PTO horsepower

[oosik]

A question. Why is PTO horsepower alway rated less than engine horsepower? Any other system of gears or pulleys that goes thru an approximate 4 to 1 reduction will increase useable power by a factor of 4.

Example: My Kubota M6040 is rated at 62 engine HP and 55 pto HP. My engine turns approximately 2000 rpm when the pto is turning roughly 540 rpm. Almost a 4 to 1 reduction - yet, rather than the pto being rated at around 240 hp - its rated at 55 hp.

Why?

 

[RyeThomas]

The pto is a straight reduction off engine rpm. Now the same rpm goes thru a transmission to the wheels that affects wheel speed. You would need a transmission attached to the pto to speed it up, then a transfer case to increase power (in simple terms), added cost, weight, size.

Just my simple theory, not saying it’s right.

 

[ericm979]

In a gear tractor it'd be lower due to friction in the gear train to the PTO. Each set of spur gears takes 1-2%. With an HST the PTO is taken before the HST part of the transmission- that's how you can change tractor speed while the PTO rpms stay the same. The HST has pumping losses which are greater than gear friction losses in a gear tractor. I'm not sure why with an HST the PTO HP is lower than the equivalent gear model.

Remember that HP = torque * rpm. Gear up or down and the torque and rpm change but the HP is the same.

 

[oosik]

Rye - when you pose that answer - don't look at me for confirmation. I sure as heck do not know. That's why I ask. This is, no way, a trick question.

 

[RyeThomas]

That’s just my theory buddy, but I will also admit I’m not saying I’m right.

 

[oosik]

Give you another example. I have a Warn winch mounted on my Jeep. Single line - first wrap on the drum - it will pull 8000 pounds. Run the winch line out to a snatch block on that tree out there and back to the Jeep. The winch will now pull 16,000 pounds. Snatch block on the Jeep bumper - winch line out to a two pulley snatch block on the tree - back to the Jeep bumper - now the winch is pulling 32,000 pounds. Probably exceeded the capacity of the winch line but you get the idea.

 

[TerryR]

ooski,

Each wrap around a block pulley doubles the pulling force, but also halves the speed. As a result the power remains the same, less friction loss. Power is the amount of work done divided by the time it takes to do the work. So you can move twice the load, but half as fast.

The same is true of the gears driving the PTO you mention in your first post. If there is a 4 to 1 gear reduction the speed is reduced by a factor of four, while the torque is increased by a factor of four. No change in power again, except for friction losses which reduce available power a bit.

 

[RyeThomas]

That’s a different story and gets into another ball of wax of change of direction, angles, load sharing/multiplying (like old school block and tackle) making 2:1 force multipliers etc.
Now the load on the winch empty spool vs loaded spool is a gear ratio deal just like turning a 9” torque wrench to say XXX lbs vs a 14” torque wrench the same XXX lbs.
All very interesting stuff, good post.

 

[RickB]

Any other system of gears or pulleys that goes thru an approximate 4 to 1 reduction will increase useable power by a factor of 4.


Why?

To begin with, this statement is not based in fact if one considers ‘power’ to be synonymous with horsepower. It flies in the face of a basic law of physics, the conservation of energy.
To answer the underlying question, PTO HP is less than most published engine HP ratings due to parasitic friction loss in the PTO drivetrain and any engine accessories present and operational when PTO HP is measured that may not be present when engine HP is measured. Additionally, engine and PTO HP figures may be observed at different engine RPM values.

 

[LouNY]

A question. Why is PTO horsepower alway rated less than engine horsepower? Any other system of gears or pulleys that goes thru an approximate 4 to 1 reduction will increase useable power by a factor of 4.

Example: My Kubota M6040 is rated at 62 engine HP and 55 pto HP. My engine turns approximately 2000 rpm when the pto is turning roughly 540 rpm. Almost a 4 to 1 reduction - yet, rather than the pto being rated at around 240 hp - its rated at 55 hp.

Why?

The engine hp is the max for the engine with no loss such as the alternator and hydraulic pumps and any drivetrain loss.
If you look at the torque to horsepower it should correspond to this;
HP = T * N /5252
Where
T = Torque (lbft)
N = Speed (rpm)
which will rearrange to;
Torque = (HP * 5252) / RPM
Torque = (62 * 5252) / 2000
Torque = 325,624 / 2000

Torque = 162.812 foot pounds of torque gross torque from the engine however;
(55*5252) / 2000 = 288,860 / 2000 = 144.43 foot pounds of torque going to the PTO
at the PTO you will have approximately 535 foot pounds of torque at 540 rpm.
HP = T* N / 5252
55 = (535 x 540) / 5252.

If you had a PTO dynamometer attached to your tractors pto it should indicate 535 foot pounds of torque at 540 rpm and full throttle,
which would be 55 HP.

 

[VroomVroom]

You will never increase your HP through gearing. The engine, combusting a specific fuel will provide "X" amount of Horsepower. How that horsepower is used or geared will change the torque values for you. The lower HP rating at the PTO is through efficiency loss. The only thing you can multiply in this case would be the torque. 540 is a lot slower in revolutions then say 2400 engine RPM. So comparing the torque at the crank and at the PTO shaft you will see the difference.

Basically, you got what the engine is going to give you and that's it. From there, its up to how its transmitted to the PTO how much you may have left after parasitic losses. How its geared along the way will dictate how much torque you have back there.

 

[ruffdog]

Torque is the twisting force. HP is a measure of the work that can be done. You can have all the torque in the world but if it is only turning 1 rpm, not much "work" is getting done. If you had the same amount of torque but at 5000 rpm, now you can get much more work done. That means more HP.

 

[Texasmark]

".......Any other system of gears or pulleys that goes thru an approximate 4 to 1 reduction will increase useable power by a factor of 4." Say what?


Power is the consumption of energy, referred to as "work". HP is [torque (in ft-lbs) x RPMS]/5252 for a simplistic equation usually relating to engines and gear trains.

What happens in a tractor is the engine HP "being applied" is transmitted to the drive train via a geared system of shafts. The ratio of the gears between the source and load determines the speed and torque of the final drive, not necessarily 4 and at the same HP less losses in the system.

The gears change the RPMS of the applied HP and in doing so change the torque; usually at a higher ratio which reduces the RPMS of the power while multiplying the torque which is necessary for moving heavy loads. The losses are in the transfer mechanisms due to friction.

The transferred HP is still what it was, it just went different places and changed characteristics!

 

[Opti-Mist]

You will never increase your HP through gearing. The engine, combusting a specific fuel will provide "X" amount of Horsepower. How that horsepower is used or geared will change the torque values for you. The lower HP rating at the PTO is through efficiency loss. The only thing you can multiply in this case would be the torque. 540 is a lot slower in revolutions then say 2400 engine RPM. So comparing the torque at the crank and at the PTO shaft you will see the difference.

Basically, you got what the engine is going to give you and that's it. From there, its up to how its transmitted to the PTO how much you may have left after parasitic losses. How its geared along the way will dictate how much torque you have back there.

Parasitic losses, mostly due to friction associated with the numerous mating surfaces involved in the transfer of power. A 12% loss between engine and PTO is about what one can expect.

 

[CincyFlyer]

A question. Why is PTO horsepower alway rated less than engine horsepower? Any other system of gears or pulleys that goes thru an approximate 4 to 1 reduction will increase useable power by a factor of 4.

Example: My Kubota M6040 is rated at 62 engine HP and 55 pto HP. My engine turns approximately 2000 rpm when the pto is turning roughly 540 rpm. Almost a 4 to 1 reduction - yet, rather than the pto being rated at around 240 hp - its rated at 55 hp.

Why?

Power is force times speed. If you get 1/4 the speed, you get 4 times the force (usually in the form of twisting motion, torque), so the math always comes out to be the same horsepower.
Or else we'd all have tiny little electric motors in our cars, geared up to make a thousand horsepower.

 

[oosik]

Sincerely - thanks to all who posted so far. I think I have a better idea of the answer to my question. I've wondered about that situation for a long time.