no B sharp or E sharp

B#=C
E#=F

Now there is a pitch between B and B# but we don't use that in western music. You have to understand that our system of having twelve notes in an octave is something our western culture came up with, you could divide an octave in as many ways as you want. For example, some African and Asian music does use some pitches that do not exist according to our music theory.

To keep your analogy of 'cycles'. It would be like this (just for the sake of understanding it, this does NOT equal actual frequencies!):

A=400
A#=450
B=500
C=550
D#=600
E=650
F=700
F#=750
G=800
G#=850

As you can see the step from B to C is as big as from A to A#. Now there is a pitch for every number of cycles but we only use those above. Any others are perceived by us as severely de-tuned. Now the reason why we divide it like we do involves a fair bit of math which I can't be bothered to think about right now. ;)

And welcome to GN!

Each pitch is related by the value 1.05946309436. By multiplying the pitch of any note by this number, you get the frequency of the next higher note. So A=440, A#=466.2, B=493.9 and C=523.3. There is no note between B & C.
By dividing a pitch, by 1.05946309436, you get the next lower semitone.
There is a B#, which is enharmonic to C.

Thanks Arien and Greybeard for the replies. I'll put the info in my brainpan and let it simmer for awhile. Thanks for the welcome.

If you take a piece of string, stretch it and pluck it, you will get a note. Shorten the string by one third, and pluck it again. Repeat 12 times. You end up with a note that is the same as the first note but 7 octaves higher. Shrink them all down to size by halving the frequency (doubling the length of the string) until they all fit into a single range without any gaps, and you'll find that what you have is just the 12 notes we all know and love. And you'll notice that there is no note between B and C or between E and F.

Next session - why C# is not really the same as Db; or "how big is a semi-tone" and "what's a comma"

Best,

A :-)

And, of course, how long is a piece of string?

why C# is not really the same as Db

Go on then, enlighten me - because I really don't get it.

What Alan is alluding to is the 'Pythagorean Comma'.

It's easier to illustrate than explain. A perfect fifth has a 3:2 relationship - if you start with A440, E should be 3x2x440 = E660. And the octave has a 2:1 relationship - sp the next A should be A880.

You can combine those relationships to fill in the tones in an octave - if you go up a fifth from E660, you get B990, and dividing that by 2 tells you the B right over A440 will be B495.

So far so good, right? Now for the comma.

If you go up twelve fifths, you've gone through the entire circle of fifths, and you're back at the starting point. You should be exactly 7 octaves over your starting point. To keep the math simple, let's say the starting tone is 100. Your 3:2 tones will be:

100-150-225-337.5-506.25-759.38-1139.06-1708.59-2562.89-3844.37-5766.5-8649.76-12974.63

But moving up seven octaves gives you this:

100-200-400-800-1600-3200-6400-12800

Finding the octave by using fifths ends up giving you a tone that's a bit sharp. That difference is called the 'Pythatgorean comman'.

Today we 'split the difference' by using equal tempered tuning... but we didn't always do that. Earlier tuning methods tried to make fifths, thirds, or some other interval as perfectly tuned as possible. And depending on what starting interval you used, some intervals sounded REALLY bad (they were called 'wolf' intervals). But where they fell depended on where you started... and on what interval was your tuning standard. This made modulation between keys impossible.... unless the 'black keys' were tuned to different frequencies. So in a perfect world, C# is not the same as Db - the sharps would come from the 3:2 ratio fifths, making the C# above A440 C#556.875; the flats would come from the 2:3 ratio fourths, making the Db549.38.