Hydraulic disc brakes - total friction as a function of rotor size I have a hydraulic disc road bike with a 160mm rotor on the front and a 140mm on the rear.  The setup is exactly the same front to rear, except for rotor size. I'd like to know the disparity between the force I would have to apply at the brake lever for the total friction or stopping power to be equal front to rear, for a given speed over a given distance, to a complete stop, i.e. total work done by each to be exactly equal.  Since the 140mm rotor is 7/8th's the size of the 160mm, I'd be tempted to say that an equal lever pull would produce 7/8th's the friction on the rear, but I'm sure there is more to it than that. 

So, for example, If I'm riding at 10mph and come to a stop over 100ft, how hard would I have to pull each lever (as a ratio of front brake/rear brake) to ensure that my momentum is converted to exactly the same amount of heat by each rotor respectively?
Don't worry about equal torque.  The front wheel will do as much as ninety percent of the work on a maximum stop.  You will learn to modulate the rear to prevent lock-up.  Equal pressure will be fine for normal stops.