These are the rotors I bought
http://cgi.ebay.com/ebaymotors/ws/eBayISAPI.dll?ViewItem&item=220328885601&sspagename=STRK%3AMEWNX%3AIT&viewitem=#ht_7800wt_1448Just drilled and slotted Cordoba rotors. A little fancier than stock, but nothing all that special. I bought my caliper brackets for $85 from a guy on A-bodies.
Actually, the braking area for the 11.75" rotors is bigger...
If you figure the outside of the rotor is 11.75" and the width of the braking surface is 1.8" ( I measured it),
then the total braking area is π (11.75/2)^2 - π(9.95/2)^2 = 30.67 square inches for the 11.75" rotors
For the 10.98 rotors the braking area is π(10.98/2)^2 - π(9.18/2)^2 = 28.5 square inches
The pad is the same size, but because the outer diameter is larger, so is the circumference of the rotor. So at any given moment the pad area in contact with the rotor is the same, but the pad "sweeps out" a larger area on the rotor because it travels a longer distance in doing so. This doesn't directly improve braking force, but because the pad is acting over a larger area the brakes are dispersing the energy of braking over a larger area. Also, there is more surface area to disperse the heat energy once the braking is done.
Long story short, the larger rotors will heat up slower than the smaller ones. They would also cool down faster if they had the same mass as the smaller rotor. I'm guessing that even with the drilling and slotting they still weigh more, so its a toss up if the greater surface area wins out over the greater mass in cooling the brakes. And I really don't feel like pulling out my thermodynamics book to figure it out
, or pulling my 11.75" disks to weigh them.
As HP2 mentioned, the braking force itself is only greater because of the greater leverage, but thats a direct multiplier in the force equation, so a little leverage distance makes a big difference. I can feel the braking difference with the new rotors.