## Re: need formula for stretch-tuning

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• Thanks Allan! But... I still don t understand how to use this information. I m using Microsoft Excel to compute the values. exp means e raised to the
Message 1 of 8 , Oct 1, 2000
Thanks Allan! But... I still don't understand how to use this
information. I'm using Microsoft Excel to compute the values.

'exp' means 'e' raised to the powers 0.085 to -0.062, right?

Can you please explain how the numbers '-48' to '48' fit into
the formula, and how I use the range from '0.085' to '-0.062'?
Does that decrement in some way?

When I use the Excel formula EXP(x), where 'x' is the decremented
range from 0.085 to -0.062, I get ratios with cents-values that
are quite a bit larger than those on the graph, and the the
graph of these values is linear, which is not the case with the
values given in the Rhodes manual. And, the 'horizontal axis'
numbers 48 to -48 are not part of the calculation.

???

Thanks.

-monz

--- In tuning@egroups.com, Allan Myhara <amyhara@m...> wrote:
> http://www.egroups.com/message/tuning/13863
>
> Hello!
> A rare opportunity for me to contribute here! If you
> re-label the horizontal axis as -48, -36, -24, -12, 0, 12, 24,
> 36, 48 then you can use exponential functions and an accurate
> formulation is exp^0.085 - exp^-0.062. I hope this is useful
> to you!
>
> " Monz" wrote:
> > > The following values are taken from the stretch-tuning chart
> > > in the _Rhodes Keyboard Instruments USA Service Manual_,
> > > (c) 1979, which came with my Rhodes electric piano.
> >
> >
> > Figuring that someone familiar with calculus might recognize
> > the key to the formula quickly by seeing a graph of the values,
> >
> > http://www.egroups.com/files/tuning/monz/rhodes.jpg

-monz
http://www.ixpres.com/interval/monzo/homepage.html
• From: Monz ... I just looked at what I sent. Sorry, the formula should have been e^0.085x - e^-0.062x. I neglected to type the variable, x.
Message 2 of 8 , Oct 1, 2000
From: " Monz" <MONZ@...>

> Thanks Allan! But... I still don't understand how to use this
> information. I'm using Microsoft Excel to compute the values.
>
> 'exp' means 'e' raised to the powers 0.085 to -0.062, right?

I just looked at what I sent. Sorry, the formula should have been
e^0.085x - e^-0.062x. I neglected to type the variable, x. Yes, e raised
to those powers, the products of the constants and the variable. I
didn't mean to imply a range of values. There are two exponentials in
the function: e^0.085x minus e^-0.062x. Simply feed it the numbers -48
to 48.

> Can you please explain how the numbers '-48' to '48' fit into
> the formula, and how I use the range from '0.085' to '-0.062'?
> Does that decrement in some way?
>
> When I use the Excel formula EXP(x), where 'x' is the decremented
> range from 0.085 to -0.062, I get ratios with cents-values that
> are quite a bit larger than those on the graph, and the the
> graph of these values is linear, which is not the case with the
> values given in the Rhodes manual. And, the 'horizontal axis'
> numbers 48 to -48 are not part of the calculation.
>
> ???
--
Bye for now

Allan Myhara
• ... Thanks, Allan! It fits the Rhodes numbers like a glove!! If one numbers the variable _x_ from 1 to 88, as piano keys, then the precise formula in Excel
Message 3 of 8 , Oct 1, 2000
--- In tuning@egroups.com, Allan Myhara <amyhara@m...> wrote:
> http://www.egroups.com/message/tuning/13879
>
> I just looked at what I sent. Sorry, the formula should have
> been e^0.085x - e^-0.062x. I neglected to type the variable, x.
> Yes, e raised to those powers, the products of the constants
> and the variable. I didn't mean to imply a range of values.
> There are two exponentials in the function: e^0.085x minus
> e^-0.062x. Simply feed it the numbers -48 to 48.

Thanks, Allan! It fits the Rhodes numbers like a glove!!

If one numbers the variable _x_ from 1 to 88, as piano keys,
then the precise formula in Excel format is:

=EXP(0.085*(x-48))-EXP(-0.062*(x-48))

In Excel, 'EXP()' means 'e^'.

Would you mind explaining to us how you derived it, step by
step? I need it in 'remedial calculus 001'.

In case anyone's wondering why I wanted this: be on the lookout
for a new retuning of an old piece I did originally in 12-tET.
It features a piano timbre, so it's going to be retuned in a
stretch pseudo-JI.

I need the formula so that after I calculate the amount of
MIDI pitch-bend to add or subtract from the 12-tET MIDI-note
number, I can simply plug these in as the variables in Allan's
formula to get the amount of additional pitch-bend needed for
the proper amount of stretch.

-monz
http://www.ixpres.com/interval/monzo/homepage.html
• ... I thought it should do quite well. ... I really didn t have to apply any calculus (although exponentials are quite common in calculus). At first I thought
Message 4 of 8 , Oct 1, 2000
Monz wrote:

> > I just looked at what I sent. Sorry, the formula should have
> > been e^0.085x - e^-0.062x. I neglected to type the variable, x.
> > Yes, e raised to those powers, the products of the constants
> > and the variable. I didn't mean to imply a range of values.
> > There are two exponentials in the function: e^0.085x minus
> > e^-0.062x. Simply feed it the numbers -48 to 48.

> Thanks, Allan! It fits the Rhodes numbers like a glove!!

I thought it should do quite well.

> If one numbers the variable _x_ from 1 to 88, as piano keys,
> then the precise formula in Excel format is:
>
> =EXP(0.085*(x-48))-EXP(-0.062*(x-48))
>
> In Excel, 'EXP()' means 'e^'.
>
> Would you mind explaining to us how you derived it, step by
> step? I need it in 'remedial calculus 001'.

I really didn't have to apply any calculus (although exponentials are
quite common in calculus). At first I thought of approximating it with
polynomials, but I soon realized that the horizontal axis was labelled
arbitrarily and that the shape of the graph was an exponential function
on both sides of the graph from the zero crossing point from which the
graph trends negative to the left and positive to the right. An
exponential function, on its own, never decreases to zero nor is ever
negative, it can only get really small as its exponent gets negatively
big. (But it can be forced negative by multiplying it by -1, in other
words by putting a minus sign in front of it.) So, setting the variable
to -48, the positive term, with its negative exponent, could be ignored
as I adjusted the constant in the exponent of the negative term, with
its positive exponent, to make the negative term agree with the left end
of the graph. Then, setting the variable to +40, I followed the same
procedure with the positive term, ignoring the contribution of the
negative term. After finding suitable constants, I plugged in the
different variable values of -36 through +36 into the two term function
to see if it matched the numbers indicated by the graph, and lo and
behold, it looked nice. Of course this is a just-so-story, things like
this are not reasoned out quite so smoothly!

> In case anyone's wondering why I wanted this: be on the lookout
> for a new retuning of an old piece I did originally in 12-tET.
> It features a piano timbre, so it's going to be retuned in a
> stretch pseudo-JI.
>
> I need the formula so that after I calculate the amount of
> MIDI pitch-bend to add or subtract from the 12-tET MIDI-note
> number, I can simply plug these in as the variables in Allan's
> formula to get the amount of additional pitch-bend needed for
> the proper amount of stretch.

--
Bye for now

Allan Myhara