing. My experience is that the loco's transmission efficiency normally runs in the 80% range. This means that if the physics of the train, grade, and speed dictate X HP then you really need X / .80 HP. If the physics say 12,000 HP then you really need 12,000 /.80 which is 15,000 HP. Another full unit!
Put another way....The 15,000 ton coal train going up a 1% grade at 10 mph requires 9564 HP. That is 8442 HP for the speed up that grade and 1122 Hp for the rolling resistance at that speed. (We'll get to rolling resistance in a minute). But that assumes 100% efficiency. At 80% efficiency this train would need 9564 Hp / .80 which is 11,995 Hp. SURPRISE! That is three 4,000Hp SD70MACs or four 3,000Hp SD40-2s to get a 15,000 ton coal train up a 1% grade at 10 mph. Sound familiar?

The Rolling Resistance of the Train.

Now we'll look at rolling resistance. Assume the same train as above, ie., 15,000 tons plus 840 tons of locos (4 SD40-2s) rolling at 10 mph on a 1.0% grade. Using the well known Davis formula we get the following values:

Resistance	Pull	HP
Grade	316,800 LBs	8447 HP
Rolling	41,880 LBs	1116 HP
Total	358,680 LBs	9563 HP

The calculated Hp required is 9563 Hp. Since our locos are only about 80% efficient this means we need a Hp rating of 12,000 to actually deliver the required 9563 Hp.

Put the train on a 2000 ft long 3 degree curve and you get:

Resistance	Pull	HP
Grade	316,800 LBs	8447 HP
Rolling	41,880 LBs	1116 HP
Curve	15,840 LBs	422 HP
Total	374,520 LBs	9985 HP

Using a loco efficiency of 80% the required 9985 Hp becomes 12,481 Hp. The 12,000 Hp of four SD40s is not going to be able to pull this train up the hill and around the curve at 10 mph. The speed will drop until the rolling resistance and grade Hp drops enough that the actual Hp required equals 80% of 12,000 Hp (9600 Hp). That speed is 9.6 mph.

At 9.6 mph we get the following values:

Resistance	Pull	HP
Grade	316,800 LBs	8109 HP
Rolling	41,557 LBs	1063 HP
Curve	15,840 LBs	405 HP
Total	374,197 LBs	9577 HP

 

Accelleration.

Lets look at accellerating trains. 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. That force adds to the drawbar pull account of the grade alone and it can break the train in two. 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 you may end up in two pieces.

Lets say we want to accelerate a 15,000 ton train at a rate of 1 mph per minute. In other words we want to be going 1 mph faster at the end of one minute than we are going now. To accellerate at that rate requires a steady force of 23,450 lbs. Note that it doesn't matter whether we are going up hill or on the level. We need to supply an additional 23,450 LBs of drawbar pull to accellerate at 1 mph per minute. Horsepower is pull times speed. Since the force to accellerate this train at 1mph/min is a constant, the HP required to accelerate the train varies according to speed. At 10 mph the HP needed is 625 Hp, at 40 mph the Hp needed is 2500 Hp.

An accelleration of 1 mph/minute is slow. It would take a train 60 minutes to go from 0 to 60 mph. But if we want to accelerate at 10 mph per minute it requires 10 times the force and 10 times the Hp at each speed. At an accelleration rate of 10 mph per minute the drawbar force needed is 234,520 lbs. At 10 mph that requires 6,250 Hp. At 20 mph it requires 12,500 Hp. At 40 mph it requires a whopping 25,000 Hp. Note that if we have only 12,000 Hp then we run out of Hp before we reach 20 mph. We can no longer accelerate at 10 mph per minute and will fall back to lower and lower acceleration rate as speed increases.

Keep in mind that these values of drawbar pull and Hp are ONLY for acceleration. You still need to supply the normal pull and Hp for any grade and rolling resistance. Lets look at that. A 15,000 ton train on a 1% grade going 8 mph requires 357,104 lbs of pull and 7617 Hp. If we have 4 SD40-2s we have 12,000 Hp times 80% efficiency = 9600 Hp available. 7620 Hp is what you get in throttle 7. We have one more throttle notch and 1980 Hp (9600-7620) remaining that we can use for acceleration. At the stated 8 mph that equates to 97,345 lbs of additional drawbar pull available. This additional force will accelerate the train at a rate of almost 4 mph per minute. Yeehaw! Put 'em in number 8 throttle and we'll be doing 12 mph at the end of the next minute.

Well not quite. A few problems crop up with that assumption. One is that as the train speed increases so does the horsepower required for both the grade and the rolling resistance. So as the speed begins to increase we have less "left over HP" for acceleration. The rate of acceleration will drop, we cannot maintain that 4 mph / minute rate we started with. In fact when we reach 10 mph all of the loco's HP is being used to pull the train and none is left over for acceleration. The speed will level out at 10 mph and stay there. Ain't physics neat?

The second problem is that you just broke the train in two so you are actually stopped. Why? Because the train traveling at 8 mph required 357,104 lbs of drawbar pull to maintain that speed on this grade. When you opened the throttle from notch 7 to notch 8 to accelerate you just put the additional 92,602 lbs of available loco tractive effort into that same drawbar. 357,104 lbs + 92,602 lbs equals a total of 449,706 lbs. Since drawbars are only good for about 390,000 lbs you just pulled one in two. Moral: When you are moving slowly you'd better handle that throttle gently if you want the train to remain in one piece. Acceleration can break trains in two.

Now for the purists, those 4 SD40-2s are not going to develope that 454,449 lbs of TE. That would mean an adhesion factor of 27% and SD40-2s can rarely if ever achieve that. They would most likely slip. But they can develope the 390,00 lb rating of the drawbars either continuously or by slipping and jerking. So either way the train is going to be in two pieces.

Note that the above train theoretically can go up this hill at 10 mph based upon Hp, efficiency, grade, and drawbar strength. Whether it actually can or not is in doubt. If we rolled onto this hill at a speed higher than 10 mph then all would be OK. As the train rolled onto the grade in number 8 throttle it would simply slow down to 10 mph and proceed up the hill. The drawbar force would be that 357,104 lb figure. Well within the rating of the drawbars. But if the train had stopped on this grade or had started from a stop on a lesser grade and was not yet up to 10 mph then we may be in trouble. Under these circumstance we may find ourselves in the situation above where we are only going 8 mph when the entire train is on the hill. We cannot go from notch 7 to notch 8 because the drawbar force will exceed their rating. Thus we cannot get to 10 mph. The only recourse is to slug it out all the way up the hill in the lower throttle position and a lower speed. It is very annoying to know you have the HP to go faster but you can't use it. If you are a real good engineer and really know what you are doing you can get around this obsticle in some cases. How? By applying some independent brakes to the locomotives drivers. Those brakes will absorb some of the extra Hp you get when you go from #7 to #8. Therefore that amount of the extra Hp and its attendent TE never reaches the train's drawbars. In that manner you can keep the total drawbar force lower than the drawbar rating. As the speed increases you feather off more and more of the independent brake until finally you are at the 10 mph physical limit and the brake is fully released. But make one mistake during that process, fail to coordinate the independent brake just right with the increasing Hp as the locos rev up and increase their load....or feather it off too quickly.......and Kapow! You are in two pieces. You have let enough extra TE reach the drawbars that their rating was exceded. Going 10 mph in #8 vs 8 mph in #7 saves you 15 minutes on the hour. But if you break it in two attempting to reach 10 mph then you are delayed 2 hours while you chain up a car and set it out and double the hill. If you are not sure of your expertise maybe it is better to just go up the hill at 8 mph in number 7 instead of trying for 10 mph in number 8.

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!

<="" a="">

<="" a="">Definitions.

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as used in this document

<="" a="">

·         <="" a=""><="" a="">Axle - Two wheels and an axle with a traction motor geared to it. All "axles" are powered, there are no idler axles.

·         <="" a="">GP40 - A 4 axle locomotive of 3,000 Hp that weighs 280,000 lbs.

·         <="" a="">SD40 - A 6 axle locomotive of 3,000 Hp that weighs 420,000 lbs.

·         <="" a="">Slug - A 4 axle unit that has traction motors but no diesel engine. Its traction motors get their electrical power from adjacent units. A concrete weight ballasts the slug to 280,000 lbs.

·         <="" a="">C44- A 6 axle locomotive of 4,400 Hp that weighs 420,000 lbs. Actual model designation is Dash 9-44CW.

·         <="" a="">SD70MAC - A 6 axle locomotive of 4,000 Hp that weighs 420,000 lbs and has AC traction motors.

·         <="" a="">SD90 - A 6 axle locomotive of 6,000 Hp that weighs 420,000 lbs and has AC traction motors.

·         <="" a="">Grade Pull or Grade Resistance - The force required on a grade to prevent a train from roling back down the hill. It is expressed as 20 lbs per ton per percent of grade.

·         <="" a="">Adhesion - The ability of the steel wheels of a locomotive to "stick" to the steel rails to prevent spinning or sliding of the wheels. The amount of force required to slide the wheels of a locomotive.

·         <="" a="">Adhesion Factor - The ratio of the adhesion to the weight of a locomotive. A good ballpark figure for steel wheels on steel rails is 30%. IE, it requires a force equal to 30% of the loco's weight to slide its wheels.

·         <="" a="">Tractive Effort - The pull developed by a locomotive. The maximum tractive effort value is directly proportional to the weight on drivers and the adhesion.

·         <="" a="">Horsepower - Any combinaton of pull and speed that equals 550 lb-ft per second. Examples: Pulling with a force of 1 pound for 550 feet and accomplishing that in one second. Pulling with a force of 550 lbs for 1 foot and accomplishing that in 1 second. Pulling with a force of 225 lbs for 2 feet and accomplishing it in 1 second. Etc.

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[BradenD]

Like the title says I have little understanding of how HSP and tractive effort translate into cars being pulled. I looked up the metrics for a Tier 4 Gevo and it has a TE of 166,000 lbs while the Big boy only has 135,000 lbs of TE. In real life the Big boy has the pulling power of multiple Gevos, right? If so how do manufacturers like GE know how many cars their designs will pull? In a sense I guess I'm asking for the larger equation of pulling power besides just tractive effort or horsepower.