sci.math.research SUNA25, Repeatable Elements =========================== Least Squares Finite Elements are produced in our theory by taking the square of Finite Difference like equations: E(i,j) = A(i).A(j) Here A(i) are the coefficients + (y2 - y1).u3 - (x2 - x1).v3 + ... = 0 of the following equation system: + (x2 - x1).u3 + (y2 - y1).v3 + ... = 0 Suppose the problem is transformed by an orthogonal transformation, for example a rotation: ^^^^^^^^^^^ x = + cos(phi).x + sin(phi).y ; u = + cos(phi).u + sin(phi).v y = - sin(phi).x + cos(phi).y ; v = - sin(phi).u + cos(phi).v Then the first equation of the above two becomes something like: [ - (x2 - x1).sin(phi) + (y2 - y1).cos(phi) ].[ + u3.cos(phi) + v3.sin(phi) ] + [ - (x2 - x1).cos(phi) - (y2 - y1).sin(phi) ].[ - u3.sin(phi) + v3.cos(phi) ] = + (y2 - y1).u3.[ cos^2 + sin^2 ] - (x2 - x1).v3.[ sin^2 + cos^2 ] + + (x2 - x1).u3.[ - sin.cos + cos.sin ] + (y2 - y1).v3.[ cos.sin - sin.cos ] = + (y2 - y1).u3 - (x2 - x1).v3 And the second: [ + (x2 - x1).cos(phi) + (y2 - y1).sin(phi) ].[ + u3.cos(phi) + v3.sin(phi) ] + [ - (x2 - x1).sin(phi) + (y2 - y1).cos(phi) ].[ - u3.sin(phi) + v3.cos(phi) ] = + (x2 - x1).u3.[ cos^2 + sin^2 ] + (y2 - y1).v3.[ sin^2 + cos^2 ] + + (x2 - x1).v3.[ cos.sin - sin.cos ] + (y2 - y1).u3.[ sin.cos - cos.sin ] = + (x2 - x1).u3 + (y2 - y1).v3 Conclusion: _both_ equations of the F.D. system are _invariant_ for rotations. Now it is trivial that they are also invariant for translations (: differences of all coordinates). So both F.D. equations are independent of the observer's coordinate system as a whole. Now we know from the "Weight Tracing" poster that L.S. elements allways can be normed. So multiplying the A(i) with one and the same factor has no influence whatsoever upon the element-matrix. But coefficients A(i) depend only upon coordinate-differences. Consequently, the element-matrix must remain the same if these distances are all multiplied with a constant factor. Therefore, two element-matrices mus be the same if their element geometries only differ by a scaling factor. In case of the Labrujere problem, this can be accomplished as follows. Start with the circular cylinder at radial position R0; suppose the mesh ends at radial position Rn. Then we have the little theorem: | | A mesh consisting of repeatable elements is given by radial positions: | ^^^^^^^^^^ | Ri = R0.(1 + dR/R0)^i where: dR = R0.(Rn/R0)^(1/n) - 1 | Proof of the first relation: R(i+1) - R(i) = R0.(1 + dR/R0)^i.(1 + dR/R0 - 1) = = Ri.dR/R0 . The other sides of the quadrilateral are: 2.Ri.sin(phi/2) and 2.R(i+1).sin(phi/2) = 2.Ri.sin(phi/2).dR/R0 . Thus all sides scale up with only one variable factor Ri . Hence all quadrilateral (super)elements have the same geometric form. The second relation follows from: Rn = R0.(1 + dR/R0)^n . QED . The little theorem is BASICally implemented by the following program fragment. Here it is assumed that the circular cylinder where the flow goes around has a radius R = R0 = 1 . The outer boundary of the mesh is limited at R = Rn = 10 . 10 REM The Labrujere Problem 20 REM ===================== 30 REM Series on Unified Numerical Approximations 40 REM 50 REM MESH GENERATOR 60 REM -------------- 70 INPUT "Number of radius meshlines (7) : ",NDR 80 INPUT "Number of angle meshlines (5) : ",NPH 90 IF NDR<=0 THEN NDR=7 100 IF NPH<=0 THEN NPH=5 110 REM --------------------------- 230 PRINT "Coordinates ..." 240 OPEN "R",#1,"coord31.bin",8 250 FIELD #1, 4 AS X\$, 4 AS Y\$ 260 PI=4*ATN(1) : DPHI=.5*PI/(NPH-1) 275 R=10 : P=R^(1/(NDR-1))-1 : REM Last meshline at R 280 FOR I=1 TO NDR 290 R=(1+P)^(I-1) : REM Gives equivalent elements 300 FOR J=1 TO NPH 310 PHI=(J-1)*DPHI : K=(I-1)*NPH+J 320 X=R*COS(PHI) : Y=R*SIN(PHI) 330 LSET X\$=MKS\$(X) : LSET Y\$=MKS\$(Y) 340 PUT #1,K : NEXT J : NEXT I 350 CLOSE #1 360 REM It should hardly be defended that repeatable elements have the immence advance that their element-matrix must be evaluated only _once_. Especially at a slow PC, where I do my homework on, this saves me a whole lot of coffee & beer ;-). 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 * E-mail: Han.deBruijn@RC.TUDelft.NL --| Fax: +31 15 78 37 87 ----* Blood