9

Let's take a 440Hz A pitch and the 880Hz A an octave higher.
If we divide space between 440Hz and 880Hz into 12 equal parts, we would have:

[440Hz, 476.6, 513.2 ... 880Hz.]

And this looks equally divided. Why do we say equally divided if the differences between notes are 12ths of 2?

  • 12
    Because 'equal' refers to a geometric progression, not an arithmetic progression. – user207421 Oct 13 at 22:48
  • 1
    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 Oct 14 at 12:58
  • A geometric progression is based on multiplication, not addition. So consecutive keys don't differ by the addition of 1/12 the frequency, but by multiplication by the 12th root of 2. – Lee Daniel Crocker Oct 14 at 20:50
  • There is a graph of pitch vs. frequency on the following answer: music.stackexchange.com/questions/39992/… – Caleb Hines Oct 15 at 2:21

10 Answers 10

38

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 (almost) the exact ratio between two half steps.

  • 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 Oct 11 at 17:55
  • 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 Oct 13 at 5:35
28

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.

  • 8
    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 Oct 12 at 17:50
  • 3
    external link with audio examples: what does a linearly spaced sequence sound like vs. a logarithmic one – Cee McSharpface - it Oct 13 at 10:14
  • IMHO the top answer and this one together make perfect explanation – Tukkan Oct 17 at 13:19
20

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).

  • 4
    @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 Oct 11 at 17:51
  • 1
    @Tim : Does this mean I get incompatible step sizes if I take my octaves to be C-to-C? – Eric Towers Oct 12 at 19:23
  • 2
    Negative frequency means that the music is played backwards, revealing hidden Satanic messages. – dan04 Oct 14 at 13:03
  • going down in equal steps to get zero or negative: really clear way to explain it will not work that way, nice compliment to @SagebusherGardener's answer – Michael Curtis Oct 14 at 20:58
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.

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.

  • 6
    @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 Oct 12 at 16:05
  • I agree, but it is a good analogy and shows that the differences of the steps aren't continuous. – Albrecht Hügli Oct 12 at 16:12
4

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.

  • 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 Oct 13 at 1:38
  • 2
    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 Oct 13 at 4:21
  • 3
    @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 Oct 13 at 7:10
  • @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 Oct 13 at 15:35
  • @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 Oct 13 at 16:14
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.

  • Stackexchange bug, is anyone else seeing this? The contents of the table disappear after page load. – whatsisname Oct 13 at 5:45
  • @whatsisname yes, I'm seeing it inconsistently, too. I'll try modifying the table to see if I can work around it. – phoog Oct 13 at 15:37
1

If an Octave is defined by this:

  • doubling of frequency
  • 12 steps

Why should the way to move from one key to the next be governed by a different rule (i.e. move along a different curve in a X,Y diagram) than moving about 12 keys, which is nothig else but applying the rule from key-to-key 12 times? There is a function which dictates how to move from one key to the next, which is defined by above terms. What you want to do is move linearly from key to key, which contradicts above definition. The curve is not a line. It's not defined as an addition of something, but a doubling (multiplication) over a certain number of keys (12). An octave above 110 Hz is 220. But an octave above that is 440, not 330 - you don't add a number (which would get equal steps), you multiply (the linear step size increases as you go higher).

Hence, if x is the multiplication step from one key to the next, f is the start frequency, and 2*f is one octave above:

f * x * x * ... * x = 2*f  | 12 steps, i.e. 1 (multiplication) step applied 12 times
f * x^12 = 2*f  | divide by f
x^12 = 2  | solve for x
x = 2 ^ (1/12)

i.e. 12th root of 2. See the image below: The orange curve follows that rule from 110 Hz to 880 Hz, with all semi tone steps in between. The blue curve is what would happen if you tried to satisfy both of your requirements: Doubling of frequency per octave, but also going in equal steps (i.e. linearly) from one octave to the next. Both curves meet at each octave: 110, 220, 440, 880. See how that blue line does not follow one smooth function but rather is pieced together of linear segments? I don't think you'd expect this to sound natural and even, going up with the frequency that way for semi tones ;) To move up smoothly and satisfy the "doubling the frequency per octave", your semi tones need to be on that orange curve (and sub-semitones like cents, too, of course, i.e. 100 cents are also not spaced equi-distantly)

logarithmic (musical) curve vs. linear pieces as the OP liked to do

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?

0

Going up an octave does not mean adding 440 Hz; rather it means multiplying by 2. Every time you go up half a tone, you multply by the same amount; you don't add the same amount.

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