Re: Chord Inversion Problem.

[pls]

Hi Alan,

c/e (which is actually the first inversion) doesn't work correctly in 
chordmode. This is a known issue. But there is a workaround. I use two 
different variables: one for the chords and one for the chord names (see 
below). And I use this construct: c1:3.5.8^1 in chordmode. This way any 
inversion can be defined, e.g.: c:maj7/e = c:3.5.7+.8^1

hth
patrick

\version "2.15.37"
accompaniment = {\easyHeadsOn \chordmode{
   c'1 c'/g %c'/e
  c1:3.5.8^1 } <e' g' c''>
}

chordSymbols = \chordmode{
  c1 c/g c/e c/e 
}

 \simultaneous{
 \context ChordNames {
   \chordSymbols
 }
 \context Staff {
    {\accompaniment}
 }
}



Am 17.07.2012 um 18:06 schrieb Phil Holmes:

> ----- Original Message ----- From: "AlanRobertClark" <address@hidden>
> To: <address@hidden>
> Sent: Tuesday, July 17, 2012 4:25 PM
> Subject: Chord Inversion Problem.
> 
> 
>> Dear All.
>> 
>> I am trying to put together a simple chord (triad) inversion exercise 
>> (piano).
>> 
>> Snippet:
>> \version "2.12.3"
>> accompaniment = {\easyHeadsOn \chordmode{
>>   c'1 c'/g c'/e} <e' g' c''>
>> }
>> \simultaneous{
>> \context ChordNames {
>>     \accompaniment
>> }
>> \context Staff {
>>    {\accompaniment}
>> }
>> }
>> 
>> This produces the first two chords, as expected, but c'/e ends up producing 
>> <c'
>> c'' g''>, which is not in the way the second inversion would ordinarily be
>> played. Entering what I want produces (correctly an Eminor with a flattened 
>> 6th,
>> instead of the more usual C/E notation.
>> 
>> Any help will be appreciated.
>> Alan.
> 
> I think you'll find this is the same as 
> http://code.google.com/p/lilypond/issues/detail?id=2617, which I realise 
> doesn't help you much....
> 
> --
> Phil Holmes 
> 
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