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Old May 14, 2012, 01:22 PM
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Originally Posted by richard hanson View Post
no-- different ballgame.
We are working with a symetrical wing tho it could have a different curve -
The 'claim" is th the craft could be trimmed for steady speed, level flight - then simply rolled inverted and with NO trim inputs or speed change continue on in steady level flight
If there were no gravity -it would be easier but in real world the lift oprovided by the wing has to reverse sides and that requires some force(s) to change and hold the wing in the new position.
I am pretty good at trimming but I can't see how this is possible.
Sorry I confused you by addressing two of your claims in one post.

So, l'll stick with this one. With symetrical sections and with the cg in just the right spot it would invert and require no trim change if it were not for the drag of verical fin that is causing a pitching moment. Or, no matter where the cg is, inverting the plane would have the exact opposite decalage for trim if it not for the pitching moment from fin drag.
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Old May 14, 2012, 01:25 PM
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There've been any number of lateral/vertical area equalized pattern planes which needed trim changes upright and inverted, if memory serves.
If there was no trim change needed, these planes would dominate the field.
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Old May 14, 2012, 02:09 PM
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Originally Posted by Sparky Paul View Post
There've been any number of lateral/vertical area equalized pattern planes which needed trim changes upright and inverted, if memory serves.
If there was no trim change needed, these planes would dominate the field.
Read my last sentence again....

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Originally Posted by Steve Anderson View Post
Or, no matter where the cg is, inverting the plane would have the exact opposite decalage for trim if it not for the pitching moment from fin drag.
Exact opposite decalage may be a trim change! Without a verticle fin, or some other asymmetry, inverted would have the exact opposite trim. It could be many degrees, or none at all, depending on the cg location.
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Old May 14, 2012, 03:28 PM
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Well Steve - how exactly does the required AOA re establish it's self?
Try it -setup a cg where downforce is positive- zero -or positive - I really don't care which
any setup has to fight gravity and when craft is turned upside it is then reversed in it's effect.
The wing will NOT establish correct AOA all on it's own-it must be held at correct AOA by some force.
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Old May 14, 2012, 05:34 PM
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Originally Posted by richard hanson View Post
Well Steve - how exactly does the required AOA re establish it's self?
Try it -setup a cg where downforce is positive- zero -or positive - I really don't care which
any setup has to fight gravity and when craft is turned upside it is then reversed in it's effect.
The wing will NOT establish correct AOA all on it's own-it must be held at correct AOA by some force.
The exact same way it is established right side up. If everything is symmetrical and trimmed and you roll it over why would it require anything but an equally opposite trim setting?

If the cg is maintaining a given AOA for level flight without any decalage as it rolls to 90 gravity is no longer influencing the airplane in pitch and so the wing is at 0 AOA. From there it does not matter if you go inverted or back to right side up, the AOA is going right back to what it was before. Why wouldn't it?
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Old May 14, 2012, 05:37 PM
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Quote:
Originally Posted by richard hanson View Post
Well Steve - how exactly does the required AOA re establish it's self?
Try it -setup a cg where downforce is positive- zero -or positive - I really don't care which
any setup has to fight gravity and when craft is turned upside it is then reversed in it's effect.
The wing will NOT establish correct AOA all on it's own-it must be held at correct AOA by some force.
I'm curious as to what level of fidelity would be required to demonstrate this sufficiently. I made a 5g free-flight balsa glider that seems to show that you can achieve upright and inverted trim without changing anything. My specs are:

wing root chord: 50 mm
wing tip chord: 50 mm
wing semi-span: 100 mm
wing sweep: 0

tail root chord: 30 mm
tail tip chord: 30 mm
tail semi-span: 50 mm
tail sweep: 0

wing LE to tail LE: 70 mm
wing/tail incidence: 0/0

CG: approx. 18 mm from wing LE

vertical tail chord: 25 mm
vertical tail span: 32 mm

Flat (unshaped) balsa sheeting for all surfaces.

I can throw this and it will achieve a stable glide. I turn it upside down and throw it at about the same speed. It achieves a stable glide that looks identical (to my eye) as when I threw it "upright".
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Old May 14, 2012, 05:39 PM
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Because there must be some effective AOA on the wing to establish a pressure difference top /bottom
No pressure difference = no lift.
however
Your example would work in vertical travel .
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Old May 14, 2012, 05:41 PM
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Originally Posted by ShoeDLG View Post
I'm curious as to what level of fidelity would be required to demonstrate this sufficiently. I made a 5g free-flight balsa glider that seems to show that you can achieve upright and inverted trim without changing anything. My specs are:

wing root chord: 50 mm
wing tip chord: 50 mm
wing semi-span: 100 mm
wing sweep: 0

tail root chord: 30 mm
tail tip chord: 30 mm
tail semi-span: 50 mm
tail sweep: 0

wing LE to tail LE: 70 mm
wing/tail incidence: 0/0

CG: approx. 18 mm from wing LE

vertical tail chord: 25 mm
vertical tail span: 32 mm

Flat (unshaped) balsa sheeting for all surfaces.

I can throw this and it will achieve a stable glide. I turn it upside down and throw it at about the same speed. It achieves a stable glide that looks identical (to my eye) as when I threw it "upright".
Not the same thing .
But I don't doubt your results. Your model is sinking - I am looking at sustained level flight as a goal.. That requires power
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Old May 14, 2012, 06:22 PM
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Quote:
Originally Posted by richard hanson View Post
Because there must be some effective AOA on the wing to establish a pressure difference top /bottom
No pressure difference = no lift.
however
Your example would work in vertical travel .
Like I wrote...
Quote:
Originally Posted by Steve Anderson View Post
If the cg is maintaining a given AOA for level flight without any decalage
I think "given AOA for level flight" means the same thing as effective AOA for level flight. In other words, it has sufficient lift for level flight. Does it not?

You do understand that the AOA of an airplane can be controlled with the elevator fixed in place, with or without decalage, simply by moving the cg fore and aft?
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Old May 14, 2012, 06:47 PM
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Originally Posted by Steve Anderson View Post
Like I wrote...


I think "given AOA for level flight" means the same thing as effective AOA for level flight. In other words, it has sufficient lift for level flight. Does it not?

You do understand that the AOA of an airplane can be controlled with the elevator fixed in place, with or without decalage, simply by moving the cg fore and aft?
absolutely -I do it all the time.
You are proposing a arrangement where the stab produces NO net downforce- correct?
Have you tried this?
powered model - held level hands off flight? upright and inverted?
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Old May 15, 2012, 04:11 AM
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Quote:
Originally Posted by richard hanson View Post
Well Steve - how exactly does the required AOA re establish it's self?
Try it -setup a cg where downforce is positive- zero -or positive - I really don't care which
any setup has to fight gravity and when craft is turned upside it is then reversed in it's effect.
The wing will NOT establish correct AOA all on it's own-it must be held at correct AOA by some force.
Richard,

A free-flight glider that glides steadily both upright and inverted with no configuration change demonstrates that the required AOA will reestablish itself. If the glider maintained the same AOA when inverted it would accelerate into the ground in a progressively steeper dive... It doesn't do that. The wing establishes one AOA upright and a different AOA inverted.
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Old May 15, 2012, 06:58 AM
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Not the same thing-
You are describing a model -powered by gravity which
"possibly" if balanced with NO downforce on the tail, could roll inverted then reestablish a positive AOA on the wing.
This not the same as trimming for level ,constant speed - then rolling to inverted and reestablishing positive AOA.
How do you re establish the AOA needed ?
Also - the craft must be powered to even attempt this
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Old May 15, 2012, 09:08 AM
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Originally Posted by richard hanson View Post
absolutely -I do it all the time.
Great. Then you already know your next question is wrong.

Quote:
Originally Posted by richard hanson View Post
You are proposing a arrangement where the stab produces NO net downforce- correct?
No. If the stab/elevator is fixed and the AOA is adjusted by moving the cg fore and aft, the load between the wing and tail changes as well.

Quote:
Originally Posted by richard hanson View Post
Have you tried this?
No. But I have tried that. Does that count?

Quote:
Originally Posted by richard hanson View Post
powered model - held level hands off flight? upright and inverted?
It makes no difference if it is gliding, powered level flight or climbing, conventional, tandem or canard, trim is trim. If everything is symmetrical the equal opposite trim setting while inverted will result in the exact same AOA and lift. If everything is symmetrical how can anything be different?
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Old May 15, 2012, 09:35 AM
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Also - the craft must be powered to even attempt this
The power is a red herring to the issue at hand, but I'm sure this is also possible.
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Old May 15, 2012, 10:44 AM
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[QUOTE=Steve Anderson;21618506...


It makes no difference if it is gliding, powered level flight or climbing, conventional, tandem or canard, trim is trim. If everything is symmetrical the equal opposite trim setting while inverted will result in the exact same AOA and lift. If everything is symmetrical how can anything be different?[/QUOTE]
.
Other than a symmetrical airfoil having no lift at zero AOA?
And the horizontal at zero AOA?
There must be longitudinal dihedral for stable flight, upright.
Either the wing will have (must have) incidence if symmetrical, or the tail is deflected from zero to give the download the plane will need for stable flight.
With the surfaces fixed, it can not fly stably inverted, it will dive.
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