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Question: Iterative scheme to get the catenary

Question:Iterative scheme to get the catenary

Maplehelpee123 55 Maple
iterator
May 10 2013
1

Hi everyone,

I recently posted about a problem I had with an iterative scheme in 2 dimensions.. someone answered that and it worked perfectly:)

I've now moved on to a slightly more complicated version. It's in 3d but only the r and z values will be used in the iteration (trying to get the catenary)

so it's not as hard as the main problem that I'm building towards.

Basically, I'm having trouble converting the points to polar coordinates and assigning them each a [j] value, even though I've solved a similar problem at a different time.  I can't spot the problem, so if anyone could tell me why this isn't working I'd be very thankful.

Thanks,

Rach

 

Trying_to_get_the_Ca.mw

restart

with(plots):NULL

G := (1/2)*RR[j-1]*(RR[j]-RR[j-1])^2/(ZZ[j]-ZZ[j-1])+(1/2)*RR[j]*(RR[j+1]-RR[j])^2/(ZZ[j+1]-ZZ[j]):

g := proc (x, y) options operator, arrow; x^3-3*x+y^3-3*y end proc:

IP;

  

SortedPts := sort(IP[1 .. -1], proc (A, B) options operator, arrow; evalb(A[1] < B[1] or A[1] = B[1] and A[2] < B[2] or A[1 .. 2] = B[1 .. 2] and A[3] < B[3]) end proc):

CylPts := proc (IP) options operator, arrow; [sqrt(P[1]^2+P[2]^2), arctan(P[2], P[1]), P[3]] end proc:

R := Vector(n+1, proc (j) options operator, arrow; CylPts[j][1] end proc):
 

GR := unapply(Gp, [RR, ZZ, j]); GZ := unapply(Gq, [RR, ZZ, j]); GRR := unapply(Gpp, [RR, ZZ, j]); GZZ := unapply(Gqq, [RR, ZZ, j])

ErrR := Vector(n+1, [0, 1]); ErrZ := Vector(n+1, [0, 1]); rho := 1.2; epsilon := 10.^(-5); while LinearAlgebra:-Norm(`<,>`(ErrR, ErrZ)) >= epsilon do for k from 2 to n do Rk := R[k]-rho*GR(R, Z, k)/GRR(R, Z, k); ErrR[k] := Rk-R[k]; R[k] := Rk; Zk := Z[k]-rho*GZ(R, Z, k)/GZZ(R, Z, k); ErrZ[k] := Zk-Z[k]; Z[k] := Zk end do end do

FinalPlot := pointplot3d({seq([R[l], T[l], Z[l]], l = 1 .. n+1)}, color = red); display(InitialPlot, FinalPlot)

NULL



Download Trying_to_get_the_Ca.mw

 

 

 

 
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