visiblity?

Am I correct to say that you're looking for an algorithm to find the distance between a rectangle and a point in 2D?

I'm not sure how to tackle a problem like that. If you'd check the distance between the view location and the 4 corner points and then check the closest distance then it might get flagged as too far away while it isn't because the distance to one edge is smaller than the distance to one corner.

And if you'd calculate the distances to each edge you'd still have to figure out which edge-distance to take.

My guess is that you should determine the distance to an edge and then check if this point is actually on the edge. I hope this picture explains it:


So the distance you're looking for is the smallest distance to an edge that is actually on an edge. These distances can be found with a dotproduct I guess and to check if it's on the edge itself you could use the found distance and the edge's normal.

I hope this made any sense.

Greets, Dirk


Edited by - Scarab0 on January 22, 2001 3:26:58 AM

After two weeks working on the problem and not being able to figure it out I finally ralized something I did wrong! I had an error in my dot product code! one wrong letter and the results were wrong! I''ve been retrying what I was doing before and I''m getting closer!
This is the idea that I''m working on...
compute the angles between the four vectors from the view location and the points of the region. Then compare the largest of those agenst the angle for one pixel.
So far it seams to be working now that I found that error in the dot product code! I should remember to never assume existing code is flawless!

pixelsonscreen = bs.radius / zDistance;
pixelsonscreen = pixelsonscreen.squared
pixelsonscreen *= (screenHeight* screenWidth)

this is approx how many pixels on screen the object will take remeber theres a VERY FINE LINE between 0 and 1 pixels yet mathmatically there is a IMPOSSIBLE FINE LINE between the 2

http://members.xoom.com/myBollux