Here's how I see it

Method A is fine until you send the "extra" torque to only one wheel. It goes to both equally.

On the slipping wheel it goes into wheel angular acceleration (a lot).

On the gripping wheel it goes into wheel angular acceleration(not a lot cause it doesn't accelerate a lot) and matching vehicule acceleration.

So why does the car does not accelerate then? It does...just not very much because that extra torque is very small no matter what.

You are on throttle, not much torque is going in any of the wheels. So where does all the torque the engine would "normally" produce under grip go ? Accelerating the engine itself(inertia). Soon hitting the rev limiter

Now on how to determine that ExtraTorque...I'm guessing it's directly related to the slipping wheel inertia.(In my mind it has to) but I'd have to think about it.

Quote from Shotglass :

im not sure what the first one is trying to achieve but from what i can see both are wrong since they result in different torques on the lest and right drive shaft

There is a difference between the torque you try to apply, and the torque you could read if you were to somehow get a torque wrench into the system.

If a single wheel (of 10m radius) is connected to an electric motor which creates 1000Nm of torque, but that wheel is in the air, then 1000Nm of torque goes into spinning the wheel, and 0Nm when be read at the wrench.

If the tyre is on the ground and able to produce 750N of force, then you will read 750Nm at the wrench, and the other 250Nm increases the angular velocity of the wheel.

Likewise when the tyre can produce 1500N of force, then all 1000Nm can be read on the wrench, and the wheel does not accelerate.

It's this figure, at the wrench, that is actually felt by the diff, and should remain equal on both shafts during steady state cornering. Hence why if you just try to send an equal amount of torque to each shaft, the actual torque that could be read will differ, and you don't have an open diff.

Or at least, that's the conclusion I've come up with.

I suspect that, although method A doesn't seem to work as well, it is just more prone to integration error and oscilations, as I think the tyre which produced more force alternates each frame, so on average it might actually be correct. I might try to investigate a bit deeper later on, but for now it seems to work, so I'll stick with what I've got, unless somebody here can come up with something better.

Quote from Bob Smith :

It's this figure, at the wrench, that is actually felt by the diff, and should remain equal on both shafts during steady state cornering.

uhm... no
it doesnt matter whether the torque on the shaft goes into pushing against the road or into pushing against the wheels inertia
ther torque is right there acting on the driveshaft and whatever torque is acting on one driveshaft must also act on the other one with an open diff regardless of what reaction the torque causes

just think of torque in terms of force (the equvalent to torque in linear motion) and maybe that helps make sense of the thing

Quote from Shotglass :

uhm... no

So why is it then that keeping the felt torques equal creates an open diff, and sending equal torque to each wheel creates a LSD?

Quote from Bob Smith :

So why is it then that keeping the felt torques equal creates an open diff, and sending equal torque to each wheel creates a LSD?

Sending equal torque under all conditions is a locked diff, not an LSD.
The open diff simply distributes the torque of the highest magnitude (and that's not necessarily coming from the engine) between the other two outputs, in relation to their resistance and possible counteracting torque.

Here's a little something from a German book, makes understanding the open diff way easier than any explanation could

A representing the "input", that is the torque being applied to the differential from the engine, C and D the outputs to the wheels.
If C and D are of equal mass and have no additional forces other than gravity applied to them, pulling A upwards will pull both C and D over the same distance in the same time.

In this illustration, however, D has a weight attached to it. Think of this weight as all the combined forces counteracting at one wheel, but not the other, including friction, inertia, braking torque, etc.

Can't provide you with a mathematical solution, but it's going to be fairly complex if you want to factor in internal resistance, material fatigue, inertia, etc. You've got me infected though, I wanna work this out

Quote from morpha :

Sending equal torque under all conditions is a locked diff, not an LSD.

No. A locked diff is capable of sending infinite torque to one wheel and 0 to the other. See Torque Bias Ratio (TBR).

Take a car with a locked diff. Pivot it around one of the drive wheels on a surface with non-zero friction. That wheel sees 0 longitudinal force. The other wheel sees a lot. Likewise, lift one wheel and leave the other in contact with the surface. The wheel in the air sees zero torque, while the one on the surface sees all of the input torque.

This is all verbal diff basics, though. Again, Bob's looking for the math behind it.

Quote from Bob Smith :

So why is it then that keeping the felt torques equal creates an open diff, and sending equal torque to each wheel creates a LSD?

LSD is not about sending equal torque to each wheel. Its about limiting the angular velocity differences between the, two driven by the diff, wheels.

That's a viscous LSD. A clutch pack LSD deals only with torque, with changes in angular velocity as a side-effect.

Quote from Bob Smith :

So why is it then that keeping the felt torques equal creates an open diff, and sending equal torque to each wheel creates a LSD?

Just means the modelling sending equal torque is wrong. Keeping felt torques equal will create a close-to but not really how-it-should-behave open diff.

Perfect open diff. Two wheels in the air. Your finger holding one.

Constant engine speed: Your finger is feeling only the other side wheel bearing friction. The only resistance there is after the diff.

Slam the throttle : Your finger will be feeling wheel bearing friction + the exact same amount of torque that is accelerating the other wheel.

The thing is that torque is REALLY small unless you have a big ass heavy wheel. The real heavy torque is going through the diff ONLY if there is resistance on the other side of it. Without resistance it just never gets out of the engine.

WHat do you do when you want to check if your clutch is slipping ? Jack the rear and slam the throttle? No. You get in 5th gear, where there is a shitload of resistance and then apply torque. Torque that now gets out of the engine and is able slip the clutch.

The real challenge of this modelisation is determining how much torque ACTUALLY goes into the diff. Like I said its probably related the the wheels inertia. And since the wheel speed matches engine speed, probably engine inertia plays a role too.

Quote from Forbin :

That's a viscous LSD. A clutch pack LSD deals only with torque, with changes in angular velocity as a side-effect.

indeed. I just focused at the end result of any limited slip differential.

Anyway still even a clutch pack LSD or any torque sensitive LSD is not about keeping equal torque in both wheels. The point is always to distribute the torque towards the tire that has more traction.

Quote from Bob Smith :

So why is it then that keeping the felt torques equal creates an open diff, and sending equal torque to each wheel creates a LSD?

how exactly do you judge which kind of diff youve created?

also im guessing your open diff model doesnt handle excess input torque correctly

Quote from PhilS13 :

Slam the throttle : Your finger will be feeling wheel bearing friction + the exact same amount of torque that is accelerating the other wheel.

Quote from Shotglass :

how exactly do you judge which kind of diff youve created?

Compare tyre forces, if they are equal at all times*, then diff is open.

*Steady state, of course

With diff model B stated above, tyre forces stay within 5% of one another, which is pretty good given the massive integration error involved at the time step.

Quote from Shotglass :

also im guessing your open diff model doesnt handle excess input torque correctly

One wheel will spin more than the other, unless resistances are exactly equal, in which case they will both accelerate at the same rate, which is only ever going to happen when they are in the air.

Quote from Bob Smith :

One wheel will spin more than the other

yes but neither of your models apply the torque required to change the angular velocity of the freely spinning wheel to the other wheel that still has grip