The Other Noble Fifth

Is there a name for the noble hypermeantone fifth that makes the chromatic
semitone phi times wider than the diatonic semitone? It's the hyper-meantone
analog to the golden fifth. To find the golden fifth on the Stern-Brocot
tree you start at 4/7, then go to 7/12, then alternate directions yielding
4/7,7/12,11/19,18/31,29/50,47/81,76/131...; to find the fifth I am talking
about you start at 3/5 and do the same for
3/5,7/12,10/17,17/29,27/46,44/75,71/121... (The denominators of these are
obviously MOS.). The actual value of the golden fifth is 2 ^ ( (4 + 3 *
(phi) ) / (7 + 5 * (phi) ) ), while the actual value of my fifth is 2 ^ ( (3
+ 4 * (phi) ) / (5 + 7 * (phi) ) ); all the rational and phi-related
coefficients are swapped (The fourths, which some people prefer for good
reason, are 2 ^ ( (3 + 2 * (phi) ) / (7 + 5 * (phi) ) ) and 2 ^ ( (3 + 2 *
(phi) ) / (7 + 5 * (phi) ) ).). If it doesn't have a name yet, it should
obviously be called the silver fifth.

Keenan P.

On Sun, 10 Sep 2000, Keenan Pepper wrote:

> Is there a name for the noble hypermeantone fifth that makes the chromatic
> semitone phi times wider than the diatonic semitone? It's the hyper-meantone
> analog to the golden fifth.

I have another analog. You set the ratio of the octave to the fourth to be the
same as that of the fourth to the major tone. This gives the scale exact self
symmetry. If you take 29 notes from the spiral of fifths, then the fourth of
that scale looks like the octave of a 12-note fifth-generated scale.

In the golden meantone, the ratio of a minor third to a tone is the same as
that of a tone to a diatonic semitone, diatonic to chromatic semitone, etc.
However, I the scale does not show self symmetry in the same way that the
one described above does.

So, it happens that the perfect fourth is sqrt(2)-1 oct or 497.1 cents. The
perfect fifth is 2-sqrt(2) oct or 702.9 cents.

I don't think these intervals have any musical significance. The fifth is a
bit wide for a schismic scale, by my tastes.

> To find the golden fifth on the Stern-Brocot
> tree you start at 4/7, then go to 7/12, then alternate directions yielding
> 4/7,7/12,11/19,18/31,29/50,47/81,76/131...; to find the fifth I am talking
> about you start at 3/5 and do the same for
> 3/5,7/12,10/17,17/29,27/46,44/75,71/121... (The denominators of these are
> obviously MOS.). The actual value of the golden fifth is 2 ^ ( (4 + 3 *
> (phi) ) / (7 + 5 * (phi) ) ), while the actual value of my fifth is 2 ^ ( (3
> + 4 * (phi) ) / (5 + 7 * (phi) ) ); all the rational and phi-related
> coefficients are swapped (The fourths, which some people prefer for good
> reason, are 2 ^ ( (3 + 2 * (phi) ) / (7 + 5 * (phi) ) ) and 2 ^ ( (3 + 2 *
> (phi) ) / (7 + 5 * (phi) ) ).). If it doesn't have a name yet, it should
> obviously be called the silver fifth.

I make this fifth 704.1 cents. That sounds close to something Margo Schulter
mentioned in one of her epic posts, but I don't have it to hand.

Hang on, aren't your fifths the wrong way round?

Graham

"I toss therefore I am" -- Sartre

--- In tuning@egroups.com, Graham Breed <
graham@m...> wrote:
> So, it happens that the perfect fourth is sqrt(2)-1 oct or 497.1
cents. The
> perfect fifth is 2-sqrt(2) oct or 702.9 cents.

Neat! THIS COULD be called the silver fifth, since
the continued fraction expansion ends in all 2's.