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Box stability calculation

  1. Dec 13, 2014 #1
    Ok I know this should be easy but it's been a few years since my physics lessons at college and I'm stumped.

    I work in packaging. I'm working on a tool that will tell me if a box will fall over when it is subjected to an edge drop test. That means that a block is placed under one edge of a box / pallet and then yanked out after which the box falls to a vertical position. If the center of gravity is in the middle of the box, there's no way it can topple over after hitting the floor, otherwise it would already have toppled when placing the block for the edge drop test. (Right?) But if the center of gravity is close to the lifted edge of the box the situation is different. My common sense tells me that there's a risk of toppling, but I'm stumped as to how to calculate it.

    See picture for what I have so far. Any pointers?
     

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  2. jcsd
  3. Dec 13, 2014 #2

    Bystander

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    "R1" = distance from C.O.M. to corner about which the box rotates as the test begins. "R2" = distance from C.O.M. to corner being dropped. "I1, I2" the moments of inertia about the two axes.

    Getting ideas yet?
     
  4. Dec 14, 2014 #3
    Hi Bystander,

    Thanks for helping me! While I can see how those parameters are relevant, I don't quite see what to do with them. As I said, long time since I've had physics class. I made an attempt, mostly based on what my common sense tells me.

    At the moment of impact, the CoM trajectory is perpendicular to r1. I can split this vector in its horizontal and vertical components.
    The horizontal component is what pushes the CoM forward and possibly makes the box topple.
    2014-12-14 12.13.35.jpg

    I figured that I would convert the kinetic energy in the horizontal component to potential energy and see how far it could lift the box CoM. If that distance is smaller than the value delta h in the sketch below, the box could not fall over.
    2014-12-14 12.13.44.jpg

    This is all grossly simplified :) But with that in mind: is there any merit to what I've done here?
     
  5. Dec 14, 2014 #4

    Bystander

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    What you've got works --- "KISS" beats doing a master's on it.
     
  6. Dec 14, 2014 #5
    Awesome. Thanks! So in this case you can just assume that the vertical component of the impact vector will dissipate on impact? I mean, I know I'm the one deciding how far I want to simplify this. But I want to end up with something that has a foothold in reality. Would you say that this is realistic or oversimplified?

    I'm thinking I could take 20-30% of the vertical impact component and invert that, subtracting the resulting gained rise of the CoM from delta h before starting the horizontal component calculation as described above.
     
  7. Dec 14, 2014 #6

    Bystander

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    What you're doing with the previous approach is essentially translating the angular momentum about the first corner ("hinge" for the drop test) to angular momentum about the second corner (the "hinge" you'll be tipping around). If the box crushes, there'll be losses, but there'll also be added instabilities, so I wouldn't apply any corrections/"fudge factors" to your "lift" simplification.
     
  8. Dec 15, 2014 #7
    I'm still missing something. Right now I'm calculating v(impact) as in this pic, based on the fact that the CoG lowers 123 mm vertically during the edge drop test.
    image1.JPG
    The values I use in the calculation come from a situation where the CoG is in the exact middle of the product, so I'd expect a maximum CoG rise that is also 123 mm. So I'm doing something wrong calculating v(impact), but what?
     
  9. Dec 15, 2014 #8

    Bystander

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    0.155 not 1.55.
     
  10. Dec 17, 2014 #9
    I've toyed with this on and off in between "real" work, and by now I've come to the conclusion that what I did was not correct.

    I assumed the vertical component of the impact vector to be based on the distance the CoG will drop during the test.
    Then I calculated the horizontal component of the impact vector based on the length of vertical and the angle perpendicular to the arc the CoG is making
    Then I assumed that this horizontal vector would lift the CoG by making the box tilt over its dropped edge.

    But the energy that goes into this system (potential energy of dropped CoG delta h) is way smaller than what comes out (kinetic energy of horizontal component impact vector). That way I'm creating free energy. I'm pretty sure I went wrong by assuming the vertical component of the impact vector is based on delta h.
     
  11. Dec 17, 2014 #10

    Bystander

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    You've got a practical problem, and we've been playing with the kinematics, making it harder than it has to be. Let me double-check myself on a short-cut and I'll get back to you.
     
  12. Dec 17, 2014 #11

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    Back to the original problem statement: quick test is to do the lift on the opposite side at the same distance from the COM as the corner you're using for the original drop. Set a sling, or stop of some sort to catch the box so it doesn't actually topple, and if we're both terribly bored we can work on the kinematics of rolling square wheels at our leisure.
     
  13. Dec 17, 2014 #12
    Ha :) I like the pragmatic approach, even though the purpose of the tool would be to have a guideline during 3D modelling before the physical packaging exists.
    I'll play around with this when I have time and maybe some new bright flash of insight will hit me down the road. Thanks for your help so far!
     
  14. Dec 29, 2014 #13

    OldEngr63

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    I think I have a satisfactory analysis of this problem as shown in the attached pdf file. Any comments would be much appreciated.
     

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  15. Dec 31, 2014 #14
    Hello OldEngr63,

    Wow. This looks fantastic! Did you have this analysis lying around by coincidence or did you write it up especially for this thread? Either way: I am very happy for your contribution. As soon as I'm back from holidays I will look into this in depth.
     
  16. Dec 31, 2014 #15

    OldEngr63

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    HI, boerminat0r,

    I wrote it up especially for you, but I will keep it as I collect interesting mechanics problems.

    The problem really bugged me, and I worked on it longer than I would like to admit, spending a lot of time looking into impact, coefficient of restitution, etc. before this energy approach came to me in the wee hours of the morning a few days ago. I got up about 3 am, drew the figures, and wrote it up to post. I hope that it is useful to you.

    If you find any errors in it, by all means let me know. As far as I can see, it is correct.
     
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