5674Re: number of images at non-zero pitch?
- Jan 3, 2007erik, many thanx for your help!
--- In PanoToolsNG@yahoogroups.com, "Erik Krause" <erik.krause@...> wrote:
> 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
> 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 ;-)
yes, of course... ;-) i too was assuming rectilinear, and should have
> 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
this is just what i needed! i guessed there should be a cosine
somewhere, but didn't quite know where. oh, and subtracting half the
vertical fov is a very thoughtful touch, but may not be strictly
necessary. with it, we get the minimum overlap, without, an average
overlap (i think). at any rate, a first rate reply, as usual!
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