Ok, since I've been called out for lack of "Data", here's all the calculations for braking force. This uses a 15/16" master cylinder for both system, the stock pedal ratio of 6.5, a modest foot input of 70 lbs, the same brake pad coefficient and the same wheel diameter. The effective diameter for the rotor is slightly different because of the 10.95 vs 11" rotors, but I did include it. You can clearly see which system generates more clamping force, and which has more total force at the wheels.
View attachment 1715195810
Now, this calculator assumes 100% of the force generated at the pistons is transferred to the rotor. That never happens, and as I have said previously the Wilwood multi-piston, fixed caliper will be more efficient- it will not lose as much force to flex in the caliper, pads, etc as the Mopar single piston sliding design.
The question is, how much less efficient is the Mopar caliper? Do you think it loses 744 pounds of clamping force in flex? I don't. Now, that last part is my opinion, and you'd need a brake dyno to figure out how much force is actually being applied to the rotor by each caliper. But quite frankly, there's no way it's losing that much force to flex.
Here's the source of my calculator, all I did was input the Wilwood and Mopar numbers for the calculations. The spreadsheet is set up to calculate front and rear, I changed the header titles to match the brake systems I was comparing, and added the "bold" to the clamping force calculation results.
http://norcal-cobras.com/misc/brake-formula-mod.xls
For those of you curious about the equations used in the spreadsheet, there's a really awesome article by StopTech that includes all of the equations, as well as a pretty good amount of brake theory. It should be attached at the bottom.
So, there are the numbers. Make your choices, spend your money.
