effect of power on climbing speed

Similar to the last post, the next question is: what is the effect of power on climbing speed? Answering this is a fairly trivial extension of the previous calculation.

Again consider a power-speed model which, simplified and neglecting wind speed and acceleration, is expressed as:

p = fm m s + fw

where terms are as described last time. Then we can calculate the effect of fractional changes in speed on fractional changes in power required:

(s / p) (∂ p / ∂ s) =

(s / p) ( fm m + 3 fw s² )

What we want, however, is the inverse of this: the effect of fractional changes in power on fractional changes in speed:

(p / s) / ( fm m + 3 fw s² ).

Expanding p / s yields:

( fm m + fw s² ) / ( fm m + 3 fw s² )

which can be simplified in terms of the ratio of wind resistance to mass-proportional power used in the last blog entry:

(1 + r) / (1 + 3r).

Again, if what we know isn't the ratio of wind resistance power to mass-proportional power, but rather the ratio of wind resistance power to total power, using r = f / (1 - f) and multiplying numerator and denominator by 1 - f:

1 / (1 + 2 f).

In contrast to the previous analysis, in this one the f form is the simpler one. So if wind resistance constitutes 10% of total power, and I put out 1% additional power on a climb, I get 1% / 1.2 = 0.83% more speed on the hill. Compare this to the case for saving 1% of mass, which increased speed by 0.75%.

Consider, then, the case where you gain 1% of body mass, starting from your bike + equipment weighing 12% of body mass and climbing with 10% of power going into wind resistance. How much power do you need to gain to keep climbing speed the same? Well, if you're addicted to "watts/kg", the answer is obvious: 1%. But that's not quite right, assuming wind resistance was unaffected by the mass gain. Total mass increase = 1% / 1.12 = 0.89%, decreasing speed by 0.67%, requiring a 0.80% increase in power to compensate. Of course there is typically some increase in cross-sectional area with any mass increase, so the actual number may be a bit higher.


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