Relationship between axle torque and wheel torque

[sprashanth]

Knowing the torque transmitted from the engine onto an axle of a vehicle, how do you find out the torque on each of the wheels? Is it the same as the axle torque or is it half or it?

Assuming of course, two wheels on the axle - one on each side

 

[CapriRacer]

well, it's been 2 days and no one has responded, so I'll take a stab at it.

If a vehicle is moving forward in a straight line, then the force transmitted to the ground has to be the same for each tire's contact patches - otherwise the vehicle would tend to turn. Of course, a little steering action can transmit an opposing force to offset what the drive portion is doing.

In a RWD this is easy to see, but in a FWD, where both the steer and the drive are the same, it's a lot harder. But "Torque Steer" is a good illustration - this is where the axle is longer on one side than the other and some of the torque is used to twist the longer shaft.

Needless to say, when the car is turning the torque split is slightly different.

Torque in = torque out + torque lost, so discounting the loss of torque due to friction, the differential will evenly split the torque between the 2 tires when the vehicle is going in a straight line and bias the torque when the vehicle is turning. And the sum of the torque out will equal the torque - once you account for the gear ratio.

So the torque on each tire is about half times the gear ratio.

 

[JohnLear]

"But "Torque Steer" is a good illustration - this is where the axle is longer on one side than the other and some of the torque is used to twist the longer shaft."

I've heard this for years, but could never understand why any diference in axle 'wind up' wouldn't be accounted for in the differential, resulting in zero torque steer. Yet torque steer is a real effect.

My understanding is that torque steer is caused by angular differences in the CV joints (i.e. left outer CV vs right outer CV) that result from the articulated drive shafts being of different length as a result of the differential being laterally offset.

Differentials are still commonly offset on FWD cars (Subaru aside), but these days it's usual for the articulated drive shafts to be of equal length due to the common use of a non articulated intermediate shaft on the side that has the greatest distance from differential to hub.

So, with a modern well engineered FWD drive train,(that isn't caused by an outside factor such as road camber, unequal alignment or tyre problem) should be non existant...?

 

[JohnLear]

Sorry, should read:
"So, with a modern well engineered FWD drive train, torque steer (that isn't caused by an outside factor such as road camber, unequal alignment or tyre problem) should be non existant...?

 

[NormPeterson]

I think we at least need to know what sort of differential is involved (open, limited slip, locker, etc.), and perhaps where the question is leading.


Norm

 

[GregLocock]

A conventional differential is a load balancing device, and so in the very simplest set of assumptions the torque exerted on each wheel is identical.

No torque is lost in driveshaft (aka halfshaft) twist in the steady state.


Torque steer is a whole different thing.







Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[CapriRacer]

OK, I'll take another stab at it:

"So, with a modern well engineered FWD drive train, torque steer (that isn't caused by an outside factor such as road camber, unequal alignment or tyre problem) should be non existant...?

But the shaft between the two half shafts is still there, and while strengthening that shaft will reduce the amount of twist, it will not eliminate it. So the best you can hope for is to minimize it.

 

[JohnLear]

So assuming equal torsional stiffness per unit of shaft length, the longer drive shaft (including everything from the hub to the differential) will twist slightly more than the shorter shaft (this additional motion taking a fraction of a second to occur...). Won't the differential planetary gears also move in compensation? (until the load is balanced, which will also occur very quickly, not over a period of time needed to feel a consistant torque steer).

So why is more force passing to one wheel than the other? I still don't think it is (assuming all else to be equal). I can see a momentary 'jerk' occuring at the steering wheel (as force is applied through the drive train) that might be caused by an initial difference in axle twist, but I also see the differential almost immediately 'cancelling out' this affect.

It's a long time since I expereinced consistant torque steer in a FWD car, and it's a long time since I drove a FWD car that did not have an intermediate shaft on the longer drive shaft that allowed all CV joints to run at equal angles.

The last example I can recall of a FWD car I've driven with obvious and significant torque steer would be a Holden Camira (which from menmry had no intermediate shaft), and before that various BMC Minis, Mokes etc.

None of these had an intermediate shaft in the drivetrain to account for the offset differential, and as a result the vertical CV angles were at least somewhat different on each side (or were so with any significant change in front ride height as might occur under hard acceleration or braking).

 

[GregLocock]

Well as an engineer you should be able to work out how long it takes the longer shaft to twist more, and to compare that time with the reaction time of the driver or vehicle. Anyway, if you want to talk about torque steer dig up the old threads and start a new one, this thread is about axles not torque steer.

Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[hashanw007]

is the Axle Torque = Engine Torque multiplied by the inverse of primary reduction ratio, gear ratio and the final drive ratio?

 

[patprimmer]

If you multiplied by the inverse of the gear ratio you would have the lowest torque in low or 1st gear and highest torque in top gear or overdrive.

Regards
Pat
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