please bear with me as I'm not a mathematician and this is difficult to word properly. :]

I need the equation for a logarithmic spiral (let's call it $S(\theta)$) that meets certain constraints for a music visualizer I'm working on. Let's call the arc-length of the spiral $A(\theta)$. I'm looking for a spiral that meets the following requirements:






I'm basically wanting the arc-length of this spiral to correspond to a frequency of a musical note. The requirements I gave basically plot the octaves of "A" notes. (http://www.phy.mtu.edu/~suits/notefreqs.html). This way I can create a "directional" spectrogram. All "A" notes will point toward $\theta=0,2\pi,4\pi,...$, all "E" notes will point roughly in the direction of $\theta=\pi,3\pi,...$.

I spent some time grinding out rough approximations of $a$ and $b$ values that give me proper values between arclengths of 27.5 and 14080.

The closes constant values I've landed on are:



I'm looking for a way to generate values of expressions for $a$ and $b$ that will produce very accurate arclength results for the extent of the average human hearing range (20Hz-20KHz). If someone could explain the process I would need to go through to get these values I would be very grateful. Let me know if something doesn't make sense. My mind is a horribly tangled place.

  • $\begingroup$ What a cool idea. I'd be interested in seeing the result $\endgroup$ – Kopper Feb 4 '11 at 4:12
  • $\begingroup$ Sounds fascinating. Please post a link to the first visualization you post online, even if the first version doesn't look pretty to you. $\endgroup$ – David Cary Apr 6 '11 at 23:53

The arc-length of a logarithmic spiral is given here. You seem to want $A(\theta + 2\pi) = 2 A(\theta)$. This gives you $b = \ln 2/(2\pi)$. Use the value you want for $A(0)$ to find $a$.

  • $\begingroup$ @Ross, thanks for the correction. $\endgroup$ – lhf May 14 '11 at 0:03

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