Turbulence penetration - light vs heavy aircraft.
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Turbulence penetration - light vs heavy aircraft.
Just trying to get my head around some of the factors that influence the effects of turbulence on different aircraft.
For example, if asked the question which aircraft would you prefer to fly through a mature CB in -
1) Large transport aircraft (737...) DLL=2.5g
or
2) Light utility aircraft / aero (warrior etc) DLL=4.4g / 6g
Looking at the two aircraft side by side and faced with a large black towering CB I would be tempted to say the large aircraft, it certainly looks stronger! However in terms of g-loading it is not, it cannot withstand as much g before it breaks! There must be other factors such as effect of weight on g-increment, wing loading etc..
Any thoughts ?
Thanks
For example, if asked the question which aircraft would you prefer to fly through a mature CB in -
1) Large transport aircraft (737...) DLL=2.5g
or
2) Light utility aircraft / aero (warrior etc) DLL=4.4g / 6g
Looking at the two aircraft side by side and faced with a large black towering CB I would be tempted to say the large aircraft, it certainly looks stronger! However in terms of g-loading it is not, it cannot withstand as much g before it breaks! There must be other factors such as effect of weight on g-increment, wing loading etc..
Any thoughts ?
Thanks
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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 )
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 )
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Just trying to get my head around some of the factors that influence the effects of turbulence on different aircraft.
For example, if asked the question which aircraft would you prefer to fly through a mature CB in -
1) Large transport aircraft (737...) DLL=2.5g
or
2) Light utility aircraft / aero (warrior etc) DLL=4.4g / 6g
Looking at the two aircraft side by side and faced with a large black towering CB I would be tempted to say the large aircraft, it certainly looks stronger! However in terms of g-loading it is not, it cannot withstand as much g before it breaks! There must be other factors such as effect of weight on g-increment, wing loading etc..
Any thoughts ?
Thanks
For example, if asked the question which aircraft would you prefer to fly through a mature CB in -
1) Large transport aircraft (737...) DLL=2.5g
or
2) Light utility aircraft / aero (warrior etc) DLL=4.4g / 6g
Looking at the two aircraft side by side and faced with a large black towering CB I would be tempted to say the large aircraft, it certainly looks stronger! However in terms of g-loading it is not, it cannot withstand as much g before it breaks! There must be other factors such as effect of weight on g-increment, wing loading etc..
Any thoughts ?
Thanks
Also, I suspect that aircraft with low L/D ratio would be more resistant to turbulence - if they hit an updraught, the wing would stall before it breaks.
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I've been in a (little) CB in a PA28 and I can absolutely assure you, that whatever the numbers may say, you do not want to do it.
I believe that we were lucky to get out in one piece, as the VSI pegged one end of the scale and then the other and the airspeed fluctuated between 40 (bottom stop) and 120 knots.
Fortunately we did not encounter ice, otherwise I reckon that would have been that.
I believe that we were lucky to get out in one piece, as the VSI pegged one end of the scale and then the other and the airspeed fluctuated between 40 (bottom stop) and 120 knots.
Fortunately we did not encounter ice, otherwise I reckon that would have been that.
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About these 2 statements :
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.
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.
Last edited by Luc Lion; 27th Sep 2006 at 16:18. Reason: cured an obscure sentence...
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g-loading rate of change due to variation of airspeed is 2.g / v
and g-loading rate of change due to changes of AoA is (g.δCL/δv) / CL
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Turbulence is also going to do nasty things in other axes; empennages may be 'designed' by gust load cases, as may engine pods. Consider that you'll usually be stressed to at least +/- 1'g' increments in the vertical axis, but 1'g' lateral load is a HUGE amount for most civil aircraft. (Even though the ability of the aircraft to generate lateral accel is of course lower)
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Normally, the limitation of the engine pods to acceleration forces is included in the g-loading limitation (usually close to the range [-2g, +4g]).
Also, it is true that the discussion above only considered gusts or windshear in the x and z axis (longitudinal and vertical)
and did not consider the effect of gusts in the lateral axis for which the plane also has structural limitation.
Also, it is true that the discussion above only considered gusts or windshear in the x and z axis (longitudinal and vertical)
and did not consider the effect of gusts in the lateral axis for which the plane also has structural limitation.
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Tailwind
As a plane is designed to tolerate some negative g-s, it would seem that tailwind would be unlikely to exceed the g limits... After all, airspeed zero would just unload the wing and put the craft to 0g - well within the limits. What would a wing do if it has negative airspeed, or 180 degrees AoA?
Whereas an updraught could cause considerable aerodynamic forces...
Whereas an updraught could cause considerable aerodynamic forces...
