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You are confusing the issue. It is possible that people like EBVT3 tuning (or some other unequal temperament) but this may have nothing to do with supposed beat (equal or unequal) cancellation/masking.

Please note the word "may" in the previous sentence. I keep an open mind, but I haven't seen any evidence for equal beat cancellation or unequal beat masking in my own experiments on my piano or in any recordings by others.

If this really worked it should be easy to demonstrate by showing a chord with and without equal beats for comparison.

Of course in a chord individual beats are always less prominent. For example play F4A4 and hear it beating. Now put your forearm on that octave: you can't hear F4A4 beating any more.

Kees

I fooled around a bit with this idea, modified to suit my own interest.

Temperament range F3-C#5 so you have all possible M3/M6 tests. Then pick a random M3/M6 test and adjust the upper note to make them equal beating. Unless it's an A. This does not work: it gets worse at every iteration. Picking the lower note is no better.

Not sure why.

Kees

Re-listen to your recording. Try to confirm what other people are saying. Re-record with better equipment so that you can confirm that what you hear in person, is there on the recording. Don't give them any opportunity to say what you know is there, is not there.

Kees,

I have discovered another equality test. (No one has every told me about it nor have I read of it anywhere, but that doesn't mean anything) But I thought you might be able to prove it mathematically. I call it the m7b5 equality. Example: FA# = BD#.

I know it works in my sequence, but is there any theoretical proof that it is valid?

FA# = BD#? Typo?

Kees

Whoops. Yes.

FG# = BD#

Thanks.

The M3/M6 test is (for example)
F3A3 = D3#C4.
However the M6 is really a tiny bit faster theoretically.
Your test, when you invert the m3 is
F3A3 = D3B3
and here the M6 is a little bit slower, but not a tiny bit.

So it may be a good bracketing test: the M3 has to beat at a rate in-between those two, but closer to D#3C4.

Kees

There is no "mathematical proof". It all depends on how much you stretch (contract) the 6:3 octaves.

If you have a pure 6:3 octave F#3F#4 then F#3A3=A3F#4 and with the inside M3 outside M6, we have B3D#4=A3F#4=F#3A3

But if you tune slightly narrow 6:3 octaves then you can have B3D#4=F3G#3
Or even B3D#4=E3G3.

(With "putting the forearm on that octave" Kees means striking all the notes together within that octave)
At least one example, where you confirm that beat masking is possible. We have a good starting point here, more later...

There is a certain degree of beat masking blending in after some time, caused by the fact that the intervals are NOT equally beating. Maybe Kees can provide us the beat rate differences of your examples?

Thank you Kees.

You say the M6 is faster in the dom7 theoretically. Is there a mathematical proof? Thanks.

I use that relational logic to prove the m7b5 relationship but thought there might be a numerical one.

I think your relationship logic assumed a pure 6:3, but I couldn't completely follow it because, while you say my test's m3 inverted is a D3B3, that would refer to B3D4 and that equality is B3D3=F4A4 so I'm completely confused. I'm hoping a few more instructions will clear it up for me. Thanks for your time.

Like I said, I'm looking for a mathematical proof. Here's my relational proof, with assumptions.

Gadzar, I'm always tuning wide 4:2 and narrow 6:3 in the temperament octave.

The m7b5 Proof.

Assumptions:
Dom7 equality. GB=FD and all transpositions within and near the temperament octave, but not over the break.
Narrow 6:3 octaves given by F3G#3>G#3F4 and all transpositions within the temperament octave and nearby, where > means "every so slightly faster; barely noticeable."
The difference in beat speed between the m3 and M6 in a narrow 6:3, is equal to the difference between chromatic M6's in ET.

(These assumptions are valid for the level of accuracy I am going for, which is high compared to some other popular tests, especially since I am using "bracketing" exclusively in my temperament sequence, which seems to produce good precision. My sequence requires that you bisect all the windows equally when setting beat speeds.)

Relational Proof

F3G#3>G#3F4 (narrow 6:3)
A3F#4>G#3F4 (ET relationship)

Therefore,
F3G#3=A3F#4 (assuming beat speed differences are equal)

A3F#4=B3D4 (Dom7 equality)

Therefore,
F3G#3=B3D#4

I'm tuning high accuracy ET temperaments (increasing RBI's and P4's) using these relationships to produce windows into which other beat speeds must bisect (that's the beauty, the bisecting may not be 100% accurate, but it seems to be way more accurate than any other one sided tests like F3A3<F3D4 for the P4, for example. Just how much faster is F3D4 supposed to be? I know some people know the answer intuitively, but how to explain that to a beginner? The windows sequence I've developed gives a solid relationship that students have to reproduce, and even if the formulas and bisecting are not 100% accurate, the process of constantly fitting beat speeds into such small windows develops aural beat speed difference sensitivity, which is the most important skill for tuning aurally using RBI, IMHO. The tolerances on the windows I've developed are 3x more precise than CM3's)

Sorry for the long winded thesis, but that's what this is; my thesis. Your help and criticism is invaluable to helping me produce a method that other technicians and students will find credible.

As always, your objective and scientific criticism is appreciated by me and many others on this forum.

(Addendum. Gadzar, your post was much more concise than mine and proved the relationship nicely, given that the 6:3 was narrow as you say. Nice job.)

Why? It may be greater or lesser, depending on how much the 6:3 octaves are stretched, in this case contracted.

There is no mathematical proof here.

P.S. In the treble we tune narrow 6:3 octaves. In the tenor we tune still narrow 6:3 octaves but less narrow than in the treble. In the bass we tune pure 6:3 octaves. And in the low bass, in some pianos, we even tune wide 6:3 octaves. So your equality will only work on a particular region of the scale where the 6:3 octaves have the required size.

The accuracy of the assumption allows me to tune extremely high accuracy, compared to the way I used to tune. (Don't say it!) You need to see the sequence to get it.

Anyway, we don't tune 6:3 octaves in the temperament, unless you're going for some excessive stretch like pure 19ths.

The purpose if this analysis is to explore a sequence that gets good accuracy and precision by placing beat speeds evenly within a window, so that beginners can get decent results and improve their beat speed difference sensitivity.

In the end, it is that sensitivity, and stability, that allow a tuner to accurately make decisions on which way to adjust a note.

The only thing a good sequence can do is get you there faster with less refining.

A bad sequence allows you to get through without showing you the errors, thereby allowing you to move on with a false sense of accomplishment, if you choose not to refine.

Before anybody gets their knickers in a twist, I am NOT saying my sequence is the only one that can do that, or even the best one. I'm just making generalizations on what I think a good sequence should allow you to do.


Anyway, I see what you are saying. Because of the variable size of the 6:3 octave, we can't have a mathematical proof. I'm just wondering how accurate a sequence can be if we assume the m7b5 relationship, when the octave stretch really only confirms FG#=CE? Kees?

Agreed. I only use it in the temperament.

I was just being simple minded and looked at the beat rates for zero IH and 2^(1/12) tuning. Then it is just a numerical coincidence that M3 and M6 beat about the same in the 3d inversion of dom 7th chord. They are not exactly the same though, M6 is a touch faster.

When there is IH and octave stretch I don't know what happens.

Regarding your proof attempt:

If x>y and z>y it does not follow that x=z unless you assume (as you did in your post) that x-y = z-y, but that is the same as assuming x=z so all you prove is that x=z if x=z.

Generally I like very much the idea of bracketing beatrates between two other beatrates.

Kees

What makes the relationship unique is that x>y for a different reason that z>y. That's why it is not as trivial as you say. Does that make sense?

Not really. No matter what the reason, it still doesn't follow that x=z unless you assume x=z (i.e., the beat rate differences (x-y and x-z are the same).

Why would they be the same in your example?

Somehow you have to mention that not only x>y and z>y, but also that both x and z are very close to y, in math speak |x-y|<=epsilon and |z-y|<=epsilon with epsilon the typical beat rate difference of your example.

In that case you can conclude |x-z|<epsilon, i.e., x is closer to z than the maximum of |x-y| and |z-y| but x could be larger or smaller than z.

To tighten the bound you need another assumption, namely that the difference between the beat rate differences d=||x-y|-|z-y|| is small, smaller than epsilon. (In your writeup you assumed this to be zero.) Suppose you know that d<q*epsilon with 0<=q<1.

The equations are now:
x>y
z>y
x-y<=epsilon
z-y<=epsilon
|(x-y)-(z-y)|<q*epsilon

Then from the last equation it follows that
|x-z|<q*epsilon

Kees

I have written a program for this. It computes a temperament range with IH data from any piano, using tunelab's model. I tune octaves as 4:2 or 6:3 or in-between equal beating. Then I compute F3A3C#4F4A4 to be progressive, and interpolate the notes in between assuming the stretch does not change over a M3.

Below the beat rates computed for 4 IH models.

If we take the usual M3/M6 test on A3 as an example, we should
see that the M3 on A3 beats the same as the M6 on G3. For ih=0 we see the M6 is 0.2 bps is faster, whereas the M6 on F#3 is 0.3 slower. So it would be more accurate to say M3 on A3 is in-between M6's on F#3 and G3.

On the Steinway D the Cerisano M3/M6 test (A3C#4=F#3D#4) is already more accurate than the usual test (A3C#4=G3E4): M3 on A3 is closer to M6 on F#3 than on G3. On the uprights the Cerisano M3/M6 test is the more accurate one.

So it seems the usual M3/M6 test should be replaced by the Cerisano M3/M6 test (or bracketing). I ran the simulation also with 4:2 and 6:3 octaves with the same results.

Did I earn a complementary copy of your book?

Kees
--
octave: equal beating 4:2 and 6:3
ih=0

M3 M6 m3 m6 P4 P5
C#3 5.5 6.3 -7.5 -8.7 0.6 -0.5
D3 5.8 6.7 -7.9 -9.3 0.7 -0.5
D#3 6.2 7.1 -8.4 -9.8 0.7 -0.5
E3 6.5 7.5 -8.9 -10.4 0.8 -0.6
F3 6.9 7.9 -9.4 -11.0 0.8 -0.6
F#3 7.3 8.4 -10.0 -11.7 0.8 -0.6
G3 7.8 8.9 -10.6 -12.4 0.9 -0.7
G#3 8.2 9.4 -11.2 -13.1 0.9 -0.7
A3 8.7 10.0 -11.9 -13.9 1.0 -0.7
A#3 9.3 10.6 -12.6 -14.7 1.1 -0.8
B3 9.8 11.2 -13.3 -15.6 1.1 -0.8
C4 10.4 11.9 -14.1 -16.5 1.2 -0.9
C#4 11.0 -- -15.0 -17.5 1.3 -0.9
D4 11.7 -- -15.9 -- 1.3 -1.0
D#4 12.4 -- -16.8 -- 1.4 --
E4 13.1 -- -17.8 -- 1.5 --
F4 13.9 -- -18.9 -- -- --
F#4 -- -- -20.0 -- -- --

ih = Steinway D

M3 M6 m3 m6 P4 P5
C#3 5.5 6.3 -7.8 -9.1 0.7 -0.5
D3 5.8 6.7 -8.3 -9.7 0.7 -0.5
D#3 6.2 7.1 -8.9 -10.3 0.7 -0.5
E3 6.5 7.5 -9.4 -10.9 0.8 -0.6
F3 6.9 8.0 -10.1 -11.6 0.8 -0.6
F#3 7.3 8.5 -10.7 -12.4 0.9 -0.6
G3 7.8 9.0 -11.4 -13.2 0.9 -0.7
G#3 8.2 9.5 -12.2 -14.1 1.0 -0.7
A3 8.7 10.1 -13.1 -15.0 1.1 -0.7
A#3 9.2 10.7 -13.9 -16.0 1.1 -0.8
B3 9.8 11.3 -14.9 -17.1 1.2 -0.8
C4 10.3 12.0 -16.0 -18.2 1.3 -0.9
C#4 10.9 -- -17.1 -19.5 1.4 -0.9
D4 11.6 -- -18.3 -- 1.5 -1.0
D#4 12.3 -- -19.7 -- 1.6 --
E4 13.0 -- -21.1 -- 1.7 --
F4 13.8 -- -22.7 -- -- --
F#4 -- -- -24.5 -- -- --

ih = HellasHelsinki
M3 M6 m3 m6 P4 P5
C#3 5.4 6.3 -7.9 -10.8 0.7 -0.3
D3 5.7 6.7 -8.4 -11.4 0.8 -0.3
D#3 6.1 7.1 -8.9 -12.0 0.8 -0.3
E3 6.4 7.5 -9.5 -12.7 0.9 -0.3
F3 6.8 8.0 -10.1 -13.4 0.9 -0.4
F#3 7.2 8.4 -10.8 -14.2 1.0 -0.4
G3 7.7 9.0 -11.5 -15.1 1.0 -0.4
G#3 8.1 9.5 -12.2 -16.0 1.1 -0.5
A3 8.6 10.1 -13.1 -17.0 1.2 -0.5
A#3 9.1 10.7 -13.9 -18.0 1.2 -0.6
B3 9.6 11.3 -14.8 -19.1 1.3 -0.6
C4 10.2 12.0 -15.8 -20.4 1.4 -0.6
C#4 10.8 -- -16.9 -21.7 1.5 -0.7
D4 11.4 -- -18.1 -- 1.6 -0.7
D#4 12.1 -- -19.3 -- 1.7 --
E4 12.8 -- -20.6 -- 1.8 --
F4 13.6 -- -22.1 -- -- --
F#4 -- -- -23.7 -- -- --

ih = large Heintzmann upright

M3 M6 m3 m6 P4 P5
C#3 5.4 6.3 -8.2 -10.6 0.7 -0.3
D3 5.7 6.7 -8.7 -11.2 0.8 -0.3
D#3 6.1 7.1 -9.3 -11.9 0.8 -0.4
E3 6.4 7.6 -9.9 -12.6 0.9 -0.4
F3 6.8 8.0 -10.6 -13.5 0.9 -0.4
F#3 7.2 8.5 -11.4 -14.3 1.0 -0.5
G3 7.7 9.0 -12.2 -15.3 1.1 -0.5
G#3 8.1 9.5 -13.0 -16.3 1.1 -0.5
A3 8.6 10.1 -13.9 -17.5 1.2 -0.6
A#3 9.1 10.8 -15.0 -18.7 1.3 -0.6
B3 9.6 11.4 -16.1 -20.0 1.4 -0.6
C4 10.2 12.1 -17.3 -21.5 1.4 -0.7
C#4 10.7 -- -18.6 -23.1 1.6 -0.7
D4 11.4 -- -20.0 -- 1.7 -0.7
D#4 12.1 -- -21.5 -- 1.8 --
E4 12.8 -- -23.2 -- 1.9 --
F4 13.5 -- -25.0 -- -- --
F#4 -- -- -27.1 -- -- --

Nice work Kees.

But whether you get a free copy depends on the accuracy of your inharmonicity model.

Did you measure all the ih values? Or what equation are you using for the ih values? Did you verify the accuracy of your ih model?