Spinning - fun question for you
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Spinning - fun question for you
Whilst chatting in the club the other day the following question was posed; you are in an unrecoverable spin, could you generate enough lift from the flying wing by applying into spin rudder so that it reduced your decent enough to make it survivable?
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proberly not.
If you could reduce the rate of decent you would have the wing flying again and thus unstalled so not spinning.
I have once been in a spin in a C152 that wouldn't recover using the normal methods.
The FIC instructor gave it a huge blast of power to get some air going over the rudder and that sorted it.
If you could reduce the rate of decent you would have the wing flying again and thus unstalled so not spinning.
I have once been in a spin in a C152 that wouldn't recover using the normal methods.
The FIC instructor gave it a huge blast of power to get some air going over the rudder and that sorted it.
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The "flying wing"? As in not stalled..? That's a new one... You might need to read up on what defines a spin.
You clearly lack anyone who flies aeros or has much of an idea at your club!
Are you suggesting that you can somehow flatten the spin and accelerate it to the extent that it works like a sycamore seed and the outer wing achieves an AoA that causes it to generate lift? I don't think it's possible unless you had rocket pods installed on each wing tip firing in different directions! Why not face the inside wing backwards too and create some kind of bizarre helicopter?!
You clearly lack anyone who flies aeros or has much of an idea at your club!
Are you suggesting that you can somehow flatten the spin and accelerate it to the extent that it works like a sycamore seed and the outer wing achieves an AoA that causes it to generate lift? I don't think it's possible unless you had rocket pods installed on each wing tip firing in different directions! Why not face the inside wing backwards too and create some kind of bizarre helicopter?!
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Hi Mad Jock, in the aero modeling world there is a manouvre called 'The Blender', where full power is applied whilst in a spin. It stops the rate of descent and the plane just rotates very rapidly about its vertical axis.
Maybe the full size C152 would not have the power to weight ratio to do that however.
Maybe the full size C152 would not have the power to weight ratio to do that however.
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phiggsbroadband - interesting. And the recovery is the opposite action presumably?
sapperkenno - read up on the meaning of the word 'fun', then have a nice lie down.
sapperkenno - read up on the meaning of the word 'fun', then have a nice lie down.
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Whilst chatting in the club the other day the following question was posed; you are in an unrecoverable spin, could you generate enough lift from the flying wing by applying into spin rudder so that it reduced your decent enough to make it survivable?
The fact that you are in an "unrecoverable spin" means that the control surfaces, and anything else you can control such as the engine power and propeller pitch, are not capable of arresting the spin. So you can't (as described above with regard to a Cessna 15x) apply power to give the rudder and tail enough effect to arrest the spin.
Since both wings are stalled, neither is generating lift, so you are basically a rock.
If you were to decrease the stall significantly in the lesser stalled wing, to the point that it was creating lift, which would slow your descent, you would, by definition have arrested the spin. But you're in an unrecoverable spin, so that's not possible.
So the answer to your question is no, you can not make it 'survivable' because you can't create lift in a stalled wing unless you un-stall the wing, and that, again, would mean that you're no longer in a spin.
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In a fully developed spin they do generate nearly the weight of the aircraft as lift+drag (the rest is generated by fuselage and tailplane), otherwise you would accelerate indefinitely...
In a really flat spin the only generate lift
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Those definitions only work for lamina flow once your past that into seperated flow and stokes flow your into a whole different area of aerodynamics of a none linear nature.
In which purely increasing the angle of attack will not give a predictable responce for what you would expect. It may very well increase the drag then reduce it again etc etc so just getting the plane flat won't garantee anything.
If you could stop the spinning and have it flat falling I will give you that its going to be more.
The only way to find out would be wind tunnel testing to find out whats the best nose attitude to spin at to create most drag in the vertical plane ie the direction your dropping.
In which purely increasing the angle of attack will not give a predictable responce for what you would expect. It may very well increase the drag then reduce it again etc etc so just getting the plane flat won't garantee anything.
If you could stop the spinning and have it flat falling I will give you that its going to be more.
The only way to find out would be wind tunnel testing to find out whats the best nose attitude to spin at to create most drag in the vertical plane ie the direction your dropping.
From http://en.wikipedia.org/wiki/Lift_(force)
Consider an extreme case of a very flat spin where the angles of attack are approaching 90 deg. Not a lot of lift at all. A lot of drag and it is acting pretty much vertically upwards. Flat plate Cd around 1.0 from memory so you can roughly calculate the vertical velocity needed for drag to equal weight.
Some-one can then consider whether that vertical velocity is survivable. Tiger Moth per abgd, apparently, with its low wing loading.
A fluid flowing past the surface of a body exerts surface force on it. Lift is the component of this force that is perpendicular to the oncoming flow direction.[1] It contrasts with the drag force, which is the component of the surface force parallel to the flow direction
Some-one can then consider whether that vertical velocity is survivable. Tiger Moth per abgd, apparently, with its low wing loading.
Yep, whatever the flow mechanism there is a resultant force - lift and drag are simply defined as the component normal to, and parallel to the relative airflow.
I agree, it is a bluff body so you can determine the drag of that and go from there.
I agree, it is a bluff body so you can determine the drag of that and go from there.
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Flat plate Cd around 1.0 from memory so you can roughly calculate the vertical velocity needed for drag to equal weight.
Also the flow would be incredibly turbulent, so I doubt that your 90 degrees would be reliably maintained. Especially after the wings come off!
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Ain't going to work in 3D and it will depend on the reynolds number whats going to happen. There is all sorts of vortexes coming into play and where they produce there max drag who knows only can be found out in a wind tunnel because its so none linear.
Also as well you don;t loose all your lift after stalling there is still a fair bit left up to half until 45degrees angle of attack.
And if its bluff body with no spin the tail is offset so will lift with the drag and the nose will drop and we are flying again.
Oh and a foil has a Cd that get up to 1.5-1.7 at 90 degrees angle of attack which is why they don't park the wind turbines with blades flat.
I reckon its going to be about 45 to 50 degrees angle of attack which is going to give you max combined lift and drag post stall.
Also as well you don;t loose all your lift after stalling there is still a fair bit left up to half until 45degrees angle of attack.
And if its bluff body with no spin the tail is offset so will lift with the drag and the nose will drop and we are flying again.
Oh and a foil has a Cd that get up to 1.5-1.7 at 90 degrees angle of attack which is why they don't park the wind turbines with blades flat.
I reckon its going to be about 45 to 50 degrees angle of attack which is going to give you max combined lift and drag post stall.
Last edited by mad_jock; 11th Dec 2012 at 19:36.
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Re the model aircraft 'blender'; if it's no longer decending, it's no longer spinning (if it ever was!).
What I saw in that vid was a series of down vertical aileron rolls, followed by an inverted spiral dive (aeroplane inverted and rotating, with down elevator applied).
The clue that it wasn't spinning was that the aeroplane recovers by the elevators making the tail go down even more and presenting the inverted wings at a greater AoA to pitch the nose up (if it had been spinning inverted it'd have just mushed into the ground as more down elevator is applied - just like more up elevator would cause in an errect spin).
Once nose-up, excess engine power pulls it into a climb.
What I saw in that vid was a series of down vertical aileron rolls, followed by an inverted spiral dive (aeroplane inverted and rotating, with down elevator applied).
The clue that it wasn't spinning was that the aeroplane recovers by the elevators making the tail go down even more and presenting the inverted wings at a greater AoA to pitch the nose up (if it had been spinning inverted it'd have just mushed into the ground as more down elevator is applied - just like more up elevator would cause in an errect spin).
Once nose-up, excess engine power pulls it into a climb.
Last edited by Shaggy Sheep Driver; 11th Dec 2012 at 19:47.
Yep. I spent some time working with NASA's spin tunnel (with free spinning models and rotary balance) so I have seen lots of data.
Somewhere online is a video of NASA flight tests of a Grumman Trainer which spins extremely flat with no power - and is generally unrecoverable.
NASA also has lots of model spin tunnel data on the same aeroplane and you should be able to find that on the NASA Langley technical data server - you may also find a copy of the same stuff at Cranfield Uni's online library.
Regardless of how you want to define lift or drag, and I agree it is non-linear, you will find that the resultant aerodynamic force acting vertically upwards will generally continue to increase (post-stall) until approx 90 deg AoA.
Depending on the balance of all the aerodynamic and inertia forces and moments - driven by controls and power as appropriate - will determine how flat a specific aeroplane will spin.
I should have mentioned Reynolds Number - you will find a NASA report on that subject too wrt comparison of model spin tests with full scale spin tests.
Some of us also played around with miniature spin tunnels and very small but dynamically scaled models which also had reasonable correlation with full scale. At one time there was one in NASA Langley's Visitor Centre with a Grumman Trainer happily spinning quite flat.
Somewhere online is a video of NASA flight tests of a Grumman Trainer which spins extremely flat with no power - and is generally unrecoverable.
NASA also has lots of model spin tunnel data on the same aeroplane and you should be able to find that on the NASA Langley technical data server - you may also find a copy of the same stuff at Cranfield Uni's online library.
Regardless of how you want to define lift or drag, and I agree it is non-linear, you will find that the resultant aerodynamic force acting vertically upwards will generally continue to increase (post-stall) until approx 90 deg AoA.
Depending on the balance of all the aerodynamic and inertia forces and moments - driven by controls and power as appropriate - will determine how flat a specific aeroplane will spin.
I should have mentioned Reynolds Number - you will find a NASA report on that subject too wrt comparison of model spin tests with full scale spin tests.
Some of us also played around with miniature spin tunnels and very small but dynamically scaled models which also had reasonable correlation with full scale. At one time there was one in NASA Langley's Visitor Centre with a Grumman Trainer happily spinning quite flat.
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I am not disputing max drag will be at 90 degrees
Just that prior to that (45-50 degrees) the lift will still be significant so much so that the drag compnent in the vertical plane with the addition of that lift will be greater than just pure drag component flat.
This will give a lower rate of decent than a pure flat plate effect.
NASA has done tons of work on turbine blades in the last 10 years from a quick search.
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I needed to work out the max moment for the base support for a FEA contract job years ago which was when I found out that taking the max force at flat wasn't a good idea.
Just that prior to that (45-50 degrees) the lift will still be significant so much so that the drag compnent in the vertical plane with the addition of that lift will be greater than just pure drag component flat.
This will give a lower rate of decent than a pure flat plate effect.
NASA has done tons of work on turbine blades in the last 10 years from a quick search.
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I needed to work out the max moment for the base support for a FEA contract job years ago which was when I found out that taking the max force at flat wasn't a good idea.