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In article <fvJ2i.193$dS1.175@newsfe03.lga>,
N114RW <qwerty_265@charter.net> wrote:
> Can someone splain this to me? I've no idea if I've it right, or not.
>
> As I understand it, maneuvering speed is the speed above which the
> aircraft may be damaged due to full control movements. Below that
> speed, the plane will stall before damage occurs. I also understand
> that maximum lift changes as the square of the speed, while stall speed
> changes as the square root of the weight.
>
> So, if I've an aircraft traveling at twice stall speed, it will have 4
> gšs acceleration prior to stalling. If I double the weight, the stall
> speed increases by 1.41, so maneuvering speed also increases by 1.41
> assuming that the aircraft is rated a 4 gšs at both weights.
> Or, to put it conversely, the maneuvering speed varies inversely
> proportionately to the square root of the weight change if I double
> the weight, the maneuvering speed increases by 41%.
> Do I've that right?
No. At higher weights the airplane will stall at a lower G loading than
at light weights, since the wing stalls at the same total lift generated.
It takes more lift to carry the higher weight. Other structural
components also carry load, such as engine mounts, tail components, seat
brackets, etc. Those components may actually be the determinant for the
max G loading, rather than the wing.
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Re: Manovering Speed
