- Jun 5, 2007
The energy lost is proportional to the torque squared (T^2). Power is a function of the torque and the cadence, and is not directly proportional to the energy loss. If you have the same torque in two examples but in the second example the cadence is doubled, then the power in the second case will be doubled. The energy loss in both cases should be the same however due to the torques being the same.TheDarkLord said:Ok, I see what you mean.
First point about why we stand up at harder gear: the reason has to do with the torque applied to the rear wheel (which is not the torque applied to the pedals). What is torque? It is the cross product of the radius vector and the force vector. Assuming that the two vectors are perpendicular, it becomes a simple multiplication. Let us assume that this is the case (it won't be the case if there is significant cross-chaining - i.e. low toothed gear at front and back). Ok. Now, when you are in a hard gear in the rear, the radius from the wheel center to the point where you are applying the torque through the chain is small. Consequently, the force on the chain required to provide the same torque to the rear wheel has to be accordingly larger. To apply this larger force (which translates to larger force on the pedal), you have to stand up. Note that this torque matters most when you are accelerating. When you are actually moving at constant speed, you have to supply a much smaller torque to counter the deceleration due to air drag, rolling friction, etc. Hence, the hard gear does not matter so much when you are moving at constant speed.
OK. Now let us consider the frame. Depending on the frame properties, a certain fraction of the force applied to the pedal goes into flexing the frame. At very high cadence, again one can imagine that there is power lost to the frame at the bottom of the pedal stroke. So, yes, there will be greater amount of power lost to the frame at higher power. But what about the fraction of power transferred to the frame? Is there a linear relation between the input power and the power lost to frame? I don't know the answer to that.
It should be noted that the torque applied to the BB that propels the bike forward has a moment arm which is the crank arm length. The torque that causes lateral torsion however has a moment arm which is the lateral distance of the centre of the pedal to the center of the bike. In reality though, the torsion axis changes as the position of the pedal changes relative to the BB. The actual axis of torsion at peak torque is perpendicular to the line from the pedal at 3 o'clock on the stroke and the centre of the BB. Which just makes the calculations harder because you need to find the spring constant/stiffness modulus of the frame in this axis.
Best way to empirically find the torsional stiffness constant/modulus would probably be to apply incremental forces to the pedals in a locked state at 3 o'clock and measure the angular deflection. From there you can determine the spring constant (k). Then you can unlock the pedals and apply forces to them, this time in normal bicycle motion with the drivetrain and wheels on normal or simulated pavement. The deflections in the frame will allow you to calculate the power absorption/loss.
All these explanations are hard without the use of diagrams.