How do you get a larger swept area if the pads and calipers are not larger as well ? Might get better cooling due to heat dissipation but that's all . Or maybe not , can you explain please .
I did. The brake rotor has a larger diameter, which means that although the brake pad is the same size, the brake area is larger.
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 of the rotor 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. 30.67 vs. 28.5 doesn't sound like a big difference, but that's a 7.6% increase in braking surface.
But the biggest advantage is still the increase in braking force. The larger diameter of the rotor means that although you have the same force applied by the brake pads, the force acting at the hub is larger. The larger diameter means the force applied by the pads has a larger "lever arm". Just like using a ratchet with a longer handle, the force applied is multiplied directly by the length of the lever arm. Longer ratchet, more power. Larger diameter rotor, more braking power.
For a 11.75" disk, the center of the brake pad is roughly 4.975" away from the center of the rotor. For a 10.98" disk, that distance is 4.59". So, if you take 4.975/4.59, you'll find that the braking force is going to increase by 8.3%, even though nothing but the rotor diameter changed.