RE: GEOMETRY OF TRIANGULAR PLOTS

For those on the harmonic entropy list -- this time, I'll do this right.

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 120° 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 . . .

Hope that helps, Jon . . .

There's one more subtlety I want to bring up . . . later . . .