How Does This Work? Voodoo Engineering?

[Cahaba Valley Farm]

I was comparing the specs between a John Deere 3038E and a Kubota LX3310 and I'm not understanding how these numbers work. If anybody has any ideas please share them.

John Deere 3038E:
Engine Torque at rated speed: 77.2 ft-lbs
Rated Speed/RPM: 2500
Engine horsepower: 37.2
PTO horsepower: 30
Weight: 2,222 lbs

Kubota LX3310:
Engine Torque at rated speed: 77.7 ft-lbs
Rated Speed/RPM: 2500
Engine horsepower: 30.8
PTO horsepower: 27
Weight: 1,918 lbs

 

[Midniteoyl]

I was comparing the specs between a John Deere 3038E and a Kubota LX3310 and I'm not understanding how these numbers work. If anybody has any ideas please share them.

John Deere 3038E:
Engine Torque at rated speed: 77.2 ft-lbs
Engine horsepower: 37.2
PTO horsepower: 30
Weight: 2,222 lbs

Kubota LX3310:
Engine Torque at rated speed: 77.7 ft-lbs
Engine horsepower: 30.8
PTO horsepower: 27
Weight: 1,918 lbs

The JD's torque is 88.5lb-ft according to Tractor Data

TractorData.com John Deere 3038E tractor engine information

www.tractordata.com

But 77.2 according to JD themselves, so...

 

[LouNY]

One very important figure is left;
what rpm did they get these numbers at.

 

[4570Man]

You need rpm and torque to establish HP. More RPM at the same torque equals more hp.

 

[deezler]

the formula is:

HP = Torque x RPM / 5252

For torque in lbf-ft. So you can have torque and hp peaks at different engine rpm depending on specs snd tuning.

why are you looking at sub 2000 pound tractors now? After all that ranting in your other thread, lol.

 

[jaydee325]

HP = Torque x RPM / 5252

Here is the derivation of this formula.

LINK

 

[Doughknob]

I was comparing the specs between a John Deere 3038E and a Kubota LX3310 and I'm not understanding how these numbers work. If anybody has any ideas please share them.

John Deere 3038E:
Engine Torque at rated speed: 77.2 ft-lbs
Engine horsepower: 37.2
PTO horsepower: 30
Weight: 2,222 lbs

Kubota LX3310:
Engine Torque at rated speed: 77.7 ft-lbs
Engine horsepower: 30.8
PTO horsepower: 27
Weight: 1,918 lbs

How they work? What do u mean?
Ideas? About what

Your signature phrase there suggest you know what torque is. So what are you asking?

 

[CobyRupert]

If “rated speed” is engine rpms when the PTO is turning at a standardized 540 rpm that most attachments operate at, it means the Deere’s engine is turning at a faster rpm to get the 540 rpm at the PTO.

Like others say:
(Horse) power = Torque x RPM
(don’t worry about the other numbers in the formula. They’re just conversion factors for keeping the units (Hp, foot-pounds, minutes, seconds, etc..) straight.

So if 2 tractors have the same torque, but one outputs more horsepower, it’s engine rpms must be higher at “rated speed”.

“Rated speed” is the undefined, confusing term in the above info.

Example your weedwacker turning at 12,000 rpm may have the same horsepower rating as an electric motor turning 1200 rpms, but it’s going to have 1/10th the torque.

 

[Jchonline]

What part of it are you having problems with?

 

[rScotty]

I was comparing the specs between a John Deere 3038E and a Kubota LX3310 and I'm not understanding how these numbers work. If anybody has any ideas please share them.

John Deere 3038E:
Engine Torque at rated speed: 77.2 ft-lbs
Engine horsepower: 37.2
PTO horsepower: 30
Weight: 2,222 lbs

Kubota LX3310:
Engine Torque at rated speed: 77.7 ft-lbs
Engine horsepower: 30.8
PTO horsepower: 27
Weight: 1,918 lbs

I think the problem is that you may be expecting that JD's "rated speed" is the same as Kubota's. It may be that both use the same RPM for rated speed, but it doesn't have to be the same and if it was it would be a coincidence.
In this case it looks to me like JD's engine is set up to produce their torque figure at a rated speed that is slightly higher than the Kubota's. There could be as much as a couple hundred RPMs difference.

If so, and the torque is the same, then when we calculate HP using HP = torque times RPM we get different HP.

Does that help?
rScotty