RE: Linear to K5 equation

>Leslie, in a previous e-mail you provided the following equation for
>converting linear spectra to K5.  Thank you.  You wrote:
>Log2(149.5744446 * partial) * 10 / step
>
>I assume that Log2 means "Log base 2","partial" is a linear value
>ranging from 0 to 1, and that "step" is dB/K5 step = 72.24719896/99.

Yup, that's it.

I got some nice confirmation that this equation works correctly today. I 
recorded the K5 playing a single partial at various amplitudes. Then I 
looked at the waveforms in a wave editor program and compared the linear 
amplitude of the waveforms to the K5 amplitudes the partial was set to (hope 
this makes sense). They matched up with what the equation said they should 
be.

>I played with this and I think I got it to work.

Cool!

>With that said, I came up with the K5 values for the Waldorf waves
>(from the 300Waves.pdf) number 10, 15, 84, 181, 385 and 415 – they
>looked interesting.  By looking at Waldorf wave 101 (pure sine wave) I
>conclude that the first line shown on the graph is actually the
>fundamental rather than the  1st harmonic. (Is that correct, is
>frequency of the 1st harmonic actually twice the fundamental or I am I
>confused on the nomenclature?)

That's my understanding. You have the fundamental, 1st harmonic, 2nd 
harmonic, etc. A lot of graphs seem to get this wrong.

Hey, I was looking at one of the links you posted, and found that some 
interesting graphs based on the Waldorf waves. Here's an example:

http://www.xs4all.nl/~hkwad/waldorf/fft000.htm

If you place the mouse over the red dots, a tooltip with appear with the 
exact amplitude of the harmonic. Would this save you some work?

I'll check out your Excel spreadsheet.

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