yeahyeah wrote:
> Dan Connelly wrote:
>
>>Mark Fennell wrote:
>>
>>
>>>OK, homework assignment complete. In a 53x12 up a 6% slope, a SE
>>>practitioner ascends ~ 0.57 m per crank revolution, or ~0.28 m per
>>>downstroke. Now I'll postulate that most of the downstroke force occurs
>>>+/-45 degrees about horizontal crank arms which means the rider is pushing
>>>his/her body and bike mass up 0.28 m with a downstroke distance of ~.24 m.
>>>So I'll concede that the low rpm SE workout is like climbing big steps in
>>>slow motion with 9 kg of dead weight on your back. Yep, that sounds about
>>>right.
>>>
>>>Mark
>>>
>>>
>>
>>peak force per pedal (F) = M * gear * effective rolling circumference * grade / (crank length * 2 alpha)
>>where alpha ~ the efficiency of the application of the pedal force. 0.5 seems a reasonable
>>estimate. 1 would be uniform in the direction of pedal motion.
>>
>>Result:
>>(F/M) = 53/12 * 0.06 * 2.1m / 0.1725 m = 3.2 m/s^2
>>
>>compare with gravity of 9.8 m/s^2
>>
>>so the force applied at the feet when pedaling is approximately 1/3 as much
>>as when stair climbing w/o additional load.
>>
>>Cadence doesn't enter the equation -- it affects power, not force.
>>
>>Dan
>
>
> So Andy et al, is there a difference between doing your LT power at 75
> rpm and 95 rpm? Is this training different physiological systems? Or
> is the only difference the gear you're riding in?
> I realize that everyone's optimal cadence is different, and one person
> might be more efficient at 75 rpm than 95rpm and vice versa, but is
> there any specific reason to train above or below your optimal cadence?
> Will your heart rate be higher at a less efficient cadence? And is
> that a good thing?
>
It was pointed out to me I missed a factor of g, and I just realized
I missed another factor of 2 pi, in my calculation, completely screwing it up.
Revised:
(F/Mg) = 53/12 * 0.06 * 2.1m / (2 pi * 0.1725 m) = 0.51
In other words, the torque-effective force on the pedals is 0.51 times body weight,
not 0.33.
Humbly submitted,
Dan
