there is a debate going on in the land rover forum about how much torque a locker can apply to each axleshaft when locked.

i say that when in 4WD, with a locked diff (spool, welded, ARB) the diff evenly distributes 25% of the engines torque (50% of the applied torque) to each axleshaft, regardless of traction of either wheel.

others say that when one wheel is off the ground, for example, that a full 50% of the engines torque (100% of the applied torque) goes to the wheel and axleshaft that has traction.

this does not make sense to me.

which is the correct theory and why?

With a locker the torque to each wheel is the same. Depending on traction the torque could be up to 100% at a wheel.:confused:

100% to a single wheel on the ground, minus frictional losses of course.

If it was 25% to each wheel regardless, then where would the rest of the torque go?

Originally posted by reddwarf
100% to a single wheel on the ground, minus frictional losses of course.

If it was 25% to each wheel regardless, then where would the rest of the torque go?

split 50/50 in the transfer case, then 50/50 at each differentials, yielding a total of 25% of the engines torque at each wheel.

Originally posted by 9V
split 50/50 in the transfer case, then 50/50 at each differentials, yielding a total of 25% of the engines torque at each wheel.

If you have no resistance, you (technically) have no torque.
If you have three wheels off the ground, 100% of the available torque, will go to the remaining one on the ground--but chances are, the resistance available (traction) will be severely limited in that situation too.
If you have two wheels on the ground with identical traction available, those two wheels will split the available torque 50/50. If traction is different between them, they'll split the available torque based on the available traction, but up to 100% of available torque, can theoretically be applied to each axleshaft, depending on how much load the others take off of it.

Depending on what kind of locker, you can (in some rare situations) actually apply more than 100% of available engine torque to one or two axleshafts, too... envision slow parkinglot circles with two ARB's locked up. Eventually, either tires will slip, or axles will break--even if the engine doesn't have enough power to slip a tire or break an axle. Subaru BRAT's had a "thing" for locking up solid if you put them in 4L and drove in circles--the engine didn't have enough power to overcome the drivetrain bind and slip tires. But you could back up (in circles) out of the situation. :)

still doesnt really make sense to me. if you have a continuous shaft (same as two axleshafts locked together with a spool), and you are applying rotational torque to the shaft from the center (ring gear). if one side of the shaft is bound up, it may have a lot more RESISTANCE to the torque being applied, but the same amount of torque that is going to the bound side will also go to the side in the air (since it is the same shaft, different ends).

right? or wrong?

Wrong. As said above, no resistence equals no torque. All of the resistence on one end of the solid axle means that all of the torque will be concentrated there.

Originally posted by billj
Wrong. As said above, no resistence equals no torque. All of the resistence on one end of the solid axle means that all of the torque will be concentrated there.

Think of it this way... you have a 500 ft-lb engine.
You have a 4:1 final drive gear.
You have no frictional losses in the system.
You have a solid-shaft rear axle--ring gear on one axleshaft, that runs wheel to wheel.
The hypothetical vehicle has infinite traction (assuming it's on the ground) and it's chained to an immovable object.
I don't want to hear about how absurd this scenario is--I already know. :)

Stomp the gas, and you have 4000 ft-lb split among two tires. Both axleshafts see equal torque (2000 ft-lb each) and the engine stalls ('cause the vehicle can't move or spin the tires) or both sides of the axle break.

Now lift one tire off the ground and stomp the gas. One tire sees no torque (no resistance--it's in the air, and not spinning--you can't overcome infinite traction) and the other gets all of the 4000 ft-lb available. You either break that side of the axle, or stall the engine again.

Now lift both off the ground and stomp the gas. Both spin freely (zero traction). You can't stomp the gas hard enough to break both axles, because they don't have any resistance (other than accelerating the rotating weight of the tires up to speed) to spinning.

Am I helping at all, or muddying the matter worse?

if you have a continuous shaft (same as two axleshafts locked together with a spool), and you are applying rotational torque to the shaft from the center (ring gear).

You just answered your own question (or argued the opposite side). Yes, you can think of it as a single axle, so....100% of the torque sent to that end of course goes to that "single" axle, and the only end it counts at is the one with traction (resistance). Thus, if you have one wheel in the air, you have 100% of the torque to that axle going to the other wheel.

9V: You're wrong. Wether you undersand it or not, you're still wrong :flipoff2:

If you had 3 tires in the air, and one on a high-traction surface, the motor would bog down. Why?

If it only has 25% of the torque, then the motor shouldn't have any trouble turning it.

Said another way... ALL OF THE POWER of the motor is going into turning that one wheel that's touching.

(Sure, some power is eaten by frictional losses in the drivetrain and by spinning 150lbs. of tire/wheel at the other 3 corners, but not a lot).

If you were in a wal-mart parking lot sitting on flat ground, you'd be right. 25% to each tire. But lift 3 in the air and the fully-locked rig will still be able to move it's self without a problem. Why? Because the motor can put down all of it's power through the one remaining tire.

Originally posted by TNToy
If you were in a wal-mart parking lot sitting on flat ground, you'd be right.......

I think you just figured out why he doesn't get it :flipoff2:

9V: torque can't just flow out into the air. It has to be transferred somewhere. If a tire isn't touching the ground it cannot transfer any torque.

Again, all torque goes to the wheel on the ground (except for the torque used up by friction, almost nil)

like the way they test engines for tourqu


you cant tell how much tourcqu it puts out by looking at it you have to try and stop it by well restince or friction or tiers on the ground or what ever you want to call it

if it helps at all remember that some...(a very small amount)...of torque is being used to spint the wheels in the air. but that is all that is needed. the onw wheel on the ground can "use" more torque by applying it to the ground.

So if a vehicle is on blocks with no wheels touching the ground and the engine is revved to 2500 rpm then by the above theories then the engine is producing no torque( assuming that there is no friction in the drivetrain).

I dont believe that this is true because energy is never lost, it is transfered or changed to a different form.Torque is a measure of mechanical energy. That enegy is generated by the conversion of chemical energy (gas) to mechanical energy via the engine.

The transfer case and diffs distribute the available mechanical energy equally to the four wheels reguardless of whether or not they have traction or not.

if no energy is ever created or destroyed then the engine never really makes torque does it? (There is just no substantial load being put on the engine in your situation.

Originally posted by ColoYJ
So if a vehicle is on blocks with no wheels touching the ground and the engine is revved to 2500 rpm then by the above theories then the engine is producing no torque( assuming that there is no friction in the drivetrain).You're right. You're also off topic. He wants to know how much torque is given to each wheel. He's says its 1/4 to each no matter where. We say it depends.

Go read his question again. We're right, and we know it. Even if we're wrong. :flipoff2:

Ok, last try.

Pretend you are starting with a motorcycle (1wd). This is equivalent to a 4x4 with 3 wheels off the ground, agreed? now, add 3 tires to the wheel, but make them just small enough they don't touch the ground..... They're sticking up in the air....do they now get 75% of the torque? OF COURSE NOT.

Tires in the air on a 4x4 is the same priciple

Another way to think of it is consider a torque wrench. Set it to 100 lb-ft and tighten a lug nut. Recall the amount of effort you used with your arm to hit make the wrench click.

Now loosen the nut until it is nearly off the stud and apply that same amount of effort to the nut. The wrench does not click, even though you are applying the same amount of effort. You are just able to spin the wrench faster. Ooh, that defines horsepower... but that is another topic..

Resistance is the only way torque can be applied.

alright i guess i get it now... thanks for the explanations everyone, they helped :D:D:D

Guys,

Check this link out:

http://www.4x4abc.com/4WD101/why_4WD.html

read the page, then for more info on torque, click the "differential locks" link, followed by "How differential locks work" link.

Very interesting reading.

It appears that 9V was right all along, assuming that this page is correct, and I understand what I'm reading! ;-)

As for the (hypothetical or otherwise) fully locked vehicle with "three tyres in the air and one on the ground" example... the tyres in the air do have 25% of the torque each. They just don't have traction. Just reach out and grab any one of the three tyres in the air and you'll be flung round and round until you let go or drop off. The torque is there.

Also, the one tyre on the ground will have a better chance of breaking a half-shaft with its share of 25% torque simply because its moving 100% of the vehicle mass... not happily sharing the work with the other three half-shafts as it was probably designed for. That's why we have heavy duty aftermarket shafts for sale!

My 2 cents.

Regards,

Steve.

That page made very little sense.
A tire beginning to slip isn't going to make the torque increase on that side. It would decrease, as the slipping tire relieves the torque. The torque to the other side thats not slipping will always be the same as the one slipping, assuming an open diff.

The open diff ALWAYS splits the torque evenly. If one tire has no traction (tire in air), theres no load= no torque. Consequently, no torque to drive the wheel still on the ground. Make sense? read it a couple times, it will.

The locker on the other hand, does NOT split torque evenly, as several have already stated.
A locked axle with one tire off the ground will certainly have all the torque going through the shaft to the tire still on the ground.

That article was written at a very "low" tech level and only represents an ideal condition where all four tires have equal traction.