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

missed?

Trausti

On Sat, May 29, 2010 at 2:26 PM, 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

>

> PS

> The exact formula is

> h = width^2 * lambda/(4*pi^2).

> Probably someone has calculated that before, but I did find no reference.

>

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