OS

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An OS, or otonal sequence, is a kind of arithmetic and monotonic tuning.

An OS is a specific (rational) type of AFS.

(n-)OSp: (n pitches of an) otonal sequence adding by p

OS(1/n) as "1 out of every n harmonics of the harmonic series" (starting with harmonic 1).

That way I could just say AFS4 and that gives me harmonics 1 5 9 13 ... (though since that ones rational we'd prefer IOS4 for iso-otonal sequence) If you wanted the iso-otonal sequence going by 4's but starting on 3 instead, you'd just need AFS(4/3). (Technically that'd give you 3/3 7/3 11/3 15/3...)


Consider AFS(3/4). That means start on 4/3 and move by 1's, so 4/3, 1+(4/3), 2+(4/3), 3+(4/3), etc. which becomes 4/4, 7/4, 10/4, 13/4... In other words, move by 3's, start on 4.

thats the way it works for 7OD3 1+(1/7*2)=~1.28571 1+(2/7*2)=~1.57143

For example, if you vary the overtone series to have a step size of 3/4 instead of 1, then you get the tuning [math]\displaystyle{ 1, 1\frac 34, 2\frac 24, 3\frac14 }[/math], which is equivalent to [math]\displaystyle{ \frac 44, \frac 74, \frac{10}{4}, \frac{13}{4} }[/math], or in other words, a class iii isoharmonic tuning with starting position of 4. We call this the otonal sequence of 3 over 4, or OS3/4.

OS and AFS are equivalent to taking an overtone series and adding (or subtracting) a constant amount of frequency. By doing this, the step sizes remain equal in frequency, but their relationship in pitch changes. For a detailed explanation of this, see the later section on the derivation of OS.

example: 8-OS(3/4)
quantity (0) 1 2 3 4 5 6 7 8
frequency (f) (4/4) 7/4 10/4 13/4 16/4 19/4 22/4 25/4 28/4
pitch (log₂f) (0) 0.81 1.32 1.70 2.00 2.25 2.46 2.64 2.81
length (1/f) (1/1) 4/7 2/5 4/13 1/4 4/19 2/11 4/25 1/7

Derivation

The tuning OS3/4 is the sequence [math]\displaystyle{ \frac 44, \frac 74, \frac{10}{4}, \frac{13}{4}... }[/math] and so on. Any OS is equivalent to shifting the overtone series by a constant amount of frequency. In the case of OS3/4, it is a shift by [math]\displaystyle{ \frac 13 }[/math]. Let's show how.

Begin with the overtone series:

[math]\displaystyle{ 1, 2, 3, 4... }[/math]

Shift it by [math]\displaystyle{ \frac 13 }[/math]:

[math]\displaystyle{ 1\frac 13, 2\frac 13, 3\frac 13, 4\frac 13... \\ }[/math]

Convert to improper fractions by first expanding the whole number:

[math]\displaystyle{ \frac 33 + \frac 13, \frac 63 + \frac 13, \frac 93 + \frac 13, \frac {12}{3} + \frac 13... \\ }[/math]

...then consolidating numerators:

[math]\displaystyle{ \frac 43, \frac 73, \frac{10}{3}, \frac{13}{3}... }[/math]

Resize to start at [math]\displaystyle{ \frac 11 }[/math] by multiplying every term by the reciprocal of the first term, [math]\displaystyle{ \frac 43 }[/math], which is [math]\displaystyle{ \frac 34 }[/math]:

[math]\displaystyle{ \frac 43 \cdot \frac 34, \frac 73 \cdot \frac 34, \frac{10}{3} \cdot \frac 34, \frac{13}{3} \cdot \frac 34... }[/math]

Cancel out:

[math]\displaystyle{ \frac{4}{\cancel{3}} \cdot \frac{\cancel{3}}{4}, \frac{7}{\cancel{3}} \cdot \frac{\cancel{3}}{4}, \frac{10}{\cancel{3}} \cdot \frac{\cancel{3}}{4}, \frac{13}{\cancel{3}} \cdot \frac{\cancel{3}}{4}... }[/math]

And we've arrived:

[math]\displaystyle{ \frac 44, \frac 74, \frac{10}{4}, \frac{13}{4}... }[/math]

So we can see that [math]\displaystyle{ \frac 13 }[/math] was the right amount to shift by because it is the delta from the starting position [math]\displaystyle{ 1 }[/math] to [math]\displaystyle{ \frac 43 }[/math], the latter of which is the reciprocal of the target step size [math]\displaystyle{ \frac 34 }[/math] and therefore the value that we need the starting position to equal in order to be sent back to [math]\displaystyle{ 1 }[/math] when we resize all steps from 1 to the target step size by multiplying everything by it.