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When one can and does tune a pure 4:2/pure 6:3, it does sound very clean to me. But more importnatly, a pure 6:3 allows for pure 19ths to be tuned as well as pure 11/12/22.
Example:
When tuning pure 4:2 and pure 12, we have
M3 = M10 < M17 = M6
Maybe it's me but I can't follow this and the rest. Which intervals are all these M3 m17 m6below etc? For the last one I quoted, for C3C4G4 I understand you mean G#2C3 = G#2C4 < D#2G4 = D#2C3 but the inequality is wrong as D#2C3 beats slower than G#2C3? Kees




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Kees,
It's easy. But I don't know where in the piano's scale it can sound right.
Take for example the octave A3A4 and the test note F3
pure 4:2 octave A3A4, M3=M10, F3A3=F3A4
wide 4:3 fourth A3D4, M3<M6, F3A3<F3D4
pure 3:1 octave plus fifth or 12th D4A5, M6=M17, F3D4=F3A5
wide 2:1 octave A4A5, M10<M17, F3A4<F3A5
wide 4:1 double octave or 15th, M3<M17, F3A3<F3A5
pure 8:3 octave plus fourth or 11th A2D4, m6down=M6, A2F3=F3D4
pure 8:1 triple octave or 22th A2A5, m6down=M17, A2F3=F3A5
pure 6:3 octave D3D4, m3down=M6, D3F3=F3D4
pure 6:1 double octave plus fifth or 19th D3A5, m3down=M17, D3F3=F3A5
This is perfectly tunable in any piano! But the question is: does it sound good?
1. Tune A4
2. Tune A3 to A4 as a pure 4:2 octave. Test M3=M10, F3A3=F3A4
3. Tune D4 to A3 as a wide 4:3 fourth (slightly more tempered than usual because we are going to tune pure 12ths). Test M3<M6, F3A3<F3D4
4. Tune A5 to D4 as a pure 3:1 twelve. Test M6=M17, F3D4=FA5
Up to this point you have a wide 4:1 double octave A3A5, M3<M17, F3A3<F3A5. You have also a wide 2:1 octave A4A5, M10<M17, F3A4<F3A5
That's where M3=M10 < M17=M6 comes from.
5. Tune A2 to D4 as a 8:3 pure 11th (octave+fourth). Test m6down=M6, A2F3=F3D4
The test note is F3, m6down means a m6 below F3, which is A2.
To understand this test think of a wide 4:3 A3D4 fourth. You have to compare the fourth partial of A3 to the third partial of D4, to make this comparison you use the fifth partial of F3. The test is M3<M6, 5:4<5:3, F3A3<F3D4. Now to test the pure octave plus fourth A2D4 you have to compare the 8th partial of A2 to the 3th partial of D4, the test will be A2F3=F3D4, 8:5=5:3, m6down=M6.
So far we have M3=M10 < M17=M6=m6down
Notice that we have a pure 8:3 eleventh A2D4 and a pure 3:1 twelve D4A5, this builds up a pure 8:1 triple octave (22th) A2A5. The test is m6down=M17, A2F3=F3A5.
How good sounds the octave A2A3? Who knows? We know that it is a narrow 8:4 octave, because we have a wide 4:1 within a pure 8:1, so 8:4 is narrow, but does it sound good? Maybe yes, maybe not, it depends on the scaling of the piano.
6. Tune D3 to D4 as a pure 6:3 octave. Test m3=M6 (or m3down=M6), D3F3=F3D4. And we know that F3D4=F3A5 so we have D3F3=F3A5 or m3down=M17, this is the pure double octave plus fifth or 19th D3A5.
How good this octave D3D4 is going to sound tuned as a 6:3?
How good is the fourth A2D3? and the fifth D3A3?
Even if these pure P8, P11, P12, P19, P22 and the resulting, not explicitly tuned, intervals sound right here at A3A4 or at any other spot, what warants you that going higher or lower in the scale they will continue to sound good?
As Jeff said, it would be FANTASTIC!
I just measured the octave spread of the Centennial Steinway D I have in my shop an the result is:
pure 4:2 0.75 cents wide 2:1 0.4 cents narrow 6:3
If you tune a 0.2 cents wide 4:2 with a 0.2 narrow 6:3 you'll have a 0.95 cents wide 2:1. Is this the best sounding octave for this piano? maybe yes, maybe not, but no matter how you tune it it won't be beatless.
My August FÃ¶rster studio
pure 4:2 0.1 cents wide 2:1 2.4 cents narrow 6:3
A Hamilton (Baldwin) Studio I am repairing
pure 4:2 0.2 cents wide 2:1 2.5 cents narrow 6:3
So where can I find a piano that can be tuned with a beatless A3A4 octave? Not even a high quality concert grand can have it beatless!




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...
So where can I find a piano that can be tuned with a beatless A3A4 octave? Not even a high quality concert grand can have it beatless!
Yes, let alone a piano with wide 6:3 with narrow 4:2. A while back Kent Swafford demonstrated such an octave on an SSB that he had. I remember Del saying that they had some well known scaling errors but are now considered a feature instead of a fault. As far as the example that Mark provided and Kees verified, I played with the numbers and could not come up with iH values that would produce those results, although I may be missing something. The ratio in iH would have to be WAY more than 1:4. I suspect there are wild partials on one or both notes. I asked what piano the audio was from and whether wound strings were involved, but there was no reply.
Jeff Deutschle PartTime Tuner Who taught the first chicken how to peck?




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...
I get:
C#3F3 = 8.6 C#3F4 = 8.9 F3G#3 = 8.2 G#3F4 = 9.7
So indeed nearly pure 4:2 and wide 6:3. So the IH of F4 must be more than 4 times the IH of F3. You could check that with tunelab.
Kees
With a freshly caffeinated mind I came up with the following possibilities for an F3F4 octave with a 4:2 beating +0.3 bps and a 6:3 beating +1.5 bps, which is what the above RBI beatrates translate to. I used Young's equation rather than Tunelab's tabulated values. Would someone please check my math/arithmetic? F3: iH 0.10, offset 0.0 cents F4: iH 0.75, offset 0.0 cents 4:2 +0.3 bps 6:3 +1.5 bps iH ratio 1:7.5F3: iH 0.20, offset 0.3 cents F4: iH 1.15, offset 0.0 cents 4:2 +0.3 bps 6:3 +1.5 bps iH ratio 1:5.75So we can't say for sure what the iH ratio might be in this case without knowing what one of the two iH values is first. Seems the higher the iH value, the less of a ratio is needed. So I expect this would happen more often in pianos with higher iH values. The problem is, pianos with higher values in the middle tend to have flatter, not steeper, iH slopes. So once again I have to wonder if this isn't a case of wild partials. Mark, again, can you tell us what piano the recording was from and if wound strings were involved?
Jeff Deutschle PartTime Tuner Who taught the first chicken how to peck?




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As Kees alluded to, he performed a mathematical proof that shows that one can have a piano with pure 4:2/pure 6:3 if the iH of the top note is 4 times that of the bottom note. According to my calculations, that is only possible if the inharmonicity of the upper note is 6.40 times as high as that of the lower note. I have a PDF of those calculations I could post or send to anyone who is interested. (Is there a way to post a PDF on Piano World?) Whether it is 4 or 6.4, it is more than any oneoctave jump in inharmonicity I have seen.




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When one can and does tune a pure 4:2/pure 6:3, it does sound very clean to me. But more importnatly, a pure 6:3 allows for pure 19ths to be tuned as well as pure 11/12/22.
Example:
When tuning pure 4:2 and pure 12, we have
M3 = M10 < M17 = M6
Maybe it's me but I can't follow this and the rest. Which intervals are all these M3 m17 m6below etc? For the last one I quoted, for C3C4G4 I understand you mean G#2C3 = G#2C4 < D#2G4 = D#2C3 but the inequality is wrong as D#2C3 beats slower than G#2C3? Kees The intervals are all from a common check note. That is why I list some as m3down, m6down. You quoted: "G#2C3 = G#2C4 < D#2G4 = D#2C3" but the intervals are all from G#2. I don't know how you got D#2 So the notes would be: G#2C3 = G#2C4 < G#2C5 = G#2F3 The pure 4:2 is C3C4 The wide double octave is C3C5 The pure 12th is F3C5 The wide 2:1 is C4C5 The wide P4 is C3F3 The narrow fifth is F3C4 All these intervals sizes are confirmed in this window. Add the m6down (C2G#3) to confirm the pure 22 C2C5, and there are a bunch more interval sizes confirmed like the pure 11. Add the m3down, if the 6:3 is pure (F2G#2 = G#2F3), and you get a pure 19 F2C5, and a bunch more intervals confirmed like the pure 8:6 P4 Thanks for the question Kees. You are always fair and polite and it is a pleasure conversing with you. You really could have made me feel stupid for the mistake I made when questioning your assumption in the proof you so generously did for me (in a quarter of a page when I had already written three pages and wasn't even done!) but you didn't. You are truly a gentleman and I want everyone to know. So thanks again.




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As Kees alluded to, he performed a mathematical proof that shows that one can have a piano with pure 4:2/pure 6:3 if the iH of the top note is 4 times that of the bottom note. According to my calculations, that is only possible if the inharmonicity of the upper note is 6.40 times as high as that of the lower note. I have a PDF of those calculations I could post or send to anyone who is interested. (Is there a way to post a PDF on Piano World?) Whether it is 4 or 6.4, it is more than any oneoctave jump in inharmonicity I have seen. Thanks for chiming in, Mr. Scott. You are considered to be an authority on the subject. Is it likely, or have you seen a situation where a wild partial, perhaps from a wound string, will make it seem there is such a large jump in inharmonicity?
Jeff Deutschle PartTime Tuner Who taught the first chicken how to peck?




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As Kees alluded to, he performed a mathematical proof that shows that one can have a piano with pure 4:2/pure 6:3 if the iH of the top note is 4 times that of the bottom note. According to my calculations, that is only possible if the inharmonicity of the upper note is 6.40 times as high as that of the lower note. I have a PDF of those calculations I could post or send to anyone who is interested. (Is there a way to post a PDF on Piano World?) Whether it is 4 or 6.4, it is more than any oneoctave jump in inharmonicity I have seen. Below I used Young's model, but I think if you plug in the tunelab IH model it will agree with you. Partials (n=1,2,..) are for A3 f3* n* (1+B3(n^21)) and for A4 f4* n* (1+B4(n^21))
4:2 pure: f3*4*(1+15*B3) = f4*2*(1+3*B4) â†’ f4 = 2*(1+15*B3)/(1+3*B4) * f3
6:3 wide: f3*6*(1+35*B3) < f4*3*(1+8*B4)
So f3*6*(1+35*B3) < 2*(1+15*B3)/(1+3*B4) * f3 *3*(1+8*B4)
Simplify:
(1+35*B3)*(1+3*B4) < (1+15*B3)*(1+8*B4) (K1)
We can solve K1 for B4 in terms of B3 for equality,
B4 = 4*B3/(3*B3 + 1) ~ 4*B3
then if we increase B4 from that value the right hand side of K1 increases faster because of factor 8*B4 on the right and 3*B4 on the left so we get wide 6:3 pure 4:2.
Kees




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Thanks Raphael and Mark for explaining those checks. I thought you guys were talking about the P12 on top of a bottom note.
Kees




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For what it's worth, here's pretty much the best I could do with the F3F4 octave. F3F4.mp3 The recording is F3 then F4 followed by the part where they're together filtered at 2:1, then 4:2 and then 6:3 using a bandpass filter set at 375Hz, 700Hz and 1050Hz. The fundamental of F3 is at about 175Hz. Both are single strings. Piano is an 1887 Steinway A1 recently restored. Strings are Roslau. Paul.




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As Kees alluded to, he performed a mathematical proof that shows that one can have a piano with pure 4:2/pure 6:3 if the iH of the top note is 4 times that of the bottom note. According to my calculations, that is only possible if the inharmonicity of the upper note is 6.40 times as high as that of the lower note. I have a PDF of those calculations I could post or send to anyone who is interested. (Is there a way to post a PDF on Piano World?) Whether it is 4 or 6.4, it is more than any oneoctave jump in inharmonicity I have seen. Below I used Young's model, but I think if you plug in the tunelab IH model it will agree with you. Partials (n=1,2,..) are for A3 f3* n* (1+B3(n^21)) and for A4 f4* n* (1+B4(n^21))
4:2 pure: f3*4*(1+15*B3) = f4*2*(1+3*B4) â†’ f4 = 2*(1+15*B3)/(1+3*B4) * f3
6:3 wide: f3*6*(1+35*B3) < f4*3*(1+8*B4)
So f3*6*(1+35*B3) < 2*(1+15*B3)/(1+3*B4) * f3 *3*(1+8*B4)
Simplify:
(1+35*B3)*(1+3*B4) < (1+15*B3)*(1+8*B4) (K1)
We can solve K1 for B4 in terms of B3 for equality,
B4 = 4*B3/(3*B3 + 1) ~ 4*B3
then if we increase B4 from that value the right hand side of K1 increases faster because of factor 8*B4 on the right and 3*B4 on the left so we get wide 6:3 pure 4:2.
Kees Actually I get 3.2 for tunelab model. One of us is missing a factor 2. Kees




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Either way, it makes me wonder about the validity of any of the RBI interval checks, especially those that involve wider intervals or in areas with jumpy iH scales.




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Partials (n=1,2,..) are for A3 f3* n* (1+B3(n^21)) and for A4 f4* n* (1+B4(n^21))
4:2 pure: f3*4*(1+15*B3) = f4*2*(1+3*B4) â†’ f4 = 2*(1+15*B3)/(1+3*B4) * f3
6:3 wide: f3*6*(1+35*B3) < f4*3*(1+8*B4)
So f3*6*(1+35*B3) < 2*(1+15*B3)/(1+3*B4) * f3 *3*(1+8*B4)
Simplify:
(1+35*B3)*(1+3*B4) < (1+15*B3)*(1+8*B4) (K1)
We can solve K1 for B4 in terms of B3 for equality,
B4 = 4*B3/(3*B3 + 1) ~ 4*B3
then if we increase B4 from that value the right hand side of K1 increases faster because of factor 8*B4 on the right and 3*B4 on the left so we get wide 6:3 pure 4:2.
Kees I don't see how you can write formulas for frequency in terms of inharmonicity that do not involve logs or exponentials. The inharmonicity constant applies to cents offset, which is essentially a logarithm. It does not apply linearly to frequency. Send me your email by PM and I will email you my PDF of the calculations. (I used Open Office Equations to write it).




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Partials (n=1,2,..) are for A3 f3* n* (1+B3(n^21)) and for A4 f4* n* (1+B4(n^21))
4:2 pure: f3*4*(1+15*B3) = f4*2*(1+3*B4) â†’ f4 = 2*(1+15*B3)/(1+3*B4) * f3
6:3 wide: f3*6*(1+35*B3) < f4*3*(1+8*B4)
So f3*6*(1+35*B3) < 2*(1+15*B3)/(1+3*B4) * f3 *3*(1+8*B4)
Simplify:
(1+35*B3)*(1+3*B4) < (1+15*B3)*(1+8*B4) (K1)
We can solve K1 for B4 in terms of B3 for equality,
B4 = 4*B3/(3*B3 + 1) ~ 4*B3
then if we increase B4 from that value the right hand side of K1 increases faster because of factor 8*B4 on the right and 3*B4 on the left so we get wide 6:3 pure 4:2.
Kees I don't see how you can write formulas for frequency in terms of inharmonicity that do not involve logs or exponentials. The inharmonicity constant applies to cents offset, which is essentially a logarithm. It does not apply linearly to frequency. Send me your email by PM and I will email you my PDF of the calculations. (I used Open Office Equations to write it). These are linearly related for small inharmonicity as follows. Partials n of note with fundamental f0 are f(n) = f0*n*(1+B*(n^21)) In cent the offset from harmonic (f0*n) is 1200/log(2)*log(f(n)/f0*n) = 1200/log(2)*log(1+B*(n^21)) ~ 1200/log(2)*B*(n^21) as B is small. So your B in cents is my 1200/log(2)*B. You can find my email under my profile. Kees




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So...the numbers are all "interesting", but how does that translate into what you hear? In other words, let's say you start out tuning a pure 4:2 A4A3 octave (with your ETD so you KNOW it's right and you confirm it with RBI tests). Now you listen to it. What are you listening FOR to decide whether you feel that's the "best" octave width or not?
Suppose you now decide to widen it a bit, perhaps to a 6:3 octave and you play it and listen, then you slightly narrow that and listen...at what point, and what criteria are you using to decide that THAT (whatever it is) octave width is "the best"? What are you listening for?
Anyone want to venture a comment or two?
Pwg




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Hi Peter, I have done a lot of research into this. I created an online survey to see what sizes tuners prefer. They choose octaves without knowing what size they were. The following criteria covers about 90% of the pianos I tune. I use this to tune A3A4 and F3F4 and from that, I can determine the m3/M3 equality that I use for my temperament sequence. (If I didn't need to know the m3/M3 equality, I don't think I would be as interested in octave sizes.) So, here is the procedure: 1) Tune a pure 4:2 2) Test the 6:3  If it's pure, leave it  If it's barely narrow, tune the octave as a wide 4:2/narrow 6:3  If it's very narrow, leave the octave as a pure 4:2/very narrow 6:3 There are other steps that I go farther into in this flow chart, but that's basically what I do. http://howtotunepianos.com/tuningbeatlessoctaves/




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Partials (n=1,2,..) are for A3 f3* n* (1+B3(n^21)) and for A4 f4* n* (1+B4(n^21))
4:2 pure: f3*4*(1+15*B3) = f4*2*(1+3*B4) â†’ f4 = 2*(1+15*B3)/(1+3*B4) * f3
6:3 wide: f3*6*(1+35*B3) < f4*3*(1+8*B4)
So f3*6*(1+35*B3) < 2*(1+15*B3)/(1+3*B4) * f3 *3*(1+8*B4)
Simplify:
(1+35*B3)*(1+3*B4) < (1+15*B3)*(1+8*B4) (K1)
We can solve K1 for B4 in terms of B3 for equality,
B4 = 4*B3/(3*B3 + 1) ~ 4*B3
then if we increase B4 from that value the right hand side of K1 increases faster because of factor 8*B4 on the right and 3*B4 on the left so we get wide 6:3 pure 4:2.
Kees I don't see how you can write formulas for frequency in terms of inharmonicity that do not involve logs or exponentials. The inharmonicity constant applies to cents offset, which is essentially a logarithm. It does not apply linearly to frequency. Send me your email by PM and I will email you my PDF of the calculations. (I used Open Office Equations to write it). DoelKees found an arithmetic error in my calculations. It turns out the right ratio is exactly 4, as he said.




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Thanks to Kees I was able to graph a beatless 4:2/6:3 octave.
I also used Parsons IH graph to find that a ratio of about 4.67 can be fit into the acceptable range for B values for F3F4
Here are the values for the differences at the partials for A3 = 219.9013, B = 0.0003, A4 = 440, B = 0.0012 (4 times)
2:1 = 0.001Hz narrow (0.002 cents) 4:2 = 0.000 Hz (0.001 cents) 6:3 = 0.001 Hz (0.002 cents) 8:4 = 0.005 Hz (0.005 cents)
I don't think anyone can hear a beat at any of those partials. At the 8:4 partial, it would take 3 minutes and 20 seconds for one beat!
That's a beatless octave with inharmonicity that is within the acceptable range according to Parsons and Sanderson.




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So...the numbers are all "interesting", but how does that translate into what you hear? In other words, let's say you start out tuning a pure 4:2 A4A3 octave (with your ETD so you KNOW it's right and you confirm it with RBI tests). Now you listen to it. What are you listening FOR to decide whether you feel that's the "best" octave width or not?
Suppose you now decide to widen it a bit, perhaps to a 6:3 octave and you play it and listen, then you slightly narrow that and listen...at what point, and what criteria are you using to decide that THAT (whatever it is) octave width is "the best"? What are you listening for?
Anyone want to venture a comment or two?
Pwg Good questions! I wonder if anyone could extra the beat rates out of the F3F4 octave I posted. To my ears, there's barely a beat in 2:1 4:2 or 6:3, but I don't have the tools to measure the actual rates. Paul.




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Partials (n=1,2,..) are for A3 f3* n* (1+B3(n^21)) and for A4 f4* n* (1+B4(n^21))
4:2 pure: f3*4*(1+15*B3) = f4*2*(1+3*B4) â†’ f4 = 2*(1+15*B3)/(1+3*B4) * f3
6:3 wide: f3*6*(1+35*B3) < f4*3*(1+8*B4)
So f3*6*(1+35*B3) < 2*(1+15*B3)/(1+3*B4) * f3 *3*(1+8*B4)
Simplify:
(1+35*B3)*(1+3*B4) < (1+15*B3)*(1+8*B4) (K1)
We can solve K1 for B4 in terms of B3 for equality,
B4 = 4*B3/(3*B3 + 1) ~ 4*B3
then if we increase B4 from that value the right hand side of K1 increases faster because of factor 8*B4 on the right and 3*B4 on the left so we get wide 6:3 pure 4:2.
Kees I don't see how you can write formulas for frequency in terms of inharmonicity that do not involve logs or exponentials. The inharmonicity constant applies to cents offset, which is essentially a logarithm. It does not apply linearly to frequency. Send me your email by PM and I will email you my PDF of the calculations. (I used Open Office Equations to write it). DoelKees found an arithmetic error in my calculations. It turns out the right ratio is exactly 4, as he said. I switched my MS Access application simulator over to Tunelab's values and found that a 1:3.2 iH ratio in an octave will result in a pure 4:2 and when 6:3 is also pure, but not 2:1 or 8:4 partial matches.
Jeff Deutschle PartTime Tuner Who taught the first chicken how to peck?



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