Tuesday, November 3, 2009

Effect of Wind and Bike Speed Fluctuations on Climbing Power: pt 3

Last time on this topic, I ended with the following equation:

<pw>fw [ s0 ( s0sw0 )² + ( 3 s0 ‒ 2 sw0 ) <Δs²> + s0 <Δsw²> + ( 2 sw0 ‒ 4 s0 ) <Δs Δsw> ] .

Up to this point, there's nothing climbing-specific about this analysis: it applies as much to descending or riding on the flats as it does to climbing. But the goal of this post is to simplify it a bit, using the result of the last analysis, the effect of headwinds on bike speed.

In that analysis, which was corrected on 31 Jan 2013, it was derived that for an absolute wind speed which is relatively small in comparison to bike speed, and for a relatively small (say 20% or less) of the power going into wind resistance, then a change in wind speed change dsw (positive = tail wind, negative = head wind) changes bike speed by ds ≈ 2 dsw f, where f is the fraction of power going into wind resistance. Here's the first "climb approximation". Obviously on a descent the assumption that wind resistance power is a small fraction of total is a bad one. On a descent, in fact, wind resistance power is likely greater than the total power.

Given this relationship, we can simplify that first equation with the relationship (corrected 31-Jan-13)

<Δs Δsw> ≈ 2<Δsw²> f.

The assumption here is that the only correlated change between s and sw is that when sw changes, the rider attempts to maintain the same power, and the speed increases (if sw increases) or decreases (if sw decreases). Of course this depends on behavior: a rider may try to maintain the same speed and apply a lesser or greater power. But we're assuming constant power for the climb.

There's also uncorrelated variations in s. These may be due to variations in grade, variations in rolling resistance due to rougher or smoother pavement, or simply variations in rider power not correlated with wind changes. Nobody can maintain perfectly steady power, and typically power fades over the course of a climb due to fatigue, but then is increased approaching the top of the climb.

Anyway, these assumptions yield (corrected 31-Jan-13):

<pw>fw { s0 ( s0sw0 )² + ( 3 s0 ‒ 2 sw0 ) <Δs²> + [ s0 + 2 ( 2 sw0 ‒ 4 s0 ) f ] <Δsw²> },

or slightly simplifying (corrected 01-Jan-13):

<pw> ‒ pw0≈ fw {( 3 s0 ‒ 2 sw0 ) <Δs²> +
[ ( 1 ‒ 8 f ) s0 + 4 f sw0 ] <Δsw²> },

where pw0 is the wind resistance power calculated using the average bike speed and the average wind speed.

A quick warning: remember the power-speed relationship we're using here yields a sign error for pw if sw > s (wind is pushing the rider or the rider is moving the opposite direction). So in this instance you'd need to carry that sign difference through; this result is not applicable.

A simplified case: if sw0 = 0 and f is sufficiently small, the result can be approximated:

<pw>pw0fw s0 ( 3 <Δs²> + <Δsw²> ),

which clearly shows bike speed fluctuations are more important than wind speed fluctuations. If you prefer unitless quantities, as I do, this can be rewritten:

<pw> / pw0 ‒ 1 ≈ ( 3 <Δs²> + <Δsw²> ) / s0².

Anyway, let's run some numbers: from the Soda Springs climb analysis:

fw = 0.20 W/(m/sec)³,
sw0 = 0,
s0 = 10 mph = 4.5 meters/sec.

Then let's assume my speed differed by around 20% from the mean typically, and the wind speed was typically around ± 1 mph, then:

<Δs²> = 0.80 meters²/second²,
<Δsw²> = 0.20 meters²/second²,
f = 0.068,

yielding (result unaffected to listed precision by revisions of 31-Jan-13):

<pw>pw0 ≈ 2.3 watts.

Okay, not a huge amount: less than 1% of total power. But it's perhaps surprising the result is even this big for such small fluctuations in bike and wind speed on a steep climb where over 90% of the power is going into either gravity of rolling resistance.

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