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Pulling power

Started by Geeraider, March 20, 2020, 11:52:37 AM

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Hunt

Drawbar pull

One way to calculate  a locomotive drawbar pull.

Multiply the engine weight, in ounces, by the maximum grade which it will climb, expressed as a decimal.

Example, if a 16 oz. locomotive will climb a 25% grade --  multiply 16oz. x 0.25=4oz. drawbar pull on level track.

Trainman203

Misread.  Car WILL roll down a 2%. But will it roll down a 1.9999999%?

😂😂😂😂😂

Len

Hunt - I just turned 18 for the 4th time a few months ago myself.

I copied all that math stuff into a text file in my "MRR Tech Info" folder some time ago for my own use. Then never actually used it for anything, just going with the "pulling power = 25% of the locos weight" rule of thumb instead. At least until I got a digital fish scale.

Len
If at first you don't succeed, throw it in the spare parts box.

jward

#18
No disrespect to you Hunt. You're usually spot on. But a locomotive will have a hard time pulling itself up a 25% grade let alone anything else.

Information I have, from an old Atlas track plan book, indicates the following:

On a 1% grade pilling power is 57% of what it is on level track.
1.4% it's 50%
2% it's 38%
2.8% it's 25%
4% it's 17% or 1/6 of level track
5.6% it's 8%

Note the odd percentages are derived from rise per standard track section, i.e. 2.8% equates to 1/4" rise per track section. Standard 18r and 22r curves are very close in length to a standard 9" straight.
Jeffery S Ward Sr
Pittsburgh, PA

Hunt

Quote from: jward on March 23, 2020, 06:29:28 AM

No disrespect to you Hunt. You're usually spot on. But a locomotive will have a hard time pulling itself up a 25% grade let alone anything else.

. . .

jward,

You having trouble remembering who wrote what and/or misreading what I wrote.

I used 25% in the example just to keep the math simple

"Example, if a 16 oz. locomotive will climb a 25% grade --  multiply 16oz. x 0.25=4oz. drawbar pull on level track."