Originally Posted by michaelmc
*F = mg(sin(theta)) where m = the mass of you + your bike, g = acceleration due to gravity, and theta = angle of incline. F is the force from gravity parallel to the incline; that is the force that you have to overcome to climb a hill. The power necessary to do that is F times your desired velocity, v (P = Fv). Therefore, m is directly proportional to P.
Originally Posted by terrymorse
Not quite. Climbing speed is not directly related to weight, since there is also rolling resistance, drive train losses, and wind resistance to overcome.
But these effects are small compared to overcoming gravity. If you drop 10% off total bike+rider weight, you can expect maybe an 8-9% improvement when climbing a typical grade.
Rolling resistance is also directly proportional to your weight. One component is the normal force, which is mass*gravity*cos(theta), where theta is the angle of the incline (making it 1 on a flat). You're right about drive train losses and wind resistance, though you should be getting more aerodynamic as you're getting thinner! So it's about proportional, but not perfectly so.
