wind speed and gas consumption

[suziest]

Where can I find out how wind speed affects gas consumption? For example, each 1 mph increase in wind speed increased gas consumption by "x" amount. Thanks

[paulsdecarli]

I don't think you can find an answer to the general question. Every car will be different, depending on its shape and weight. You may find studies of gas mileage at different speeds. Wind resistance is the big factor here. a car traveling at 20 miles per hour into a headwind of 20 miles per hour will probably use about as much gas as a car going 40 miles per hour in still air.

[suziest]

It doesn't have to be exact. I just need an average number increase per mph wind speed for a mid-sized car at 60 and 70 miles per hour.

[JanelleSchlossberger]

Perhaps coefficient of drag values might play a part in the gas consumption versus wind speed. Some of these web pages show the effect of a change in the mileage as a result of changing the coeffient of drag. Perhaps you could find a formula that relates mileage versus wind speed for differing values of coefficient of drag.

Check some of these sites for some info and ideas.


http://www.fordf150.net/misc/tonneau-co ... rticle.php

http://www.hybridcars.com/aerodynamic-design.html

http://www.aerodyn.org/Drag/tables-cd-level-2.html

http://www.aerodyn.org/Drag/tables.html

Good luck !

[deleted-71576]

I think you will find that there is no single answer. Each car will have a different frontal area - hence a different coefficient of drag. Also, the rear profile of the car will affect the total drag due to the turbulence related at the boundary layer.

Basically, a Buick will be affected differently than a Hummer, and also differently than a Lamborghini.

Many cars have their coefficients of drag as published values. If you narrowed your project down to a type of car with a known coefficient of drag, you would likely get much farther.

An interesting discussion of this, as it relates to roller coasters is in this link:

http://www.newton.dep.anl.gov/askasci/p ... y00444.htm

Once you calculate the increased force that the car has to overcome due to the wind, you would need to look at the gas consumption required to produce that extra power. This relates to the efficiency of the engine (MPG it gets at certain speeds.)

Another discussion as it relates to wind speed and bicycles is in this link:
http://damonrinard.com/aero/formulas.htm

A discussion of the phenomenon regarding a runner running into a headwind (with a sample calculation) is in this link:
http://experts.about.com/q/Physics-1358 ... unning.htm

Needless to say, this is a difficult project. Perhaps doing this with a scale model and a wind tunnel / fan would be a better approach.

[deleted-71555]

The other way to see it is: instead of let the wind come to you, let the car get the wind. That means: when the car is moving in 40mi/hour, it is equivalent to have a 40mi per hour wind on you 9assume no wind that day. Of course, you have to deduct the power you have used for the car during the driving distance.

Let's fix one car: it’s your car, drive on same shift (D4, automatic), no wind (or very small wind day), drive in speed of 40, 60, and 80 mile (check the police�)

• The gas consumption C = function of power, let’s say C = k*P (P is power needed)
  Power P = to conquer the friction of road + to conquer the air drag
  
  Power need to conquer the friction of road R = function of car weight and road friction rate
  Power need to conquer the friction air drag A = function of speed and drag factor

Then, assume the following functions:

• Drive at 40: R=W*mu, A= rho*(V1^2), 200miles, you consume 2 gallon (example, NOT real number)
  Drive at 60: R=W*mu, A= rho*(V2^2), 200 miles, you consume 3 gallon (example, NOT real number)
  Drive at 80: R=W*mu, A= rho*(V3^2), 200 miles, you consume 4 gallon (example, NOT real number)

…

You can get you rho value after solving the equation group. Then you can draw the chart of gas consumption C vs. speed V. It’s not linear. The slope at any V value can give you the rate of the gas consumption rate at that car speed; in this case, the wind speed.

The key to this problem is to proper define it, or confine it. For exact numbers you need to practice it. In the real world, the wind direction also counts, e.g., the wind behind you can push you forward and save the gas.

This link may give you more values be useful for reference. http://experts.about.com/q/Physics-1358 ... unning.htm