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Percentage error of a guess at a position, and the actual position.

  1. May 31, 2014 #1
    How do I calculate the percentage error between a guess of a position in 3D space, and the actual position. For example, if I am allowed an error of 100mm in the x, y, and z directions, in a 5m x 4m x 3m volume, what is the maximum percentage error allowed?

    Can I just do:

    [itex]\frac{100*10^{-3}^3}{5*4*3}=1.67*10^{-5}%[/itex]


    Is this correct? It just seems a bit small to me.

    Thanks!

    __________________________________________
    If the Math isn't displaying properly, it says:

    (0.1^3)/(5*4*3)=0.0000167%
     
    Last edited: May 31, 2014
  2. jcsd
  3. May 31, 2014 #2

    mathman

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    max errors:
    x direction 2%
    y direction 2.5%
    z direction 3.333...%

    What is the formula you used to get your answer?
    You need to further clarify percentage of what?
     
  4. May 31, 2014 #3
    Sorry I wasn't very clear in the question. I have a particle that can be anywhere within a 5m*4m*3m volume. I want to pinpoint (localise) the position of this particle down to 0.1m.

    e.g

    The particle is at position (4.13,2.10,2.31). If I guess the position is at (4.23,2.20,2.31), then what is the percentage error between my guess and the actual position.

    I'm allowed a maximum deviation of 0.1m from the actual particle position, when the particle can be anywhere within a 5m*4m*3m room. What is this expressed as a percentage?

    In my attempt I used the volume that the particle could be in 0.1m*0.1m*0.1m that satisfies the localisation, divided by the total volume of the space that the particle is inside.
     
    Last edited: May 31, 2014
  5. May 31, 2014 #4
    e.g: I can pinpoint the position of a oxygen particle inside a balloon within a margin of error of 1%. What is the max (x,y,z) deviation between my guess of the particle and the actual particle if the balloon has a volume of:
    a) 0.1m^3
    b)1m^3

    This is just an example question to try and explain what i'm asking.

    I actually want to know the formula for the margin of error if i'm allowed a deviation of 0.1m,0.1m,0.1m in the x, y and z positions, for a particle within a 5m*4m*3m volume.
     
  6. Jun 1, 2014 #5

    mathman

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    You keep asking for percentage of error, where a formula would be 100x/y. x is the error, y is what?
     
  7. Jun 1, 2014 #6
    Margin of error might be a better term.

    I guess in this case y would be the total volume of the container.

    Back to my balloon example. If I had a 1% margin of error in a 1m^3 baloon, The particle could be anwhere within a 0.01m^3 volume, within this balloon. If I had a 1% margin of error in a 3m^3 baloon, The particle could be anwhere within a 0.03m^3 volume, within this balloon. If I know for certainty that the particle is at [0.5[itex]\pm[/itex]0.1 , 0.7[itex]\pm[/itex]0.1 , 0.3[itex]\pm[/itex]0.1] when it is inside a 5m x 4m x 3m = 60m^3 volume, then what is the margin of error of the localisation expressed as a percentage. Can I just do 0.001/60 = 0.0000167%

    Sorry if this question isn't making much sense. I might be talking complete mathematical BS at the moment. I'm just looking for a way to quantify the accuracy of the localisation, independantly from the size of the container. This is because If i could localise the position of a particle down to [itex]\pm[/itex]0.1m within a 100m^3 area, then this is very impressive. If I can localise the position of a particle down to [itex]\pm[/itex]0.1m within a 0.1m^3 area, then this isn't impressive. In the latter case I have just said, the particle is inside this 0.1m^3 container.
     
  8. Jun 2, 2014 #7

    mathman

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    If I understand you correctly, you want to consider a sphere centered around the point of interest with a radius of 100 mm. The error term you want seems to be the ratio of the volume of the sphere to the volume of the container. I am not sure what is the significance of this ratio.
     
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