Where does all the extra torque go-to in a mu-split situation?

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In a μ-slip situation, an engine producing 100Nm of torque can achieve a maximum of 450Nm at the driveshaft, but only 150Nm can be transferred to each wheel when one is on ice and the other on tarmac. The extra torque, which amounts to 150Nm, is primarily used to accelerate the drivetrain and the spinning wheel on ice, rather than being effectively utilized for traction. An open differential tends to send most torque to the wheel with less traction, resulting in minimal torque being transferred to the wheel on tarmac. This leads to a scenario where the spinning wheel requires very little torque to maintain speed, causing the engine to rev higher until it reaches a limit. Ultimately, the system's inefficiency in torque distribution highlights the challenges of traction in slippery conditions.
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Lets say my engine is producing 100Nm of torque and after torque multiplication with the 1st gear and drivetrain,I achieve my maximum possible torque potential in the driveshaft as 450Nm.

In a μ-slip situation(left wheel on Ice and right wheel on tarmac),the max. possible torque I can transfer on Ice would be calculated to then be 150Nm. The theory says that I would have 50:50 torque distribution in a open differential.This would hence mean my Tarmac side would also have a 150Nm torque.

Original torque on driveshaft =450Nm
Torque at the wheel on Ice = 150Nm
Torque at the wheel on Tarmac= 150Nm

Torque disappeared somewhere=450Nm-(150Nm+150Nm)=150Nm

Where does this extra torque disappear?
I'm thinking all this extra torque goes into spinning the pinion gears in the open differential.Is this correct?
 
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If you have it floored, the extra torque goes into accelerating the engine and drivetrain. If you have one wheel on tarmac and one on ice, this means you will spin the wheel on ice.
 
In addition to the previus post, the torque is also being used to accelerate the spinning tire.
 
Keep in mind that the torque at the wheels that the tyre has traction on a surface is lessened greatly when the tyre starts to slip. So, if the traction limit on ice 150Nm, then once the wheel has started to spin, the traction that tyre now has may be as low as 15% of the limit, 22.5Nm! That means you only have 22.5Nm to either wheel. Once the wheel on ice is free spinning, the 'extra' torque you have available accelerates the engine and drivetrain to the point where the extra torque is being accounted for, or the engine hits the rev limiter, or the throttle position becomes a restriction to the amount of power(torque) the engine produces, reducing it or limiting engine rpm to where it is producing enough torque to keep the drivetrain and spinning wheel at speed.
Damo
 
Gosh those were good explanations.Didn't know that a spinning wheel would take-up so much of the torque! 150Nm-22.5Nm=127.5Nm of torque lost in just spinning the wheel on ice. Guess the engine would stall if it just had to produce 22.5Nm+22.5Nm of torque.
 
An open differential on ice is more like a 99/1% split differential. It will send the majority of torque to the wheel with the least traction. A 50/50 split is a locked/posi rear end.
 
marellasunny said:
Gosh those were good explanations.Didn't know that a spinning wheel would take-up so much of the torque!
To be clear, most of the torque will be used to *accelerate* the whole drivetrain (including the wheel on the ice, but not the one on the tarmac), not to keep it spinning. To keep a wheel spinning at constant speed on ice requires very little torque (due to the very low friction), so anything additional you give to it will just accelerate the whole system. The system will continue to accelerate (RPMs rising) until the torque is no longer enough to accelerate the system further. This will likely be when you hit the rev limiter or the tyre comes off the ice onto something more grippy.
 
are we talking about normal car? usual gear ratios for small car are around 4 for first gear, and than around 4 for final gear. so it is 1600Nm going to the wheels!

next thing, as Highspeed says. with one wheel on ice open diff can transfer only tiny amount of torque to the wheel on tarmac (due to internal friction and stuff). not enough to move the car at all
what you describing is how limited slip differential Torsen works. basically it can transfer to tarmac side only the same torque the ice side can transfer.
if you have locked diff (or strong viscous lsd), it locks the both sides together, so it can transfer 100% of torque to wheel with traction.

any torque over that is going to result in unloaded wheel/engine spinning faster. that requires very little torque, so your car can be standing still with 15% throttle and reving 4000rpm. at full throttle the engine would be bouncing of the rev limiter and you will be standing still with one wheel spinning wildly.
 
Highspeed said:
An open differential on ice is more like a 99/1% split differential. It will send the majority of torque to the wheel with the least traction. A 50/50 split is a locked/posi rear end.

You aren't entirely wrong with that description, but, with an open diff you will only ever have the same percentage of torque being applied to the wheel with traction as the amount of torque which is making the spinning wheel spin!
If it takes 10Nm to make an open diff spin one wheel, the other wheel will only have 10Nm to propel the car regardless of the instantaneous torque the engine/drivetrain is generating, whether it be 15Nm or 500Nm. So an open diff is a 50/50 torque splitting device, but not a 50/50 power proportioning device.:smile:
 

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