sci.math.research
SUNA27, Reduced Quads again
===========================
A patch test with four triangles was carried out for the Labrujere Problem. OK.
Together these four triangles form a quadrilateral superelement. The following
equations were found for this superelement. Here we adapt the numbering and we
replace the "global" (u,v) by "local" variables (U,V):
3
o U4  U1 + V3  V2 = 0 (1)
/  \ V4  V1  U3 + U2 = 0 (2)
/  \
1 o  *  o 4 2.U* = U1 + U4 = U2 + U3 (3 & 4)
\  / 2.V* = V1 + V4 = V2 + V3 (5 & 6)
\  /
o Consider the diagonals (1)(4) and (2)(3) of a
2 distorted quadrilateral as a local coordinate system.
Then global velocities u,v are associated with the
local U,V according to: The inverse transformation reads:
u = (x4  x1).U + (x3  x2).V ; J.U = + (y3  y2).u  (x3  x2).v
v = (y4  y1).U + (y3  y2).V ; J.V =  (y4  y1).u + (x4  x1).v
Where J = (x4  x1).(y3  y2)  (y4  y1).(x3  x2) <> 0 .
Substitute the inverse transform into (1) and (2), giving:
+ (y3  y2).(u4  u1)  (x3  x2).(v4  v1)
 (y4  y1).(u3  u2) + (x4  x1).(v3  v2) = 0
 (y4  y1).(u4  u1) + (x4  x1).(v4  v1)
 (y3  y2).(u3  u2) + (x3  x2).(v3  v2) = 0
And in (3 & 4) & (5 & 6) , giving:
+ (y3  y2).(u1 + u4  u2  u3)  (x3  x2).(v1 + v4  v2  v3) = 0
 (y4  y1).(u1 + u4  u2  u3) + (x4  x1).(v1 + v4  v2  v3) = 0
From J <> 0 it follows that: u1 + u4 = u2 + u3 & v1 + V4 = v2 + v3 .
Surely the linearity requirement is _independent_ of any coordinate system.
It is remarked that the diagonals of the quadrilateral patch must be _straight_
lines, in order for the above transformation to be valid. This puts geometrical
restrictions upon the triangles in the patch. Especially node (*) must be at
the intersection of these diagonals.
Let's rotate now the whole patch over an angle of 45 degrees, and remember the
poster "Reduced Quadrilaterals". The accompanying system
3 o  o 4 of equations is then derived by substituting in the above
  x1 = x3 = y1 = y2 = 0 , x2 = y3 = x4 = y4 = 1 , giving:
 * 
  (u4  u1) + (v4  v1)  (u3  u2) + (v3  v2) = 0
1 o  o 2  (u4  u1) + (v4  v1)  (u3  u2)  (v3  v2) = 0
Or: [ (u2 + u4)  (u1 + u3) ] + [ (v3 + v4)  (v1 + v2) ] = 0
[ (v2 + v4)  (v1 + v3) ]  [ (u3 + u4)  (u1 + u2) ] = 0
These are the same equations as derived before in "Reduced Quadrilaterals".
In that poster we have seen, however, that such elements don't work properly.
Reduced Quads give rise to a global system which interconnects only horizontal
or vertical components of the velocities exclusively. And it interconnects only
even or uneven numbered vertices exclusively. The Least Squares Finite Element
matrix associated with the Cauchy equations is repeated from this poster:
u1 u2 u3 u4 v1 v2 v3 v4
 1 1   1 +1 1 +1 1 1 +1 +1   + o o  o +  o 
 +1 1   1 1 +1 +1 +1 1 +1 1  =  o +  o  o o +  u
 1 +1   o  + o + o o  
 +1 +1  Here the elements of the Least   o o + o  + o _____
 1 +1  Squares matrix are also apart  o  + o + o o  
 1 1  from a constant factor (= 2).  + o o  o +  o 
 +1 +1  Furthermore: + = +1 ,  = 1   o o + o  + o  v
 +1 1  and o = 0 . Symmetric matrix:  o +  o  o o + 
But wait! The situation is slightly different here. We also have two equations
which impose linearity: u1 + u4 = u2 + u3 and v1 + v4 = v2 + v3 .
u1 u2 u3 u4 v1 v2 v3 v4
 +1 0   +1 1 1 +1 0 0 0 0   +   + o o o o 
 1 0   0 0 0 0 +1 1 1 +1  =   + +  o o o o  u
 1 0    + +  o o o o 
 +1 0   +   + o o o o _____
 0 +1   o o o o +   + 
 0 1   o o o o  + +  
 0 1   o o o o  + +   v
 0 +1   o o o o +   + 
To be continued ...

* Han de Bruijn; Applications&Graphics  "A little bit of Physics * No
* TUD Computing Centre; P.O. Box 354  would be NO idleness in * Oil
* 2600 AJ Delft; The Netherlands.  Mathematics" (HdB). * for
* Email: Han.deBruijn@RC.TUDelft.NL  Fax: +31 15 78 37 87 * Blood