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5674Re: number of images at non-zero pitch?

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  • pedro_silva58
    Jan 3, 2007
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      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
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