## Tuesday, November 3, 2009

### Effect of Wind Speed on Riding Speed

Recall in the last post there was a term in the result representing the correlation between wind speed and bike speed. To estimate this term, a relationship between wind speed and bike speed is needed. This is a relationship interesting on its own merit, anyway.

For this I start, as always, with a bike speed-power relationship, in this case without the usual assumption of zero wind. There are two components to power considered here: one from wind resistance and independent of mass, the other mass-proportional and due to either rolling resistance or to altitude change. Drivetrain frictional losses are assumed proportional to these other two components, and thus don't enter into this analysis.

First, the wind resistance component is:

pw = fw (ssws,

where pw is the power due to wind resistance, s is the bike speed, and sw is the speed of the wind in the direction of the bike (a head wind would be negative).

Then, we have a mass-proportional component, which we'll simply model as:

pm = fm m s,

where pm is the mass-proportional power, fm is a the sum of road grade and coefficient of rolling resistance times the acceleration of gravity, m is the total mass of the rider and bike and all equipment, and s is again the bike speed.

Then to get total power I add these components:

p = pw + pm

I'm neglecting the energy associated with acceleration in this analysis because I assume speeds change slowly. Thus I am making a "quasi-static" approximation.

So the question I need to answer is: if I change sw, how must I change s to get back the original value of p?

One approach would be to analytically solve s as a function of p and sw, then take the partial derivative with respect to sw with p fixed. That would work, but to do that I'd need to solve a cubic equation. Quadratic equations are straightforward enough, but cubic equations are yucky. Better to sidestep the need to solve for s.

1. Change sw by the differential increment dsw, and see what the resulting change in power is dp.
2. To restore p to its original value, change it by the differential increment ‒dp, and see what the resulting change ds is in s.
This sort of problem is the bread and butter of partial derivatives. Partial derivatives are essentially the answer to the question, "if I change one thing and leave some other things unchanged, what's the effect on a quantity of interest?"

So for the first step, we want the partial derivative of p with respect to sw. Since only pw and not pm depends on sw, we need only differentiate the equation for pw. The result is simple enough:

p / ∂ sw = ‒2 pw / ( ssw ).

From this, I can calculate:

dp = dsw ( ∂ p / ∂ sw ).

Okay, so then the next step: I want to restore p to its original value, so need to figure out how much to shift s (by ds) in order to do this. This is easily answered:

ds = ‒dp / ( ∂ p / ∂ s ).

So we need to solve for ∂ p / ∂ s. This is again easily done:

p / ∂ s = 2 pw / ( ssw ) + p / s.

So we have all the parts we need.

d s / d sw = 2 pw / [ 2 pp (1 + sw / s ) ]

I prefer writing these equations in terms of a parameter f, which is the fraction of power which goes into wind resistance (although my past analysis has assumed zero headwind, and I'm not assuming that here). Plugging in that parameter f yields:

d s / d sw = 2 f / [2  f + 1 ‒ sw / s ]

The first thing to do when deriving a formula is to check the units. That's not an issue here: all units cancel, as they should, since we're looking for a ratio of values with the same units. The next thing is to try some simple cases.

First consider the case where all resistance is from the wind resistance, and there is no wind. Then in this case, we get a ratio 2 / 3. This is correct: wind resistance force comes from wind speed and wind momentum, both proportional to relative wind speed, and then for power there is an additional component from rider speed.  So two parts decrease with increased tailwind, while three parts increase with increasing rider speed, so I balance 3 parts wind speed with 2 parts rider speed.

Then you can ask what happens if f is small and sw a small fraction of s? In this case, the small headwind reduces the speed by 2 f times the headwind speed. That's a nice easy-to-remember result.

So that's it. We can use this result to finish the analysis of wind speed fluctuations.

#### 1 comment:

djconnel said...

Looking back on this, deriving the result numerically, I noticed a big error in the original derivation. It was good until the final few steps, then it all came to tears. I fixed it.