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Vertical jump height from force plate

  1. Mar 24, 2014 #1
    Hi guys


    I've a problem calculating the vertical jump from ground reaction force. We tested a person with a force plate, and we got this curve as a result:
    2vudngx.png



    2. Relevant equations
    These are the equations, sorry for the danish language, but just look at the equations and not the text.
    900sxh.png


    3. The attempt at a solution
    4lpnhh.png
    fg is 9.82, but maybe it should be the persons weight in newton?
    deltat= 0.04, but maybe it should be (2.44-1.34) because that is the time for the entire jump?
    Either way, i don't get a useful result.


    Thanks for looking at it, feel free to ask if you need additional info :-)
     
  2. jcsd
  3. Mar 24, 2014 #2

    BvU

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    Hello Creamcheese, and welcome to PF.

    The picture I get from the graph is someone of around 80 kg bending his/her knees and then -- immediately and without waiting until the scale would show fg again -- stretching and jumping off the plate. (Pity the landing was not on the plate -- would have given a good check!)
    I suppose one horizontal division is 2 s ?

    Then: the ttnn terms contain a Δt in the form of e.g. (2.04 - 2) , but it looks as if there is another Δt in the form of a factor deltat ??

    And: I would say the time for the entire jump is not from 1.34 but from the moment the force is greater than mg ( I am not 100% sure about this -- it is related to the question if the force plate contributes to the FΔt or not. You decide). That way the junp height comes out a little higher.

    And, eh, you did do something to estimate the actual junp height when the experiment was carried out, I hope? With a camera (or just looking from aside) and a ruler you can get a reasonable check fro the calculations outcome! If not, maybe ou can still do that afterwards....
     
  4. Apr 2, 2014 #3
    Not sure what you mean about the horizontal division i 2 s?
    It is correct that there is two forms of Δt. Where i think "deltat" i the time for the entire jump.
    I've tried starting the calculations from where the force is greater than mg, but it really didn't make a difference.

    no, we didn't make a video, but logic tells us the jump is around 0.5m :-)

    Maybe this can help me out a bit: http://people.brunel.ac.uk/~spstnpl/Publications/VerticalJump(Linthorne).pdf
     
  5. Apr 2, 2014 #4

    haruspex

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    Wouldn't it be easier using momentum? There's a constant gravitational force. The difference between that and the measured force is the net force on the person. Integral over time gives the momentum.
     
  6. Apr 2, 2014 #5

    I thought it was momentum i was already using. Can you give an example of how to calculate it?
    From where at the graf should i start finding the integral, and after finding the integral. What then?
     
  7. Apr 2, 2014 #6

    BvU

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    You are using momentum. Your second relevant equation says mv = ##\Sigma## F ##\Delta##t

    I still don't understand the tt expressions.

    I maintain that e.g. tt18 should read (2.04-2) (1900 - fg)/m (which has the dimension m/s)
    and not (2.04-2) (1900 - fg)(deltat) /m
    and certainly not ( (2.04-2) 1900 - fg)(deltat) /m
    (as it reads now, with kgm/s minus kgm/s2 in the first subtraction.
    I certainly hope that that rendering is a mistake ! )


    Assume tt118 = (2.04-2) (1900 - fg)(deltat) /m) = 0.03309 in your figure

    From tt28 - tt18 = 0.03 I then conclude that 0.04 * 1500 deltat / m = 0.03 and with m = 80 kg (from the first figure in post 1) a deltat = 120 msec that shouldn't be there in my opinion.

    It occurs as a square in the summation (0.00718). If I correct that I get 0.5 meter as you seem to expect.

    When I reproduce the experiment, I find that even 0.3 meter is already quite a big effort ! :smile:
     
  8. Apr 2, 2014 #7

    haruspex

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    Apologies - I saw the first "relevant equation" and stopped there.
    You want the integral of the net force, i.e. the force on the plate minus the gravitational force. So you can start anywhere along the initial flat line; the net force is zero there. You should stop when the person leaves the plate, i.e. when the recorded force drops to zero.
     
  9. Apr 3, 2014 #8
    But force development is achieving an amount of force over time. So should there be a Δt when i need to calculate the jump height?
     
  10. Apr 3, 2014 #9
    Thanks for the answer!

    I'm not sure i fully understand, but tt18(and all the other tt's) should look like this: 2w536mg.png

    And then how do i get further with my calculations?

    EDIT: Got the final result to 0.47m
    For a gymnast, ain't that possible?
     
    Last edited: Apr 3, 2014
  11. Apr 3, 2014 #10

    BvU

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    Actually, that is quite possible. Think of volleyball: players can normally just reach the top of the net, but when they attack, some of them have almost an arm's length of reach available, even if starting from standstill.
     
  12. Apr 3, 2014 #11

    BvU

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    By the way, setting g=10 is a bit coarse in this context :smile:
     
  13. Apr 3, 2014 #12

    haruspex

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    As I wrote in post #7, you should start the integration anywhere along the initial flat line. Starting from the peak force is wrong. It may well be that at peak force the momentum is downwards (indeed, it looks like that from the graph).
     
  14. Apr 4, 2014 #13

    BvU

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    @Haru: I think ost is doing OK, judging from the summation expression which starts at tt1 all the way to tt28
    the earlier tt just didn't end up in the post.

    So now I am wondering if all tt < 0 contributions are really desirable in calculating vtakeoff...
    When I have more time, I might read the Linthorne link more attentively.
     
  15. Apr 4, 2014 #14

    haruspex

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    Yes, you must use all the momentum contributions, including the net negative ones.
    Rejeost, would you mind posting all the data points read from the graph? I can only see the ones from tt18 onwards, and it's a pain trying to read them off a graph with no grid.
    What are you getting for the takeoff speed? From a crude estimate from the graph it looks something like 3m/s. That would indeed give about half a metre for the jump height. Bear in mind that this height should be interpreted as the change in altitude of the centre of mass from the instant the force drops to zero. Assuming the body is straight and vertical from then, that's the same as the height the feet reach.
     
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