Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

This is a student inquiry sparked by sheer curiosity.

Wikipedia states the drag equation, $F = 1/2v^2pC_dA_c$.

(p = mass density of fluid/gas, v = velocity, c_d = drag coefficient, a_c = cross sectional area perpendicular to velocity.)

Wikipedia claims that this equation is only accurate under certain conditions:

The objects must have a blunt form factor and the fluid must have a large enough Reynolds number to produce turbulence behind the object.

Does there exist a more general equation that can accurately measure the drag of all objects, at all speeds/Reynolds numbers and considering all physical properties affecting the drag?

share|improve this question
    
I do not know how accurate this model is, but I found this model developed by N.J. Giordano which quantifies the drag force acting on a baseball. –  fibonatic Feb 5 at 21:03
add comment

2 Answers

up vote 4 down vote accepted

The only way to determine the dynamics of the system (Newtonian fluid exerting a drag on a rigid object) in full generality, for all geometries and Reynolds numbers, is to solve the Navier-Stokes equations with the appropriate boundary conditions. These equations are at heart nothing more than local expressions for conservation of mass, momentum and energy underlying all of classical mechanics.

However, this is often more time-consuming than is warranted, since there exist far simpler expressions for drag that apply in certain geometrical and viscous limits. Even when your system does not precisely match the appropriate limiting conditions, you can often get a useful approximation by modeling your system as such. But if you need more accuracy, you can't avoid the full problem.

You've already mentioned one useful limit in your question. Another applies to the low-Reynolds number case, and is referred to as Stokes drag. Notice that in this case the drag is linearly proportional to the velocity, rather than proportional to the square of the velocity as in the high Reynolds number limit.

Given these two limits, one useful approach could be to write your drag force as $C_1 v$ + $C_2 v^2$ and then perform an empirical fit to find $C_1$ and $C_2$. You'll have to be careful, though, if you are working with non-steady flow, since $C_1$ and $C_2$ might then depend on time (note that this has already been pointed out in D.W.'s answer, but hopefully it is now more clear why this is often effective).

Caveat: If your fluid is non-Newtonian, then the situation can be even more complicated, since the simplest notion of viscous drag no longer applies.

share|improve this answer
    
So ultimately the Navier-Stokes method is the best general approximation for a fluid-object system (setting aside that it's extremely tedious)? (Bear with me as I'm a community college freshman.) –  Armend Veseli Feb 5 at 23:48
    
Yes, at least for Newtonian fluids. The object provides boundary conditions for the motion of the fluid, which is governed by the N-S equations. These equations can be quite challenging to solve in general, and often require computers. For this reason, it is useful to start with a simple model to approximate the system in question, and then decide if more accuracy is needed. –  kleingordon Feb 5 at 23:51
1  
@ArmendVeseli To put the Navier-Stokes equations into perspective -- if you had a grid with $N$ points in it, to solve the Navier-Stokes equations directly (no models, called Direct Numerical Simulation, or DNS) then $N \propto Re^{9/4}$. So if you wanted to calculate the flow over a wing at flight conditions where $Re \approx 1e6$, you would need something on the order of $1e13$ points. If all you did was store the 5 variables (density, 3 velocity, energy) this would require approximately 72 TB of memory. And that doesn't even include the grid (which makes it 8 variables at each point)! –  tpg2114 Feb 6 at 2:14
    
So it's not that it's extremely tedious to solve -- it's impossible with current technology to come anywhere close to real problems! –  tpg2114 Feb 6 at 2:15
add comment

I don't know ALL that much about this, but I can offer some more insight in to drag if you'd like it. And seeing as how you asked out of general curiosity, putting in my two cents, however broad, doesn't seem too useless.

The normal differential equation for acceleration in one dimension looks like this:

$\ddot{x} = C$,

where $\ddot{x}$ is the double derivative of position with respect to time, and C can be any constant (0 for no acceleration, -g for a falling object, etc.)

Now, if there is drag on an object, there is some factor that is working against the acceleration of the object to slow it down. We know one basic thing about this drag, it is proportional to velocity in some way. In undergraduate Classical Mechanics, we learn that there are basically two different forms for this drag - linear and quadratic drag. Here's what those terms mean:

Linear Drag:

$\ddot{x} - \alpha \dot{x} = C$

Here the drag force is proportional to velocity. The faster the object moves, the more drag it experiences. It turns out that this model of drag works well for certain objects/systems. However, the force that you mentioned for drag is quadratic with respect to velocity. Let's talk about that next.

Linear AND Quadratic Drag:

$\ddot{x} - \alpha \dot{x} - \beta \dot{x}^{2} = C$

In this case there are two different drag related effects on the object, one in proportion to velocity and one in proportion to velocity squared.

Now, it turns out that for most objects, the drag, that the object experiences can be accurately modeled by choosing the correct values for $\alpha$ and $\beta$. The differences result from many factors including the cross sectional area, shape, etc, of the object.

It's much more complicated than this obviously, but that's the small piece I know about drag.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.