I don't think that such a computation of a theoretical limit of accuracy is possible. There are several sources of uncertainty in weather models:
initial and boundary data,
parameterizations,
numerical instability, rounding and approximation errors of the numerical scheme employed to solve the Navier-Stokes equations for the atmosphere.
The term "parameterization" refers to the approximation of all subgrid processes, these are all processes/influences that happen at a scale that is smaller than the length of a grid cell. This includes effects from the topography, or the local albedo. More sophisticated approximations to subgrid processes can actually lead to less predictivity of a model, because the needed more detailed initial and boundary data are not available.
The Navier-Stokes equations themselves are usually approximated up to a minimum length scale that is way larger than the length scales that would be necessary to resolve turbulent flows, these kinds of approximations are called large eddy simulations.
The accuracy of this truncation depends critically on the kind of flow and turbulence.
While I don't think that it is possible to derive a theoretical limit, what people do instead is performing ensemble runs, where the results of a weather model are compared that are calculated with slightly perturbed initial and boundary data.
An example of such a "twin" experiment can be found here:
(The result is that the error becomes significant after ca. 15 days of simulated time.)
I think @Killercam is right, I'll try to explain the same thing a little more elaborately.
Firstly. in the case considered, since the fluid and the cylinder is chosen, increase in velocity directly translates to increase in the Reynolds number as $R_e = \frac{\rho V D}{\mu}$.
Before considering flow in the range $250 < R_e < 2\times 10^5$ , lets first observe what happens in the region where the viscous force dominates over inertial forces i.e $R_e <<1$. The fluid slowly "crawls" over the surface of the cylinder. There are 2 stagnation points on the leftmost and rightmost parts of the cylinder.
From the solution for inviscid flow over cylinder (superposition) we can note that tangential velocity is maximum at mid-section and decreases as and decreases as we proceed "downhill"
This can be extended to viscous flow and two important things are to be noted here:
- The shear stress is maximum at the mid-section, which is implicit as a higher velocity gradient is created because of the larger value of tangential velocity.
- The static pressure starts to increase after the mid-section. i.e the pressure is increasing in the direction of flow $\frac{\partial P}{\partial x}>0$ which is called as an adverse pressure gradient
As $R_e$ is increased (i.e velocity is increased), the inertial forces start to dominate over viscous forces.The flow velocity is zero at the surface and the particles very close to the boundary have a very low momentum since they experience very strong viscous forces. On the right part of the cylinder, the fluid particles close to the cylinder not only experience strong viscous force, but also adverse pressure gradient which eventually forces the fluid particles to stop/reversed, causing the neighboring particles o move away from the surface. This is called as flow separation. It results in the creation of a free shear layer which ultimately rolls up to form a vortex.
@Killercam said:
The velocity of the flow divided by the diameter of the cylinder is the typical crossing time of the fluid, hence is directly related to the frequency of the observed oscillations for a specific Reynolds number.
After a vortex is shed, the fluid particles behind have to undergo the same process i.e it takes the same distance to come to rest and then causing the neighboring particles to separate. Since this distance is a small part of the cylinder, $dist \propto D$ and hence the time interval between 2 vortices shed from the same side(top/bottom) of the cylinder $time \propto \frac{dist}{V}$ i.e $time \propto \frac{D}{V}$.
The time interval is exactly the time period of vortex shedding and hence the frequency of shedding is
$f=\frac{1}{T}\propto \frac{V}{D}$
Best Answer
If you're looking for the average wind speed, you might use the Strouhal number: $St=\frac{f L}{v}$
where $f$ is the shedding frequency, $L$ is a characteristic length scale, and $v$ is velocity. You can find charts comparing Strouhal numbers to Reynolds numbers ($Re=\frac{vL}{\nu}$). I realize that both depend on velocity (v), but you know all the other variables ($L$, $\nu$, $f$). An iterative approach will quickly converge on a velocity:
if you're looking for other components ($u$) of velocity due to the shed vortices, i'm sure there are papers that will help you once you determine the average velocity $v$.