Distance between triads

...and now for the subtlety.

I derived the formula for the Cartesian coordinates (x,y) for a chord with
intervals x' and y':

x = x'
y = (2y'+x')/sqrt(3)

BUT, look again at the graph:

i2 |`-. i3
\ |60°`-. /
\ | `-. /
\ | 90° `-.c
\ | /90°`-.
\ | / `-.
\ a| / 60°`-.
\ | / ,-'`-.
\ | / ,-' `-.
\ |30°/ ,-' . `-.
\ | / ,-'b .
\ | /30° ,-' 500¢
\ |/ ,-' .
\|,-'________________._____i1
`-. .
`-. .
`-. .
`-. .
`-.
`-.

According to the formalism so far, the distance between the origin and the
center of side C is 500¢. So that says that the distance between the chord 0
0 0 and the chord 0 250 500 is 500¢. Huh? Meanwhile, the distance between
the origin and the upper corner of the triangle is 500*2/sqrt(3) = 577¢.
That says that the distance between the chord 0 0 0 and the chord 0 0 500 is
577¢. Wha???

So clearly, we should build a scaling factor of sqrt(3)/2 into our
transformation so that the triadic distances are meaningful. So the FINAL
formula for the Cartesian coordinates (x,y) for a chord with intervals x'
and y' is:

x = x'*sqrt(3)/2
y = y'+x'/2

The distance function that this leads to has the property that if two chords
differ only in the tuning of one note, the distance between them will be
equal to the amount that that note moved (hence in the picture I made with
the red (JI major triad) and blue (JI minor triad) points, each triad is
closest to its "partner" in the same inversion, at a distance of 71¢). For
chords that differ in two notes, it's less intuivite what a distance
function would mean -- the example 0 0 0 and 0 250 500 would now have a
distance of 433¢ -- but at least it makes sense as a sort of "weighted
average" of how the intervals change, and is of course the only option that
seems consistent with a geometrical picture of triads -- in other words,
it's the only possible _Euclidean_ distance function for triads.