On Tuesday, January 02, 2007 at 11:28, pedro_silva58 wrote:

> 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

best regards

--

Erik Krause

Resources, not only for panorama creation:

http://www.erik-krause.de/