# Car Drag Formula Varies with Speed?

1. Apr 20, 2014

### Vdrela

Hi,

I've read in a website that the drag (air resistance) formula for a car would be different:

-> for velocities under 86 km/h
Drag = K(coefficient) * V

-> for velocities over 86 km/h and under 1200 km/h
Drag = K(coefficient) * V^2

I've researched a little bit and only one formula is generically used to describe drag, the one that considers V^2:

D = 1/2.p.V^2.C.A

D-drag
p-density of fluid
C-drag coefficient
A-area

I understand that this is only one of the accepted representations, and that it is only applicable under certain conditions. Nevertheless, I am looking to understand very simplistic if the claim that under roughly 86km/h we should use only V, and after 86km/h we should use V^2 is valid.

The truth is that I'm finding it a little bit difficult to understand why the formula would change from 86 km/h onwards. The only reason I can conceive the formula to change would be empirical/experimental testing. But even so, what could happen at roughly 86 km/s to suddenly change the drag formula to consider V^2 instead of V? Is this some sort of mystery?

* I've also heard about the velocity being 100km/s or 60mph, instead of 86 km/h, but again, I would just like to understand if the formula changes and why would that be.

Thanks

2. Apr 20, 2014

### Staff: Mentor

The different formulas correspond to laminar flow (low velocity) and turbulent flow (high velocity). The difference between both can happen quite suddenly, but not at the same speed for all car models and not even for all sides of the car. The 86km/h are probably some experimental result to approximate the transition region. The scaling with v for slow cars and with v^2 for fast cars is physical and comes from two different drag mechanisms.

3. Apr 20, 2014

I agree that the result is apparently empirically derived, but it has nothing to do with laminar versus turbulent flow. Using the empirical drag formula, laminar and turbulent flows still show a $v^2$ dependence. Stokes flow is an example where there is a dependence on $v$, but that doesn't apply here.

4. Apr 20, 2014

### AlephZero

Can you give a link to the website where you read this?

The break point of 86 km/h doesn't seem right for typical road cars, and claiming that a formula is correct for a car traveling at 1200 km/h is hypothetical, unless they are talking about the http://en.wikipedia.org/wiki/ThrustSSC.

5. Apr 21, 2014

### Staff: Mentor

Where does the proportionality to v come from then?

6. Apr 21, 2014

### 256bits

Reserved for low Reynolds number for a small object in a viscous medium such as when encountered during sinking, or perhaps the flow of air over a slightly heated wire. The actual physics would not be all that applicable for a racecar.

explanation in next post.

7. Apr 21, 2014

### Staff: Mentor

Sure, that's the physical process. But if that process is not relevant for a car*, why does that approximation exist for the drag?

*okay, it is really not relevant, I checked the numbers. cm/s...

8. Apr 21, 2014

### 256bits

No. It is not a mystery.
No doubt your website has used curve fitting to match the data they obtained and would be applicable for only a certain type of car.

You should know that linear equations are easier to use then any other type, and engineers have a lot of linear equations they use to model a situation. Depending upon the amount of error one can be tolerant of for the analysis, if the linear model gives 5% or 10% error over using a cumbersome exponential or power equation, then that could be acceptable. If one needs more precision or accuracy then the cumbersome equation is the one to use.

You found this equation for drag:
which is similar to these equations depending upon the velocity:
and
.

What they do is group the other terms in the first equation into a value for K. I doubt if K has the same value for both of the second equations.

http://www.princeton.edu/~asmits/Bicycle_web/blunt.html
In this picture, you can see that Cd ( for equation D = 1/2.p.V^2.C.A ) is not a constant but depends itself upon the Reynolds number, and thus velocity of the object through the viscous medium.

You can see that if you want the really complicated approach then calculate Reynolds number, and match that to the coefficient of drag for every velocity the object goes through.

A less complicated approach is to pick a representative Cd for the object at a representative velocity and use that for the calculation, with some error of course at velocities outside of the picked Cd.

An even less complicated approach is to use the simplified linear equation from your site for velocities below a certain picked value of velocity, with most likely some acceptable error.

As a side note, one can approximate any part of a curve with a linear equation , y=mx + b.
As an exercise for you, try an approximation of the the sine curve along its less curvy part, up to a certain value of x. And then try to approximate the top curvy part with a curve of y = mx^2 +b.

Last edited: Apr 21, 2014
9. Apr 21, 2014

It is hard to say with certainty without seeing the OP's source, but most likely it is simply an artifact of trying to represent some seriously complex physics with an oversimplified linear equation, as alluded to by 256bits. Most likely, below that speed for the specific vehicle in question, there is something else going on, like a moving separation point along the surface of the car or something else that in this case has a dependence on $v$ so that essentially you have the drag following some rough law of $K_1 v^2$ and some other attenuating phenomenon following $K_2 v$. When you combine the two effects and let $K = K_1/K_2$, you have something like $K v$.

Again, though, without seeing the source, it is difficult to say.

10. Apr 21, 2014

### Staff: Mentor

I guess the point is that the drag coefficient CD is a function of the Reynolds Number, and the Reynolds Number is proportional to the velocity, so the drag coefficient is a function of the velocity.

Chet

11. Apr 21, 2014

Well that's true, but that Reynolds number dependence is accounted for in the drag equation for situations in which it applies such that you only need to know CD at the current conditions to get the "correct" drag. For most simple shapes at reasonable Reynolds numbers, the drag force scales (roughly) with $v^2$ (or really with the dynamic pressure). For more complex shapes like a car, the approximation simply isn't a good one for the simple drag equation.

Now, it is probably that in the OP's source, the people simply used their own equation, finding a coefficient that scales directly with velocity rather than the dynamic pressure (thus rolling the drag coefficient and density up into a single paramter, $F$). Most likely they just found that for that particular car, the drag scaled nicely with dynamic pressure as is typically the case, and for lower speeds, there were simply more complicated physics taking place than could accurately be modeled by something like the drag coefficient and it scaled better with just the velocity.

12. Apr 24, 2014

### Vdrela

Thank you for your quick replies.

The source is in Portuguese(Brazilian), maybe you'll understand if you use i.e. Google Translate.
http://www.fisicaevestibular.com.br/Dinamica10a.htm

I got there randomly because I heard that the drag formula would change from V to V^2 after a given velocity, let's say 60 mph, and I was trying to understand why would the formula change if there is no change in shape, viscosity, or density of air...

I can understand that maybe at 1200 km/h (approximated speed of sound) the formula V^2 no longer applies. Don't know if it is true or not, but at least it could make sense to me if it was true. However I can't really understand what would make the formula change at 60 mph.

And yes, in the formula Drag = K * V or Drag = K * V^2, I believe K stands for 1/2.p.C.A, which means that the following formulas are equivalent:

Drag = K * V^2 <=> Drag = 1/2.p.C.A.V^2

So very simplistically the question is still there for me. Does the following formula apply to all speeds?
Drag = 1/2.p.C.A.V^2

13. Apr 24, 2014

It can apply to all speeds, but doesn't necessarily. Depending on the Reynolds number, which is a nondimensional combination of density, velocity, size and viscosity, the drag may be proportional to $v$ or $v^2$. For very low Reynolds numbers ($Re \ll 1$), the drag is quite simple and is proportional to $v$. For pretty much any other situation, it is proportional to $v^2$.