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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 . . .

Hope that helps, Jon . . .

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

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