G force; physics question - help
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G force; physics question - help
Folks, I need some help. Here's the question;
I have a 1600KG crate. The "shockwatch" indicator shows 5G. How many meters (or feet) would I have to drop those 1600KG in order to get 5G's worth of "shock". I know G is a measure of acceleration.
I've got as far as 5 x 1600 = 8000Nm Force or 1600x(5Gx9.81Nm)/s=49ms/s/s Acceleration
Time must be the missing link, but how the heck to get it?
(PS 20 years since my last physics lesson - was a miserable student).
Thanks in advance
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rgds Rat
I have a 1600KG crate. The "shockwatch" indicator shows 5G. How many meters (or feet) would I have to drop those 1600KG in order to get 5G's worth of "shock". I know G is a measure of acceleration.
I've got as far as 5 x 1600 = 8000Nm Force or 1600x(5Gx9.81Nm)/s=49ms/s/s Acceleration
Time must be the missing link, but how the heck to get it?
(PS 20 years since my last physics lesson - was a miserable student).
Thanks in advance
------------------
rgds Rat
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Anything higher than about a millimetre!'
This is because you need to know a couple of things, mainly the compressibility of the surface that the crate is being dropped on to. If you consider a concrete floor to be totally rigid, then the deceleration will be instantaneous.
For 5 G's though, you need the crate so slow down at the rate of 49 m/sec/sec. So, if the crate is being dropped at 49 metres per second, then it'll have to stop in one second to make 5 G's.
(might be wrong on that though.)
One of the only physics formula's I remember from school, also about 20 years ago, is -
S = UT + 1/2 A T^2.
S = distance
UT = initial speed & distance
1/2 A = 4.9 m/s/s
T^2 = time squared.
Hope it helps.
This is because you need to know a couple of things, mainly the compressibility of the surface that the crate is being dropped on to. If you consider a concrete floor to be totally rigid, then the deceleration will be instantaneous.
For 5 G's though, you need the crate so slow down at the rate of 49 m/sec/sec. So, if the crate is being dropped at 49 metres per second, then it'll have to stop in one second to make 5 G's.
(might be wrong on that though.)
One of the only physics formula's I remember from school, also about 20 years ago, is -
S = UT + 1/2 A T^2.
S = distance
UT = initial speed & distance
1/2 A = 4.9 m/s/s
T^2 = time squared.
Hope it helps.
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A couple of times a year somebody gives me a new undercarriage to certify, where obviously the landing shock is important to know. The way I handle it is as follows: -
(1) Slowly load the undercarriage up until we get fed up, measuring how far the top of the undercarriage / aeroplane / something easy to measure goes down with load.
(2) Plot a graph of deflection versus static load.
(3) Integrate (usually graphically, counting squares is quite therapeutic on a bad Friday) this graph to turn it into energy stored versus static load.
(4) Using my magic book of formulae, I work out the speed at which the undercarriage will be hitting the ground. (For aircraft there are standard formulae which are generally a function of wing loading).
(5) From mass of the aircraft and the impact speed I work out the kinetic energy when it hits.
(6) I plot this energy as a line on the graph at (3) to give me mean impact force.
(7) Divide this by weight to give me the g-loading at impact.
There are a few weaknesses in my method. If the u/c uses any kind of fluid-based shock absorbers (as opposed to springs) then their characteristics depend upon the RATE of application, and so I've no choice but to actually drop it. Also, when I use this test to calculate the various static load tests I have to do to certify the undercarriage, I need to remember that much higher shock loads than static loads are needed to break metallic or composite structures. But this is in the safe sense, and so long as it passes static, there's no problem.
If you can find a formula for descent terminal velocity on your airdrop loads and then apply (1) - (4) above, you should get a meaningful answer.
Finally, this test takes a lot of mass, usually sand or gravel bags. I always advise people planning such tests to first find somebody who either wants a kids sandpit or to gravel their drive!
G
(1) Slowly load the undercarriage up until we get fed up, measuring how far the top of the undercarriage / aeroplane / something easy to measure goes down with load.
(2) Plot a graph of deflection versus static load.
(3) Integrate (usually graphically, counting squares is quite therapeutic on a bad Friday) this graph to turn it into energy stored versus static load.
(4) Using my magic book of formulae, I work out the speed at which the undercarriage will be hitting the ground. (For aircraft there are standard formulae which are generally a function of wing loading).
(5) From mass of the aircraft and the impact speed I work out the kinetic energy when it hits.
(6) I plot this energy as a line on the graph at (3) to give me mean impact force.
(7) Divide this by weight to give me the g-loading at impact.
There are a few weaknesses in my method. If the u/c uses any kind of fluid-based shock absorbers (as opposed to springs) then their characteristics depend upon the RATE of application, and so I've no choice but to actually drop it. Also, when I use this test to calculate the various static load tests I have to do to certify the undercarriage, I need to remember that much higher shock loads than static loads are needed to break metallic or composite structures. But this is in the safe sense, and so long as it passes static, there's no problem.
If you can find a formula for descent terminal velocity on your airdrop loads and then apply (1) - (4) above, you should get a meaningful answer.
Finally, this test takes a lot of mass, usually sand or gravel bags. I always advise people planning such tests to first find somebody who either wants a kids sandpit or to gravel their drive!
G
Guest
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You can get a 5g pulse for a very short period of time without much trouble just driving along a road without suspension on your trailer. It would be a ransient "spike".
Piston aircraft engine mounts see in excess of 5g when they are started.
Can I ask you to clarify your question....
a) do you have a device which has recorded that your 1600kg has exceeed 5g (sometime)
b) or do you want to have a the 1600kg see a continious 5 g.
c) you want to induce a 5g pulse accelleration/decelleration into the 1600 kg ?
Z
Piston aircraft engine mounts see in excess of 5g when they are started.
Can I ask you to clarify your question....
a) do you have a device which has recorded that your 1600kg has exceeed 5g (sometime)
b) or do you want to have a the 1600kg see a continious 5 g.
c) you want to induce a 5g pulse accelleration/decelleration into the 1600 kg ?
Z
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Thanks Zeke,
The shockwatch indicator showed readings from 5G up to 25G!! Was a cargo pallet with a machine on it. I've got all the data from the FDR which (obviously) shows that the aircraft never went anywhere near these values.
I'm assuming it was dropped somewhere between leaving the factory and arriving at destination.
It would have to be one heck of a drop though; I was wondering if it was possible to calculate exactly how far, based on the given weight/shockwatch indicator reading.
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rgds Rat
The shockwatch indicator showed readings from 5G up to 25G!! Was a cargo pallet with a machine on it. I've got all the data from the FDR which (obviously) shows that the aircraft never went anywhere near these values.
I'm assuming it was dropped somewhere between leaving the factory and arriving at destination.
It would have to be one heck of a drop though; I was wondering if it was possible to calculate exactly how far, based on the given weight/shockwatch indicator reading.
------------------
rgds Rat
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As everyone has said, your answer depends on how fast it was going, and over what time or distance the thing took to stop. Or more precisely, the stopping distance for the shockwatch (as the part to which it is fixed may have had some spring in it).
We assume that the pallet was dropped cleanly, rather than sliding off something, or tipping over. We also assume that the pallet came to a damped (squishy) stop, rather stopping on a trampoline or springy floor.
Given all that, (plus a few more unstated but less important assumptions) the g force depends on the velocity of the shockwatch at contact, and the distance over which it came to rest. And the velocity on contact depends on the height from which it was dropped.
The velocity V attained by a free-falling thing is given by V squared = U squared + 2 * g * H .....(1)
where g = 9.807 metres/sec/sec; V is metres/sec; H is height in metres and U is the initial velocity, in our case zero.
The same formula describes the stopping process, except the deceleration comes from your shockwatch, and the height is the stopping distance. So v squared = V squared + 2 * G * g * h ....(2)
where v is the final velocity (zero), V comes from above (the initial velocity on impact), G is the shockwatch (and negative); h is the stopping distance.
Now if you substitute V squared from (1) into (2) (because they are the same thing) and simplify, you get
H = G * h. That is, drop height = shockmeter * stopping distance
and when you think about it, this simply is a definition of G.
So stopping distance 1cm; Drop height 5 cm for 5G; 25 cm for 25G
Stopping distance 5cm; Drop height 25 cm for 5G; 75 cm for 25G.
Doesn't take much to generate Gs.
A hell of a lot depends on where the shockwatch is placed. If it is on the top of a deformable crate, but that crate has a machine at the bottom with only the pallet between it and the concrete to deform in taking the shock, then the machine sees more G than the shockwatch.
ET
We assume that the pallet was dropped cleanly, rather than sliding off something, or tipping over. We also assume that the pallet came to a damped (squishy) stop, rather stopping on a trampoline or springy floor.
Given all that, (plus a few more unstated but less important assumptions) the g force depends on the velocity of the shockwatch at contact, and the distance over which it came to rest. And the velocity on contact depends on the height from which it was dropped.
The velocity V attained by a free-falling thing is given by V squared = U squared + 2 * g * H .....(1)
where g = 9.807 metres/sec/sec; V is metres/sec; H is height in metres and U is the initial velocity, in our case zero.
The same formula describes the stopping process, except the deceleration comes from your shockwatch, and the height is the stopping distance. So v squared = V squared + 2 * G * g * h ....(2)
where v is the final velocity (zero), V comes from above (the initial velocity on impact), G is the shockwatch (and negative); h is the stopping distance.
Now if you substitute V squared from (1) into (2) (because they are the same thing) and simplify, you get
H = G * h. That is, drop height = shockmeter * stopping distance
and when you think about it, this simply is a definition of G.
So stopping distance 1cm; Drop height 5 cm for 5G; 25 cm for 25G
Stopping distance 5cm; Drop height 25 cm for 5G; 75 cm for 25G.
Doesn't take much to generate Gs.
A hell of a lot depends on where the shockwatch is placed. If it is on the top of a deformable crate, but that crate has a machine at the bottom with only the pallet between it and the concrete to deform in taking the shock, then the machine sees more G than the shockwatch.
ET
