Tractive Effort of Locomotives?

[Anonymous]

The term "tractive effort" is used when discribing a locomotive. I would like to know how I can calculate the value. I have found out it has a lot to do with the wieght of the locomotive. What else and does anyone know the formula needed to do the calculation? JL.

[Anonymous]

Here's a formula for the tractive effort of a steam engine:
Traditional formula for figuring tractive effort of a 2 cylinder steam locomotive:

(BP X .85) X (cb X cb) X cs
___________________________ = TE

dd

BP is boiler pressure
cb is cylinder bore
cs is cylinder stroke
dd is driver wheel diameter
TE is tractive effort

Example for K-36:

(.85 X 195) X (20 X 20) X 24
____________________________ = 36,136.3636 lbs.

This is a theoretical value. Hope this helps.

[Anonymous]

Thanks Joe! This is exactly what I've been looking for. I guess in todays railroading the BP and cb and cs of the steam engine all get combined into an amperage value or horse power rating for dissels? And the value would be derated like the BP value say at 85% level again? I was suprised to see total wieght of the engine was not in the calculation at all. I thought it had more to do with the tractive effort or is that tractive force. I guess I still have some more unanswered questions related to the topic, if you know any more formulas? thanks again. James.

[Anonymous]

There are lots of discusssions of this topic on the Web. Using the Google.com search engine, I got a lot of stuff entering the term tractive effort. That's how I got the formula I posted the other day. Here's the URL of a good discussion of steam vs. diesel performance I found using google.com:
http://www.trainweb.org/railwaytechnical/st-vs-de.html

Hope that helps.

[Anonymous]

yes it does, thanks for the link
James.