Sunday, February 3, 2013

analytic approximation to random direction wind

Last time I gave up trying to solve this integral. I wanted to watch Cyclocross Worlds so got lazy and did a numerical solution and fit. But I realized while out running after the racing that I could have done much better. So I'm back for more.

(1 / 2π) ∫ dφ exp[ (sw' / 3)2 ] exp[ −2 swx' / 3 ],

where the integral is over the full circle and φ is the angle of the wind relative to the rider (0 = pure tail wind).

The key here is to recognize that this can be well-approximated by a Gaussian for sw' to at least 1. This isn't the solution of the integral, but it's a good approximation, so once I recognize the analytic form of the solution I can get away just matching derivatives with respect to sw' = the ratio of the wind speed to the zero-wind rider speed.

So my solution will be the following, where I must solve for K:

exp[ (sw' / K)2 ].

I recognize that for every value of positive swx', there is a corresponding negative value, and so I can combine them in the integral and then integrate over the half-circle for which swx' are positive (from −π/2 to +π/2):

(1 / π) ∫ dφ exp[ (sw' / 3)2 ] cosh[ 2 swx' / 3 ].

I can then do a low-order expansion on these exponentials to generate a linear equation: exp(x) ≈ 1 + x, and cos(x) ≈ 1 + x2/2. The integral thus becomes the following approximation:

(1 / π) ∫ dφ [1 + (sw' / 3)2] [ 1 + 2 (swx' / 3)2 ].

Now I plug in the formula for the tail wind component swx':

(1 / π) ∫ dφ [1 + sw' / 3)2] [ 1 + 2 (sw' / 3)2 sin2φ ].

I can omit terms proportional to sw'4:

(1 / π) ∫ dφ [1 + (1 + 2 sin2φ) (sw' / 3)2].

Recognizing the average value of sin2φ = 1/2, the integral is trivial:

1 + 2 (sw' / 3)2 .

The low-order expansion of the Gaussian approximation is:

1 + (sw' / K)2.

So I conclude K = sqrt(9/2). This is essentially what I got numerically.

So written more elegantly, the approximation to the average speed of a rider slowed down by a constant wind coming from wind distributed evenly from all possible directions during a closed course is:

<v0> = v00 exp[− 2 (sw / 3 v00)2],

where <v0> is the average flat-road speed and v00 is the zero-wind flat-road speed.

The interesting thing about this result is the average speed reduction, including the headwind and tailwind sections, is twice the speed reduction in the cross-wind sections, for relatively small speed reductions. So if the cross-winds slow me down by 1%, over the whole course I'll be slowed around 2%, considering the effects of the head-winds, tail-winds, and partial cross-winds.

It's interesting to compare this to the result for an out-and-back with direct head and tail wind. There the average time per unit distance is scaled by cosh[ 2 sw' / 3 ] exp[ (sw' / 3)2 ] ≈ exp[3 (sw' / 3)2]. So the result in wind equally distributed over all angles equally is 2/3 as much as that from a pure headwind plus a pure tailwind.

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