Banshee-I can tell you do not have an engineering background. Quick lesson in fluid dynamics. Two givens-one tc (A) has 3 times the surface area of clutch material than another (B), but both use same clutch material. If tc A has 60 lbs of applied fluid pressure and tc B has 90 lbs they should handle (all other things in trans being equal) the same amount of torque before slipping. You cannot use assumtions about the ability of a tc to hold a certain # of foot lbs of torque unless you can plug in alot of other numbers. As reported by some trans builders FACTORY transmissions vary in fluid pressure by as much as 25%. A poorly performing transmission with low line pressures with a 3 clutch tc will not work as well as one with high line pressures and a single clutch tc. Valve bodies alone will not help with low line pressures.
Bill...
It is clear from your post that you have NO idea what you are talking about. I have no engineering background you say? Funny... I have a piece of paper from NC State University that says I have a degree in Mechanical Engineering, the State of North Carolina agrees that I'm a licensed ME, and the Dept of the Navy seems to think the same thing. FWIW, I work for what is regarded as the TOP engineering office in the Navy, and I'm sure that in the last year alone I've done more design, analysis, and engineering investigation than you EVER WILL, so put that in your pipe and smoke it. As far as the transmission goes, sit back and let me teach you a little about a clutch and multidisk clutch.
For a clutch, there are three and only three things that determine a clutch's torque holding capability: apply pressure, friction material (coefficient of friction), and disk geometry. The number of disks directly multiply the torque capacity of a single disk unit when the disks are stacked.
Starting with apply pressure or clamping force, this of course comes from fluid pressure behind a piston in our TCs. The truth is, it really doesn't matter at all where they clamping force comes from... it could be spring applied, piston applied, lever applied, or you could push it together with your hands... it really doesn't matter at all. The amount of force that is applied however is directly related to a clutch's torque holding capability, and I think everyone on this board seems to understand that fact quite well.
Moving along to coefficient of friction, this also directly applies to a clutch's torque holding ability. Obviously a "stickier" material will grab better before it slips. It sounds like a hell of a concept, and low and behold it's the truth. It's the marriage of these two ideas (clamping force and Coefficient of friction) that lead to the common formula that everyone loves to quote:
Friction force = Mu * Normal Force where
Mu (the greek letter) = coefficient of friction and
Normal Force = the clamping force
This is a great little formula for a simple block sliding across the table, but it is lost when looking a clutch. The third part of the equation, disk geometry, is the other major player in clutch disk performance. Disk diamter plays the major role in determining just how much torque the disk will hold before it slips. Let's imagine a ring 1" think in two configurations - one ring has a OD of 5" and the other has an OD of 10". The lever arm for the 10" disk is twice as long, so it is seeing less torque on the force of it's disk face and thus holds more shaft torque before slipping. It's the same idea as going to a larger gear in a transmission. The formula that describes clutch disk geometry is the following:
Area = Pi/4 (Do-Di) where:
Pi = 3. 1415
Do = Outer disk Diameter
Di = Inner disk Diamter
To apply all of this to a disk you first start with the applied claping force from the piston:
F=((Pi*Pressure)/4)*(D^2-d^2)
This formula applies to the piston in the TC and shows how much apply force is generated by this piston at a given apply pressure.
The torque holding capability for the disk is found by integrating the product of the frictional force and the radius of the clutch disk. After the integral, this formula comes out to be the following:
Torque = (F*f/3)(D^3-d^3/D^2-d^2)
Where:
F= clamping force from the above equation
f= coefficient of friction of the clutch's material
D= outer disk diamter
d= inner disk diamter
It should be noted again that this equation describes a single disk only, but the value for a multidisk setup is multiplied by the number of pairs of surfaces in contact. I would like to note that these formulas come from the book Mechanical Engineering Design by Shigley and Mischke on page 645 & 646 in the Clutches, Brakes, Couplings, and Flywheels section. If you'll notice, Bill, it's not in the Fluid dynamics section... that has NOTHING to do with a clutch's torque holding capability. Furthermore, as you can see from above, the numbers in your post do not even come close to making any sense whatsoever. How can a clutch with just 50% more apply pressure but 200% less clutch area hold more?? Is this some voodoo magic you have?? Finally, I'd like you to prove this statement with a little math:
A poorly performing transmission with low line pressures with a 3 clutch tc will not work as well as one with high line pressures and a single clutch tc.
Please apply it to the TC only... I don't think you can.
Chris,
Again, I'm saying that this TC does not have to be on a truck to show that it can hold the claimed torque value. Furthermore, math and theory go a long way as long as the example is represented clearly. Clutch design and analysis has been around a long time, and it's pretty much down to a science on determining how much a setup can hold. All I can say is make it slip, and then we'll talk about it.
John