oh boy, classical physics sure helps figure things out!!
Found some interesting friction material (prolly nothing new to the clutch people)
http://www.tribco.com/drycluch.html
so, I want a single disc ETH clutch to engage smoothly, have little to no maintenance, and hold 1500 foot pounds at the flywheel.
Torque = force x moment arm
Force = Normal force x u
u = either static or dynamic coefficients of friction... . using the dynamic (kinetic) coefficient would be a good suggestion.
Torque = Normal force x u x moment arm
The average radius of a 2001 ETH clutch is 0. 514 feet
I want 1500 foot pounds
1500 = Normal force x u x . 514 feet
2919 LBF = Normal force x u(k)
u(k) = 0. 36
8108 lbf = Normal force... now we can divide this by "2" cuz there are two sides to the friction disc.
4054 lbf... .
or
if I used the Static coefficient of friction u(s) = 1. 05 the final force = 1390 lbf
Hopefully someone else will assist in how this translated to pressure plate load.
Found some interesting friction material (prolly nothing new to the clutch people)
http://www.tribco.com/drycluch.html
so, I want a single disc ETH clutch to engage smoothly, have little to no maintenance, and hold 1500 foot pounds at the flywheel.
Torque = force x moment arm
Force = Normal force x u
u = either static or dynamic coefficients of friction... . using the dynamic (kinetic) coefficient would be a good suggestion.
Torque = Normal force x u x moment arm
The average radius of a 2001 ETH clutch is 0. 514 feet
I want 1500 foot pounds
1500 = Normal force x u x . 514 feet
2919 LBF = Normal force x u(k)
u(k) = 0. 36
8108 lbf = Normal force... now we can divide this by "2" cuz there are two sides to the friction disc.
4054 lbf... .
or
if I used the Static coefficient of friction u(s) = 1. 05 the final force = 1390 lbf
Hopefully someone else will assist in how this translated to pressure plate load.
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