Okay. We’re getting there. If I might ask for your patience once more: so when we say the torque is split 50/50 front/rear, but in the case I outlined in the first post, 100% of the torque is going to the rear wheel with traction, that’s the semantic issue I find confusing. Is it only 50/50 front/rear when wheels on both axles have traction?

I understand how 4WD works. This really is just a semantic issue. I was in a theoretical discussion the other day, and I said that 100% of the torque goes to the wheel with traction, but my discussion partner insisted that only 50% could …

In 4wd where all gearing is completely fixed, no slippage or clutches etc, I would not word it as "torque" is split, more like "power is being distributed to both axles/all wheels". The actual application of the torque can only be applied where there are opposing forces. in this case, the one wheel with traction.

The best way to look at it is, fully locked 4wd will allow distribution of torque where it is able to be applied. No traction means the power cannot be distributed there. If both rear wheels have equal traction then it would be 50/50 torque applied to both rear wheels and none to the front if they are freespinning. The distribution is based on opposing forces.

If you look at audi's new awd system in the rear of the 2022+ rs3, they use electronic controlled clutches to force the torque to the rear wheel that has the most traction, to properly apply that torque.

Torque is not force. Torque is a Moment (Force * Distance) i.e. ft*lbs. or N*m. Otherwise completely agree with your statement.

For force or torque to exist there must be an opposing force. For wheels in the air, there is no opposing force. For wheels with "no traction" vs "with traction" you are looking at different frictional forces, static friction vs kinetic friction, respectively. There is still some torque applied to a spinning wheel.

Under normal driving conditions or when the vehicle is stopped tires are in static friction with the ground. The scenario where a wheel spins (or skids) happens when the force applied to the tire overcomes the static friction force available between the tire and ground surface (traction). Loss of traction means the tire and ground are in kinetic friction (spin or skid).

Under the identical conditions static friction always exceeds kinetic friction. Ie you may feel safe driving on wet or icy road but when you experience traction loss you will have considerably less traction sliding than when driving normally. I don't have data to back this up but I would estimate you can lose up to 75% or more of friction in certain conditions when sliding takes place. This is why when traction is lost you take you foot off the brakes or gas and try to regain traction so that at a minimum you can steer and ideally you can brake or accelerate. The exception here being when you drifting or are spinning muddys.

Finally! everyone is starting to understand that torque can't go somewhere when there is no resistance or opposing force.

In my explanations, I chose to ignore friction, gear losses, etc. for clarity (why muddy the water. )

It may be a semantics issue, but I would disagree with your first statement. Torque is a rotational force. We just happen to use linear terms (lb-ft etc) to measure it or specify it.

Perhaps I phrased my question poorly.

Change torque to power but I knew what you meant.

Semantics be dammed. The engineer in me will not back down from this one.

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A simple experiment will illustrate why this rule is true.

1. Jack up the car until the wheels are off the ground and can spin freely (no brakes, transmission in neutral, etc)

2. Take your trusty torque wrench and try to torque the wheel nuts. Can't be done; the wheel will just spin because there is no opposing force and the torque wrench will read zero.

In the original scenario, any wheel that is not touching the ground gets no torque and will just spin freely. The argument of torque splitting, which direction torque will go is meaningless: the tires off the ground receive no torque.

You are both correct and this argument of torque is a force or torque is a moment is mostly semantics. Or, maybe from a scientific standpoint, it is a matter of relativity.

Torque is the rotational analogue of linear force.

Linear Work = Force x Linear Displacement. Rotational Work = Torque x Rotational(Angular) Displacement.

Not all force is torque, but all torque is rotational force.

All squares are rectangles, not all rectangles are squares.

All bourbons are whiskey, not all whiskey is bourbon.

All bourbons are whiskey, not all whiskey is bourbon.[/QUOTE]


And most importantly All Bourbons are from Kentucky. End of discussion
It's Friday, I need Bourbon

Um, yeah. A delightful discussion of physics and engineering principles, but they don’t answer my question, which is: if the transfer case splits the torque 50/50 front/rear, how can does 100% of the torque get to a single rear wheel?

It might help to think this through as available torque vs actual torque and how this flows through the system. Torque is available 50/50 at the tcase but it isn’t actually felt by the front or rear dive shafts and then axels unless traction exists at each tire. The differentials complicate this because if a dif is open and one wheel doesn’t have traction then the available torque goes to the spinning wheel instead of the one with traction. This is why we have lockers.

In the scenario where 100% of the torque goes to one wheel you would need the other three wheels to have zero traction and then have the differential locked at the axel. Now that we have mechanically locked the wheel with traction it must rotate and because it has traction the available torque goes to this tire only.

Think of the scenario above but replace the driveshafts and axels with helical springs. Which springs will feel twist under load and which ones spin freely without resistance? The driveshaft and the axel upstream of the tire with traction are the only springs that twist up and experience torque under load. The other springs spin freely but dont feel twist at all. This is 100% of the torque going to the tire with traction (Ignoring drivetrain losses and waste)

For semantics of torque we are all saying the same thing but I think it’s still important to note that:
torque = force * distance = moment = “rotational force”
Torque ≠ force
Ft * lbs ≠ lbs

Both are vectors but bodies can be under both force and moment and it’s important to distinguish the two during analysis.

I agree mostly with El Dusty. Perhaps it is all in nomenclature and semantics, but torque is a rotational force; a force is a force, no matter whether your push in a strait line, or twist.

Captain Spalding, I think you are locked into the mindset that torque must be split 50/50. That is true if all tires have the same amount of traction, but not true if there are wheels without traction. Power, torque, or a push only goes some where if there is resistance: push against an object and you can feel the returning pressure on your hand; push against air and you feel little returning force.

Thanks. Never mind.

Because it's a locker.

"Torque distribution" is a term used for differentials, like subaru's center diff, our Torsen center diff, etc. Our Torsen, for example, sends 40/60 F/R normally, and close to 29:71 when turning, 53:47 when rear loses grip. Because that's the amount of torque that the differential is capable of sending, maximum 71% rear, and maximum 53% front. If center diff is unlocked on an AWD 4runner, when rear axle loses all traction, front can only get 53%, as stated above.

When you lock all 3 diff locks, the "torque distribution" is no longer determined by the vehicle, but the surface that you drive on. If you're driving perfectly straight on a perfectly flat road, with 4 tires equally worn, with same friction (μ, if you read papers like nerds do. Just kidding), your torque is indeed 25% across 4 wheels. However, because you do not have a "differential" any more, the system is able to send 100% of torque from the T-case to any of the wheels, given that wheel does not break traction, and other wheels have 0 traction (μ=0, which never happens, seriously). The torque distribution is now completely dynamic, as it'll "assign" torque to the wheel that has the most traction. That's why "torque distribution" does not apply to lockers.

The "50:50 torque distribution" statement is really just helping people to understand a center diff lock, or traditional part time 4wd. For lockers, the statement itself is misleading, as torque distribution is completely dynamic (in a good way). If a center diff always has 50:50 torque distribution, it cannot send 100% of the torque to any of the wheels, because it's actually an open diff.