Re: Panorama Resolution Limits
- --- In PanoToolsNG@yahoogroups.com, "robert" <image360@...> wrote:
>Depends on the pixelsize of the sensor. Assuming 6 micrometer I get a minimum focal length of 9.5mm for the 10000 pixel wide panorama, and 183mm for the 192000 pixel wide panorama. Sensors with smaller pixelsize get by with shorter focal lengths.
> Interesting thought. What mm lens would this mazimum imply?
> Also, I'm assuming the optical laws might not apply to non lens capture methods;Correct.
>Well, focus stacking is like stereo panoramas: in theory it does not work, but in reality you get some useful results. The problem is that blurred foreground features may cover sharp background features, even if you did everything right (keep location of entrance pupil constant, readjust magnification). Therefore, a resolution cannot be guaranteed, or theoretically predicted. The limit I am referring to does not apply.
> And would not focus stacking allow for complete DOF.
- --- In PanoToolsNG@yahoogroups.com, Trausti Hraunfjord <trausti.hraunfjord@...> wrote:
>If you're not concerend with DOF, then the only parameter determining resolution is the physical aperture size, ~2.4m for the Hubble telescope. This would be a panorama 27 million pixel wide, and the hyperfocal distance 10568km. From these results you can conclude that we have miserable wether right now preventing me from riding my bike.
> Thanks for that Helmut. I have caught the same question once upon a time,
> but since I am not into math at all, I simply accepted there being no limit
> at all.
> My thinking went: If one uses the Hubble telescope to zoom in as far as it
> can, then shoot a 360 pano, the result would surely be more than the 512
> meters. And for an even bigger pano, one could use the coming James Webb
> telescope for the same, and get many times bigger pano as a result.
> Am I completely off track in that thinking? If I am, then what is it I
- :) ouch! 27 million pixels ... that's a panorama 72 kilometers wide and 36
km high.... MOAP (Mother Of All Panoramas), miniaturizing Mt. Everest,
which is "only" a little under 9 km's above sea level where highest.
Many thanks for the numbers, quite interesting to think about.
On Sun, May 30, 2010 at 3:23 AM, hd_de_2000 <der@...> wrote:
> If you're not concerend with DOF, then the only parameter determining
> resolution is the physical aperture size, ~2.4m for the Hubble telescope.
> This would be a panorama 27 million pixel wide, and the hyperfocal distance
> 10568km. From these results you can conclude that we have miserable wether
> right now preventing me from riding my bike.
> Helmut Dersch
[Non-text portions of this message have been removed]
>:) ouch! 27 million pixels ... that's a panorama 72 kilometers wideThere's no hard-wired link between pixel count and physical size. At
300ppi (approx 118 pixels/cm) you'd be looking at a panorama of only
1.42 miles or 2.28 km wide...
...heh. Did I really say "only"?
- hey groovy, that's the width of this pano - 192000 pixels :-)
it seems that any larger gigapixels out there have been refocused to shoot each row (which i did not do, shooting this)
--- In PanoToolsNG@yahoogroups.com, "hd_de_2000" <der@...> wrote:
> While browsing some online gigapixel panoramas, I wondered what physical resolution limits exist based on fundamental optical laws. Combining diffraction limit & depth-of-field formulas yields a universal minimum hyperfocal distance h for photographing 360 degree panoramas:
> h = width^2 * 1.39 * 10^-8 m
> Some numbers: w=10000 pixel: h=1.39m
> w=192000 pixel: h=512m
> This limit is independent of lens and sensor, but may not be reached with some lens/sensor combinations.
> For close-up panoramas a similar formula for the minimum object distance a and a desired depth-of-field da may be derived:
> a/h = da/a (da << a)
> Example: if the desired depth of field is +/- 10%, then with the same numbers as above, the minimum object distance is 0.139m fo the 10000 pixel wide panorama, and 51.2m for the 192000 pixel wide panorama.
> Helmut Dersch
> The exact formula is
> h = width^2 * lambda/(4*pi^2).
> Probably someone has calculated that before, but I did find no reference.