NVIDIA Are the Industry Standard Shader Driving Force?

Discussion in 'Graphics and Semiconductor Industry' started by Dave Baumann, Jun 19, 2003.

  1. darkblu

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    see, your 'arc' comment shows you're just still thinking in terms of spherical eyes :)

    don't think in arcs, think in planar units of your 'planar-eye'. the surface of the paper being mapped onto one square unit of your planar-eye area is constant when that sheet is flat-on your eye.
     
  2. RussSchultz

    RussSchultz Professional Malcontent
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    Erg. So you're saying if 1 sq cm of the center of the surface being viewed maps into 1mm of the center of the the 'screen', then it would do the same on the outside?

    How in the hell does that end up being correct? The geometry just does not work out when you keep the 'screen' the same size, keep the object the same size, but push the 'screen' closer to the object (and still encompass the object) Or, in other words, increase the FOV toward 180 degrees.

    (Or is it not supposed to?)
     
  3. darkblu

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    yes, russ, exactly.
    because your planar eye 'samples' the image by planar units (i.e. square units, or area), not by arc-units. this means that starting from the center of your eye you don't advance in degrees (or any arc-units) but in planar units. i intended to post a sketch image last night but my hosting abilities are rather limited. so i'll try ascii. let's start with a scheme of what you think (i.e. eye samples by arc-units), and then we'll see how things actually are.

    Code:
    p2|...
      |   ...
      |      ...
      |         ...
      |            ...
      |               ...
    p1|.......           ...
      |       .........     ...
      |                ...........|
    p0|...........................|..
    

    there are 3 rays cast from your eye to the paper sheet: one central to the eye (let's call it ray0) and two advancing to the periphery (ray1 and ray2, respectively). the essential property of those rays is that each of them offsets from the previous by a constant arc-magnitude, i.e. ang(ray0, ray1) = ang(ray1, ray2). now let each of those rays intersect the paper sheet at points p0, p1 and p2, repspectively. what we can say about those, as you have rightfully pointed out, is that |p0, p1| < |p1, p2|.

    now, what i've been trying to tell you, OTH, is that your planar eye 'sampling' does not advance by arc-units starting from the center, but by planar units. this means that the property of rays 0 through 2 which says there is a constant arc-magnitude between two neighbouring rays actually does not hold, i.e. ang(ray0, ray1) != ang(ray1, ray2), but rather another property holds in this case: let the rays intersect our eye-plane at e0, e1 & e2, respectively, then |e0, e1| = |e1, e2|. i.e. our rays advance to the periphery of the eye by a constant planar magnitude, measured in the eye-plane. do you see now why i told you you kept thinking in spherical terms? once you realise that |e0, e1| = |e1, e2| from that it naturally follows that for the points of ray-paper intersection p0 through 2 now another property holds true, namely |p0, p1| = |p1, p2|.
    which proves what i told you, namely that if 1 sq cm of the center of the surface being viewed maps into 1mm of the center of the the 'screen', then it would do the same on the outside.
     
  4. RaolinDarksbane

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    I haven't tested this yet on my 9800, but it seems that the eyelashes are fixed with the new opengl wrapper for Dawn.


    http://www.rage3d.com/articles/atidawning/

     
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