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

-----------

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)

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.

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)

thus

thus

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

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 ?

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.

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...

* * *

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)

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

……………

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

( (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

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 valueSo 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.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…

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

.....

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 EDOWhen 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….

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

…..

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

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

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

....

....

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

.....

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

....

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

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.

* * *

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 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, 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

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

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

......

‘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, 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

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

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

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)

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:

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.

* * *

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)

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)

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

.....

* * * * *

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 :-__