Linear to K5 equation

Ok,
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. 
I played with this and I think I got it to work.  

For example, if the Linear value is 1, the K5 value should be 99.  
Log2(149.5744446*1)*10*99/72.24719896 = 99.

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?) 

Anyway, I will post an Excel file with the result.  What I did was
simply copy the spectrum from the Waldorf .pdf (I first zoomed to
500%), and put it in the spread sheet, sized it and then just
eyeballed the values and put them in the cells right below the graph.
 I then used the above equation to get the K5 values.  This excel
sheet works like a template, so all you have to do is copy a spectra,
resize the graph, enter the linear values in ROW 50 starting at B50
and the K5 values will show up in column "BS" (no pun intended)
automatically  (more or less).  Once you get the hang of it you can
decode a wave in less than 5 minutes. 

I will post this Excel file in the same place as the .pdfs (Hard Copy
Spectra).  




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