Could one by using only 3 coefficients of air resistance, one for each axis $x$, $y$ and $z$, model the resistance on an object from any angle? So for example if one has a square, could one summarise air resistance by having a vector for each face of the square? I do not think it would work exactly as described, but is there any way one could calculate drag on an object from any angle?
$\begingroup$
$\endgroup$
1
-
2$\begingroup$ Pedantry: you can model any physical phenomenon in any way you want. The real question is whether your model actually corresponds to the real world. $\endgroup$– Michael SeifertCommented Jun 29, 2017 at 13:16
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
1
I don't expect that to work well.
The coefficient for drag is based on the geometry of the object, and is experimentally determined (and can potentially change if flow conditions are outside of the scope of how the value was experimentally determined).
In the simple example of a cube, we know that the drag equation is $$F_d = \frac {C_d \rho u^2 A} 2$$ The problem will be the value of $C_d$ (drag coefficient). As you can see here the drag coefficient for a cube face first into the flow is $\approx 1.05$, yet when it is from the side it is $\approx 0.80$. I do not see any direct analytical correlation between the two, including when the changes in size are accounted for.
It would be even worse for more complicated shapes where opposite faces in the same axis could have completely different geometry (one side flat, the other side rounded for example), which would mean you would at least need 6 coefficients, one for every face, not just for each axis.
But yet again, that should not work in practice, as the drag coefficient is experimentally determined (I remember I once tried to find an equation to determine how drag coefficient changed with respect to angle; my luck was extremely limited and the results were questionable).
-
$\begingroup$ Sadly enough that was what I expected. I'll leave the question open for one more day. $\endgroup$– YadesesCommented Jun 29, 2017 at 13:51
$\begingroup$
$\endgroup$
For laminar flows around rigid objects, where there would be some hope of the problem being approximately linear even if the shape is complicated, I'd expect you would have to specify at least six components of some tensor to relate the object's orientation (three degrees of freedom) to the direction of the drag force (also three degrees of freedom). For example, consider an airfoil where the force imposed by airflow is not antiparallel to the incident velocity --- don't be confused by the mathematical tool of breaking this force into components and labeling them "lift" and "drag" as if those were different things.
For turbulent flows, I'd expect the problem to be nonlinear. You might be able to construct a "drag tensor" at one velocity, but expect something different at a different velocity.
Now consider an object where the aspect ratio is very large, like a sheet of paper, moving at modest velocity. Turn it one way, like a flag flying, and you have laminar flow. Turn it a different way, like a parachute, and you have turbulent flow. At some angle between these you go through the chaotic transition to turbulence. This is not the sort of thing you can model effectively with three or six or nine coefficients.
