# Need help understanding the double sharp turn in Chopin's prelude in e minor [duplicate] I need help understanding why the notes in this turn are B-A#-Gx-A#.

I don't understand how the double sharp affects the turn. I would assume that instead of A# being the "base note" (for lack of a better word), it would be C because of the double sharp. Then the turn would be D-C-A#-C.

I think the problem for me is the turn, but It's my understanding that turns go up one whole tone from the base, the base, down one whole tone from the base, then base again.

If A# is the "base note" then I would think that the turn is C-A#-G#-A#.

Every answer that I have found online gives the notes, but not a very good explanation of how they got those notes. I don't doubt that the notes are B-A#-Gx-A#, but i don't know why. Please help me understand what part of this double sharp turn makes it so different than what I would expect.

## 1 Answer

There is no rule that says the turn has to go up one whole tone from the note you call the base. It can be a half tone or a whole tone. The turn follows the key signature unless something else is indicated.

So in this case what you call the base note is A♯. If you follow the key signature then the next tone up from A♯ is B, so it should be B. And regarding the tone below A♯. Here the double sharp sign below the turn sign indicates the tone below A♯ must be Gx.

Thus B-A♯-Gx-A♯ is the solution.

• Thanks, it took a little bit more research, buy I now think I have a full understanding of what is going on here. The double sharp only modifies to note below the "base note". I.E. if the accidental k is below the turn symbol, only the note below the "base" is modified. – Jared L. Jun 16 '19 at 20:41
• Yes, that is how it works. – Lars Peter Schultz Jun 16 '19 at 20:54