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December 9

New musical tuning system calculation, math question.

Sound is frequency. In the way we usually tune instruments we start with an frequency of 440hz. Thats A4. A5 is A4 * 2 = 880hz, A6 is A5 *2 = 1760hz...... Between A4 and A5, we have B4, C4..., but this is not needed to my question, so I will not explain it.

The question.

I, was thinking about how, some "logical" tuning system would work. And need your help.

This logical tuning would start with 20 hz (lowest frequency we can hear), and would continue increasing following some formula/calculation/whateaver, until it goes to 440hz. Then it would continue to increase, using THE SAME formula/calculation/whateaver as before until it reach 20000hz (highest frequency we can hear). The amount of steps this tuning needs to go from 20 to 440hz, NEED to be the same amount of steps it would take to go from 440 to 20000

Anyway how this formula could be?. 201.78.126.14 (talk) 02:11, 9 December 2012 (UTC)[reply]

Well, you basically have 3 points on a graph, say (0,20), (50,440), and (100,20000). You can run a circle through any 3 non-collinear points like this. StuRat (talk) 04:11, 9 December 2012 (UTC)[reply]

... or you can fit a quadratic function. Or almost any other formula that involves 3 independent parameters. And you have even more choice if you don't fix the number of steps at 100. Of course there are physical reasons why tuning involves doubling of frequencies, so replacing this with some other convention chosen at random will be less logical, not more logical. Gandalf61 (talk) 05:00, 9 December 2012 (UTC)[reply]

Agreed. If you tried to tune your instrument like that, it would be all forked up. StuRat (talk) 05:03, 9 December 2012 (UTC) [reply]

Yeah, who cares about logic, it would produce hideous dissonance. Looie496 (talk) 16:36, 9 December 2012 (UTC)[reply]

Although other scales are possible, see scale (music). StuRat (talk) 16:40, 9 December 2012 (UTC)[reply]

Why should the number of steps on each side of 440Hz be the same? Why is that frequency special in your tuning system? Before defining a "logical" system, you should look into why instruments are tuned the way they are, and explain what is illogical about them and why. Then you can start working on designing something that solves the problems you've identified. 209.131.76.183 (talk) 19:30, 10 December 2012 (UTC)[reply]

Indeed, 440Hz is a pretty random choice, as can be seen from the history of concert pitch. And the 20Hz and 20000Hz endpoints are fairly random as well, the actual range of hearing varies substantially between people, these are just ballpark figures neatly expressible with the quite randomly chosen unit of frequency and semi-randomly chosen numerical notation system we happen to be using. Trying to base anything “logical” on these three points is rather ridiculous.—Emil J. 20:40, 10 December 2012 (UTC)[reply]

OP HERE: Because 440hz would be the middle point, and so you would have to have the same amount of steps to get from the starting point (when humans start to listen to sounds) to 440hz as you would need to go from 440hz to the endpoint(when humans stop being able to listen to it). About my stuff being logical or not, the idea is that, if 440 is the middle point (or that would be middle C 261.626 Hz?) the amount of steps between the starting point (when we start to listen) and 440 need to be the same as the amount of steps between the middle point and the end point (when we stop listening to sounds). .201.78.166.79 (talk) 22:27, 11 December 2012 (UTC)[reply]

Well I was taught that "middle C" was 256 Hz but either choice cannot be the "middle" of a geometric progression ranging from 20Hz up to 20 kHz. The musical "middle" would need to be the geometric mean of the endpoints. You could make an equal-tempered scale starting at 20 Hz with a middle of 440 Hz if you take your top note to be 9680 Hz (no-one plays notes higher than that, though some overtones go higher). If you require n steps in each half, then each interval ratio would be the nth root of 22. If you try n = 53 or 54 you will get intervals that are approximate semitones. For quarter tones (perhaps the smallest interval clearly heard by most people), try n = 107. By selecting notes from the n = 107 scale, you could get very close to Arab, Indian and Western music (the difference would be noticed only by a highly-trained ear).

Alternatively, for exact semitones, take your lowest note to be 23.542 Hz and your highest note to be 8223.6 Hz and n = 50 (or n = 100 for quarter tones). You might be interested in the article: Microtonal music, but I think all such scales are based on the octave (doubling the frequency), and your exact original specification would not produce a musical scale as we understand music because tunes would sound different depending on the point on the scale at which you started, and you could never play an exact octave. This doesn't mean that you couldn't play "music" of some kind on your musically uneven scale, but it would sound very strange and dissonant unless you selected just certain notes that roughly corresponded to conventional intervals. Dbfirs 18:38, 12 December 2012 (UTC)[reply]

WIkipedia says middle C is 261.626.

You said: "all such scales are based on the octave (doubling the frequency), and your exact original specification would not produce a musical scale as we understand music because tunes would sound different depending on the point on the scale at which you started, and you could never play an exact octave.", yes I know that and I dont care if it the end result is not based on doubling the frequency. By starting with an frequency of 20hz and doubling it, you would go to 40hz, then 80hz, 160, 320, 640. So it would be impossible to go to middle c or 440hz by just doubling the frequency.Also, YES, it needs to "start" with 20hz and "ends" with 20000hz.Also the number of steps not being divisible by 12 is not an problem.201.78.197.147 (talk) 22:47, 12 December 2012 (UTC)[reply]

I wasn't doubting Wikipedia, just reminding you that "middle" was different in the past and is just an arbitrary convention. It is impossible to make a "musical" scale (based on consistent frequency ratios) starting at 20 Hz with 440 Hz in the "middle" and ending at 20 000 Hz, but you could construct an infinite number of different "mathematical" scales with this property (though the simple quadratic and cubic fits go into negative frequency, the circle fit doesn't give a single-valued function, and my attempt at a simple exponential fit doesn't give good results). I suppose the simplest solution is to construct two half-scales with a change in ratios each side of 440 Hz, but you wanted a single formula. An inelegant example of a function that fits your requirement (using 100 intervals, where f is the frequency and n is your note in the scale) is f = 18.0473 plus (4.5334 times n) plus (1.952862 times ten to the power of (n over 25)). You can adjust the constants to get a better fit if you wish, but you will not be able to hear the difference. Some intervals are rather less than a semitone and some are a bit more (but all are less than two semitones except for a few notes below 50Hz). When you create some music using this scale, perhaps you will post your composition so that we can drive away stray cats with it! Perhaps someone more numerate and functionate than I am can come up with a simpler Monotone increasing function through (0,20) (50,440) and (100,20000) with intervals closer to a GP? Dbfirs 09:05, 13 December 2012 (UTC)[reply]

Yes this middle is an arbitrary convention, I knew that before posting the question. The thing about my question/idea is: if this is the middle, the, the number of steps between starting position and ending position should be the same, or it wouldnt be the middle. Also I dont care if is not "musical" (thats subjective anyway). Anyway, I will check the math you posted.189.115.206.17 (talk) 10:30, 14 December 2012 (UTC)[reply]

Tested your formula using wolfram math. With n equal to 50, f=440.004 and not 440, with n=0 f=20.00002 not 20. Anyway thanks, its pretty close to what I wanted.201.78.219.30 (talk) 21:01, 14 December 2012 (UTC)[reply]

Yes, as I said, you can adjust the three constants to get any degree of accuracy you wish, but I stopped there because it is impossible to tune any instrument or analogue sound generator anywhere near this degree of accuracy. Digital sound generators might theoretically generate an exact number of cycles per second, but they cheat, and the result will not be a pure frequency. I'd be interested to hear what your scale sounds like if you compose some music using it. Dbfirs 09:06, 15 December 2012 (UTC)[reply]

"Maybe they're all wrong" ensembles

Do any ensemble learning methods account for correlated errors that can recognize a disagreement between two inner models as evidence against both? An example would be the Bayesian posterior probabilities given that inner classifier X predicted class A, classifier Y independently predicted class B, and both X and Y had the following confusion matrix:

Neon Merlin 04:01, 9 December 2012 (UTC)[reply]

What would be the point of using multiple models if you require them all to agree with each other? Looie496 (talk) 16:30, 9 December 2012 (UTC)[reply]

We don't -- that's the point. In this case, assuming errors are uncorrelated within actual classes (which is probably safe given low noise and heterogeneous inner algorithms), explaining why one predicts A and the other predicts B leads to the conclusion that C is the most probable. Neon Merlin 16:59, 9 December 2012 (UTC)[reply]

Points determining an ellipse or sphere

3 points in the plane (a) are on a circle, (b) which is unique, (c) iff a condition on their relative positions is satisfied: that they not be collinear.

(1) Does there exist n such that n points in the plane (a) are on a conic section, (b) which is unique, (c) iff some condition on their relative positions is satisfied? (Note n>3 since 3 points on an ellipse are also on a circle, making the conic for those 3 points not unique.)

(2) Does there exist k such that k points in R³ (a) are on a sphere, (b) which is unique, (c) iff some condition on their relative positions is satisfied? Duoduoduo (talk) 15:56, 9 December 2012 (UTC)[reply]

Do our five points determine a conic and cubic function articles help ? I don't think R³ can lie on the surface of a sphere, since it's both planar and infinite, and a sphere is neither. StuRat (talk) 16:26, 9 December 2012 (UTC)[reply]

Any 4 points on R³ are on a unique sphere iff there is no plane containing all of them. --77.125.75.226 (talk) 16:52, 9 December 2012 (UTC)[reply]

I see. So those 4 points are on the sphere, while the rest of R³ is not. StuRat (talk) 16:55, 9 December 2012 (UTC)[reply]

(ec)Off the top of my head, your initial statement extends to any number of dimensions (this generalizes the answer to your question (2)): a (k − 2)-sphere is determined by k points. Your condition would be that for k = 4 points, they not be coplanar to determine a (2-)sphere (and the number of Euclidean dimensions does not have to be limited, e.g. 3 points determine a circle in Rⁿ, n ≥ 2). In general, the condition becomes that the k points must not be co-(k − 1)-hyperplanar to determine a (k − 2)-sphere.

A variation is to treat lines, planes etc. as circles, spheres and so on (which they conformally are), which allows the condition to be relaxed. The condition amounts to something like: every point must be separate from the lower-dimensional figure determined by the previous points. E.g., given one point (no constraint), the second point must not be coincident (this determines a 0-sphere, being two points). The third point must be away from the 0-sphere (i.e. not coincident with either of the first two points), and determines a 1-sphere (meaning a circle or line). A fourth point not on the 1-sphere determines a 2-sphere (meaning a sphere or plane). And so on. — Quondum 17:02, 9 December 2012 (UTC)[reply]

Thanks all, for the link and the explanations. Quondum, you say "Your condition would be that for k = 4 points, they not be coplanar to determine a (2-)sphere" and also say "In general, the condition becomes that the k points must not be co-(k − 1)-hyperplanar to determine a (k − 2)-sphere." But it seems to me (maybe I'm missing something) that the latter says that the k=4 points determining a sphere must not be 3-hyperplanar, contrasting with "not be coplanar". Should "not be co-(k − 1)-hyperplanar" be "not be co-(k − 2)-hyperplanar"? Duoduoduo (talk) 17:28, 9 December 2012 (UTC)[reply]

Oops. Yes. (*blush*) — Quondum 17:35, 9 December 2012 (UTC)[reply]