Clutchpack diffs - why more clutches?

I'm in the process of building a diff to handle my car's power and I want to rebuild the LSD core. There are options where I can take out material from the carrier and add more clutches. The supplier claims that you add lockup, from 25% (2), 40% (3) to 60%(4). I can't really understand why this is so and why it's desirable? The latter of the 3 options is especially painful as it requires machining both the cap of the carrier and the other end inside the barrel.

I mean, I start off with 2 discs, they take a certain amount of force to break loose. How does adding another set of clutch/dog prevent the original discs from breaking loose? Each individual disc still has the same surface area, same friction coefficient... so if the same torque as before is applied, it should still break loose just the same, no? If you add more preload sure, but you don't need to add clutches to get that.

The only benefit I can logically understand is that it would prolong the life of the clutches, as doubling the number of clutches, probably doubles the lifespan of the clutches. I can also understand changing the ramp angle. But why more clutches?

If you keep stacking discs in there, the pressure per surface area remains similar (since they're stacked, and not spreading across a larger area: If I stand on a peanut butter and jelly sandwich, the pressure is "one ransom/peanut butter and jelly sandwich". If I stack two sandwiches, I'm applying my full pressure to the top sandwich, and it passes that full pressure to the bottom sandwich; the result is that I have two sandwiches, each being under a pressure of "one ransom/peanut butter and jelly sandwich").

Another exceedingly silly physical example: Give David and Tim each a string tied to a tile of linoleum. Stack Tim's on top of David's, and put a bucket of paint on top to weight them down, and then have Tim and David pull opposite directions. Not much effort, probably.

Now give them each fifteen pieces of linoleum, each with a string. Stack them alternating Tim/David/Tim/David etc... Then put the same bucket of paint on top. Ignoring the (admittedly significant) weight of the linoleum itself, it'll be much harder for David and Tim to pull the stack apart, won't it?

The bottom line isn't that the new discs prevent the old discs from breaking loose, it's that the force trying to move the discs in opposite directions are applied simultaneously across all the discs. All the outer-toothed discs are going one way, and all the inner-toothed discs the other. Because they're stacked, every interface has the same load per surface area, but the surface area keeps getting larger.

Each friction:steel contact has a certain amount of static friction associated with it, they add up. Like breaking two sticks in half rather than one, and then dragging both through the mud(dynamic friction), rather than one. Friction is the force times the friction coefficient, and the force stays constant here if spring preload is kept at the same level.

ransom: So how does adding more clutches help? If it's the same load per surface area, then any single clutch still breaks loose at the same torque as before. Why does it matter what the other clutches do if one is slipping?

Regarding the Tim and David example, it will be easier for them to do it with the second example if you ignore the weight of the sheets, since the paint bucket creates more normal force per surface area and hence more friction. (EDIT: sorry confused myself with the same problem... if you ignore the weight of the sheets then the pressure per area is the same)

Kenny_McCormic wrote: Each friction:steel contact has a certain amount of static friction associated with it, they add up. Like breaking two sticks in half rather than one, and then dragging both through the mud(dynamic friction), rather than one. Friction is the force times the friction coefficient, and the force stays constant here if spring preload is kept at the same level.

That's just it though, it's not like breaking two sticks in half, it's like two identical people breaking two identical sticks. The additional clutch doesn't make the first clutch any stronger. The fact that there is another steel:friction pair doesn't affect the static friction of any single clutch, so it's liable to overcome said static friction at the same torque as before...

Sorry if I'm being dense here... but if you isolate any given clutch, I don't see how the additional clutches change that clutch's behavior in any way and since they are in series, if one breaks loose, it doesn't matter how many you have in total, you're slipping.

You are being dense here. Unless you happen to understand 3d vector mechanics up to moments, you're probably going to need an engineer actually in front of you to explain this.

You aren't isolating any clutch, it works as a system, the force is distributed.

I think the problem is that you're thinking of breaking it down to one clutch, when you have to consider all of the clutches.

Here: http://www.eng.auburn.edu/~marghitu/MECH2110/C_6.pdf Page 15, equation 6.23

Kenny_McCormic wrote: You are being dense here. Unless you happen to understand 3d vector mechanics up to moments, you're probably going to need an engineer actually in front of you to explain this. You aren't isolating any clutch, it works as a system, the force is distributed.

Why can't you isolate a clutch? Until the clutches begin to slip, the force on any one of them is identical. And if the static friction is also unchanged, then the force at which they overcome the static friction is also unchanged. And since static friction is greater than dynamic, once one clutch begins to slip, the others aren't likely to and just become part of the fixed assembly of the carrier...

If you can explain it, please do. If I don't get it, I can ask for clarification, here or elsewhere. You don't have to get into low level details. Anything can be explained at a high enough level to get the idea across.

BigD wrote: ransom: So how does adding more clutches help? If it's the same load per surface area, then any single clutch still breaks loose at the same torque as before. Why does it matter what the other clutches do if one is slipping?

If it takes one pound to move a particular interface, and I have ten interfaces all working together, and it is only possible to move them simultaneously (which is how this works), it will take ten pounds to move the "interface pack".

You are aware that the alternating discs are tied to the input and output by the teeth on the inside or outside edges, respectively, and that the force is not being transmitted through the stack of clutches?

BigD wrote: Regarding the Tim and David example, it will be easier for them to do it with the second example if you ignore the weight of the sheets, since the paint bucket creates more normal force per surface area and hence more friction.

You've got this half right-ish (second half), but the first part I think you've either got wrong or you've phrased it in a fashion that isn't making sense to me.

The second example is the 30 total sheets. The bucket is the same, so the normal force is the same. The area is 29 times as great (the original two sheets had a one-square interface area; if we have 30 sheets, there are 29 interfaces). Why would you say it would be easier for them to pull it apart when you also state that there is more friction? Incidentally, the normal force per surface area remains the same, but the interface surface area is vastly (29 times) greater.

ransom wrote: You are aware that the alternating discs are tied to the input and output by the teeth on the inside or outside edges, respectively, and that the force is not being transmitted *through* the stack of clutches?

And the last horse crosses the finish line! I get it now. I was only thinking about it as a friction stack and the force being transferred purely through shear. The shaft teeth was the missing link in my thinking.

Thanks guys!

Excellent!

The clutches act in parallel, not in series- the fact that they are all next to eachother is most likely what is throwing you off.

EDIT- You got it already, I was too slow!

NONACK wrote: The clutches act in parallel, not in series- the fact that they are all next to eachother is most likely what is throwing you off. EDIT- You got it already, I was too slow!

Hehe, you weren't the slow one. But yeah exactly, I was only thinking of the clutches as a friction stack between the spider gear barrel and the carrier body but of course that's not how it works.

BigD wrote:

ransom wrote: You are aware that the alternating discs are tied to the input and output by the teeth on the inside or outside edges, respectively, and that the force is not being transmitted *through* the stack of clutches?

And the last horse crosses the finish line! I get it now. I was only thinking about it as a friction stack and the force being transferred purely through shear. The shaft teeth was the missing link in my thinking. Thanks guys!

Ah, too small of a frame of reference.