From: Dave Rusin <rusin@math.niu.edu>
Subject: Re:  Stewart platform
Date: Wed, 22 Sep 1999 03:23:50 -0500 (CDT)
Newsgroups: [missing]
To: TriciaRz@aol.com
Keywords: Sample numerical calculations showing non-unique solutions possible

The last time I dealt with Stewart platforms I learned that there is a fairly
substantial bit of work already done, so I don't know if I should give you
my (amateur) conclusions. I like to think about these things geometrically
but in fact it's all a fairly simple (by today's standards) algebraic
calculation. I don't know about unique solutions with restricted values
of the variables but it appears there can be multiple solutions. In fact
I just ran some calculations on a platform with these characteristics:
base runs from (0,0) to (10,0) to (5,8) in the plane (i.e. nearly equilateral);
platform has sides of length 9, 10, 11 (i.e. not quite equilateral or even
isoceles); piston lengths  6 & 12, 7 & 11, 9 & 13 at the three platform
vertices, respectively. I fed this into the computer algebra package  Magma
in the following format:

Q:=RationalField();
R<x1,y1,z1,x2,y2,z2,x3,y3,z3>:=PolynomialRing(Q,9);
a:=10;b:=5;c:=8;l1:=9;l2:=10;l3:=11;p12:=6;p13:=12;p23:=7;p21:=11;p31:=9;p32:=13;
G:=ideal<R|
(x1-x2)^2+(y1-y2)^2+(z1-z2)^2-l3^2,
(x1-x3)^2+(y1-y3)^2+(z1-z3)^2-l2^2,
(x2-x3)^2+(y2-y3)^2+(z2-z3)^2-l1^2,
x1^2+y1^2+z1^2-p12^2,
(x1-a)^2+y1^2+z1^2-p13^2,
(x2-a)^2+y2^2+z2^2-p23^2,
(x2-b)^2+(y2-c)^2+z2^2-p21^2,
(x3-b)^2+(y3-c)^2+z3^2-p31^2,
x3^2+y3^2+z3^2-p32^2>;
//ignore... EliminationIdeal(G,{x1,y1,z1});
Groebner(G);

I won't show the value of  G  at this point because it's many pages long,
but it shows polynomials which are equivalent to the given set of  9  but
of the very special form
   x1 + 2/5,
   y1 - [bignumber] z3^14 + ...
   z1 + [more  z3 stuff]
[likewise for  x2, y2, z2, x3, y3, and then finally ]

z3^16 + 
233520143876274504789/242177763763360000*z3^14 + 
7255016240942576514134370463827/34830199726657445888000000*z3^12 - 
6135855545804589799118335658802661631/445826556501215307366400000000*z3^10 -
815801522057339831032333444255677424613211/365221115085795579794554880000000*z3^8 + 
2893043016821964206280419999396021687312902961/584353784137272927671287808000000000*z3^6 + 
9429558429072446997426234034876744407099094545398667/1495945687391418694838496788480000000000*z3^4 + 
1487785383664482494153160877129482075894250036230791553/7659241919444063717573103557017600000000*z3^2 + 
329576025910924583342728864397438051105463712280733369140161/196076593137768031169871451059650560000000000

This means there is one solution  (x1,y1,...,y3,z3)  for each of the roots
of this last polynomial, of which there can be at most  16. As it happens, this
polynomial has only 4 real roots (two pos/neg pairs), but that means there
are exactly 2 orientations of the platform having z3 positive. (The values
of  z3  are   8.998519356  and  8.953804466). If I've done the calculations
correctly, the three points are then either
 [-0.4,  0.070818136749, 5.98623293829],
 [10.48920749452, 1.36825468408, 6.84752182521], 
 [4.84097632462, 8.03688979710, 8.998519356]
OR
 [-0.4, 5.85896219001, 1.22986261669],
 [ 6.69211212987, -1.00492991882,   6.08670630940],
 [ 5.74287843157, 7.47320098026, 8.95380446674]
As you can see these are two rather different solutions.

I don't claim this situation is representative; I made up numbers rather
randomly.

The calculations with Magma take just seconds.

dave