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• ## Re: Number of Images in a Multi-row Panorama

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• greetings! back in january, erik krause kindly and expertly replied to a similar question of mine (generic, not tied to any lens in particular). you should be
Message 1 of 16 , Sep 18, 2007
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greetings!
back in january, erik krause kindly and expertly replied to a similar
question of mine (generic, not tied to any lens in particular). you
should be able to apply it to your case.
i quote...

<quote>
> we all know the formulas to compute the number of images required at
> the horizon, given pano field of view and image overlap, but i would
> like to be able to calculate the numbers of images at non-zero
> pitches. simple reasoning indicates they must be less than at the
> horizon, but i can neither find nor deduce the formulas. can anyone
help?

Some trigonomtry should help - lets see:
I assume rectilinear images since they have the same vertical and
horizontal FoV throughout the whole image - for fisheyes it would be
too hard to calculate exactly since fullframe ones have a far larger
FoV along the edges than through the center and for cuircular ones
you probably won't need to calculate ;-)

Rectilinear images will have least overlap at the side which is
nearest to the horizon. Hence you need to know the pitch angle of
that side. You get this by subtracting half the vertical FoV from the
pitch angle of the image (center). If you get a negative value here
(this is the case if the image crosses the horizon) you are done.
Same formula as for horizontal shooting applies.

For a positive value the circumference which is to cover is by
cosinus of that angle smaller than the horizon. If you for example
pitch the camera 60° up and your image has 30° vertical F0V the lower
side will be at 45° and hence need to cover cos(45) = 0.7071 times
the horizon. Given a 20° horizontal FoV and 20% overlap you need
cos(45)*360/(20*0.8)=15.9 -> 16 images (instead of 23 for the
horizon). If you have the exact (floating point) number of images for
the horizon you can use the factor directly: 22.5 * 0.7071 = 15.9
</quote>

hope this helps!
cheers,
pedro

--- In PanoToolsNG@yahoogroups.com, "Wayne Heil" <kwheil@...> wrote:
>
> Can anyone tell me the number of rows and images per row required to
take an
> ultra high-res multi-row panorama using a 60mm lens with a Nikon D200
> camera?
>
> I have used online calculators to determine that 52 images are
required in
> portrait mode with a 30% overlap. However, I'm have trouble
determining how
> many images should be in each of the other required rows. From the
> calculators it seems I may need as many as 17 rows, 1 horizontal
with 8 rows
> up and 8 rows down. I would assume I need fewer images as I proceed
from the
> horizontal row, but how many fewer?
>
...