"Differential action" here should read "the action of that specific slip-limiting action." In other words, helical and clutch types operate differently, but similar cicumstances trigger their function. Sloppy writing on my part. I apologize for the confusion.

Seriously, man, please stop with the point-by-point quoting and knee-jerk refutation. You have been ignoring the influence of reaction torque at each axle on the behavior of the LSD. The paper I linked discussing the operation of the Torsen design makes this as clear as it can be. Or are you disputing their white paper? The linked paper discussing the Eaton E-LSD makes clear the clutch transfer torque to that axle in addition to the torque transmitted by the differential gears. Or do you dispute that paper?

The position that the clutch stacks on both axles are always bound with the same force might be valid for a particular design, but it isn't valid universally. IOW, you can be correct for certain designs, I'm not saying you aren't. But you need to refer to those designs specifically, rather than insist on being correct in all cases and any other position is invalid.

The Torsen paper makes clear that the design is capable of delivering different amounts of torque to each axle based on differences in reaction torque.

I do want your input on the Torsen paper: http://torsen.com/files/Traction_Control_Article.pdf

When discussing its behavior during a turn on page 5, (numbered 79), it states:
That statement clearly states that the outside wheel, the faster rotating wheel, sees more torque than the slower wheel. It is ambiguous given that when turning, it is not a guarantee that the inside wheel has a greater resistance to rotation.

The paper then states that the Torsen design transfers a greater proportion of torque to the inside wheel due to its resistance to transfers of torque between drive wheels in proportion to the amount of torque applied to the carrier. A greater proportion than what? A greater proportion of torque than is applied to the outside wheel? That can't be, because they just said that the slower wheel transfers torque to the outside wheel. So what is the alternative situation? An open diff? Some other LSD design?

It claims that the Torsen's "traction management abilities" stem from the fact that "since torque is already distributed in increased proportion to the inside drive wheel, it is exceedingly unlikely the outside wheel will 'spin up'. Alternatively, should the torque of the inside drive wheel exceed available traction, it is equally unlikely to 'spin up' since such a 'spin up' would require a difference in traction between drive wheels which exceeds the bias ratio."

Also, "in all directions of travel, the Torsen differential will resist 'spin up' of either drive wheel by instantly dividing torque between the drive axles in proportions up to the bias ratio to match prevailing traction conditions."

Assuming a bias ratio of 5:1, this would imply that even when cornering, the Torsen could transfer to the outside wheel five times the torque being transferred to the inside wheel. It does also say that when the inside wheel slips, torque is divided more evenly based on the maximum amount of torque that can be applied by the inside wheel. So in a case where the inside wheel slips, the outside wheel could only see the amount of torque being transferred to the inside wheel. Which the paper states earlier can only be the maximum amount of reactive torque, i.e. the amount of available traction. Now when the inside wheel slips, it will only rotate at the same speed as the outside wheel, the degree of slip is minimal, and it can still transfer torque. The amount of torque transferred to the outside wheel will drop to the amount supported by the inside wheel.

For me the ambiguity stems from the claim that the faster wheel will receive torque transferred from the slower wheel up to the bias ratio, yet the further discussion states that the transfer of torque is increasingly resisted as input torque at the carrier increases, implying that the more torque applied to the carrier, the smaller the torque split until the point both wheels see the same amount of torque being transferred, i.e. the point when the inside wheel begins to slip. It would appear that a Torsen transfers more torque to the outside wheel as long as the inside wheel has traction. Is that how you read it?

 

Both of you need to try going outside instead of rambling about differentials on a forum.

 

Why? I completely disagree with Nunco as it relates to this thread. However, if I were calling him names or if he was telling me less than flattering things about my lineage I would agree with you. The fact is we are having a reasonably civil discussion about a car related topic. I see no reason why we need to go "outside". You are of course welcome to contribute to the topic.

Nunco, I don't have time to reply now but I will try to get back to your post later tonight or this weekend.

 

So is that the thing that causes the reaction torque in the diff or do you mean the relative motion of the axles? The two aren't always the same. I'm still not sure which you mean.

Point by point is a good way to dive into the details. I'm often frustrated with your responses because you don't actually refute details. Instead you post generalities which I have already addressed in detail. However, rather than explaining why the details are wrong you simply repost the general statements which I was refuting.

The Torsen paper and what I have said do not conflict. I have never ignored the reaction torque on the axles. The sum of the reaction torques equals the torque applied to the diff by the motor. The torque bias ratio is the maximum ratio of torque between the left and right wheel. What else is there to say? Nothing I have ever posted has disagreed with that. I have also covered all those points already. If you disagree then it is up to you to say why what I just said was wrong. If you repeat that I am wrong without providing details I can only assume you don't know what is going on but your knee jerk reaction is to assume I'm wrong. If you are in a position to say I'm wrong then you must be able to say why just as I say why I feel your statements are wrong. This is also why I have challenged you to put your views in math terms.

I have always said the total torque to an axle is that which is transferred via the gears +/- that which comes from the clutch/friction system. The papers agree with that and their equations match my equations. What makes you think they don't? Again this is a time when you NEED to offer a detailed answer. Yet another claim that my facts conflict with something is meaningless if you don't provide the details that back your views.

You are right, it's not universal. The diffs used in the later Miatas to the best of my knowledge do not use ramps to clamp the clutches. The thing the paper showed is that even if the clutch is loaded by just one set of the gears the forces it creates are used by both sets.

The difference is ONLY true in the case where both wheels spin at the same speed. I have already shown how it isn't the case when one wheel spins faster. Yet again, you either need to relinquish that point or in mathematical terms explain why I am wrong. This is a point where the papers all agree with me.

I will read it. Is there any part you think conflicts with my previous statements.
BTW, you should check out section 3.2. They include some math equations describing torque transfer while the wheels are moving at different speeds. Funny, they are the same as my equations! It appears that the guys who make the Torsen ALSO agree with my claim that the slower wheel gets the greater torque!

Ummm yeah. When you turn why would the inside wheel go slower? Well the ground under that wheel moves slower. Why is that? Well the reactions of the chassis spinning make it harder to spin the inside tire faster unless it were to slip. The inside wheel spins slower because it would take more force to make it spin faster than the ground would allow it to spin!

Not at all. You need to read the next paragraph!
So even the Torsen paper agrees with me, the inside wheel gets more torque!

Yes! You did read the paragraph! But you didn't understand it. The paragraph says the inside wheel gets more torque BECAUSE it's the slower wheel! So do you know what that all means? Can you tell us how this is different that what I have said? None of it conflicts with what I have said.

Yes... but only when the inside wheel spins up with respect to the outside wheel.

Yes, because when the inside wheel slips it speeds up and we are closer to equal speeds. However, so long as it is the slower wheel the inside wheel gets more torque. IF you read the equations on page 78, right you will see that they agree with me, T_1=(Trg+Td)/2 while the other wheel is T_2=(Trg-Td)/2. So if Td is adding to one it's taking away from the other.
I'm not sure you are understanding what the paper is saying but in different words it is again saying the slower wheel gets the torque and the extra torque is Td and Td is proportional to Trg. That is the same thing I have been saying all along.

That's not how I would read it. They are using rather confusing language to say that Td is proportional to Trg. So when one wheel slips while the other maintains grip the total torque on the two outputs drops. Since Trg=Tleft+Tright, Trg drops. When Trg drops, Td has to decrease because it's proportional to Trg. So when Td drops the wheels are more able to move freely.

Again, what they have said is rather a different phrasing of what I have said but it still agree. The key thing is to look at their equations. Note that their fundimental equations which I have quoted are the same as the equations in the Eaton papers and are equivalent to mine. If that all is the same then the behavior of the diffs must be the same to the extent that we understand what drives Td. In a clutch diff it's the same as in a Torsen (largely the same but some detail differences exist). In the case of a viscous diff Td=f(relative wheel velocities). In the eLSD Td=f(computer code). In the end we get Td in different ways but once we have it the other equations hold true.