turbolarry
06-19-2009, 09:42 AM
Just kinda curious if anyone has tried a six puck sprung disk with a stock pressure plate?
I know the clamp load for the stocker is weak at approx. 2000 lbs, but matching it with a more aggressive disk should increase it's torque rating, right?
I know some guys daily drive with a six puck and claim no problems, and even the four puck Exide.
It's simple math according to ACT;
T = N x F x P x R
T = torque capacity
N = number of disk surfaces (ususally 2)
F = coefficient of friction (.25 for organic/.32 for race disk)
P = pressure /lbs. of clamp force
R = radius of gyration in feet (What is this?)
Working the formula backwards from known ACT clutch torque capacities I'm getting R = .36028 (approx.), but I really don't know what this means.
If I go ahead and try and calculate a torque capacity for a stock pressure plate with an ACT six puck I'm using these values;
N = 2 (but does this still apply to pucked disks?)
F= .30 (somewhere between organic and race, right?)
P = 2000 (stock pressure in lbs.)
R = .36028 (again derived from known clutch torque capacities)
2 x .30 x 2000 x .36028 = 432.3
Am I doing this right?Â* Would this combo really be a 430 ft. lbs. of torque clutch?Â* I know I have some approximations in there, but I used the lesser of the values.
Light pedal feel, but with harsh engagement?
I don't even need a clutch, I'm just thinking.
I know the clamp load for the stocker is weak at approx. 2000 lbs, but matching it with a more aggressive disk should increase it's torque rating, right?
I know some guys daily drive with a six puck and claim no problems, and even the four puck Exide.
It's simple math according to ACT;
T = N x F x P x R
T = torque capacity
N = number of disk surfaces (ususally 2)
F = coefficient of friction (.25 for organic/.32 for race disk)
P = pressure /lbs. of clamp force
R = radius of gyration in feet (What is this?)
Working the formula backwards from known ACT clutch torque capacities I'm getting R = .36028 (approx.), but I really don't know what this means.
If I go ahead and try and calculate a torque capacity for a stock pressure plate with an ACT six puck I'm using these values;
N = 2 (but does this still apply to pucked disks?)
F= .30 (somewhere between organic and race, right?)
P = 2000 (stock pressure in lbs.)
R = .36028 (again derived from known clutch torque capacities)
2 x .30 x 2000 x .36028 = 432.3
Am I doing this right?Â* Would this combo really be a 430 ft. lbs. of torque clutch?Â* I know I have some approximations in there, but I used the lesser of the values.
Light pedal feel, but with harsh engagement?
I don't even need a clutch, I'm just thinking.