First generation diesel- tractive effort?

I'm no expert on this but I think that tractive effort implies the amount of force needed to start a train in motion.  Afterwards you have the mass and momentum that keeps the train rolling with less effort from the diesel engines.  Obviously grades and curves need to be taken into account.

By your calculations a 100 car train moving at 40mph would require nearly 20 diesels.

@PRR Man posted:

In my pursuit for more accurate model operation. I'd like to ask the trainmen here to comment or verify my findings.

In determining the amount of weight a single unit engine can handle on the road. My research came up with these figures:

Using 1750hp as found in a F7 or GP9 (567c engine) the stated tractive effort (TE) is 64,750 ftlbs. Using a 72% efficiency rating the TE at the rails is about 46,620. I have found specs that state it takes 2-5 lbs of TE per ton to start a train. Since I model the mid-50s with a lot of friction bearing rolling stock, I'll use the high number. Add +10 lbs of TE per ton to accelerate the train, and +20 lbs per ton for a 1% grade. That's a total of 35 lbs TE per ton of train.

That provides 1332 tons of TE. At nominally 50 tons per car, that's 26.6 cars?! It doesn't sound correct. Then I realized that TE is reduced when speed increases. So at 25 mph the figure becomes 9.2 cars. At 40 mph it becomes 5.8 cars.

Am I anywhere near right in where I'm going with this? Thanks for your input.

First, drag, or friction, is highest when the train is stopped, until it gets to a speed where wind resistance takes over, the speed of which is probably quite high for a train.

If you pull a car with a constant force, you will accelerate until the drag increases to match the pulling force.

I think there are two problems with your assertion. You state it takes more TE to accelerate a static load. This is not true. You also estimate quite a high number for “friction bearings”. I doubt that they are the 100-200% increase in friction you estimated.

PRR Man: EMD’s F7 had a 567B as a prime mover, which produced 1500 hp like the GP7. 😉

Tractive effort is highest at starting or at a relatively low speed - usually under 10 miles-per-hour. And, unless I'm mistaken, tractive effort is the actual force applied at a locomotive's drawbar or coupler - no factor relating to efficiency is relevant. Power applied to a train is the product of the tractive effort and the speed. Therefore, if an engine produces approximately constant power at the crankshaft and the generator and electric motors have constant efficiency, the tractive effort will decrease as the speed increases. In starting a train, tractive effort is needed to overcome static friction and also to accelerate the train. The acceleration is typically largest at starting and very low speeds. As the speed increases, the acceleration decreases and the tractive effort works only against the internal friction of the trucks, rolling resistance, and aerodynamic drag.

MELGAR

For starters, the starting tractive effort of any locomotive is generally calculated by multiplying the weight on drivers by the maximum available adhesion, i.e. a nominal 27% (back in the early diesel and steam days, but now with AC traction and 420,000 pound 6 axle diesel units, available adhesion now approaches 46% providing starting tractive efforts of over 180,000 pounds). Thus, the 1750HP GP9, with a working weight of about 240,000 pounds, gives a starting tractive effort of a bit over 64,000 pounds.

Also, remember the Tractive STARTS the train, while horse power ACCELERATES the train. Diesel electric locomotives are essentially CONSTANT HP, variable torque (tractive effort) machines ( the slower a diesel electric goes, the more the tractive effort increases), while steam Locomotives are just the opposite ; being CONSTANT TORQUE, variable HP (the faster the steam locomotive goes, to a point, the more the HP increases).

Jack, so I can revise my figures using the higher 64,000 ft lbs tractive effort AT THE RAIL. Which already allowed for the parasitic effect of the take-off machinery of the prime mower-generator. Can you comment on the load figures I found for acceleration and grade?

To reiterate, acceleration will happen as long as the force at the coupler pulling faces is greater than the parasitic drag of the friction force and wind resistance.

Friction forces go down immediately after the train starts to roll. The train will continue to accelerate until as long as a force greater than the drag is present.

An increase in tractive effort isn’t required as long as the above is true.

I disagree about acceleration. If it takes X amount of TE to start a train. Speed will become constant once friction is overcome for a given amount of TE. It is not a case of ever increasing speed unless there's an attendant increase in TE acceleration.

@PRR Man posted:

In my pursuit for more accurate model operation. I'd like to ask the trainmen here to comment or verify my findings.

In determining the amount of weight a single unit engine can handle on the road. My research came up with these figures:

Using 1750hp as found in a F7 or GP9 (567c engine) the stated tractive effort (TE) is 64,750 ftlbs. Using a 72% efficiency rating the TE at the rails is about 46,620. I have found specs that state it takes 2-5 lbs of TE per ton to start a train. Since I model the mid-50s with a lot of friction bearing rolling stock, I'll use the high number. Add +10 lbs of TE per ton to accelerate the train, and +20 lbs per ton for a 1% grade. That's a total of 35 lbs TE per ton of train.

That provides 1332 tons of TE. At nominally 50 tons per car, that's 26.6 cars?! It doesn't sound correct. Then I realized that TE is reduced when speed increases. So at 25 mph the figure becomes 9.2 cars. At 40 mph it becomes 5.8 cars.

Am I anywhere near right in where I'm going with this? Thanks for your input.

My question to you is what are you trying to ACTUALLY determine?

I am an Mechanical Design Engineer, I have spent my life dealing with calculations.  There are many ways to solve a problem.  The point is I can give you a textbook or Webpage full of calculations.  But without knowing  when and how to apply them they are useless.

So is it the amount of F-Units to pull a given train?

I would be happy to help you.  But trying to determine a "magical calculation" for a general locomotive type under all conditions is not the way it works.  The way motive power for a given train was/is determined is based on specifics for a particular "run".

Let me just touch on some of the issues.  All F-7's were not identical from a NET power DELIVERY standpoint.  For instance they could be geared differently.

Operating factors also come into play.  This is a very complex problem with a lot of factors.

If you study the situation in depth you will find that motive power was/is lent to RR's prior to purchase orders being signed for this reason.  The so called "Demo Units" provided by Locomotive Manufacturers were not just rolling Billboards in fancy paint schemes, they served a purpose.

Hope this helps!

@PRR Man posted:

I disagree about acceleration. If it takes X amount of TE to start a train. Speed will become constant once friction is overcome for a given amount of TE. It is not a case of ever increasing speed unless there's an attendant increase in TE acceleration.

It may seem counterintuitive, but that’s how the physics works out. This is all within the limits of increasing friction and wind resistance, and any grade of course.

Relating to the starting friction, here’s a great resource. You can see in the graph they use as an example, starting friction increases until motion starts, then drops immediately after.

http://hyperphysics.phy-astr.g...du/hbase/frict2.html

Here is a good summary with detailed calculations:

http://evilgeniustech.com/idio...erAndTractiveEffort/

Chris, from years in the right hand seat, I can tell you this:  The amount of tonnage you can start depends on the weight on the drivers and on the horsepower of the engine.  Here's a comparison from a 1984 Santa Fe timetable Special Instruction, three classes of locomotives from the same builder with identical traction motors and the same gear ratio.

SD39  2300 HP  Weight 391,500 lbs   Tractive Effort 82,284 lbs

SD40  3000 HP  Weight 391,500 lbs   Tractive effort 70,067 lbs

SD45  3600 HP  Weight 391,500 lbs   Tractive effort 72,286 lbs

It may be a surprise that the lowest horsepower locomotive has the highest tractive effort.  This is starting tractive effort.  The higher horsepower locomotives produced enough horsepower to lose adhesion, whereas the humble SD39 was able to use its lower horsepower more efficiently in starting a train.  Notice though, that the SD40 and SD45, both within the high horsepower category of that age, have tractive effort comparable to their respective horsepower.  I used EMD locomotives because the three classes were ballasted to the same weight, just for simplicity.  Starting tractive effort would have been higher on comparable GE locomotives due, in part, to their more robust traction motors, but the Santa Fe GE units were ballasted differently and the EMD comparison was less complex.

Once you get the train moving, the horsepower makes all the difference.  Power, as indicated by traction motor current, decreases as speed increases.  

For example, on gently undulating territory with a 7,000 ton train, if you have 4 SD39's at 2300 Horsepower apiece, then you will find it difficult to maintain a speed exceeding 50-55 MPH.  However, if your 4 units are SD45's, at 3600 Horsepower each, you can keep a steady speed of 70 MPH.  At 55 MPH the quartet of SD39's will only be able to produce about 200 amps to each traction motor, but the SD45's will be producing 500-600 amps.  The SD39's and the SD45's each weigh the same, but the horsepower makes the difference in speed.  Even on an ascending mountain grade, where the SD39's could maintain a continuous speed of around 10-12 MPH, the SD45's would be making 15-18 MPH.

Weight on drivers is important.  A 2250 HP E8 cannot start as much tonnage as a 2300 horsepower SD39 .  The E8 only has two thirds of the locomotive weight on its drivers because the center idler axle of each truck supports a third of the weight but produces no tractive effort.  The SD39, though, has all axles powered and all the weight of the locomotive is on drivers.

So, to return as closely as possible to your question about what an F7 can pull, I suggest you look at photos from the 1950s and see how many cars your favorite railroad had them pulling,  That's much easier than trying to figure a tonnage rating.

For an understandable, technical explanation, I suggest you read Hudson 5432's excellent post which follows.

BTW to further illustrate what I am getting at with my prior post.  I saw a picture (can't find it now) of a WP GS-64 on the point (added as a helper) of the California Zephyr.  The caption said it was added, because the train was running behind schedule and there were adverse conditions ahead, so they were doing it to help make up some time.

When motive power was calculated for arguably the Pinnacle of Streamlined Domeliners, that was never the intention.

1) The tractive effort quoted for diesels is STARTING tractive effort. Unless a RR has a chassis dynamometer like at DOT Pueblo or like PRR once had at Altoona, the starting tractive effort is a CALCULATION based on the horsepower of the diesel engine at Notch 8, the size and characteristics of the traction generator (or alternator) and traction motors. Available tractive effort (at all speeds) is dependent on the adhesion level that can be achieved by the locomotive (or locomotives).

2) The input horsepower of the diesel engine into the main generator or traction alternator is dependent on temperature, altitude, the high heat value of the fuel, and of course the condition of the engine.

3) Every "link" in this chain of devices has an efficiency lower than 100%, and they are compounded in the calculations.

4) The force exerted at the coupler is DRAWBAR PULL, not tractive effort. The difference between tractive effort and drawbar pull is the energy required to move the locomotive (or locomotives).

5) As speed increases, tractive effort and drawbar pull decline in a curve with a hyperbolic shape. This decline is not a direct relationship since, as speed increases, voltage increases and electrical amperage declines, so apparatus heating (and losses) are reduced.

6) The starting resistance of a train "of journal boxes or bearings" is usually estimated at about 3 lb/ton. Tests have shown that an adequately lubricated journal box has slightly lower starting resistance than a roller bearing journal. This difference is small and usually disregarded.

7) The resistance of the train increases mainly due to wind resistance, until the train reaches a "balancing speed". This point is the intersection of the declining drawbar pull curve and the increasing resistance of the train drawn as a curve. Wind resistance is not much of a factor below about 40 mph.

8) Grade resistance in lb/trailing ton is 20 lb/ton/1% grade. So if a grade is 1.5%. the resistance is 30 lb/ton IN ADDITION TO the force required to move the train on level track.

9) Everywhere there are grades, there are curves. A one degree curve ADDS 0.8 lb/ton to the train resistance.

This is a complicated subject. Recommend you google "Davis Locomotive and Train Resistance" or Railroad "AAR" resistance for formulae. In my opinion, trying to downsize this calculation for a basement model is not possible if you want a correct answer that mimics real RR operations.

@Number 90 posted:

So, to return as closely as possible to your question about what an F7 can pull, I suggest you look at photos from the 1950s and see how many cars your favorite railroad had them pulling,  That's much easier than trying to figure a tonnage rating.

Thanks for your insight!  "I was just shooting from the hip".  Being a Mechanical Engineer not a REAL Engineer.

Strangely we came to the same conclusion! 

@Hot Water posted:

Also, remember the Tractive STARTS the train, while horse power ACCELERATES the train. Diesel electric locomotives are essentially CONSTANT HP, variable torque (tractive effort) machines ( the slower a diesel electric goes, the more the tractive effort increases), while steam Locomotives are just the opposite ; being CONSTANT TORQUE, variable HP (the faster the steam locomotive goes, to a point, the more the HP increases).

I don't know, perhaps railroads have their own unique physics that is applicable to only railroads and nowhere else in the universe.

The folks in physics seem to believe that it is force that accelerates according to Isaac Newton's second law of motion, force = mass times acceleration (F=ma).

Also the good folks in physics seem to think that horsepower, or power in general, is the rate at which work is being done.

For example, 1 horsepower is work being done at the rate of 550 ft lb/sec.

Gentlemen; what I hoped to gain from this exercise was to determine the number of cars a single engine can handle. Based on the particular conditions of it's run. For me it is to enhance my hobby with realistic train handling. Sure I can hitch 20 cars to one F3, but that's not realistic. After doing the theoretical research. Which did include that 'evil genius' site. I wanted to hear about real world experience out on the road.

You have all been generous with your knowledge and experience. I have learned what I was after. Thank you!

@PRR Man posted:

...Sure I can hitch 20 cars to one F3, but that's not realistic...

It depends.

How many cars any locomotive can pull has a lot to do with the terrain of the railroad they have to operate on. Here's a first-hand example, on a train I ran many times on the Ohio Central.

We had a line running south out of Youngstown, Ohio that had a 2.02% grade for the first 2.2 miles out of the yard. A single GP9 was good for 10 cars on that grade. We could get up the grade with the throttle in Run 8 at 10-11 mph and just touch the 30 minute short time ratings on the traction motors as we crested the top of the hill.

We assigned four GP9s (mechanical almost identical to an F7) to those trains and made sure that the loaded trains on that run were always 40 cars. I had to have all four units on line to get up that grade, with all the units wide open at 10-11 mph. However, once we topped that grade, I would isolate the lead two units (take them off line so they just idled) and run with only two GP9s on those 40 cars for the rest of the run. That saved fuel and made it a lot quieter for me up in the lead unit. The rest of the run on that line had milder and shorter grades, and two GP9s could handle those 40 cars with ease.

So it is entirely prototypical and realistic to put 20 cars behind an F7, if you are running on level track.

@Hudson5432 posted:

5) As speed increases, tractive effort and drawbar pull decline in a curve with a hyperbolic shape. This decline is not a direct relationship since, as speed increases, voltage increases and electrical amperage declines, so apparatus heating (and losses) are reduced.

This calculation is actually quite close to the traction curve of an ES44AC

https://ogrforum.com/...s44ac-traction-curve

WBC,

Yes that is how it is supposed to be.

However,

The system efficiency of the locomotive OVER ITS SPEED RANGE changes the "tractive effort-speed curve" from a "pure" hyperbola. In the early diesel days and with the use of main generators instead of traction alternators and DC traction motors and not AC traction motors, the system efficiency was significant. For example, the "375" horsepower constant in the English system was not used by EMD. They used "308". If you divide 308 by 375, this results in a system efficiency of 82.1 percent OVER THE SPEED RANGE OF THE LOCOMOTIVE, which was about average for the actual "rail horsepower" of an F3, F7, GP7, GP9, etc. (Or at least EMD thought so...)

So if you take the 1750 HP "input to the generator for traction" (UNDER STANDARD CONDITIONS OF AMBIENT TEMP, ELEVATION, AND FUEL HEAT VALUE*) and multiply by main generator efficiency, and then multiply by the four traction motors IN SERIES (ie low speed, high amperage), per the EMD ratio the AVERAGE RAIL HP is 1437, which is 1750 x .821. At higher speeds than the continuous speed (MCS) where the motors are reconnected electrically and operated in full parallel (vs series or series parallel), the system efficiency will be higher and the rail horsepower will also be higher.

Note this is calculated tractive effort and Rail Horsepower, NOT drawbar pull/drawbar HP which as posted prior will be lower.....

*STANDARD conditions= 60 degr F, less than 1000 ft elevation, and a fuel high heat value of 18850 (all from my memory). Turbo engines have an advantage to several thousand feet elevation.

None of the above violates any rules of physics, it just recognizes that several factors are in play that yield a different result...

This might be considered as "splitting hairs", but consider in the early diesel days units were operated with 2 to 4 in MU, and these differences were large enough to affect train schedules, performance on grades, balancing speed, tonnage ratings, and other things that are very important for railroads (and Dispatchers) to know.

Cool, thanks for the explanation which makes sense. I believe that the constants such as 375 mile lb/s are for the point at which work is being done.   If output is measured at the end of the crankshaft or at the generator, that is not at the point of work being done. That output has to power all the support systems for the locomotive and be reduced due to electrical resistance before it hits the wheels. Thus, it makes sense that at the wheels a constant that takes all that into account is used and that constant (looks like 308 mile lb/s) would be reduced.

At the crankshaft it would be 375 and then after subtracting off all the support systems and electrical resistance it is 308 at the wheel. 

At the crankshaft of the diesel engine it would be measured (or calculated if the engine was too potent to measure the force by mechanical means) in foot-lb of torque, converted to HP by using "5252", etc.

The two major engine builders use carefully calibrated large alternators to "back into" the HP. They measure the alternator volts and amps, and they know very precisely the alternator efficiency in order to accurately measure the HP of the diesel engine. Then an adjustment is made for temperature, etc. Obviously, the GROSS input HP of the diesel engine has to be higher since the engine, by either mechanical or electrical means, has to develop HP to run the air compressor, the dynamic brake blowers, the radiator fan(s), etc.

The 82.1%, the former EMD number, is for ONLY the losses in the transmission system (ie the traction generator and either four or six traction motors including cabling) and not the power required to operate the auxiliaries. Instead of also deducting the power required for the auxiliaries, the diesel engine builders just set the engine for a higher gross HP.

There is a slight reduction in system efficiency for a six motor vs a four motor locomotive. The newest locomotives with modern traction alternators, efficient power electronics, and efficient AC traction motors, have measurably higher efficiencies than 1940-1960 era equipment.

I was an engineer, what tractive effort meant to me was how much power I could put into the traction motors I had available before they started to unload for wheel slip. All the above goes out the window when you have less than optimal wheel/rail conditions, and the sanders are not working.

I agree. The realizable adhesion trumps everything, and that is why the builders all work hard on that.