Single engine Cessna tailplane lift

[photofly]

@Colonel Sanders:

I know you're worried about the point where the tail force switches sign.

Here's a system where the tail force also switches sign; just like an airplane. Just like an airplane, nothing very significant happens at that point:

seesaw.png (11.88 KiB) Viewed 4926 times

You can move the box to the left, or the right, and move the cg to the left or the right of the fulcrum. Or put it directly in the middle. By doing so you make the spring on the right (the "tail") either compress or extend, or take up its natural length. But apart from a change in deck angle there's not really any difference to the behaviour of the system in either case.

If you want to extend the diagram to make it a little more aircraft-like, consider that by slowing down and speeding up the pilot is moving the fulcrum back and forth (because of the way the wing's centre of pressure moves back and forth depends on angle of attack).

A note: this system doesn't have the same stability criteria as an airplane. (It would if you changed the stiffness of the spring as the box - or fulcrum - moved, but then it wouldn't be so simple any more.)

Therein lies the rub: it's not the direction of the stretch or compression that interests us, it's how stiff it is, i.e. how the restoring force grows or shrinks as the tilt varies. If you put your mathematical hat on for a second, it's the gradient of the force around the equilibrium that counts (and specifically whether the gradient is positive or negative) and not whether, at equilibrium, the force itself is positive, or negative, or neither.

You raise some interesting points about knife-edge flight. I'd like to address them after I've written to PilotDAR.

[photofly]

So in my effort to either understand or refute this really hard to accept concept, I'm trying to relate it to the real world of flying planes. It does not relate for me yet...

Thanks for being patient with me, and giving me the opportunity to marshal some thoughts and write about them in a concise cogent manner.

Takeoffs, and particularly water takeoffs involve some strange combinations of forces, because of the pitching moment of the water contact, and the changes in airflow as you enter/leave ground effect.

Firstly, let's not worry about control forces. You simply cannot tell the force on the tail from the direction of the control force (*). I can have a tail-down force with either a pull or a push required on the yoke by adjusting the trim. Control forces are important to the pilot but they're determined by taking moments around the elevator hinge and they're determined by the trim tab setting at the time.

In your analysis, when you talk about the direction of the change of forces on the tail for a pitch up or pitch down I believe you're mixing together dynamic effects with the changes at the new equilibrium position. This distinction is really important to get right.

Let's talk through a very simple manoeuver. I think you'll be surprised by the conclusions.



Let's start by flying straight and level at 100 knots. Then let's maintain level flight and slow to 90kts.

We ignore contributions to pitching moment from the engine, fuselage, undercarriage etc. They're important in real life but they don't change the essence of what's going on. It's long, complicated and very detailed; but the devil, as always is in the detail. Stick with it.



1. The pilot makes a sudden step-change to the yoke position - back - and then fixes the yoke.


2. This instantly makes the angle of attack of the elevator more negative. For anyone who finds this hard to imagine, think of an aircraft with a stabilator, where the whole stabilizer rotates, instead of having an elevator tab.


3. Because of this change in tail AoA The force at the tail becomes more downward. (Either the tail lift decreases, or the down-force increases.) Nothing else has changed (yet) so the aircraft is suddenly out of equilibrium. A pitch-up acceleration is now present because the forces are unbalanced.


4. The acceleration persists, causing a nose up rotational motion to start. The pitch starts to increase.


As the pitch increases the following things happen:


5. The AoA of the wing starts to increase. Therefore the coefficient of lift of of the wing grows. To maintain the steady lift required for level flight the speed must decrease (we adjust power).


6. As the Cl of the wing increases the centre of pressure of the wing moves forward. (You can see this on the graph that dr.aero posted, from Aerodynamics for Naval Aviators - high Cl means forward cp.) This means that the wing is trying to pitch the aircraft nose up more, as it slows down. The wing's natural tendency is to amplify any pitching up.


7. Because the whole aircraft is rotating in pitch, at the same time as the AoA of the wing is increasing, the AoA of the horizontal stabilizer is increasing at the same rate.


8a. The increase in AoA at the tail increases the lift (or decreases the down-force) on the tail. We added a fixed amount of tail force by moving the elevator, but now the aircraft is rotating, subtracting from that change. The change in tail force works counter to the pitching-up motion. The horizontal stabilizer's natural tendency is to work against any pitch change. That's why it's called the stabilizer.


8b. The tail downforce (or lift) decreases in magnitude because the aircraft is slowing.


9. The twisting moment of the wing and the stabilizer set the rotational acceleration of the airplane. It's still rotating, nose up. It will continue to rotate nose up faster and faster until the extra contribution from the wing (6) is exactly balanced out by the changing force on the tail (8 and 8b). When the forces balance the acceleration stops.


10. Just because the acceleration has stopped, doesn't mean the rotation has stopped. The aircraft is still rotating nose up, and keeps doing so. In fact it overshoots the point where the forces balance. It rotates nose up too much. Now the tail is lifting (or not pushing down) more than the change in pitching moment from the wing. It will start to accelerate nose down.


11. We set up an oscillation. This is called the longitudinal short period mode. It's easily predicted by working out the relevant equations. I will spare you. You can go and demonstrate and measure it in any aircraft.


12. We rely on pitch damping to damp out the oscillations. All flyable aircraft have plenty of pitch damping. If you want to know how, we can talk about that another time.


13. When the oscillation has stopped, the aircraft has reached a new equilibrium. The equilibrium point was actually the pitch attitude that was reached in (9). That's where the moments balanced again, which means that the rotational acceleration is zero. The pitch damping lets the rotational velocity go to zero, and zero net moment at this pitch angle means it stays at zero.



14. The new equilibrium differs from where we started in the following ways:

i) We're flying slower.
ii) the nose is higher, therefore ...
iii) the angle of attack of the wing is higher but ...
iv) the lift is the same, so ...
iv) the wing is flying at a higher Cl...
v) which means a more forward centre of pressure...
vi) ...which means a more nose-up moment created between the wing and the weight acting at the centre of gravity ...
vii) ...which means more lift (or less downforce) from the tail.


That is: to fly slower, pull back on the elevator to reduce the angle of attack of the tail, initially. By the time a new equilibrium is established, the whole aircraft has rotated so that the tail is providing more lift (or less downforce.)

This is counterintuitive. It's also true.

Although the lift on the tail increases (downforce decreases) I'm going to hold off making too strong an assertion on where the tail AoA ends up, because the change in speed affects it.

Notes:

A) For an aircraft to reach a stable pitch angle after an elevator input, the change from the wing in (6) must be overtaken by the change from the tail in (8). You can quite quickly work out when that's not the case, in which case the aircraft will be unflyable. In fact It happens when the cg is too far rearwards for the size and position of the tail. Since the pilot can't change the tail geometry he only has to worry about the location of the cg.

B) None of this depends on which direction the tail force goes. In every case I said "tail lift increases or downforce decreases" or vice versa. It works just as well either way around.

C) The fact that the wing tries to twist nose up more when it slows is an unfortunate feature of a regular wing. It's why you need a tail in the first place. If you design a wing carefully you can make it try to twist nose down when it slows. The feature of such a wing is a reflexed trailing edge. That's what aircraft without horizontal stabilizers have: think Concorde, or delta wings. In fact, if you take a delta wing and cut off the reflexed trailing edge, and stick it at the back of a long fuselage extension then it becomes a regular tail.

D) the requirement for a tail to stabilize a regular wing was worked out by a guy called Alphonse Pénaud around 1870.
http://en.wikipedia.org/wiki/Alphonse_P%C3%A9naud (see especially the comment about the stability of his "Planophore")

E) The fuselage has a significant pitch destabilizing effect too (just like the wing) - as you fly slower, it tries to pitch up more. You can (and aerodynamicists do) add the effects together when working out what size and shape the tail has to be.



Enough for now. Fire away with any faults, questions, etc.


(*) Control forces are an important clue to the pilot to tell him how far he or she is from equilibrium (how far out of trim) and the stability margin (soft or stiff yoke) but those are effects that are deliberately designed in to the aircraft's trim system. Yoke feel doesn't affect the natural stability of the aircraft, only the way the pilot handles it.

[photofly]

@CS:

Here's an aircraft in knife-edge flight. Ferocious angle of attack on the fuselage which is now acting as a lifting body, of very small aspect ratio. The rudder is a now a cross between a regular tail and the edge of a reflexed wing (along with the fuselage).

You wouldn't know what the force on the rudder is unless you know the centre of pressure of the fuselage at this angle of attack, and where the cg is.

knife.png (161.82 KiB) Viewed 4895 times

[bezerker]

It appears from this image that the 1/4 chord C of L is aft of the most rearward allowable C of G.

[photofly]

I went through the algebra of the equilibrium situation again, this time to look at the angle of attack of the tail for different airspeeds, at equilibrium.

Using the aerodynamic angle of attack (i.e. angle of airflow relative to the zero-lift line) for both tail and wing, and again, ignoring the fuselage contribution. Also ignoring for the moment the downwash contribution to the tail angle of attack.

Starting again with:

M0 the pitching moment of the wing
M'0 the pitching moment of the tail
Mf the pitching moment of the fuselage
Mp the pitching moment of the propellor thrust
L the lift of the wing
L' the lift of the tail (both lifts positive when lifting upwards)
x_a the horizontal distance from the cg to the aerodynamic centre (ac) of the wing
x'_a the horizontal distance from the cg to the ac of the tail
M the total moment
W the weight of the aircraft

and

Q they dynamic pressure
Cl coefficient of lift of the wing
Cl' the coefficient of lift of the tail
Cm0 the pitching moment coefficient of the wing
c the mean aerodynamic chord of the wing
S the area of the wing
S' the area of the tail

(We follow convention and say that positive moments are in the nose-up direction.)

M = M0 + M'0 + x_a.L - x'_a.L' + Mp + Mf = 0 for equilibrium

Again we ignore Mp, Mf and note that M'0, the pitching moment of the tail is zero because it's a symmetric airfoil.

This time write in terms of the coefficients, by diving by QS:

c.S.Cm0 + S.x_a.Cl - x'_a.S'.Cl' = 0

we write the coefficient of lifts as kα and k'α' for wing and tail respectively (α and α' relative to zero lift line, so strictly proportional relationship)

c.S.Cm0 + S.x_a.k.α - x'_a.S'.k'.α' = 0 --(1)

The lift equation is that L + L' = W

therefore

S.k.α + S'.k'.α' = W / Q

which gives

k.α = W / Q - S'.k'.α' --(2)

Substitute (2) in to (1) and rearrange to get:

α' = [ c.S.Cm0 + x_a . (W/Q) ] / [S'k' (x_a + x_a') ]

The numerator becomes zero when the tail force is zero, as before.
The term (x_a + x_a') is the distance between the two aerodynamic centres, i.e. the distance between the chord/4 of the wing and the chord/4 of the tail.

Using the following data:

c = 1.75m - chord
S = 16.2 m^2 - wing area
S' = 3.5 m^2 - tail area
W = 2950lbs = 1340kg * 9.81 m/s^2
x_a + x_a' = 5.47m
ρ = 1.229 kg/m^3
Q = 0.5 ρ v^2
k' is the gradient of the tail cl vs. α' curve. A good estimate for k' is
0.1 / (1 + 2/aspect ratio) (Von Mises, p149).
The aspect ratio of the tail is about 3, so we use k' = 0.06, and for the main wing k=0.075.

And taking an example cg station of 43" making x_a = 43 -37" = 6" = 0.15m

α' = [-1.28 + 3286 / v^2] / 1.15

with v in m/s

or

α' = [-1.28 + 12416 / v^2] / 1.15

with v in knots.

This is what that looks like:

tail_angle.png (27.04 KiB) Viewed 4884 times

Comments: You can see how the angle of attack of the tail increases, the slower the airspeed. You can also see that with this choice of cg etc. it crosses from negative (down force) to positive (lift) around 100kts.



Now that we have α', we can write down an expression for α from (2):

α = ( W / Q - S'.k'.α' ) / S.k

The first term in the numerator is the component required to support just the weight of the aircraft on the wing; the second component is the adjustment to take into account the tail force (either up or down.)

With v in knots again this works out as:

α = ( 86320 / v^2 ) - 0.173 α'

Let's plot both these quantities on a graph:

tail_angle2.png (37.21 KiB) Viewed 4887 times

Comments: The blue line doesn't cross the green line at the stalling AoA. But we used a linear relationship for Cl: Cl = kα. This is no longer true as the AoA approaches the stall. So I'm not too worried about it. We've guessed various parameters, and estimated some others, so I wouldn't read too much into the numerical values on the graph. But the broad behaviour should be correct.

Secondly: the difference between the mauve and blue lines represents the elevator setting, i.e. the yoke position. The bigger the distance between the two lines the further back the yoke, because the pilot has rotated the elevator more negatively, to provoke a pitch-up. We can see that by doing so, the equilibrium AoA of both tail and main wing has increased - the wing by more than the tail - and the trim speed of the aircraft has decreased.

You can also see that the distance the elevator needs to move change airspeed by 10 knots increases as the airspeed decreases. Small changes in elevator position make bigger differences in airspeed at high speed, but when you fly slowly you need to make much coarser movements to change speeds.

[photofly]

It appears from this image that the 1/4 chord C of L is aft of the most rearward allowable C of G.

That's a C172.

Leading edge at station 28" (approx)
Chord 64"
Quarter chord therefore at 28+16 = 44".
CG range 35 - 47"

Seeing as the cp is always somewhere aft of the ac (chord/4) it's possible that for the C172 the cg never goes aft of the cp, in which case the tail always provides a downforce.

The 182 has somewhat different geometry though.

[digits_]

taildragger takeoffs

I think that it's pretty obvious that during a taildragger
takeoff with the mains supporting the weight of the
aircraft, with sufficient power and forward stick you
can raise the tail before you even start moving.

Obviously the tail is providing "upwards" lift - relative
to the pilot's frame of reference - at that point.

This is not necessary. While the tail might move up, it is possible that his happens with no lift or negative lift. While you are on the ground, with the brakes on the weel, you are rotating around the point where the wheels hit the concrete. This is a completely different balance situation compared to an "in flight" balance, where you rotate around the center of gravity.

The situation you get on the ground is similar to the following: imagine you pull a box around, over a flat floor , with a rope tied +- on half it height. This will go fluently, the box will move. Now, if someone puts a few roks in front of the path (= friction), the bottom part of the rox may get stuck. If you continue to pull, the box will rotate, even though it generates no lift anywhere.

Watch this video and try to tell me that the tail isn't
providing upwards lift during both takeoff and landing:

Note that only half of the planes raise their tail before takeoff. Note that with those planes, the tail starts dropping as soon as the airplane starts to move. When the brakes are released, the force disappears (partially only, we still have some acceleration) which causes the tail to fall down again. I admit this last part might happen partially because the pilot might move the stick (I couldn't see the elevator movement during this phase of the flight clearly on the video).


My point is that you can't deduct the lift/downforce from watching the airplanes fly. You have to measure or calculate it.

[Colonel Sanders]

Sigh. Ok, let's consider the case with no power on -
a tailwheel landing.

Let's make it a wheel landing. After touchdown, the
aircraft is in a level pitch attitude. In the aircraft that
I fly, the horizontal stab is rigged to be perfectly aligned
with the longitudinal axis of the aircraft. So, when the
aircraft is level, so the the horiztontal stab.

Additionally, the horizontal stab is symmetrical. Therefore
is it providing no lift in any direction. It has no angle of
attack, when the aircraft is in a level attitude.

Now, to keep the tail up after performing a wheel landing,
I progressively apply full forward stick which points the
elevator full down, and the aircraft in a level attitude:



Talk to me about what aerodynamic forces result.

I'm not sure what experience you have wheel landing
a tailwheel aircraft - or landing any aircraft - but if you
apply full forward stick immediately after landing, you
can get the tail high enough that the prop hits the runway,
the spinner digs into the pavement, and the aircraft
will nose over inverted.

Back to the aircraft, which has symmetrical wings, again
with no angle of incidence. Therefore, in a level attitude
in a wheel landing, they are providing no lift.

The aircraft's weight is mostly supported by the main
gear. There can be no disputing that the C of G is behind
that. To maintain that level attitude, about 50 lbs of force
needs to be applied vertically to the tail. I know this,
because I have jacked it up (and lifted it) many times
in the hangar during maintenance.

Now, where is this lifting force coming from, during a power
off wheel landing?

I did a quick google search and found this slightly tail-low wheel landing:



Just as in straight-and-level inverted flight (see previous
pictures with vectors) that tail is providing "up" lift, relative
to the pilot's frame of reference.

[cgzro]

Not sure it makes much difference but your quarter cord line appears to be only correct for the inner 1/3 rd of the wing. The double tapered outer 2/3 line seems wrong because your line is closer to the 1/5 chord at the tip. The average cp would therefore be a tiny bit aft of what you drew depending on how you average the contributions. Hard to tell from an iphone screen and likely not marerial to your point but since you are being precise..

[photofly]

Colonel:

You didn't account for a few things.

Biplanes have a special advantage when it comes to stability. They have decalage between the upper and lower wings.

The wings each have a symmetrical profile, but they're not rigged at the same angle (to my eyes). And, just like the tail, the lower wing flies in the downwash of the upper wing (and the upper wing flies in the upwash of the lower wing.) This is important. Combine this with the fact that the bottom wing is to the rear of the top wing, and you have positive decalage. That gives the wing stability even without the tail. If you like, the bottom wing is acting as a stabilizer for the top wing.

In practical terms that means that as long as the aircraft is moving the two wings between them are generating a nose up pitching moment (it doesn't depend on there being any overall lift) which you have to account for.

You effectively have an upper wing, and two tail surfaces, one large one just under and to the rear of the main wing (with massive downwash effect) and the second smaller tail (with less of a downwash effect.)

Unless you include the effects of all three wings and their interaction, you won't get the tail forces right.

I'm not saying you got them wrong, but ...

[Colonel Sanders]

The wings each have a symmetrical profile, but they're not rigged at the same angle

According to Mr Pitts, they are. This is important in terms
of establishing vertical uplines and downlines.

My apologies for making things too complicated, which is a
real problem in this thread. Concentrate on this diagram
for now:



What forces result?

[digits_]

Sigh. Ok, let's consider the case with no power on -
a tailwheel landing.

Let's make it a wheel landing. After touchdown, the
aircraft is in a level pitch attitude. In the aircraft that
I fly, the horizontal stab is rigged to be perfectly aligned
with the longitudinal axis of the aircraft. So, when the
aircraft is level, so the the horiztontal stab.

Additionally, the horizontal stab is symmetrical. Therefore
is it providing no lift in any direction. It has no angle of
attack, when the aircraft is in a level attitude.

Now, to keep the tail up after performing a wheel landing,
I progressively apply full forward stick which points the
elevator full down, and the aircraft in a level attitude:


Talk to me about what aerodynamic forces result.

If everything is perfectly symmetrical (including the wings as you specified below), then I would guess that what you say is correct (ignoring the moment from the airframe etc, like we did in the previous calculations). Note that this would be an airplane in free fall. Like a brick.

I'm not sure what experience you have wheel landing
a tailwheel aircraft - or landing any aircraft - but if you
apply full forward stick immediately after landing, you
can get the tail high enough that the prop hits the runway,
the spinner digs into the pavement, and the aircraft
will nose over inverted.

No tailwheel experience. I don't think that's relevant though. I am not disputing the stick forces you feel.
The above effect can, in general, also be created by an elevator with a downforce. As long as the lifting force by the main wings is big enough to rotate the airplane around his main wheel.

Back to the aircraft, which has symmetrical wings, again
with no angle of incidence. Therefore, in a level attitude
in a wheel landing, they are providing no lift.

The aircraft's weight is mostly supported by the main
gear. There can be no disputing that the C of G is behind
that. To maintain that level attitude, about 50 lbs of force
needs to be applied vertically to the tail. I know this,
because I have jacked it up (and lifted it) many times
in the hangar during maintenance.

Now, where is this lifting force coming from, during a power
off wheel landing?

If this is still about the perfectly symmetrical airplane, then would think it would be generated by the tail then yes.
However, from this you can't conclude the following

I did a quick google search and found this slightly tail-low wheel landing:

Just as in straight-and-level inverted flight (see previous
pictures with vectors) that tail is providing "up" lift, relative
to the pilot's frame of reference.

In this case, even with a symmetrial airplane, you are flying straight and level, and thus the wings generate lift. This will create a moment around the CoG. Depending on the positions of the CoG and CoL etc (see previous posts), the tailplane will either lift positively or negatively.

[photofly]

The wings each have a symmetrical profile, but they're not rigged at the same angle

According to Mr Pitts, they are.

Yes. I ran the photo through a graphics package, and I couldn't detect any difference in the rigging angle. The difference in downwash/upwash between the wings is a real effect though. It's one obvious asymmetry: in inverted flight the top wing is behind the lower wing. Maybe Mr. Pitts was extremely clever and set the lower wing rearwards by just the right amount to exactly overcome the pitching moment of the fuselage and undercarriage.

I like your videos of the short takeoffs. The tail is lifting, at least at the start, when the airspeed is lowest. Which is what you'd expect from looking at the pitching moments, as I did.

One caveat though: the direction of the force needed to change the attitude of the aircraft isn't the same as the change in force achieved when the speed reaches equilibrium. That is, don't judge the equilibrium forces from what you see happen while the aircraft is rotating in pitch.

[digits_]

I like your videos of the short takeoffs. The tail is lifting, at least at the start, when the airspeed is lowest.

How can you be sure of that ? Why can't the tail lifting be caused by the generated engine power / thrust ?

[photofly]

I like your videos of the short takeoffs. The tail is lifting, at least at the start, when the airspeed is lowest.

How can you be sure of that ? Why can't the tail lifting be caused by the generated engine power / thrust ?

Of course you're right, you can't be 100% sure. But once the wheels are off the ground, the pitching moment from the engine isn't very large, because the thrust line passes fairly close to the cg.

[bezerker]

I can't see the difference...can you see the difference?

Again, C of L is nicely behind most aft C of G. And using 28 as leading edge of chord is ambitious (as is 1/4 chord for C of L for entire aircraft).

[photofly]

I'm not sure what point you're making.

The chord/4 for the C182 is at a station of 37". The cg range is 33 to 48.5". Which bit do you think I have wrong?

Which bit of your diagram shows where the leading and trailing edges are?

as is 1/4 chord for C of L for entire aircraft

... Yeah... Steve Pomroy pointed that out to me somewhere early on. Thereafter I made the proper calculations using M0 and Cm0.

[Colonel Sanders]

(Mr Pitts put) the top wing (ahead of) the lower wing

The reason he did that was so that you could
get in and out of the airplane! The original version
of the airplane had flat-bottom wings. The round
wings came much later.

I'm not it matters much, frankly. The Beech Staggerwing
did it the other way 'round (top wing behind bottom wing)
and it flew just fine.

The tail is lifting, at least at the start, when the airspeed is lowest

That's not surprising. During a takeoff with a powerful engine -
my understanding is they use nitrous oxide for more power -
there is considerable slipstream generated, and it will cause
the tail to produce a lot of lift.

[pdw]

[quote/]

What forces result?

Strong lift above this horizontal stab (amount depending on actual airflow speed) given a sharp down-deflection angle of the elevator surface into the relative airflow as shown.

[bezerker]

Photofly, the point I was making was that only for the last inch or two of the C of G range is behind the aerodynamic centre of lift (which was not obvious from your first post).

If you include the pitching moment about the aerodynamic centre of lift (which may have been covered, too lazy to read it all again), there is no question that the aircraft (even with a max aft C of G) will be pitching forward, needing a downward force on the elevator to counteract it.

Again, too lazy to do math to compute rotational forces about centre of aerodynamic centre.

[photofly]

Photofly, the point I was making was that only for the last inch or two of the C of G range is behind the aerodynamic centre of lift (which was not obvious from your first post).

Oh, ok. If that's your point, it's wrong.

Again, too lazy to do math to compute rotational forces about centre of aerodynamic centre.

Thanks for your contribution.

Let me say that I'm not too lazy, and I have done the maths. I posted it in this thread. You're welcome to check it and see if I made an error, or work through it your own. In fact I'd be grateful if you did; then if you get a different result we can check why. But given that I have done the maths, and you're too lazy to do so, I'm not too lazy to say that you're wrong:

If you include the pitching moment about the aerodynamic centre of lift (which may have been covered, too lazy to read it all again),

It was indeed covered. How much effort does it take to read through a thread before posting?

http://www.avcanada.ca/forums2/viewtopi ... 76#p810376
http://www.avcanada.ca/forums2/viewtopi ... 34#p810434
http://www.avcanada.ca/forums2/viewtopi ... 39#p810939

there is no question that the aircraft (even with a max aft C of G) will be pitching forward, needing a downward force on the elevator to counteract it.

If you stop being too lazy, and actually do the maths, then you'll find exactly the opposite.

[photofly]

(Mr Pitts put) the top wing (ahead of) the lower wing

The reason he did that was so that you could
get in and out of the airplane! The original version
of the airplane had flat-bottom wings. The round
wings came much later.

I'm not it matters much, frankly. The Beech Staggerwing
did it the other way 'round (top wing behind bottom wing)
and it flew just fine.

It's important when designing the aircraft to be stable; the Staggerwing proves the point that you can make the stability up elsewhere, but it makes a big difference. You might as well say that it's not very important that an aircraft has a tail, after all Concorde didn't have one and it flew just fine.

[Colonel Sanders]

There's all sort of nonsense in aircraft design, and
very little of it was engineering driven.

For example, back in the 1950's, Cessna had a
straight tail. Wonderfully effective. But then the
marketing suits got involved, and decided it would
look at lot racier if the tail was swept. So, we ended
up with the crappy configuration of today, with limited
rudder effectiveness.

Cessna aren't the only styling idiots. Piper decided in
the 1970's that their Cherokee 6 would look at lot
zoomier with a T-tail. Now, that required another 100+
lbs of metal to support the structure, and they ended
up with an inferior aircraft because the elevator was
now out of the slipstream. It was so bad, they eventually
scrapped it and went back to the conventional tail.

My point is that aircraft are not styled and designed by
Albert Einstein, which is the definite impression one could
get by reading AvCan.

I might remind the people here that Curtis Pitts inspected
welds for the USN during WWII, and B. D. Maule was a
mechanic. I am fairly certain that neither of these very
successful aircraft designers and manufacturers could
keep up with the conversation here, having met them both.

I can assure you that neither of them were Albert Einstien,
or even Richard Feynman.

[Colonel Sanders]

On the subject of stick vs chalk ...

Has Denker (or anyone else here) ever flown an
ultralight Challenger? What did you think of it's
flight characteristics?

[Chuck Ellsworth]

Has Denker (or anyone else here) ever flown an
ultralight Challenger? What did you think of it's
flight characteristics?

Oh dear God that was one of the most delightful aircraft I ever flew.

Such well balanced flight control response, made a Pitts feel like a dump truck with flat tires.