My point is a bit different. As braking continues, the rotor tends to run cooler than the pad (pad surface in your no-pad-conductivity example) though both are exposed to the same “friction interface” temperature. Not only is rotor a better heat conductor but only a relatively small portion of the swept area is exposed to the friction heat source at any given time, i.e. a lower duty cycle. Thus the rotor would have not only a greater conductive coefficient h, but would also see a greater ΔT relative to the “friction interface” area that would be neither pad nor rotor but the common area there between.riff_raff wrote:Do you mean heat transfer and temperature, rather than heat energy and temperature? I would agree with your comment that the pad is subject to a 100% "duty cycle" during braking, while the rotor is not. But consider this hypothetical: Let's assume we have a pad material that conducts absolutely no heat, and a rotor material that has a very high thermal conductivity. At the instant the brakes are applied and there is friction heat generated at the pad/rotor interface, the relative pad and rotor surfaces would assume a state of thermal equilibrium based on how efficiently the rotor conducts the thermal energy away from the contact interface. Since the pad material does not conduct heat, while the pad outer surface will become hot, there will be no conductive heating of the pad body. And all of the heat load produced at the friction interface will be forced into the rotor body.olefud wrote:Very true. But it’s important to distinguish between heat energy and temperature. During braking the pads have a 100% duty cycle while any particular point on the rotor has a rather low duty cycle. Thus the rotor will tend to run hotter than the rotor which, as your point out, tends to divert heat energy to the lower temperature rotor
No conductivity is a tricky concept. I’m thinking it would be like very low conductivity ceramic heat shields that still show surface heating, i.e. space craft reentry heat shields.