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Uniformity in polar coordinates

Imagine a polar web of similar quadrilaterals, as shown in the picture below:

Start with an innermost circle at a radial position R0 ; suppose that the mesh ends at a radial position RN . Let the vertex points of the quadrilaterals in the mesh be given by: xi,j=Ricos(αj);yi,j=Risin(αj) Where αj=j/M2π with j=0,,M1 and i=0,,N . Then we have the following
Theorem. The radial positions in a mesh consisting of similar quadrilaterals are given by: Ri=R0.(RNR0)i/Nwhere:i=0,,N Proof. R0=(RN/R0)0/N and RN=(RN/R0)N/N to begin with.
In general, there are two edges of the quadrilaterals which have the same length, equal to: Ri+1Ri=R0.[(RNR0)(i+1)/N(RNR0)i/N]=Ri.[(RNR0)1/N1] Now αj+1αj=2π/M ; let c=(RN/R0)1/N for short; then: Ri+1Ri=Ri(c1) .

The lengths of the two other edges of the quadrilaterals are (see above figure): Ri.2.sin(π/M)andRi+1.2.sin(π/M)=Ri.c.2.sin(π/M) Thus all edges scale up with only one variable factor Ri . Edges counterclockwise: Ri×[2.sin(π/M),(c1),c.2.sin(π/M),(c1)] Therefore all quadrilaterals have the same geometric shape. With other words: they are all similar. But we are not finished yet.

What we furthermore need is the following Lemma.
wa(xA)2+wb(xB)2=(wa+wb)(xwaA+wbBwa+wb)2+wawbwa+wb(AB)2 Where x,A,B and wa,wb>0 are real numbers.
Proof. wa(xA)2+wb(xB)2=wa(x22Ax+A2)+wb(x22Bx+B2)=(wa+wb)x22(waA+wbB)x+(waA2+wbB2)=(wa+wb)[x22waA+wbBwa+wb+(waA+wbBwa+wb)2](wa+wb)(waA+wbBwa+wb)2+(waA2+wbB2)=(wa+wb)(xwaA+wbBwa+wb)2+(waA+wbB)2+(wa+wb)(waA2+wbB2)wa+wb=(wa+wb)(xwaA+wbBwa+wb)2+(w2aA2+2wawbAB+w2bB2)+w2aA2+w2bB2+wawbA2+wawbB2wa+wb=(wa+wb)(xwaA+wbBwa+wb)2+wawb(A22AB+B2)wa+wb=(wa+wb)(xwaA+wbBwa+wb)2+wawbwa+wb(AB)2 Without Loss Of Generality (WLOG) we can put for the weighting factors: w1+w2=1 . Then we have: wa(xA)2+wb(xB)2=[x(waA+wbB)]2+wawb(AB)2 In particular when w1=w2=1/2 : 12[(xA)2+(xB)2]=(xA+B2)2+(AB2)2 Application. We have found for the edges: Ri×[2.sin(π/M),(c1),c.2.sin(π/M),(c1)] What we want now is to determine the constant c in such a way that the following approximate equalities hold as good as possible: 2.sin(π/M)c1andc.2.sin(π/M)c1 Meaning that the quadrilaterals are similar to squares as good as possible. Formulated in a Least Squares sense, this in turn means that: 12[(xA)2+(xB)2]=minimum(c) with the substitutions x=c1 , A=2.sin(π/M) and B=c.2.sin(π/M) . The minimum is obviously attained when x=(A+B)/2 or: c1=sin(π/M)+csin(π/M)c[1sin(π/M)]=1+sin(π/M)1+sin(π/M)1sin(π/M)=c=(RNR0)1/NRNR0=(1+sin(π/M)1sin(π/M))N If the outer mesh radius is normed to one, ipse est RN=1, then we have for the inner mesh radius: R0=(1sin(π/M)1+sin(π/M))N