HEALpix: where are the equal areas ??

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Laurent Jacques
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HEALpix: where are the equal areas ??

Post by Laurent Jacques » October 21 2004

Hi,

If there are some persons specialized in HEALpix on this forum, can somebody explain me why this "Hierarchical Equal Area isoLatitude Pixelisation" contains the two words "Equal Area" ?

IMHO, there is no truly equal area in this grid, excepted if you define a new definition of area on the sphere, as it is defined in the HEALpix original paper, i.e.
Dz * Dphi !

For me, this equal area concept is just a rough approximation. What do you think ?

Best,
Laurent.

Anthony Banday
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Post by Anthony Banday » October 21 2004

I think you are wrong.

Please take a look at the recent paper available on the astro-ph
archive. You will find a prescription for the pixel boundaries there.
Formal integration will demonstrate that the pixels are exactly equal area.
This was a driving requirement in the development of HEALPix, and
great care was taken to ensure this condition was met.

Tony Banday

Laurent Jacques
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I'm not convinced

Post by Laurent Jacques » October 22 2004

If you speak about the paper astro-ph/0409513

Title: HEALPix -- a Framework for High Resolution Discretization, and Fast Analysis of Data Distributed on the Sphere. K. M. Gorski, E. Hivon, A. J. Banday, B. D. Wandelt, F. K. Hansen, M. Reinecke, M. Bartelman.

I didn't find any clues about the integration realized: Is-it done in (z=cos(theta),phi) coordinate or truly in (theta,phi) ? The former is not an area computing on the sphere by on the cylinder even if we take into account the non-geodesic pixel boundaries equation.

Do you know a paper where I can find a true proof of this "equal area" theorem ?
astro-ph/0409513 is too much summarized for that.

Best,
Laurent

Antony Lewis
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Re: I'm not convinced

Post by Antony Lewis » October 22 2004

HealPix is constructed by division of the sphere into 12 equal large pixels, then recursiverly diving each pixel into 4 equal size parts. So you get 12, 12 x 4, 12 x 4 x 4, 12 x 4x 4x 4, ... pixels at each higher devision, and by construction they are equal area. Just look at Fig 4 in the paper of how they are constructing higher resolutions.

Eric Hivon
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Post by Eric Hivon » October 22 2004

Hi Laurent,

what is your problem with the dz * dphi integration ?
You know that the integration element on the sphere is
d omega = sin (theta) * d theta * d phi,
which is equal to d omega = d z * d phi if you define z = cos (theta).
Since -1 < z < 1, the total area available in (z,phi) is 4 Pi, llike on the sphere.

In other terms, the mapping from the sphere (theta, phi) to cylindrical coordinates (z, phi) preserves the area.
Therefore, any area measured in (z,phi) will be the same on the sphere.

As for the exact calculation, in the equatorial region,
the pixel boundaries have the form
zp = 2/3 - a*k + b*phi,
zm = 2/3 - a*k' - b*phi,
where a=4/3/Nside and b=8/3/Pi and k and k' are integers
(see http://arxiv.org/abs/astro-ph/0409513 Eq 22) .
The pixel boundaries are therefore straight lines (in (z,phi)) which are either increasing (zp) or decreasing (zm) with phi, and the pixel is a diamond, with its diagonals along
the North-South axis and East-West axis respectively (ie, crossing at a right angle).

A pixel enclosed in the lines zp and zm with the parameters k, k+1 and k', k'+1 respectively,
will have a distance Dz = a between its North and South corners,
and a distance Dphi = a/b between its East and West corners.

As you know, the surface of a diamond with orthogonal diagonals
if a half of the products of its diagonal. Therefore
S = Dz * Dphi / 2 = a * a / b / 2 = 2 Pi / (3 * Nside * Nside) = 4 Pi / Npix

In the polar regions, the pixel boundary equations are slightly more complicated
(too much for an e-mail),
requiring the integration to be done in details (it is easier to divide the pixel in 4 quadrants
along its diagonals, compute the 4 integrals and add them) to find a surface of
4 Pi / Npix per pixel.

Hope this helps,

Eric

Laurent Jacques
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Re: I'm not convinced

Post by Laurent Jacques » October 24 2004

Antony Lewis wrote:HealPix is constructed by division of the sphere into 12 equal large pixels, then recursiverly diving each pixel into 4 equal size parts. So you get 12, 12 x 4, 12 x 4 x 4, 12 x 4x 4x 4, ... pixels at each higher devision, and by construction they are equal area. Just look at Fig 4 in the paper of how they are constructing higher resolutions.
This doesn't explain the equal area property, it explains just the subdivision. It is not clear that you truly divide each pixel in 4 equal size part without to deal with equation.

However, Eric Hivon's answer explains me what I didn't understand.

Thanks.

Best,
Laurent.

Laurent Jacques
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Post by Laurent Jacques » October 24 2004

Eric Hivon wrote:Hi Laurent,

what is your problem with the dz * dphi integration ?
You know that the integration element on the sphere is
d omega = sin (theta) * d theta * d phi,
which is equal to d omega = d z * d phi if you define z = cos (theta).
Since -1 < z < 1, the total area available in (z,phi) is 4 Pi, llike on the sphere.

In other terms, the mapping from the sphere (theta, phi) to cylindrical coordinates (z, phi) preserves the area.
Therefore, any area measured in (z,phi) will be the same on the sphere.
Indeed.
As for the exact calculation, in the equatorial region,
the pixel boundaries have the form
zp = 2/3 - a*k + b*phi,
zm = 2/3 - a*k' - b*phi,
where a=4/3/Nside and b=8/3/Pi and k and k' are integers
(see http://arxiv.org/abs/astro-ph/0409513 Eq 22) .
The pixel boundaries are therefore straight lines (in (z,phi)) which are either increasing (zp) or decreasing (zm) with phi, and the pixel is a diamond, with its diagonals along
the North-South axis and East-West axis respectively (ie, crossing at a right angle).

A pixel enclosed in the lines zp and zm with the parameters k, k+1 and k', k'+1 respectively,
will have a distance Dz = a between its North and South corners,
and a distance Dphi = a/b between its East and West corners.

As you know, the surface of a diamond with orthogonal diagonals
if a half of the products of its diagonal. Therefore
S = Dz * Dphi / 2 = a * a / b / 2 = 2 Pi / (3 * Nside * Nside) = 4 Pi / Npix
Perfect.
In the polar regions, the pixel boundary equations are slightly more complicated
(too much for an e-mail),
requiring the integration to be done in details (it is easier to divide the pixel in 4 quadrants
along its diagonals, compute the 4 integrals and add them) to find a surface of
4 Pi / Npix per pixel.
OK. IMHO, these exact integrals, the point which was for me the most unclear, have to be written somewhere in further papers describing HEALpix ! It's important for rigorous reading.
Hope this helps,

Eric

Thanks for your very detailed answer.

Best regards,
Laurent.

Boud Roukema
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Re: HEALpix: where are the equal areas ??

Post by Boud Roukema » February 24 2005

Laurent Jacques wrote:Hi,

If there are some persons specialized in HEALpix on this forum, can somebody explain me why this "Hierarchical Equal Area isoLatitude Pixelisation" contains the two words "Equal Area" ?

IMHO, there is no truly equal area in this grid, excepted if you define a new definition of area on the sphere, as it is defined in the HEALpix original paper, i.e.
Dz * Dphi !

For me, this equal area concept is just a rough approximation. What do you think ?

Best,
Laurent.
The healpix pixelisation is really two parts: (1) a spherical projection and (2) a pixelisation.

See Fig. 2 of astro-ph/0409533 for a picture of (1) the projection. Each of the 1/12 parts of the sphere are projected onto squares of equal area.

Part (2), the pixelisation, is then obvious, since equal-area pixelisation of a square is obvious. You don't really need a lot of integration to prove that equal-area pixelisation of the square is possible, IMHO.

There's also a more general paper on this subject: astro-ph/0412607.

Boud Roukema
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Joined: February 24 2005
Affiliation: Torun Centre for Astronomy, University of Nicolaus Copernicus
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HEALpix: where are the equal areas ??

Post by Boud Roukema » February 24 2005

i guess i forgot to say: there's still the question of whether or not the spherical projection is an equal area projection - e.g. the Mercator projection is clearly not an equal area projection.

In (1) the spherical projection for healpix:

- In the equatorial area, this is done by using sin(theta) instead of theta e.g. see eq. (7) of astro-ph/0409533 .

- In the polar area, you need to check that the partial derivatives give the right values - eqs. (29)-(31) give this in astro-ph/0409533 .

hope this helps.

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