Parallax and Perspective

OK, I'm clearly out of my depth running on just memory here. Let me try and explain, instead, things as I would look at them.

I'll be working off the following picture:


Here, I have the viewing plane aligned to the origin of the coordinate system, such that it's perpendicular to the z axis and in-line with the x and y axes. I've further defined a two-dimensional coordinate system on the plane with an origin where z intersects the plane and with x' and y' coordinate axes (which I forgot to note down, oops!) parallel to the x and y axes of the space. From here on out, I picked a point P defined by its central vector (e.i. defined as P(x,y,z)), and have marked out the physical representations of its coordinates as down on the chart. Red is z, blue is y and green is X. The A, B, A' and B' points I'm using just so that I can call things more easily, and despite having noted the angle between the Z axis and the image of the O->P vector in the xz plane, I probably won't call it by name.

Oh, and before I forget, I've positioned the distance of the viewing plane away from the origin as 1 unit, just to simplify the calculations I'm doing by hand.

Here's what I have in mind: First let's find the image of the O->P vector in the xz plane. That would be OB (noting it as a fixed-length line, lacking a better English term) as B is the image of P on xz. Furthermore, the image of point B onto the z axis is point A. This gives us a triangle OAB, with a right (90 degree) angle at point A. It may not look like that's where it is, but remember - both OA and BA are images of the coordinate axes, which are perpendicular between each other.

By virtue of this being a right angle triangle, I can calculate the tangent of angle BOA, which comes to AB/OA (again, defined just as length values, not as vectors). But because AB = x and OA = z, the tangent of the angle comes up to x/z. Now, because the viewing plane is perpendicular to the x and y axes, and A' and B' are intersection points of OA and OB with the plane, the OA'B' triangle is, in fact, similar to OAB, and because it shares an angle to which we already know the tangent, we can recalculate. Specifically, within triangle OA'B', the tangent of angle B'OA' can be calculated as A'B'/OA'. However, A'B' is actually the x' value of point P by the coordinate system of the viewing plane, and OA' is the distance to the viewing plane, which I defined as 1. This gives us a tan(B'OA') = x'/1 = x'. What's more, because we're looking at the same tangent, that gives us x' = x/z.

I can do an identical process and find out that y' = y/z. I don't want to bore you with it, it's nothing interesting to describe.

This gives me the 3D to 2D transformation of
|x' = x/z
|y' = y/z
For a coordinate system Ox'y' as defined in the viewing plane.

This also tells me one easy thing: A line parallel to the viewing direction will be directed towards the middle of the "screen," crossing it in infinity. The easy answer as to why is that both horizontal and vertical coordinates will decrease, as x and y remain constant, but z (the denominator) increases, causing each subsequent point on the line to converge on the centre as the distance increases. You can also look at it as the angle decreasing, inclining its tangent off centre towards zero, though that's a less interesting way of putting it.

Also, because y' is not affected by the x coordinate of the point, a line parallel to the horizon (where y and z are constant) will transcribe into a line "on screen." Moreover, x' will increase linearly, giving me the function of a line, which I don't feel like bothering to define. It isn't relevant either way.

Basically, that disproves my original claim that lines parallel to the horizon would converge in infinity left and right, necessitating that they be drawn curved, but it does so at the cost of redefining my original model which required refracting through a sphere (and a rather more complicated calculation, involving square roots and variables squared).

Nevertheless, this "flat" model still produces noticeable distortion as you progress away from its centre, only its distortion is angular, rather than curved. I'm not sure which would be more appropriate to describing human eyesight as perceived by the brain, but from the look of things, the flat model is remarkably easier to calculate, which would make it remarkably more suitable for real-time rendering.

*edit*
And again - I graduated from degree a few years ago, and this particular field I studied at the beginning of the course, so it's quite possible I've forgotten a lot, or single-handedly bastardised analysis and geometry. Possibly both.

Quote:

Originally Posted by Arcanaville Samuel_Tow is the only poster that makes me want to punch him in the head more often when I'm agreeing with him than when I'm disagreeing with him.