Yahoo Tuning Groups Ultimate Backup tuning Woolhouse's derivation of 7/26-comma 'optimal' meantone

This is an explanation of how Woolhouse (and Paul Erlich
independently) derived 7/26-comma meantone as the one
with the least error from JI, by the RMS (= 'root mean
square) method.

I've added it to the Woolhouse webpage, which is now complete:
http://www.ixpres.com/interval/monzo/woolhouse/essay.htm

Thanks to Paul for his invaluable help at one crucial spot.

=======================================

Woolhouse's derivation of 7/26-comma 'optimal' meantone
-------------------------------------------------------

First, here are the variables used in Woolhouse's equation.
I'm substituting Roman letters for Woolhouse's Greek ones,
which are lacking in ASCII, to keep everything as simple as
possible.

Woolhouse Monzo name ratio

variables not known

tau T tempered tone ?
sigma S tempered semitone ?

variables known

t t just major-tone 9/8
t, t, just minor-tone 10/9
theta s just diatonic semitone 16/15
c c syntonic comma 81/80

Some of the just relations can be expressed:

t = t, + c major-tone = minor-tone + comma
t, = t - c minor-tone = major-tone + comma
c = t - t, comma = major-tone - minor-tone

Woolhouse states that the 5L,2s 'octave' must = 1:2 ratio
thus, 'octave' == unison.

So let's examine the composition of the 'octave' in terms
of these basic-interval variables.

The 'octave' is broken down into these basic steps:

degree: I II III IV V VI VII (I)
\ / \ / \ / \ / \ / \ / \ /
tempered T T S T T T S
just t t, s t t, t s

Therefore,

'octave'
= 5T + 2S tempered intervals
= 3t + 2t, + 2s just intervals
= 3(t, + c) + 2t, + 2s just intervals
= 5t, + 2s + 3c just intervals

Subtracting the just from the tempered:

5T + 2s tempered intervals
- 5t, + 2s + 3c just intervals
----------------------------
= 5(T - t,) + 2(S - s) - 3c
= 0

We can eliminate the variables S and s by solving for (S - s).
We get:

5(T - t,) + 2(S - s) - 3c = 0
= 2(S - s) - 3c = -5(T - t,)
= 2(S - s) = -5(T - t,) + 3c
= (S - s) = (-(5/2)*(T-t,))+((3/2)*c)

Now let's look at our three basic intervals.

Minor 3rd
---------

tempered minor 3rd = T + S

just minor 3rd = t + s
= (t, + c) + s
= t, + s + c

Subracting the JI minor 3rd from the tempered minor 3rd,
we find the 'error':

T + S tempered minor 3rd
- t, + s + c just minor 3rd
-------------------------
= (T - t,) + (S - s) - c minor 3rd error

Substituting what we found above for (S - s):

(T - t,) + (S - s) - c
= (T - t,) + (-(5/2)*(T-t,))+((3/2)*c) - c
= (-(3/2) * (T - t,)) + ((1/2) * c)

Major 3rd
---------

tempered major 3rd = 2T

just major 3rd = t + t,
= (t, + c) + t,
= 2t, + c

Subracting the JI major 3rd from the tempered major 3rd,
we find the 'error':

2T tempered major 3rd
- 2t, + c just major 3rd
----------------
= 2(T - t,) - c major 3rd error

This has no need of further simplification.

4th
----

tempered 4th = 2T + S

just 4th = t + t, + s
= (t, + c) + t, + s
= 2t, + s + c

Subracting the JI 4th from the tempered 4th,
we find the 'error':

2T + S tempered 4th
- 2t, + s + c just 4th
--------------------------
= 2(T - t,) + (S - s) - c 4th error

Substituting what we found above for (S - s):

2(T - t,) + (S - s) - c
= 2(T - t,) + (-(5/2)*(T-t,))+((3/2)*c) - c
= (-(1/2) * (T - t,)) + ((1/2) * c)

Now all three errors are in the same form:

minor 3rd error = (-(3/2) * (T - t,)) + ((1/2) * c)
major 3rd error = 2 * (T - t,) - c
4th error = (-(1/2) * (T - t,)) + ((1/2) * c)

Now to find to RMS [= 'root mean square'] total error of
the three intervals,

> [Woolhouse 1835, p 44]
>
> To determine the value of (T - t,), so that the
> sum of the squares of these three errors shall be
> the least possible, multiply them by the respective
> coefficients -(3/2), +2, -(1/2), and the sum of
> the products will be
>
> (13/2)(T - t,) - 3c.
>
> By putting this = 0, we find
>
> (T - t,) = (6/13)c,
>
> which used in the above value of (S - s), gives also
>
> (S - s) = (9/26)c
>

Paul Erlich explained this crucial step to me in more detail:

> [Paul Erlich, private communication]
>
>
> You can't just use algebra to solve an optimization
> problem You need calculus.
>
> Let's look at the expression for the errors:
>
>> minor 3rd error = (-(3/2) * (T - t,)) + ((1/2) * c)
>> major 3rd error = 2 * (T - t,) - c
>> 4th error = (-(1/2) * (T - t,)) + ((1/2) * c)
>
> Let's simplify the notation by calling (T - t,) "x":
>
> minor 3rd error = (-(3/2) * x) + ((1/2) * c)
> major 3rd error = 2 * x - c
> 4th error = (-(1/2) * x) + ((1/2) * c)
>
> The sum of the squared error is thus:
>
> ((-(3/2) * x) + ((1/2) * c))^2
> + (2 * x - c )^2
> + ((-(1/2) * x) + ((1/2) * c))^2
>
> To minimize this sum-of-squared errors as a function of x,
> you set its "derivative" or rate of change with respect to x
> equal to zero. That means that the rate at which the
> sum-of-squared error is changing when x changes is zero,
> which can only happen at a local minimum or a local maximum.
> In this case we know it will be a local minimum.
>
> The derivative of the sum-of-squared error with respect to x is:
>
> 2*((-(3/2) * x) + ((1/2) * c))*(-(3/2))
> + 2*(2 * x - c )*2
> + 2*((-(1/2) * x) + ((1/2) * c))*(-(1/2))
>
> Since we are setting this equal to 0, we can eliminate the
> three "2*"'s:
>
> ((-(3/2) * x) + ((1/2) * c))*(-(3/2))
> + (2 * x - c )*2
> + ((-(1/2) * x) + ((1/2) * c))*(-(1/2))
> = 0
>
> Simplifying:
>
> (9/4)*x - (3/4)*c + 4*x - 2*c + (1/4)*x - (1/4)*c = 0
>
> or
>
> (13/2)*x - 3*c = 0
>

= (13/2) * (T - t,) - 3c = 0

> [Paul Erlich, private communication]
>
> What the equation above actually represents is not
> the sum of the squared errors, but a relationship
> between the amount of tempering and the comma when
> the amount of tempering is such as to minimize the
> sum of the squares of the errors. And that relationship
> comes from setting the _derivative_ (or rate of change)
> of the sum-of-squared error, with respect to the amount
> of tempering, equal to zero, because that is a condition
> that only occurs at the minimum.

Now we can calculate the amount of temperament for the tone.

Solving for (T - t,) gives:

(13/2) * (T - t,) - 3c = 0
(13/2) * (T - t,) = 3c
13 * (T - t,) = 6c
(T - t,) = (6/13) * c

There's our error, or the amount of tempering, for the tone.

So the tempered tone is:

T = (t, + ((6/13) * c))

tempered tone = just minor-tone + 6/13 comma

For the semitone:

Solving for (S - s) gives:

(S - s)
= (-(5/2) * (T - t,) ) + ((3/2) * c)
= (-(5/2) * ((6/13) * c)) + ((3/2) * c)
= (-(15/13) * c ) + ((3/2) * c)
= (-(30/26) * c ) + ((39/26) * c)
= (9/26) * c

There's our error, or the amount of tempering, for the semitone.

S = (s + ((9/26) * c))
tempered semitone = just diatonic semitone + 9/26 comma

In addition to knowing the tempering from the 'minor-tone' as
calculated above, it will be convenient to know the amount of
tempering for the tone from the 'major-tone':

t = (t, + ((13/13) * c))
T = (t, + (( 6/13) * c))
T = (t - (( 7/13) * c))

tempered tone = just major-tone - 7/13 comma

Now, proving that this temperament is indeed Paul Erlich's
7/26-comma meantone, let's calculate the tempered '5th':

'5th'
= 3T + s
= 2(t - ( 7/13) * c) + (t, + ( 6/13) * c) + (s + (9/26) * c)
= 2(t - (14/26) * c) + (t, + (12/26) * c) + (s + (9/26) * c)
= 2t - ((28/26) * c) + t, + (12/26) * c) + s + (9/26) * c)
= 2t - t, + s - ((7/26) * c)
= just '5th' - ((7/26) * c)
= just '5th' - 7/26 comma

Which in numbers, of course, is:

(3/2) / ( (81/80)^(7/26) ) = ~696.165 cents

There it is.

I asked Paul Erlich to factor that as far as
could be done, and his answer was:

((2^-1)*(3^1)) / (((2^-4)*(3^4)*(5^-1))^(7/26))

= ((2^-1)*(3^1)) / (((2^-(28/26))*(3^(28/26))*(5^-(7/26))))

= 2^(2/26)*3^(-2/26)*5^(7/26)

= 2^(1/13)*3^(-1/13)*5^(7/26)

-monz

Joseph L. Monzo Philadelphia monz@juno.com
http://www.ixpres.com/interval/monzo/homepage.html
|"...I had broken thru the lattice barrier..."|
| - Erv Wilson |
--------------------------------------------------

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Manuel and Wim:

Thanks, Manuel. The tuning you gave Wim is fine for
experimenting on a 12-tone keyboard.

But you're missing a very important point of Woolhouse's
theory.

He plainly advocated 12-EDO for the 'common keyboard' and
other 'usual' instruments. The reason he went thru the
trouble to calculate the 7/26-comma meantone and to advocate
its close EDO approximations of 50-EDO and 19-EDO, was to
be able to have *two* different sizes of semitone, one
'diatonic' and one 'chromatic', to simulate characteristic
5-limit JI harmonic/melodic gestures, while at the same time
having only one size for the 'tone' to prevent commatic drift.

Woolhouse apparently didn't feel that the 7/26-comma meantone
was practical enough for actual musical use, because he
doesn't give any example of a complete (or even partial) scale,
but jumps immediately into describing 50-EDO as a substitute,
and in pretty good detail. As you can see on my webpage,
his 'basic scale' consists of 21 pitches.

Since Wim's guitar has frets which can be moved anywhere on
the fingerboard, he can fret his guitar precisely to Woolhouse's
complete scale, at least for one fingerboard-'octave'.

Here it is:

cents notation 12-EDO cents adjustment

1200.000 C C 0.000
1153.978 B# C - 46.022
1126.846 Cb B + 26.846
1080.824 B B - 19.176
1007.670 Bb A#/Bb + 7.670
961.648 A# A#/Bb - 38.352
888.495 A A - 11.505
815.341 Ab G#/Ab + 15.341
769.319 G# G#/Ab - 30.681
696.165 G G - 3.835
623.011 Gb F#/Gb + 23.011
576.989 F# F#/Gb - 23.011
503.835 F F + 3.835
457.813 E# F - 42.187
430.681 Fb E + 30.681
384.659 E E - 15.341
311.505 Eb D#/Eb - 11.505
265.484 D# D#/Eb - 34.516
192.330 D D - 7.670
119.176 Db C#/Db + 19.176
73.154 C# C#/Db - 26.846
0.000 C C 0.000

Note that the 'accidentals' which fall between the steps in
the diatonic scale which are already semitones, break the
usual pattern: Fb is *lower* than E# and Cb is *lower* than
B#.

-monz

Woolhouse's derivation of 7/26-comma 'optimal' meantone

🔗Joe Monzo <monz@xxxx.xxxx>

🔗manuel.op.de.coul@xxx.xxx

🔗Joe Monzo <monz@juno.com>