Headwind when riding in a loop
If the effect of wind resistance was linear with respect to speed, then yes, the forces would balance out. That is to say: if pedaling into a 20 mph wind was only twice as hard as pedaling into a 10 mph wind, the forces would balance.

But this is not the case, as @Daniel mentions. The effort to overcome the 20mph wind is more than twice that needed for the 10mph wind. If we simplify your loop into an out-and-back (half directly into the wind, half with it at your back), say the wind is a constant 10 mph, and that you cycle to maintain 20mph, you can get an intuitive idea of how this works.

With those assumptions, for half your trip you would be fighting a 30mph headwind (10 wind + 20 from your own speed). The second half the "headwind" would be only 10mph (20 from your own speed - 10 from the wind). This is compared to a day with no wind, in which the whole trip would have a headwind of 20mph. The jump from 20mph to 30 requires much more energy (what is the last time you hit 30mph?) than the energy saved from dropping from 20 to 10mph.

Drag varies as the square of velocity. This means that velocity is the dominant factor in the equation (as opposed to drag coefficient).

From wikipedia:


  For high velocities (or more precisely, at high Reynolds number) drag will vary as the square of velocity. Thus, the resultant power needed to overcome this drag will vary as the cube of velocity. The standard equation for drag is one half the coefficient of drag multiplied by the fluid mass density, the cross sectional area of the specified item, and the square of the velocity.


For an excellent treatment of the forces affecting a cyclist, see Rainer Pivit's article. The graphs are especially helpful.