Last update of all pages 17.8.2014
In a previous article entitled 'A Use Of Quantum Numbers With Beat Frequencies in Psychoacoustics' there is an expression for Secondary order beat frequency = δ for an interval = (b+δ)/a where a & b are integers & δ is a small change.
We will briefly recall this now:
In a previous article entitled 'A Use Of Quantum Numbers With Beat Frequencies in Psychoacoustics' there is an expression for Secondary order beat frequency = δ for an interval = (b+δ)/a where a & b are integers & δ is a small change.
We will briefly recall this now:
So as an example Considering two component frequencies being close to a numerical integer ratio like 8/5
Letting bottom frequency, fa = 5 and top frequency fb = 8, and then increment fb = 8 to (8+δ) where δ is a small change
The primary difference frequency (or conjunction alignment frequency) = (8+δ)-5 = (3+δ)
As shown in the picture before,
the primary angle swept out = 5/(3+δ) which is fa/(fb-fa+δ)
So how many conjunctions will it take to have a conjunction point rotated round a number of full turns to be back close to the starting point?
Well to get the conjunction angle=5/(3+δ) back to near the start angle, can multiply by 3 to give close to 5 complete full rotations.
To find the angle of displacement θ as shown in the picture:
θ = 5-[3*(5/(3+δ))]
or
θ = fa-[(fb-fa)*(fa/(fb-fa+δ))]
Simplifying equation gives :
θ = fa*δ/(fb+δ-fa)
The secondary order time period, Tβ = (primary order time period)/ θ
SO if primary order time period = Tα = (1/fo)*[fa/((fb+δ)-fa)] {because timeperiod = 1/frequency}{f0 = fundamental base frequency in relation to frequency component , fa}
Then expression for Tβ = (1/fo)*[fa/(fb+δ-fa)]*[(fb+δ-fa)/(fa*δ)]
Terms cancel giving :
Tβ = 1/(δ*fo)
By definition, frequency = 1/timeperiod
So for secondary order beat frequency,Fβ:
-----------
Fβ = δ*fo
-----------
Finding the Peakβ EDO:
EDO stands for ‘Equal Divisions Per Octave’ which is the same as the alternative term , ET = ‘Equal Temperament’
So using a geometrical analogy for the next picture, interval (4/3) = 2^Θ Angle= Θ
Letting bottom frequency, fa = 5 and top frequency fb = 8, and then increment fb = 8 to (8+δ) where δ is a small change
The primary difference frequency (or conjunction alignment frequency) = (8+δ)-5 = (3+δ)
As shown in the picture before,
the primary angle swept out = 5/(3+δ) which is fa/(fb-fa+δ)
So how many conjunctions will it take to have a conjunction point rotated round a number of full turns to be back close to the starting point?
Well to get the conjunction angle=5/(3+δ) back to near the start angle, can multiply by 3 to give close to 5 complete full rotations.
To find the angle of displacement θ as shown in the picture:
θ = 5-[3*(5/(3+δ))]
or
θ = fa-[(fb-fa)*(fa/(fb-fa+δ))]
Simplifying equation gives :
θ = fa*δ/(fb+δ-fa)
The secondary order time period, Tβ = (primary order time period)/ θ
SO if primary order time period = Tα = (1/fo)*[fa/((fb+δ)-fa)] {because timeperiod = 1/frequency}{f0 = fundamental base frequency in relation to frequency component , fa}
Then expression for Tβ = (1/fo)*[fa/(fb+δ-fa)]*[(fb+δ-fa)/(fa*δ)]
Terms cancel giving :
Tβ = 1/(δ*fo)
By definition, frequency = 1/timeperiod
So for secondary order beat frequency,Fβ:
-----------
Fβ = δ*fo
-----------
Finding the Peakβ EDO:
EDO stands for ‘Equal Divisions Per Octave’ which is the same as the alternative term , ET = ‘Equal Temperament’
So using a geometrical analogy for the next picture, interval (4/3) = 2^Θ Angle= Θ
Taking natural logarithms, ln:
ln(4/3)/ln(2) =Θ
Angle turned, Θ , can be expressed as it’s reasonably low form which is about 2/5
estimating this angle = 2/(5+ δ):
In terms of
|fb+δ | = proposed upper integer of geometry ratio in picture
|fa | = lower integer in geometry ratio
|fb+δ-fa| = conjunction difference frequency
Then we have:
|fa | =|2 |
|fb+δ-fa | =|5+δ|
Letting Integer 2 = a
And Integer 5 = c
So:
ln(I)/ln(2) = a/(c+ δ) ln(4/3)/ln(2) = 2/(5+δ)
δ = [a*ln(2)/ln(I)] – c in this example, interval, I=(4/3)
δ = [2*ln(2)/ln(4/3)] – 5 = 0.18115832….
And from equation found above: Fβ = δ*fo
then:
1/(Peakβ EDO) = δ*2/(5+δ) 1/(Peakβ EDO) = δ*a/(c+δ)
Peakβ EDO = (c+δ)/δ*a
Peakβ EDO = (5+δ)/δ*2 = -13.3001...
This is where the exact zero angle is reached but after a number of angular turns we have to pick an EDO closest to this value for the peak. In the picture above showing angular turn , Θ = ln(4/3)/ln(2) , The first angle turn is shown as a line drawn from top Zero angle to lower right, marked ‘1’EDO
Then the next angular turn is shown as a line to the top left marked ‘2’ EDO
And the next angular turn is shown as a line from ‘2’ to the middle right marked ‘3’ EDO
Etc…
Observing just above right along the point ‘2’ , there is 5 angular turns made to make the mark ‘7’EDO
And another 5 angular turns from ‘7’ is the point further right at the top marked ‘12’
This ‘12’ is closest to the found Peakβ EDO = (5+δ)/δ*2 = -13.3001...
And occurs very near the Zero angle at the top of the picture Wheel.
This next part was derived from intuition of the geometry:
We have defined angle integers, (a/c), in this example (a/c) = (2/5)
The first EDO in a series of EDOs to be derived we let = q, an integer
If δ is negative, Peakβ EDO and the successive series of EDO’s drawn with ‘c’ angular steps will be rotating clockwise :
If δ is postive, Peakβ EDO and the successive series of EDO’s drawn with ‘c’ angular steps will be rotating anti-clockwise
(or opposite way round depending on perspective view)
When dealing with a negative Peakβ EDO value,
( (q*a)+1)/c = N where N is the lowest whole number found
thus
q = ((N*c)-1)/a
When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
thus
q = ((N*c)+1)/a
And the Series of associated EDOs will be of the form:
EDO = q+(k*c) where k is an integer incremented from 0 to however many.
In this example where angle integers , (a/c) = 2/5
We notice that the Peakβ EDO = (5+δ)/δ*2 = -13.3001... is a negative value
So using the found equation : q = ((N*c)-1)/a
Trying N = 1 initially
q = ((1*5)-1)/2 = 2, and it is a whole number and therefore the valid value
So series for EDO = q+(k*c) = 2+(k*5)
2EDO
7EDO close
12EDO closest to the Peakβ EDO = -13.3001
17EDO close
22EDO further
27EDO
....We can rearrange the previous equation to find the value of k that corresponds to an EDO being close to the Peakβ EDO:
k = (1/c)(| Peakβ| - q) where |Peakβ| is the modulus positive value of Peakβ EDO
In this example, k = (1/5)(13.3001-2) = 2.26002
So we can choose some rounded integer values of ‘k close to this irrational value that will correspond with EDOs in the region at the top of the Wheel picture near the Zero angle.
So there is a choice of
k = 2 for 12EDO
and k = 3 for 17EDO
So we could choice a set of values of ‘k’ close to its irrational value will be near to the zero angle, therefore “close” to the Just intonation interval ‘I’ in some way.
But what is going to be this limiting angle ?
Imposing a limiting angle = γ
In the series of EDOs we produced, we have a range of EDOs in the region of the close Peakβ EDO , but as the EDOs in the series get further away from the Peak, the angle they are placed at in a geometrical representation gets further away from the Zero angle with Each angular Turn.
The exact angular turn we know as ,Θ=ln(I)/ln(2) where I = interval
For a selected EDO from the series it's angle will =
EDO*Θ = EDO*ln(I)/ln(2)
Now we need to round off this value of the EDO angle to the nearest whole number (i.e. the zero angle) subtracted from its irrational true value, and the modulus of the result is imposed to be less than the limiting angle, γ.
'ROUND' is defined to mean rounded to the nearest integer
We have
| ROUND[EDO*ln(I)/ln(2)]-(EDO*ln(I)/ln(2)) | < Imposing a limiting angle = γ
In the series of EDOs we produced, we have a rangle of EDOs in the region of the close Peakβ EDO , but as the EDOs in the series get further away from the Peak, the angle they are placed at in a geometrical representation gets further away from the Zero angle with Each angular Turn.
The exact angular turn we know as ,Θ=ln(I)/ln(2) where I = interval
For a selected EDO from the series it's angle will =
EDO*Θ = EDO*ln(I)/ln(2)
Now we need to round off this value of the EDO angle to the nearest whole number (i.e. the zero angle) subtracted from its irrational true value, and the modulus of the result is imposed to be less than the limiting angle, γ.
'ROUND' is defined to mean rounded to the nearest integer
We have
| ROUND[EDO*ln(I)/ln(2)]-(EDO*ln(I)/ln(2)) | < γ
So we could have EDOs part of a limiting angle , γ Set
Some groups may alternatively wish to find EDOs that are opposite the Zero angle at half a full turn limited by an angle γ also. Perhaps they will have a set of what could be called Anti-EDOs.
Example set of intervals with limiting angle parameter γ
Setting limiting angle γ = 1/4 a full turn
and finding ETs that occur near the zero angle < γ so < 1/4 or the top half of the circle
and using Just intonation intervals in the temperament being 16/15, 6/5, 5/4, 45/32 ,3/2 Other intervals of these notes produced by the ET will be downwards 15/8,5/3,8/5,64/45,4/3 included in the total scale of course.
Using a simple program (that we've checked to a good degree and very sure works!)
the set of γ angular Limit ET's are:
γ<1/4
ETs:
12,22,31,41,53,65,87,106,118,130,140,152,159,171,183,193,205,224,236,248,258,270,277,289,301,311,323,335,342,354,376,388,407,419,429......
Meaning 12ET,22ET,31ET,41ET,53ET,65ET, etc…
γ<1/6
ETs: 12,53,65,106,118,171,183,236,258,289,323,354,376,388,429,441,472,494,506,547,559,612,624,665,677,718,730,783,795,817,848,870.....
γ<1/4
ETs:
53,65,118,171,236,354,376,441,494,506,547,559,612,665,677,730,783,795,848.....
γ<1/12
ETs: 53,118,236,494,559,612,665,677,730.......
γ<1/16
ETs: 118,494,559,612,730....
!Now for anti EDOs where we search for EDOs close to opposition of zero angle = ± ½ turn.
Using intervals 3/2,5/4,6/5 & reverse 4/3,8/5,5/3 only
@ ± ½ turn & γ<1/4
ETs: 20,32,33,52,85,86,119,138,139,150,151,170,203,204,237,257,290,322,355,375,408,421,440,442,461,462,473,474,493,526,527,560,579,580,591,592,611,644,645,678,697,698,731,750,762..........
@ ± ½ turn & γ<1/5
ETs: 33,86,119,204,322,408,526,645,763,816,849,934,967,1138,1375,1579,1664,1697,1750,1987,2105,2191,2309,2546,2599,2632,2717,2750,2921,3158,3329,3362,3447,3480,3533,3651......
@ ± ½ turn & γ<1/6
We have nothing
Next a mixture of 2 Up zero harmonious close to just regions for intervals (5/4) & (6/5) and 1 down Bad region for interval (3/2)
interval = (3/2) @ ± ½ turn & γ<1/4
interval = (5/4) @ zero angle & γ<1/6
interval = (6/5) @ zero angle & γ<1/6
For these 3 conditioned intervals only
ETs: 103,221,391,509,832,833,951,1121.....
Changing limiting angle parameter, γ
interval = (3/2) @ ± ½ turn & γ<1/4
interval = (5/4) @ zero angle & γ<1/7
interval = (6/5) @ zero angle & γ<1/7
For these 3 conditioned intervals only
ETs: 509,1121,2292,3175,3787,4805,5417,5688,6588.....{room for error}
Changing limiting angle parameter, γ
interval = (3/2) @ ± ½ turn & γ<1/4
interval = (5/4) @ zero angle & γ<1/8
interval = (6/5) @ zero angle & γ<1/8
For these 3 conditioned intervals only
ETs: ...nothing...
* * *
Calculating the fret or step number of an EDO that
corresponds close to interval, I
To find out what the numbered step like a fret on an instrument is connected with this EDO that should sound close to interval, in this example interval ,I= (4/3), then we know that, I = 2^(n/EDO) so:
step or fret ‘n’=EDO*ln(I)/ln(2) rounded to the nearest integer.
and in this example, n = 12*ln(4/3)/ln(2) = 4.98045….. = 5 rounded to closest integer.
Continuing from earlier....
A more accurate angle turned can be expressed as it's reasonably low form which is alternatively about 5/12, estimating this angle = 5/(12+ δ):
In terms of
Angle integers (a/c) = 5/12
| 5 |
|12+δ|
So:
ln(4/3)/ln(2) = 5/(12+δ)
δ = [5*ln(2)/ln(4/3)] – 12 = 0.0471042…
And then:
1/(Peakβ EDO) = δ*5/(12+δ)
Peakβ EDO = (12+δ)/δ*5 = +51.1508...This is where the exact zero angle is reached but after a number of angular turns we have to pick the EDO closest to this value for the peak.
Recalling from before,
When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
Thus: q = ((N*c)+1)/a
Trying N = 1 initially: then q = (12+1)/5 = 13/5, not a whole number so invalid.
Trying N = 2 next: then q = ((2*12)+1)/5 = 25/5 = 5, a whole number so therefore valid.
So q = 5 and the series to be derived is from before : EDO = q+(k*c)
We have
q = 5ET
5+12 = 17ET further
17+12 = 29ET further
29+12 = 41ET close
41+12 = 53ET closest to peakβ EDO
53+12 = 65ET close
65+12 = 77ET further
77+12 = 89ET further
……………
Another example, (5/4) = 2^Θ
ln(4/3)/ln(2) =Θ
Angle turned, Θ , can be expressed as it’s reasonably low form which is about 2/5
estimating this angle = 2/(5+ δ):
In terms of
|fb+δ | = proposed upper integer of geometry ratio in picture
|fa | = lower integer in geometry ratio
|fb+δ-fa| = conjunction difference frequency
Then we have:
|fa | =|2 |
|fb+δ-fa | =|5+δ|
Letting Integer 2 = a
And Integer 5 = c
So:
ln(I)/ln(2) = a/(c+ δ) ln(4/3)/ln(2) = 2/(5+δ)
δ = [a*ln(2)/ln(I)] – c in this example, interval, I=(4/3)
δ = [2*ln(2)/ln(4/3)] – 5 = 0.18115832….
And from equation found above: Fβ = δ*fo
then:
1/(Peakβ EDO) = δ*2/(5+δ) 1/(Peakβ EDO) = δ*a/(c+δ)
Peakβ EDO = (c+δ)/δ*a
Peakβ EDO = (5+δ)/δ*2 = -13.3001...
This is where the exact zero angle is reached but after a number of angular turns we have to pick an EDO closest to this value for the peak. In the picture above showing angular turn , Θ = ln(4/3)/ln(2) , The first angle turn is shown as a line drawn from top Zero angle to lower right, marked ‘1’EDO
Then the next angular turn is shown as a line to the top left marked ‘2’ EDO
And the next angular turn is shown as a line from ‘2’ to the middle right marked ‘3’ EDO
Etc…
Observing just above right along the point ‘2’ , there is 5 angular turns made to make the mark ‘7’EDO
And another 5 angular turns from ‘7’ is the point further right at the top marked ‘12’
This ‘12’ is closest to the found Peakβ EDO = (5+δ)/δ*2 = -13.3001...
And occurs very near the Zero angle at the top of the picture Wheel.
This next part was derived from intuition of the geometry:
We have defined angle integers, (a/c), in this example (a/c) = (2/5)
The first EDO in a series of EDOs to be derived we let = q, an integer
If δ is negative, Peakβ EDO and the successive series of EDO’s drawn with ‘c’ angular steps will be rotating clockwise :
If δ is postive, Peakβ EDO and the successive series of EDO’s drawn with ‘c’ angular steps will be rotating anti-clockwise
(or opposite way round depending on perspective view)
When dealing with a negative Peakβ EDO value,
( (q*a)+1)/c = N where N is the lowest whole number found
thus
q = ((N*c)-1)/a
When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
thus
q = ((N*c)+1)/a
And the Series of associated EDOs will be of the form:
EDO = q+(k*c) where k is an integer incremented from 0 to however many.
In this example where angle integers , (a/c) = 2/5
We notice that the Peakβ EDO = (5+δ)/δ*2 = -13.3001... is a negative value
So using the found equation : q = ((N*c)-1)/a
Trying N = 1 initially
q = ((1*5)-1)/2 = 2, and it is a whole number and therefore the valid value
So series for EDO = q+(k*c) = 2+(k*5)
2EDO
7EDO close
12EDO closest to the Peakβ EDO = -13.3001
17EDO close
22EDO further
27EDO
....We can rearrange the previous equation to find the value of k that corresponds to an EDO being close to the Peakβ EDO:
k = (1/c)(| Peakβ| - q) where |Peakβ| is the modulus positive value of Peakβ EDO
In this example, k = (1/5)(13.3001-2) = 2.26002
So we can choose some rounded integer values of ‘k close to this irrational value that will correspond with EDOs in the region at the top of the Wheel picture near the Zero angle.
So there is a choice of
k = 2 for 12EDO
and k = 3 for 17EDO
So we could choice a set of values of ‘k’ close to its irrational value will be near to the zero angle, therefore “close” to the Just intonation interval ‘I’ in some way.
But what is going to be this limiting angle ?
Imposing a limiting angle = γ
In the series of EDOs we produced, we have a range of EDOs in the region of the close Peakβ EDO , but as the EDOs in the series get further away from the Peak, the angle they are placed at in a geometrical representation gets further away from the Zero angle with Each angular Turn.
The exact angular turn we know as ,Θ=ln(I)/ln(2) where I = interval
For a selected EDO from the series it's angle will =
EDO*Θ = EDO*ln(I)/ln(2)
Now we need to round off this value of the EDO angle to the nearest whole number (i.e. the zero angle) subtracted from its irrational true value, and the modulus of the result is imposed to be less than the limiting angle, γ.
'ROUND' is defined to mean rounded to the nearest integer
We have
| ROUND[EDO*ln(I)/ln(2)]-(EDO*ln(I)/ln(2)) | < Imposing a limiting angle = γ
In the series of EDOs we produced, we have a rangle of EDOs in the region of the close Peakβ EDO , but as the EDOs in the series get further away from the Peak, the angle they are placed at in a geometrical representation gets further away from the Zero angle with Each angular Turn.
The exact angular turn we know as ,Θ=ln(I)/ln(2) where I = interval
For a selected EDO from the series it's angle will =
EDO*Θ = EDO*ln(I)/ln(2)
Now we need to round off this value of the EDO angle to the nearest whole number (i.e. the zero angle) subtracted from its irrational true value, and the modulus of the result is imposed to be less than the limiting angle, γ.
'ROUND' is defined to mean rounded to the nearest integer
We have
| ROUND[EDO*ln(I)/ln(2)]-(EDO*ln(I)/ln(2)) | < γ
So we could have EDOs part of a limiting angle , γ Set
Some groups may alternatively wish to find EDOs that are opposite the Zero angle at half a full turn limited by an angle γ also. Perhaps they will have a set of what could be called Anti-EDOs.
Example set of intervals with limiting angle parameter γ
Setting limiting angle γ = 1/4 a full turn
and finding ETs that occur near the zero angle < γ so < 1/4 or the top half of the circle
and using Just intonation intervals in the temperament being 16/15, 6/5, 5/4, 45/32 ,3/2 Other intervals of these notes produced by the ET will be downwards 15/8,5/3,8/5,64/45,4/3 included in the total scale of course.
Using a simple program (that we've checked to a good degree and very sure works!)
the set of γ angular Limit ET's are:
γ<1/4
ETs:
12,22,31,41,53,65,87,106,118,130,140,152,159,171,183,193,205,224,236,248,258,270,277,289,301,311,323,335,342,354,376,388,407,419,429......
Meaning 12ET,22ET,31ET,41ET,53ET,65ET, etc…
γ<1/6
ETs: 12,53,65,106,118,171,183,236,258,289,323,354,376,388,429,441,472,494,506,547,559,612,624,665,677,718,730,783,795,817,848,870.....
γ<1/4
ETs:
53,65,118,171,236,354,376,441,494,506,547,559,612,665,677,730,783,795,848.....
γ<1/12
ETs: 53,118,236,494,559,612,665,677,730.......
γ<1/16
ETs: 118,494,559,612,730....
!Now for anti EDOs where we search for EDOs close to opposition of zero angle = ± ½ turn.
Using intervals 3/2,5/4,6/5 & reverse 4/3,8/5,5/3 only
@ ± ½ turn & γ<1/4
ETs: 20,32,33,52,85,86,119,138,139,150,151,170,203,204,237,257,290,322,355,375,408,421,440,442,461,462,473,474,493,526,527,560,579,580,591,592,611,644,645,678,697,698,731,750,762..........
@ ± ½ turn & γ<1/5
ETs: 33,86,119,204,322,408,526,645,763,816,849,934,967,1138,1375,1579,1664,1697,1750,1987,2105,2191,2309,2546,2599,2632,2717,2750,2921,3158,3329,3362,3447,3480,3533,3651......
@ ± ½ turn & γ<1/6
We have nothing
Next a mixture of 2 Up zero harmonious close to just regions for intervals (5/4) & (6/5) and 1 down Bad region for interval (3/2)
interval = (3/2) @ ± ½ turn & γ<1/4
interval = (5/4) @ zero angle & γ<1/6
interval = (6/5) @ zero angle & γ<1/6
For these 3 conditioned intervals only
ETs: 103,221,391,509,832,833,951,1121.....
Changing limiting angle parameter, γ
interval = (3/2) @ ± ½ turn & γ<1/4
interval = (5/4) @ zero angle & γ<1/7
interval = (6/5) @ zero angle & γ<1/7
For these 3 conditioned intervals only
ETs: 509,1121,2292,3175,3787,4805,5417,5688,6588.....{room for error}
Changing limiting angle parameter, γ
interval = (3/2) @ ± ½ turn & γ<1/4
interval = (5/4) @ zero angle & γ<1/8
interval = (6/5) @ zero angle & γ<1/8
For these 3 conditioned intervals only
ETs: ...nothing...
* * *
Calculating the fret or step number of an EDO that
corresponds close to interval, I
To find out what the numbered step like a fret on an instrument is connected with this EDO that should sound close to interval, in this example interval ,I= (4/3), then we know that, I = 2^(n/EDO) so:
step or fret ‘n’=EDO*ln(I)/ln(2) rounded to the nearest integer.
and in this example, n = 12*ln(4/3)/ln(2) = 4.98045….. = 5 rounded to closest integer.
Continuing from earlier....
A more accurate angle turned can be expressed as it's reasonably low form which is alternatively about 5/12, estimating this angle = 5/(12+ δ):
In terms of
Angle integers (a/c) = 5/12
| 5 |
|12+δ|
So:
ln(4/3)/ln(2) = 5/(12+δ)
δ = [5*ln(2)/ln(4/3)] – 12 = 0.0471042…
And then:
1/(Peakβ EDO) = δ*5/(12+δ)
Peakβ EDO = (12+δ)/δ*5 = +51.1508...This is where the exact zero angle is reached but after a number of angular turns we have to pick the EDO closest to this value for the peak.
Recalling from before,
When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
Thus: q = ((N*c)+1)/a
Trying N = 1 initially: then q = (12+1)/5 = 13/5, not a whole number so invalid.
Trying N = 2 next: then q = ((2*12)+1)/5 = 25/5 = 5, a whole number so therefore valid.
So q = 5 and the series to be derived is from before : EDO = q+(k*c)
We have
q = 5ET
5+12 = 17ET further
17+12 = 29ET further
29+12 = 41ET close
41+12 = 53ET closest to peakβ EDO
53+12 = 65ET close
65+12 = 77ET further
77+12 = 89ET further
……………
Another example, (5/4) = 2^Θ
Basic expanded triangle so let angle = 1/(3+ δ)
a = 1
c = 3
| 1 |
|3+δ|
δ = [1*ln(2)/ln(5/4)] – 3 = 0.1062837…
Peakβ EDO = (3+δ)/δ*1 = +29.2263362... Remembering this is the exact zero angle so we need to find close EDO
When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
So
q = ((N*c)+1)/a
Trying N = 1, q = (3+1)/1 = 4, a whole number therefore valid.
So q = 4 and the EDO series to be derived is from before : EDO = q+(k*c)
4EDO
4+3 = 7EDO
7+3 = 10EDO
10+3 = 13EDO
13+3 = 16EDO
16+3 = 19EDO
19+3 = 22 EDO Getting closer
22+3 = 25EDO close
25+3 = 28EDO Next to peakβ EDO
28+3 = 31EDO Next to Peakβ EDO
31+3 = 34EDO close
34+3 = 37EDO further
.....
Another example (9/8) = 2^Θ
a = 1
c = 3
| 1 |
|3+δ|
δ = [1*ln(2)/ln(5/4)] – 3 = 0.1062837…
Peakβ EDO = (3+δ)/δ*1 = +29.2263362... Remembering this is the exact zero angle so we need to find close EDO
When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
So
q = ((N*c)+1)/a
Trying N = 1, q = (3+1)/1 = 4, a whole number therefore valid.
So q = 4 and the EDO series to be derived is from before : EDO = q+(k*c)
4EDO
4+3 = 7EDO
7+3 = 10EDO
10+3 = 13EDO
13+3 = 16EDO
16+3 = 19EDO
19+3 = 22 EDO Getting closer
22+3 = 25EDO close
25+3 = 28EDO Next to peakβ EDO
28+3 = 31EDO Next to Peakβ EDO
31+3 = 34EDO close
34+3 = 37EDO further
.....
Another example (9/8) = 2^Θ
A contracted hexagon So let angle = 1/(6+δ)
a = 1
c = 6
| 1 |
|6+δ|
δ = [1*ln(2)/ln(9/8)] – 6 = -.1150508….
Peakβ EDO = (6+δ)/δ*1 = -51.1509...
As before, When dealing with a negative Peakβ EDO value,
( (q*a)+1)/c = N where N is the lowest whole number found
So
q = ((N*c)-1)/a
Trying N = 1 initially, q = (6-1)/1 = 5, a whole number therefore valid.
So q = 5 and the EDO series to be derived is from before : EDO = q+(k*c)
5ET
5+6 = 11ET
11+6 = 17ET
17+6 = 23ET
23+6 = 29ET
29+6 = 35ET
35+6 = 41ET close
41+6 = 47ET closer
47+6 = 53ET closest to peakβ ET
53+6 = 59ET closer
59+6 = 65ET close
…..
2nd ring, 3rd Ring, integer ‘M’th Ring for higher EDOs
But as more rotations happen we can go to a higher EDO for what may be called the 2nd ring
a = 1
c = 6
| 1 |
|6+δ|
δ = [1*ln(2)/ln(9/8)] – 6 = -.1150508….
Peakβ EDO = (6+δ)/δ*1 = -51.1509...
As before, When dealing with a negative Peakβ EDO value,
( (q*a)+1)/c = N where N is the lowest whole number found
So
q = ((N*c)-1)/a
Trying N = 1 initially, q = (6-1)/1 = 5, a whole number therefore valid.
So q = 5 and the EDO series to be derived is from before : EDO = q+(k*c)
5ET
5+6 = 11ET
11+6 = 17ET
17+6 = 23ET
23+6 = 29ET
29+6 = 35ET
35+6 = 41ET close
41+6 = 47ET closer
47+6 = 53ET closest to peakβ ET
53+6 = 59ET closer
59+6 = 65ET close
…..
2nd ring, 3rd Ring, integer ‘M’th Ring for higher EDOs
But as more rotations happen we can go to a higher EDO for what may be called the 2nd ring
in the case of (4/3) = 2^Θ
2ndRing Peakβ EDO = 2*Peakβ EDO = 51.1508*2 = 102.302EDO
From
a = 5
c = 12
| 5 |
|12+δ|
Angular turn = 5/(12+δ)
From before ,it was said When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
In this example we found, q = 5
But instead of our next series starting on 5 which is the point a certain able to the right of the zero angle at the top of the wheel in the picture representation, We can move to the next point before which must be twice 5 =10 for dealing with the 2nd ring.
So starting value in EDO series = q*M+k*c, where M is the integer of Order of Ring.
=10
10+12 = 22
22+12 = 34
34+12 = 46
46+12 = 58
58+12 = 70
70+12 = 82
82+12 = 94 close
94+12 = 106ET Closest to 2nd ring Peak EDO
106+12 = 118ET close
118+12 = 230ET
And how about a 3rd ring?
3rd ring Peakβ EDO = 3*Peakβ EDO = 51.1508*3 = 153.452...
instead of starting next series on 0, 5 or 10, we choose the next one along which is 3 angular turns of point 5, so M = 3, and ‘q’ is still = 5
So starting value in EDO series = q*M+k*c, where M is the integer of Order of Ring.
15ET
15+12 = 27ET
27+12 = 39ET
39+ 12 = 51ET
etc....
....
111+12 = 123ET
123+12 = 135ET further
135+12 = 147ET Close to 3rd ring Peakβ EDO
147+12 = 159 ET Close to 3rd ring Peakβ EDO
159+12 = 171ET further
For the same interval =(4/3) we will revert back to a lower order approximation
From angle = 2/(5+δ):
a = 2
c = 5
| 2 |
|5+δ|
In this case, from before, Peakβ EDO = (5+δ)/δ*2 = -13.3001...
2nd Ring Peakβ EDO = 2*-13.3001... = -26.6002...
From before, we found q = 2 and this is the 2nd ring so ‘M’th ring, M = 2
Starting value in EDO series = q*M+k*c, where M is the integer of Order of Ring
4ET
4+5 = 9ET
9+5 = 14ET
14+5 = 19ET Close
19+5 = 24ET Closest to 2nd Ring Peakβ EDO
24+9 = 33ET Close
3rd ring Peakβ EDO = 3*-13.3001... = -39.9003...
‘M’th ring, and M = 3, ‘q’ remains =2
6ET
6+5 = 11ET
11+5 = 16ET
16+5 = 21ET
21+5 = 26ET
26+5 = 31ET close
31+5 = 36ET Closest to 3rd Ring Peakβ EDO
36+5 = 41ET Close
.....
4th ring Peakβ EDO = 4*-13.3001... = -53.2004...
‘M’th ring, and M = 4, ‘q’ remains =2
8
8+5 = 13
13+5 = 18
18+5 = 23
23+5 = 28
28+5 = 33
33+5 = 38
38+5 = 43
43+5 = 48ET Close
48+5 = 53ET Closest to 4th ring Peakβ EDO
53+5 = 58ET close
58+5 = 63ET
...
We see that this matches with the 53EDO found before when the angle = 5/(12+ δ), the higher order approximation for the geometry.
In the previous example (4/3) = 2^angle, Simplifying angle to lowest order angle = 1/(2+δ) instead of better approximations angle =2/(5+δ) or better still angle = 5/(12+δ)
So we have
a = 1
c = 2
|1 |
|2+δ|
So δ = ln(4/3)/ln(2) -2 = +0.4094208...
Peakβ EDO = (2+δ)/δ*1 = +5.884949....
When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
So
q = ((N*c)+1)/a
Trying N = 1 initially: then q = (2+1)/1 = 3, a whole number so valid.
So q = 3 and the EDO series to be derived is from before : EDO = q+(k*c)
So
3ET
3+2=5ET closest to Peakβ EDO
5+2 = 7ET
2nd ring Peakβ EDO = 2*5.884949 = 11.769898...
‘M’th ring so M = 2, still ‘q’ = 3
Starting value in EDO series = q*M+k*c
6ET
6+2 = 8ET
8+2 = 10ET
10+2 = 12ET closest
12+2 = 14ET
3rd ring Peakβ EDO = 3*5.884949 = 17.654847…
‘M’th ring so M = 3, still ‘q’ = 3
9ET
9+2 = 11ET
11+2 = 13ET
13+2 = 15ET
15+2 = 17ET closest
17+2 = 19
4th ring Peakβ EDO = 4*5.884949 = 23.5398...
M = 4, still ‘q’ = 3
12ET
12+2 = 14ET
14+2 = 16ET
16+2 = 18ET
18+2 = 20ET
20+2 = 22ET
22+2 = 24ET closest
24+2 = 26ET
5th ring Peakβ EDO = 5*5.884949..= 29.424745....
M = 5 still ‘q’ = 3
15ET
15+2 = 17ET
17+2=19ET
…
…
25+2 = 27ET close
27+2 = 29ET closest
29+2 = 31ET close
6th ring Peakβ EDO = 6*5.884949....= 35.30969....
M = 6, still ‘q’ = 3
18ET
18+2 = 20ET
20+2 = 22ET
……
……
32+2 = 34 far
34+2 = 36ET closest
36+2 = 38 far
7th ring Peakβ EDO = 7*5.884949....= 41.194643....
M = 7, still ‘q’ = 3
21ET
21+2 = 23ET
23+2 = 25ET
…..
…..
37+2 = 39 far
39+2 = 41ET closest
41+2 = 43 far
8th ring Peakβ EDO = 8*5.884949....= 47.07959....
M = 8, still ‘q’ = 3
24ET
24+2 =26ET
26+2 = 28ET
….
….
44+2 = 46ET close
46+2 = 48ET Close
9th ring Peakβ EDO = 9*5.884949....= 52.964541....
M = 9, still ‘q’ = 3
27ET
27+2 = 29ET
29+2 = 31ET
…..
…..
49+2 = 51
51+2 = 53ET closest
53 = 2+ 55
.....
So even for the low order reduced integer approximation for the angle yields identical matching results so far.
In a previous example we can now investigate abstract geometry of (5/4) = 2^ angle:
A good approximation for the angle is (1/(3+δ))
But this gives:
δ = [1*ln(2)/ln(5/4)] - 3
Peakβ EDO = (3+δ)/δ*1 = +29.2263362...A bit too large for more uses
so what about reduced approximating the angle to 1/(2+δ) instead?
a = 1
c = 2
this gives:
δ = [1*ln(2)/ln(5/4)] - 2
Peakβ EDO = (2+δ)/δ*1 =+2.807854499…
As stated before, When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
So
q = ((N*c)+1)/a
Trying N = 1 initially: then q = (2+1)/1 = 3, a whole number so valid.
So q = 3 and the EDO series to be derived is from before :
EDO = q+(k*c)
So starting on q = 3 for series:
3ET closest to Peakβ EDO
2nd ring Peakβ EDO = 2*2.807854 = +5.615709
For ‘M’th ring, M =2 for this and still ‘q’ = 3
6ET closest
3rd ring Peakβ EDO= 3*2.807545= +8.42356
M = 3, still , ‘q’ = 3
9ET closest
4th ring Peakβ EDO = 11.2314
M = 4, still ‘q’ = 3
12ET closest
5th ring Peakβ EDO = 14.0393
M = 5, still ‘q’ = 3
Starting value in EDO series = q*M+k*c
15ET close
But we see that for incrementing values in the series, the ET will be getting further away, so we need to go backwards in the series, that is let ,’k’ = negative integer values too.
So also if k = -1, another value in the series is 15-c = 13ET close also
Other ET’s in this series get further away.
6th ring Peakβ EDO = 16.84713
M = 6, still ‘q’ = 3
Starting value in EDO series = q*M+k*c
So we have 18EDO close
And as again we need to go backward in series
so letting k = -1 : we have 18-c = 16EDO closer
7th ring Peakβ EDO = 19.655
M = 7, still ‘q’ = 3
Starting value in EDO series = q*M+k*c
21EDO, but again this has gone too far so we need to go backwards in series
So also 21-c = 19EDO closest
8th ring Peakβ EDO = 22.4628
M = 8, still ‘q’ = 3
Starting value in EDO series = q*M+k*c
24EDO, but again too far, so need to go backwards in series, k = -1, EDO = 24-c=
22 closest
9th ring Peakβ EDO = 25.2707...
M = 9, still ‘q’ = 3
Starting value in EDO series = q*M+k*c
27EDO, too far, we need to go backwards in series, k = -1, EDO = 27-c=
25EDO closest
10th ring Peakβ EDO = 28.0785
M = 10, still ‘q’ = 3
Starting value in EDO series = q*M+k*c
30EDO, but again this has gone too far so we need to go backwards , k = -1, EDO =30-c=
28edo closest
11th ring Peakβ EDO = 30.8864
M = 11, still ‘q’ = 3
Starting value in EDO series = q*M+k*c
33EDO, but again too far so we going backwards in series,k = -1, EDO = 33-c=
31EDO closest
Another example (8/7) = 2^Θ
Distorted pentagon so angle = 1/(5+ δ):
a = 1
c = 5
| 1 |
|5+δ|
δ = [1*ln(2)/ln(8/7)] - 5
Peakβ EDO = (5+δ)/δ*1 = +27.1927...
When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
So
q = ((N*c)+1)/a
Trying N = 1 initially, q =(5+1)/1 = 6, a whole number therefore valid.
6EDO
6+5 = 11EDO
11+5 = 16EDO
16+ 5 = 21EDO close
21+5 = 26 EDO closest to peakβ
26+5 = 31EDO close
31+5 = 36EDO further
36+5 = 41EDO further
2nd Ring Peakβ EDO =2*27.1927 = +54.3854
M = 2, still ‘q’ = 6
Starting value in EDO series = q*M+k*c
12ET
12+5 = 17EDO
17+5 = 22EDO
22+5 = 27EDO
27+5 = 32EDO
32+5 = 37EDO
37+5 = 42EDO
42+5 = 47EDO Close
47+5 = 52EDO Closest to 2nd Ring Peakβ EDO
52+5 = 57EDO Close
57+2 = 62
....
....
Another example (9/7) = 2^Θ
a = 1
c = 5
| 1 |
|5+δ|
δ = [1*ln(2)/ln(8/7)] - 5
Peakβ EDO = (5+δ)/δ*1 = +27.1927...
When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
So
q = ((N*c)+1)/a
Trying N = 1 initially, q =(5+1)/1 = 6, a whole number therefore valid.
6EDO
6+5 = 11EDO
11+5 = 16EDO
16+ 5 = 21EDO close
21+5 = 26 EDO closest to peakβ
26+5 = 31EDO close
31+5 = 36EDO further
36+5 = 41EDO further
2nd Ring Peakβ EDO =2*27.1927 = +54.3854
M = 2, still ‘q’ = 6
Starting value in EDO series = q*M+k*c
12ET
12+5 = 17EDO
17+5 = 22EDO
22+5 = 27EDO
27+5 = 32EDO
32+5 = 37EDO
37+5 = 42EDO
42+5 = 47EDO Close
47+5 = 52EDO Closest to 2nd Ring Peakβ EDO
52+5 = 57EDO Close
57+2 = 62
....
....
Another example (9/7) = 2^Θ
Can be thought of as a contracted triangle form or perhaps 11-pointed star form
We will try the contracted triangle first
a = 1
c = 3
| 1 |
|3+δ|
δ = [1*ln(2)/ln(9/7)] - 3
Peakβ EDO = (3+δ)/δ*1 = -11.4012...
When dealing with a negative Peakβ EDO value,
( (q*a)+1)/c = N where N is the lowest whole number found
So
q = ((N*c)-1)/a
Trying N = 1 initially, q = (3-1)/1 = 2, a whole number, therefore valid
And as stated before, the series of associated EDOs will be of the form:
EDO =( M*q)+(k*c) where ‘k’ is an integer incremented from 0 to however many but now we know ‘k’ can have negative integer values also. ‘M’ is we recall the integer of the ‘M’th Ring, in this first case, for 1st ring, M = 1
So incrementing, k, from 0 to however many needed, in this example , EDO =
2ET
2+3 = 5ET
5+3 = 8ET
8+3 = 11ET closest to Peakβ EDO
11+3 = 14ET
14+3 = 17
.....
2nd Ring Peakβ EDO = 2*-11.4012 = -22.802…
M = 2, still, ‘q’=2,
EDO =( M*q)+(k*c)
4ET
4+3 = 7ET
7+3 = 10ET
10+3 =13ET
13+3 = 16ET
16+3 = 19ET close
19+3 = 22ET closest to 2nd Ring Peakβ EDO
22+3 = 25ET
....
3rd Ring Peakβ EDO = 3*-11.4012 = -34.2036
M = 3, still, ‘q’=2
EDO =( M*q)+(k*c)
6EDO
6+3 = 9EDO
9+3 = 12EDO
……..
……..
27+3 = 30EDO
30+3 = 33EDO Closest to 3rd Ring Peakβ EDO
33+3 = 36EDO Close
……..
So instead of the geometry integer approximation for a contracted triangle, we will try the 11 pointed star with angle, 4/(11+ δ)
a = 4
c = 11
| 4 |
|11+δ|
δ = [4*ln(2)/ln(9/7)] -11
Peakβ EDO = (11+δ)/δ*4 = 85.258…a much higher value
S stated, When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
q = ((N*c)+1)/a
Trying N = 1 initially, q = (11+1)/4 = 12/4=3, a whole number therefore valid.
So q = 3 and the EDO series to be derived is from before : EDO =(M*q)+(k*c)
M = 1 since this is 1st ring of ‘M’th ring.
EDO = 3
3+11 = 14
14+11 = 25
25+11 = 36
36+11 = 47
47+11 = 58
58+11 = 69
69+11 = 80EDO closest to Peakβ EDO
80+11 = 91EDO close too
We will now revert back to the lower order approximation for the geometry before for contracted triangle
7th Ring Peakβ EDO = 7*-11.4012 = -79.808
we deduced ‘q’ = 2, and M = 7 for 7th ring
from before : ET =(M*q)+(k*c)
14ET
14+3 = 17ET
17+3 = 20ET
…..
…..
71+3 = 74
74+3 = 77ET close
77+3 = 80ET closest to 7th Ring Peakβ EDO
80+3 = 83ET close
83+3 = 86
So 80EDO seems to appear for both forms of integer geometrical approximations in this case.
* * *
Example (6/5) = 2^Θ
We will try the contracted triangle first
a = 1
c = 3
| 1 |
|3+δ|
δ = [1*ln(2)/ln(9/7)] - 3
Peakβ EDO = (3+δ)/δ*1 = -11.4012...
When dealing with a negative Peakβ EDO value,
( (q*a)+1)/c = N where N is the lowest whole number found
So
q = ((N*c)-1)/a
Trying N = 1 initially, q = (3-1)/1 = 2, a whole number, therefore valid
And as stated before, the series of associated EDOs will be of the form:
EDO =( M*q)+(k*c) where ‘k’ is an integer incremented from 0 to however many but now we know ‘k’ can have negative integer values also. ‘M’ is we recall the integer of the ‘M’th Ring, in this first case, for 1st ring, M = 1
So incrementing, k, from 0 to however many needed, in this example , EDO =
2ET
2+3 = 5ET
5+3 = 8ET
8+3 = 11ET closest to Peakβ EDO
11+3 = 14ET
14+3 = 17
.....
2nd Ring Peakβ EDO = 2*-11.4012 = -22.802…
M = 2, still, ‘q’=2,
EDO =( M*q)+(k*c)
4ET
4+3 = 7ET
7+3 = 10ET
10+3 =13ET
13+3 = 16ET
16+3 = 19ET close
19+3 = 22ET closest to 2nd Ring Peakβ EDO
22+3 = 25ET
....
3rd Ring Peakβ EDO = 3*-11.4012 = -34.2036
M = 3, still, ‘q’=2
EDO =( M*q)+(k*c)
6EDO
6+3 = 9EDO
9+3 = 12EDO
……..
……..
27+3 = 30EDO
30+3 = 33EDO Closest to 3rd Ring Peakβ EDO
33+3 = 36EDO Close
……..
So instead of the geometry integer approximation for a contracted triangle, we will try the 11 pointed star with angle, 4/(11+ δ)
a = 4
c = 11
| 4 |
|11+δ|
δ = [4*ln(2)/ln(9/7)] -11
Peakβ EDO = (11+δ)/δ*4 = 85.258…a much higher value
S stated, When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
q = ((N*c)+1)/a
Trying N = 1 initially, q = (11+1)/4 = 12/4=3, a whole number therefore valid.
So q = 3 and the EDO series to be derived is from before : EDO =(M*q)+(k*c)
M = 1 since this is 1st ring of ‘M’th ring.
EDO = 3
3+11 = 14
14+11 = 25
25+11 = 36
36+11 = 47
47+11 = 58
58+11 = 69
69+11 = 80EDO closest to Peakβ EDO
80+11 = 91EDO close too
We will now revert back to the lower order approximation for the geometry before for contracted triangle
7th Ring Peakβ EDO = 7*-11.4012 = -79.808
we deduced ‘q’ = 2, and M = 7 for 7th ring
from before : ET =(M*q)+(k*c)
14ET
14+3 = 17ET
17+3 = 20ET
…..
…..
71+3 = 74
74+3 = 77ET close
77+3 = 80ET closest to 7th Ring Peakβ EDO
80+3 = 83ET close
83+3 = 86
So 80EDO seems to appear for both forms of integer geometrical approximations in this case.
* * *
Example (6/5) = 2^Θ
Angle turned appears like a contracted square form so
a = 1
c = 4
| 1 |
|4+δ|
δ = [1*ln(2)/ln(6/5)] - 4
Peakβ EDO = (4+δ)/δ*1 = -19.180007...
When dealing with a negative Peakβ EDO value,
( (q*a)+1)/c = N where N is the lowest whole number found
So
q = ((N*c)-1)/a
Trying N = 1 initially, q = (4-1)/1 = 3, a whole number, therefore valid.
3ET
3+4 = 7ET
7+4 = 11ET
11+4 = 15ET Close
15+4 = 19ET Closest to peakβ
19+4 = 25ET Close
25+4 = 29ET
2nd Ring Peakβ EDO = 2*-19.180007 = -38.36014
2nd ring so M = 2, still q = 3
6EDO
6+4 = 10
10+4 = 14
14+4 = 18
18+4 = 22
22+4 = 26
26+4 = 30
30+4 = 34EDO close
34+4 = 38EDO closest to 2nd Ring Peakβ EDO
38+4 = 42EDO close
42+4 = 46
3rd Ring Peakβ EDO = 3*-19.180007 = -57.540
3rd ring, so M = 3, still q = 3
9ET
9+4 = 13ET
13+4 = 17ET
……
.......
41+4 = 45
45+4 = 49
49+4 = 53EDO close
53+4 = 57EDO Closest to 3rd Ring Peakβ EDO
57+4 = 61EDO close
61+4 = 65
Example (7/5) = 2^Θ
a = 1
c = 4
| 1 |
|4+δ|
δ = [1*ln(2)/ln(6/5)] - 4
Peakβ EDO = (4+δ)/δ*1 = -19.180007...
When dealing with a negative Peakβ EDO value,
( (q*a)+1)/c = N where N is the lowest whole number found
So
q = ((N*c)-1)/a
Trying N = 1 initially, q = (4-1)/1 = 3, a whole number, therefore valid.
3ET
3+4 = 7ET
7+4 = 11ET
11+4 = 15ET Close
15+4 = 19ET Closest to peakβ
19+4 = 25ET Close
25+4 = 29ET
2nd Ring Peakβ EDO = 2*-19.180007 = -38.36014
2nd ring so M = 2, still q = 3
6EDO
6+4 = 10
10+4 = 14
14+4 = 18
18+4 = 22
22+4 = 26
26+4 = 30
30+4 = 34EDO close
34+4 = 38EDO closest to 2nd Ring Peakβ EDO
38+4 = 42EDO close
42+4 = 46
3rd Ring Peakβ EDO = 3*-19.180007 = -57.540
3rd ring, so M = 3, still q = 3
9ET
9+4 = 13ET
13+4 = 17ET
……
.......
41+4 = 45
45+4 = 49
49+4 = 53EDO close
53+4 = 57EDO Closest to 3rd Ring Peakβ EDO
57+4 = 61EDO close
61+4 = 65
Example (7/5) = 2^Θ
Spliced diameter so:
a = 1
c = 2
| 1 |
|2+δ|
δ = [1*ln(2)/ln(7/5)] - 2
Peakβ EDO = (2+δ)/δ*1 = +34.3096...
When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
q = ((N*c)+1)/a
Trying N = 1 initially, q =(2+1)/1 = 3, a whole number, therefore valid.
3EDO
3+2 = 5EDO
5+2 = 7EDO
.....
.....
29+2 = 31EDO
31+2 = 33EDO Next to peak
33+2 = 35EDO Next closest to peak
35+2 = 37EDO
Trying for interval (11/8) = 2^Θ
a = 1
c = 2
| 1 |
|2+δ|
δ = [1*ln(2)/ln(7/5)] - 2
Peakβ EDO = (2+δ)/δ*1 = +34.3096...
When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
q = ((N*c)+1)/a
Trying N = 1 initially, q =(2+1)/1 = 3, a whole number, therefore valid.
3EDO
3+2 = 5EDO
5+2 = 7EDO
.....
.....
29+2 = 31EDO
31+2 = 33EDO Next to peak
33+2 = 35EDO Next closest to peak
35+2 = 37EDO
Trying for interval (11/8) = 2^Θ
A kind of expanded spliced diameter geometry
So
a = 1
c = 2
| 1 |
|2+δ|
δ = [1*ln(2)/ln(11/8)] - 2
Peakβ EDO = (2+δ)/δ*1 = +12.3249
When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
So
q = ((N*c)+1)/a
Trying, N = 1 initially, q = (2+1)/1 = 3, valid
3
3+2 = 5
5+2 = 7
7+2 = 9
9+2 = 11ET close to Peakβ EDO
11+2 = 13ET closest to Peakβ EDO
13+2 = 15
......
2nd Ring Peakβ EDO = 2*12.3249 = +24.6498
‘M’th ring, M=2, still q = 3
6ET
6+2 = 8
8+2 = 10
10+2 = 12
12+2 = 14
14+2 = 16
16+2 = 18
18+2 = 20
20+2 = 22ET near
22+2 = 24ET closest to 2nd ring Peakβ EDO
24+2 = 26ET near
3rd Ring peakβ EDO = 3*12.3249 = 36.9747
3rd ring, so M = 3
3ET
3+2 = 5
5+2 = 7
7+2 = 9
....
....
31+2 = 35ET
33+2 = 35ET near
35+2 = 37ET Closest 3rd ring Peakβ EDO
37+2 = 39ET near
39+2 = 41 further
Next Testing For (22/17)
So
a = 1
c = 2
| 1 |
|2+δ|
δ = [1*ln(2)/ln(11/8)] - 2
Peakβ EDO = (2+δ)/δ*1 = +12.3249
When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
So
q = ((N*c)+1)/a
Trying, N = 1 initially, q = (2+1)/1 = 3, valid
3
3+2 = 5
5+2 = 7
7+2 = 9
9+2 = 11ET close to Peakβ EDO
11+2 = 13ET closest to Peakβ EDO
13+2 = 15
......
2nd Ring Peakβ EDO = 2*12.3249 = +24.6498
‘M’th ring, M=2, still q = 3
6ET
6+2 = 8
8+2 = 10
10+2 = 12
12+2 = 14
14+2 = 16
16+2 = 18
18+2 = 20
20+2 = 22ET near
22+2 = 24ET closest to 2nd ring Peakβ EDO
24+2 = 26ET near
3rd Ring peakβ EDO = 3*12.3249 = 36.9747
3rd ring, so M = 3
3ET
3+2 = 5
5+2 = 7
7+2 = 9
....
....
31+2 = 35ET
33+2 = 35ET near
35+2 = 37ET Closest 3rd ring Peakβ EDO
37+2 = 39ET near
39+2 = 41 further
Next Testing For (22/17)
A kind of an 8shaped star (3/(8+δ)) angle swept out.
So
a = 3
c = 8
| 3 |
|8+δ|
δ = [3*ln(2)/ln(22/17)] - 8
Peakβ EDO = (8+δ)/δ*3 = +41.23748599
When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
q = ((N*c)+1)/a
Trying N = 1 initially, q =(8+1)/3 = 3, a whole number, therefore valid.
3EDO
3+8 = 11
11+8 =19
19+8 = 27
27+8 = 35
35+8 = 43EDO closest to Peakβ EDO
And Testing For (27/20)
So
a = 3
c = 8
| 3 |
|8+δ|
δ = [3*ln(2)/ln(22/17)] - 8
Peakβ EDO = (8+δ)/δ*3 = +41.23748599
When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
q = ((N*c)+1)/a
Trying N = 1 initially, q =(8+1)/3 = 3, a whole number, therefore valid.
3EDO
3+8 = 11
11+8 =19
19+8 = 27
27+8 = 35
35+8 = 43EDO closest to Peakβ EDO
And Testing For (27/20)
A kind of an 7shaped star (3/(7+δ)) angle swept out.
So
a = 3
c = 7
| 3 |
|7+δ|
δ = [3*ln(2)/ln(27/20)] - 7
Peakβ EDO = (7+δ)/δ*3 = -32.55648044…
When dealing with a negative Peakβ EDO value,
( (q*a)+1)/c = N where N is the lowest whole number found
So
q = ((N*c)-1)/a
Trying N = 1 initially, q = (7-1)/3 = 2, a whole number therefore valid
2ET
2+7=9ET
9+7 = 16ET
16+7 = 23ET
23+7 = 30ET closest to Peakβ EDO
43+8 = 51
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Measure of Closeness, μ of EDO to PeakβEDO
So we want to find a measure , μ of how close the found EDO is to the just ratio Interval, I
So if in the geometry, assigning the angle swept out to be (a/(c+δ)) where a & c are integers & δ is a small change
We know interval, I = 2^(a/( c+δ)) and also , I = 2^(n/EDO) where n is the nth step in the EDO that approximates for close to Just intonation Interval, I.
We can say that : 2^(n/EDO) = I*(2^μ)
So
ln(2)*(n/EDO)=ln(I)+ μ*ln(2)
μ = (n/EDO)-ln(I)/ln(2)
Formula from before:
δ = [a*ln(2)/ln(I)] - c
Peakβ EDO = (c+δ)/(δ*a)
=1/[a-(c*ln(I)/ln(2))]
This can be rearranged as ln(I) = (a-(1/ Peakβ))*(ln(2))/c
Just intonation interval, I = 2^[(1/c)*(a-(1/Peakβ))]
So substituting ln(I) in formula before for μ gives, after rearranging:
μ = (n/EDO)-(a/c)+(1/(c*Peakβ))
The closer this value of μ to zero, the more accurate and close the Just intonation interval, I is to the deduced EDO and the nth step.
To deduce the deviation measured in cents, from the just ratio interval and the nth step for the EDO, then:
deviation in cents = 1200* μ
We now need to test this equation for when the measure of closeness, μ is zero.
That should be when the predicted close EDO in the series previously shown derived from the geometry, equals the PeakβEDO
Let's include the m'th Ring PeakβEDO where m is the integer corresponding to the ring order.
So using previous equations:
m*Peakβ = m/[a-(c*ln(2)/ln(I))]
closeness, μ = (n/EDO)-(a/c)+(1/(c*m*Peakβ))
Since for this special case, just intonation interval, I=2^(n/EDO)
then
n/(EDO) = ln(I)/ln(2)
Substituting in the equation for μ gives:
μ = [ln(I)/ln(2)] - (a/c) + [(1/(c.m))*(a-c*ln(I)/ln(2))]
Rearranging this equation results in:
μ = [1 - (1/m)] * [(ln(I)/ln(2)) - (a/c)]
But since μ is zero then either [ ] bracket factor = 0
So either [(ln(I)/ln(2)) - (a/c)] = 0
or [1 - (1/m)] = 0
So (a/c) = ln(I)/ln(2)
that is:
Just intonation interval I = 2^(a/c)
or same as (n/EDO) = (a/c)
Of course, it takes an infinite valued EDO for Perfect just intonation integer fraction intervals to ocuur. So a & c are infinitely high integers for this exact event to exist.
But this result seems to show the retained balance of the equation and no mathematical laws in Logic are broken. So when the geometrically predicted EDO = Peakβ, then closeness, μ = 0
And when m = 1 for the first ring PeakβEDO, closeness μ = 0 so this theory seems valid.
* * *
The Set of Cascading EDOs
Here is an investigation into what cascade EDOs exist for already found Basic low numbered EDOs, by using the angle in the geometry to be equal to n/EDO , that is the step number divided by the EDO, and then finding a new associated PeakβEDO and finding close range EDO’s near to the new PeakβEDO.
We know the definition that interval, I = 2^(n/EDO)
Fret number, same as step number, n = EDO * ln(I)/ln(2)
using the example interval, I = (4/3)
and using the calculated EDOs from examples before:
We had a close EDO = 12 step, n = 5
And converting, a = n & c = EDO since integer fraction (a/c) is the approximated angle in the geometry we found the next stage , Peakβ EDO = +51.1508...
And the closest equal temperament in a series = 53ET
So in 53ET, the step size, n = 22 and (22/53) will be a more accurate approximation of the angle turned in the geometry.
New ‘a’ = 22,
New ‘c’ = 53
As before:
| 22 |
| 53+ δ|
So:
ln(4/3)/ln(2) = 22/(53+δ)
δ = [22*ln(2)/ln(4/3)] – 53 = 0.007258473
and as the process earlier:
1/(Peakβ EDO) = δ*22/(53+δ)
Peakβ EDO = (53+δ)/δ*22 = +344.92505….
Recalling from before,
When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
Thus: q = ((N*c)+1)/a
Trying N = 1 initially: then q = (53+1)/22 = 54/22=27/11, not a whole number so invalid.
Trying N = 2 next: then q = (106+1)/22 = 107/22 , not a whole number so invalid.
Trying N = 3 next: then q = (159+1)/22 = 160/22=80/11 , not a whole number so invalid.
Trying N = 4 next: then q = (212+1)/22 = 213/22 , not a whole number so invalid.
Trying N = 5 next: then q = (265+1)/22 = 266/22 =133/11, not a whole number
Trying N = 6 next: then q = (318+1)/22 = 319/22 , not a whole number
Trying N = 7 next: then q = (371+1)/22 = 372/22 = 186/11, not a whole number
Trying N = 8 next: then q = (424+1)/22 = 425/22 , not a whole number
Trying N = 9 next: then q = (477+1)/22 = 478/22 =239/11 , not a whole number
Trying N = 10 next: then q = (530+1)/22 = 531/22 , not a whole number
Trying N = 11 next: then q = (583+1)/22 = 584/22 , not a whole number
Trying N = 12 next: then q = (636+1)/22 = 637/22 ,not a whole number
Trying N = 13 next: then q = (689+1)/22 = 690/22 not a whole number
This is getting a bit tedious, so leaving the calculations to a computer algorithm…
We find when N = 39:then ‘q’ = ((39*53)+1)/22 = 94
And the ET series to be deduced is of the form, ET = (q*M)+(k*c)
This is the 1st ring, so M = 1
series is
94ET
94+53 = 147
147+53=200
200+53=253
253+53=306ET close
306+53=359ET closest to Peakβ EDO
Fret number, same as step number, n = EDO * ln(I)/ln(2)
For 359ET, n = 148.99846…. = 149 rounded to nearest integer
Continuing this cascading process we let new approximate angle in geometry,
(a/c) = (149/359)
| 149 |
| 359+ δ|
So:
ln(4/3)/ln(2) = 149/(359+δ)
δ = [149*ln(2)/ln(4/3)] – 359 = 0.003705114
and as the process earlier:
1/(Peakβ EDO) = δ*149/(359+δ)
Peakβ EDO = (359+δ)/δ*149 = +650.296007….
Recalling from before,
When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
Thus: q = ((N*c)+1)/a
Trying N = 1 initially: then q = (359+1)/149 =360/149, not a whole number so invalid.
Speeding up this process by an electronic algorithm we find, with increasing integer, N:
When N =276, then q = 665
And the ET series to be deduced is of the form, ET = (q*M)+(k*c)
1st ring so M = 1
Series is
665ET closest to Peakβ EDO
665+359 =1024ET close
And going backwards by letting k = negative integer -1,-2,-3… we have:
665-359 = 306ET
306-359=-53ET
All close in the region of the Peak β ET
Fret number, same as step number, n = EDO * ln(I)/ln(2)
For 665ET, n = 275.999937…. = 276 rounded to nearest integer
Yet again, Continuing this cascading process we let new approximate angle in geometry,
(a/c) = (276/665)
| 276 |
| 665+ δ|
So:
ln(4/3)/ln(2) = 276/(665+δ)
δ = [276*ln(2)/ln(4/3)] – 665 = 0.000151754
and as the process earlier:
1/(Peakβ EDO) = δ*276/(665+δ)
Peakβ EDO = (665+δ)/δ*276 = +15877.1488….
Recalling from before,
When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
Thus: q = ((N*c)+1)/a
Trying when N = 1, q = not a whole number therefore invalid.
Speeding up this process by a mechanical algorithm we find, with increasing integer, N:
When N =403, then q = 971
And the ET series to be deduced is of the form, ET = (q*M)+(k*c)
In this case , dealing with 1st ring so M = 1
Series is
971ET
+665=1636ET
+665=2301ET
+665=2966ET
....
....
+665=14271
+665=14936ET close
+665=15601ET closest
+665=16266ET close
*
For this same interval, I = (4/3) a number of close EDOs were found from the series made before. Some significant close Equal temperaments found were :
Temperament deviation=
1200*ln(I)/ln(2)-1200*ln(a/c)
53ET -0.07cents
41ET +0.48cents
29ET +1.49cents
12ET -1.95cents
46ET +2.40cents
17ET +3.93cents
31ET -5.18cents
22ET +7.14cents
19ET -7.22cents
7ET -16.24cents
5ET +18.04cents
Using 41ET, the fret step number is as before found from, interval I = 2^(n/ET)
n = ET * ln(I)/ln(2),
thus, n = 17 so letting (a/c) = (17/41) and finding the Peakβ ET :
Continuing this cascading process we let new approximate angle in geometry,
a = 17
c = 41
| 17 |
| 41+ δ|
So:
ln(4/3)/ln(2) = 17/(41+δ)
δ = [17*ln(2)/ln(4/3)] –41 =- 0.039845725…..
and as the process earlier:
1/(Peakβ EDO) = δ*17/(41+δ)
Peakβ EDO = (41+δ)/δ*17 = -60.46874091….
As before, When dealing with a negative Peakβ EDO value,
( (q*a)+1)/c = N where N is the lowest whole number found
So
q = ((N*c)-1)/a
Trying N = 1 initially, q = (41-1)/17 =40/17 not a whole number, therefore invalid.
Next N = 2 initially, q = ((2*41)-1)/17 = 81/17 not a whole number, therefore invalid.
Next N = 3 initially, q = ((3*41)-1)/17 = 122/17 not a whole number, therefore invalid.
Continuing, we find When N = 22, q = 53
And the ET series to be deduced is of the form, ET = (q*M)+(k*c)
1st ring so M = 1
Series is
53ET closest to Peakβ ET
53+41=94ET close
94+41=135ET
And going backwards, letting k have negative integer values instead,
53-41 = 12ET
12-41 = -29ET
.....
* * * * *
Appendix :-
So
a = 3
c = 7
| 3 |
|7+δ|
δ = [3*ln(2)/ln(27/20)] - 7
Peakβ EDO = (7+δ)/δ*3 = -32.55648044…
When dealing with a negative Peakβ EDO value,
( (q*a)+1)/c = N where N is the lowest whole number found
So
q = ((N*c)-1)/a
Trying N = 1 initially, q = (7-1)/3 = 2, a whole number therefore valid
2ET
2+7=9ET
9+7 = 16ET
16+7 = 23ET
23+7 = 30ET closest to Peakβ EDO
43+8 = 51
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Measure of Closeness, μ of EDO to PeakβEDO
So we want to find a measure , μ of how close the found EDO is to the just ratio Interval, I
So if in the geometry, assigning the angle swept out to be (a/(c+δ)) where a & c are integers & δ is a small change
We know interval, I = 2^(a/( c+δ)) and also , I = 2^(n/EDO) where n is the nth step in the EDO that approximates for close to Just intonation Interval, I.
We can say that : 2^(n/EDO) = I*(2^μ)
So
ln(2)*(n/EDO)=ln(I)+ μ*ln(2)
μ = (n/EDO)-ln(I)/ln(2)
Formula from before:
δ = [a*ln(2)/ln(I)] - c
Peakβ EDO = (c+δ)/(δ*a)
=1/[a-(c*ln(I)/ln(2))]
This can be rearranged as ln(I) = (a-(1/ Peakβ))*(ln(2))/c
Just intonation interval, I = 2^[(1/c)*(a-(1/Peakβ))]
So substituting ln(I) in formula before for μ gives, after rearranging:
μ = (n/EDO)-(a/c)+(1/(c*Peakβ))
The closer this value of μ to zero, the more accurate and close the Just intonation interval, I is to the deduced EDO and the nth step.
To deduce the deviation measured in cents, from the just ratio interval and the nth step for the EDO, then:
deviation in cents = 1200* μ
We now need to test this equation for when the measure of closeness, μ is zero.
That should be when the predicted close EDO in the series previously shown derived from the geometry, equals the PeakβEDO
Let's include the m'th Ring PeakβEDO where m is the integer corresponding to the ring order.
So using previous equations:
m*Peakβ = m/[a-(c*ln(2)/ln(I))]
closeness, μ = (n/EDO)-(a/c)+(1/(c*m*Peakβ))
Since for this special case, just intonation interval, I=2^(n/EDO)
then
n/(EDO) = ln(I)/ln(2)
Substituting in the equation for μ gives:
μ = [ln(I)/ln(2)] - (a/c) + [(1/(c.m))*(a-c*ln(I)/ln(2))]
Rearranging this equation results in:
μ = [1 - (1/m)] * [(ln(I)/ln(2)) - (a/c)]
But since μ is zero then either [ ] bracket factor = 0
So either [(ln(I)/ln(2)) - (a/c)] = 0
or [1 - (1/m)] = 0
So (a/c) = ln(I)/ln(2)
that is:
Just intonation interval I = 2^(a/c)
or same as (n/EDO) = (a/c)
Of course, it takes an infinite valued EDO for Perfect just intonation integer fraction intervals to ocuur. So a & c are infinitely high integers for this exact event to exist.
But this result seems to show the retained balance of the equation and no mathematical laws in Logic are broken. So when the geometrically predicted EDO = Peakβ, then closeness, μ = 0
And when m = 1 for the first ring PeakβEDO, closeness μ = 0 so this theory seems valid.
* * *
The Set of Cascading EDOs
Here is an investigation into what cascade EDOs exist for already found Basic low numbered EDOs, by using the angle in the geometry to be equal to n/EDO , that is the step number divided by the EDO, and then finding a new associated PeakβEDO and finding close range EDO’s near to the new PeakβEDO.
We know the definition that interval, I = 2^(n/EDO)
Fret number, same as step number, n = EDO * ln(I)/ln(2)
using the example interval, I = (4/3)
and using the calculated EDOs from examples before:
We had a close EDO = 12 step, n = 5
And converting, a = n & c = EDO since integer fraction (a/c) is the approximated angle in the geometry we found the next stage , Peakβ EDO = +51.1508...
And the closest equal temperament in a series = 53ET
So in 53ET, the step size, n = 22 and (22/53) will be a more accurate approximation of the angle turned in the geometry.
New ‘a’ = 22,
New ‘c’ = 53
As before:
| 22 |
| 53+ δ|
So:
ln(4/3)/ln(2) = 22/(53+δ)
δ = [22*ln(2)/ln(4/3)] – 53 = 0.007258473
and as the process earlier:
1/(Peakβ EDO) = δ*22/(53+δ)
Peakβ EDO = (53+δ)/δ*22 = +344.92505….
Recalling from before,
When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
Thus: q = ((N*c)+1)/a
Trying N = 1 initially: then q = (53+1)/22 = 54/22=27/11, not a whole number so invalid.
Trying N = 2 next: then q = (106+1)/22 = 107/22 , not a whole number so invalid.
Trying N = 3 next: then q = (159+1)/22 = 160/22=80/11 , not a whole number so invalid.
Trying N = 4 next: then q = (212+1)/22 = 213/22 , not a whole number so invalid.
Trying N = 5 next: then q = (265+1)/22 = 266/22 =133/11, not a whole number
Trying N = 6 next: then q = (318+1)/22 = 319/22 , not a whole number
Trying N = 7 next: then q = (371+1)/22 = 372/22 = 186/11, not a whole number
Trying N = 8 next: then q = (424+1)/22 = 425/22 , not a whole number
Trying N = 9 next: then q = (477+1)/22 = 478/22 =239/11 , not a whole number
Trying N = 10 next: then q = (530+1)/22 = 531/22 , not a whole number
Trying N = 11 next: then q = (583+1)/22 = 584/22 , not a whole number
Trying N = 12 next: then q = (636+1)/22 = 637/22 ,not a whole number
Trying N = 13 next: then q = (689+1)/22 = 690/22 not a whole number
This is getting a bit tedious, so leaving the calculations to a computer algorithm…
We find when N = 39:then ‘q’ = ((39*53)+1)/22 = 94
And the ET series to be deduced is of the form, ET = (q*M)+(k*c)
This is the 1st ring, so M = 1
series is
94ET
94+53 = 147
147+53=200
200+53=253
253+53=306ET close
306+53=359ET closest to Peakβ EDO
Fret number, same as step number, n = EDO * ln(I)/ln(2)
For 359ET, n = 148.99846…. = 149 rounded to nearest integer
Continuing this cascading process we let new approximate angle in geometry,
(a/c) = (149/359)
| 149 |
| 359+ δ|
So:
ln(4/3)/ln(2) = 149/(359+δ)
δ = [149*ln(2)/ln(4/3)] – 359 = 0.003705114
and as the process earlier:
1/(Peakβ EDO) = δ*149/(359+δ)
Peakβ EDO = (359+δ)/δ*149 = +650.296007….
Recalling from before,
When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
Thus: q = ((N*c)+1)/a
Trying N = 1 initially: then q = (359+1)/149 =360/149, not a whole number so invalid.
Speeding up this process by an electronic algorithm we find, with increasing integer, N:
When N =276, then q = 665
And the ET series to be deduced is of the form, ET = (q*M)+(k*c)
1st ring so M = 1
Series is
665ET closest to Peakβ EDO
665+359 =1024ET close
And going backwards by letting k = negative integer -1,-2,-3… we have:
665-359 = 306ET
306-359=-53ET
All close in the region of the Peak β ET
Fret number, same as step number, n = EDO * ln(I)/ln(2)
For 665ET, n = 275.999937…. = 276 rounded to nearest integer
Yet again, Continuing this cascading process we let new approximate angle in geometry,
(a/c) = (276/665)
| 276 |
| 665+ δ|
So:
ln(4/3)/ln(2) = 276/(665+δ)
δ = [276*ln(2)/ln(4/3)] – 665 = 0.000151754
and as the process earlier:
1/(Peakβ EDO) = δ*276/(665+δ)
Peakβ EDO = (665+δ)/δ*276 = +15877.1488….
Recalling from before,
When dealing with a positive Peakβ EDO value,
( (q*a)-1)/c = N where N is the lowest whole number found
Thus: q = ((N*c)+1)/a
Trying when N = 1, q = not a whole number therefore invalid.
Speeding up this process by a mechanical algorithm we find, with increasing integer, N:
When N =403, then q = 971
And the ET series to be deduced is of the form, ET = (q*M)+(k*c)
In this case , dealing with 1st ring so M = 1
Series is
971ET
+665=1636ET
+665=2301ET
+665=2966ET
....
....
+665=14271
+665=14936ET close
+665=15601ET closest
+665=16266ET close
*
For this same interval, I = (4/3) a number of close EDOs were found from the series made before. Some significant close Equal temperaments found were :
Temperament deviation=
1200*ln(I)/ln(2)-1200*ln(a/c)
53ET -0.07cents
41ET +0.48cents
29ET +1.49cents
12ET -1.95cents
46ET +2.40cents
17ET +3.93cents
31ET -5.18cents
22ET +7.14cents
19ET -7.22cents
7ET -16.24cents
5ET +18.04cents
Using 41ET, the fret step number is as before found from, interval I = 2^(n/ET)
n = ET * ln(I)/ln(2),
thus, n = 17 so letting (a/c) = (17/41) and finding the Peakβ ET :
Continuing this cascading process we let new approximate angle in geometry,
a = 17
c = 41
| 17 |
| 41+ δ|
So:
ln(4/3)/ln(2) = 17/(41+δ)
δ = [17*ln(2)/ln(4/3)] –41 =- 0.039845725…..
and as the process earlier:
1/(Peakβ EDO) = δ*17/(41+δ)
Peakβ EDO = (41+δ)/δ*17 = -60.46874091….
As before, When dealing with a negative Peakβ EDO value,
( (q*a)+1)/c = N where N is the lowest whole number found
So
q = ((N*c)-1)/a
Trying N = 1 initially, q = (41-1)/17 =40/17 not a whole number, therefore invalid.
Next N = 2 initially, q = ((2*41)-1)/17 = 81/17 not a whole number, therefore invalid.
Next N = 3 initially, q = ((3*41)-1)/17 = 122/17 not a whole number, therefore invalid.
Continuing, we find When N = 22, q = 53
And the ET series to be deduced is of the form, ET = (q*M)+(k*c)
1st ring so M = 1
Series is
53ET closest to Peakβ ET
53+41=94ET close
94+41=135ET
And going backwards, letting k have negative integer values instead,
53-41 = 12ET
12-41 = -29ET
.....
* * * * *
Appendix :-