Best technique when driving in strong head/tailwinds

[Peakster]

Originally Posted by Lug_Nut

The 40/60 will not average to 50. For the sake of argument (and easier math) make the one way distance 60 miles. It will take 1.5 hours out at 40 and 1 hour back at 60 for 2.5 hours total for the 120 miles, an average of 48 mph.
To average some desired speed at some speed out and a different speed back it is necessary to use ratios. At 40 mph out (4/5 of 50 mph) the speed back will need to be 62.5 (5/4 of 50 mph) to average 50 mph. 62.5 mph is 1.5625 times faster than 40 mph. The air drag increase will be that speed increase squared or 2.44 times the air drag.
A 10 mph tail wind at 62.5 will make the "air speed" 52.5. A 10 mph headwind at 40 mph will make an "air speed" of 50 mph, almost an equal trade in air drag.

Very good information! Thanks a bunch!

Edit: I'm not that great at math problems, so I'm going to throw this one out there:
Today I went on a flat empty highway that goes SE/NW. Temperature was -15*C with SE winds @ 41km/h.

Going SE, I accelerated like a bat out of hell to 100km/h within 0.4kms. Let's say I got 15mpg for that 0.4km stretch. I then shut off the engine and was able to coast to a complete stop in 0.7kms.

I then turned around, heading NW and once again accelerated like a BOH to 100km/h. This time I only needed 0.3km to reach 100km/h and lets say I got the same 15mpg within the 0.3km. I then shut off the engine and was amazed that I coasted to a stop after 1.6kms.

Question is this: What was my MPG(US) in each direction and what was my combined MPG(US) for the 'round trip (not including the fuel used to turn the car around)?

Edit #2 - This the the answer I came up with:
Going into wind: 62.7ml of fuel used, 41.25 mpg for the 1.1km total
Going away from wind: 47.0ml of fuel used, 95.00 mpg for the 1.9km total
Both trips: 109.7 ml used for 3.0km = 64.29 mpg