Best technique when driving in strong head/tailwinds

[brucepick]

Because the penalty for higher air speed is so high you're better off minimizing the "hit" by driving slower into the wind. When the wind is helping you from behind, you can drive a bit faster and not pay the usual penalty for that road speed.

Driving with the wind coming from behind, the net air speed is your speed minus the wind speed. If you're going 60 and the wind is 20 then the air resistance is what you would have if driving 40 in no wind. 60-20 = 40

Driving into the wind you have to add wind speed to car speed. 60 mph + 20 wind = 80 mph. So your wind / air resistance is the same as if you were driving 80 mph in no wind. Big drag penalty there.

[Peakster]

Originally Posted by brucepick

Because the penalty for higher air speed is so high you're better off minimizing the "hit" by driving slower into the wind. When the wind is helping you from behind, you can drive a bit faster and not pay the usual penalty for that road speed.

Driving with the wind coming from behind, the net air speed is your speed minus the wind speed. If you're going 60 and the wind is 20 then the air resistance is what you would have if driving 40 in no wind. 60-20 = 40

Driving into the wind you have to add wind speed to car speed. 60 mph + 20 wind = 80 mph. So your wind / air resistance is the same as if you were driving 80 mph in no wind. Big drag penalty there.

Thanks for the pointer. I drove on rural roads today and tried this technique (windy again) and it's looking good on the Scangauge.

[brucepick]

Its the cube of the airspeed.
From wikipedia: http://en.wikipedia.org/wiki/Drag_%2..._high_velocity

Power
The power required to overcome the aerodynamic drag is given by:


Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, four times a [the] work in half the time requires eight times the power.

It should be emphasized here that the drag equation is an approximation, and does not necessarily give a close approximation in every instance. Thus one should be careful when making assumptions using these equations.

[end of Wikipedia quote]

[Lug_Nut]

Originally Posted by brucepick

Its the cube of the airspeed.

With a doubling of speed the drag (force) quadruples per the formula.

So two times faster increases air drag four times? That's squared, not cubed. Two times two.

Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, four times a [the] work in half the time requires eight times the power.

Four times the force already figures in the double speed, why is the double speed considered twice? If you are keeping a constant time, then the distance is double and thus the power exerted over the same time increases by the cube.

80 mph for one hour uses 8 times (2x2x2) the power of 40 mph for one hour. That's a cube, but the distance (displacement) is double.
80 mph for 40 miles uses 8 times (2x2x2) the power of 40 mph for 40 miles (a fixed displacement or distance), but for half as long. 2x2x2x.5 is the same as 2x2. 2x2 is a square, not a cube.
To me that's logical. If I'm wrong, why does it make sense to me?

[GasSavers_BluEyes]

Remember power is work over time, so it can be reduced to a limit expression and calculated on an instantaneous basis. Basically, it doesn't matter if you are going 80mph for one hour or one minute, it still uses 8x the power as going at 40mph, but you are only doing 4x the work. (work=force x distance)

I'm a mechanical mngineering student and can attest that the formula that Bruce posted is correct.

To get FE from all this, you just need some engine data on BSFC - that's pounds of fuel used per hour per hp generated. Of course you'd need the BSFC number at the RPM and throttle position you cruise at (most dyno data is at WOT unless you know someone at an OEM) but then you can calculate lb/hr of fuel used and use the density of fuel and your speed to calculate mpg.

[Lug_Nut]

And that's where I was making my error. The work increases with the INVERSE (not the product) of the time change.
Doubling the speed cuts the time in half, thus again doubling the squared drag resistance increase. (2x2)/.5 is the same as 2x2x2.