A Level 1 AOA clarification

On Tue, 28 Dec 2004 at 16:55:05 in message
, Peter Duniho
wrote:
It does this as long as the lift is slightly less and the speed drops to
produce _less_ drag and lift, leaving more engine power and thrust to
climb.

At an airspeed just above stall, a reduction in speed results in MORE drag.
There is a reduction in parasitic drag, but there is a greater increase in
induced drag, with a net increase in total drag (and that's ignoring drag
caused by the rudder and any other control surfaces that require a change in
position).

HI Peter. How easy it is to get slightly confused on Usenet! What I was
trying to say is that if you maintain the same AoA then the lift drag
ratio remains the same, but because the lift required when climbing is
less than when flying level you can climb at a reduced speed but with
less drag. 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..

There is a maximum lift drag ration at a modest angle of attack. Above
_and_ below that angle of attack that ratio worsens.

When climbing extra work must be done against gravity. That extra work can
come from increasing power or from reducing speed and therefore drag.

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. 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?

A reduction in speed
is only guaranteed to produce a net increase in power available if the new
airspeed is higher than Vbg. It can sometimes also produce a net increase,
if the old airspeed was sufficiently higher than Vbg, and the new airspeed
is close enough to Vbg, even if less than, but you need to know more about
the old and new airspeeds in that case to say for sure what happens. More
importantly, a reduction in speed is guaranteed to produce a net decrease of
power available if the OLD airspeed is lower than Vbg (as it is when just
above stall speed).

I think that is another way of saying what I have just said? I cannot
remember if we started off with an assumption that the aircraft was only
just above stall speed? If so then you are correct of course.

Even in a steady glide the required lift is less than that needed in
level flight! That is easier to see because the drag vector helps the
lift match the gravity vector.

--
David CL Francis

"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.

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.

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.

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".

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.

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.

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).

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.

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.

But when you ask the question about HOW MUCH it does this, the answer is
"not enough to change the fundamentals". Not for the airplanes we fly, and
I think probably not for any airplane.

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.

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.

I think that is another way of saying what I have just said?

Well, since I seem to disagree with what you said, I sure hope not.

I cannot remember if we started off with an assumption that the aircraft
was only just above stall speed? If so then you are correct of course.

Yes, this part of the discussion was entirely about the regime of flight
near the stalling speed and AOA. Reducing drag by pitching up while above
L/Dmax airspeed is uninteresting, since that's a direct consequence of
slowing to an airspeed closer to L/Dmax airspeed.

Even in a steady glide the required lift is less than that needed in level
flight! That is easier to see because the drag vector helps the lift match
the gravity vector.

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.

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.

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

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.

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.

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.

Pete

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%.

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

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

Although I agree that a small increase in AOA would not contribute
enough vertical component of lift to overcome the initial increase in
induced drag, there are ways to get into this regime of flight. 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.

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. The only component of
thrust that accelerates the airplane is that parallel to the flight
path. If the angle of climb remains the same, then increasing thrust
will obviously accelerate the airplane. However, as thrust increases,
the angle of climb increases up to the point that the component of
aircraft weight along the flight path is equal to the increase in
thrust.


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.

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

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

Given that in more than 100 years of study, this concept has never
shown up

I doubt you're familiar with even 1% of the 100 years of aerodynamic
research and thought. I'm certainly not.

You should realize that "I've never heard of it so it must be false"
is a weak argument.

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.

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.

This is based on your mistaken notion above.

Since the lift vector points slightly aft in level flight,

Only because of induced drag.

However, in gliding flight, the vector is pointed forward, helping
counteract the contribution drag makes to lift.

No, the lift vector is perpendicular to the local relative wind
causing it. There is a rearward component (called "induced drag"),
but there is no component forward along the flight path.

"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 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

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 21:23:58 in message
, Peter Duniho
wrote:
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?

Peter,

After posting another long article with a correction I am a bit
reluctant to step in again here. However the explanation is fairly
simple.

Induced drag and wing tip vortices are almost one and the same thing.
Very roughly induced drag is proportional to

(Lift coefficient)^2/(aspect ratio)

The higher the aspect ratio then the smaller is the induced drag until
when the Aspect Ratio is infinite it is zero. Airfoil sections always
used to be tested in wing tunnels by taking them right across the tunnel
so that there are no tip effects. A correction has to be made for the
tunnel wall but the effect is to test a two dimensional section. Of
course there is still the parasitic drag component to be measured and
that is the Cd that is quoted for the particular wing section.

You can try to measure induced drag but you need a large tunnel and a 3d
model of the aircraft sufficiently far from the walls so that the flow
is not too distorted.
--
David CL Francis

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