What makes you think I'd want to respond to your OP? Narcissism? You've asked a question that I've answered dozens and dozens of times, you've not put any new angle on it, you aren't smart enough after over 350 posts to use the forum search function, you still haven't found the flat earth wiki and you are completely devoid of humour, empathy or manners.
Like you I come to this site for a bit of cerebral stimulation and hopefully some light entertainment. Can you imagine my face after reading that boring tripe that you've tried to dress up as a fascinating question? I will happily avoid your threads.
OK I'm sorry if I was mean. It wasn't appropriate for me to mention your name and your post on laminar flow over the earth was very productive and interesting. I've taken your name out of the OP, I hope you won't stay mad

Having said that, I still think that what Thork has suggested in that thread is completely implausible, and since this is also related to the discussion at hand I'll go through it here:
(This is all rather rough so if you spot a mistake please let me know)
At first glance, I would guess the Reynolds of the UA would have to be in the order of at least in the billions based on our observations of the turbulence of atmospheric flows on earth. Take the earth, with a radius of 6,371 km. This gives it a projection area of according to a round earth model of;
A = pi * 6371^2 = 1.28e14 square metres (let's assume this RE convention to be correct for now, we can vary this number by an order of magnitude later to gauge its impact on the result)
Assuming a coefficient of friction (C) for a sphere of ~0.5, we have a force exerted of
F = 0.5 * rho * C * v^2
where v is the velocity of the aether and C is the coefficient of friction:
F = 0.5 * rho *0.5 * v^2
Assuming that the flat earth weighs "about" as much as the round earth (so in the order of 6e24 kg), this gives (let's assume gravity is 10m/s^2 for simplicity):
F = ma = 6e24 * 10 = 0.5 * rho *0.5 * v^2
This gives 2.4e26 = rho*v^2
Now taking a simple rule-of-thumb reynolds number calculation
Re = rho * v * L / mu
We know that rho*v = 2.4e26/v (from the above result), so that gives:
Re = 2.4e26*L / (mu * v)
Taking the diameter of the earth (around 10 million metres) as a reference length;
Re = 2.4e33 / (mu*v)
Assuming that the aether can't be travelling faster than the speed of light:
Re = 2.4e33 / (mu*c) = 2.4e33 / (mu*3e8) = 8e24 / mu;
Assuming that the UA is as viscous as water:
Re = 8e24 / 1e-3 = 8e27
That is an absolutely ridonculously large number. Given the scale of this number there is little point in testing it for the effect of small changes to the assumptions (such as the projection area of the earth).
There is no way that flow is laminar, even if the UA was made of solid titanium (however that would work). The above is assuming that the UA is travelling at the speed of light relative to the earth, is as viscous as water and the earth has a mass equal to the RE model. The earth would literally have to weigh less than a bacterium for that flow to be laminar. I haven't taken the lorentz transformation for the velocity of the flow into account.
Flows above 300,000 reynolds number tend to become turbulent if they are very viscous, but anything above 10 million is almost certainly turbulent. I don't think there is any substance that can remain laminar above 50 million, not to mention 1 billion. I don't even know what 10^27, but it's certainly not laminar!
Based on this, and my own intuition as an engineer specialised in transition flow hypersonics, there is absolutely no way that flow would not be turbulent. Anything above us should not be affected by the UA based on our understanding of turbulent flow and the possible materials the UA could be made of.