Linearising Eqn for Fractal dimension

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Homework Help Overview

The discussion revolves around an experiment aimed at determining the fractal dimension of hand compressed aluminium spheres, with a focus on linearizing an equation related to the relationship between the radius of the spheres and the length of the foil used in the experiment.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to derive a linearized form of the equation relating the radius of the spheres to the length of the foil, questioning the correctness of their approach. Some participants suggest alternative forms that would allow for plotting without knowing the fractal dimension directly.

Discussion Status

The discussion is active, with participants providing guidance on how to reformulate the equation for linearization. There is a recognition of the need to express the relationship in a way that separates the variables, allowing for a linear plot.

Contextual Notes

Participants are working under the constraint of needing to linearize the equation without prior knowledge of the fractal dimension, which influences the direction of the discussion.

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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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