________________________________________________________________________
Date: Thu, 5 Oct 2000.
Subject: RE: GEOMETRY OF TRIANGULAR PLOTS
I'm posting this to the tuning list as well since Jon Wild and others
there have expressed interest in Chalmers' triangular plots. For that
purpose you have to rotate Chalmers' triangle 90°, so that one side is
vertical. Or you could take the result below and simply switch x and y.
The axes are perpendicular to the sides of the triangle, so they are at
120° angles.
Let's see how to get the Cartesian coordinates (x,y) for any chord. Let
x' and y' be two intervals in the chord. The i1 and i2 axes are at a
120° angle. The point representing the chord is x' units from the origin
in the direction of the i1 axis and y' units along the origin in the
direction of the i2 axis:
i2
\
\
\
\
\
\ (x,y)
\ ,+--------------------
\ ,-' | .
\ ,-' 60°| .
\ ,-' | .
\ ,-' | .
\ ,-| | .
\ ,-' | | .
\,-' | | .
,-'\90° | | .
,-' \ | | .
,-' \ \ | | .
,-' \ \ | | .
\ \ \ | | .
\ \ \ | | .
\ y' \ |......x'.....| (y'+2x')*2/sqrt(3)
\ \ \| | .
\ \ \_______________________________.______________i1
\ \ ,-'\60° 90°| . .
\ ,-' \ | . .
\ ,-' \ \ | . .
\ \ \ | . .
\ \ \ | . .
\ \ \ | . .
y'+2x' \ \ | x'*sqrt(3) .
\ 2x' \ | . .
\ \ \ | . .
\ \ \ | . .
\ \ 30°| . .
\ \ \ | . .
\ \ \|____._______________
\ \ ,-'
\ ,-'
\ ,-'
,-'
Clearly x = x'
But y = (y'+2x')*2/sqrt(3) - x'*sqrt(3)
= (2y'+4x'-3x')/sqrt(3)
= (2y'+x')/sqrt(3).
Anyhow, the trick behind the triangular plot is that there is a third
axis (call it i3) at 60° angles to the other two, along which the point
is x'+y' units from the origin. Let's show this:
i3
i2 /
\ -------,-+.--------/----------------
\ . ,-' | `-. / .
\ .-' |60° `-./ .
\ ,-'. | 90°/`-. .
\ ,-' . | / `-. .
\ ,-' . | / .`-. .
\ ,-' . |30°/ . .
\ ,-' c | / b (2y'+x')/sqrt(3)
\,-' . | / . .
,-'\ . |/ . .
,-' \ ---------/-------.-------- .
\ \ /|`-. . . .
\ \ / | `-. . .
\ \ / | `-. . .
\ \ /30°| `-. . .
\ \ / | . . .
\ \ / | . . .
y' \ / | . x'*sqrt(3) .
\ \ / | . . .
\ \ |../.x'.....| 2x' . .
\ \ | / | . . .
\ \ |/60° 90°| . . .
\ \|___________|__________________________i1
\ ,-'`-. .
,-' `-. .
`-. .
`-.
c = (2y'+x')/sqrt(3)-x'*sqrt(3)
= (2y'+x'-3x')/sqrt(3)
= (2y'-2x')/sqrt(3)
b = c*sqrt(3)/2 = y'-x'
So the length along the i3 axis is y'-x' + 2x' = y'+x'.
THAT'S THE MAGIC THAT MAKES THE TRIANGULAR PLOT WORK!
How does this apply to the Chalmers plot?
Let's orient the triangle like this:
.
|`-.
| `-.
| `-.
| `-.c
| `-.
| `-.
| `-.
a| `-.
| ,-'
| ,-'
| ,-'
| ,-'
| ,-'b
| ,-'
| ,-'
|,-'
'
Placing it relative to the axes above:
i2 |`-. i3
\ |60°`-. /
\ | `-. /
\ | 90° `-.c
\ | /90°`-.
\ | / `-.
\ a| / 60°`-.
\ | / ,-'`-.
\ | / ,-' `-.
\ |30°/ ,-' . `-.
\ | / ,-'b .
\ | /30° ,-' 500¢
\ |/ ,-' .
\|,-'________________._____i1
`-. .
`-. .
`-. .
`-. .
`-.
`-.
So you can see that the distance from side a is the first interval, x';
the distance from side b is the second interval, y'; and the distance
from side c is 500¢ minus (x'+y'), which is of course the third interval
in the tetrachord . . .
~~~
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
intuitive 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.
________________________________________________________________________