How can I bin the polar angles of a unit sphere for non-equal bin widths?

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Homework Help Overview

The original poster is exploring how to bin polar angles on a unit sphere to achieve non-equal bin widths, aiming for approximately equal area distribution across the sphere. The discussion involves binning the azimuthal angle with equal widths while adjusting the polar angle bins to have larger widths near the poles and smaller widths near the equator.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Some participants suggest using equal spacing for the cosine of the polar angle as a method for binning. Others provide explanations regarding the surface area differential on a sphere and how it relates to the polar angle.

Discussion Status

The discussion includes attempts to clarify the reasoning behind the proposed methods for binning polar angles. Some participants have provided additional explanations, and there appears to be a productive exchange of ideas regarding the mathematical principles involved.

Contextual Notes

The original poster is seeking assistance within the constraints of homework guidelines, which may limit the type of solutions or methods that can be directly provided.

susantha
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hello,
I am trying to bin the unit sphere so that each bin has approximately equal area on the sphere. I hope binning the azimuthal angle (0...2*pi) to equal bin widths. Then i need to bin the polar angle(0...pi) so that at poles(angles close to 0 and pi) bin widths are large and close to equator(angles close to pi/2) the bin widths are small. I would appreciate any kind of help for binning the polar angles for non-equal bin widths.
Thanks in advance.
Susantha
 
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Use equal spacing for the cosine of the polar angle.
 
Last edited:
mathman said:
Use equal spacing for the cosine of the polar angle.

Could you please give little more explanation.
Thanks
 
It just comes from the fact that the surface area differential on a sphere is given by: dA=sinφdφdθ, where φ is the polar angle (0,π) and θ is the azimuthal angle (0,2π). Integrate dA over some range in φ results in the cosine difference.
 
Further explanation: Unit sphere - then the radius of a small circle at angle φ is sinφ. A circular strip of width dφ would have an area 2πsinφdφ.
 
Now i got it. Thank you very much for your reply.
 

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