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Springs for cushioning a fall

  1. Feb 25, 2016 #1
    Lets say there is a rigid metal box. It is 1meter wide, 2meters long and an undetermined height. The thickness of the box's metal walls are also undetermined. The entire box weighs 2 metric tons.

    There is a person lying face down or face up in this box, and inside the box, the bottom and top of the box is covered in springs. The human weighs 88kg

    If the box is dropped from a cliff, assuming the box lands with its largest side hitting the ground, how high can the cliff be without the human inside sustaining injury?

    M - metal
    S - spring
    P - padding (so the springs don't pierce the human)
    H - human

    MMMMM
    SSSSS
    PPPPP
    HHHHH
    PPPPP
    SSSSS
    MMMMM

    This is just a basic model to give me an introduction to calculating springs as a cushioning device. I learn the fastest when i have an example i can dissect so a simple example would be greatly appreciated. I plan on picking random springs I can acquire and calculate all of them to see which one is best for my purposes.
     
  2. jcsd
  3. Feb 25, 2016 #2
    The undetermined height of box means you can't answer the question. The answer will all be around the G Force you put on the body. If your box is not much larger than the human, then no spring will be good enough for even the smallest thing you could call a cliff. If the box is HUGE, then you could get away with any size of cliff, as the box will reach a terminal velocity and be able to provide enough damping to keep the g-force less than lethal (where HUGE = "sufficient height to provide non-lethal amount of damping at terminal velocity :-)

    PS - I really hope this is a thought experiment.... stay away from the box!
     
  4. Feb 25, 2016 #3
    Yes this is a thought experiment. Don't worry, I'm not gonna hurt myself in the name of physics ^_^. Springs are required in my construct and I am just curious if they can double as a cushion.

    I set the height as undetermined because of the reasons you stated. I haven't decided on the spring I want to use therefore I can't set a height for the box but I guess maybe I should pick a height.

    Lets say the box is 0.5m high, and the human is a box that is 0.4m wide, 2m long, and 0.15m high. So i guess the springs are 0.175m high.

    Oh and the springs are designed for compression.
     
  5. Feb 26, 2016 #4

    Baluncore

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    Science Advisor

    Firstly, springs do not absorb energy, they reflect it. You need something that will decelerate the human body to a velocity of zero over the longest possible time. That calls for a crumple zone that will absorb the kinetic energy with the minimum spring action.

    The survivability of a body in a fall is usually specified as an acceleration, (or deceleration), measured in Gs. That is relative to the acceleration due to gravity at the Earth's surface.

    A very useful relationship is that; If a body is accelerated by gravity for ta seconds and then decelerates back to zero velocity in td seconds, the deceleration will be ( ta / td ) G.

    You know the hight of the drop and so can calculate the time of fall and the speed on impact. Once you decide on a survivability acceleration, say about 20 G, you can calculate the time to stop and therefore the length of the crumple zone needed.
     
  6. Feb 27, 2016 #5
    Well, you have springs on either side of the human in the box, so compression/expansion - doesn't really matter - they all can be modeled as performing ideally, by the spring formula F= kx.

    What you will end up with when the box hits the ground, is the human oscillating forever from the middle position as a simple harmonic oscillator.
    https://en.wikipedia.org/wiki/Harmonic_oscillator
    Section : simple harmonic oscillator
    if there is no damping to dissipate the energy ( of the oscillating human from the fall ). You will need some kind of friction device so that the human eventually comes to rest. See the little picture of the moving ball - that's what the human will be doing with only springs.

    When the box hits the ground - you can initially assume that the box and ground are absolutely rigid as a first assumption, and then go on to some deformation of the box and/or ground to absorb some of the energy - the box and the human will be travelling at a particular velocity. The box immediately stops, but the human continues with the same velocity. The human will slow down to stop, reverse direction, travel through the middle point again, slow down again to a stop, reverse direction, etc.

    You are given the formula x(t) for displacement. A derivative gives the velocity v(t). Another derivative gives the acceleration a(t) of the human.

    You also are given the formula for w, frequency of oscillation.
    Knowing the mass, m, of the human, and the maximum force a human can withstand, you can put it all together and solve for you spring constant.

    If you decide to iterate, try using a spreadsheet ( MS Excel ), to see what different values of k do for you human volunteer in a box. Or even different heights of cliffs.

    Have fun.
    ( I might try that myself just for fun if and when I get the time.
     
  7. Mar 1, 2016 #6

    CWatters

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    Homework Helper

    The length of the springs determines the effective "stopping" distance (if we assume the person must stop before he hits the bottom of the box). If you know the maximum tolerable g you can calculate the minimum length of the spring and the spring constant.

    Sure the person won't actually stop but subsequent "bounces" will be at reducing g so hopefully we can ignore them.
     
  8. Mar 1, 2016 #7
    Baluncore's points are pretty straight on. Not that these numbers will directly assist in your research but there has been several good studies done on related subjects. One of them was done by the Royal Air Force. There determination was that Human bones when subjected to rapid deceleration start to break at very close to 12 Kn in a well supporting harness. Most recovery systems use this number as a base line. Knowing the Kinect energy of your human subject that has developed due to time accelerating one can than calculate the time of the impulse required to decelerate again to 0.
    The dynamic ropes that are used in mountaineering and rock climbing are also built around this number. The ropes are designed to have a high elongation with a initial yield point that varies around 7Kn. This normally results in an elongation of the rope of approximately 30%. The structure due to internal friction also provides a significant dampening effect.
    When selecting ropes in those cases calculations are made to determine what they refer to as " Fall Factor." This is a ratio of the unrestrained verticle fall compared to the length of rope used to arrest. A 10 meter fall arrested with 5 meters of rope would be a ff of 2. The lower yield point provides a slower initial deceleration. In that type of use a 5 m fall arrested by 2.5 m of rope will have the same deceleration as a 30 m fall arrested by 15 m of rope. The variable is the time of the impulse.

    In your spring example many of the same principles apply with the exception that I do not see dampening force to prevent the springs from rebounding.
     
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