• Tractive effort

  • General discussion about locomotives, rolling stock, and equipment
General discussion about locomotives, rolling stock, and equipment

Moderator: John_Perkowski

  by N. Todd
Most of us know that for first and second-generation locomotives, maximum tractive effort is simply 25% of the unit's total weight.
Now how do you figure out continuous tractive effort? Weight aside, I suppose this revolves around the minimum continuous speed and maximum speed. The following were taking from locomotive databooks-assume all have a maximum speed of 70 mph:
Example 1 weighs 268850 lbs. and exerts 57200 lbs. of con't t/e at 13.8 mph.
Example 2 weighs 348000 lbs. and exerts 79500 lbs. of con't t/e at 8.5 mph.
Example 3 weighs 391500 lbs. and exerts 90600 lbs. of con't t/e at 9.9 mph.

So is there a formula for figuring this out?
  by Jay Potter
In theory, the easiest way to determine a unit's "continuous tractive effort" is to enter the tractive effort curve (tractive effort as a function of speed) with the minimum continuous speed for that model unit. The amount of tractive effort for that speed is the unit's continuous tractive effort.

However, as a practical matter, the "mcs" concept is somewhat outdated because it doesn't really apply to AC-traction units or to DC-traction units that are configured with protective circuits that automatically reduce the current to traction motors when necessary to prevent them from overheating from extended operation beneath the unit's mcs. The mcs concept is even less applicable to AC-traction units because their traction motors can operate at extremely low speeds (one mph or even less) for hours without overheating.

Because of this, the term "nominal continuous tractive effort" is oftentimes used instead of "continuous tractive effort". A unit's nominal continuous tractive effort is the tractive effort that the equipment manufacturer has determined that the unit can be relied upon to produce under most operating conditions. For example, an AC4400CW has a nominal continuous tractive effort of 145,000 pounds. Under most operating conditions, it will begin producing that level of tractive effort when its speed falls to 9.8 mph. So 9.8 mph is the "nominal continuous speed" for an AC4400CW. However that has nothing to do with traction-motor overheating because, as indicated previously, the unit can operate for extended periods at much lower speeds without the risk of overheating its traction motors.

  by N. Todd
Thanks for clearing that up a bit.
However I am not concerned with AC or the DC units with all the microprocessor wheelslip systems- that's all too complex. I was just referring to older examples like aN SD40, C-636, U30C, etc.
  by Jay Potter
If you want a less "complex" approach, you could use the following formula: Continuous tractive effort equals (horsepower) times (308) divided by (mcs). However my personal view of that approach is that it's too general to be really useful. I'll use the SD40 as an example.

According to the operator's manual that I use, the SD40 was offered with three different gear ratios, each of which had a different mcs. These were 11.1, 11.9, and 12.9 mph. EMD considered the SD40's standard weight to be approximately 368,000 pounds.

When the formula is applied to each mcs, the respective continuous tractive efforts are 83,200 pounds, 77,600 pounds, and 71,600 pounds.

Using Chessie System SD40s as an example, their weights ranged from 390,000 pounds to 420,000 pounds. The railroad assigned all of them an mcs of 11.1 mph and a continuous tractive effort rating of 82,100.

So if I wanted to cite the continuous tractive effort for "an SD40", what value should I use?

  by Typewriters
I'll just note a couple of things here, as I've researched this very thing for a long time and arrived at a simple solution --- which is that the only way to do this properly is to discover actual official manufacturer-supplied data.

The "308" so ofted quoted is only a very rough approximation, which seems best to apply to an early generation of EMD units. In fact, when you find actual information and plug it into the formula and solve for this conversion factor, you can find that it's a low as the mid or high 290's and up as high as the low 310 plus range. It's really a derivation of another formula relating efficiency of electrical machines. Just using 308 all the time gives gibberish -- you have to find the actual data.

Second, note that traction motors and electrical equipment often was changed during production of a given locomotive. For example, early ALCO C-628 units are rated at 79,500 lbs continuous tractive effort, while later ones are rated 85,800. Just the same, early ALCO / MLW 3000 HP C-630 units were rated 85,800 lbs continuously, with later units being 90,600 lbs. You have to find the actual data for the unit you're looking at, and I know of nowhere yet that you can find the exact break for every make and model. It's a tedious thing to search out the data, rather than calculate it, but it yields the correct answers. ( I have lots of this kind of actual hard data if you care to ask for specifics!)

The last thing to consider is whether or not the locomotive data you are looking at includes the use of power matching equipment. Generally, this is electrical equipment that reduces power at lower locomotive speeds -- and you can see that it's used in the variously differing minimum continuous speeds that you'll find for various GE U30B units, as a good example. Also, Chessie adjusted its GP-40 units' Performance Control for a lower-than-standard minimum continuous speed. This must be considered when comparing, for example, the data for NYC U30B units and for IC U30B units which have different minimum speeds listed with roughly similar continuous efforts. Notable too is the comparison between the roughly 17 MPH minimum speed of the PC/CR U33B units and the roughly 12 MPH MCS of the CR U36B units. That's adjustment of the Automatic Power Matching System at work.

-Will Davis