Tractor H,P./ torque

[Anonymous-0]

My brother's tiller is rated for up to 45 hp. My 400 will exceed that by a few. The rated torque in the owners manual is 472 lbs. Does anyone know how to figure out the torque delivered at the PTO. I do not want to shear a shaft again.

It is a 6 foot New Holland 105A. At 540 PTO RPMs the tines are turning 210 RPMs.

I intend to keep the tractor at a lower engine speed to not overpower the little bugger.

Also, I reversed the right wheel, so that the tiller will cover the tire track. Will this cause a significant amount of traction loss when I plow snow this winter? I do have two weights on each wheel.

Thank you
SDE

 

[Bob]

Doesn't the silly tiller have a safety slip clutch?

torque = HP X 5252 / RPM

For instance,

50HP x 5252 / 540 = 486.3 ft/lbs
45HP X 5252 / 540 = 437.7 ft/lbs

 

[RAB]

Bob is right but you may well be over-torquing at lower speeds. The safety bolt should shear, of course, as well as the slip clutch option. Operate it at the manufacturer's recommended speed. Those ratings will be appliccable at those particular values. The shaft will be rated at that particular speed as well.
RAB

 

[Midwest redneck]

Bob is right, also the torque at the rear wheels is multipled through gear reduction. So if you have 400ft lbs and the torque is multipled by 20 times the torque is 8000ft lbs, speed is reduced though.

 

[Anonymous-0]

Yes it does, but it allowed us to shear a $650 shaft. I will make sure it is going to slip before I use it again.

It is recommended that at the beginning of each tillage season that the tension is reduced and the clutch allowed to slip for 1 to 2 seconds and then reset the spring tension to the proper setting. We learned the hard way.
SDE

 

[Jerry/MT]

At constant horsepower, the product of torque x rpm = a constant i.e Tengine x Nengine = Tpto x Npto. this is an ideal relation shipt that assume no loss through the gears. So if the engine torque is 472lb-ft at 2000 rpm and the pto is running at 540 rpm when the engine is running at 2000 rpm then the pto torque is:
T2 = T1 x Ne/Npto

472lb-ft x 2000/540 = 1748 lb-ft

If the gear train has an efficency of say 90 % than Tpto actual =efficiency x Tpto i =0.9 x 1748 = 1573 lb-ft.