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The Physics of Multi-Plate Clutches

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Old May 15, 2009 | 10:58 PM
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The Physics of Multi-Plate Clutches

Yeah, I'm a geek.

1) Most manufacturers appear rate their multiplate clutches to have significantly more torque-bearing capacity than their single plate clutches.

2) People who have installed said multiplate clutches have had positive experiences with respect to holding power.

If the clamping force between a single plate and a multiplate clutch is the same, how can the frictional force be different?

In other words, frictional force = normal force x coefficient of friction. This is true of static or dynamic situations (the coefficient may be different between static & dynamic situations) and is an accepted formula in physics; hopefully we can safely avoid any debate on the correctness of the formula.

What's the point? Nowhere within this formula is the surface area of the two (or more) surfaces mentioned.

Given this background, several questions arise:

a) Why are multi-plate clutches capable of holding more torque than single plate clutches? Assuming all else (clamping load, friction material, etc.) the same.

b) Why do wider tires with a larger contact patch provide for more traction? Assuming all else (vehicle weight, surface conditions, etc.) the same.

http://en.wikipedia.org/wiki/Frictional_force
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Old May 15, 2009 | 11:04 PM
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This is you sir! Not in a bad way at all. It's people like you who help me understand the true physics of the automotive industry and thus gaining more knowledge to help pass to others.
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Old May 15, 2009 | 11:49 PM
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Originally Posted by BoostLover99
Yeah, I'm a geek.


a) Why are multi-plate clutches capable of holding more torque than single plate clutches? Assuming all else (clamping load, friction material, etc.) the same.
More surface area.
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Old May 16, 2009 | 12:38 AM
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Originally Posted by crispeed
More surface area.
this.




Last edited by dxbtune; May 16, 2009 at 12:43 AM.
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Old May 16, 2009 | 12:48 AM
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Originally Posted by crispeed
More surface area.
Way more surface area.
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Old May 16, 2009 | 01:09 AM
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that equation is for one clutch disk. each cluch disk will have its own normal force to it and also its own coeffient of friction.
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Old May 16, 2009 | 01:23 AM
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One pressure plate will exert X amount of force and the torque capacity of the clutch is proportional to the combined surface area and the force exerted by the pressure plate.

The same X force applied on 2 plates will be the same however the surface area will change allowing for a larger torque bearing capacity.

Thing is, if you increase the surface area in order to have more strength, you also need to increase the force of the pressure plate otherwise you will slip no matter how much surface area you have....

power,surface area,force of pressure plate are proportional in their own respects

Thats the way I see it I could be off my rocker though... anyone else concur?

Last edited by dxbtune; May 16, 2009 at 01:28 AM.
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Old May 16, 2009 | 01:57 AM
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Sorry to be dense, but where is surface area in the frictional force equation?

F = mu x N

mu = frictional coefficient between the surfaces

N = normal force applied to the surfaces

Where does surface area enter into this? Intuitively, yes, more surface area => more friction. But I'm asking about the physics.
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Old May 16, 2009 | 03:42 AM
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From: I'm Canadian living and working in Dubai/Kuwait/Bahrain
In my point of view it can be broken down in simple terms and I could be completely off

Without taking wear, tear and stress into this:

Say you have a 2 puck clutch and you ask for it to withstand 500 NM, in order for you to be able to do so, you will require the pressure plate to exert 1500N of force ' large amount of force' to be able to hold up the required torque.

Now let's say you have a six puck clutch ( lets say all pucks are the same size), you wont need to have the same amount of force exerted because you have a larger surface area applied say 500N in a perfect world as an example
with the coefficient of friction you can devise the exact pressure or force the pressure plate needs to exert according to the surface area covered thus allowing the pressure plate to be of a lighter material as it does not have to be so robust.

I dunno man just throwing ideas around...

Last edited by dxbtune; May 16, 2009 at 03:45 AM.
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Old May 16, 2009 | 08:54 AM
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The thing you aren't taking into account is that the Coefficient of Friction (COF) is not a static number, it is an empirical number discovered for each and every instance of friction. The reason for this is that friction is not a MATERIAL property, but a SYSTEM property. Each system must be looked at differently.

For example, stick a piece of tape to a plate of steel. Attempt to pull the tape across the surface. This COF is pretty high (above 1). Now take your blow torch and heat the back of the steel plate. Pull the same piece of tape across the surface with the same amount of force. It will move much easier. The only thing we have changed from the 2 instances is the COF.

The differences between multiple-plate clutches might be that the materials/compounds used work better in that instance or a multitude of other factors that the engineers took into account.
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Old May 16, 2009 | 10:09 AM
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Originally Posted by M4Lki3r
The thing you aren't taking into account is that the Coefficient of Friction (COF) is not a static number, it is an empirical number discovered for each and every instance of friction. The reason for this is that friction is not a MATERIAL property, but a SYSTEM property. Each system must be looked at differently.

For example, stick a piece of tape to a plate of steel. Attempt to pull the tape across the surface. This COF is pretty high (above 1). Now take your blow torch and heat the back of the steel plate. Pull the same piece of tape across the surface with the same amount of force. It will move much easier. The only thing we have changed from the 2 instances is the COF.

The differences between multiple-plate clutches might be that the materials/compounds used work better in that instance or a multitude of other factors that the engineers took into account.
Ahh, now we're getting somewhere!

Good point RE the COF being empirical & a system property. So suffice to say that the COF of a multiplate system *must* be significantly higher than that of a single plate system, then?

This makes perfect sense that COF would encompass the surface area of the *SYSTEM*.

Thank you!
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Old May 16, 2009 | 12:37 PM
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From: The Last Sorta Free State in the US
Originally Posted by M4Lki3r
The thing you aren't taking into account is that the Coefficient of Friction (COF) is not a static number, it is an empirical number discovered for each and every instance of friction. The reason for this is that friction is not a MATERIAL property, but a SYSTEM property. Each system must be looked at differently.

For example, stick a piece of tape to a plate of steel. Attempt to pull the tape across the surface. This COF is pretty high (above 1). Now take your blow torch and heat the back of the steel plate. Pull the same piece of tape across the surface with the same amount of force. It will move much easier. The only thing we have changed from the 2 instances is the COF.

The differences between multiple-plate clutches might be that the materials/compounds used work better in that instance or a multitude of other factors that the engineers took into account.
I thought about this a bit more and unfortunately, it still doesn't make sense.

Check out the COFs for various surfaces. http://en.wikipedia.org/wiki/Frictio...ts_of_friction

Notice that COF (mu) is expressed as a dimensionless CONSTANT. It is *not* expressed as a function of surface area. For example, mu for dry & clean cast iron against zinc is 0.85, regardless of surface area. Mu for dry & clean cast iron against zinc is NOT 0.85 x surface area.

So now we're back to the same question - why do multi-plate clutches hold more torque?
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Old May 16, 2009 | 01:36 PM
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ok generally, a twin-plate clutch has roughly twice the friction threshold as a single plate ceteris paribus. so in essence, the equation

Force = COF x Normal force

would be doubled in value due to the second clutch and would be tripled if the clutch was a triple plate...

to answer your question about it being a dimensionless constant, yes that's true but the force value as a whole is doubled because there is twice the amount a surface area. the coefficient of friction is a constant number but it is the total value that is doubled. there will be twice the normal force and twice the kinetic force.

look at it this way...put a brick down on a slight incline and say it begins to slide down at an extremely slow rate. holding the weight constant, double the surface area of the brick and it'll stop moving correct? more surface area means more frictional threshold.

it's almost the same as surface tension on water. there are spiders that literally stand in one place on a body of water because they increase their surface area...go jump in a pool and ball up. what happens? you sink...spread yourself out and you can float on the top of the water. it's almost the general concept as friction but it's kind of a different topic all together at the same time

hopefully that clears things up a bit
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Old May 16, 2009 | 01:59 PM
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Ok - I got it. The block analogy helped.

Two blocks sliding down the same slope have twice the frictional force as one block sliding down the same slope.

Thanks!
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Old May 16, 2009 | 02:16 PM
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From: SLC
Originally Posted by BoostLover99
Ok - I got it. The block analogy helped.

Two blocks sliding down the same slope have twice the frictional force as one block sliding down the same slope.

Thanks!
well yes and no...two blocks will have twice the net frictional force. the force of each block is the same regardless but the net (or total) frictional force is doubled and that's what counts

and the same is for the tire question

no problem...i always enjoy intellectual discussions like this...makes for a better community

EDIT: in a more technical sense, there is a more complicated equation people use to determine torque capacity and it is as the following...

Torque capacity = N (number of friction surfaces) x F (clamping force of the pressure plate) x COF x Rg (radial gyration)

that's specifically geared toward centripetal dynamics....F = COF x N still works but it's way more simplistic

Last edited by Babs; May 16, 2009 at 02:22 PM.
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