# Linearising Eqn for Fractal dimension

1. Nov 25, 2012

### CAF123

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 $y =( \frac{r}{a})^{d_f}, x = L^2, c = 0$

2. Nov 25, 2012

### haruspex

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.

3. Nov 25, 2012

### CAF123

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

4. Nov 25, 2012

Yes.