HEALpix: where are the equal areas ??

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 Joined: October 20 2004
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HEALpix: where are the equal areas ??
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.
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.

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 Joined: October 01 2004
 Affiliation: CESR
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I think you are wrong.
Please take a look at the recent paper available on the astroph
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
Please take a look at the recent paper available on the astroph
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

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 Joined: October 20 2004
 Affiliation: FYMA/PHYS/UCL
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I'm not convinced
If you speak about the paper astroph/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: Isit 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 nongeodesic pixel boundaries equation.
Do you know a paper where I can find a true proof of this "equal area" theorem ?
astroph/0409513 is too much summarized for that.
Best,
Laurent
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: Isit 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 nongeodesic pixel boundaries equation.
Do you know a paper where I can find a true proof of this "equal area" theorem ?
astroph/0409513 is too much summarized for that.
Best,
Laurent

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 Joined: September 23 2004
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Re: I'm not convinced
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.

 Posts: 7
 Joined: September 25 2004
 Affiliation: IAP
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/astroph/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 NorthSouth axis and EastWest 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 email),
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
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/astroph/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 NorthSouth axis and EastWest 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 email),
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

 Posts: 7
 Joined: October 20 2004
 Affiliation: FYMA/PHYS/UCL
 Contact:
Re: I'm not convinced
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.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.
However, Eric Hivon's answer explains me what I didn't understand.
Thanks.
Best,
Laurent.

 Posts: 7
 Joined: October 20 2004
 Affiliation: FYMA/PHYS/UCL
 Contact:
Indeed.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.
Perfect.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/astroph/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 NorthSouth axis and EastWest 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
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.In the polar regions, the pixel boundary equations are slightly more complicated
(too much for an email),
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
Thanks for your very detailed answer.
Best regards,
Laurent.

 Posts: 87
 Joined: February 24 2005
 Affiliation: Institute of Astronomy, Nicolaus Copernicus University
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Re: HEALpix: where are the equal areas ??
The healpix pixelisation is really two parts: (1) a spherical projection and (2) a pixelisation.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.
See Fig. 2 of astroph/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 equalarea pixelisation of a square is obvious. You don't really need a lot of integration to prove that equalarea pixelisation of the square is possible, IMHO.
There's also a more general paper on this subject: astroph/0412607.

 Posts: 87
 Joined: February 24 2005
 Affiliation: Institute of Astronomy, Nicolaus Copernicus University
 Contact:
HEALpix: where are the equal areas ??
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 astroph/0409533 .
 In the polar area, you need to check that the partial derivatives give the right values  eqs. (29)(31) give this in astroph/0409533 .
hope this helps.
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 astroph/0409533 .
 In the polar area, you need to check that the partial derivatives give the right values  eqs. (29)(31) give this in astroph/0409533 .
hope this helps.