Fabio,
I have been following the "Gigapixel" correspondence with increasing
dismay. Your idea of using the resolution around the circumference
of the sphere that forms the spherical projection is simple, logical,
and seems to me to provide a sensible and (dare I say it) "intuitive"
measure of how many useful, meaningful pixels there are in the whole
panorama. There is a one-to-one correspondence between the area of
the sphere and the amount of visual information (pixels) in the image.
This does leave the problem that there really is no point in including
lots of sky in a very large panorama (there isn't always a lot of
point in including sky in your typical spherical panorama, either, but
we usually do it).
For a rectilinear image the area of the rectangle (measured in pixels
along the sides) is an equally intuitive and unambiguous measure of
the useful visual data in the image. It's the equirectangular image
that is misleading. You do NOT get a useful handle on the amount of
visual information by multiplying its height by its width...
That leave the problem of the cylindrical panorama (which many of
these wannabe gigapixel images are). But the number of pixels is
always going to be less than those of the corresponding sphere and the
appropriate ratio is very simply calculated. The calculation is left
to the reader as an exercise. <grin>
Roger W
On Sat, 23 Oct 2010 01:47:08 +0900, Fabio Bustamante
<#removed#> wrote:
,,,,,
> I was thinking about Mark's suggestion right before he posted it. It
> seems a good approach to the problem, to use original pixels as
> reference. IMO much better than taking an arbitrary planar projection
> and multiplying lengh by width. However even in the source images,
> pixels in theory have different relation with the sphere. We must keep
> in mind that the images we get from the camera to make the pano have
> arbitrary projection themselves. So in the case of circular fisheye
> lenses (which in theory have an f-theta projection) a square of, say,
> 20x20 pixels at the center of the image represent a very different real
> world angular area than a square of 20x20 pixels at the border of the
> circle.
>
> I thought about a different approach. Take an equirectangular image of
> 6000x3000. In theory the least distorted part of the pano is the central
> 6000px line, the horizon. The more you go up or down this 6000 wide line
> starts representing less information, up to the extreme of the top and
> bottom lines having 6000 pixels to represent a single point. But we do
> have a horizon of "pure", unstretched information.
>
> My idea is to take the horizon as a reference unit. In this case we
> would have a horizon 6000 long (not pixels, just... "units of
> information"). Then we use this number as the circunference of the
> sphere and go after the area. For this particular sphere the area would
> be of 11.459.156 (and not the 18M we get from 6K x 3K). So if this
> approach is right, the panorama represented by a 6000x3000 image would
> be roughly the equivalent of a 11.4mp image with the sharpness found on
> it's horizon.
--
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