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If equal temperament divides octave into 12 equal parts, why hertz differences are not the same but element 12th of two?


Why are pianos traditionally tuned “out of tune” at the extremes?In C major, if the third note is 14 cents sharp from JI, does the V/VI have a third that is 28 cents sharp?Do capable harmony singers sing in just intonation or tempered tuning?Are sharp keys “bright” and flat keys “dark”?Two people listening to the same song, yet they hear two different guitar chords that sound right played. Why?How do I spell a note when none of the usual rules make a difference?In meantone temperaments, do A sharp and B flat have different frequencies?Need help understanding harmonic series and intervalsWhat is the difference between 2/4 and 4/4 when it comes the accented beats?






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8

















Let's take A sound 440Hz
If we divide space between 440Hz and 880Hz into 12 equal parts, we would have:
440Hz, 476.6, 513.2 ... 880Hz. And this for me looks like equally divided.
Why we say equally divided if its element 12th of 2?









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  • 9





    Because 'equal' refers to a geometric progression, not an arithmetic progression.

    – user207421
    18 hours ago












  • The frequency ratio of ANY 2 notes N notes apart is the same. This is because all notes are based on a ratio of frequencies. If you used equal magnitude spacings the ratios between notes would vary continually.

    – Russell McMahon
    4 hours ago

















8

















Let's take A sound 440Hz
If we divide space between 440Hz and 880Hz into 12 equal parts, we would have:
440Hz, 476.6, 513.2 ... 880Hz. And this for me looks like equally divided.
Why we say equally divided if its element 12th of 2?









share









New contributor



Tukkan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • 9





    Because 'equal' refers to a geometric progression, not an arithmetic progression.

    – user207421
    18 hours ago












  • The frequency ratio of ANY 2 notes N notes apart is the same. This is because all notes are based on a ratio of frequencies. If you used equal magnitude spacings the ratios between notes would vary continually.

    – Russell McMahon
    4 hours ago













8












8








8


4






Let's take A sound 440Hz
If we divide space between 440Hz and 880Hz into 12 equal parts, we would have:
440Hz, 476.6, 513.2 ... 880Hz. And this for me looks like equally divided.
Why we say equally divided if its element 12th of 2?









share









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Tukkan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












Let's take A sound 440Hz
If we divide space between 440Hz and 880Hz into 12 equal parts, we would have:
440Hz, 476.6, 513.2 ... 880Hz. And this for me looks like equally divided.
Why we say equally divided if its element 12th of 2?







theory harmony tuning intervals temperament





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share









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asked Oct 11 at 15:08









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  • 9





    Because 'equal' refers to a geometric progression, not an arithmetic progression.

    – user207421
    18 hours ago












  • The frequency ratio of ANY 2 notes N notes apart is the same. This is because all notes are based on a ratio of frequencies. If you used equal magnitude spacings the ratios between notes would vary continually.

    – Russell McMahon
    4 hours ago












  • 9





    Because 'equal' refers to a geometric progression, not an arithmetic progression.

    – user207421
    18 hours ago












  • The frequency ratio of ANY 2 notes N notes apart is the same. This is because all notes are based on a ratio of frequencies. If you used equal magnitude spacings the ratios between notes would vary continually.

    – Russell McMahon
    4 hours ago







9




9





Because 'equal' refers to a geometric progression, not an arithmetic progression.

– user207421
18 hours ago






Because 'equal' refers to a geometric progression, not an arithmetic progression.

– user207421
18 hours ago














The frequency ratio of ANY 2 notes N notes apart is the same. This is because all notes are based on a ratio of frequencies. If you used equal magnitude spacings the ratios between notes would vary continually.

– Russell McMahon
4 hours ago





The frequency ratio of ANY 2 notes N notes apart is the same. This is because all notes are based on a ratio of frequencies. If you used equal magnitude spacings the ratios between notes would vary continually.

– Russell McMahon
4 hours ago










8 Answers
8






active

oldest

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32


















The intervals between notes are "equal" not in the sense that the difference in Hz between them is the same, but the ratio a between them is the same. Let's say g is one semitone higher than f, then g = a f.



Note Hz Ratio a to previous note, rounded to 3 decimal places
A4 440.00
A#4 466.16 1.059 (466.16 / 440.0 = 1.059, and so on down the column)
B4 493.88 1.059
C5 523.25 1.059
C#5 554.37 1.059
D5 587.33 1.059
D#5 622.25 1.059
E5 659.25 1.059
F5 698.46 1.059
F#5 739.99 1.059
G5 783.99 1.059
G#5 830.61 1.059
A5 880.00 1.059


It might be easier to understand when you think of the frequency of the octaves. The number of Hz between octaves is different (220, 440, 880, 1760, etc.), but the ratio of 2:1 is always the same. The same concept applies to the notes in the scale.



Mathematically, what we are doing is dividing an octave (2:1 ratio) into 12 equal steps (equal in ratio, i.e. a^12=2). Using a scientific calculator, we can solve for a=2^(1/12) = 1.0594630943592952645618252949463, which is the exact ratio between two half steps.






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  • 2





    I think the OP already knows it's the twelfth root of two. He says that in the subject line, "why hertz differences are not the same but element 12th of two?"

    – piiperi
    2 days ago






  • 4





    You have not written the exact ratio. 2^(1/12) is irrational. You have written the initial segment of its decimal expansion, but it continues non-repetitively forever. The next 20 digits are ... 41700779204317494185 ....

    – Eric Towers
    yesterday






  • 2





    I think this answer is spot-on. This is the reason. But I do think the phrase "the difference is the same ratio" sounds a little clumsy. May I suggest "the ratio is the same" as an improvement?

    – luser droog
    yesterday


















25


















The division of notes has to do with human perception and psychoacoustics. One description of human perception is the Weber-Fechner law, where a human will perceive equal changes in some sensory input, such as sound level or sound pitch, not by absolute level or value difference, but by the ratio of the change. e.g. larger values need a proportionately larger change for the change be perceived (if small) or perceived as about the same, within some reasonable range (e.g. audible, but not causing ear damage, etc.)



Thus, for a semitone (forth, fifth, etc.) interval to sound the same, no matter which base note one starts from, in the equal temperament scale, the notes have to differ not by equal absolute frequency differences (as would be created by equal Hz deltas between notes), but by equal ratio differences (the 12th root of 2, so that twelve equal multiplications will equal one octave).



e.g. the "equality" in equal division has to be in equal in ratio, not additive absolute value.






share|improve this answer























  • 7





    This answer nails the error in the thought behind the question — intervals are defined by human perception, and human perception of intervals is logarithmic as to frequency, not linear. Octave above 440 is 880. Octave below 440 is 220. Same with all the intervals — equal interval means same ratio of frequencies. Where it gets interesting is that perfect 4ths and 5ths are not equal to 5 and 7 well-tempered half-steps, respectively.

    – Jeff Y
    yesterday






  • 3





    external link with audio examples: what does a linearly spaced sequence sound like vs. a logarithmic one

    – Cee McSharpface - it
    yesterday


















17


















What happens if you go down by the same steps:



  • 440Hz

  • 1 step down : 403.33Hz

  • 2 steps down : 366.67Hz

  • 3 steps down : 330.Hz

  • ...

  • 11 steps down : 36.67Hz

  • 12 steps down : 0Hz

  • 13 steps down : -36.67Hz

So, using your "equally divided" logic, we are at zero Hz after 12 steps, and the next step beyond that is minus 37 Hz! What does that even mean? But ok, let's follow your logic a little bit ... what's the frequency exactly in the middle of the octave 440 - 880 Hz, that would be 660 Hz. What's an octave above that? That would be 2 * 660 Hz = 1320 Hz. What would be the steps in that octave - 660 Hz / 12 = 55 Hz? Ok, then let's take one step up from 660 Hz, that's 660 Hz + 55 Hz = 715 Hz. But wait ... the step was supposed to be 37 Hz, not 55 Hz??? Does your step size depend on the start and end points of the octave? Or does it take a sudden jump at 880 Hz - steps below 880 would be 440 / 12, but above 880 they would be 880 / 12? Where does such a divider come from, is it embedded in nature? I thought A = 440 Hz was only an agree-upon convention, not a law of nature.



Where did you get the 880Hz? By multiplying by 2, i.e. one octave higher. I guess the same has to apply for any frequency, not only 440Hz? For example, one octave higher from 880Hz has to be 880Hz * 2? And any other frequency like 1000Hz... one octave above that must be 2000Hz. If the interval of an octave is calculated with multiplication, how could other intervals be calculated with addition?



So, ask yourself: if F1 and F2 are the frequencies of two consecutive semitones, what is the relationship between F1 and F2, if (F1 * 2) and (F2 * 2) has to have the same relationship?



You're looking for a function f(F) such that f
applied 12 times gives 2*F.




f(f(f(f(f(f(f(f(f(f(f(f(F)))))))))))) = 2 * F


If you step up one semitone from F, you get a frequency f(F). The frequency one octave higher from that is 2 * f(F).



If you first step up an octave, you get F*2. And if you step up one semitone from that, you get f(F*2), which should be the same frequency, so:




2 * f(F) = f(2 * F)


What might function f be like?



From the subject line "why hertz differences are not the same but element 12th of two?" I assume that you already know that consecutive semitones have a ratio of 2^(1/12).






share|improve this answer




























  • Not quite qhat OP means! 440>880 = 440, so each of the 12 steps should be 36.66Hz. In the lower octave, between 220 and 440, there's only 220 Hz difference, so each of the 12 steps should be 18.33 Hz.

    – Tim
    Oct 11 at 16:29






  • 3





    @Tim I think this is what the OP means. He means that he's trying to figure out things and wants help doing that. Looking at the subject line, he already has the solution, and he wants to find the right perspective for understanding why it is like it is. Which is what I'm trying to provide. YMMV but I was able to figure this out in high-school when writing a music player program, having only the information that an octave higher is multiplying by two, and applying the same relationship 12 times brings you there, an octave higher. :)

    – piiperi
    2 days ago












  • @Tim : Does this mean I get incompatible step sizes if I take my octaves to be C-to-C?

    – Eric Towers
    yesterday











  • Negative frequency means that the music is played backwards, revealing hidden Satanic messages.

    – dan04
    4 hours ago


















6


















A simple way is to look at ratios as suggested above. One can divide an interval equally arithmetically such that the length (size, or more technically "measure") of each subinterval is identical. Dividing an interval arithmetically in 12 pieces (I can explain the 12 but it takes more math.) yields, 1=12/12, 13/12, 14/12, 15/12, 16/12, 17/12, 18/12, 19/12, 20/12, 21/12, 22/12, 22/12, 24/12=2. However people's hearing seems (experimentally) to distinguish ratios of frequencies rather than differences as being more identical. For example (Taking A=440cps), the fifth above A is E at 660cps not 19/12*440=696.666....



If we want equal ratios for each half-step, instead of (2-1)/12, we 2^(1/12). The point is that the ratio of G to C is constant for all fifths (A-D, C-F, etc.). Since antiquity, the ratio of a fifth is 3:2 (or 3/2 times the frequency of the lower note.) This comports with dividing a string into intervals and listening to the frequency of the two shorter pieces. (Aside: Vincenzo Galilei suggested using 18/17 as a approximation to the twelfth root of two; it's remarkably good.)



However: for computational work, we can use logarithms; the logarithm of a ratio is the difference in the logarithms of that ratio's constituents. One divides the octave into 1200 cents (the 1200th root of 2) and assigns 100 cents to the equal-tempered semitone. This allows one to easily (at least when using pencil and paper instead of a calculator) compute interval sized for varying tunings.



So, even though our ears hear by ratio (experimentally), we can calculate by ratios or addition. Wiki has a bunch of articles qG (quod Google in analogy with qv) which give a more comprehensive explanation.






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    4


















    Possibly a simple way to look at it is to look at a guitar neck. An octave there is divided into 12 parts - equal as far as each fret is a semitone away from its neighbour. But looking carefully, it's fairly obvious that each fret isn't the same size. In fact, the eleventh fret is very nearly half the size of the first one, from nut to fret 1. Go further, and the 12th (octave) is actually half the size of the first.



    Your hypothesis is that they would all be the same size - one twelfth of the half length of the open string? Were that the case, what woud happen at fret 13? And apart, each fret would produce a note that was out of tune. So there needs to be a ratio of each fret against its neighbour, as pointed out in other good answers.






    share|improve this answer





















    • 4





      @AlbrechtHügli The frets on a guitar neck illustrate the physical relationship between the notes of an equal-tempered scale, but they don't explain it. The explanation goes in the opposite direction: The scale explains the spacing of the frets.

      – Solomon Slow
      2 days ago











    • I agree, but it is a good analogy and shows that the differences of the steps aren't continuous.

      – Albrecht Hügli
      2 days ago


















    4


















    Start by considering the equal division of octaves into one part. That is, think about changing pitch by octaves only.



    If we start with A1=55 Hz, we have the following pitches:




    Pitch Frequency
    ----------------
    A1 55 Hz
    A2 110 Hz
    A3 220 Hz
    A4 440 Hz
    A5 880 Hz
    ...


    You can see that when you increase the pitch by an equal additive amount, you increase the frequency by an equal multiplicative factor. That is, every time you increase the pitch by one octave, you double the frequency. This means that the relationship between pitch and frequency is logarithmic.



    From there, it is fairly easy to reach the conclusion that to divide the octave into some number of equal parts, you need to find the factor that, when multiplied by itself that number times, yields 2. In other words, the frequency factor corresponding to a division of the octave into n parts is the nth root of 2.






    share|improve this answer




























    • Stackexchange bug, is anyone else seeing this? The contents of the table disappear after page load.

      – whatsisname
      yesterday











    • @whatsisname yes, I'm seeing it inconsistently, too. I'll try modifying the table to see if I can work around it.

      – phoog
      yesterday


















    3


















    Our note system is a logarithmic scale for frequency. A logarithmic scale turns equal fractions into equal distances. You can define equal temperament as a constant step size of 1/12 on the log_2 scale of frequency.



    Going back to the linear scale, this means that a semitone translates into a factor of 2^(1/12) (the twelfth root of two).




    The reason for this is, that the sound of an interval depends on how the overtone spectra of the two nodes match up.



    The octave has the unique feature that all harmonics of the higher note match up with some harmonic of the lower note. Likewise, if you have a perfect fifth (factor 3/2), every second harmonic of the upper note coincides with every third harmonic of the lower note. Similar relations hold for the perfect fourth (factor 4/3), the mayor third (5/4) and the mayor sixth (5/3). And so on, and so forth. The pattern of how the harmonics match up defines the sound of the interval, and the harmonics are defined by factors of frequency.



    Thus, only a logarithmic scale can be used to describe intervals well (our note system). And by consequence, equal temperament must be defined on the logarithmic scale.






    share|improve this answer





















    • 2





      Good idea to talk about log scale. You could even show one graph with equal spacing on a log scale and one with uneven spacing on a linear scale, like a guitar neck.

      – Eric Duminil
      yesterday






    • 1





      The pitch-class equivalence of octaves is not related to overtones. It is apparent even in sine waves, which have no overtones. Conversely, a "real life" pitch of 220 Hz has an overtone at 660 Hz, yet 660 Hz isn't an equivalent pitch class to 220 Hz.

      – phoog
      yesterday






    • 2





      @phoog 660 Hz is one octave + one fifth above 220 Hz, and it does blend excellently. You even have a stop for this interval in many pipe organs because it blends so well. The organist uses this stop to change the sound of the fundamental stop, not to get a transposition. Also, did you ever try to tune two sines to an octave? You can do it if you run the signal through a distorting guitar amp (at the very least this adds the frequencies f1-f0 and f0+f1 to the signal), but I know that I won't be able to do it precisely without technological help.

      – cmaster
      yesterday











    • @cmaster of course it blends excellently, but it isn't equivalent in the way that 440 and 880 are. They're both A, but 660 and 1980 are E and B. Three instruments playing pitches in parallel separated by factors of two sound more unified than three instruments separated by factors of three, even if the instruments are sine wave generators (relatively easy to achieve with waveform synthesis or Hammond organs). Pipe organ stops for non-octave harmonics only blend in certain registrations. A 5 1/3' stop with a single 8' stop is likely to sound like parallel fifths rather than a richer tone.

      – phoog
      yesterday












    • @phoog For a 5 1/3' stop you need the 16' stop to blend. And, yes, a 3 1/5' stop and a 5 1/3' stop blend perfectly with a 16' stop. What makes it difficult to use that combination, is that stops are usually either on or off with nothing in between, and the high stops are just too loud in the combination. I once played an electronic organ that allowed partly pulling a stop, allowing me to reduce the volume of the high stops, and I used that combination quite frequently for its nice sound. With only on/off stops, you'd need to pull some more 16' stops to get the right relative volume.

      – cmaster
      yesterday


















    0


















    This is another answer trying to help understanding also the question to people who can't cope with ratios and other abstract terms:



    Imagine you have a tone of 12 Hz frequency (a string waving 12 times/second).
    How must the 12 half steps between the octava (24 Hz) be tuned, so that the differences between all half steps are equal?



    The question implies: If the range between the octava is 12 Hz, why is the difference between the 12 half steps not always just 1 Hz?



    root=12Hz



    minor second 13Hz



    major second 14Hz



    .



    .



    .



    .



    perfect fifth 18Hz



    .



    .



    .



    major seventh: 23Hz



    octave: 24



    We can see that the difference between the first half 12Hz and 13Hz is just 1/10 of 12Hz (10% of the whole octava), while the additional difference between the octava 24Hz and the precedent half tone (23Hz) would have been almost only a 1/20 (=5%) of the difference between the next upper half tone above the octave will be 2Hz more - because this must be a 1/10 of the next octava of 48Hz, as the difference between of ocatava' (24Hz) and octava'' (48Hz) is 24Hz! (48-24=24) and a half step of 1/12 between octava' and octava'' will be 2?



    From this we can derive that the differences between the half steps are not additional of 1/12 but proportional by multiplying each half step with 1/12.



    Hope this is not droning and confusing. TLDR?






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      8 Answers
      8






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      8 Answers
      8






      active

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      32


















      The intervals between notes are "equal" not in the sense that the difference in Hz between them is the same, but the ratio a between them is the same. Let's say g is one semitone higher than f, then g = a f.



      Note Hz Ratio a to previous note, rounded to 3 decimal places
      A4 440.00
      A#4 466.16 1.059 (466.16 / 440.0 = 1.059, and so on down the column)
      B4 493.88 1.059
      C5 523.25 1.059
      C#5 554.37 1.059
      D5 587.33 1.059
      D#5 622.25 1.059
      E5 659.25 1.059
      F5 698.46 1.059
      F#5 739.99 1.059
      G5 783.99 1.059
      G#5 830.61 1.059
      A5 880.00 1.059


      It might be easier to understand when you think of the frequency of the octaves. The number of Hz between octaves is different (220, 440, 880, 1760, etc.), but the ratio of 2:1 is always the same. The same concept applies to the notes in the scale.



      Mathematically, what we are doing is dividing an octave (2:1 ratio) into 12 equal steps (equal in ratio, i.e. a^12=2). Using a scientific calculator, we can solve for a=2^(1/12) = 1.0594630943592952645618252949463, which is the exact ratio between two half steps.






      share|improve this answer











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      • 2





        I think the OP already knows it's the twelfth root of two. He says that in the subject line, "why hertz differences are not the same but element 12th of two?"

        – piiperi
        2 days ago






      • 4





        You have not written the exact ratio. 2^(1/12) is irrational. You have written the initial segment of its decimal expansion, but it continues non-repetitively forever. The next 20 digits are ... 41700779204317494185 ....

        – Eric Towers
        yesterday






      • 2





        I think this answer is spot-on. This is the reason. But I do think the phrase "the difference is the same ratio" sounds a little clumsy. May I suggest "the ratio is the same" as an improvement?

        – luser droog
        yesterday















      32


















      The intervals between notes are "equal" not in the sense that the difference in Hz between them is the same, but the ratio a between them is the same. Let's say g is one semitone higher than f, then g = a f.



      Note Hz Ratio a to previous note, rounded to 3 decimal places
      A4 440.00
      A#4 466.16 1.059 (466.16 / 440.0 = 1.059, and so on down the column)
      B4 493.88 1.059
      C5 523.25 1.059
      C#5 554.37 1.059
      D5 587.33 1.059
      D#5 622.25 1.059
      E5 659.25 1.059
      F5 698.46 1.059
      F#5 739.99 1.059
      G5 783.99 1.059
      G#5 830.61 1.059
      A5 880.00 1.059


      It might be easier to understand when you think of the frequency of the octaves. The number of Hz between octaves is different (220, 440, 880, 1760, etc.), but the ratio of 2:1 is always the same. The same concept applies to the notes in the scale.



      Mathematically, what we are doing is dividing an octave (2:1 ratio) into 12 equal steps (equal in ratio, i.e. a^12=2). Using a scientific calculator, we can solve for a=2^(1/12) = 1.0594630943592952645618252949463, which is the exact ratio between two half steps.






      share|improve this answer











      New contributor



      Sagebrush Gardener is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.
















      • 2





        I think the OP already knows it's the twelfth root of two. He says that in the subject line, "why hertz differences are not the same but element 12th of two?"

        – piiperi
        2 days ago






      • 4





        You have not written the exact ratio. 2^(1/12) is irrational. You have written the initial segment of its decimal expansion, but it continues non-repetitively forever. The next 20 digits are ... 41700779204317494185 ....

        – Eric Towers
        yesterday






      • 2





        I think this answer is spot-on. This is the reason. But I do think the phrase "the difference is the same ratio" sounds a little clumsy. May I suggest "the ratio is the same" as an improvement?

        – luser droog
        yesterday













      32














      32










      32









      The intervals between notes are "equal" not in the sense that the difference in Hz between them is the same, but the ratio a between them is the same. Let's say g is one semitone higher than f, then g = a f.



      Note Hz Ratio a to previous note, rounded to 3 decimal places
      A4 440.00
      A#4 466.16 1.059 (466.16 / 440.0 = 1.059, and so on down the column)
      B4 493.88 1.059
      C5 523.25 1.059
      C#5 554.37 1.059
      D5 587.33 1.059
      D#5 622.25 1.059
      E5 659.25 1.059
      F5 698.46 1.059
      F#5 739.99 1.059
      G5 783.99 1.059
      G#5 830.61 1.059
      A5 880.00 1.059


      It might be easier to understand when you think of the frequency of the octaves. The number of Hz between octaves is different (220, 440, 880, 1760, etc.), but the ratio of 2:1 is always the same. The same concept applies to the notes in the scale.



      Mathematically, what we are doing is dividing an octave (2:1 ratio) into 12 equal steps (equal in ratio, i.e. a^12=2). Using a scientific calculator, we can solve for a=2^(1/12) = 1.0594630943592952645618252949463, which is the exact ratio between two half steps.






      share|improve this answer











      New contributor



      Sagebrush Gardener is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      The intervals between notes are "equal" not in the sense that the difference in Hz between them is the same, but the ratio a between them is the same. Let's say g is one semitone higher than f, then g = a f.



      Note Hz Ratio a to previous note, rounded to 3 decimal places
      A4 440.00
      A#4 466.16 1.059 (466.16 / 440.0 = 1.059, and so on down the column)
      B4 493.88 1.059
      C5 523.25 1.059
      C#5 554.37 1.059
      D5 587.33 1.059
      D#5 622.25 1.059
      E5 659.25 1.059
      F5 698.46 1.059
      F#5 739.99 1.059
      G5 783.99 1.059
      G#5 830.61 1.059
      A5 880.00 1.059


      It might be easier to understand when you think of the frequency of the octaves. The number of Hz between octaves is different (220, 440, 880, 1760, etc.), but the ratio of 2:1 is always the same. The same concept applies to the notes in the scale.



      Mathematically, what we are doing is dividing an octave (2:1 ratio) into 12 equal steps (equal in ratio, i.e. a^12=2). Using a scientific calculator, we can solve for a=2^(1/12) = 1.0594630943592952645618252949463, which is the exact ratio between two half steps.







      share|improve this answer











      New contributor



      Sagebrush Gardener is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.








      share|improve this answer




      share|improve this answer








      edited 21 hours ago









      M. Stern

      1714 bronze badges




      1714 bronze badges






      New contributor



      Sagebrush Gardener is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      answered Oct 11 at 17:26









      Sagebrush GardenerSagebrush Gardener

      4211 silver badge3 bronze badges




      4211 silver badge3 bronze badges




      New contributor



      Sagebrush Gardener is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.




      New contributor




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      Check out our Code of Conduct.












      • 2





        I think the OP already knows it's the twelfth root of two. He says that in the subject line, "why hertz differences are not the same but element 12th of two?"

        – piiperi
        2 days ago






      • 4





        You have not written the exact ratio. 2^(1/12) is irrational. You have written the initial segment of its decimal expansion, but it continues non-repetitively forever. The next 20 digits are ... 41700779204317494185 ....

        – Eric Towers
        yesterday






      • 2





        I think this answer is spot-on. This is the reason. But I do think the phrase "the difference is the same ratio" sounds a little clumsy. May I suggest "the ratio is the same" as an improvement?

        – luser droog
        yesterday












      • 2





        I think the OP already knows it's the twelfth root of two. He says that in the subject line, "why hertz differences are not the same but element 12th of two?"

        – piiperi
        2 days ago






      • 4





        You have not written the exact ratio. 2^(1/12) is irrational. You have written the initial segment of its decimal expansion, but it continues non-repetitively forever. The next 20 digits are ... 41700779204317494185 ....

        – Eric Towers
        yesterday






      • 2





        I think this answer is spot-on. This is the reason. But I do think the phrase "the difference is the same ratio" sounds a little clumsy. May I suggest "the ratio is the same" as an improvement?

        – luser droog
        yesterday







      2




      2





      I think the OP already knows it's the twelfth root of two. He says that in the subject line, "why hertz differences are not the same but element 12th of two?"

      – piiperi
      2 days ago





      I think the OP already knows it's the twelfth root of two. He says that in the subject line, "why hertz differences are not the same but element 12th of two?"

      – piiperi
      2 days ago




      4




      4





      You have not written the exact ratio. 2^(1/12) is irrational. You have written the initial segment of its decimal expansion, but it continues non-repetitively forever. The next 20 digits are ... 41700779204317494185 ....

      – Eric Towers
      yesterday





      You have not written the exact ratio. 2^(1/12) is irrational. You have written the initial segment of its decimal expansion, but it continues non-repetitively forever. The next 20 digits are ... 41700779204317494185 ....

      – Eric Towers
      yesterday




      2




      2





      I think this answer is spot-on. This is the reason. But I do think the phrase "the difference is the same ratio" sounds a little clumsy. May I suggest "the ratio is the same" as an improvement?

      – luser droog
      yesterday





      I think this answer is spot-on. This is the reason. But I do think the phrase "the difference is the same ratio" sounds a little clumsy. May I suggest "the ratio is the same" as an improvement?

      – luser droog
      yesterday













      25


















      The division of notes has to do with human perception and psychoacoustics. One description of human perception is the Weber-Fechner law, where a human will perceive equal changes in some sensory input, such as sound level or sound pitch, not by absolute level or value difference, but by the ratio of the change. e.g. larger values need a proportionately larger change for the change be perceived (if small) or perceived as about the same, within some reasonable range (e.g. audible, but not causing ear damage, etc.)



      Thus, for a semitone (forth, fifth, etc.) interval to sound the same, no matter which base note one starts from, in the equal temperament scale, the notes have to differ not by equal absolute frequency differences (as would be created by equal Hz deltas between notes), but by equal ratio differences (the 12th root of 2, so that twelve equal multiplications will equal one octave).



      e.g. the "equality" in equal division has to be in equal in ratio, not additive absolute value.






      share|improve this answer























      • 7





        This answer nails the error in the thought behind the question — intervals are defined by human perception, and human perception of intervals is logarithmic as to frequency, not linear. Octave above 440 is 880. Octave below 440 is 220. Same with all the intervals — equal interval means same ratio of frequencies. Where it gets interesting is that perfect 4ths and 5ths are not equal to 5 and 7 well-tempered half-steps, respectively.

        – Jeff Y
        yesterday






      • 3





        external link with audio examples: what does a linearly spaced sequence sound like vs. a logarithmic one

        – Cee McSharpface - it
        yesterday















      25


















      The division of notes has to do with human perception and psychoacoustics. One description of human perception is the Weber-Fechner law, where a human will perceive equal changes in some sensory input, such as sound level or sound pitch, not by absolute level or value difference, but by the ratio of the change. e.g. larger values need a proportionately larger change for the change be perceived (if small) or perceived as about the same, within some reasonable range (e.g. audible, but not causing ear damage, etc.)



      Thus, for a semitone (forth, fifth, etc.) interval to sound the same, no matter which base note one starts from, in the equal temperament scale, the notes have to differ not by equal absolute frequency differences (as would be created by equal Hz deltas between notes), but by equal ratio differences (the 12th root of 2, so that twelve equal multiplications will equal one octave).



      e.g. the "equality" in equal division has to be in equal in ratio, not additive absolute value.






      share|improve this answer























      • 7





        This answer nails the error in the thought behind the question — intervals are defined by human perception, and human perception of intervals is logarithmic as to frequency, not linear. Octave above 440 is 880. Octave below 440 is 220. Same with all the intervals — equal interval means same ratio of frequencies. Where it gets interesting is that perfect 4ths and 5ths are not equal to 5 and 7 well-tempered half-steps, respectively.

        – Jeff Y
        yesterday






      • 3





        external link with audio examples: what does a linearly spaced sequence sound like vs. a logarithmic one

        – Cee McSharpface - it
        yesterday













      25














      25










      25









      The division of notes has to do with human perception and psychoacoustics. One description of human perception is the Weber-Fechner law, where a human will perceive equal changes in some sensory input, such as sound level or sound pitch, not by absolute level or value difference, but by the ratio of the change. e.g. larger values need a proportionately larger change for the change be perceived (if small) or perceived as about the same, within some reasonable range (e.g. audible, but not causing ear damage, etc.)



      Thus, for a semitone (forth, fifth, etc.) interval to sound the same, no matter which base note one starts from, in the equal temperament scale, the notes have to differ not by equal absolute frequency differences (as would be created by equal Hz deltas between notes), but by equal ratio differences (the 12th root of 2, so that twelve equal multiplications will equal one octave).



      e.g. the "equality" in equal division has to be in equal in ratio, not additive absolute value.






      share|improve this answer
















      The division of notes has to do with human perception and psychoacoustics. One description of human perception is the Weber-Fechner law, where a human will perceive equal changes in some sensory input, such as sound level or sound pitch, not by absolute level or value difference, but by the ratio of the change. e.g. larger values need a proportionately larger change for the change be perceived (if small) or perceived as about the same, within some reasonable range (e.g. audible, but not causing ear damage, etc.)



      Thus, for a semitone (forth, fifth, etc.) interval to sound the same, no matter which base note one starts from, in the equal temperament scale, the notes have to differ not by equal absolute frequency differences (as would be created by equal Hz deltas between notes), but by equal ratio differences (the 12th root of 2, so that twelve equal multiplications will equal one octave).



      e.g. the "equality" in equal division has to be in equal in ratio, not additive absolute value.







      share|improve this answer















      share|improve this answer




      share|improve this answer








      edited 2 days ago

























      answered 2 days ago









      hotpaw2hotpaw2

      1,2272 gold badges15 silver badges22 bronze badges




      1,2272 gold badges15 silver badges22 bronze badges










      • 7





        This answer nails the error in the thought behind the question — intervals are defined by human perception, and human perception of intervals is logarithmic as to frequency, not linear. Octave above 440 is 880. Octave below 440 is 220. Same with all the intervals — equal interval means same ratio of frequencies. Where it gets interesting is that perfect 4ths and 5ths are not equal to 5 and 7 well-tempered half-steps, respectively.

        – Jeff Y
        yesterday






      • 3





        external link with audio examples: what does a linearly spaced sequence sound like vs. a logarithmic one

        – Cee McSharpface - it
        yesterday












      • 7





        This answer nails the error in the thought behind the question — intervals are defined by human perception, and human perception of intervals is logarithmic as to frequency, not linear. Octave above 440 is 880. Octave below 440 is 220. Same with all the intervals — equal interval means same ratio of frequencies. Where it gets interesting is that perfect 4ths and 5ths are not equal to 5 and 7 well-tempered half-steps, respectively.

        – Jeff Y
        yesterday






      • 3





        external link with audio examples: what does a linearly spaced sequence sound like vs. a logarithmic one

        – Cee McSharpface - it
        yesterday







      7




      7





      This answer nails the error in the thought behind the question — intervals are defined by human perception, and human perception of intervals is logarithmic as to frequency, not linear. Octave above 440 is 880. Octave below 440 is 220. Same with all the intervals — equal interval means same ratio of frequencies. Where it gets interesting is that perfect 4ths and 5ths are not equal to 5 and 7 well-tempered half-steps, respectively.

      – Jeff Y
      yesterday





      This answer nails the error in the thought behind the question — intervals are defined by human perception, and human perception of intervals is logarithmic as to frequency, not linear. Octave above 440 is 880. Octave below 440 is 220. Same with all the intervals — equal interval means same ratio of frequencies. Where it gets interesting is that perfect 4ths and 5ths are not equal to 5 and 7 well-tempered half-steps, respectively.

      – Jeff Y
      yesterday




      3




      3





      external link with audio examples: what does a linearly spaced sequence sound like vs. a logarithmic one

      – Cee McSharpface - it
      yesterday





      external link with audio examples: what does a linearly spaced sequence sound like vs. a logarithmic one

      – Cee McSharpface - it
      yesterday











      17


















      What happens if you go down by the same steps:



      • 440Hz

      • 1 step down : 403.33Hz

      • 2 steps down : 366.67Hz

      • 3 steps down : 330.Hz

      • ...

      • 11 steps down : 36.67Hz

      • 12 steps down : 0Hz

      • 13 steps down : -36.67Hz

      So, using your "equally divided" logic, we are at zero Hz after 12 steps, and the next step beyond that is minus 37 Hz! What does that even mean? But ok, let's follow your logic a little bit ... what's the frequency exactly in the middle of the octave 440 - 880 Hz, that would be 660 Hz. What's an octave above that? That would be 2 * 660 Hz = 1320 Hz. What would be the steps in that octave - 660 Hz / 12 = 55 Hz? Ok, then let's take one step up from 660 Hz, that's 660 Hz + 55 Hz = 715 Hz. But wait ... the step was supposed to be 37 Hz, not 55 Hz??? Does your step size depend on the start and end points of the octave? Or does it take a sudden jump at 880 Hz - steps below 880 would be 440 / 12, but above 880 they would be 880 / 12? Where does such a divider come from, is it embedded in nature? I thought A = 440 Hz was only an agree-upon convention, not a law of nature.



      Where did you get the 880Hz? By multiplying by 2, i.e. one octave higher. I guess the same has to apply for any frequency, not only 440Hz? For example, one octave higher from 880Hz has to be 880Hz * 2? And any other frequency like 1000Hz... one octave above that must be 2000Hz. If the interval of an octave is calculated with multiplication, how could other intervals be calculated with addition?



      So, ask yourself: if F1 and F2 are the frequencies of two consecutive semitones, what is the relationship between F1 and F2, if (F1 * 2) and (F2 * 2) has to have the same relationship?



      You're looking for a function f(F) such that f
      applied 12 times gives 2*F.




      f(f(f(f(f(f(f(f(f(f(f(f(F)))))))))))) = 2 * F


      If you step up one semitone from F, you get a frequency f(F). The frequency one octave higher from that is 2 * f(F).



      If you first step up an octave, you get F*2. And if you step up one semitone from that, you get f(F*2), which should be the same frequency, so:




      2 * f(F) = f(2 * F)


      What might function f be like?



      From the subject line "why hertz differences are not the same but element 12th of two?" I assume that you already know that consecutive semitones have a ratio of 2^(1/12).






      share|improve this answer




























      • Not quite qhat OP means! 440>880 = 440, so each of the 12 steps should be 36.66Hz. In the lower octave, between 220 and 440, there's only 220 Hz difference, so each of the 12 steps should be 18.33 Hz.

        – Tim
        Oct 11 at 16:29






      • 3





        @Tim I think this is what the OP means. He means that he's trying to figure out things and wants help doing that. Looking at the subject line, he already has the solution, and he wants to find the right perspective for understanding why it is like it is. Which is what I'm trying to provide. YMMV but I was able to figure this out in high-school when writing a music player program, having only the information that an octave higher is multiplying by two, and applying the same relationship 12 times brings you there, an octave higher. :)

        – piiperi
        2 days ago












      • @Tim : Does this mean I get incompatible step sizes if I take my octaves to be C-to-C?

        – Eric Towers
        yesterday











      • Negative frequency means that the music is played backwards, revealing hidden Satanic messages.

        – dan04
        4 hours ago















      17


















      What happens if you go down by the same steps:



      • 440Hz

      • 1 step down : 403.33Hz

      • 2 steps down : 366.67Hz

      • 3 steps down : 330.Hz

      • ...

      • 11 steps down : 36.67Hz

      • 12 steps down : 0Hz

      • 13 steps down : -36.67Hz

      So, using your "equally divided" logic, we are at zero Hz after 12 steps, and the next step beyond that is minus 37 Hz! What does that even mean? But ok, let's follow your logic a little bit ... what's the frequency exactly in the middle of the octave 440 - 880 Hz, that would be 660 Hz. What's an octave above that? That would be 2 * 660 Hz = 1320 Hz. What would be the steps in that octave - 660 Hz / 12 = 55 Hz? Ok, then let's take one step up from 660 Hz, that's 660 Hz + 55 Hz = 715 Hz. But wait ... the step was supposed to be 37 Hz, not 55 Hz??? Does your step size depend on the start and end points of the octave? Or does it take a sudden jump at 880 Hz - steps below 880 would be 440 / 12, but above 880 they would be 880 / 12? Where does such a divider come from, is it embedded in nature? I thought A = 440 Hz was only an agree-upon convention, not a law of nature.



      Where did you get the 880Hz? By multiplying by 2, i.e. one octave higher. I guess the same has to apply for any frequency, not only 440Hz? For example, one octave higher from 880Hz has to be 880Hz * 2? And any other frequency like 1000Hz... one octave above that must be 2000Hz. If the interval of an octave is calculated with multiplication, how could other intervals be calculated with addition?



      So, ask yourself: if F1 and F2 are the frequencies of two consecutive semitones, what is the relationship between F1 and F2, if (F1 * 2) and (F2 * 2) has to have the same relationship?



      You're looking for a function f(F) such that f
      applied 12 times gives 2*F.




      f(f(f(f(f(f(f(f(f(f(f(f(F)))))))))))) = 2 * F


      If you step up one semitone from F, you get a frequency f(F). The frequency one octave higher from that is 2 * f(F).



      If you first step up an octave, you get F*2. And if you step up one semitone from that, you get f(F*2), which should be the same frequency, so:




      2 * f(F) = f(2 * F)


      What might function f be like?



      From the subject line "why hertz differences are not the same but element 12th of two?" I assume that you already know that consecutive semitones have a ratio of 2^(1/12).






      share|improve this answer




























      • Not quite qhat OP means! 440>880 = 440, so each of the 12 steps should be 36.66Hz. In the lower octave, between 220 and 440, there's only 220 Hz difference, so each of the 12 steps should be 18.33 Hz.

        – Tim
        Oct 11 at 16:29






      • 3





        @Tim I think this is what the OP means. He means that he's trying to figure out things and wants help doing that. Looking at the subject line, he already has the solution, and he wants to find the right perspective for understanding why it is like it is. Which is what I'm trying to provide. YMMV but I was able to figure this out in high-school when writing a music player program, having only the information that an octave higher is multiplying by two, and applying the same relationship 12 times brings you there, an octave higher. :)

        – piiperi
        2 days ago












      • @Tim : Does this mean I get incompatible step sizes if I take my octaves to be C-to-C?

        – Eric Towers
        yesterday











      • Negative frequency means that the music is played backwards, revealing hidden Satanic messages.

        – dan04
        4 hours ago













      17














      17










      17









      What happens if you go down by the same steps:



      • 440Hz

      • 1 step down : 403.33Hz

      • 2 steps down : 366.67Hz

      • 3 steps down : 330.Hz

      • ...

      • 11 steps down : 36.67Hz

      • 12 steps down : 0Hz

      • 13 steps down : -36.67Hz

      So, using your "equally divided" logic, we are at zero Hz after 12 steps, and the next step beyond that is minus 37 Hz! What does that even mean? But ok, let's follow your logic a little bit ... what's the frequency exactly in the middle of the octave 440 - 880 Hz, that would be 660 Hz. What's an octave above that? That would be 2 * 660 Hz = 1320 Hz. What would be the steps in that octave - 660 Hz / 12 = 55 Hz? Ok, then let's take one step up from 660 Hz, that's 660 Hz + 55 Hz = 715 Hz. But wait ... the step was supposed to be 37 Hz, not 55 Hz??? Does your step size depend on the start and end points of the octave? Or does it take a sudden jump at 880 Hz - steps below 880 would be 440 / 12, but above 880 they would be 880 / 12? Where does such a divider come from, is it embedded in nature? I thought A = 440 Hz was only an agree-upon convention, not a law of nature.



      Where did you get the 880Hz? By multiplying by 2, i.e. one octave higher. I guess the same has to apply for any frequency, not only 440Hz? For example, one octave higher from 880Hz has to be 880Hz * 2? And any other frequency like 1000Hz... one octave above that must be 2000Hz. If the interval of an octave is calculated with multiplication, how could other intervals be calculated with addition?



      So, ask yourself: if F1 and F2 are the frequencies of two consecutive semitones, what is the relationship between F1 and F2, if (F1 * 2) and (F2 * 2) has to have the same relationship?



      You're looking for a function f(F) such that f
      applied 12 times gives 2*F.




      f(f(f(f(f(f(f(f(f(f(f(f(F)))))))))))) = 2 * F


      If you step up one semitone from F, you get a frequency f(F). The frequency one octave higher from that is 2 * f(F).



      If you first step up an octave, you get F*2. And if you step up one semitone from that, you get f(F*2), which should be the same frequency, so:




      2 * f(F) = f(2 * F)


      What might function f be like?



      From the subject line "why hertz differences are not the same but element 12th of two?" I assume that you already know that consecutive semitones have a ratio of 2^(1/12).






      share|improve this answer
















      What happens if you go down by the same steps:



      • 440Hz

      • 1 step down : 403.33Hz

      • 2 steps down : 366.67Hz

      • 3 steps down : 330.Hz

      • ...

      • 11 steps down : 36.67Hz

      • 12 steps down : 0Hz

      • 13 steps down : -36.67Hz

      So, using your "equally divided" logic, we are at zero Hz after 12 steps, and the next step beyond that is minus 37 Hz! What does that even mean? But ok, let's follow your logic a little bit ... what's the frequency exactly in the middle of the octave 440 - 880 Hz, that would be 660 Hz. What's an octave above that? That would be 2 * 660 Hz = 1320 Hz. What would be the steps in that octave - 660 Hz / 12 = 55 Hz? Ok, then let's take one step up from 660 Hz, that's 660 Hz + 55 Hz = 715 Hz. But wait ... the step was supposed to be 37 Hz, not 55 Hz??? Does your step size depend on the start and end points of the octave? Or does it take a sudden jump at 880 Hz - steps below 880 would be 440 / 12, but above 880 they would be 880 / 12? Where does such a divider come from, is it embedded in nature? I thought A = 440 Hz was only an agree-upon convention, not a law of nature.



      Where did you get the 880Hz? By multiplying by 2, i.e. one octave higher. I guess the same has to apply for any frequency, not only 440Hz? For example, one octave higher from 880Hz has to be 880Hz * 2? And any other frequency like 1000Hz... one octave above that must be 2000Hz. If the interval of an octave is calculated with multiplication, how could other intervals be calculated with addition?



      So, ask yourself: if F1 and F2 are the frequencies of two consecutive semitones, what is the relationship between F1 and F2, if (F1 * 2) and (F2 * 2) has to have the same relationship?



      You're looking for a function f(F) such that f
      applied 12 times gives 2*F.




      f(f(f(f(f(f(f(f(f(f(f(f(F)))))))))))) = 2 * F


      If you step up one semitone from F, you get a frequency f(F). The frequency one octave higher from that is 2 * f(F).



      If you first step up an octave, you get F*2. And if you step up one semitone from that, you get f(F*2), which should be the same frequency, so:




      2 * f(F) = f(2 * F)


      What might function f be like?



      From the subject line "why hertz differences are not the same but element 12th of two?" I assume that you already know that consecutive semitones have a ratio of 2^(1/12).







      share|improve this answer















      share|improve this answer




      share|improve this answer








      edited yesterday

























      answered Oct 11 at 15:40









      piiperipiiperi

      6,6031 gold badge9 silver badges26 bronze badges




      6,6031 gold badge9 silver badges26 bronze badges















      • Not quite qhat OP means! 440>880 = 440, so each of the 12 steps should be 36.66Hz. In the lower octave, between 220 and 440, there's only 220 Hz difference, so each of the 12 steps should be 18.33 Hz.

        – Tim
        Oct 11 at 16:29






      • 3





        @Tim I think this is what the OP means. He means that he's trying to figure out things and wants help doing that. Looking at the subject line, he already has the solution, and he wants to find the right perspective for understanding why it is like it is. Which is what I'm trying to provide. YMMV but I was able to figure this out in high-school when writing a music player program, having only the information that an octave higher is multiplying by two, and applying the same relationship 12 times brings you there, an octave higher. :)

        – piiperi
        2 days ago












      • @Tim : Does this mean I get incompatible step sizes if I take my octaves to be C-to-C?

        – Eric Towers
        yesterday











      • Negative frequency means that the music is played backwards, revealing hidden Satanic messages.

        – dan04
        4 hours ago

















      • Not quite qhat OP means! 440>880 = 440, so each of the 12 steps should be 36.66Hz. In the lower octave, between 220 and 440, there's only 220 Hz difference, so each of the 12 steps should be 18.33 Hz.

        – Tim
        Oct 11 at 16:29






      • 3





        @Tim I think this is what the OP means. He means that he's trying to figure out things and wants help doing that. Looking at the subject line, he already has the solution, and he wants to find the right perspective for understanding why it is like it is. Which is what I'm trying to provide. YMMV but I was able to figure this out in high-school when writing a music player program, having only the information that an octave higher is multiplying by two, and applying the same relationship 12 times brings you there, an octave higher. :)

        – piiperi
        2 days ago












      • @Tim : Does this mean I get incompatible step sizes if I take my octaves to be C-to-C?

        – Eric Towers
        yesterday











      • Negative frequency means that the music is played backwards, revealing hidden Satanic messages.

        – dan04
        4 hours ago
















      Not quite qhat OP means! 440>880 = 440, so each of the 12 steps should be 36.66Hz. In the lower octave, between 220 and 440, there's only 220 Hz difference, so each of the 12 steps should be 18.33 Hz.

      – Tim
      Oct 11 at 16:29





      Not quite qhat OP means! 440>880 = 440, so each of the 12 steps should be 36.66Hz. In the lower octave, between 220 and 440, there's only 220 Hz difference, so each of the 12 steps should be 18.33 Hz.

      – Tim
      Oct 11 at 16:29




      3




      3





      @Tim I think this is what the OP means. He means that he's trying to figure out things and wants help doing that. Looking at the subject line, he already has the solution, and he wants to find the right perspective for understanding why it is like it is. Which is what I'm trying to provide. YMMV but I was able to figure this out in high-school when writing a music player program, having only the information that an octave higher is multiplying by two, and applying the same relationship 12 times brings you there, an octave higher. :)

      – piiperi
      2 days ago






      @Tim I think this is what the OP means. He means that he's trying to figure out things and wants help doing that. Looking at the subject line, he already has the solution, and he wants to find the right perspective for understanding why it is like it is. Which is what I'm trying to provide. YMMV but I was able to figure this out in high-school when writing a music player program, having only the information that an octave higher is multiplying by two, and applying the same relationship 12 times brings you there, an octave higher. :)

      – piiperi
      2 days ago














      @Tim : Does this mean I get incompatible step sizes if I take my octaves to be C-to-C?

      – Eric Towers
      yesterday





      @Tim : Does this mean I get incompatible step sizes if I take my octaves to be C-to-C?

      – Eric Towers
      yesterday













      Negative frequency means that the music is played backwards, revealing hidden Satanic messages.

      – dan04
      4 hours ago





      Negative frequency means that the music is played backwards, revealing hidden Satanic messages.

      – dan04
      4 hours ago











      6


















      A simple way is to look at ratios as suggested above. One can divide an interval equally arithmetically such that the length (size, or more technically "measure") of each subinterval is identical. Dividing an interval arithmetically in 12 pieces (I can explain the 12 but it takes more math.) yields, 1=12/12, 13/12, 14/12, 15/12, 16/12, 17/12, 18/12, 19/12, 20/12, 21/12, 22/12, 22/12, 24/12=2. However people's hearing seems (experimentally) to distinguish ratios of frequencies rather than differences as being more identical. For example (Taking A=440cps), the fifth above A is E at 660cps not 19/12*440=696.666....



      If we want equal ratios for each half-step, instead of (2-1)/12, we 2^(1/12). The point is that the ratio of G to C is constant for all fifths (A-D, C-F, etc.). Since antiquity, the ratio of a fifth is 3:2 (or 3/2 times the frequency of the lower note.) This comports with dividing a string into intervals and listening to the frequency of the two shorter pieces. (Aside: Vincenzo Galilei suggested using 18/17 as a approximation to the twelfth root of two; it's remarkably good.)



      However: for computational work, we can use logarithms; the logarithm of a ratio is the difference in the logarithms of that ratio's constituents. One divides the octave into 1200 cents (the 1200th root of 2) and assigns 100 cents to the equal-tempered semitone. This allows one to easily (at least when using pencil and paper instead of a calculator) compute interval sized for varying tunings.



      So, even though our ears hear by ratio (experimentally), we can calculate by ratios or addition. Wiki has a bunch of articles qG (quod Google in analogy with qv) which give a more comprehensive explanation.






      share|improve this answer






























        6


















        A simple way is to look at ratios as suggested above. One can divide an interval equally arithmetically such that the length (size, or more technically "measure") of each subinterval is identical. Dividing an interval arithmetically in 12 pieces (I can explain the 12 but it takes more math.) yields, 1=12/12, 13/12, 14/12, 15/12, 16/12, 17/12, 18/12, 19/12, 20/12, 21/12, 22/12, 22/12, 24/12=2. However people's hearing seems (experimentally) to distinguish ratios of frequencies rather than differences as being more identical. For example (Taking A=440cps), the fifth above A is E at 660cps not 19/12*440=696.666....



        If we want equal ratios for each half-step, instead of (2-1)/12, we 2^(1/12). The point is that the ratio of G to C is constant for all fifths (A-D, C-F, etc.). Since antiquity, the ratio of a fifth is 3:2 (or 3/2 times the frequency of the lower note.) This comports with dividing a string into intervals and listening to the frequency of the two shorter pieces. (Aside: Vincenzo Galilei suggested using 18/17 as a approximation to the twelfth root of two; it's remarkably good.)



        However: for computational work, we can use logarithms; the logarithm of a ratio is the difference in the logarithms of that ratio's constituents. One divides the octave into 1200 cents (the 1200th root of 2) and assigns 100 cents to the equal-tempered semitone. This allows one to easily (at least when using pencil and paper instead of a calculator) compute interval sized for varying tunings.



        So, even though our ears hear by ratio (experimentally), we can calculate by ratios or addition. Wiki has a bunch of articles qG (quod Google in analogy with qv) which give a more comprehensive explanation.






        share|improve this answer




























          6














          6










          6









          A simple way is to look at ratios as suggested above. One can divide an interval equally arithmetically such that the length (size, or more technically "measure") of each subinterval is identical. Dividing an interval arithmetically in 12 pieces (I can explain the 12 but it takes more math.) yields, 1=12/12, 13/12, 14/12, 15/12, 16/12, 17/12, 18/12, 19/12, 20/12, 21/12, 22/12, 22/12, 24/12=2. However people's hearing seems (experimentally) to distinguish ratios of frequencies rather than differences as being more identical. For example (Taking A=440cps), the fifth above A is E at 660cps not 19/12*440=696.666....



          If we want equal ratios for each half-step, instead of (2-1)/12, we 2^(1/12). The point is that the ratio of G to C is constant for all fifths (A-D, C-F, etc.). Since antiquity, the ratio of a fifth is 3:2 (or 3/2 times the frequency of the lower note.) This comports with dividing a string into intervals and listening to the frequency of the two shorter pieces. (Aside: Vincenzo Galilei suggested using 18/17 as a approximation to the twelfth root of two; it's remarkably good.)



          However: for computational work, we can use logarithms; the logarithm of a ratio is the difference in the logarithms of that ratio's constituents. One divides the octave into 1200 cents (the 1200th root of 2) and assigns 100 cents to the equal-tempered semitone. This allows one to easily (at least when using pencil and paper instead of a calculator) compute interval sized for varying tunings.



          So, even though our ears hear by ratio (experimentally), we can calculate by ratios or addition. Wiki has a bunch of articles qG (quod Google in analogy with qv) which give a more comprehensive explanation.






          share|improve this answer














          A simple way is to look at ratios as suggested above. One can divide an interval equally arithmetically such that the length (size, or more technically "measure") of each subinterval is identical. Dividing an interval arithmetically in 12 pieces (I can explain the 12 but it takes more math.) yields, 1=12/12, 13/12, 14/12, 15/12, 16/12, 17/12, 18/12, 19/12, 20/12, 21/12, 22/12, 22/12, 24/12=2. However people's hearing seems (experimentally) to distinguish ratios of frequencies rather than differences as being more identical. For example (Taking A=440cps), the fifth above A is E at 660cps not 19/12*440=696.666....



          If we want equal ratios for each half-step, instead of (2-1)/12, we 2^(1/12). The point is that the ratio of G to C is constant for all fifths (A-D, C-F, etc.). Since antiquity, the ratio of a fifth is 3:2 (or 3/2 times the frequency of the lower note.) This comports with dividing a string into intervals and listening to the frequency of the two shorter pieces. (Aside: Vincenzo Galilei suggested using 18/17 as a approximation to the twelfth root of two; it's remarkably good.)



          However: for computational work, we can use logarithms; the logarithm of a ratio is the difference in the logarithms of that ratio's constituents. One divides the octave into 1200 cents (the 1200th root of 2) and assigns 100 cents to the equal-tempered semitone. This allows one to easily (at least when using pencil and paper instead of a calculator) compute interval sized for varying tunings.



          So, even though our ears hear by ratio (experimentally), we can calculate by ratios or addition. Wiki has a bunch of articles qG (quod Google in analogy with qv) which give a more comprehensive explanation.







          share|improve this answer













          share|improve this answer




          share|improve this answer










          answered 2 days ago









          ttwttw

          11.8k13 silver badges42 bronze badges




          11.8k13 silver badges42 bronze badges
























              4


















              Possibly a simple way to look at it is to look at a guitar neck. An octave there is divided into 12 parts - equal as far as each fret is a semitone away from its neighbour. But looking carefully, it's fairly obvious that each fret isn't the same size. In fact, the eleventh fret is very nearly half the size of the first one, from nut to fret 1. Go further, and the 12th (octave) is actually half the size of the first.



              Your hypothesis is that they would all be the same size - one twelfth of the half length of the open string? Were that the case, what woud happen at fret 13? And apart, each fret would produce a note that was out of tune. So there needs to be a ratio of each fret against its neighbour, as pointed out in other good answers.






              share|improve this answer





















              • 4





                @AlbrechtHügli The frets on a guitar neck illustrate the physical relationship between the notes of an equal-tempered scale, but they don't explain it. The explanation goes in the opposite direction: The scale explains the spacing of the frets.

                – Solomon Slow
                2 days ago











              • I agree, but it is a good analogy and shows that the differences of the steps aren't continuous.

                – Albrecht Hügli
                2 days ago















              4


















              Possibly a simple way to look at it is to look at a guitar neck. An octave there is divided into 12 parts - equal as far as each fret is a semitone away from its neighbour. But looking carefully, it's fairly obvious that each fret isn't the same size. In fact, the eleventh fret is very nearly half the size of the first one, from nut to fret 1. Go further, and the 12th (octave) is actually half the size of the first.



              Your hypothesis is that they would all be the same size - one twelfth of the half length of the open string? Were that the case, what woud happen at fret 13? And apart, each fret would produce a note that was out of tune. So there needs to be a ratio of each fret against its neighbour, as pointed out in other good answers.






              share|improve this answer





















              • 4





                @AlbrechtHügli The frets on a guitar neck illustrate the physical relationship between the notes of an equal-tempered scale, but they don't explain it. The explanation goes in the opposite direction: The scale explains the spacing of the frets.

                – Solomon Slow
                2 days ago











              • I agree, but it is a good analogy and shows that the differences of the steps aren't continuous.

                – Albrecht Hügli
                2 days ago













              4














              4










              4









              Possibly a simple way to look at it is to look at a guitar neck. An octave there is divided into 12 parts - equal as far as each fret is a semitone away from its neighbour. But looking carefully, it's fairly obvious that each fret isn't the same size. In fact, the eleventh fret is very nearly half the size of the first one, from nut to fret 1. Go further, and the 12th (octave) is actually half the size of the first.



              Your hypothesis is that they would all be the same size - one twelfth of the half length of the open string? Were that the case, what woud happen at fret 13? And apart, each fret would produce a note that was out of tune. So there needs to be a ratio of each fret against its neighbour, as pointed out in other good answers.






              share|improve this answer














              Possibly a simple way to look at it is to look at a guitar neck. An octave there is divided into 12 parts - equal as far as each fret is a semitone away from its neighbour. But looking carefully, it's fairly obvious that each fret isn't the same size. In fact, the eleventh fret is very nearly half the size of the first one, from nut to fret 1. Go further, and the 12th (octave) is actually half the size of the first.



              Your hypothesis is that they would all be the same size - one twelfth of the half length of the open string? Were that the case, what woud happen at fret 13? And apart, each fret would produce a note that was out of tune. So there needs to be a ratio of each fret against its neighbour, as pointed out in other good answers.







              share|improve this answer













              share|improve this answer




              share|improve this answer










              answered 2 days ago









              TimTim

              116k12 gold badges113 silver badges290 bronze badges




              116k12 gold badges113 silver badges290 bronze badges










              • 4





                @AlbrechtHügli The frets on a guitar neck illustrate the physical relationship between the notes of an equal-tempered scale, but they don't explain it. The explanation goes in the opposite direction: The scale explains the spacing of the frets.

                – Solomon Slow
                2 days ago











              • I agree, but it is a good analogy and shows that the differences of the steps aren't continuous.

                – Albrecht Hügli
                2 days ago












              • 4





                @AlbrechtHügli The frets on a guitar neck illustrate the physical relationship between the notes of an equal-tempered scale, but they don't explain it. The explanation goes in the opposite direction: The scale explains the spacing of the frets.

                – Solomon Slow
                2 days ago











              • I agree, but it is a good analogy and shows that the differences of the steps aren't continuous.

                – Albrecht Hügli
                2 days ago







              4




              4





              @AlbrechtHügli The frets on a guitar neck illustrate the physical relationship between the notes of an equal-tempered scale, but they don't explain it. The explanation goes in the opposite direction: The scale explains the spacing of the frets.

              – Solomon Slow
              2 days ago





              @AlbrechtHügli The frets on a guitar neck illustrate the physical relationship between the notes of an equal-tempered scale, but they don't explain it. The explanation goes in the opposite direction: The scale explains the spacing of the frets.

              – Solomon Slow
              2 days ago













              I agree, but it is a good analogy and shows that the differences of the steps aren't continuous.

              – Albrecht Hügli
              2 days ago





              I agree, but it is a good analogy and shows that the differences of the steps aren't continuous.

              – Albrecht Hügli
              2 days ago











              4


















              Start by considering the equal division of octaves into one part. That is, think about changing pitch by octaves only.



              If we start with A1=55 Hz, we have the following pitches:




              Pitch Frequency
              ----------------
              A1 55 Hz
              A2 110 Hz
              A3 220 Hz
              A4 440 Hz
              A5 880 Hz
              ...


              You can see that when you increase the pitch by an equal additive amount, you increase the frequency by an equal multiplicative factor. That is, every time you increase the pitch by one octave, you double the frequency. This means that the relationship between pitch and frequency is logarithmic.



              From there, it is fairly easy to reach the conclusion that to divide the octave into some number of equal parts, you need to find the factor that, when multiplied by itself that number times, yields 2. In other words, the frequency factor corresponding to a division of the octave into n parts is the nth root of 2.






              share|improve this answer




























              • Stackexchange bug, is anyone else seeing this? The contents of the table disappear after page load.

                – whatsisname
                yesterday











              • @whatsisname yes, I'm seeing it inconsistently, too. I'll try modifying the table to see if I can work around it.

                – phoog
                yesterday















              4


















              Start by considering the equal division of octaves into one part. That is, think about changing pitch by octaves only.



              If we start with A1=55 Hz, we have the following pitches:




              Pitch Frequency
              ----------------
              A1 55 Hz
              A2 110 Hz
              A3 220 Hz
              A4 440 Hz
              A5 880 Hz
              ...


              You can see that when you increase the pitch by an equal additive amount, you increase the frequency by an equal multiplicative factor. That is, every time you increase the pitch by one octave, you double the frequency. This means that the relationship between pitch and frequency is logarithmic.



              From there, it is fairly easy to reach the conclusion that to divide the octave into some number of equal parts, you need to find the factor that, when multiplied by itself that number times, yields 2. In other words, the frequency factor corresponding to a division of the octave into n parts is the nth root of 2.






              share|improve this answer




























              • Stackexchange bug, is anyone else seeing this? The contents of the table disappear after page load.

                – whatsisname
                yesterday











              • @whatsisname yes, I'm seeing it inconsistently, too. I'll try modifying the table to see if I can work around it.

                – phoog
                yesterday













              4














              4










              4









              Start by considering the equal division of octaves into one part. That is, think about changing pitch by octaves only.



              If we start with A1=55 Hz, we have the following pitches:




              Pitch Frequency
              ----------------
              A1 55 Hz
              A2 110 Hz
              A3 220 Hz
              A4 440 Hz
              A5 880 Hz
              ...


              You can see that when you increase the pitch by an equal additive amount, you increase the frequency by an equal multiplicative factor. That is, every time you increase the pitch by one octave, you double the frequency. This means that the relationship between pitch and frequency is logarithmic.



              From there, it is fairly easy to reach the conclusion that to divide the octave into some number of equal parts, you need to find the factor that, when multiplied by itself that number times, yields 2. In other words, the frequency factor corresponding to a division of the octave into n parts is the nth root of 2.






              share|improve this answer
















              Start by considering the equal division of octaves into one part. That is, think about changing pitch by octaves only.



              If we start with A1=55 Hz, we have the following pitches:




              Pitch Frequency
              ----------------
              A1 55 Hz
              A2 110 Hz
              A3 220 Hz
              A4 440 Hz
              A5 880 Hz
              ...


              You can see that when you increase the pitch by an equal additive amount, you increase the frequency by an equal multiplicative factor. That is, every time you increase the pitch by one octave, you double the frequency. This means that the relationship between pitch and frequency is logarithmic.



              From there, it is fairly easy to reach the conclusion that to divide the octave into some number of equal parts, you need to find the factor that, when multiplied by itself that number times, yields 2. In other words, the frequency factor corresponding to a division of the octave into n parts is the nth root of 2.







              share|improve this answer















              share|improve this answer




              share|improve this answer








              edited 20 hours ago









              Dom

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              38.5k20 gold badges117 silver badges238 bronze badges










              answered yesterday









              phoogphoog

              2,0608 silver badges12 bronze badges




              2,0608 silver badges12 bronze badges















              • Stackexchange bug, is anyone else seeing this? The contents of the table disappear after page load.

                – whatsisname
                yesterday











              • @whatsisname yes, I'm seeing it inconsistently, too. I'll try modifying the table to see if I can work around it.

                – phoog
                yesterday

















              • Stackexchange bug, is anyone else seeing this? The contents of the table disappear after page load.

                – whatsisname
                yesterday











              • @whatsisname yes, I'm seeing it inconsistently, too. I'll try modifying the table to see if I can work around it.

                – phoog
                yesterday
















              Stackexchange bug, is anyone else seeing this? The contents of the table disappear after page load.

              – whatsisname
              yesterday





              Stackexchange bug, is anyone else seeing this? The contents of the table disappear after page load.

              – whatsisname
              yesterday













              @whatsisname yes, I'm seeing it inconsistently, too. I'll try modifying the table to see if I can work around it.

              – phoog
              yesterday





              @whatsisname yes, I'm seeing it inconsistently, too. I'll try modifying the table to see if I can work around it.

              – phoog
              yesterday











              3


















              Our note system is a logarithmic scale for frequency. A logarithmic scale turns equal fractions into equal distances. You can define equal temperament as a constant step size of 1/12 on the log_2 scale of frequency.



              Going back to the linear scale, this means that a semitone translates into a factor of 2^(1/12) (the twelfth root of two).




              The reason for this is, that the sound of an interval depends on how the overtone spectra of the two nodes match up.



              The octave has the unique feature that all harmonics of the higher note match up with some harmonic of the lower note. Likewise, if you have a perfect fifth (factor 3/2), every second harmonic of the upper note coincides with every third harmonic of the lower note. Similar relations hold for the perfect fourth (factor 4/3), the mayor third (5/4) and the mayor sixth (5/3). And so on, and so forth. The pattern of how the harmonics match up defines the sound of the interval, and the harmonics are defined by factors of frequency.



              Thus, only a logarithmic scale can be used to describe intervals well (our note system). And by consequence, equal temperament must be defined on the logarithmic scale.






              share|improve this answer





















              • 2





                Good idea to talk about log scale. You could even show one graph with equal spacing on a log scale and one with uneven spacing on a linear scale, like a guitar neck.

                – Eric Duminil
                yesterday






              • 1





                The pitch-class equivalence of octaves is not related to overtones. It is apparent even in sine waves, which have no overtones. Conversely, a "real life" pitch of 220 Hz has an overtone at 660 Hz, yet 660 Hz isn't an equivalent pitch class to 220 Hz.

                – phoog
                yesterday






              • 2





                @phoog 660 Hz is one octave + one fifth above 220 Hz, and it does blend excellently. You even have a stop for this interval in many pipe organs because it blends so well. The organist uses this stop to change the sound of the fundamental stop, not to get a transposition. Also, did you ever try to tune two sines to an octave? You can do it if you run the signal through a distorting guitar amp (at the very least this adds the frequencies f1-f0 and f0+f1 to the signal), but I know that I won't be able to do it precisely without technological help.

                – cmaster
                yesterday











              • @cmaster of course it blends excellently, but it isn't equivalent in the way that 440 and 880 are. They're both A, but 660 and 1980 are E and B. Three instruments playing pitches in parallel separated by factors of two sound more unified than three instruments separated by factors of three, even if the instruments are sine wave generators (relatively easy to achieve with waveform synthesis or Hammond organs). Pipe organ stops for non-octave harmonics only blend in certain registrations. A 5 1/3' stop with a single 8' stop is likely to sound like parallel fifths rather than a richer tone.

                – phoog
                yesterday












              • @phoog For a 5 1/3' stop you need the 16' stop to blend. And, yes, a 3 1/5' stop and a 5 1/3' stop blend perfectly with a 16' stop. What makes it difficult to use that combination, is that stops are usually either on or off with nothing in between, and the high stops are just too loud in the combination. I once played an electronic organ that allowed partly pulling a stop, allowing me to reduce the volume of the high stops, and I used that combination quite frequently for its nice sound. With only on/off stops, you'd need to pull some more 16' stops to get the right relative volume.

                – cmaster
                yesterday















              3


















              Our note system is a logarithmic scale for frequency. A logarithmic scale turns equal fractions into equal distances. You can define equal temperament as a constant step size of 1/12 on the log_2 scale of frequency.



              Going back to the linear scale, this means that a semitone translates into a factor of 2^(1/12) (the twelfth root of two).




              The reason for this is, that the sound of an interval depends on how the overtone spectra of the two nodes match up.



              The octave has the unique feature that all harmonics of the higher note match up with some harmonic of the lower note. Likewise, if you have a perfect fifth (factor 3/2), every second harmonic of the upper note coincides with every third harmonic of the lower note. Similar relations hold for the perfect fourth (factor 4/3), the mayor third (5/4) and the mayor sixth (5/3). And so on, and so forth. The pattern of how the harmonics match up defines the sound of the interval, and the harmonics are defined by factors of frequency.



              Thus, only a logarithmic scale can be used to describe intervals well (our note system). And by consequence, equal temperament must be defined on the logarithmic scale.






              share|improve this answer





















              • 2





                Good idea to talk about log scale. You could even show one graph with equal spacing on a log scale and one with uneven spacing on a linear scale, like a guitar neck.

                – Eric Duminil
                yesterday






              • 1





                The pitch-class equivalence of octaves is not related to overtones. It is apparent even in sine waves, which have no overtones. Conversely, a "real life" pitch of 220 Hz has an overtone at 660 Hz, yet 660 Hz isn't an equivalent pitch class to 220 Hz.

                – phoog
                yesterday






              • 2





                @phoog 660 Hz is one octave + one fifth above 220 Hz, and it does blend excellently. You even have a stop for this interval in many pipe organs because it blends so well. The organist uses this stop to change the sound of the fundamental stop, not to get a transposition. Also, did you ever try to tune two sines to an octave? You can do it if you run the signal through a distorting guitar amp (at the very least this adds the frequencies f1-f0 and f0+f1 to the signal), but I know that I won't be able to do it precisely without technological help.

                – cmaster
                yesterday











              • @cmaster of course it blends excellently, but it isn't equivalent in the way that 440 and 880 are. They're both A, but 660 and 1980 are E and B. Three instruments playing pitches in parallel separated by factors of two sound more unified than three instruments separated by factors of three, even if the instruments are sine wave generators (relatively easy to achieve with waveform synthesis or Hammond organs). Pipe organ stops for non-octave harmonics only blend in certain registrations. A 5 1/3' stop with a single 8' stop is likely to sound like parallel fifths rather than a richer tone.

                – phoog
                yesterday












              • @phoog For a 5 1/3' stop you need the 16' stop to blend. And, yes, a 3 1/5' stop and a 5 1/3' stop blend perfectly with a 16' stop. What makes it difficult to use that combination, is that stops are usually either on or off with nothing in between, and the high stops are just too loud in the combination. I once played an electronic organ that allowed partly pulling a stop, allowing me to reduce the volume of the high stops, and I used that combination quite frequently for its nice sound. With only on/off stops, you'd need to pull some more 16' stops to get the right relative volume.

                – cmaster
                yesterday













              3














              3










              3









              Our note system is a logarithmic scale for frequency. A logarithmic scale turns equal fractions into equal distances. You can define equal temperament as a constant step size of 1/12 on the log_2 scale of frequency.



              Going back to the linear scale, this means that a semitone translates into a factor of 2^(1/12) (the twelfth root of two).




              The reason for this is, that the sound of an interval depends on how the overtone spectra of the two nodes match up.



              The octave has the unique feature that all harmonics of the higher note match up with some harmonic of the lower note. Likewise, if you have a perfect fifth (factor 3/2), every second harmonic of the upper note coincides with every third harmonic of the lower note. Similar relations hold for the perfect fourth (factor 4/3), the mayor third (5/4) and the mayor sixth (5/3). And so on, and so forth. The pattern of how the harmonics match up defines the sound of the interval, and the harmonics are defined by factors of frequency.



              Thus, only a logarithmic scale can be used to describe intervals well (our note system). And by consequence, equal temperament must be defined on the logarithmic scale.






              share|improve this answer














              Our note system is a logarithmic scale for frequency. A logarithmic scale turns equal fractions into equal distances. You can define equal temperament as a constant step size of 1/12 on the log_2 scale of frequency.



              Going back to the linear scale, this means that a semitone translates into a factor of 2^(1/12) (the twelfth root of two).




              The reason for this is, that the sound of an interval depends on how the overtone spectra of the two nodes match up.



              The octave has the unique feature that all harmonics of the higher note match up with some harmonic of the lower note. Likewise, if you have a perfect fifth (factor 3/2), every second harmonic of the upper note coincides with every third harmonic of the lower note. Similar relations hold for the perfect fourth (factor 4/3), the mayor third (5/4) and the mayor sixth (5/3). And so on, and so forth. The pattern of how the harmonics match up defines the sound of the interval, and the harmonics are defined by factors of frequency.



              Thus, only a logarithmic scale can be used to describe intervals well (our note system). And by consequence, equal temperament must be defined on the logarithmic scale.







              share|improve this answer













              share|improve this answer




              share|improve this answer










              answered yesterday









              cmastercmaster

              2475 bronze badges




              2475 bronze badges










              • 2





                Good idea to talk about log scale. You could even show one graph with equal spacing on a log scale and one with uneven spacing on a linear scale, like a guitar neck.

                – Eric Duminil
                yesterday






              • 1





                The pitch-class equivalence of octaves is not related to overtones. It is apparent even in sine waves, which have no overtones. Conversely, a "real life" pitch of 220 Hz has an overtone at 660 Hz, yet 660 Hz isn't an equivalent pitch class to 220 Hz.

                – phoog
                yesterday






              • 2





                @phoog 660 Hz is one octave + one fifth above 220 Hz, and it does blend excellently. You even have a stop for this interval in many pipe organs because it blends so well. The organist uses this stop to change the sound of the fundamental stop, not to get a transposition. Also, did you ever try to tune two sines to an octave? You can do it if you run the signal through a distorting guitar amp (at the very least this adds the frequencies f1-f0 and f0+f1 to the signal), but I know that I won't be able to do it precisely without technological help.

                – cmaster
                yesterday











              • @cmaster of course it blends excellently, but it isn't equivalent in the way that 440 and 880 are. They're both A, but 660 and 1980 are E and B. Three instruments playing pitches in parallel separated by factors of two sound more unified than three instruments separated by factors of three, even if the instruments are sine wave generators (relatively easy to achieve with waveform synthesis or Hammond organs). Pipe organ stops for non-octave harmonics only blend in certain registrations. A 5 1/3' stop with a single 8' stop is likely to sound like parallel fifths rather than a richer tone.

                – phoog
                yesterday












              • @phoog For a 5 1/3' stop you need the 16' stop to blend. And, yes, a 3 1/5' stop and a 5 1/3' stop blend perfectly with a 16' stop. What makes it difficult to use that combination, is that stops are usually either on or off with nothing in between, and the high stops are just too loud in the combination. I once played an electronic organ that allowed partly pulling a stop, allowing me to reduce the volume of the high stops, and I used that combination quite frequently for its nice sound. With only on/off stops, you'd need to pull some more 16' stops to get the right relative volume.

                – cmaster
                yesterday












              • 2





                Good idea to talk about log scale. You could even show one graph with equal spacing on a log scale and one with uneven spacing on a linear scale, like a guitar neck.

                – Eric Duminil
                yesterday






              • 1





                The pitch-class equivalence of octaves is not related to overtones. It is apparent even in sine waves, which have no overtones. Conversely, a "real life" pitch of 220 Hz has an overtone at 660 Hz, yet 660 Hz isn't an equivalent pitch class to 220 Hz.

                – phoog
                yesterday






              • 2





                @phoog 660 Hz is one octave + one fifth above 220 Hz, and it does blend excellently. You even have a stop for this interval in many pipe organs because it blends so well. The organist uses this stop to change the sound of the fundamental stop, not to get a transposition. Also, did you ever try to tune two sines to an octave? You can do it if you run the signal through a distorting guitar amp (at the very least this adds the frequencies f1-f0 and f0+f1 to the signal), but I know that I won't be able to do it precisely without technological help.

                – cmaster
                yesterday











              • @cmaster of course it blends excellently, but it isn't equivalent in the way that 440 and 880 are. They're both A, but 660 and 1980 are E and B. Three instruments playing pitches in parallel separated by factors of two sound more unified than three instruments separated by factors of three, even if the instruments are sine wave generators (relatively easy to achieve with waveform synthesis or Hammond organs). Pipe organ stops for non-octave harmonics only blend in certain registrations. A 5 1/3' stop with a single 8' stop is likely to sound like parallel fifths rather than a richer tone.

                – phoog
                yesterday












              • @phoog For a 5 1/3' stop you need the 16' stop to blend. And, yes, a 3 1/5' stop and a 5 1/3' stop blend perfectly with a 16' stop. What makes it difficult to use that combination, is that stops are usually either on or off with nothing in between, and the high stops are just too loud in the combination. I once played an electronic organ that allowed partly pulling a stop, allowing me to reduce the volume of the high stops, and I used that combination quite frequently for its nice sound. With only on/off stops, you'd need to pull some more 16' stops to get the right relative volume.

                – cmaster
                yesterday







              2




              2





              Good idea to talk about log scale. You could even show one graph with equal spacing on a log scale and one with uneven spacing on a linear scale, like a guitar neck.

              – Eric Duminil
              yesterday





              Good idea to talk about log scale. You could even show one graph with equal spacing on a log scale and one with uneven spacing on a linear scale, like a guitar neck.

              – Eric Duminil
              yesterday




              1




              1





              The pitch-class equivalence of octaves is not related to overtones. It is apparent even in sine waves, which have no overtones. Conversely, a "real life" pitch of 220 Hz has an overtone at 660 Hz, yet 660 Hz isn't an equivalent pitch class to 220 Hz.

              – phoog
              yesterday





              The pitch-class equivalence of octaves is not related to overtones. It is apparent even in sine waves, which have no overtones. Conversely, a "real life" pitch of 220 Hz has an overtone at 660 Hz, yet 660 Hz isn't an equivalent pitch class to 220 Hz.

              – phoog
              yesterday




              2




              2





              @phoog 660 Hz is one octave + one fifth above 220 Hz, and it does blend excellently. You even have a stop for this interval in many pipe organs because it blends so well. The organist uses this stop to change the sound of the fundamental stop, not to get a transposition. Also, did you ever try to tune two sines to an octave? You can do it if you run the signal through a distorting guitar amp (at the very least this adds the frequencies f1-f0 and f0+f1 to the signal), but I know that I won't be able to do it precisely without technological help.

              – cmaster
              yesterday





              @phoog 660 Hz is one octave + one fifth above 220 Hz, and it does blend excellently. You even have a stop for this interval in many pipe organs because it blends so well. The organist uses this stop to change the sound of the fundamental stop, not to get a transposition. Also, did you ever try to tune two sines to an octave? You can do it if you run the signal through a distorting guitar amp (at the very least this adds the frequencies f1-f0 and f0+f1 to the signal), but I know that I won't be able to do it precisely without technological help.

              – cmaster
              yesterday













              @cmaster of course it blends excellently, but it isn't equivalent in the way that 440 and 880 are. They're both A, but 660 and 1980 are E and B. Three instruments playing pitches in parallel separated by factors of two sound more unified than three instruments separated by factors of three, even if the instruments are sine wave generators (relatively easy to achieve with waveform synthesis or Hammond organs). Pipe organ stops for non-octave harmonics only blend in certain registrations. A 5 1/3' stop with a single 8' stop is likely to sound like parallel fifths rather than a richer tone.

              – phoog
              yesterday






              @cmaster of course it blends excellently, but it isn't equivalent in the way that 440 and 880 are. They're both A, but 660 and 1980 are E and B. Three instruments playing pitches in parallel separated by factors of two sound more unified than three instruments separated by factors of three, even if the instruments are sine wave generators (relatively easy to achieve with waveform synthesis or Hammond organs). Pipe organ stops for non-octave harmonics only blend in certain registrations. A 5 1/3' stop with a single 8' stop is likely to sound like parallel fifths rather than a richer tone.

              – phoog
              yesterday














              @phoog For a 5 1/3' stop you need the 16' stop to blend. And, yes, a 3 1/5' stop and a 5 1/3' stop blend perfectly with a 16' stop. What makes it difficult to use that combination, is that stops are usually either on or off with nothing in between, and the high stops are just too loud in the combination. I once played an electronic organ that allowed partly pulling a stop, allowing me to reduce the volume of the high stops, and I used that combination quite frequently for its nice sound. With only on/off stops, you'd need to pull some more 16' stops to get the right relative volume.

              – cmaster
              yesterday





              @phoog For a 5 1/3' stop you need the 16' stop to blend. And, yes, a 3 1/5' stop and a 5 1/3' stop blend perfectly with a 16' stop. What makes it difficult to use that combination, is that stops are usually either on or off with nothing in between, and the high stops are just too loud in the combination. I once played an electronic organ that allowed partly pulling a stop, allowing me to reduce the volume of the high stops, and I used that combination quite frequently for its nice sound. With only on/off stops, you'd need to pull some more 16' stops to get the right relative volume.

              – cmaster
              yesterday











              0


















              This is another answer trying to help understanding also the question to people who can't cope with ratios and other abstract terms:



              Imagine you have a tone of 12 Hz frequency (a string waving 12 times/second).
              How must the 12 half steps between the octava (24 Hz) be tuned, so that the differences between all half steps are equal?



              The question implies: If the range between the octava is 12 Hz, why is the difference between the 12 half steps not always just 1 Hz?



              root=12Hz



              minor second 13Hz



              major second 14Hz



              .



              .



              .



              .



              perfect fifth 18Hz



              .



              .



              .



              major seventh: 23Hz



              octave: 24



              We can see that the difference between the first half 12Hz and 13Hz is just 1/10 of 12Hz (10% of the whole octava), while the additional difference between the octava 24Hz and the precedent half tone (23Hz) would have been almost only a 1/20 (=5%) of the difference between the next upper half tone above the octave will be 2Hz more - because this must be a 1/10 of the next octava of 48Hz, as the difference between of ocatava' (24Hz) and octava'' (48Hz) is 24Hz! (48-24=24) and a half step of 1/12 between octava' and octava'' will be 2?



              From this we can derive that the differences between the half steps are not additional of 1/12 but proportional by multiplying each half step with 1/12.



              Hope this is not droning and confusing. TLDR?






              share|improve this answer






























                0


















                This is another answer trying to help understanding also the question to people who can't cope with ratios and other abstract terms:



                Imagine you have a tone of 12 Hz frequency (a string waving 12 times/second).
                How must the 12 half steps between the octava (24 Hz) be tuned, so that the differences between all half steps are equal?



                The question implies: If the range between the octava is 12 Hz, why is the difference between the 12 half steps not always just 1 Hz?



                root=12Hz



                minor second 13Hz



                major second 14Hz



                .



                .



                .



                .



                perfect fifth 18Hz



                .



                .



                .



                major seventh: 23Hz



                octave: 24



                We can see that the difference between the first half 12Hz and 13Hz is just 1/10 of 12Hz (10% of the whole octava), while the additional difference between the octava 24Hz and the precedent half tone (23Hz) would have been almost only a 1/20 (=5%) of the difference between the next upper half tone above the octave will be 2Hz more - because this must be a 1/10 of the next octava of 48Hz, as the difference between of ocatava' (24Hz) and octava'' (48Hz) is 24Hz! (48-24=24) and a half step of 1/12 between octava' and octava'' will be 2?



                From this we can derive that the differences between the half steps are not additional of 1/12 but proportional by multiplying each half step with 1/12.



                Hope this is not droning and confusing. TLDR?






                share|improve this answer




























                  0














                  0










                  0









                  This is another answer trying to help understanding also the question to people who can't cope with ratios and other abstract terms:



                  Imagine you have a tone of 12 Hz frequency (a string waving 12 times/second).
                  How must the 12 half steps between the octava (24 Hz) be tuned, so that the differences between all half steps are equal?



                  The question implies: If the range between the octava is 12 Hz, why is the difference between the 12 half steps not always just 1 Hz?



                  root=12Hz



                  minor second 13Hz



                  major second 14Hz



                  .



                  .



                  .



                  .



                  perfect fifth 18Hz



                  .



                  .



                  .



                  major seventh: 23Hz



                  octave: 24



                  We can see that the difference between the first half 12Hz and 13Hz is just 1/10 of 12Hz (10% of the whole octava), while the additional difference between the octava 24Hz and the precedent half tone (23Hz) would have been almost only a 1/20 (=5%) of the difference between the next upper half tone above the octave will be 2Hz more - because this must be a 1/10 of the next octava of 48Hz, as the difference between of ocatava' (24Hz) and octava'' (48Hz) is 24Hz! (48-24=24) and a half step of 1/12 between octava' and octava'' will be 2?



                  From this we can derive that the differences between the half steps are not additional of 1/12 but proportional by multiplying each half step with 1/12.



                  Hope this is not droning and confusing. TLDR?






                  share|improve this answer














                  This is another answer trying to help understanding also the question to people who can't cope with ratios and other abstract terms:



                  Imagine you have a tone of 12 Hz frequency (a string waving 12 times/second).
                  How must the 12 half steps between the octava (24 Hz) be tuned, so that the differences between all half steps are equal?



                  The question implies: If the range between the octava is 12 Hz, why is the difference between the 12 half steps not always just 1 Hz?



                  root=12Hz



                  minor second 13Hz



                  major second 14Hz



                  .



                  .



                  .



                  .



                  perfect fifth 18Hz



                  .



                  .



                  .



                  major seventh: 23Hz



                  octave: 24



                  We can see that the difference between the first half 12Hz and 13Hz is just 1/10 of 12Hz (10% of the whole octava), while the additional difference between the octava 24Hz and the precedent half tone (23Hz) would have been almost only a 1/20 (=5%) of the difference between the next upper half tone above the octave will be 2Hz more - because this must be a 1/10 of the next octava of 48Hz, as the difference between of ocatava' (24Hz) and octava'' (48Hz) is 24Hz! (48-24=24) and a half step of 1/12 between octava' and octava'' will be 2?



                  From this we can derive that the differences between the half steps are not additional of 1/12 but proportional by multiplying each half step with 1/12.



                  Hope this is not droning and confusing. TLDR?







                  share|improve this answer













                  share|improve this answer




                  share|improve this answer










                  answered 8 hours ago









                  Albrecht HügliAlbrecht Hügli

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                  9,3961 gold badge10 silver badges31 bronze badges
























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