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Linearising Eqn for Fractal dimension

  1. Nov 25, 2012 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    I am doing an experiment to determine the fractal dimension of hand compressed aluminium spheres. I cut a square of foil of some length ##L## and known thickness, ##t##. I do this a few times, varying ##L##. The radius of the hand compressed spheres, $$r = aL^{\frac{2}{d_f}}, a\, \text{some constant}$$ where ##d_f## is the fractal dimension sought after. Linearise this eqn so that the data can be plotted linearly.

    3. The attempt at a solution
    I suppose they would have got to the given eqn by saying $$\frac{4}{3}\pi r^{d_f} = L^2t,$$ and solving for ##r##, with ##a = (\frac{3t}{4\pi})^{1/d_f}##?
    When they say 'data', I presume that means my values of ##L, r## that I measure using a ruler or Vernier callipers.
    Now to linearise: I said a linearised form would be $$( \frac{r}{a})^{d_f} = L^2.$$ Is this correct? I have ##y =( \frac{r}{a})^{d_f}, x = L^2, c = 0##
     
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  3. Nov 25, 2012 #2

    haruspex

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    That doesn't help because you don't know df, so can't plot y. You need to get it in the form g(r) = m(df)h(L) + c, for some functions g, h and m, and some constant c, where g and h are not dependent on df. That would allow you to plot g(r) against h(L) and extract m(df) as the slope.
     
  4. Nov 25, 2012 #3

    CAF123

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    Ok, thanks. So I should have $$ ln(\frac{r}{a}) = \frac{2}{d_f} ln(L)? $$
     
  5. Nov 25, 2012 #4

    haruspex

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