4 Fifths and a Major ThirdThere is a mathematical relationship between 4 Fifths and a Major Third. This relationship creates an irreconcilable dilemma.
Have you wondered about why Pure Octave Equal Temperament contains heavily tempered Major Thirds?
This happens because its Fifths are only slightly tempered.
An image of Pure Octave Equal Temperament:
http://rollingball.com/equal.jpgHave you wondered about why 1/4 Syntonic Meantone Temperament contains many Pure Major Thirds?
This happens because its Fifths are heavily tempered.
An image of 1/4 Syntonic Meantone Temperament:
http://rollingball.com/images/Aaron.gifThese two tunings highlight the following irreconcilable dilemma:If all the Fifths are either pure or slightly tempered, all the Major Thirds will become heavily tempered. If all the Fifths are heavily tempered, most of the Major Thirds will become either pure or slightly tempered.Roshan Kakiya's Idealised Young I: Reconciling the Irreconcilable Dilemma to the Most Possible ExtentAn image of Roshan Kakiya's Idealised Young I:
http://rollingball.com/images/Kakiya-Young.pngThe main thread of Roshan Kakiya's Idealised Young I:
https://my.ptg.org/communities/community-home/digestviewer/viewthread?GroupId=43&MessageKey=531b7c8c-0ed8-44f6-b17a-345dc9be27d8&CommunityKey=6265a40b-9fd2-4152-a628-bd7c7d770cbfUnequal temperaments are compromises that reconcile the irreconcilable dilemma that has been mentioned above to different extents. This irreconcilable dilemma can only be reconciled to the most possible extent. It cannot be completely reconciled because it is irreconcilable. The way in which it can be reconciled to the most possible extent is as follows.
The relationship between 4 Fifths and a Major Third is important because the size of a Major Third is determined by 4 consecutive Fifths around the Circle of Fifths.
The size of C-E is determined by the size of C-G, G-D, D-A and A-E. C-E will be the only Major Third that is closest to Just if C-G, G-D, D-A and A-E are the most tempered Fifths around the Circle of Fifths.
The maximum limit must be established before continuing any further. There must only be a maximum of 4 consecutive Pure Fifths around the Circle of Fifths in order to ensure that there is only one Pythagorean Major Third. If C-E is the only Major Third that is closest to Just, the most logical thing to do is to make F#-A# the only Major Third that is furthest from Just. This is because F#-C# is opposite C-G in the Circle of Fifths.
If F#-C#, C#-G#, G#-D# and D#-A# are all Pure Fifths, F#-A# will become a Pythagorean Major Third. 4 out of 12 Fifths have now been established. The Pythagorean comma has not been distributed at all by doing this.
The Pythagorean comma must be distributed across C-G, G-D, D-A, A-E, E-B, B-F#, A#-F and F-C.
The most logical thing to do is to equally temper C-G, G-D, D-A and A-E because their counterparts opposite the Circle of Fifths, F#-C#, C#-G#, G#-D# and D#-A#, are equally pure. C-G, G-D, D-A and A-E must be the most tempered Fifths in order to ensure that C-E is the only Major Third that is closest to Just. 8 out of 12 Fifths have now been established.
The remaining 4 Fifths, E-B, B-F#, A#-F and F-C, must be equally tempered in order to eliminate the Pythagorean comma in its entirety. They must be equally tempered because C-G, G-D, D-A and A-E are all equally tempered and F#-C#, C#-G#, G#-D# and D#-A# are all equally pure. E-B, B-F#, A#-F and F-C must all be less tempered than C-G, G-D, D-A and A-E in order to ensure that C-E is the only Major Third that is closest to Just. 12 out of 12 Fifths have now been established.
The optimum way of distributing the Pythagorean comma is as follows:
Pythagorean comma = −1.
a = The amount of the Pythagorean comma by which C-G, G-D, D-A and A-E must each be narrower than Just.
b = The amount of the Pythagorean comma by which E-B, B-F#, A#-F and F-C must each be narrower than Just.
c = The amount of the Pythagorean comma by which F#-C#, C#-G#, G#-D# and D#-A# must each be narrower than Just.
a = 2b.
c = 0.4a + 4b + 4c =
−1.
4a + 4b + 4
× 0 = −1.
4a + 4b =
−1.
4 × 2b + 4b = −1.
8b + 4b = −1.
12b = −1.
b = − 1/12.
a = 2 × (− 1/12).
a = − 1/6.
In summary,
C-G, G-D, D-A and A-E must each be narrower than Just by 1/6 Pythagorean comma.
E-B, B-F#, A#-F and F-C must each be narrower than Just by 1/12 Pythagorean comma.
F#-C#, C#-G#, G#-D# and D#-A# must all be Pure Fifths.
C-E will be the only Major Third that is closest to Just.
F#-A# will be the only Major Third that is furthest from Just.
In conclusion, Roshan Kakiya's Idealised Young I reconciles the irreconcilable dilemma that has been mentioned above to the most possible extent.------------------------------
Roshan Kakiya
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