Coming late to the party here, I've been in Tampa in simulator training I'm trading in the DC-6 for the Herc. Anyway, against my better judgment I'll weigh in here.
First off, I don't take the position that there isn't "bad air" particularly down low. No doubt that with convection, wind gradients, obstacle effects on the wind, and a myriad of other things, some days are better for flying on the ragged edge than others.
That said, it appears from the reading I've done (and I admit that haven't read every post on all 10 or so pages of the 2 threads) that there are folks who believe that at say, 10,000 ft, in a steady, uniform, consistent wind, with no shear, or vertical air motion, you can turn an airplane "downwind" and stall the airplane in a turn which wouldn't stall the airplane if that same turn was performed in still, calm air.
The fallacy comes because it *seems* like the inertia of the airplane will resist the acceleration required to maintain airspeed in the downwind turn, and that the airplane requires more acceleration to maintain airspeed in the downwind turn. This fallacy is seductive, because on the face of it is *seems* so *right* and I can certainly see the appeal of that line of thinking. The trouble is, it is just a fallacy, the aircraft turning "downwind" accelerates at the identical rate as the airplane turning at the same turn rate in still air. Let’s take all the "seems like", "common sense tells you" and other fuzzy thinking out of the equation and analyze what is actually happening, using numbers. It’s not hard to do.
Acceleration is defined as change in velocity per unit of time. So what is the required acceleration?
Take an airplane flying first directly north at 50 knots, then turning at standard rate, constant altitude and airspeed turn 180 degrees to directly south..
We’ll consider only north/south winds, so we need only consider north/south acceleration. East west acceleration becomes irrelevant.
So, what is the acceleration when this is done in still air? Immediately before the turn, the velocity is 50 knots north, and immediately after the turn the velocity is 50 knots south, so the velocity change is 100 knots (168.8 ft/sec). The change takes place in exactly one minute, so the average north-south acceleration during the turn is 100 knots/minute , or 2.81 ft/sec./sec
Ok, now what happens when we make the same turn in a wind? Lets say we have a 25 knot wind out of the north. Now, I’ll consider the groundspeed here, because that is what causes the erroneous perception. In reality, it’s only the airspeed that matters, but the results won’t change if you consider the groundspeed correctly.
What is the groundspeed before the turn? 25 knots north
After the turn? 75 knots south
Net change in goundspeed? 100 knots (The velocity difference between 25 knots one way and 75 knots the opposite direction is 100 knots or 168.8 ft/sec)
Time to turn 180 degrees in a standard rate turn? 1 minute.
Average acceleration ? 100 knots per minutes or 2.81 ft/sec/sec. The average acceleration through the turn, after you add a wind, is identical, right out to however many decimal places you want to carry it out to.
OK, how about a really big wind, surely if we use enough wind the airplane will have to accelerate faster to keep flying speed right? Well let’s take a look. Let’s say we had a 200 knot wind out of the north.
Groundspeed before the turn 150 knots south (50 knots in a 200 knot head wind).
Groundspeed after the turn 250 knots south. (50 knots plus a 200 knot tail wind)
Change in groundspeed 100 knots (168.8 ft/sec)
Time to turn 180 degrees in a standard rate turn? Still 1 minute.
Average acceleration needed to maintain 50 kt. airspeed? Still 100 knots in one minute, or 2.81 ft/sec/sec.
I know what some of you are thinking. You’re saying, well a standard rate turn isn’t a much of a turn, it’s such a gradual turn that the airplane has enough time to accelerate downwind, you need a faster turn to get the "wind blowing backward over your wing causing you to fall out of the sky" effect.
OK, let’s use a faster turn, see how that changes things. Lets use a 45 degree bank. Lets say that our airplane can bank 45 degrees at 50 knots without stalling, and that it has sufficient power to circle indefinitely at 45 degrees bank and 50 knots in a level turn without loosing airspeed. And lets further say that 50 kt is hte slowest this airplane can fly in a 45 degree bank, 49.5 knots and it falls out of the sky. 45 degrees bank at 50 knots will give a turn rate of approximately 22 degrees per second (from the chart on pg. 179 of Aerodynamics for Naval Aviators) That means that the 180 degree turn will take about 8.2 seconds.
So what is the acceleration in still air?
50 knots north to 50 knots south is still a change of 100 knots. The 180 degree turn now takes 8.2 seconds, so the average acceleration needed is 100 knots/8.2 seconds, or 20.6 ft/sec/sec.
Now, let’s see what the acceleration is in the dreaded downwind turn. Let’s skip right to a really big wind and not piddle around with insignificant 25 knot winds. Let’s use the same 200 knot north wind we used before with a standard rate turn.
Groundspeed before the turn? 150 knots south
Groundspeed after the turn? Must be 250 knots south, 249.4 knots across the ground (49.5 kt. airspeed) and the airplane stalls
Total velocity change required. 100 knots (168.8 ft/sec)
Time for velocity change 8.2 seconds
Average acceleration needed to maintain 50 kt airspeed? 168.8/8.2 = 20.6 ft/sec/sec
Huh, turns out even in a really steep bank, really low airspeed, high turn rate and ridiculously huge wind, the acceleration needed to maintain airspeed in the turn with a 200 kt. wind is identical to the same turn in still air.
And that is where the rubber meets the road. All the downwind turn theories depend on "inertia causing the airplane to not accelerate fast enough to maintain airspeed" with the fundamental flawed belief that the acceleration *must* be greater if you turn downwind or you lose airspeed.
However, when you actually analyze what the required acceleration is, by taking the velocity before, the velocity after, and the time of the turn, (change in velocity divided by time is the only valid way you can consider acceleration, because that is the definition of acceleration) we find that the wind makes absolutely no difference in the acceleration required through the turn in order to come out the other side with 50 knots of airspeed.
Now some of you are probably thinking, yes but that is the *average* acceleration, the acceleration varies during the turn. Well, yes, the acceleration in hte north/south direction does vary, but just like the *average* acceleration remains identical out the nth decimal place, regardless of the wind, so does the acceleration at any particular point during the turn. Think about it, the acceleration during the turn changes significantly, but the *average* acceleration remains the same? Not likely. For those of you who are unswayed by the unlikeness of that, we can take a more analytical look at the situation.
Yes, it is true that the acceleration during the turn changes, more specifically, the acceleration in the North/South direction (which is what we’re interested in) changes, the total acceleration during a constant turn remains the same. The north south component of the acceleration will be greatest at the 90 degree point of the turn and the least at the 0 degree and 180 degree points in the turn. So lets look at the point where the acceleration is the greatest, that’s where we’ll fall out of the sky right?
First off, what is the lateral (across the ground) acceleration? Well, it we stick with the 45 degree bank, the lateral acceleration is conveniently, 1 g , or 32.2 ft/sec/sec. I see some of you saying, no, a 45 degree turn is more than 1 g, well yes and no. It gives a "load factor factor" of 1.414 but that’s 1 g of gravity vertically and 1 g of horizontal acceleration, add Pythagorean theorem and that gives you 1.414 g aligned with your butt in the seat. So, 1 g lateral acceleration for a 45 degree banked turn. Now in a zero wind situation, what is our groundspeed (or airspeed, no difference if there’s no wind) at the exact 90 degree point in the turn? Zero, right? That’s the precise point when our groundspeed (and airspeed) is reversing from just a little in the north direction to just a little in the south direction. And right at the reversal, north/south groundspeed (and airspeed) is zero. So let’s examine that one second bracketing the precise 90 degree point. One half second after the reversal, the north/south groundspeed is 16.1 ft sec (9.5 knots) to the south, Acceleration is 32.2 ft/sec/sec, and we’ve accelerated at that rate for 1 half second from zero north/south groundspeed, (32.2 * 0.5 = 16.1 ft/sec). Similarly, at one half second before the 90 degree point the north/south groundspeed (and airspeed) is 9.5 knots in the north direction. So in that second the groundspeed and airspeed changes from 16.1 ft/sec (9.5 kt.) north to 16.1 ft/sec (9.5 kt) south for a total velocity change of 32.2 ft/sec in one second which is an acceleration of 32.2 ft/sec/sec or one g.
Well what if we throw a wind in there? Again, we’ll use the 200 knot wind because that ought to magnify any small effects. So, in a 200 kt wind out of the north, what is the airplane’s north/south groundspeed at the exact 90 degree point? As we saw in the still wind case the north south airspeed goes to zero, so the groundspeed must be 200 kt. (337.6 ft/sec) to the south. A half second before the 90 degree point, groundspeed is 190.5 kts. (321.5 ft/sec) south (200 knots wind to the south plus 9.5 knots airspeed to the north = 190.5 knots to the south) one half second after the 90 degree point the groundspeed is 209.5 kt (353.6 ft/sec) to the south (200 knots wind to the south plus 9.5 knots airspeed to the south = 209.5). So, over that one second centered on the 90 degree point in the turn the north/south groundspeed goes from 190.5 knots south to 209.5 knots south, a change of 19 knots (32 ft/second) in one second when is 32 ft/second/second, which is the same acceleration we had in the no wind condition.
I could go on, and do the same analysis for the 45 and 135 degree points for the turn, or any other point, but the math will start to get complicated, and the answer will always be the same. A constant, unaccelerated wind will not change the acceleration of the plane in the turn, one iota, nor will it change the acceleration required to maintain airspeed as you turn down wind.
... then again maybe I will be perpetually wrong 
