About these 2 statements :
Originally Posted by
Ka8 Flyer
G is an acceleration and not a force -> F = m * a
So it takes a lot more force to accelerate a B-whatever to 2,5G's than for a 2t SEP to 6 G's. So I'd rather be in a B744 in a CB (actually I'd rather not be in a CB
at all )
Originally Posted by
chornedsnorkack
I suppose that it would be an issue of wing loading multiplied by g-loading.
they tend to state that different planes develop different g-loading when confronted to the same windshear.
The first post investigate the mass and the second investigate the wing loading.
Actually, the plane which best resists to turbulence is the one that has the smallest ratio
between the rate at which g-loading is varying with the windshear and its g-loading margin.
Physically, this ratio is expressed by
(δg/δv) / Δg
where Δg is (g-loading max - 1).
If I am not mistaken with my formulas, minimising the ratio comes down to 2 terms.
The first term corresponds to minimising the possibility of increasing the g-loading
through an increase of the lift factor CL and it matches the recommendation of flying below the maneuvering speed.
The second term corresponds to minimising 1 / (Δg * Vs1)
And so the plane which best resists turbulence is the plane with the highest product between g-load margin and stall speed.
For a light aircraft (a P28A - non-utility) , I compute 64kt * 2.5g = 160 g*kt
For a light aircraft (a P28A - utility version) , I compute 64kt * 3.4g = 218 g*kt
For a 737, I compute 130 kt * 1.2g = 156 g*kt (not sure of the parameters)
So, there is not much difference between the Piper (non-utility) and the 737 because the 737 lower g-load sensitivity to windshear
is compensated by the Piper extra g-load margin and the Piper utility version would be safer due to its supplementary g-load margin.
Now, that's not so simple. I made these example calculations only with the positive load factor margin.
Actually, the plane may also break up with negative load factor.
So, one should choose the smallest g-load margin, the positive one being (g-load max - 1) and the negative one being (1 - g-load min).
For instance, if your g-load limits are -1.2 and +3.5, your margins are 2.2 and 2.5 .
Another simplification I made is that I took Vs1 as the lowest speed one can fly with flaps up,
but the real speed one should elect to use is the maneuvering speed minus a margin to compensate for gusts.
I put here the derivative formula I used for the math addicts and to allow for corrections in case I made some stupid mistakes.
(δg/δv) / Δg = (ρ.S / m.Δg) . (v².δCL/δv + 2.CL.v)
= (g.δCL/δv) / (CL.Δg) + 2.g / (v. Δg)
with g being the acceleration in smooth flying conditions (g-load = 1) where it matches the equation:
g = CL.ρ.S.v² / m
Coming back on the quoting of the two earlier posts, I don't see how the resistance of a plane to turbulence
can be expressed as a factor of the mass of the plane or as a factor of the wing loading.
I don't say that there is no possibility of a relationship, but simply I can't see a way to express it.