In rec.aviation.student WingFlaps wrote:
On Apr 27, 8:22?am, Michael Ash wrote:
In rec.aviation.student WingFlaps wrote:

On Apr 27, 6:04?am, wrote:
? ? ? ? Lots of people had the impression you were talking about the
dreaded downwind turn, with all the talk about the energy required to
accelerate to maintain airspeed. The energy required, as pointed out
in a much earlier post with several very good references, is so tiny
that it's not worth fooling with at all.

Perhaps you could put a number on that? Could you try a gliding turn
with stopwatch and altimeter and compare that to a straight glide?

In the optimal 45-degree-banked turn the load factor will be about 1.4.
Your best glide speed and min sink speed will increase by the square root
of that, or 20%. The glide angle remains the same if you increase your
airspeed appropriately, so your sink rate will also increase by 20%. So
instead of 650fpm you'll be coming down at 780fpm. At 78kts (65kts best
glide speed from previous post plus 20%) and a 45 degree bank you're
making a circle a bit over 500ft across which will take you 13 seconds to
complete half of. The extra sink rate from the turn will therefore cost
you 30 feet over what you would have experienced in a straight glide for
the same amount of time.

You'll also lose about 80 feet to accelerate from 65kts to 78kts. But
you'll gain this back at the end, so as long as the end of your turn ends
at a reasonable height it can be ignored.

The numbers will, of course, vary between aircraft but it would appear
that the extra energy loss due to the turn itself isn't all that
significant. If 30 feet is the difference between making it and not making
it you probably should not be turning around in the first place.

I make the turn diameter bigger than that using the formula
rad=(knots^2)/(11.26 x tan(bank)) (assuming it's right) or about
1080'?

You're right, I gave the radius, not the diameter, but worded it as though
it were the diameter. My 13 seconds is based on v/(pi*r) though so it
ought to be correct.

On a side note, these equations are generally vastly more comprehensible
if you leave out unit conversions altogether. Turn radius in any circle
caused by acceleration is v^2/a, and here the acceleration is
9.8m/s^2*tan(bank). Anyway, the result is the same, I just find it easier.
Moving on....

So, what would you consider the minimum height taking decision
time into account and a 225 degree turn followed by a 45 to line up
back on the runway?

I really haven't a clue. I don't fly these things and thus don't have the
experience to comment on this. I've heard that you need to be quite high
to be reasonably safe doing it. The physics only gives you a raw minimum.
You need a hefty safety margin on top of that, plus knowledge of your
personal ability to perform close to the ideal, obstacles which may modify
your options, and other such things. Wind, density altitude, engine
performanc, aircraft weight, and other such things will all contribute as
well to change the answer.

My personal decision height for the analogous glider launch emergency is
*usually* a shade under 200 feet. Conditions, performance, obstacles, and
landing opportunities ahead can all modify this value. The physics would
tell you that the altitude required is around 35% of your wingspan,
because the energy from the extra speed you carry is theoretically
sufficient to zoom slightly, make the turn at min sink speed, and roll out
lined up with the adjacent taxiway at the same altitude you started. (If
you don't lose altitude then you just need to be high enough to avoid
hitting your wingtip on the ground, thus the 35%.) In reality you'd have
to be completely nuts to try this maneuver starting from an altitude of
20ft at normal takeoff speeds. And as I mentioned before, this has very
little bearing on the options available to a typical powered airplane.

--
Michael Ash
Rogue Amoeba Software