Linearising Eqn for Fractal dimension

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AI Thread Summary
The discussion focuses on linearizing the equation for determining the fractal dimension of hand-compressed aluminum spheres. The equation relates the radius of the spheres to the length of the foil square and the sought fractal dimension. The proposed linearized form is $$\ln(\frac{r}{a}) = \frac{2}{d_f} \ln(L)$$, which allows for plotting the data in a linear format. This transformation enables the extraction of the fractal dimension from the slope of the plotted data. The conversation emphasizes the importance of expressing the relationship in a way that separates the variables for effective analysis.
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Homework Statement


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.

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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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.
 
Ok, thanks. So I should have $$ ln(\frac{r}{a}) = \frac{2}{d_f} ln(L)? $$
 
Yes.
 
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