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    Re: Miracle on the Hudson
    From: Gary LaPook
    Date: 2017 Jul 30, 08:53 +0000
    If that were true then airlines wouldn't have to pay for new a tires and brakes on a regular basis since all landings would take advantage of that. And nobody would ever be hurt in an emergency landing. 

    L = Cl rho V^2 S / 2
     Lift equals coefficient of lift times rho times velocity squared times wing area over two.

    Rho = air density which in the standard atmosphere at sea level and 15 C  is 0.002378 slugs per cubic foot. At colder temps rho is greater so V can be less and still produce enough Lift to keep the plane in the air.

    Velocity is in feet per second

    S = wing area in square feet

    Coefficient of lift is determined by the shape of the wing and by any high lift devices and the angle of attack. So you increase the angle of attack as you are landing as the plane slows down which maintains the lift until you reach Cl = max.and then Cl drops off if you increase angle of attack further. This is the "stall" and the airplanes starts to accelerate downwards because the lift is less than the weight of the plane.  If you wanted to land at zero velocity and zero kinetic energy then Cl would have to be infinite. 

    cl




    From: Herbert Prinz <NoReply_HerbertPrinz@fer3.com>
    To: garylapook---.net
    Sent: Friday, July 28, 2017 2:09 PM
    Subject: [NavList] Re: Miracle on the Hudson

    I don't believe what you are saying. There must be a way of dissipating energy while still in the air, by changing the angle of attack, or changing air resistance with flaps or whatever. Otherwise, how could a glider ever land safely? Gary was merely speaking about the optimum angle of attack with respect to gaining maximum distance. I see no reason why the plane could in theory not have touched the water with zero kinetic energy left.
    Dave said: Interestingly, the initial impact might have been less survivable in July, because there would have been more energy to dissipate. 


       
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