How much does wind affect the speed of a professional cycling peloton? It's a well known fact that due to air resistance increase of speed on a bike takes non linear increase of power, so the huge marginal increase of power needed to go faster while riding at around 40-45km/h has made average top speeds of the peloton stay in this range for many years. I was wondering if the air drag plays such a major role why doesn't wind affect the average speeds more? 

I'd imagine that given at these speeds air resistance is a major factor (it's almost like a wall if you look at the power curve) the cyclist relative 'airspeed' in the wind should also be a major factor but it doesn't look so judging from the single day race results like Milan-San Remo (which has constant almost straight route so should be strongly affected by the wind). You can see results here: http://www.bikeraceinfo.com/classics/Milan-San%20Remo/milan-san-remo-index.html. If I were to guess I'd expect that wind should make the differences on range of at least -/+10 km/h but clearly this is not the case (it's more like a 37-45km/h range). Why is that so? 

(There was an incredibly interesting question here about the increase (or lack thereof) of avg speed of TdF winners over the years: Why aren&#39;t Tour de France riders going any faster? I understand that in multiday event the wind effects and tactics will cancel out but what about single stage races?)
One thing worth noting is that battling a headwind of, say, 10 km/h will not entail an increase the effort as large as would be incurred in increasing you ground speed by the same amount.

The basic equation for the power required to overcome air resistance is:

P ~ 1/2*ACdV^3

where A is frontal area, Cd de coefficient of drag and V the ground speed = air speed. If now we add a headwind v:

P ~ 1/2*ACdV*(V+W)^2 = 1/2*ACd(V^3+Vv^2+2vV^2)

The net power increase due to the headwind is therefore 1/2*ACd(V*v^2+2*v*V^2).

If instead of a headwind we increase the ground speed by the same amount v, we have:

P ~ 1/2*ACd(V+v)^3 = 1/2*ACd(V^3+3*V*v^2+3*v^2*V+v^3)

The power increase in this case is 1/2*ACd(3*V*v^2+3*v^2*V+v^3). It is obviously larger than in the headwind case, by 1/2*ACd(V*v^2+v^2*V+v^3).

This is a factor in explaining why winds do not affect average speeds as much as you expected.