How much does wind affect the speed of a professional cycling peloton?
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.