Horn transposition

Discussion in 'The Rehearsal Room' started by CRIMSONDRAGON, Jul 23, 2011.



    Hi, I was hoping someone can help in clarifying how to transpose a piece written for tenor horn in Eb to Flugal horn in Bb. Is it a case of transposing everything written on the horn copy down a perfect 5th?? to sound exactly the same and how do you work out the new key signature. The horn part is originally in 2 flats.
    many thanks
  2. SteveT

    SteveT Member

    The Eb tenor horn is transposed up a major 6th from concert pitch. The Bb Flugel Horn is transposed up a major second. Therefore simply transpose down a perfect 5th from the Horn part and Bobs your uncle! You are then playing at the same pitch and in the same octave!
    Last edited: Jul 23, 2011
  3. PeterBale

    PeterBale Moderator Staff Member

    Further to the clear answer above, to amend the key signature you add 1 flat, ie written Bb becomes written Eb.

    Edit: Sorry for the brainstorm - corrected now :oops:
    Last edited: Jul 23, 2011
  4. brassneck

    brassneck Active Member

    Just transpose the horn part up a perfect 4th (and a flat is ADDED to the key signature ... Eb Horn part in Bb major/G minor = Db major/Bb minor concert pitch. Transposing key for Bb Flugel would be a major 2nd up = Eb major/C minor). ;)
  5. PeterBale

    PeterBale Moderator Staff Member

    Thanks for spotting my deliberate mistake, although you've got the flugel playing an octave higher than the horn was ;)
  6. brassneck

    brassneck Active Member

    It's the principle that counts, Peter! Unless the player has a 4-valve instrument! ;)
  7. Tam O Shanter

    Tam O Shanter Member

    The easiest way to work out the key signature change is to do the same transposition to that as you would do to the notes. So if you are transposing a horn part to Flugel and taking the notes down a 5th, then you do the same with the key signature.

    If the key signature is Bb for Horn it becomes Eb for Flugel - so it goes from 2 flats to 3 flats - down a 5th.



    Hi guys..many many thanks for all your help with this..Ihave got the transpositions done..tho found a few problems with sibelius!!! but found a way round it so it all works well
    thanks again
  9. worzel

    worzel Member

    This may not work for everyone, but I like to think of it as follows.

    n = number of sharps (or flats if negative)

    to transpose up a semitone n = n + 7
    to transpose down a semitone n = n - 7
    to convert to enharmonic equivalent (say C# to Db) n = n +/- 12

    Then from Bb (-2) down 7 semitones =

    -2 - (7 * 7) + 4 * 12 = -3

    I.e. 3 flats.

    I'm guessing that way might appeal to your sensibilities, C# man :)
    Last edited: Jul 26, 2011
  10. MoominDave

    MoominDave Well-Known Member

    Interesting way of thinking about it. It's basically clock arithmetic -

    Let k be original key signature (positive for sharps, negative for flats); let s be semitones to transpose by (can be + or -). Then your new key signature is

    (k + 7s) mod 12

    Neat. But I suspect most will complain that it's too mathsy, even in your simpler form.
  11. worzel

    worzel Member

    In your more concise form, wouldn't

    (((k + 7s) + 6) mod 12) - 6

    be better?
  12. MoominDave

    MoominDave Well-Known Member

    I was assuming that a negative number mod a positive number would be taken as -n times the positive number with a negative remainder left over rather than -(n+1) times the positive number with a positive remainder left over. You can avoid that assumption that way, as you point out. Not sure whether the choice of convention has serious arithmetical ramifications?
  13. Bass Trumpet

    Bass Trumpet Active Member

    Well, that's cleared that one up. Thanks Dave ;)
  14. worzel

    worzel Member

    I don't see how that works. If I started at C and went up three semitones, with your formula I get:

    (0 + 7*3) mod 12 = 9. I.e. 9 sharps

    whereas with mine I get:

    (((0 + 7*) + 6) mod 12) - 6) = -3. I.e. 3 flats.

    The only difference I can see is that mine comes out with the more conventional of two enharmonic answers.
  15. MoominDave

    MoominDave Well-Known Member

    Yeah, you're quite right. My apologies. Didn't think it through properly before posting.
  16. worzel

    worzel Member

    To be honest, nor did I originally. Which is why I left my +/- enharmonic adjustment was so vague :)

    If you want a real brain teaser (or maybe it's really easy and I'm just thick), how about formalising how to denote the key from the number of sharps or flats? It's got to work in general though, so 11=E#, 14 = C##, etc.
  17. mikelyons

    mikelyons Supporting Member

    After all of this wonderful edification and polishing of brain cells, if you are using Sibelius, wouldn't it just be easier to create a flugel instrument in the same score (press I) and then simply copy and paste the music across? Or am I missing something? ;-)
  18. worzel

    worzel Member

    Yeah, but I'm guessing the questioner was after more than just, "give the job to someone/something that knows how to do if for you." And anyway, you'd want to know how it's done so you could check that Sibelius has done it properly.
  19. MoominDave

    MoominDave Well-Known Member

    How about:

    Let m = semitones away from C of the key name.
    Let n = number of sharps (negative n = flats).
    Let a = actual semitones away from C of the key
    Let s = number of sharps at the end of the key name (negative s = flats)

    Then a = 7n mod 12
    and s = floor( (n+1) / 7 )

    So m = (a - s) mod 12.

    Then you match m to a note name (e.g. m = 4 => E - you'll never have a sharp or flat here due to the a-s bit) and add s sharps (or -s flats if s is negative).

    I've knocked up a bit of code to run this, and it seems robust...

    N.B. Ignore the rubbish I spouted before about mod conventions - remainders are always positive.
    Last edited: Jul 26, 2011
  20. worzel

    worzel Member

    Cool. I've just knocked up a bit of code to check it too. My code doesn't print a key letter for 12,19,24,26,31,33 (that's as far as I got) and doesn't print anything for negatives. Have I misunderstood your pseudo code?

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