RE: Horiz stab = any lift???  
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RE: Horiz stab = any lift??? - 1/25/2004 5:46:59 PM   
Ben Lanterman



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From: St. Charles, MO, USA
Status: offline 		The reason it seems counterinuitive and this is the hard part to change in your thinking........... is that most people seem to think there is a down load on the tail holding up the weight of the engine on the nose and it is balanced about the CG. They forget about the wing lift in front of the CG and that the weight of the motor along with the other airplane components is vectorially summed and becomes the CG. When you sum moments about the CG location the motor is already taken care of. You have just the aerodynamic wing and tail forces relative to the CG to take care of. You can use other locations but the math is more complex and the answer is the same.

The angle of attack of the wing (leading edge up) is 20 degrees. The elevator deflection (trailing edge up) is say 20 degrees. The angle of attack alone on the tail at zero elevator deflection would give a humonguous (tech term) lift. The elevator deflection decreases it. But the balance about the CG must be maintained. If the wing is lifting up and the airplane is in steady flight the tail must be lifting up. It is a simple balance. It is easy to work out.

10 inch wing chord, 20 pounds of lift on wing, CG is one inch aft of the wing 25% point, tail arm from CG to tail is 20 inches.

20lbs x 1in = 20in x Xlbs

X = 1 pound of lift on the tail directed up. It works for any steady state flight condition with zero camber and no flaps (assume no strange fuselage shapes)

For an airplane weight there is an angle of attack and an elevator deflection that allows the airplane to trim out and fly level. When you change elevator from that trim condition, say 5 degrees, to 10 degrees trailing edge up you are no longer aerodynamically balanced for that weight. The airplane rotates until the angle of attack of the wing and tail increase and you probably start climbing at a steady rate. In that climb the tail load is still up. With even more elevator deflection you have reduced the load on the tail enough to loop.

Imagine if you will an all moving horizontal tail and the angle is measured with respect to the wing.

If the wing is 10 degrees angle of attack and the tail is 15 deg leading edge down then the tail load is down.
If the wing is 10 degrees angle of attack and the tail is 10 deg leading edge down then the tail load is zero.
If the wing is 10 degrees angle of attack and the tail is 0 deg leading edge down then the tail load is up.
If the wing is 10 degrees angle of attack and the tail is 10 deg leading edge up then the tail load really up.

What the airplane does at these conditions depends on the airplane weight and stability parameters.

If you add flaps or camber on the wing which make a big enough nose down pitching moment then the tail must counter it and can end up with a down load. Look at how it works. The wing and tail are at say 5 degrees angle of attack and every thing is balanced out in level flight. Lower the flaps and the airplane takes a nose dive and the lift goes up. The angle of attack might even have to go to zero to maintain level flight. The net result can make enough nose down moment that the tail must create a down load. Since the angle of attack of the tail is the same as the wing (zero maybe) it is not creating any forces due to angle of attack. You get the down load from the application of up elevator.

I have carefully ignored what happens in 3 D maneuvers, snaps, etc. There the forces are highly irregular and it can only be determined by watching where the airplanes goes - too hard and too much work.


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Do this experiment - 1/25/2004 6:00:38 PM   
Ben Lanterman



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Status: offline 		Get your favorite airplane. Put it on the table and put wedges under each wing panel until the airplane balances. It should be nice and tippy in pitch. Any small push on nose or tail makes it move easily. The mass vectors are balanced.

Now put an eyescrew on the centerline of the fuselage and located at the 25% chord location and tie a string to it. The eyescrew should be in front of the CG. Ok just tape a string at the 25% location if you don't want to screw it.

Pull up (wing lift simulation) 1 pound.

Which way do you have to push on the tail to make it level again. How much did you have to push?

Pull up (wing lift simulation) 5 pounds.

Which way do you have to push on the tail to make it level again. How much did you have to push?


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Ben Lanterman

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RE: Do this experiment - 1/25/2004 11:47:25 PM   
Tall Paul



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Status: offline 		Airflow, airflow, airflow....
"Get your favorite airplane. Put it on the table and put wedges under each wing panel until the airplane balances. It should be nice and tippy in pitch. Any small push on nose or tail makes it move easily. The mass vectors are balanced."
Without that, a suspended test isn't valid.
Aerodynamic forces on the airplane AND inertial forces sum up in flight.

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I can't tell if you agree with me or not ! - 1/26/2004 1:11:54 AM   
Ben Lanterman



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Status: offline 		"Aerodynamic forces on the airplane AND inertial forces sum up in flight." Which is the reason I mention a lot about steady state trimmed flight thich effectively removes the inertial forces from the simulation. I have also omitted downwash which isn't that big of a factor (put the elevator high out of it like a Cap). The you have the total lift from wing and tail equals the weight of the airplane, the pitching moment of inertia doesn't count unless you are in accelerated flight and it leaves you with just... the moments due to aerodynamics, which in a simplified examination, are a result of two things, the wing lift and tail lift and their associated moment arms.

I could sum moments for steady state flight about the nose of the spinner. Then you have the wing lift times the distance from the nose to the quarter cord of the wing, the weight of the airplane times the distance from the nose to the CG and the tail lift times the distance from the nose to the tail.

You can get a simple answer by examination since the weight (CG) moment is greater than the wing (lift) moment you must have a lift at the tail. Neat stuff.


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RE: I can't tell if you agree with me or not ! - 1/26/2004 8:19:32 AM   
Tall Paul



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Status: offline 		As long as there's airflow over a cambered surface, there's a pitching moment.
It's not an acceleration, it's a constant force.
Its effect requires a down-load on the horizontal for level trimmed flight.
Lift in level flight is composed of both weight AND that increment needed to carry the download.

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RE: I can't tell if you agree with me or not ! - 1/26/2004 3:27:16 PM   
Oryx


 

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quote:

ORIGINAL: Tall Paul
As long as there's airflow over a cambered surface, there's a pitching moment.


I agree that with a cambered airfoil there is usually a downward pitching moment for the wing in isolation about the aerodynamic center (AC) of the wing. However, to determine whether a tail is lifting up or down, you have to do the force and moment balance about the CG of the aircraft so that you can take the weight of the aircraft out of the equation (it is of course still implicitly in the equation since the amount of lift required from the aircraft depends on the weight). The amount the tail needs to lift and whether that lift is up or down, is going to depend on the angle of attack, the size and location of the tail, pitching moment of the tail airfoil, etc. Many aircraft, even ones with a lot of camber on the wing, can have the tail lift down at cruise or high speed, while the tail lifts up at higher angles of attack. You also have to consider that although the CG is ahead of the aircrafts NP for positive stability, it can still be (and often is) behind the wings own AC. If you now resolve the forces and moments about the CG, you can see how the lift of the wing, which causes a nose-up pitching moment about the CG, may cancel out or be more than the constant downward pitching moment about the wings AC.

As an example, have a look at OHS (outboard horizontal stabilizer) designs, which I have seen papers on at a few more recent AIAA conferences. These aircraft have the tail in the upwash behind and outboard of the wing, but since they are still behind the wing the longitudinal stability works the same way as for a conventional aircraft. These designs rely on the tail ALWAYS lifting up, since that is required to gain the low induced drag advantage of the OHS configuration. That is usually achieved despite the aircraft having a large positive static margin and a conventional main wing airfoil with a negative pitching moment (again refering to the wing or airfoil AC of course). To design it that way all you need to do is to adjust the size and relative location of the tail. You can achieve the same effect with a conventional design if the tail is sized correctly and placed at the right location. It generally requires a configuration that places the NP as far as possible behind the wings AC, so that a positive static margin still places the CG relatively far behind the wings AC.

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RE: I can't tell if you agree with me or not ! - 1/26/2004 5:52:43 PM   
Tall Paul



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Status: offline 		Oryx, the OHS planes I've seen at the SAE Aero Design West competitions hardly inspire confidence in the practicality of the configuration, if only from a structural standpoint.
Placing a potential oscillator at the end of the wing, and then out on a boom is a flutter incident begging to occur.
As for the tail lifting up. Nope!
Look at these images of U.Ill's 2002 entry.. the horizontal is keeping the nose up. always...
I'd check the math on the neutral point computations.
Moving the tail outboard doesn't do the magic thing it's claimed to do.
OHS designs are IMNSHO, exercises in futility, since the exact same result can be achieved lighter, cheaper and easier with a conventional layout.
Interesting to watch, but aeronautical dead ends.

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Agreed, but....... - 1/26/2004 5:54:00 PM   
Ben Lanterman



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Status: offline 		Well written Oryx.

>As long as there's airflow over a cambered surface, there's a pitching moment.

Agreed

>It's not an acceleration, it's a constant force.

Agreed

>Its effect requires a down-load on the horizontal for level trimmed flight.


It requires an "incremental" down load. I usually do the lift on the tail with all symmetrical airfoils to remove the camber or flap pitching moment and have always said that they change the amount of lift the tail has to make. With a midly cambered airfoil there is an incremental lift needed at the tail to compensate. For a complete answer that increment plus the elevator increment is added to the basic lift at what ever the local angle of attack due to downwash is. But the final answer is usually up.

The reason my experiment with a pattern ship works is because there are no camber effects.

The other thing that messes up perceptions is when you look at a typical trainer. The tail is set negative to the wing so the assumption is that the tail is lifting down. But all that does is give you angle of attack stability. It hasn't given a down load on the tail.

>Lift in level flight is composed of both weight AND that increment needed to carry the download.

Agreed

I looked up naca report 792 on the internet. They put strain gages and pressure taps on a typical fighter and flew the beast. The tail loads were up during maneuvering except..... because of airplane moments of inertia if you give a small quick stick yank, the tail load is down initially. As the airplane accelerates in pitch the loads begin to decrease until the reverse and return to up at the new trimmed condition.


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RE: I can't tell if you agree with me or not ! - 1/26/2004 6:19:49 PM   
Oryx


 

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quote:

ORIGINAL: Tall Paul
Oryx, the OHS planes I've seen at the SAE Aero Design West competitions hardly inspire confidence in the practicality of the configuration, if only from a structural standpoint.
Placing a potential oscillator at the end of the wing, and then out on a boom is a flutter incident begging to occur.
As for the tail lifting up. Nope!
Look at these images of U.Ill's 2002 entry.. the horizontal is keeping the nose up. always...
I'd check the math on the neutral point computations.
Moving the tail outboard doesn't do the magic thing it's claimed to do.
OHS designs are IMNSHO, exercises in futility, since the exact same result can be achieved lighter, cheaper and easier with a conventional layout.
Interesting to watch, but aeronautical dead ends.


Paul, that particular airplane that you showed flew with the tail lifting up under all trimmed flight conditions. At low angle of attack (when flown with little ballast), full down elevator had to be kept in for the entire flight and it was landed by pulling back on the throttle all the way (which btw matched the simulations prior to construction exactly). The pictures are very deceptive - the horizontal is placed in a very strong upwash from the wingtips, which is the reason why they are mounted at such a nose-down incidence. However, they still lift up: the local angle of attack is always positive for trimmed flight, and even with up elevator the local CL is positive. The NP calculations on the plane and trim incidences were very close as calculated and no CG or incidence adjustments had to be made during the test flights prior to the competition. You can ask Mike Cross whether it was stable in pitch - his only complaint was roll stability due to the lack of dihedral, but otherwise he was very happy with the stability of the model.

I am not an advocate of OHS designs either, but the one in your pictures was a very interesting design project and from what I remember the configuration was chosen more out of curiosity towards that particular concept than anything else. My use of the OHS as an example in this discussion was because it is a good example of a design that relies on an upward lifting tail. It still does not mean it has to be an OHS to have an upward lifting tail - the same can be true for a conventional configuration with the right sizing and location of the tail.

< Message edited by Oryx -- 1/26/2004 6:51:07 PM >

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RE: I can't tell if you agree with me or not ! - 1/26/2004 7:55:41 PM   
Tall Paul



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Status: offline 		Thinking on it, and relating the flight characteristics of my Senior Kadet with full-flap and full power.. (it has an uncontrollable pitch-up in that configuration), the angle on the tail of the U.Ill plane very well could represent it lifting up.
That airfoil though will have an enormous Cm with that amount of flap.
Do you know it might be? (seeing as you're close to the source)

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RE: I can't tell if you agree with me or not ! - 1/26/2004 8:17:54 PM   
Oryx


 

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quote:

ORIGINAL: Tall Paul
Do you know it might be? (seeing as you're close to the source)


Yes, from what I remember it does have an enormous Cm The guy who did the airfoil design is not in town anymore, so I would have to ask around to track the data down. I think the airfoil was analyzed using MSES, but never tested in the wind tunnel. The data should still be somewhere.

The main problem with that design was the huge amount of drag associated with that airfoil at higher lift coefficients. You will remember that its final attempt with the 21lb load had it take-off in a very short distance, but it then ran out of steam due to all that drag. It climbed out to about 20 feet and was even landed again before the end of the runway was reached. In retrospective, Mike should have kept it on the ground longer so it would take off at a lower CL, but hindsite is always 20/20. Wish I had time to get involved again this year, but my thesis is consuming all my time. It is projects like these that made being a grad student again for a while so much fun

I will try to find out for you what the Cm of the airfoil was.

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RE: I can't tell if you agree with me or not ! - 1/26/2004 8:54:17 PM   
Tall Paul



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Status: offline 		I pulled these images out of the pile...
Way back when Lockheed hooked up with NASA to do a wake-turbulence alleviation program.
We mounted 8 smoke generators supplied by Frank Sanders, who made them for aerobatic airplanes, underneath the Tristar wing..
4 per side.. one immediately on the side next to the landing gear cover, one at the inboard flap outboard edge; the outboard flap outer edge, and underneath the wing tip.
The test was to fly a T-37 into the smoke trail behind the Tristar in landing configuration.. everything down..
It was discovered the turbulence at 5 miles behind the L-1011 was sufficient to kick the Tweet out of the smoke path.
Then the "alleviation" was tried, which consisted of full cycles left and right on the control wheel in the Tristar.
This had an enormous effect on the wake strength.. the T-37 could in much closer.
Of course the manuver would be completely unacceptable in a commerical plane. But it did work.
I took these images.. note how the tip flow almost goes away with down aileron...
There's a NASA report on the test, I'm sure... author should be Joe Tymcyzm.. something odd like that..
Pilot in the Tristar was Fitz-Fulton, and Bill Dana in the Tweet.
.
Model aviation tie-in: To see if the most inboard generator was -on fire-, I had a sheet of Chrome Monokote pasted to the inboard side of each wing engine. Not that we had any way putting out a fire....

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