Converting vertical descent rate to g load
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tubby,
I don't think so really. It depends on a handful of parameters, such as gear design (stroke, spring constant, non-linear behaviour), aircraft weight, whether you're talking a "hard" landing (gear bottoming out), and I'm probably forgetting a couple.
And don't take the 'g' from the FDR or QAR for gospel. It isn't sampled fast enough.
Does anybody know of a simple way of converting vertical speed to an impact g load?
And don't take the 'g' from the FDR or QAR for gospel. It isn't sampled fast enough.
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Obviously then it is complicated by the energy absorbtion of the undercarraige, the pressure of the tyres etc. But as a guide, in the 777, 600 ft/min is roughly equiv to 1.8g ( it gets its knickers in a knot at 1.98 g hard landing warning depending on what sort of a day it's having)
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Different parts of the - ahem - vehicle will see vastly different g-loads.
The pointy end stops VERY quickly, ergo by definition very high acceleration loads. The tail end is cushioned by the collapse of everything in front, so it sees much lower loads - thus they install the FDR, CVR etc back there to improve the likelihood of them surviving. (75-80 years ago the pilot might be back there too - my how far we've progressed!
)
The pointy end stops VERY quickly, ergo by definition very high acceleration loads. The tail end is cushioned by the collapse of everything in front, so it sees much lower loads - thus they install the FDR, CVR etc back there to improve the likelihood of them surviving. (75-80 years ago the pilot might be back there too - my how far we've progressed!
)
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Ah Barit you've been in one of my landings.
If you take 1"g" as 32 ft/sec/sec then all you need to know is the time it takes the undercarriage to stop the vertical movement ( this ignores wing flex etc which may be quite considerable.
So if we assume the time taken is 2 sec then 1 g would be (32*60)/ 2
or 960 ft/min rate of descent.
however, N2 over oil oleo's do not give constant decelaration so i do not think there would be an easy conversion
If you take 1"g" as 32 ft/sec/sec then all you need to know is the time it takes the undercarriage to stop the vertical movement ( this ignores wing flex etc which may be quite considerable.
So if we assume the time taken is 2 sec then 1 g would be (32*60)/ 2
or 960 ft/min rate of descent.
however, N2 over oil oleo's do not give constant decelaration so i do not think there would be an easy conversion
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Experimentation? ...
700fpm arrival on a concrete surface equates to 2.1g in a certain long thin canadian twin turboprop. According to its Flight Data Monitoring equipment. Allegedly.
... 
!!
700fpm arrival on a concrete surface equates to 2.1g in a certain long thin canadian twin turboprop. According to its Flight Data Monitoring equipment. Allegedly.
... !!
