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## Remapping question

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• I have a Pano2VR project with input of six cube faces of 2000x2000 pixels each. I want to create a transformation to an angular map little planet (circular).
Message 1 of 3 , Nov 2, 2009
I have a Pano2VR project with input of six cube faces of 2000x2000
pixels each.

I want to create a transformation to an angular map "little planet"
(circular).

How do I calculate the maximum resolution possible given the size of the
cube faces?

Many thanks,
Robert

--
Manor Photography 07890 564889
http://www.manor-photography.com
• ... Just an estimate: 2000px cube face width has the same resolution as 6282px equirectangular width. You can do little planet with a 360° fisheye mapping
Message 2 of 3 , Nov 2, 2009

> I have a Pano2VR project with input of six cube faces of 2000x2000
> pixels each.
>
> I want to create a transformation to an angular map "little planet"
> (circular).
>
> How do I calculate the maximum resolution possible given the size of the
> cube faces?

Just an estimate: 2000px cube face width has the same resolution as
6282px equirectangular width. You can do "little planet" with a 360°
fisheye mapping which would be the same width as the equirect. It very
much looks as if Pano2VR uses this mapping.

However, the better mapping for "little planet" is stereographic, which
stretches the outer regions. PTGui f.e. suggest 300° for "little planet"
with about 14000px for a 6000px wide input equirect...

Perhaps you find some more info following the links on
http://wiki.panotools.org/Stereographic_Projection.

best regards
--
Erik Krause
http://www.erik-krause.de
• ... The formula is quite simple: 2 * equirect width * tan(FoV/4) / pi The result is the stereographic image width and FoV is the angle of the desired field
Message 3 of 3 , Nov 2, 2009
Erik Krause wrote:

> However, the better mapping for "little planet" is stereographic, which
> stretches the outer regions. PTGui f.e. suggest 300° for "little planet"
> with about 14000px for a 6000px wide input equirect...

The formula is quite simple:

2 * equirect width * tan(FoV/4) / pi

The result is the stereographic image width and "FoV" is the angle of
the desired field of view of the stereographic image.

However, Pano2VR most likely doesn't use this mapping but the fisheye
one, where the output image width is the same as the equirect image width.

best regards
--
Erik Krause
http://www.erik-krause.de
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