#1
I know what an A7 is by itself, but what do the #9 and b13 do? Can someone please give me a tab or chord chart of this chord? Also, would someone be so kind as to explain what you are doing scale-wise to get a #9 and b13?
#2
A 9 is the second degree of a scale, raised by an octave (to create a ninth degree). A #9 is the nine sharpened (half a tone higher). The 13 is the same. It's the sixth scale of a degree, raised by an octave, creating a thirteenth. Then flatten it.

This would be your chord (notic it misses the fifth, the E)

8 B#
6 F
6 C#
5 G
-
5 A

Also, I think this is in the wrong forum, but not sure.
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#3
They're exactly the intervals they say they are, added to the chord.

#9 being enharmonic with a b3 (so for A, that's a C... or possibly a B#, I'm not sure) and a b13 is a #5 enharmonically (in A again that's an F).

In scalar terms... it's not from a scale, at least not one that I know of. It's just a chord that gets the sound of whatever it is the original composer wanted.
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#4
#9 => 7th + sharpened second
b13 => 7th, 9th, (2nd) 11th (4th) and b13th (6th)
Although commonly on the highest of those is played.

so in A a #9 is a B# (enharmonic to C but not actually a C)
b13 = F

=> A7#9b13 = A C# E G B# (D) F
The D is in brackets because it is technically implied but probably not used
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#5
Basically you grab your initial A scale

A - B - C# - D - E - F# - G#

you count along obviously ending at seven
But then start again so that A would be classed as 8

the sharp from number (9) B is you move it up a semi-tone / one note higher
So that the actual note will be infact B# (Which also is the same note as C, this term is known as enharmonic)

the b in front of the (13) means bring that note back a semi-tone so that the F# actually becomes an F

EDIT: **** I'm a slower typer
#6
All of the extra #s and bs are worked out basically by looking at the scale of the chord - In this case, the scale of A major. If 8th is obviously an octave, so 9th is the same as the 2nd, but an octave above. The 2nd in A Major is B, so #9 would be B#, or C.
The same applies to 13, so if you work it out, the 13th is the same as the 6th, which is F, but an octave above the root. F#b is obviously just F.
The chord would also have all of the A7 in, which you already know is A C# E G. So, overall, the chord would be A C# E G C F
I think this is right, but if not, I'm sure I will be quickly corrected

Edit: Good god, what happened here
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Last edited by HeretiK538 at May 25, 2011,
#8
To all people saying the #9 is a C in the key of A, it is not. It's a B# (which is enharmonically the same note).
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#9
Quote by RDSElite
To all people saying the #9 is a C in the key of A, it is not. It's a B# (which is enharmonically the same note).


*feels slightly smug*
R.I.P. My Signature. Lost to us in the great Signature Massacre of 2014.

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“A man who mistakes secrets for knowledge is like a man who, seeking light, hugs a candle so closely that he smothers it and burns his hand.”


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#10
Quote by Zaphod_Beeblebr
*feels slightly smug*


yeah
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