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Landing Gear
- Thread starter Fajer
- Start date
- Thread starter
- #3
the maximum deflection has been provided which = 0.3 m (Value Assumed ) For lynx 300 Helicpter
Mass of the helicopter = 5126 Kg
sink speed = 0.5m
k= F/x = 5126/0.3 = 167620.2 N/m
Need to find out damping ratio and Damping Coeficient at which the helicopter will have a deflection of 0.3 m
Mass of the helicopter = 5126 Kg
sink speed = 0.5m
k= F/x = 5126/0.3 = 167620.2 N/m
Need to find out damping ratio and Damping Coeficient at which the helicopter will have a deflection of 0.3 m
ok, now you've told us the limiting displacement, and calc'd a linear stiffness. wrongly btw (is the helicopter's mass 5126 kg or is it it's weight ?)
but damping controls how the systems responds to step changes over time ... does it bounce ?
which'll require other data
but damping controls how the systems responds to step changes over time ... does it bounce ?
which'll require other data
- Thread starter
- #5
btrueblood
Mechanical
"it should bounce because its landing on a aircraft carrier"
Do you mean it should NOT bounce?
Do you mean it should NOT bounce?
if the mass is 5126 kg, then it's weight will be 5126*9.81 = 50240 N (no?)
but i wonder if that's the right force to apply (btw, there are two skids supporting the helicopter, so each supports 1/2 the weight)
maybe an energy equation, the initial kinetic and potential energy are converted into spring energy
1/2mv^2+mgh = 1/2kx^2 ... (h = x ??)
... but i don't think that's the right spring energy ... it's not like there's a spring mounted under the fuselage, being compressed; it's more like a leaf spring.
but i wonder if that's the right force to apply (btw, there are two skids supporting the helicopter, so each supports 1/2 the weight)
maybe an energy equation, the initial kinetic and potential energy are converted into spring energy
1/2mv^2+mgh = 1/2kx^2 ... (h = x ??)
... but i don't think that's the right spring energy ... it's not like there's a spring mounted under the fuselage, being compressed; it's more like a leaf spring.
- Thread starter
- #8
ok, so if you have a oleo gear it'll compress on impact and rebound in some way. and then it'd be reasonable to assume x = h (in the previous energy balance).
still the reasonable assumption would be landing on the two main gear.
this is starting to sound suspiciously like a school problem ...
still the reasonable assumption would be landing on the two main gear.
this is starting to sound suspiciously like a school problem ...
Perhaps you could ask Prof Newman if he's still around?
Posting guidelines faq731-376 (probably not aimed specifically at you)
What is Engineering anyway: faq1088-1484
What is Engineering anyway: faq1088-1484
GregLocock
Automotive
It's like watching a train wreck in slow motion.
OK, is the load split between all three legs?
Secondly, what is the complete list of your design aims? I gather maximum deflection of 0.3m is one. Or are you saying the spring deflection at rest is 0.3m?
If you don't have any damping then the helicopter will land, compress the spring, and then bounce back into the air.
"k= F/x = 5126/0.3 = 167620.2 N/m"
wrong. Or at least, if it happens to be right it'll be because of coincidence not logic.
Cheers
Greg Locock
New here? Try reading these, they might help FAQ731-376
OK, is the load split between all three legs?
Secondly, what is the complete list of your design aims? I gather maximum deflection of 0.3m is one. Or are you saying the spring deflection at rest is 0.3m?
If you don't have any damping then the helicopter will land, compress the spring, and then bounce back into the air.
"k= F/x = 5126/0.3 = 167620.2 N/m"
wrong. Or at least, if it happens to be right it'll be because of coincidence not logic.
Cheers
Greg Locock
New here? Try reading these, they might help FAQ731-376
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