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ptadjust plugin

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  • crane@ukonline.co.uk
    hello, can somebody smarter than me tell me what is the relationship between the horizontal field of view and the width size in the ptadjust photoshop plugin
    Message 1 of 5 , Dec 2, 2008
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      hello,
      can somebody smarter than me tell me what is the relationship between the
      horizontal field of view and the width size in the ptadjust photoshop plugin
      from epaperpress when extracting a rectilinear from an equirectangular.
      I'd like to extract a 90 degree horizontal view but then have a bit extra
      either side please.

      regards

      mick






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    • Jim Watters
      ... There has been several discussions on this list as to what should be the optimal dimensions for cube faces when extracting them to make a QTVR. For 90 deg
      Message 2 of 5 , Dec 2, 2008
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        crane@... wrote:
        > hello,
        > can somebody smarter than me tell me what is the relationship between the
        > horizontal field of view and the width size in the ptadjust photoshop plugin
        > from epaperpress when extracting a rectilinear from an equirectangular.
        > I'd like to extract a 90 degree horizontal view but then have a bit extra
        > either side please.
        >
        > regards
        >
        > mick
        >
        >
        There has been several discussions on this list as to what should be the
        optimal dimensions for cube faces when extracting them to make a QTVR.
        For 90 deg use width / pi for optimal, to ensure no loss of detail width / 3

        What you want to do is keep the same (or a little better) pixels per
        degree as the original. For more details see Resolution of Various
        Panorama Formats http://www.worldserver.com/turk/quicktimevr/panores.html


        --
        Jim Watters

        jwatters @ photocreations . ca
        http://photocreations.ca
      • crane@ukonline.co.uk
        ... That doesn t explain very clearly, assumes a lot of prior understanding. eg: The focal length is defined as the distance to the imaging surface at the
        Message 3 of 5 , Dec 3, 2008
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          Quoting Jim Watters <jwatters@...>:

          > What you want to do is keep the same (or a little better) pixels per
          > degree as the original. For more details see Resolution of Various
          > Panorama Formats http://www.worldserver.com/turk/quicktimevr/panores.html

          That doesn't explain very clearly, assumes a lot of prior understanding.
          eg:
          "
          The focal length is defined as the distance to the imaging surface at the
          center of projection.

          For a cubic panorama, this is the distance to the center of one of the faces.
          "
          I have no idea how to calculate that.

          regards


          mick

          > Jim Watters


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        • crane@ukonline.co.uk
          ... yes I know i am replying to my own post. has anybody explained in terms that are readily understandable what happens to the arithmetic from real world to
          Message 4 of 5 , Dec 3, 2008
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            > Quoting Jim Watters <jwatters@...>:
            >
            > > What you want to do is keep the same (or a little better) pixels per
            > > degree as the original. For more details see Resolution of Various
            > > Panorama Formats http://www.worldserver.com/turk/quicktimevr/panores.html
            >
            > That doesn't explain very clearly, assumes a lot of prior understanding.
            > eg:
            > "
            > The focal length is defined as the distance to the imaging surface at the
            > center of projection.
            >
            > For a cubic panorama, this is the distance to the center of one of the faces.
            >
            > "
            > I have no idea how to calculate that.

            yes I know i am replying to my own post.

            has anybody explained in terms that are readily understandable what happens to
            the arithmetic from real world to sensor image to stitching to projecting onto
            the monitor , in QT/ flash etc because I realize I do not have a clue as to
            the maths involved and I would like to.

            regards


            mick

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          • Erik Krause
            ... If you take fisheye images you have a more or less linear mapping between distances on the sensor and angles of view in the world. This is: A certain pixel
            Message 5 of 5 , Dec 3, 2008
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              crane wrote:
              > has anybody explained in terms that are readily understandable what
              > happens to
              > the arithmetic from real world to sensor image to stitching to
              > projecting onto
              > the monitor , in QT/ flash etc because I realize I do not have a clue as to
              > the maths involved and I would like to.

              If you take fisheye images you have a more or less linear mapping
              between distances on the sensor and angles of view in the world. This
              is: A certain pixel distance is always a certain angle distance, no
              matter where on the sensor you measure.

              For equirectangular images it is just the same: The width represents
              360° the height 180°, that's why the aspect ratio always is 2:1. How to
              calculate this is outlined on
              http://wiki.panotools.org/DSLR_spherical_resolution

              If it comes to rectilinear it gets a bit more complicated, since the
              mapping is a tangent one. But it gets fairly easy if you do it in
              geometry. Draw a circle. Now draw a surrounding square. Where the square
              touches the circle the pixel density is equal to that one of the circle.
              One cube face is the diameter of the circle wide. The diameter of the
              circle (and hence the cube face) is circumference / pi.

              If the square would not be surrounding but inside the circle, pixel
              information would be lost, because the information formerly on the
              circle must be compressed to a smaller distance, which means to less
              pixels. If you map to equirect again (or map to an arbitrary view in the
              viewer) this information can not be re-created.

              Hence if you extract a cube face from an equirectangular and you don't
              want to loose any information you need a cube face size of equirect
              width / pi.

              However, as Hans Nyberg demonstrated recently: If you remap from a
              maximum size equirect to cube faces for a viewer (be it Flash, QT or
              whatever) the best compromise between size and quality is at about 70%
              of that size, since any bayer interpolated image can be reduced by about
              70% with negligible loss of image quality. Hence you can divide equirect
              width by about 4.4 to get the optimum cube face size.

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