Patent val

From Xenharmonic Wiki
Jump to: navigation, search

日本語


Introduction

The patent val (aka nearest edomapping) for some EDO is the val that you obtain by simply finding the closest rounded-off approximation to each prime in the tuning. For example, the patent val for 17-EDO is <17 27 39|, indicating that the closest mapping for 2/1 is 17 steps, the closest mapping for 3/1 is 27 steps, and the closest mapping for 5/1 is 39 steps. This means, if octaves are pure, that 3/2 is 706 cents, which is what you get if you round off 3/2 to the closest location in 17-equal, and that 5/4 is 353 cents, which is what you get is you round off 5/4 to the closest location in 17-equal. This val can be extended to the case where the number of steps in an octave is a real number rather than an integer; this is called a generalized patent val, or GPV. For instance the 7-limit generalized patent val for 16.9 is <17 27 39 47|, since 16.9 * log2(7) = 47.444, which rounds down to 47.

There are other vals or edomappings besides the patent or nearest one. You may prefer to use the <17 27 40| val as the 5-limit 17-equal val instead, which rather than <17 27 39| treats 424 cents as 5/4. This val has lower Tenney-Euclidean error than the 17-EDO patent val. However, while <17 27 39| may not necessarily be the "best" val for 17-equal for all purposes, it is the obvious, or "patent" val, that you get by naively rounding primes off within the EDO and taking no further considerations into account. However, <17 27 40| is the generalized patent val for 17.1, since 17.1 * log2(5) = 39.705, which rounds up to 40.

Further explanation

A p-limit val contains the number of steps it takes to get to each prime number up to p, in prime number order:

< [2/1] [3/1] [5/1] [7/1] ... [p/1] |

Given N-EDO, the equal division of the octave into N parts, we may define vals that map a specific number of N-EDO steps to these primes.

For any prime p we can find a corresponding p-limit val in a canonical manner by scalar multiplying <1 log2(3) log2(5) ... log(p)| by N and rounding to the nearest integer. In general this is not guaranteed to be the most accurate available val, but if N-edo has enough relative accuracy in the p-limit, it will be. The name patent comes from the fact that "patent" in one sense of the word is a synonym for "obvious"; the patent val may or may not be the best choice but it's the obvious choice.

One way to think of this process is to first ask, "How many 1200-cent steps (octaves) does it take to get to each prime?" It takes one full-octave step to get to 2/1, log2(3) steps to get to 3/1, log2(5) to get to 5/1, and so on. This gives us <1 1.585 2.322 2.807 3.459 ... log2(p) |.

Then ask, "How many more N-EDO steps does it take to get to the same places?" One 12-EDO step is 1/12th of an octave, by definition; therefore, you need 12 times as many steps to reach 2/1, 3/1, 5/1... Similarly, one 31-EDO step is 1/31st of an octave, so you need 31 times as many steps to reach 2/1, 3/1, 5/1...

Thus, the way to get the p-limit patent val for N-EDO is to multiply <1 1.585 2.322 2.807 ... log2(p) | by N. Then, since you can't take fractional steps in an EDO, you round the results to the nearest integers.

A 12 EDO Example

Multiplying 12 times <1 1.585 2.322 2.807 3.459|

yields <12 19.020 27.863 33.688 41.513|,

rounded to <12 19 28 34 42|,

which is the 11-limit patent val for 12edo.

An alternate and expanded example for 31 EDO

As stated above, the val contains the number of steps it takes to get to a given prime number, in prime number order:

< [2/1] [3/1] [5/1] [7/1] [etc.] |

By definition, for any EDO, the number of steps to 2/1 is the EDO division: 31 for 31 EDO. The 2-limit patent val is < 31 |.

What's the number of steps to 3/1?

The step size for 31 EDO is 38.70967742 cents.

3/1 is 1901.96 in cents.

1901.96 cents / 38.70967742 cents/step = 49.13383752 steps.

This is an EDO, so we can't take 0.13383752 steps. Instead, we round. This is clearly closer to 49 steps, so that's the "obvious" or "patent" choice. The 3-limit patent val is

< 31 49 |. Doing the same thing up through 17, and we get an 17-limit patent val of

< 31 49 72 87 107 115 127 |

To see how to extend from one limit to another, we may look at what to do for 19/1 and use that to go from the 17-limit to the 19-limit.

19/1 = 5097.51 cents, 5097.51 / 38.70967742 cents/step = 131.6857529 steps. Round to get 132. The 19-limit patent val is

< 31 49 72 87 107 115 127 132 |

Note that these are the same answers you would get if you multiplied 31 times <1 1.585 2.322 2.807 3.459 3.700 4.087 4.248 | and rounded the result.

How this defines a rank 1 temperament

A val defines a rank 1 temperament by defining the deliberate introduction of an error into one or more primes. In 12 EDO, for instance, the perfect fifth (ratio 3/2, or exactly 1.5) is mapped to 700 cents, which is actually just barely flat: a ratio of 2^(700/1200), or 1.4983070769.

As stated above, the 2/1 in the patent val is perfect. The patent val for 12 EDO, <12 19 28 34 42 (etc) |, implies that it takes 12 steps to get the octave, which is does: 12 steps * 100 cents / step = 1200 cents = 1 octave.

In the patent val for 12 EDO, the number 19 is in the second spot -- the place reserved for 3/1. That implies that it takes 19 steps to get to 3/1. We know that's not true: We rounded numbers off when we created the val in the first place, so essentially we're just acting as if 19 steps gets you to 3/1. To put it differently, we're deliberately introducing an error into 3/1. More precisely, any time we would have used 3/1 (e.g., in 9/8, 3/2, etc.), we're going to use a slightly different number instead.

We can calculate the error we're introducing into 3/1 as follows for 12 EDO:

12 EDO steps are 100.0 cents each.

19 steps of 12 EDO = 19 steps * 100.0 cents/step = 1900.0 cents.

1900.0 cents => 2^(1900/1200), or 2.9966141538. This is the value that 12 EDO uses in place of prime 3.

That means that the 12 EDO patent val substitutes 2.9966141538 any time you would have had prime 3 in a ratio. 9/8 and 3/2 will be somewhat flat. 4/3 will be somewhat sharp. (Note that, for now, the example intervals (3/2, 4/3, 9/8) deliberately avoid any primes other than 3 and 2, and 2 is pure, so the sharpness and flatness comes only from the impure 3/1 value.)

We can do the same calculations for 31 EDO.

The patent val for 31 EDO is <31 49 72 87 107 (etc) |. The 49 implies that it takes 49 steps to get to 3/1.

31 EDO steps are 38.70967742 cents each.

49 steps of 31 EDO = 49 steps * 38.70967742 cents/step = 1896.774194 cents.

1896.774194 cents => 2^(1896.774194/1200), or 2.991035765. This is what 31 EDO uses in place of prime 3.

Again, as in 12 EDO, it's less than 3, so 9/8 and 3/2 will be somewhat flat, and 4/3 will be somewhat sharp. Note that 31 EDO's prime 3 is a little farther away from 3/1 than 12 EDO's 3/1 -- i.e., it has a greater error. That means 31 EDO's 3/2 will be even flatter, and its 4/3 will be even sharper, than in 12 EDO.

That doesn't make 31 EDO better or worse than 12; it just means there's more error in the 3/1 ratio in 31 EDO than in 12 EDO. If you run these calculations for 5/1 using the patent vals for 12 EDO and 31 EDO, you'll find that 5/1 has more error in 12 EDO than in 31 EDO: 5.0396842 vs. 5.002262078, respectively. 31 EDO may therefore be preferred by people who like sweeter thirds (5/4 ratios) and are willing to have flatter fifths (3/2 ratios).

How this relates to commas

These deliberate errors ensure that certain commas get tempered out. The patent vals for both 12 EDO and 31 EDO temper out 81/80. Here are the calculations:

81 = 3*3*3*3. This can also be written as a power of a prime -- 3^4 -- or as a monzo -- | 0 4 >.

80 = 2*2*2*2*5. This can also be written as a product of powers of primes -- (2^4)*(5^1) -- or as a monzo -- | 4 0 1 >.

Substitute in the values for 81/80 in 12 EDO and get this: (2.9966141538^4) / (2^4)*(5.0396842) = 80.6349472 / 80.6349472 = 1/1.

Substitute in the values for 81/80 in 31 EDO and get this: (2.991035765^4) / (2^4)*(5.002262078) = 80.036193 / 80.036193 = 1/1.

The lesson here is that even though the errors in the primes are different for each EDO + patent val in these cases, 81/80 is still tempered out. However, that's not true for all commas; for instance, 12 EDO tempers out 128/125, the diesis, while 31 EDO does not; and 31 EDO tempers out 393216/390625, the Wuerschmidt comma, while 12 EDO does not.

By the way, there's a faster way to find out if a comma is being tempered out. Recall that multiplying by a number means adding logarithms: for instance, to go up an octave you can multiply a ratio by 2/1 or you can add 1200 to the cents value. Since the patent val shows you how many EDO steps it takes to get to a prime number, you can add that many steps instead of multiplying things out like we did above. Similarly, you can subtract if you need to divide.

81/80 = 3*3*3*3 / (2*2*2*2 * 5).

To get 81, we need to multiply by three, four times. That means we add the number of steps to 3/1 in the patent val, four times. Using the 31 EDO patent val, that's 49+49+49+49, or 4*49, or 196.

To get 80, we need to multiply by two, four times, and multiply again by five, once. So add together the number of steps to 2/1, four times, and then add the number of steps to 5/1. Using the 31 EDO patent val, that's 31+31+31+31+72, or 4*31+72, or 196.

You're dividing 81 by 80, so (assuming we're starting at zero, though it works no matter where you start) you add the steps for 81 (+196) and subtract the steps for 80 (-196). 196-196 = 0. This means that it takes zero steps to reach 81/80 -- in other words, 81/80 "vanishes".

.