s = s0 + Δs,
sw = sw0 + Δsw,
where, for <s> signifying the average of s, etc:
<s> = s0 + <Δs> = s0,
<sw> = sw0 + <Δsw> = sw0.
So these can be plugged into the wind power calculation:
pw = fw (s ‒ sw)² s.
First, a word of warning: this equation only applies if sw ≤ s, otherwise you need to flip the sign. If sw > s, then it could mean a tailwind is pushing the rider, or else the rider is moving in the negative direction. These sign issue is a big clue that this equation is not grounded in fundamental physics, and its not, it's an approximation. But given that, we'll assume the rider keeps moving forward, and continue:
= fw (s0 + Δs ‒ sw0 - Δsw)² ( s0 + Δs ).
Okay, that obviously needs to be expanded:
pw = fw ( s0² + 2 s0 Δs ‒ 2 s0 sw0 ‒ 2 s0 Δsw + Δs² ‒ 2 Δs sw0 ‒ 2 Δs Δsw + sw0² + 2 sw0 Δsw + Δsw² ) ( s0 + Δs ).
A clarification of sloppy notation: Δs² refers to the square of the difference, not the difference of the square. This is an important distinction.
Anyway, we're not done expanding yet (gulp):
pw = fw ( s0³ + 2 s0² Δs ‒ 2 s0² sw0 ‒ 2 s0² Δsw + s0 Δs² ‒ 2 s0 Δs sw0 ‒ 2 s0 Δs Δsw + s0 sw0² + 2 s0 sw0 Δsw + s0 Δsw² + Δs s0² + 2 s0 Δs² ‒ 2 Δs s0 sw0 ‒ 2 Δs s0 Δsw + Δs³ ‒ 2 Δs² sw0 ‒ 2 Δs² Δsw + Δs sw0² + 2 Δs sw0 Δsw + Δs Δsw² ).
It's downhill from here, I promise. First, it's clear there's some terms in the above listed twice (those with both s0 and Δs). These can be combined:
pw = fw ( s0³ + 3 s0² Δs ‒ 2 s0² sw0 ‒ 2 s0² Δsw + 3 s0 Δs² ‒ 4 s0 Δs sw0 ‒ 4 s0 Δs Δsw + s0 sw0² + 2 s0 sw0 Δsw + s0 Δsw² + Δs³ ‒ 2 Δs² sw0 ‒ 2 Δs² Δsw + Δs sw0² + 2 Δs sw0 Δsw + Δs Δsw² ).
Still ugly. But we care only about the average of power. For this, remember that s0 and sw0 are constants, and the average of a constant multiplied by either Δs or Δsw is zero:
<pw> = fw ( s0³ ‒ 2 s0² sw0 + 3 s0 <Δs²> ‒ 4 s0 <Δs Δsw> + s0 sw0² + s0 <Δsw²> + <Δs³> ‒ 2 <Δs²> sw0 ‒ 2 <Δs² Δsw> + 2 <Δs Δsw> sw0 + <Δs Δsw²> ).
- We assume the probability distributions of Δs and Δsw are symmetric, and therefore<Δs³> = 0. <Δsw³> doesn't appear in the equation, but it's also zero.
- Retain only a first-order correlation between Δs and Δsw, eliminating <Δs Δsw²> and <Δs² Δsw>. Perhaps this assumption should be revisited later.
Then this simplifies further:
<pw> ≈ fw ( s0³ ‒ 2 s0² sw0 + 3 s0 <Δs²> ‒ 4 s0 <Δs Δsw> + s0 sw0² + s0 <Δsw²> ‒ 2 <Δs²> sw0 + 2 <Δs Δsw> sw0 ).
Combining Δ terms and removing a common s0 from the rest yields:
<pw> ≈ fw [ s0 ( s0² ‒ 2 s0 sw0 + sw0² ) + ( 3 s0 ‒ 2 sw0 ) <Δs²> + s0 <Δsw²> + ( 2 sw0 ‒ 4 s0 ) <Δs Δsw> ].
The first term is an obvious square, so we can simplify once again, happy to see the relative wind speed back:
<pw> ≈ fw [ s0 ( s0 ‒ sw0 )² + ( 3 s0 ‒ 2 sw0 ) <Δs²> + s0 <Δsw²> + ( 2 sw0 ‒ 4 s0 ) <Δs Δsw> ].
First, a sanity check: if all of the Δ terms are zero (constant bike and wind speed), we have the first equation back. This had to be the case. If it weren't, then we'd made an error.
Then an observation: it's only true that Δs and Δsw have the same effect if s0 = sw0, implying the wind is the same speed as the rider and in the same direction. If the wind is less of a tailwind or a headwind, then fluctuations in bike speed have a stronger effect than those of wind speed. In zero wind, bike speed fluctuations have three times the effect of wind speed fluctuations of the same magnitude. Why? When wind speed increases without bike speed changing, the retarding force increases and power increases in proportion to the retarding force increase. But when bike speed increases the same amount, the retarding force increases the same amount, but further the bike is moving faster against this increased retarding force. Okay, perhaps a bit too much on this...
The remaining terms feature:
- <Δs²>: the expectation of the square of the deviation in bike speed,
- <Δsw²>: the expectation of the square of the deviation in wind speed,
- <Δs Δsw>: the expectation of the product of the deviations of bike speed and wind speed.
Now suppose that bike speed varied randomly, and wind speed varied randomly, but the variations were uncorrelated. For example, if I told you just the relative wind speed, you'd have no further information about the bike speed. This would be a convenient assumption, as then this term would average out to zero, and we could neglect it. Maybe on a very steep hill in light wind, where the small wind variations might have a very small effect on bike speed in comparison to variations in road grade, this might be an acceptible approximation. Or when riding in deep mud, where the depth of the mud results in variations in rolling resistance which strongly limit speed, it would also be a good assumption. But in general if the head wind is stronger, the bike moves slower, and vice versa. So a model is needed for the correlation of bike speed and wind speed.
Alternately this parameter could be evaluated directly, for example from iBike data. I love the iBike. It's amazing I haven't gotten one yet. Maybe when I switch to the Metrigear Vector and need an ANT+ Sport head unit, and when iBike reveals their new version in March 2010 around the time the Vector is due for release. But I digress.
Another approach is to model it. The simplest model is a quasi-static model: assume the bike has no inertia, and accelerates or decelerates instantly to the speed at which propulsive force and retarding forces cancel. This is a good approximation if the variations in bike and wind speed tend to be relatively gradual.
But this blog post is getting a bit long, and my train is almost at Palo Alto station, so it's a good time to pause. Next time I'll analyze the quasi-static effect of wind speed on bike speed. This will provide an estimate of <Δs Δsw>.