# Car drag coefficent vs fuel consumption link

• I
• Jurgen M
Cd is not inversely proportional with fuel consumption?No, change of Cd is not inversely proportional with fuel consumption. As the drag coefficient decreases, the aerodynamic force opposing the car's forward movement also decreases, resulting in a decrease in fuel consumption. However, there are other factors at play, such as the power-torque curve of the engine and selected gear reduction, which can affect fuel consumption as well. So while a lower drag coefficient may lead to a decrease in fuel consumption, it is not a direct or proportional relationship.
Jurgen M
Car has drag coeffecient of 0.35 and drink 5L/100km, if car Cd is reduced to 0.20 and everthying else stay the same(frontal area,power,weight etc),how much will car drink?

Is it possible to calcualte this or we need something more?

I am interested if exist direct link/formula between Cd and fuel consumption?

I'm pretty sure this is not a schoolwork question, so I'll page our gas mileage vs. Cd expert @jrmichler

topsquark
Jurgen M said:
Car has drag coeffecient of 0.35 and drink 5L/100km, if car Cd is reduced to 0.20 and everthying else stay the same(frontal area,power,weight etc),how much will car drink?

Is it possible to calcualte this or we need something more?

I am interested if exist direct link/formula between Cd and fuel consumption?
What is directly proportional to the Cd is the aerodynamic force opposing the forward movement of the car.
How the fuel consumption is related to that force, depends on the power-torque curve of the engine and selected gear reduction.

topsquark
Lnewqban said:
What is directly proportional to the Cd is the aerodynamic force opposing the forward movement of the car.
How the fuel consumption is related to that force, depends on the power-torque curve of the engine and selected gear reduction.

0.35 : 0.2 = 1.75, so Fd is 1.75 smaller for car with CD=0.20

5L/100km : 1.75 = 2.8L/100km

?

There isn't going to be a simple answer. The forces on the car include drag (which is a square function of speed) and various frictional resistances (which are linear with speed), and the engine and drivetrain efficiency vary with RPM and gear ratio. Or, more directly: it depends more on how you drive than the drag coefficient. If you drive mostly city/low speeds drag coefficient makes no difference. If you drive only on the highway it will make a difference.

Lnewqban
Here is how I would estimate it. I would start with the Brake Specific Fuel Consumption ##BSFC## definition:
$$BSFC = \frac{\frac{dm_f}{dt}}{P}$$
Where ##\frac{dm_f}{dt}## is the fuel mass flow rate consumed and ##P## is the power output. I would assume the ##BSFC## to be constant in your scenario. This can be replaced by the fuel volumetric flow rate ##\frac{dV_f}{dt}## when introducing the fuel density ##\rho_f##:
$$BSFC = \frac{\rho_f\frac{dV_f}{dt}}{P}$$
Introducing the vehicle velocity ##v##:
$$BSFC = \frac{\rho_f\frac{dV_f}{dt}}{P}\frac{v}{v}$$
$$BSFC = \frac{\rho_f\frac{dV_f}{dt}}{P}\frac{v}{\frac{dx}{dt}}$$
$$BSFC = \frac{\rho_f \frac{dV_f}{dx}}{\frac{P}{v}}$$
$$BSFC = \frac{\rho_f \frac{dV_f}{dx}}{F}$$
Where ##\frac{dV_f}{dx}## is the fuel consumption in volume by distance traveled and ##F## the equivalent force produced by the engine. Since ##\cancelto{}{BSFC}## and ##\cancelto{}{\rho_f}## are constant then:
$$\cancelto{}{\Delta \left(\frac{dV_f}{dx}\right) = \Delta F}$$
Or assuming scenarios ##1## and ##2##:
$$\cancelto{}{\left(\frac{dV_f}{dx}\right)_1 - \left(\frac{dV_f}{dx}\right)_2 = F_1 - F_2}$$
$$\cancelto{}{\left(\frac{dV_f}{dx}\right)_1 - \left(\frac{dV_f}{dx}\right)_2 = F_1 - \left(F_1 - \frac{1}{2}\rho \left(C_{d\ 1} - C_{d\ 2}\right) A v^2\right)}$$
$$\cancelto{}{\left(\frac{dV_f}{dx}\right)_2 = \left(\frac{dV_f}{dx}\right)_1 - \frac{1}{2}\rho \left(C_{d\ 1} - C_{d\ 2}\right) A v^2}$$

Since ##BSFC## and ##\rho_f## are constant, then assuming scenarios 1 and 2:
$$\frac{\left(\frac{dV_f}{dx}\right)_2}{F_2} = \frac{\left(\frac{dV_f}{dx}\right)_1}{F_1}$$
$$\frac{\left(\frac{dV_f}{dx}\right)_2}{\left(\frac{dV_f}{dx}\right)_1}=\frac{F_2}{F_1}$$
$$\frac{\left(\frac{dV_f}{dx}\right)_2}{\left(\frac{dV_f}{dx}\right)_1}=\frac{F_1 - \frac{1}{2}\rho \left(C_{d\ 1} - C_{d\ 2}\right) A v^2}{F_1}$$
$$\frac{\left(\frac{dV_f}{dx}\right)_2}{\left(\frac{dV_f}{dx}\right)_1} = 1 - \frac{\frac{1}{2}\rho \left(C_{d\ 1} - C_{d\ 2}\right) A v^2}{F_1}$$
$$\frac{\left(\frac{dV_f}{dx}\right)_2}{\left(\frac{dV_f}{dx}\right)_1} = 1 - \frac{\frac{1}{2}\rho \left(C_{d\ 1} - C_{d\ 2}\right) A v^3}{P_1}$$
Say in your first scenario, ##\rho = 1.23\ kg/m^3##, ##C_d = 0.35##, ##A= 0.7\ m^2##, ##v= 27.8\ m/s##, and ##P = 20\ 000\ W##, then by changing only ##C_d = 0.2## in scenario 2, you get:
$$\frac{\left(\frac{dV_f}{dx}\right)_2}{\left(\frac{dV_f}{dx}\right)_1} = 1 - \frac{\frac{1}{2}(1.23) \left(0.35 - 0.2\right) (0.7) (27.8)^3}{20\ 000} = 0.93$$
And if you were doing 5 L/100km before, you are now doing 4.65 L/100km.

Edit: Redone the awful math I did to something that made sense.

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erobz, russ_watters and Lnewqban
If you are keeping all the underlying complexity the others are mentioning out of it, you can get a sense for how the fuel consumption will change keeping everything else constant.

$$P = F \cdot v$$

Imagine you had already determined the power demand driving at some speed ##v_o##, call it ##P_o##.

Lets make the simplistic supposition that the retarding force is given by:

$$F = F_s + \beta v_o^2$$

##\beta## is proportional to ##C_d##

Then the power requirement as a function of ##\beta## would be linear:

$$P = P_o \left( \frac{F_s}{F_s + \beta_o v_o^2} + \frac{v_o^2}{F_s + \beta_o v_o^2} \beta \right)$$

thus;

$$\Delta P = \frac{P_o v_o^2}{F_s + \beta_o v_o^2} \Delta \beta$$

So even in this model you don't have the simple proportionality to ##\beta## ( or ## C_D##).

Also, the fuel consumption is going to be proportional to the to the Power input, so we should divide what is above by the drive train efficiency at ##v_o## namely ##\eta_o##

$$\Delta P_{input} = \frac{1}{\eta_o}\frac{P_o v_o^2}{F_s + \beta_o v_o^2} \Delta \beta$$

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Lnewqban
So change of Cd is not proportional with fuel consumption?
edit..

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Jurgen M said:
So change of Cd is not inversely proportional with fuel consumption?

No. If ##C_d## gets larger the power required ##P## gets larger, and if the power required is larger the fuel consumption is larger.

erobz said:
No. If ##C_d## gets larger the power required ##P## gets larger, and if the power required is larger the fuel consumption is larger.
Sorry I make mistake, I mean proportional...

Jurgen M said:
Sorry I make mistake, I mean proportional...
In the simplistic model I made its proportional ( but the constant of proportionality is likely not 1 ) The question is how realistic does it need to be?

erobz said:
In the simplistic model I made its proportional ( but the constant of proportionality is likely not 1 ) The question is how realistic does it need to be?
For rough calculation 5/(0.35/0.2)= 2.8L/100km

If Cd drops by 42% then fuel cons. drops by 42% as well..

If that is case, now I understand why electric cars chase low Cd x frontal area...

Jurgen M said:
For rough calculation 5/(0.35/0.2)= 2.8L/100km

If Cd drops by 42% then fuel cons. drops by 42% as well..

If that is case, now I understand why electric cars chase low Cd x frontal area...
No, that is using ##1## as constant of proportionality...there is no obvious indication that is anywhere near reality.

erobz said:
No, that is using ##1## as constant of proportionality...there is no obvious indication that is anywhere near reality.
How you get 1?
0.35/0.2 is 1.75

Jurgen M said:
0.35 : 0.2 = 1.75, so Fd is 1.75 smaller for car with CD=0.20

5L/100km : 1.75 = 2.8L/100km

?
Fd for more aerodynamic car is 57% smaller.
Only 57% less torque at the driving wheels is required.
When parked, engine idling, zero torque is required, but fuel consumption is not zero.
Heat losses are still huge.

https://en.m.wikipedia.org/wiki/Fuel_economy_in_automobiles#Energy_considerations

tech99, russ_watters and erobz
Jurgen M said:
How you get 1?
0.35/0.2 is 1.75
No, what I'm saying is that would imply for the formula I derrived:

$$\frac{1}{\eta_o}\frac{P_o v_o^2}{F_s + \beta_o v_o^2} = 1$$

There is no justification for that.

Oh, and there would be another constant of proportionality in there for the conversion from Power to fuel mileage. All of that would have to equal 1...

erobz said:
No, what I'm saying is that would imply for the formula I derrived:

$$\frac{1}{\eta_o}\frac{P_o v_o^2}{F_s + \beta_o v_o^2} = 1$$

There is no justification for that.

Oh, and there would be another constant of proportionality in there for the conversion from Power to fuel mileage. All of that would have to equal 1...
What is Fs in your formula?

Jurgen M said:
What is Fs in your formula?
It would be the equivalent of any (approximately) constant retarding forces. Forces that do not have a strong dependency on the vehicle's velocity.

I think I've over complicated my first approach.

The input power ##P_{in}## should simply be given by:

$$P_{in} = \frac{1}{\eta_o} \left( F_s + \beta v_o^2\right) v_o$$

That implies:

$$\Delta P_{in} = \frac{v_o^3}{\eta_o} \Delta \beta$$
If I haven't done anything egregious, it does seem that there are potentially large fuel savings potential from modest changes in ##\beta## at highway speeds (with this simple model).

The question now seems to be: How hard is it to actually make a modest change ##\beta## remembering from above that ##\beta = \frac{1}{2} C_D \rho A## ?

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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.

olivermsun, Jurgen M, russ_watters and 3 others
erobz said:
I think I've over complicated my first approach.

The input power ##P_{in}## should simply be given by:

$$P_{in} = \frac{1}{\eta_o} \left( F_s + \beta v_o^2\right) v_o$$
Still too complicated in my opinion.

If we are being asked for a percentage improvement in gas mileage then ##v_o## is irrelevant, ##\beta## is irrelevant and ##{v_o}^2## is irrelevant.

All that matters for the mileage ratio is the power ratio.

If we are holding velocity fixed, all that matters for the power ratio is the force ratio.

Per the model you have chosen, we are reducing the "drag" portion of the total force by a fraction of 4/7 and leaving the static ("s") portion unchanged.

So the mileage would improve by a ratio of: ##\frac{F_s + F_\text{drag}}{F_s + \frac{4}{7}F_\text{drag}}##

We could express the same formula a bit differently if instead of knowing both retarding forces we knew only their ratio. Call it ##S## for how "streamlined" the car is (at the selected cruising speed) with ##S = \frac{F_s}{F_\text{drag}}##.

Then the mileage improvement ratio would be ##\frac{S + 1}{S + \frac{4}{7}}##

In the limit of high ##S## (very stylish car with all the brakes locked hard on), air resistance is not even a factor and mileage improves negligibly when you reduce air resistance by a factor of 4/7.

In the limit of low ##S## (a huge sail on a cart with high pressure tires and teflon-coated ball bearings), air resistance is everything and you get a 7/4 improvement in mileage.

If ##S## = 1 then you get 27% improvement in fuel economy.

erobz
jack action said:
And if you were doing 5 L/100km before, you are now doing 4.65 L/100km.

Edit: Redone the awful math I did to something that made sense.
That seems to small. For 42% reuduction of Cd only 7% drop in f.c.

jbriggs444 said:
In the limit of high ##S## (very stylish car with all the brakes locked hard on), air resistance is not even a factor and mileage improves negligibly when you reduce air resistance by a factor of 4/7.

In the limit of low ##S## (a huge sail on a cart with high pressure tires and teflon-coated ball bearings), air resistance is everything and you get a 7/4 improvement in mileage.

If ##S## = 1 then you get 27% improvement in fuel economy.
"High S" is car with lots of drag from tyres, transmission but low aero drag?
What is most often ratio between aero drag and rolling+transmision drag in passenger car at let say 100-150km/h?

Keep in mind that 0.35 is very bad aero, this cars looks like brick
0.20 is very low, sleek cars, modern electric cars...so difference from 0.35 and 0.20 is huge

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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.
He claim he improve Cd from 0.31 to 0.17

He improve civic top speed from OEM -152km/h to 225km/h?
that seems unreal
152km/h(42m/s)

F=75kW / 42
F=1777N

Cd=1777N /0.5 x42(square) x1,84m2 x1,225
=0.89
Cd from math is too big, top speed is too small or other resistance forces are not negligible)something is wrong here, car with 75kW must have higer top speed, 152km/h is too low

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Jurgen M said:
That seems to small. For 42% reuduction of Cd only 7% drop in f.c.
I took numbers out of thin air, but the equation shows that it depends on what the 5L/100km is used for. A lot is lost in mechanical inefficiencies or other uses (like A/C for example). That is why you need to refer to the actual power output of the first scenario.

If you imagine that the entire 5L/100km of fuel is converted to fight the drag force only - no other losses - then then ##P_1 = \frac{1}{2}\rho C_{d\ 1} A v^3## and the whole equation reduces to:
$$\frac{\left(\frac{dV_f}{dx}\right)_2}{\left(\frac{dV_f}{dx}\right)_1} = 1 - \frac{C_{d\ 1} - C_{d\ 2}}{C_{d\ 1}} = \frac{C_{d\ 2}}{C_{d\ 1}}$$
Or a ratio of 0.5714 in our case, which corresponds to the 4/7 (##= \frac{C_{d\ 2}}{C_{d\ 1}}##) of @jbriggs444 in post #22.

This equation and the one from @jbriggs444 just show that your fuel consumption reduction will be somewhere between 0 and ##\left(1- \frac{C_{d\ 2}}{C_{d\ 1}}\right)## depending on the actual portion of fuel used to fight the drag force in the first scenario.
Jurgen M said:
Cd=1777N /0.5 x42(square) x1,84m2 x1,225
=0.89
Cd from math is too big, top speed is too small or other resistance forces are not negligible)
Answer: Other resistance forces are not negligible.

Your calculations just demonstrate my preceding point.

There exists a large body of research into vehicle fuel efficiency. Much of that is summarized in this online calculator: https://ecomodder.com/forum/tool-aero-rolling-resistance.php.

Engine efficiency varies with RPM and throttle opening. Engine efficiency is measured as BSFC, Brake Specific Fuel Consumption, and shown in BSFC charts. Some example BSFC charts: https://ecomodder.com/forum/showthread.php/bsfc-chart-thread-post-em-if-you-got-1466.html.

The top speed of a vehicle is dependent on the power available at the top speed. A vehicle designed for top speed will have gear ratios such that the engine will be at peak power RPM at the top speed. A vehicle designed for gas mileage will have gear ratios such that the engine will be at the best efficiency for the power needed at a particular speed.

The total drag on a vehicle can be measured by means of a coastdown test. The results can be used to calculate both aerodynamic drag (proportional to ##velocity^2##), and rolling resistance. An example of such a test: https://ecomodder.com/forum/showthread.php/coastdown-test-06-gmc-canyon-20405.html.

Jurgen M, russ_watters, Lnewqban and 2 others
jrmichler said:
The total drag on a vehicle can be measured by means of a coastdown test. The results can be used to calculate both aerodynamic drag (proportional to ##velocity^2##), and rolling resistance. An example of such a test: https://ecomodder.com/forum/showthread.php/coastdown-test-06-gmc-canyon-20405.html.
How do you get aero drag and rolling drag separetly if meassurement include both?
You can only separate them if you know Cd and frontal area of car that you meassure...

How you exclude car inertia, car that has higher mass will travel longer.
Can you post math of that calculation?

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Jurgen M said:
How do you get aero drag and rolling drag separetly if meassurement include both?
You can only separate them if you know Cd and frontal area of car that you meassure...
Your measurements of drag will be of drag versus velocity. One can model that as a quadratic function and separate out the coefficient on the squared term separately from the coefficient of the linear term and for the constant term. ##F = av^2 + bv + c##. You have a table of measured values for ##F## versus ##v## and you want to estimate ##a##, ##b## and ##c##.

That would be a reasonable task for a least squares regression to a quadratic fit. Excel can do it. The first link I Googled up (no guarantees on quality) was: https://www.statology.org/quadratic-regression-excel/

The "least squares" approach is very commonly used when you have data that you think fits a particular family of functions but you are not sure what parameters identify the one "best fit" function out of that family. It is based on a measure of "best fit" which is the sum of squared differences between the selected function and the measured data points. This choice of "best fit" is well behaved mathematically and makes it straightforward to calculate the appropriate set of parameters. [Other choices for "best" such as "sum of absolute values of the differences" are not as easily solved]

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Jurgen M said:
I know that but civic with102HP must has top speed at least 170km/h..I talk from exprience..
or his car has problem with engine..

https://www.auto-data.net/en/honda-civic-vi-hatchback-1.4i-s-90hp-12251
https://www.auto-data.net/en/honda-civic-vi-hatchback-1.5-vtec-e-114hp-12252
The 1992 USDM (US domestic market) CX model (which is the model in our case) is rated at 70 bhp. The engine was later changed to one from a VX model of an unspecified year (although 1992 was rated at 92 bhp).

Although a more representative link from your website shows a top speed of 170 km/h, this link advertises a top speed of 155 km/h for a 1993 CX model. I personally doubt that dropping maximum power from 92 bhp to 70 bhp will yield a drop from 177 km/h to 170 km/h in top speed.

Jurgen M said:
How do you get aero drag and rolling drag separetly if meassurement include both?
You can only separate them if you know Cd and frontal area of car that you meassure...

How you exclude car inertia, car that has higher mass will travel longer.
Can you post math of that calculation?
Empirical determination of coefficients for aerodynamic drag and rolling resistance

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..

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?

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.

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Lnewqban
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.

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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