Suppose you are cycling on a flat out-and-back course on a windy day. The first half of your ride is into the wind, the second half is with the wind. Is there any advantage to expending more power during the headwind half vs. the tailwind half?
This should be simple to calculate? To compare scenarios, we will specify that the time-average-power (Paverage) be a specific value. In equation form this is...
TotalTime x Paverage = P1t1 + P2t2
The time spent at P1 (into the wind), t1, will be longer than the time spent at P2 (with the wind).
The other assumption we will make is that our 'air speed', W, is high enough that most of our power is consumed in combating drag. We can write...
W1 = kP11/3
...meaning the air speed is solely a function of a drag coefficient, k, times power to the 1/3rd power. This is a fairly good assumption at air speeds over 20 mph. This relationship tells us that to bike 10% faster, we need to increase power by the cube of 1.1, or about 33%. [It also suggests that efforts to reduce k (a 'drag coefficient') can pay dividends when power increases are hard to find.]
The 'ground velocity', V, (what appears on your cyclocomputer) is W-w, where w is the wind speed.
Steady Power
Let's take the base case of steady power (P1/P2=1, so Paverage=P1=P2) and no wind (w=0, so W1=V1=kP11/3). The equation simplifies to
TotalTime/Distance = 1/(kPaverage1/3).
The inverse of that is
Vaverage = Distance/TotalTime = kPaverage1/3.
As expected, when drag forces predominate, our speed varies as power to the 1/3rd.
No wind
For the case of no wind, the equations 'simplify' to give
Paverage/P2 = (1+(P1/P2)2/3) / (1 + (P1/P2)-1/3)
As it turns out, for values of P1/P2 of interest, a distance-weighted Paverage is a very good approximation of the time-weighted Paverage. [Paverage/P2 = (P1/P2 +1)/2] when there is no wind.
That is shown graphically in the bottom curve below. The degradation in average speed is visible at all values of P1/P2 other than one, but it is almost insignificant. It suggests that steady pace is the best option in calm conditions.
Wind
Now let's looks at how many seconds we can save in a 40K race (20K into the wind and 20K with the wind) at a given ratio of P1/P2 and various wind speeds. This is shown in the graph below.
The next two curves compare two different Paverages in a 5 mph wind. Interestingly, the advantage of using a "P1/P2>1 strategy" decreases for stronger cyclists. The difference between speeds into the wind versus with the wind is less for the stronger cyclists, so they gain less from this strategy.
The top two curves are with a 10 mph wind. The strategy yields larger returns in windier conditions.
Conclusion
Having said all that, I conclude that the time savings suggested here are insignificant. The analysis ignores too many other variables and the savings are so small, that I would not implement this strategy myself.
Other Thoughts on Pacing
I have followed Alex Dowsett in his quest to set the 1-hour record (he set it in 2020 but it has since broken by others). I recall one of his comments on pacing. He said that during the first-30-minutes he focuses on his aero position (the k we talked about above). He has spent enough time in the wind tunnel that he knows what his optimal position is. It is an unnatural position and takes some energy to maintain. In the second-30-minutes his form degrades, and his power increases to maintain or increase his lap speed. This might be a good strategy for a windy day. Focus on aero position heading into the wind, then focus more on power output heading with the wind?
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Calculation Notes:
Unfortunately, these equations cannot be solved analytically. A wind speed, power ratio, and target Paverage were chosen. A P1 was guessed. That guess was used to calculate a TotalTime/Distance (EQ 1), which was used to calculate Paverage (EQ 2). The difference between the target Paverage and the calculated Paverage was used to make a new guess of P1. Usually convergence to <0.01% error occurred in less than 4 iterations.
Equations:
TotalTime/Distance = 1/V1 + 1/V2 = 1/(W1-w) + 1/(W2+w) EQ 1
and
TotalTime x Paverage = P1t1 +P2t2
[or TotalTime/Distance x Paverage = P1/V1 + P2/V2] EQ 2
where W1 = kP11/3 etc.
and V1 = W1 - w and V2 = W2 + w
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