>Ron, >Thanks. I actually saw your response to JD's post about >15 minutes after I posted this, when I had worked from e-mail >150 down to about e-mail 70. But since we're on the subject, >I believe you mentioned in that post that you use a modified >form of the Conklin formula and then you proceeded to give >the Conklin formula. In what way do you modify it, or is >that proprietary? > >Phil I diced a kinetic energy formula into Conklin's formula to see what it would do, and ended up liking it a little better. It's no big secret, but it doesn't particularly mean anything either so I haven't made it available to avoid being asked, or ordered, to defend it. Neither formula means anything exact, but is just a relative indication so you can see where you are in relation to the rest of the scale. Much like most aspects of scaling. Ron N
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