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Re: Re: M motors
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Posted by G-MAN on September 19, 2002 at 16:38:38 from (206.106.139.74):
In Reply to: Re: M motors posted by Steven@nd on September 18, 2002 at 16:49:36:
I'm not trying to start problems here, but I'm having some trouble with your math. The formula for horsepower is as follows: HP = (Torque x RPM) / 5252. A couple different sources I referenced show rated speed for the Farmall M to be 1450 rpm. According to my calculations using the formula listed above, a Farmall M would have to produce 97 horsepower at 1450 rpm to have 350 lb. ft. of torque at that RPM. I calculated that as follows - 350 ft.lb. x 1450 RPM = 507,500. 507,500 / 5252 = 96.62 HP. Using a HP level of 45 at 1450 RPM (although the Nebraska test results I saw showed around 36 HP on gasoline) torque would be 163 lb.ft., or less than half of the claimed 350. I calculated this by the following formula: HP x 5252 / 1450. I have dynoed dozens of tractors, and I can tell you first hand that as you continue to pull a tractor down past rated speed, horsepower and therefore torque will continue to increase. But, according to my math, an M would have to produce 45 HP at a scant 675 RPM to be producing 350 lb.ft. of torque. A John Deere "G" with 412.5 cubic inches and a monster stroke of 7 inches only produces 204 lb.ft. of torque at 975 RPM (according to the Nebraska test HP rating), so if you have or know of a stock Farmall M that can produce 350 at 675, I am extremely interested in purchasing it. Even a die-hard Deere man such as myself can see that such a tractor would be a force to be reckoned with in stock pulls. Once again, none of this is intended to start some big color war, but I find engine-related math to be interesting, and something didn't look quite right in your post. Feel free to correct any errors that I may have made.
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