Thread Subject:

Uniform Distribution of Points about a sphere

Hello,

I am trying to distribute points uniformly about the surface of a sphere using the spiral distribution detailed in the following paper 'Comparison pf Various Spherical Antenna Array Element Distributions'.

It details a method by Rakhmanov et al, the details as follows:
0<theta<pi,
0<phi<2*pi,
Thetak=arccos(hk),
hk=-1+2(k-1)/(N-1), 1<k<N,
Phik=(Phik(-1)+((3.6)/sqrt(N))*1/sqrt(1-hk^2))*mod(2*pi), 2<k<N-1,
Phik(1)=Phik(N)=0

I have tried to implement this using the following code but the points are non-uniform down the opposing sides of the sphere:

N=100;
k=1:N;
k=fliplr(k);
hk=-1+(2.*(k-1))./(N-1);
Thetak=acos(hk);
Thetak=fliplr(Thetak);
Phik(1)=0;
for n=2:N-1;
   Phik(n)=mod((Phik(n-1)+(3.6./sqrt(N)).*1/sqrt(1-hk(n).^2)),2*pi);
end
Phik(1)=0;
Phik(N)=2*pi;
R=ones(size(Thetak));
[x1 y1 z1]=sph2cart(Thetak,Phik,R);
plot3(x1,y1,z1,'o')
axis equal
view([90 0])

Does anybody have an idea of where I might be going wrong? Thank you for your time in advance.

What toolboxes do you have (type ver on the command line)? Do you have the Mapping toolbox? (I don't but perhaps there is something in there that might help.)

"Image Analyst" <imageanalyst@mailinator.com> wrote in message <i4tpb0$m5f$1@fred.mathworks.com>...
> What toolboxes do you have (type ver on the command line)? Do you have the Mapping toolbox? (I don't but perhaps there is something in there that might help.)

Hi, thanks for replying.

No I don't have that tool box. The code/equation derivation is supposed to just give uniformly spaced spherical coordinates. Not looking for anything to fancy just wondering if there is anything obviously wrong with my code. I was thinking it might have something to do with how I have implemented the mod function?

"Christopher " <chris_quested@yahoo.co.uk> wrote in message <i4tbps$oii$1@fred.mathworks.com>...
> Hello,
>
> I am trying to distribute points uniformly about the surface of a sphere using the spiral distribution detailed in the following paper 'Comparison pf Various Spherical Antenna Array Element Distributions'.
>
> It details a method by Rakhmanov et al, the details as follows:
> 0<theta<pi,
> 0<phi<2*pi,
> Thetak=arccos(hk),
> hk=-1+2(k-1)/(N-1), 1<k<N,
> Phik=(Phik(-1)+((3.6)/sqrt(N))*1/sqrt(1-hk^2))*mod(2*pi), 2<k<N-1,
> Phik(1)=Phik(N)=0
>
> I have tried to implement this using the following code but the points are non-uniform down the opposing sides of the sphere:
>
> N=100;
> k=1:N;
> k=fliplr(k);
> hk=-1+(2.*(k-1))./(N-1);
> Thetak=acos(hk);
> Thetak=fliplr(Thetak);
> Phik(1)=0;
> for n=2:N-1;
> Phik(n)=mod((Phik(n-1)+(3.6./sqrt(N)).*1/sqrt(1-hk(n).^2)),2*pi);
> end
> Phik(1)=0;
> Phik(N)=2*pi;
> R=ones(size(Thetak));
> [x1 y1 z1]=sph2cart(Thetak,Phik,R);
> plot3(x1,y1,z1,'o')
> axis equal
> view([90 0])
>
> Does anybody have an idea of where I might be going wrong? Thank you for your time in advance.

Just a guess, but are you sure about using the fliplr
function on k to compute hk which in turn is used on Phik?
(Thetak then gets flipped again.)

It seems to me that Phik and Thetak are not
indexed to the same k, but I could be wrong.

"Christopher " <chris_quested@yahoo.co.uk> wrote in message <i4tbps$oii$1@fred.mathworks.com>...
> Hello,
>
> I am trying to distribute points uniformly about the surface of a sphere using the spiral distribution detailed in the following paper 'Comparison pf Various Spherical Antenna Array Element Distributions'.
>
> It details a method by Rakhmanov et al, the details as follows:
> 0<theta<pi,
> 0<phi<2*pi,
> Thetak=arccos(hk),
> hk=-1+2(k-1)/(N-1), 1<k<N,
> Phik=(Phik(-1)+((3.6)/sqrt(N))*1/sqrt(1-hk^2))*mod(2*pi), 2<k<N-1,
> Phik(1)=Phik(N)=0
>
> I have tried to implement this using the following code but the points are non-uniform down the opposing sides of the sphere:
>
> N=100;
> k=1:N;
> k=fliplr(k);
> hk=-1+(2.*(k-1))./(N-1);
> Thetak=acos(hk);
> Thetak=fliplr(Thetak);
> Phik(1)=0;
> for n=2:N-1;
> Phik(n)=mod((Phik(n-1)+(3.6./sqrt(N)).*1/sqrt(1-hk(n).^2)),2*pi);
> end
> Phik(1)=0;
> Phik(N)=2*pi;
> R=ones(size(Thetak));
> [x1 y1 z1]=sph2cart(Thetak,Phik,R);
> plot3(x1,y1,z1,'o')
> axis equal
> view([90 0])
>
> Does anybody have an idea of where I might be going wrong? Thank you for your time in advance.
- - - - - - - -
  Your quoted ranges for theta and phi are:

  0<theta<pi and 0<phi<2*pi,

whereas the ranges for theta and phi in sph2cart are:

 0<=theta<=2*pi and -pi/2<=phi<=+pi/2.

This suggests to me that phi and theta are reversed from what you are assuming in using sph2cart and also that sph2cart's phi is off by pi/2. The fact that sqrt(1-hk^2)) (= sin(theta)) occurs in the denominator of the increment for Phik would also strongly imply that; it should make the increments of Phik larger near the two extremes of theta at: 0 and pi. You had better check your authors' definitions for theta and phi. There are varying definitions for spherical coordinates.

  I have not been able to figure out where the constant 3.6 comes from in the authors' formula for incrementing Phik. Do you know?

  I should point out that doing a mod(~,2*k) ought to be unnecessary if you are transforming coordinates back to cartesian. Taking sines and cosines would automatically do that for you since they have a period of 2*pi.

Roger Stafford

"Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in message <i4uiuc$9ah$1@fred.mathworks.com>...
> Your quoted ranges for theta and phi are:
>
> 0<theta<pi and 0<phi<2*pi,
>
> whereas the ranges for theta and phi in sph2cart are:
>
> 0<=theta<=2*pi and -pi/2<=phi<=+pi/2.
>
> This suggests to me that phi and theta are reversed from what you are assuming in using sph2cart and also that sph2cart's phi is off by pi/2. The fact that sqrt(1-hk^2)) (= sin(theta)) occurs in the denominator of the increment for Phik would also strongly imply that; it should make the increments of Phik larger near the two extremes of theta at: 0 and pi. You had better check your authors' definitions for theta and phi. There are varying definitions for spherical coordinates.
>
> I have not been able to figure out where the constant 3.6 comes from in the authors' formula for incrementing Phik. Do you know?
>
> I should point out that doing a mod(~,2*k) ought to be unnecessary if you are transforming coordinates back to cartesian. Taking sines and cosines would automatically do that for you since they have a period of 2*pi.
>
> Roger Stafford
- - - - - - - -
  I should have given you what I would consider the appropriate transformation to cartesian coordinates (for unit radius) in your problem.

 x = cos(Phik).*sin(Thetak);
 y = sin(Phik).*sin(Thetak);
 z = cos(Thetak);

  You will note that this causes the increments in z to be of uniform size (as I suspect they should be.)

Roger Stafford

"Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in message <i4ul1s$rc6$1@fred.mathworks.com>...
> "Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in message <i4uiuc$9ah$1@fred.mathworks.com>...
> > Your quoted ranges for theta and phi are:
> >
> > 0<theta<pi and 0<phi<2*pi,
> >
> > whereas the ranges for theta and phi in sph2cart are:
> >
> > 0<=theta<=2*pi and -pi/2<=phi<=+pi/2.
> >
> > This suggests to me that phi and theta are reversed from what you are assuming in using sph2cart and also that sph2cart's phi is off by pi/2. The fact that sqrt(1-hk^2)) (= sin(theta)) occurs in the denominator of the increment for Phik would also strongly imply that; it should make the increments of Phik larger near the two extremes of theta at: 0 and pi. You had better check your authors' definitions for theta and phi. There are varying definitions for spherical coordinates.
> >
> > I have not been able to figure out where the constant 3.6 comes from in the authors' formula for incrementing Phik. Do you know?
> >
> > I should point out that doing a mod(~,2*k) ought to be unnecessary if you are transforming coordinates back to cartesian. Taking sines and cosines would automatically do that for you since they have a period of 2*pi.
> >
> > Roger Stafford
> - - - - - - - -
> I should have given you what I would consider the appropriate transformation to cartesian coordinates (for unit radius) in your problem.
>
> x = cos(Phik).*sin(Thetak);
> y = sin(Phik).*sin(Thetak);
> z = cos(Thetak);
>
> You will note that this causes the increments in z to be of uniform size (as I suspect they should be.)
>
> Roger Stafford


Roger thank you so much for such a concise answer. Simply placing the standard spherical to cartesian transform that you detailed above has done the job. I should have tried this in the first place!

With regards to the constant 3.6, the author uses this to assign the packing density. The constant is not the optimum but is a good compromise due to the distances in the hexagonal lattice that the derivation is based on shrinking at certain points.

Many thanks for your help again.

Chris

"Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in message
> I have not been able to figure out where the constant 3.6 comes from in the authors' formula for incrementing Phik. Do you know?

Roger,

I find the reference here, section 5: http://www.math.vanderbilt.edu/~esaff/texts/155.pdf

Bruno

"Bruno Luong" <b.luong@fogale.findmycountry> wrote in message <i4uqvb$k62$1@fred.mathworks.com>...
> "Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in message
> > I have not been able to figure out where the constant 3.6 comes from in the authors' formula for incrementing Phik. Do you know?
>
> Roger,
>
> I find the reference here, section 5: http://www.math.vanderbilt.edu/~esaff/texts/155.pdf
>
> Bruno
- - - - - - - -
  Thanks, Bruno. I suspected there was something of a heuristic nature to that constant, 3.6 .

Roger Stafford