A Level 1 AOA clarification

"Greg Esres" wrote in message
...

relatively small fraction of the total weight of the airplane in the
first place, less than 10% in at least some cases, perhaps most
cases)

Lift in a 10 degree climb should be reduced about 1.5%.

Yes. So? Not relevant to the statement you quoted (which was about
thrust).

I'm not sure your theory holds up very well.

"His" theory is mentioned in a number of aerodynamics books.

Fantastic. It would have been nice of you to provide the name of one
popular (i.e. easy to find) one, so that I can read up on it.

[...] If
you had enough unused AOA left to generate a load factor, you could
change the flight path then return the AOA to its original value. The
aircraft may be able to stay on a steeper flight path due to the
reduced parasite drag and reduced effective weight. Don't forget that
thrust will increase slightly with a lower airspeed.

I admit, I didn't consider scenarios where one is taking advantage of
transient changes in drag and lift. Still, there's not much "unused AOA" in
the regime of flight we're talking about, nor did David suggest that might
be required (his implication, to my reading, was that his suggestion applied
generally, not with very specific pilot techniques and situational
characteristics).

To have a vertical component high enough to support the airplane
will require a horizontal component so high that the airplane won't
slow.

Not really clear on what you mean by that.

Yeah, I was posting pretty late. That wasn't clear at all. My point is
simply that I don't see how you can increase thrust enough to support the
airplane significantly, while still managing to slow the airplane down to
theoretically lower-drag steady state. Perhaps the zoom maneuver you
described is the answer to that.

All of the above is very vague. What I hear you say is "I don't want
to believe you." ;-)

Yes, I admit that readily. But the reason I don't want to believe is that
the proposal bears no resemblance to the behavior of any airplane I've
flown, not while I've been flying it anyway.

I agreed up front that my response is as much hand waving as anything else.
But then so is David's. I'd be more than happy to see someone step in with
some real math that shows the answer one way or the other. I don't happen
to be patient enough with the math. There's a reason that, when I was
working on my math degree, I focused on theory and stayed away from numbers.
Topology was my favorite class, differential equations my least.

There are an infinite number of steady states; every time I move the
elevator, I create a new steady state.

It seems to me that in this context, my qualification of "new steady state"
(and David's for that matter) should have been clear. That is, he's
proposing that at the same speed, there are multiple steady states that
produce different amounts of drag.

Lift is always generated perpendicular to the wing's chord.

No, for subsonic flight, it's perpendicular to the *local* relative
wind, the relative wind that is modified by wingtip vortices.

Mea culpa. Still, in a climb (or descent), lift is not being applied
entirely to counteracting weight.

If lift
were perpendicular to the chordline, you would have induced drag in a
wind tunnel, and you don't.

I understand my error regarding chord versus relative wind. Still, I'm
boggled by the lack of induced drag in a wind tunnel. If the wing's not
creating lift (0 AOA), I can see how there wouldn't be induced drag. But
this would happen in the real world too. If the wing is creating lift,
shouldn't there be a measurable force parallel to the relative wind? Even
in a wind tunnel?

You can measure lift in a wind tunnel. Why not induced drag?

Pete

Yes. So? Not relevant to the statement you quoted (which was about
thrust).

The issue under discussion was how much less lift was needed when
thrust supported some of the weight of the aircraft. The reduction in
necessary lift could accommodate a lower airspeed at the same AOA, or
a lower AOA at the same airspeed. But, as you said, the reduction in
lift is not a whole lot.

It would have been nice of you to provide the name of one popular
(i.e. easy to find) one, so that I can read up on it.

It's not always easy to find a reference to something I've read; I
often lose hours doing so. Anyway, here's one: "Introduction to
Flight", by John D. Anderson. p. 290. Quote:

"As seen in this example, for steady climbing flight, L (hence Cl) is
smaller, and thus induced drag is smaller. Consequently, total drag
for climbing flight is smaller than for level flight at the same
velocity."

Still, there's not much "unused AOA" in the regime of flight we're
talking about, nor did David suggest that might be required (his
implication, to my reading, was that his suggestion applied ??

True

But the reason I don't want to believe is that the proposal bears no
resemblance to the behavior of any airplane I've flown, not while I've
been flying it anyway.

If the effect exists, I agree that it would probably be small and
might well be lost in the wash.

I'd be more than happy to see someone step in with some real math
that shows the answer one way or the other.

Anderson shows some numbers. I hate trying to depict the math in this
forum, because it looks so ugly.

I don't happen to be patient enough with the math. There's a reason
that, when I was working on my math degree,

I only delve into math when the concepts are not clear. Putting some
numbers to the theory makes things real sometimes.

That is, he's proposing that at the same speed, there are multiple
steady states that produce different amounts of drag.

There are precedents. A banked aircraft at a given airspeed will have
a larger AOA than a non-banked one, and thus incur larger amounts of
induced drag.

I envision that a climbing airplane is essentially a lighter one,
since thrust will support a small amount of weight.

Still, I'm boggled by the lack of induced drag in a wind tunnel. If
the wing's not creating lift (0 AOA), I can see how there wouldn't be
induced drag. But this would happen in the real world too. If the
wing is creating lift, shouldn't there be a measurable force parallel
to the relative wind? Even in a wind tunnel?

I should qualify that. The great body of wing sections that NACA
tested in the early part of the last century were placed flush against
the walls; no wingtips. It is the wingtip vortices which cause the
local relative wind to be different from the "real" relative wind.
Absent that, the total aerodynamic force is perpendicular to the
incoming freestream.

They did this intentially, since the actual induced drag on a real
wing will depend on its aspect ratio. Better to make their data
"pure" and allow builders to adjust it to fit their specific
planforms. There is still drag, of course, but just not induced drag.

Greg Esres wrote:

relatively small fraction of the total weight of the airplane in the
first place, less than 10% in at least some cases, perhaps most
cases)

Lift in a 10 degree climb should be reduced about 1.5%.

How did you arrive at 1.5%?


Lift is always generated perpendicular to the wing's chord.

No, for subsonic flight, it's perpendicular to the *local* relative
wind, the relative wind that is modified by wingtip vortices. If lift
were perpendicular to the chordline, you would have induced drag in a
wind tunnel, and you don't.

Come on Greg, you're telling me that wings in wind tunnels have no induced
drag? That's ridiculous. How about if the wind tunnel was 1000 miles long
by 10 miles high - would the wings in that wind tunnel have induced drag?
(I seem to remember this same argument a few months ago)

Hilton

Hilton wrote:
Greg Esres wrote:
No, for subsonic flight, it's perpendicular to the *local* relative
wind, the relative wind that is modified by wingtip vortices. If lift
were perpendicular to the chordline, you would have induced drag in a
wind tunnel, and you don't.

Come on Greg, you're telling me that wings in wind tunnels have no induced
drag? That's ridiculous. How about if the wind tunnel was 1000 miles
long
by 10 miles high - would the wings in that wind tunnel have induced drag?
(I seem to remember this same argument a few months ago)

Just to follow-up my own post, here is a line from the NASA web site:
"During the winter, with the aid of their wind tunnel, they began to
understand the role of high induced drag on their aircraft's poor
performance."

http://www.grc.nasa.gov/WWW/Wright/airplane/drag1.html

Sorry Greg, "no induced drag in a wind tunnel" is simply not true.

Hilton

How did you arrive at 1.5%?

L = Wcos(theta)


Come on Greg, you're telling me that wings in wind tunnels have no
induced drag? That's ridiculous. How about if the wind tunnel was
1000 miles long by 10 miles high - would the wings in that wind tunnel
have induced drag? (I seem to remember this same argument a few
months ago)


Argument? No, Todd Pattist and I attempted to educate you on this
subject, but apparently failed.

The size of the wind tunnel is irrelevant. What matters is that the
wing tips are flush against the walls of the wind tunnel. This
produces 2-D airflow, rather than 3-D. In 2-D flow, there are no wing
tip vortices and thus will have no induced drag.

(If you would read something other than Aerodynamics for Naval
Aviators, you'd understand this.)

Sorry Greg, "no induced drag in a wind tunnel" is simply not true.

Once again, you don't understand.

If you put a 3-D wing in a wind tunnel, you will get induced drag.

The article you posted even documents the effect:

----------snip--------------
This drag occurs because the flow near the wing tips is distorted
spanwise as a result of the pressure difference from the top to the
bottom of the wing. Swirling vortices are formed at the wing tips, and
there is an energy associated with these vortices.
----------snip--------------


See that? Flow "near the wing tips is distorted"....vortices are
formed at the wing tips


2-D flow is achieved when the wing tips are flush against the sides of
the wind tunnel. No wing tips = no pressure leaking around the wing
tips = no wingtip vortices = no induced drag.

"Greg Esres" wrote in message
...
"As seen in this example, for steady climbing flight, L (hence Cl) is
smaller, and thus induced drag is smaller. Consequently, total drag
for climbing flight is smaller than for level flight at the same
velocity."

I'm not questioning whether thrust contributes to lift, and thus reduces the
total lift requirement. It is patently obvious to me that a force directed
at least partially downward contributes to lift. That quote says nothing
more than that. What I am questioning is whether for a given performance
scenario there are multiple drag scenarios.

That is, he's proposing that at the same speed, there are multiple
steady states that produce different amounts of drag.

There are precedents. A banked aircraft at a given airspeed will have
a larger AOA than a non-banked one, and thus incur larger amounts of
induced drag.

It's clear that I continue to fail to state my objection properly. Let me
try again...

David's post implies that for a given performance scenario (straight and
level flight, for example) you can nudge the airplane into a "new steady
state" where drag is lower. Your examples of climbing and turning don't
address that issue; they are entirely different performance scenarios (that
is, the airplane is doing something different) than the scenario to which
drag is being compared.

According to David's original post (if I read it correctly), there are
multiple drag scenarios for a given path of flight. Each time you come up
with an example, it starts out by assuming a new path of flight compared to
the "base case".

I envision that a climbing airplane is essentially a lighter one,
since thrust will support a small amount of weight.

Seems reasonable to me. But what if you don't want to climb? And in
particular, if we're talking about comparing one airplane in straight and
level flight to another in straight and level flight, introducing a climb to
the discussion doesn't help much.

Pete

Greg Esres wrote:
How did you arrive at 1.5%?

L = Wcos(theta)

*If* you assume level and climb airspeeds are the same.


[zap: induced drag disagreement]

(If you would read something other than Aerodynamics for Naval
Aviators, you'd understand this.)

Yeah, I also hate it when people use decades of research, hours of wind
tunnel testing, and accepted aerodynamic principals in these newsgroups.

Hilton

According to David's original post (if I read it correctly), there
are multiple drag scenarios for a given path of flight.

I didn't pick up on that, but if so, I agree with you. That scenario
seems unlikely.

On Fri, 31 Dec 2004 at 02:06:08 in message
, Peter Duniho
wrote:

Peter, I am sorry that this reply is a bit delayed as we have had
visitors over the last three days. I have gone back to basic equations
and looked again at them! I now have to confess that you are, at least
in part correct. I offer my apologies.

"David CL Francis" wrote in message
...
[...] So under some conditions if you just raise the nose a little you can
find a new steady state where speed is slightly reduced but with the same
thrust you can climb at the same AoA..

Your theory sounds wonderful, but I doubt it holds water in practice.


My statement above was wrong as written. It is true in the sense that
you can find a new steady state at the same AoA and a slightly reduced
speed but only by applying more thrust. You apply more thrust and if the
AoA stays the same the velocity will initially increase and the flight
path will curve upward arriving at a new steady state where there is a
climb, but at a slightly reduced speed.

Theta is the angle of climb in degrees.

If the L/D is fixed at 10 then here is a little table. I hope it comes
out not too screwed up by different fonts and line lengths:

weight Lift/ Theta Drag Thrust Lift Lift
Velocity
lb. drag deg lb. lb. lb.
% %
10000 10 0 1000 1000 10000 100 100.00
10000 10 1 1000 1174 9998 100 99.99
10000 10 2 999 1348 9994 100 99.97
10000 10 5 996 1868 9962 100 99.81
10000 10 15 966 3554 9659 97 98.28
10000 10 30 866 5866 8660 87 93.06
10000 10 45 707 7778 7071 71 84.09
10000 10 60 500 9160 5000 50 70.71
10000 10 90 0 10000 0 0 0.00

In practice aircraft cannot be controlled at zero velocity and more
thrust would be needed at the higher angles of climb to maintain a
controllable airspeed. Note that lift is reduced to zero in a vertical
climb as it must be.

I freely admit that these equations give only a simplified
demonstration. There are many refinements to be added but the basics are
generally accepted I believe.

For example at the same power setting thrust is not independent of
velocity, the thrust is not always exactly opposite to drag and control
deflections also have an effect on drag and total lift.

Weight is the gravitational force on the aircraft and points to the
centre of the earth. Thrust is defined, in this simplification as a
force along the flight path and drag is a force in the opposite
direction. Lift (you seem to have got this wrong further down in your
post) is defined, in the usual way, as acting at right angles to the
flight path.

Drag acts along the flight path in the opposite general direction to
thrust. It has two components. One is the standard parasitic drag and
the other is induced drag which can be looked at as the drag component
due to the tilting back of the force vector to allow for the deflection
of the air stream by the generation of the lift. By making a few
assumptions about the size of the airflow that is actually deflected
this can be shown to be proportional to the square of the lift
coefficient.

Thrust does not reduce the required lift by much, especially not in the
light planes we tend to fly. Steady-state pitch angles in climbs tend to be
modest, meaning a tiny fraction of the thrust vector is the downward
component. Just 17% of the total thrust, for a 10 degree pitch angle.
Given how little thrust a light plane has in the first place (only a
relatively small fraction of the total weight of the airplane in the first
place, less than 10% in at least some cases, perhaps most cases), even
taking 20% (or even 34%) of that and applying it to lift just isn't going to
help that much.

The thrust must be equal to the drag or the aircraft will not fly at
all! If the thrust is smaller than 10% of the weight then that implies
that the lift/drag ratio is better than 10 for it to even fly at best
lift/drag ratio. A motor glider requires little thrust to fly but may
not climb very well. Note that the pitch angle, depending on how it is
defined, is not normally the same as the angle of climb

Even assuming an airplane with a thrust-to-weight ratio of 1.0 (a rare
occurrance, but they do exist...some F-16s, for example), I'm not sure your
theory holds up very well. You might think that you could simply increase
thrust as you slow the airplane in order to allow a smaller AOA to suffice
to provide the remaining necessary lift. But there's a problem with that
idea.

See the table above. I think you are using what you call 'my theory' in
a strange way in the above. If an aircraft is capable of climbing
vertically and steadily then thrust must always be greater than weight
so that a velocity can be maintained sufficient to maintain control
Under those circumstances the situation requires that lift is zero.

If the AOA is kept small, then the increased thrust will prevent the
airplane from decellerating. To have a vertical component high enough to
support the airplane will require a horizontal component so high that the
airplane won't slow. If the AOA is allowed to increase, then what the
increased thrust is actually allowing is for the wing to stall without the
airplane being pulled downward by gravity (the wing WILL stall at the
appropriate AOA, regardless of airspeed). It's not demonstrating anything
about some "new steady state".


I do not understand the above and in particular your remark that the
'aircraft will not slow'. If AoA is not changed then increased thrust
will increase speed and increased speed will increase lift. That, unless
checked, will lead to the flight path curving upward and a climb
beginning until a new state is reached. The gravity component then also
plays its part in speed reduction.

The above thought experiment should also illustrate another problem with
your theory: it ignores the change in the portion of lift actually
contributing to counteracting gravity that occurs due to pitch changes. See
below for more commentary on that.


Again I am not sure what you mean there. If you mean it ignores the
effect of a climbing flight path then it does not. I have no special
theory - just an attempt to explain basics in the simplest reasonable
way using the simplest possible static equations. AFAIK there are an
infinite number of different possible steady states in the flight of an
aircraft.

Of course, to me the biggest problem intuitively with your theory is that I
am sure that aerodynamics involves only continuous functions. Given that,
if you assume more than one steady state, you are claiming that there are
multiple local minima/maxima between which are apparently "lower efficiency"
areas. And of course, if there's more than one, I see no reason to believe
that there are only two. This would then imply that the flight envelope has
numerous of these local minima/maxima points.

I certainly agree that aerodynamics is a series of continuous functions
with changes of those functions at certain values.

I am not sure about the rest of your paragraph above. An aircraft can
fly at any speed and angle within its capabilities. Change any of the
basic 4 forces and a new situation that can be maintained may or may not
be possible.

Given that in more than 100 years of study, this concept has never shown up
as a noted element of the relationship between speed, drag, and lift, I'm
inclined to believe that it's just not true (just as the idea of "cruising
on the step" is not true).

What concept are you talking about? I am just using simple balance of
forces equations - apart from when I get it wrong of course! :-(

I admit that I have not brought out the equations and proved my point
irrefutably. Someone like Julian Scarfe or Todd Pattist would probably do a
better job discussing this, since they seem to be more "math oriented" (that
is, they don't mind crunching some equations now and then ). But I still
feel reasonably confident that there's no secondary "new steady state" one
can achieve by increasing AOA and taking advantage of thrust.

I would welcome Todd's participation, he has corrected me a few times
but we have often agreed.

The extra work comes ONLY from a net surplus of power.

I agree with that. However that net surplus can come from either more
engine power or reduced drag. Because even in a modest climb the engine
thrust vector plus the lift vector combine to match the weight.

I assume by "the engine thrust vector" you really mean "the vertical
component of the engine thrust vector". Assuming that, I agree with your
statement, but I don't find it informative. The engine thrust vector has a
non-zero vertical component even during level cruise flight, and yes it does
contribute to counteracting gravity, reducing the lift required.

Fair enough and I have made a part of that point above somewhere. It is
a part of the simplification involved.

Put another way if you are flying below the maximum lift drag ratio and
you increase the AoA to the optimum whilst keeping the same power the
aircraft should climb. This is self evident if you are flying level at
high speed and at a climbing power setting. Bring the nose up increasing
the AoA and your aircraft will definitely climb. Agreed?

I never said anything to the contrary. If you are at an angle of attack
lower than the best L/D AOA (and thus at an airspeed higher than the L/Dmax
airspeed), increasing pitch angle without a change in power will result in a
climb, yes.

I am glad we agree on that!

But that has nothing to do with what happens at an airpseed near stall,
which occurs below the L/Dmax airspeed, and well above the best L/D AOA.

Agreed. You cannot decrease drag by increasing AoA near the stall. This
is part of where we were at cross purposes. I was not thinking
specifically about near the stall.

This seems like a good point at which to mention something else you've left
out and which I alluded to earlier...

Lift is always generated perpendicular to the wing's chord. Some people
like to call just the vertical component of this force "lift", but the
amount of force acting through the wing is the force perpendicular to the
chord.

That is wrong. Lift is always defined as perpendicular to the free
stream velocity of the aircraft. In the same way lift is not defined
relative to gravity either except that in steady level flight it just
happens to be exactly opposite and balancing gravity.

However you label things, you cannot avoid the fact that when you change the
angle of the lift vector, the portion of the force created by the wing used
to counteract gravity is also changed. In particular, in the low-airspeed,
power-on example, even as thrust is helping support the airplane, you are
using your lift less efficiently, which means that the wing needs to
generate more total lift just to provide the necessary vertical component.

Insofar as I understand it, I disagree still with that statement. If
thrust is providing a vector in the lift direction then in steady flight
less lift is required as my table does show.

This is similar to the required increase in lift while in a turn, but due to
redirecting the lift vector in a different way.

No, it is different to that because the increased force required in a
turn is needed to produce an acceleration which is felt as 'g'. That
does not apply in steady level, climbing or gliding flight when just the
normal 1g is felt.

Since the lift vector points slightly aft in level flight, even at high
airspeeds when the angle of attack is low, it's easier to see how this
negates at least some of whatever contribution thrust might make as the
angle of attack is increased. However, in gliding flight, the vector is
pointed forward, helping counteract the contribution drag makes to lift.

The lift vector points at right angles to the line of flight no matter
what. Lift and drag coefficients are measured like that unless things
have changed a lot since my long ago student experiments in a wind
tunnel.. The only time you consider a backward shift of the 'lift'
vector is in calculating induced drag using a simplified method. By
including a separate term for induced drag proportional to Cl^2 that
effect is allowed for. Lift and drag coefficients for an airfoil are
determined for infinite aspect ratio.

This is getting too long to deal with like this! It has got to the point
where we would need to discuss this face to face to achieve much more
and come to a common understanding of terms and simple equations.

I wrote "at least some" up there, but it should be apparent from the
disparate magnitudes of the lift and thrust vectors that you get a much more
significant change in lift than you do in thrust.

Not at small angles of climb as my correction and table indicates. Small
angles of climb mean a small change of the lift vector component in the
truly vertical direction vector.

The bottom line he even though thrust does contribute at least a little
to counteracting gravity, it does not do so in a way significant enough to
change the fact that, as you slow the airplane from any airspeed at or below
L/Dmax airspeed, you experience increased drag.

That is true and I do agree with that and apologise for the confusion.
Only if the L/D increases with AoA do you get a benefit of reduced drag.
Although the L/D of an aircraft may be 10 or more at maximum. at full
speed in level flight it will be much lower, perhaps as low as 2 or
less. At the zero lift AoA the Lift/drag ratio is also zero but you
cannot then maintain level flight!!!.

I hope that helps to clear some of the misunderstandings between us. My
only excuse is my great age!

I repeat my apology for my error.

I always enjoy your posts even if I don't quite agree!
--
David CL Francis