Patent val: Difference between revisions
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== Generalized patent val == | == Generalized patent val == | ||
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 {{val| 17 27 39 47 }}, since 16.9 | 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 {{val| 17 27 39 47 }}, since 16.9 × log<sub>2</sub>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 {{val| 17 27 40 }} val as the 5-limit 17-equal val instead, which rather than | There are other vals or edomappings besides the patent or nearest one. You may prefer to use the {{val| 17 27 40 }} val as the 5-limit 17-equal val instead, which rather than {{val| 17 27 39 }} treats 424 cents as 5/4. This val has lower Tenney-Euclidean error than the 17-EDO patent val. However, while {{val| 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, {{val| 17 27 40 }} is the generalized patent val for 17.1, since 17.1 × log<sub>2</sub>5 = 39.705, which rounds up to 40. | ||
== Further explanation == | == Further explanation == | ||
A [[p-limit]] [[ | A [[harmonic limit|''p''-limit]] [[val]] contains the number of steps it takes to get to each prime number up to ''p'', in prime number order: | ||
{{val| [2/1] [3/1] [5/1] [7/1] | {{val| [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. | 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 [ | For any prime ''p'' we can find a corresponding ''p''-limit val in a canonical manner by [[Wikipedia: Scalar multiplication|scalar multiplying]] {{val| 1 log<sub>2</sub>3 log<sub>2</sub>5 … log<sub>2</sub>''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, | 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, log<sub>2</sub>3 steps to get to 3/1, log<sub>2</sub>5 to get to 5/1, and so on. This gives us {{val| 1 1.585 2.322 2.807 3.459 … log<sub>2</sub>''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 | 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 {{val| 1 1.585 2.322 2.807 | Thus, the way to get the ''p''-limit patent val for ''N''-EDO is to multiply {{val| 1 1.585 2.322 2.807 … log<sub>2</sub>''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 == | == A 12-EDO Example == | ||
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rounded to {{val| 12 19 28 34 42 }}, | rounded to {{val| 12 19 28 34 42 }}, | ||
which is the | which is the ''11-limit patent val for [[12edo]]''. | ||
== An alternate and expanded example for 31 EDO == | == An alternate and expanded example for 31 EDO == | ||
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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 | 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 | ||
{{val| 31 49 }}. Doing the same thing up through 17, and we get a 17-limit patent val of | {{val| 31 49 }}. | ||
Doing the same thing up through 17, and we get a 17-limit patent val of | |||
{{val| 31 49 72 87 107 115 127 }} | {{val| 31 49 72 87 107 115 127 }} | ||
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== How this defines a rank-1 temperament == | == 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 | 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<sup>(700/1200)</sup>, or 1.4983070769. | ||
As stated above, the 2/1 in the patent val is perfect. The patent val for 12 EDO, {{val| 12 19 28 34 42 (etc) }}, implies that it takes 12 steps to get the octave, which is does: 12 steps | As stated above, the 2/1 in the patent val is perfect. The patent val for 12 EDO, {{val| 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 | 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: | We can calculate the error we're introducing into 3/1 as follows for 12 EDO: | ||
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12 EDO steps are 100.0 cents each. | 12 EDO steps are 100.0 cents each. | ||
19 steps of 12 EDO = 19 steps | 19 steps of 12 EDO = 19 steps × 100.0 cents/step = 1900.0 cents. | ||
1900.0 cents => 2 | 1900.0 cents => 2<sup>1900/1200</sup>, 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.) | 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.) | ||
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We can do the same calculations for 31 EDO. | We can do the same calculations for 31 EDO. | ||
The patent val for 31 EDO is | The patent val for 31 EDO is {{val| 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. | 31 EDO steps are 38.70967742 cents each. | ||
49 steps of 31 EDO = 49 steps | 49 steps of 31 EDO = 49 steps × 38.70967742 cents/step = 1896.774194 cents. | ||
1896.774194 cents => 2 | 1896.774194 cents => 2<sup>1896.774194/1200</sup>, 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 | 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). | 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). | ||
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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: | 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 = | 81 = 3×3×3×3. This can also be written as a power of a prime – 3<sup>4</sup> – or as a monzo – {{monzo| 0 4 }}. | ||
80 = | 80 = 2×2×2×2×5. This can also be written as a product of powers of primes – 2<sup>4</sup>×5<sup>1</sup> – or as a monzo – {{val| 4 0 1 }}. | ||
Substitute in the values for 81/80 in 12 EDO and get this: | Substitute in the values for 81/80 in 12 EDO and get this: 2.9966141538<sup>4</sup> / (2<sup>4</sup>×5.0396842) = 80.6349472 / 80.6349472 = 1/1. | ||
Substitute in the values for 81/80 in 31 EDO and get this: | Substitute in the values for 81/80 in 31 EDO and get this: 2.991035765<sup>4</sup> / (2<sup>4</sup>×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 | 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 Würschmidt 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. | 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 = | 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 | 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 | 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 | 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". | ||
[[Category:Theory]] | [[Category:Theory]] | ||