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?

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?