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Help- mathematical results for a banked turn

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  • Steve Uzochukwu
    replied
    Help- mathematical results for a banked turn

    Dave Smith wrote: Wasn't it you Paul (could have been BB) who worked out it was impossible to load a flexwing beyond 4G because of unloading due to airframe distortion?
    Dave
    The BHPA test rig has managed to produce figures well above 4G on some wings, but others shed load. It's not flight though as the ability of the rig to load the wing is limited mostly by engine power overcoming the drag at those high lift values.

    Think the highest figures ever sustained without failure on a test rig are close to 7G. Think this was on the US test rig in the late 1990s.

    Mark Dale ran the rig most of the time but current Tech Officer Ian Currer also has some experience on it, and used to be a flex wing manufacturer.

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  • Mick Broom
    replied
    Help- mathematical results for a banked turn

    Found this on my quest

    Web page of bank/loads

    Stall speed doubles at just over 70 degrees and trebles at just over 80 degrees which probably explains why we cannot pull High loads.

    We just cannot go fast enough

    Leave a comment:


  • Paul Dewhurst
    replied
    Help- mathematical results for a banked turn

    Depends on the individual machine and its aeroelastic coefficient. We plotted the curves from flight testing stall speed against g for the kiss 400 and Va ( the speed where you ca get 4g without stalling first) was above Vne.

    But types with a higher sail tension don't twist off so much under load and can pull more g.

    Paul

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  • Dave Smith
    replied
    Help- mathematical results for a banked turn

    Wasn't it you Paul (could have been BB) who worked out it was impossible to load a flexwing beyond 4G because of unloading due to airframe distortion?
    Dave

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  • Paul Dewhurst
    replied
    Help- mathematical results for a banked turn

    its still all down to lift - even at 90 degrees of bank - it's the lift that is producing the turn radius and the G. And in this context lift is the force the wing is making - not the vertical component - which at 90 degrees can only be zero!

    There was some weird stuff earlier in the thread about not being able to roll out of the turn at 90 degrees due to gravity pulling the trike / hanglider pilot out of Perpendicularity with the wing. That's not really relevant and the angle the trike adopts has no effect on the wing - it's roll torque applied by the pilot through his / her arms to the bar that produces roll force to the wing.

    Paul

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  • Mick Broom
    replied
    Help- mathematical results for a banked turn

    Your correct Paul,

    I am struggling to relate the maths with the facts but I now think I see what you are saying is the maths is correct but it takes the plane into a place which is impossible to do.

    I understand the G is produced by the lift but if the forces are of a greater magnitude than the static 1G downwards which can be produced by the centrifugal effect ( speed / radius ) then it acts through the mass/lift axis which could be at any angle, even upside down???

    I am afraid what my head says is that when its at 90 degrees the dominant force is due to the centrifugal load which is dependent on the speed and radius of the turn , push out more G. Obviously too stay level you would need to fly at less than 90 degrees to offset the gravity.

    Not enough brain cells left , more drink needed

    :burger:

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  • Paul Dewhurst
    replied
    Help- mathematical results for a banked turn

    Not sure why it's difficult? g is just produced by lift. The bank angle is irrelevant - unless you want to maintain level flight. Of course that's impossible to sustain for high bank angles - which is why the maths ramps the G up almost infinitely.

    It seems like you are struggling with trying to relate what you have observed / experienced with the math? Seems to me it's the level flight bit that is the trouble. The boys in the last WAG were banking 90 degrees but only pulling 4g - but that's because they weren't increasing lift enough to support level flight - but it didn't matter becuase the time scale was very short, so there was no noticeable height loss.

    Paul

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  • Mick Broom
    replied
    Help- mathematical results for a banked turn

    Hi Steve,

    We are obviously talking about a different form of flying which we would not want to go near.

    Exploring the loads and limits in the aircraft with words are something which will bring knowledge and appreciation of the complexity of problems and the reason for limits in our flying and with a better understanding of what the plane will do and what effects it must make you a better pilot in my personal opinion.

    Paul,

    The angle concerns came from the discussion, it was the G loads which interested me , mainly because given the accepted formula it appeared to be completely wrong when calculating the steeper turns , Steve appears to be working along these lines of using the formula which predicted G loads which obviously were not happening so the question had to be asked why. I now understand the reasons but still have not got a mathematical conclusion or crossover point even after simplifying it to a flat constant speed turn. I will work on it some more.

    Its obviously difficult but would be nice so we could move on to predicted stall speeds but the main motivation was to check out the data logger I am working on which in theory at least gives all the angles, accelerations and GPS positioning to maybe get enough information to start monitoring changes.

    I know from my work on the bikes that information is key and a good logger when understood is very useful.

    Thanks for the contributions

    Leave a comment:


  • Steve Uzochukwu
    replied
    Help- mathematical results for a banked turn

    Paul Dewhurst wrote: Not really Steve, with a trike you can quite precisely control the G applied. It's all directly down to bar position. Not difficult at all really.
    I was talking about wing overs or asymmetric spirals. Input has to be precise and correctly timed or the pilot is in the sail. The video link demonstrates my point...

    Leave a comment:


  • Paul Dewhurst
    replied
    Help- mathematical results for a banked turn

    Not really Steve, with a trike you can quite precisely control the G applied. It's all directly down to bar position. Not difficult at all really. Done lots of it for test flying for sectionS approval, where we measure control force against G - usually In a continuous spiral turn, but applies to any turn, - the G is directly proportional to bar position - not bank angle.

    Of course I am not advocating we all go flying 90 degree turns ( although we have an exemption from the 60 degree bank limit from CAA for the WAG team). Just pointing out that it's not the bank angle that applies the G in itself - its all down to bar position. It's also not particularly useful Mick to try and calibrate a G meter form flying bank angles - it's not so easy to precisely measure the bank angle, and just a tiny bit of climb or descent varies the G considerably.

    Paul

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  • Steve Uzochukwu
    replied
    Help- mathematical results for a banked turn

    Paul Dewhurst wrote: Of course you can be at 90 degrees or greater with any g load. It's only if you start in level flight and increase bank whilst maintaining level flight that G will rise according to the hard relationship law.

    With aerodynamic pitch control you can be at 90 degrees and have negative G if you wish.
    On a PG with no rigid structure we can be above the wing in an asymmetric spiral or even a tumble. On a flex wing with a wingover we can go beyond 90 degrees of bank. Agreed.

    But it's instantaneous, and not something you can adjust or control slowly whilst watching a G meter or ASI.

    Just warning of the dangers, as the pilot in my previous link found out when he peloyed his reserve....

    Leave a comment:


  • Paul Dewhurst
    replied
    Help- mathematical results for a banked turn

    Of course you can be at 90 degrees or greater with any g load. It's only if you start in level flight and increase bank whilst maintaining level flight that G will rise according to the hard relationship law.

    With aerodynamic pitch control you can be at 90 degrees and have negative G if you wish.

    In a flexwing doing a pylon turn ( which I think is Micks interest as the WAG is coming up and son is flying in it?) the G is governed by the pilot and how much push forward on the bar is made. The machine may briefly get to 90 degrees or greater at the turn apex, but no hard relationship with G can be assumed - the machine can be considered to maintain height 'ballistically' for the very small amount of time it is at that bank angle.

    An accelerometer is a very useful instrument so the pilot can get a feel for what load they are subjecting the machine to. I think last time at the WAG, with accelerometer fitted they found it very difficult to exceed 4g. As they were flying the Quikr solo this was equivalent to around 3/4 the limit load values.

    Paul

    Paul

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  • Steve Uzochukwu
    replied
    Help- mathematical results for a banked turn

    Dave Smith wrote: Are we saying that the G loads are so great the pilot is incapable of pulling himself up the bar? I must say that I don't see the hang strap move much out of the vertical (perpendicular) throughout the maneuver. At one point he seems to be doing the classic HG beginner error of moving across the bar, but allowing his body to twist, losing all weightshift effect. Although that is close to the point where he gives up and goes for his chest pack.
    Dave
    It's worth drawing a triangle of forces.



    As you get closer to the 90 degree point, the G forces build up, forcing you more and more to the lowest part of the arc through which you swing, plus the effect of gravity is always downwards. In extremis you have massive G away from the direction of turn plus nothing to balance gravity (any force will have a zero element at 90 degrees to it).

    If you look to the end of the video, you see an earlier successful exit from a spiral. Part of the twisting is looking for the reserve handle.

    Mick Broom wrote:
    I agree with you that manufacturers limitations should not be exceeded and it obviously initiated the problem but with the inside wing unloading the batten clips I believe it became an aerodynamic failure, not a lot to do with G loads or is it?
    It is to do with G loads. If you read the report, you will see that the extra loading on the sail caused the batten clips to break, and therefore to disengage from the sail. In the report minimum failure loads for the batten clips were specified.

    I have seen a fair few aerobatic manoeuvres on HG, and I've seen a few go wrong and parachutes get deployed.

    When you go beyond 60 degrees angle of bank (or 45 degrees with a Flash 2), you are headed into unknown territory where hospital food and wooden boxes wait for the unwary.

    From my spreadsheet, at 75 degrees angle of bank you are pulling 4G. 80 degrees is 6G. Beyond this lies AAIB report land. Your flex wing may shed load by washing out, or it may break. Not many wings go much over 7G on the test rig.

    Beyond 4G you need to have access to a reserve.

    I'm fairly sure you know what you are doing but just in case I'll have a clear conscience in stating that once you get near 80 degrees there is a fair chance you'll be packing the aircraft into a carrier bag using a dust pan and brush.

    Here's a quick example of a brief excursion outside the placards:

    https://www.youtube.com/watch?v=b0gu8UWDqL0

    PM me an e-mail address and I'll send you the reverse engineered G loading to bank angle.

    Leave a comment:


  • Mick Broom
    replied
    Help- mathematical results for a banked turn

    Hi Dave,

    That is the message coming from a few but it plainly is not happening.

    If as stated with the wing at about 90 degrees using the formula as now gives say 8G , well the pilot would have lost vision and certainly not be in a position to do anything but hang on without success plus the real possibility of the plane /wing failure.

    I believe when in a steep turn the G loads are speed and turn dependant so a bit of speed but more important a tighter radius will produce the G load but this is a lot less than 1/cos( angle) and is probably half the value.

    Now all I need is a mathematical case which supports this or someone to say its not !

    Apart from any structural or pilot limitations without an accurate prediction of the G we cannot establish where the stall would occur.

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  • Dave Smith
    replied
    Help- mathematical results for a banked turn

    Are we saying that the G loads are so great the pilot is incapable of pulling himself up the bar? I must say that I don't see the hang strap move much out of the vertical (perpendicular) throughout the maneuver. At one point he seems to be doing the classic HG beginner error of moving across the bar, but allowing his body to twist, losing all weightshift effect. Although that is close to the point where he gives up and goes for his chest pack.
    Dave

    Leave a comment:

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