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" How I Almost Destroyed a 50 million War Plane...'

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" How I Almost Destroyed a 50 million War Plane...'

Old 30th Jan 2016, 11:58
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Originally Posted by Fortissimo
Tourist, there are indeed some misunderstandings on here. You do not get -1g simply by rolling inverted. You have to remain in level flight to achieve that. But I don't have a monopoly on wisdom, so please feel free to give me the benefit of yours so I can rest easy as well.
Yes you do.
Simply rolling inverted gives you exactly -1g.

Doing anything else does not, i.e. not remaining in level flight.


I can make this easier for you if you want, but I thought the phrase "simply" was clear enough.

Here goes.

Roll inverted.
Don't do anything else.
Don't apply full left or right pedal.
Don't change the power setting.
Don't deploy speedbrakes.
Don't deploy flaps.

Just roll inverted and continue to maintain non accelerative flight with reference to the mean surface of the planet you are currently orbiting.

At this point the only acceleration you are experiencing is the acceleration due to the local gravity of the position you are in. This is generally considered to be a constant of approx. 9.81m/s2 on the surface of planet earth with obvious local variations for velocity of aircraft, position and height on planet earth and as affected by our orbit around the sun and the moons orbit around us.
We call this 1g, or in the event we are discussing, -1g due to the orientation of the aircraft




Personally I think my first statement covered the relevant facts and was more succinct.
If you want any more clarification, please get back to me.
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Old 30th Jan 2016, 13:54
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If you roll inverted and wish to maintain level flight you will require to apply more thrust - unless you have a symmetrical aerofoil with a zero angle of incidence.
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Old 30th Jan 2016, 14:08
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Originally Posted by Basil
If you roll inverted and wish to maintain level flight you will require to apply more thrust - unless you have a symmetrical aerofoil with a zero angle of incidence.
Not necessarily, but good general point.

I should have said apply the correct amount of thrust to maintain speed.
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Old 30th Jan 2016, 14:49
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To Tourist.

Not necessarily, but good general point.
Why was Basil's statement not necessarily correct Tourist? Please explain.
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Old 30th Jan 2016, 15:10
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Originally Posted by deefer dog
Why was Basil's statement not necessarily correct Tourist? Please explain.
Because depending on the aerofoil and angle of incidence, it is possible that you would have to reduce thrust.

Come on, keep up.
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Old 30th Jan 2016, 15:49
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As Tourist points out, any non-accelerative (constant-speed vector ) trajectory will give you constant pilot-relative 1g. You can then choose to apply this 1g in any pilot-relative direction you want by rotating the plane into any acceptable attitude.

Presumably vectored thrust fly by wire aircraft can maintain very strange constant-speed-vector attitudes, in the same way as powered aircraft can maintain constant-speed-vector attitudes quite different from gliders.


Edmund
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Old 30th Jan 2016, 16:24
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back to tourist

Because depending on the aerofoil and angle of incidence, it is possible that you would have to reduce thrust.
Err...but as basil pointed out, in the case of a symmetrical airfoil, with zero angle of incidence, that is not the case. So I ask again, why "not necessarily" in response to basil's post?
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Old 30th Jan 2016, 18:22
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Originally Posted by deefer dog
Err...but as basil pointed out, in the case of a symmetrical airfoil, with zero angle of incidence, that is not the case. So I ask again, why "not necessarily" in response to basil's post?
Ok, I'll try again.

It is possible, would you agree, to have an asymmetric aerofoil that has superior efficiency inverted than the right way up. It is also possible to mount a wing with an angle of incidence giving reduced drag inverted.

Under both or either of those circumstances, it is possible to have a reduced thrust requirement inverted than the right way up.

Not likely, but possible, and in the realms of this "aeronautic basic concepts shlong measuring" debate which we are all involved in, possible is plenty.

Hence "not necessarily, but good general point"

As I said, come on, keep up!
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Old 31st Jan 2016, 19:20
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There are 2 reasons I find this 'story' a load of bs,
A- no engineer would suggest this solution unless it was a one off rtb.
B- surely select the gear count to ten before going negative g would be a far smarter solution or as suggested inverted flight before gear retraction.
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Old 31st Jan 2016, 20:09
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Point A is a fair point.
Point B is another misreading. It was zero not negative they wanted.
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Old 31st Jan 2016, 21:20
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How about just doing the whole maneuver in IMC(if you decide to do the maneuver that is).
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Old 31st Jan 2016, 21:33
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Simply rolling inverted gives you exactly -1g.
No you are not.

The aerofoil (or for what matters the entire airframe) doesn't know what's up and what's down as long as the entire contraption is flown straight and lever at constant speed (for relatively short distances the said planet can be considered flat). Drag might be different but that's already another story.

+ or - are mathematical conventions, depending upon the referencing system one's choose. Since weight is a "standard" force experienced by everything on earth, it has been chosen to be positive, perpendicular to the surface and directed to the centre. Wanna go zero-g or negative? Get high the push down sharply...

PZ
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Old 31st Jan 2016, 22:15
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Question Up from Down

The aerofoil (or for what matters the entire airframe) doesn't know what's up and what's down as long as the entire contraption is flown straight and lever at constant speed
True- but what is missing from this ' erudite' discussion is the concept of up versus down.

For a standard airfoil - it is shaped such that the ( lift vector ) in normal flight is UP which allows the plane to fly level and generally at right angles to the gravity vector ( DOWN ) in normal flight. When the two vectors ( lift versus gravity or UP versus DOWN ) balance- the plane flys level at some elevation above local earth.

BUT if you turn the same wing inverted or upside down , then the LIFT VECTOR ADDS TO THE GRAVITY VECTOR, and absent some other forces the plane will go down, still in level flight -at least for a while.

Keep in mind that I am talking steady state conditions- absent any other control manipulations or forward vector changes in speed- momentum which can delay the result net DOWN for a while.
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Old 1st Feb 2016, 04:50
  #34 (permalink)  
 
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You do not get -1g simply by rolling inverted. You have to remain in level flight to achieve that
You don't have to remain in level flight. Rolling inverted and maintaining the pre roll power setting and airspeed will see the aircraft accelerate into and establish a fixed rate of descent due to increased drag (symmetrical airfoils aside). Once the steady state inverted descent is established -1 "g" will be experienced. Accelerating to the steady state descent will see >-1 "g", -.8 as an arbitrary example.
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Old 1st Feb 2016, 08:57
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Originally Posted by CONSO

BUT if you turn the same wing inverted or upside down , then the LIFT VECTOR ADDS TO THE GRAVITY VECTOR, and absent some other forces the plane will go down, still in level flight -at least for a while.
That's exactly right, and that is why when doing a quick aileron roll, you find the aircraft accelerating downwards at nearly 20 m/s2 during the inverted phase.....

....oh, wait.
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Old 1st Feb 2016, 09:06
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Originally Posted by megan
You don't have to remain in level flight. Rolling inverted and maintaining the pre roll power setting and airspeed will see the aircraft accelerate into and establish a fixed rate of descent due to increased drag (symmetrical airfoils aside). Once the steady state inverted descent is established -1 "g" will be experienced. Accelerating to the steady state descent will see >-1 "g", -.8 as an arbitrary example.
Not really true if we are using the aircraft as the basis of what is up or down re g.

As the angle of descent increases, the g relative to the plane of the aircraft's usual up or down drops to zero as the aircraft reaches the vertical.

ie gravity will be pulling directly towards the nose once vertical so zero g relative to gear going up or down.

Sorry, that is badly explained, but I hope you understand what I mean.

I am told that test pilots sometimes do test in an angle of descent for specifically this reason to get less than one g with respect to the aerodynamic surfaces etc despite being in a steady state.

Despite the drop in negative g experienced by the undercarriage due to a descent, you could bring that negative g back up to -1g by throwing in a turn.
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Old 1st Feb 2016, 10:21
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interesting ,

I always thought I'd have -1G as I am pushing the stick to stay at level while inverted.
Pushing the stick further gives -2G or more (but then I am not flying level anymore) up to e.g. an inverted loop with -3G up to -4G.

That's the way I explained the -1G inverted in the simplest terms , still correct I hope ?
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Old 1st Feb 2016, 11:46
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Originally Posted by joe two
I always thought I'd have -1G as I am pushing the stick to stay at level while inverted.
The fact that if you build a symmetric aerofoil aircraft with zero angle of incidence it is possible to have zero trim change inverted suggests that the push is not what is giving you the negative g....
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Old 1st Feb 2016, 13:09
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yep , convinced ...
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Old 1st Feb 2016, 14:42
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Missing the point?

Am I the only one who is bemused that the folks posting here seem to have missed the point of the article? I'd call it a classic I Learnt About Flying From That (ILAFFT) piece. Normalization of deviation, Human Factors etc. just saying...

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