I Car drag coefficent vs fuel consumption link

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Reducing a car's drag coefficient (Cd) from 0.35 to 0.20 can significantly impact fuel consumption, with calculations suggesting a potential decrease from 5L/100km to approximately 2.8L/100km, assuming other factors remain constant. However, the relationship between Cd and fuel consumption is complex, as it also depends on driving conditions, engine efficiency, and the power-torque curve. While a direct proportionality is suggested, real-world results indicate that fuel savings from reduced drag are often about half the decrease in aerodynamic drag. Empirical studies show that a 10% reduction in drag typically yields around a 5% improvement in gas mileage. Overall, while lowering Cd can enhance fuel efficiency, the actual gains depend on various driving and mechanical factors.
  • #31
jrmichler said:
This is a case where empirical knowledge needs more weight than physics calculations. The people at Ecomodder.com (https://ecomodder.com/) study the practical effects of aerodynamic drag ad nauseam. One of their heroes modified a Honda Civic until he was able to get 50 MPG at 90 MPH, and 90 MPG at more reasonable speeds. He built the car because he drove a lot of miles, mostly at high speed, and wanted good gas mileage. The story is at: https://aerocivic.com/ and also at: https://ecomodder.com/forum/showthread.php/aerocivic-how-drop-your-cd-0-31-0-a-290.html.

The short version is that gas mileage improvement in the real world is about half the decrease in aerodynamic drag. A 10% reduction in drag typically yields about a 5% improvement in gas mileage. That will vary depending on many variables other than aerodynamic drag. Also, reductions in aerodynamic drag require other changes in order to realize the full benefit.

The best book on automobile aerodynamics is the book by Hucho: https://www.amazon.com/dp/1483108414/?tag=pfamazon01-20. Highly recommended if you are interested in the research on the subject. If you want to learn practical techniques on how to modify your car to reduce drag, the best book that I know of is the book by Julian Edgar: https://www.amazon.com/dp/1787112837/?tag=pfamazon01-20. Highly recommended if you want to modify your car and do it right.

One person, who shall not be named because that would be bragging, made a number of modifications and documented the effects on gas mileage. Link: https://ecomodder.com/forum/showthread.php/modding-06-gmc-canyon-17070.html. In that case, the gas mileage of a 2006 GMC Canyon crew cab truck was improved from 21 MPG in winter and 27 MPG in summer to 32 MPG in winter and 38 MPG in summer.
Do you think it is realistic that civic increase top speed from 150 to 225km/h?
This is huge difference..
 
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  • #32
quote:
"The ecological result of sharply focused aerodynamic work is demonstrated clearly, said Fecker, with an improvement of Cd by a factor of 0.04 cutting fuel consumption of a car cruising on an autobahn at 130 km/h (81 mph) by 0.5 L/100 km, which equates to approximately 13 g/km of CO2.

“Engineers would need to find a weight saving of 35 kg in the chassis structure to manage a similar drop in CO2 emissions,” according to Mercedes."


What is math meanning of " improvement of Cd by factor 0.04 "?

Lets say orignal Cd is 0.26, if divide by 0.04, get 6.5... this is cleary wrong number..
Do they mean 0.26 - 0.04 = 0.22 (that make sense) ?

Isnt in english "by factor" mean same as times?
 
  • #33
Jurgen M said:
quote:
"The ecological result of sharply focused aerodynamic work is demonstrated clearly, said Fecker, with an improvement of Cd by a factor of 0.04 cutting fuel consumption of a car cruising on an autobahn at 130 km/h (81 mph) by 0.5 L/100 km, which equates to approximately 13 g/km of CO2.

“Engineers would need to find a weight saving of 35 kg in the chassis structure to manage a similar drop in CO2 emissions,” according to Mercedes."


What is math meanning of " improvement of Cd by factor 0.04 "?

Lets say orignal Cd is 0.26, if divide by 0.04, get 6.5... this is cleary wrong number..
Do they mean 0.26 - 0.04 = 0.22 (that make sense) ?

Isnt in english "by factor" mean same as times?
It is poor wording at best.

Yes, the natural usage of "an improvement by a factor of x" is that some figure of merit is multiplied by x. Big numbers are good. Small numbers are bad. Improving something by a factor of 0.04 means making that thing 25 times worse.

This is not a plausible reading. Without more context, it is impossible to know the intended meaning. If we had a reference to the article where this claim is made, it might be possible to disambiguate.
 
Last edited:
  • #34
Jurgen M said:
... Do they mean 0.26 - 0.04 = 0.22 (that make sense) ?
It seems to me that it does not make sense for most commercial cars.
According to the equation of post #3 above, the reduction in drag force or drag work should be directly proportional to the reduction in fuel consumption at same car's velocity.
If that is true, 15% less Cd would imply that 0.5 liters/100 km would be 15% of consumption with Cd=0.26.
That value would then be 3.3 liters/100 km or 78 miles/gallon.
 
  • #35
Lnewqban said:
It seems to me that it does not make sense for most commercial cars.
According to the equation of post #3 above, the reduction in drag force or drag work should be directly proportional to the reduction in fuel consumption at same car's velocity.
If that is true, 15% less Cd would imply that 0.5 liters/100 km would be 15% of consumption with Cd=0.26.
That value would then be 3.3 liters/100 km or 78 miles/gallon.
But according to post #15, it seems that a lot of fuel energy is lost in the combustion process and the friction losses. Even idling, a car uses a lot of fuel.

I doubt the fuel reduction would be proportional to the drag reduction. If I refer to a modified version of the equation found in post #6:
$$\frac{\left(\frac{dV_f}{dx}\right)_2}{\left(\frac{dV_f}{dx}\right)_1} = 1 - \frac{\frac{1}{2}\rho \left(C_{d\ 1} - C_{d\ 2}\right) A v^3}{P_1}$$
Or:
$$\frac{\left(\frac{dV_f}{dx}\right)_2}{\left(\frac{dV_f}{dx}\right)_1} = 1 - \frac{1}{1 + \frac{P_{other}}{\frac{1}{2}\rho C_{d\ 1} A v^3}} \frac{C_{d\ 1} - C_{d\ 2}}{C_{d\ 1}}$$
Where ##P_1 = \frac{1}{2}\rho C_{d\ 1} A v^3 + P_{other}##, and ##P_{other}## is the power loss through other mechanisms than aerodynamic drag.

According to this equation, if you assume that the drag power accounts for only half the total power loss (##\frac{1}{2}\rho C_{d\ 1} A v^3 = P_{other}##), a drop from 0.26 to 0.22 will result in 0.5 L/100km reduction if the initial fuel consumption was 6.5 L/100km.
 
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  • #36
jack action said:
According to this equation, if you assume that the drag power accounts for only half the total power loss (##\frac{1}{2}\rho C_{d\ 1} A v^3 = P_{other}##), a drop from 0.26 to 0.22 will result in 0.5 L/100km reduction if the initial fuel consumption was 6.5 L/100km.
If your calculation is correct then they mean minus 0.04 ..
 
  • #37
I find this two text for same car:

"Every reduction of one thousandth in wind resistance pays: with a higher speed and lower fuel consumption. Even a 0.01 reduction in the Cd value represents a saving of 0.1 litres per 100 kilometres. "
and this text

"Good flow characteristics are tremendously important in this context, because if the Cd value can be cut by one tenthousandth, fuel consumption for the average customer is reduced by a tenth of a litre, and at very fast motorway speeds by as much as 0.4 litres per 100 kilometres."

one tenthousandth is 0.0001, that can be reduce consumption by 0.1Liter, something is wrong here..
 

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