On Fri, 25 Apr 2008 00:48:45 +0200, Stefan
wrote in :

WingFlaps schrieb:

Try reading the statement again, here it is:

"Now we add in the energy losses from having to accelerate with the
wind and to glide speed."

Now perhaps you would like to revise some physics and try to critcise
it for us?

It's the "having to accelerate with the wind" part which is complete BS
unless I completely misunderstand what you are trying to say.


Here's what a physics professor, pilot and author of Performance of
Light Aircraft* had to say on the subject of downwind turns when he
was active on Usenet:



http://groups.google.com/group/rec.a...41feb8649bbe12
Nov 22 1996, 1:00 am

The Kinetic Energy "Paradox"

Several recent rec.aviation.xx threads have featured or involved the
apparent paradox that an aircraft which turns downwind thereby
undergoes a large gain of kinetic energy (KE) w.r.t. (with respect
to) the earth but no gain of kinetic energy w.r.t. the air mass.
Initial
(incorrect) physical reasoning is often along these lines: there is
only a constant velocity (that of the wind) difference between CSs
(coordinate systems) fixed w.r.t. the earth and w.r.t. the air. Hence
time rates of change of velocities (accelerations) are the same in
the two systems (the derivative of a constant being zero); hence so
are the forces the same in both (the mass never varies). Since it
takes a difference in forces to produce a difference in kinetic
energies, there SHOULD be no such KE difference. But there is.

The error is in the seemingly innocuous statement that two
identical forces F(t), t the time, produce the same KE difference.
The solution is to see that even though the forces, as seen in the
two CSs, are identical, the DISTANCES moved under those
forces, and the parallelism between force and displacement, are
NOT the same. Delta_KE = Int(F dot dr) = Int(F dot v)dt=F dot
vdelta_t; Int' stands for integral, F and dr and v are vectors and
dot' is that of the dot or scalar product, the product of the two
flankers times the cosine of the angle between them. Angular
brackets, as X , stand for the average of quantity X. If one keeps
track of the displacements, and their orientation with respect to the
forces, which is what vector analysis does, then the kinetic energy
difference computes properly and, as always, there is NO
PARADOX. The truth is that the forces ARE the same, but the
kinetic energy changes w.r.t. the two systems are NOT the same.

Here's a simple example. A jet is heading North, unit vector i, at
speed V w.r.t. the earth. The stewardess, Linda, of mass m, is
walking South at velocity -si w.r.t. the jet. V might be 600 mph, s
might be 4 mph. She hands a copy of Mountain Pilot magazine to a
passenger, reverses course, and walks North at velocity +si.

W.r.t. the jet, Linda has undergone a momentum change delta_p =
2msi. And, also w.r.t. the jet, a KE change delta_KE = 0.

W.r.t. the earth, Linda has undergone a momentum change delta_p
= 2msi (the same as w.r.t. the jet). And, also w.r.t. the earth, a KE
change delta_KE = (m/2)*(V+s)^2 - (m/2)*(V-s)^2 = 2mVs. Not at
all the same as w.r.t. the jet.

The force of the deck carpeting on Linda's pumps, F(t), is in the +i
direction during her turnabout maneuver; some sort of Gaussian
peak over a relatively small time interval delta_t. Ignoring the
small walking speed difference, her speed w.r.t. the earth
throughout that maneuver was essentially V. (A relatively tiny bit
less before she got stopped, a relatively tiny bit more afterwards.)
So F dot V = FV (they're in the same direction). Now her
momentum change is Int(F dt) = F delta_t = 2msi. We can use
this fact to evaluate Fdelta_t, plug it in the average of the
integral formulation in the second paragraph, and find that indeed
her delta_KE = 2msV, just as calculated by taking simple
differences.

The short time intervals, integrals approximated as averages, etc.,
are only incidental features which have no bearing on the essential
argument. (Not any more than accelerations at the end and
beginning of one twin's journey at 0.9c has any real bearing on the
twin "paradox" in special relativity.) The only trouble with treating
the original airplane case, which I've done, is that you get two and
a half pages of fairly dense vector calculus and the Internet is not
yet sophisticated enough to let us put that across uniformly in this
kind of forum.

The result: though the forces are the same in the two coordinate
systems (air mass and earth), the greater distance travelled in the
earth-based system, and the fact that those displacements in space
are not perpendicular to the forces (as they always are in the air
mass based system), means that indeed kinetic energy IS gained in
a downwind turn w.r.t. the earth. Here's the final result:
Delta_KE(w.r.t. earth) = -m*omega*R*Vw*(cos(omega*tf)-1),
where omega is the angular speed (yaw rate) in radians/sec, R is
the radius of the turn w.r.t. the air mass, Vw is the wind speed, and
tf is the time of flight. When you do a 180-degree turn, tf = T/2,
where T is the time needed to make a circle w.r.t. the air mass.
Then delta_KE = 2m*omega*R*Vw. If you continue around to
make a 360, you find, as intuition suggests, that delta_KE(w.r.t.
earth) = 0.

NO paradox. There never is. There's only (hopefully, temporary)
CONFUSION. Physics is wonderful.
-------------------------------------


http://groups.google.com/group/rec.a...c4f35133062386
Jun 25 1996, 12:00 am

Here's a posting I came up with this afternoon which *SHOULD* excite
some discussion. At the moment, however, it is also
my genuine opinion.
John

You Do Lose Airspeed in a Downwind Turn

John T. Lowry, PhD
Flight Physics
Billings, Montana
(406) 248 2606

June 1996

I came across a side reference to the infamous "downwind turn
controversy" in an Internet user group posting. Then I located a
reference to that conundrum, in Joe Christy's "Good Takeoffs
and Good Landings," 2nd edition. On page 11 Mr. Christy writes:

...there is no difference between downwind, upwind, or
crosswind turns in flight. They are all the same to the
airplane.

It seems that beginning flight students think that they will
need to add power or angle of attack when turning from upwind
to downwind because of the loss of airspeed inherent in the
reversal of direction. Later on they learn the more sophisti-
cated "truth." As expressed by Joe Christy farther down the
same page:

Remember, when airborne, you are carried *with* a moving
air mass independently of your movement *through* it.

His italics shown with *XX*. So the conventional wisdom is that
there is no difference, except for the name, between a turn from
downwind to upwind and one from upwind to downwind. Let's leave
it at that for the moment.

And let's change the subject; to that of horizontal wind shear,
or gusts. Say you're heading into the wind at 100 KTAS and at
first (Wind A) the wind is 40 knots. Suddenly (Wind B) there is
a slackening gust to only 10 knots. (All wind speeds are ground
speeds, speeds with respect to the earth just below them.) What
happens? Your airspeed suddenly drops 30 knots to 70 KTAS. (For
the moment, your 60 knot ground speed stays the same.) Let's say
you take no immediate pilot control action. If your airplane is
properly trimmed, the lower airspeed results in the airplane
both losing altitude and nosing over. Airspeed begins to
recover. When the original 100 KTAS is regained, you're back
into trim. Albeit at a lower altitude. The point is: your air
speed really did diminish. In fact there's a caveat to the
conventional wisdom in a figure caption on that same page of
Mr. Christy's book:

To an airplane in flight, there is no difference between
turning downwind, crosswind, or upwind, *except when
flying through a wind shear*.

This time, my (**) italics. So at this point we have it that
turning towards downwind does not result in loss of air speed
but flying into a slackening gust does result in loss of
airspeed.

But here's the kicker. *Turning to downwind is the same thing
as a slow wind shear*. Say you start a coordinated turn from
upwind towards downwind at standard rate, 3ø per second. At the
start, say you have a 30 knot headwind. A second later you have
a 30*cos(3ø) = 30*0.9986 = 29.96 knot headwind. It's not much
less, but it *is* less. After turning for 10 seconds you've
turned 30ø and the headwind has slacked off to "only"
30*cos(30ø) = 30*0.8660 = 25.98 knots. That's getting to be at
least marginally perceptible, but remember the effect has been
continuously spread over ten seconds. After a quarter turn from
headwind to crosswind, 90ø, which took 30 seconds, your relative
headwind has changed by 30 knots.

That 30 knots is almost precisely the size (50 ft/sec) of the
FAA's standard gust. But the FAA takes it that the gust develops
its full force over only one quarter second! There's the
difference. The same wind change, spread over a time interval
30/0.25 = 120 times as long and at a time while you've already
increased angle of attack to counteract your banked and
therefore off-vertical lift vector, and possibly also increased
power a little to counteract increased induced drag and
increased trim drag in the turn, is essentially imperceptible.
The time scales are vastly different.

So you do lose airspeed in a downwind turn. Only, normally, not
much.

I can imagine that students and instructors of aviation have
infused their hangar flying sessions on this subject with
several additional scientific red herrings. The Internet
posting I mentioned even got into general relativity. Here are
the more likely possibilities I can think of.

Possible Red Herrings in the Downwind Turn Controversy

1. The air mass in which the airplane is embedded is an inertial
frame and so . . . .

If the air mass is steady, that's so. But if you fly from Wind A
into slower moving Wind B, you're talking about two *different*
inertial frames. Your air speed is your speed through the air in
which you're currently flying, not your speed relative to the
air a mile back. An airplane can change air speed quite rapidly
because the speeds of air molecules can vary considerably from
one place to another place fairly near by. That is, gusts exist.

2. Air speed is air speed, so . . . .

Air speed is speed through that air *in the direction the
airplane is pointed*. Say I told you an airplane was progressing
northward against a 30 knot wind out of the north and its ground
speed was northward at 70 knots. You'd probably say its air
speed was 100 knots. But what if I told you, in addition, that
that airplane's nose was pointed directly to the *south*. (The
ultimate skidded flat turn.) Tilt! Unfair! Now the air speed has
suddenly become -100 knots. The concept of air speed takes a bit
more refinement than most of us give it.

3. The earth isn't actually an inertial frame of reference
anyway . . . .

True. But the non-inertial terms (due to rotation about its axis
and revolution about the sun) are known. And quite small. So for
all practical intents and short term purposes, the earth is an
inertial frame. Therefore, a force is necessary to change the
speed of the airplane relative to the earth. Even if the
airplane flew into a vacuum, it would *initially* still move at
the same speed in the same direction, just as does a tennis ball
you drop out the window of a speeding car. Then, as the changed
and unbalanced forces took over, the airplane would accelerate.

In summary, my answer to the downwind turn controversy is: You
do lose air speed in a downwind turn, but in a "noisy"
environment and only over a time span that renders that loss
almost imperceptible. Still, I think those beginning students
were right.

COMMENTS? CONTRARY VIEWS?

-------------------------------------

http://groups.google.com/group/sci.p...37f136c766e351
Feb 10 1999, 1:00 am

There is a problem similar to the ball/train/kinetic energy problem in
aviation. When turning from heading upwind to heading downwind, the
airplane apparently has a large gain of kinetic energy (as measured
with respect to the Earth underneath). Even though the forces acting
on the airplane during the turn (aerodynamic + weight) are the same in
both the Air Mass- and the Earth-based systems (uniform motion between
them), the kinetic energies are not. It's because Delta_KE is the
integral of Force dot dr and the space increments dr are NOT the same
in the two systems.

Rather than go into this with pilots I usually simplify to a flight
attendant handing a passenger a magazine, then turning on her pumps
and walking back the way she came from. Similar to the bouncing ball.

John.

John T. Lowry, PhD
Flight Physics; Box 20919; Billings MT 59104
Voice: 406-248-2606
----------------------



http://groups.google.com/group/aus.a...e03975757e040d
Mar 16 1999, 1:00 am

Gavan Cook asked me to address this issue. It's a repost of an older
response on the same subject in rec.aviation.piloting.

For me, the convincing reason there is NO DIFFERENCE
between a downwind turn and an upwind turn is: **all the relevant
FORCES in the turn (except gravity, which doesn't change) are
aerodynamic.** With those forces being only between the airframe
or control surfaces or propeller and the AIR, what possible effect
could anything else have? It's only forces which make
accelerations which are changes in velocity. This of course
assumes the exact same air mass motion (no gusts or wind shear),
the same control inputs, etc., throughout the two (to downwind or
to upwind) turn maneuvers.

Now the airplane DOES slow down (in air speed) in a turn,
due to increased induced drag due to larger lift due to bank, but it
doesn't slow down any MORE in a downwind turn than in an
upwind one.

Phil's insistence on a **coordinated** turn is very
important in a practical sense. Say you turned by kicking hard
rudder, with a real big rudder, so that you skidded around 180
degrees in (say) two seconds. (Strange to say, one airline pilot told
me another airline pilot told him that this slewing-around skidding
turn is just the thing to do!) THEN you've got a problem all right.
You've then got (almost) your original 50 knots of ground speed,
in the same direction (now backwards), plus a 20 knot tailwind.
Seventy knots air speed, but a NEGATIVE 70 knots air speed with
reference to the direction the airplane is pointed. You'd "fall out of
the sky." However, there is no difference in this foolish maneuver
whether it's a turn to downwind or upwind. In the other case,
where you'd initially have 70 knots air speed and 90 knots ground
speed, an instantaneous skidded reversal would leave you at 90
knots ground speed backwards and a 20 knot headwind, net result
still the exact same negative 70 knots air speed with reference to
the direction the airplane is pointed. So still a real bad idea to
skid around, but the SAME bad idea whether to downwind or upwind.

There's no AERODYNAMIC difference between turns to
dowwind or to upwind. The only differences are between their final
ground speeds, and in the perception of the pilot who unwittingly
slows his air spseed when he notices, by looking at the earth,
increased ground speed during and after his turn to downwind.

Hope this clarifies. Sometime later on I'll address the
question "Where does the additional kinetic energy (with respect to
the earth, an (approximate) inertial frame of reference), after a
downwind turn, come from?" It DOES come, even though all the relevant
forces, as argued above, are the same. It's a subtler issue.

John Lowry.
--
John T. Lowry, PhD
Flight Physics; Box 20919; Billings MT 59104
Voice: 406-248-2606

---------------------

*
http://www.bookfinder.com/search/?ac...t%2520aircraft