AEM 2pc rotors

AZmike

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not sure if i agree with this.. the braking surface is still the same as tht is determined by the caliper pistons.. how can brake torque increase with just the relocation of the caliper?

It's not the area of the pad, it's the forces and where they are applied that makes the difference.

The master cylinder/caliper combination squeezes the pads against the rotor with a force roughly proportional to the force on the brake pedal. I'll use some numbers to illustrate:
Assume 1" master cylinder bore, 2" caliper bore, 5:1 pedal to master cylinder stroke ratio. I'll assumes these aren't power brakes to simply things a little.
Let's say the driver presses on the brake pedal with 50 lbf. The lever arm of the pedal produces 250 lbf at the master cylinder. The difference in bore area produces a 1000 lbf clamp force at the caliper (ratio of areas is 2^2=4). None of these figures depend on the diameter of the rotor.

The job of the brakes is the resist the rotation of the wheel--to produce a torque. Just like the extra length of a breaker bar helps to free stuck fasteners, moving the clamping force of the caliper outward increases the brake torque. To continue the example, assume a pad to rotor friction coefficient of 0.4 and two rotor sizes: one with the pad centered at 5" from the wheel axis and another with the pad centered at 6" from the wheel axis.

Regardless of rotor size, the caliper is squeezing each side of the rotor with 1000 lbf, producing a 800 lbf friction force on the turning rotor from the pads (1000*0.4=400 lbf from each pad). The smaller rotor with the 5" effective radius makes a brake torque of 800*5=3000 in-lbf (333 ft-lbf). The larger rotor has a brake torque of 800*6=2400 in-lbf (400 ft-lbf); same pedal force, same caliper, more brake torque.
 

Darianjr1

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It's not the area of the pad, it's the forces and where they are applied that makes the difference.

The master cylinder/caliper combination squeezes the pads against the rotor with a force roughly proportional to the force on the brake pedal. I'll use some numbers to illustrate:
Assume 1" master cylinder bore, 2" caliper bore, 5:1 pedal to master cylinder stroke ratio. I'll assumes these aren't power brakes to simply things a little.
Let's say the driver presses on the brake pedal with 50 lbf. The lever arm of the pedal produces 250 lbf at the master cylinder. The difference in bore area produces a 1000 lbf clamp force at the caliper (ratio of areas is 2^2=4). None of these figures depend on the diameter of the rotor.

The job of the brakes is the resist the rotation of the wheel--to produce a torque. Just like the extra length of a breaker bar helps to free stuck fasteners, moving the clamping force of the caliper outward increases the brake torque. To continue the example, assume a pad to rotor friction coefficient of 0.4 and two rotor sizes: one with the pad centered at 5" from the wheel axis and another with the pad centered at 6" from the wheel axis.

Regardless of rotor size, the caliper is squeezing each side of the rotor with 1000 lbf, producing a 800 lbf friction force on the turning rotor from the pads (1000*0.4=400 lbf from each pad). The smaller rotor with the 5" effective radius makes a brake torque of 800*5=3000 in-lbf (333 ft-lbf). The larger rotor has a brake torque of 800*6=2400 in-lbf (400 ft-lbf); same pedal force, same caliper, more brake torque.

You definitely got some rep from me. That is some great info that needs to be stickied and posted somewhere for everybody to see. :thumbsup:
 

DarkSideAccord

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It's not the area of the pad, it's the forces and where they are applied that makes the difference.

The master cylinder/caliper combination squeezes the pads against the rotor with a force roughly proportional to the force on the brake pedal. I'll use some numbers to illustrate:
Assume 1" master cylinder bore, 2" caliper bore, 5:1 pedal to master cylinder stroke ratio. I'll assumes these aren't power brakes to simply things a little.
Let's say the driver presses on the brake pedal with 50 lbf. The lever arm of the pedal produces 250 lbf at the master cylinder. The difference in bore area produces a 1000 lbf clamp force at the caliper (ratio of areas is 2^2=4). None of these figures depend on the diameter of the rotor.

The job of the brakes is the resist the rotation of the wheel--to produce a torque. Just like the extra length of a breaker bar helps to free stuck fasteners, moving the clamping force of the caliper outward increases the brake torque. To continue the example, assume a pad to rotor friction coefficient of 0.4 and two rotor sizes: one with the pad centered at 5" from the wheel axis and another with the pad centered at 6" from the wheel axis.

Regardless of rotor size, the caliper is squeezing each side of the rotor with 1000 lbf, producing a 800 lbf friction force on the turning rotor from the pads (1000*0.4=400 lbf from each pad). The smaller rotor with the 5" effective radius makes a brake torque of 800*5=3000 in-lbf (333 ft-lbf). The larger rotor has a brake torque of 800*6=2400 in-lbf (400 ft-lbf); same pedal force, same caliper, more brake torque.

nice write up for sure. But for the physics retards, it basically means that rotor size does not matter. The only thing that was improving the braking was the relocation of the caliper further from the wheel axis.
 
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