From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Repost: Help! Coordinate transformation problem
Date: 14 Jun 1997 16:54:52 GMT

In article <339ED8D3.41C6@aptnames.austin.ibm.com>,
Hal Fulton  <hal9000@aptnames.austin.ibm.com> wrote:

> Consider these constraints:
>    0 <= x <= 1
>    0 <= y <= 1
>    0 <= z <= 1
>    x + y + z = 1
> 
> We see that the set of possible points (x,y,z) is the triangular
> region nestled by the origin in one octant of three-space, the
> vertices being the points K(0,0,1), L(0,1,0), and M(1,0,0).
>            z                         B
> 
>            |                         |   K
>          K .                         |   /\
>            |                         |  /  \
>            |     L                   | /    \
>            |_____.__    y            |/      \
>           /                          .________._____    A
>          /
>       M .                           M          L
>        /
> 
>       x
> 
> Now suppose we pick another coordinate plane with A and B axes
> as follows: Origin is point M, positive A axis is in the direction
> of L, and so on. Obviously the triangular region lies within this
> AB plane, and the AB coordinates of point K would be (I think)
> (sqrt(2)/2,sqrt(3)/2).
[not quite; see below]

> I am trying to find the relatively simple functions that will
> transform (x,y,z) to (A,B), that is
>    A(x,y,z) = ...
>    B(x,y,z) = ...

You want affine functions: A  and  B  will have the form
	ax+by+cz+d
Determine  a b c d  by stating where the points should go.
	M: (sqrt(2), 0, 0) -> (0,0)
	L: (0, sqrt(2), 0) -> (sqrt(2),0)
	K: (0, 0, sqrt(2)) -> (sqrt(2)/2, w)
(Here you know the distance from  K  to  M  so  2 = (1/2) + w^2 implies
w = sqrt(6)/2.) Similarly I assume you want a rigid motion, so the
origin should get sent to some place of distance  1  from  M, L, and K.
If this point is  (x0,y0,z0) then you get 3 equations to determine these
three coordinates. I'll save  you the trouble: they solve to
x0=sqrt(2)/2, y0=sqrt(6)/6.

So for the first coordinate function  A  we have
a*sqrt(2) + d = 0, b*sqrt(2) + d = sqrt(2), c*sqrt(2) + d = sqrt(2)/2,
and  d = sqrt(2)/2. That is, 
	A(x,y,z) = (-1/2)x + (1/2)y + 0 z + sqrt(2)/2
Likewise
	B(x,y,z) = (-sqrt(3)/6)x+(-sqrt(3)/6)y+(sqrt(3)/3)z+(sqrt(6)/6)
if I've done the arithmetic correctly.

dave