If you don't know the torque of the engine, (I've never seen torques
published for mil engines?) then you can get a good guess by doing the
calculations based upon the max speed of the engine. Chances are that
the engine produces max hp at max rpm. So to figure torque, use this
formula hp = rpm* torque(ft-lb)/5252 and turn it around.
Hence, torque= hp*5252/rpm.
Then you can guess that max torque is probably about 120% of torque at
the max rpm and that should be a good guess.
To get max torque at axles, multiple max engine torque * reduction ratio
to axles in a certain gear.... ie 300 lb-ft * 5 (transmission) * 2
(transfer case) * 6.00 (axle) = ??
To get ground traction (max pull assuming no slip) divide final torque by
radius of tires in ft.
Dave
On Fri, 2 Apr 1999 17:15:01 -0600 "Ken" <kweiss@wans.net> writes:
>List,
>
>Does anyone know the foot pounds of torque that the m211 GMC engine
>puts out
>at the crank?
>Any guess about foot pounds of torque to the ground in both 4 wheel
>and 6
>wheel drive, if there is any difference?
>How about any gasoline powered deuce and a half? (I.e., CCKW, DUKW,
>M135,
>M35)
>
>Just wondering
>
>Thanks
>Ken Weiss
>1953 XM211 2 1/2 ton troop/cargo truck "Deuces are Wild"
>1954 XM105E3 1 1/2 ton cargo trailer
>Secretary, Newsletter Editor, and Webmaster
>Lone Star MVPA and "The Transfer Case"
>Check out our new web site
>http://www.geocities.com/motorcity/pit/4393/index.html
>
>
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