erik, 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

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 ;-)

yes, of course... ;-) i too was assuming rectilinear, and should have

been explicit.

> 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!

thanx again,

pedro