**comp.graphics.algorithms**

## Subject: **Re: Basic perspective/normal q**

"Hans-Bernhard Broeker"

news:4bkt8rF123r6gU1@news.dfncis.de...

> [jongware]

>

> > I know it's no good -- that's why I get substandard results. Can you

> > substantiate your comment with the required math? All inputs are as

> > mentioned: a number of x,y,z coordinates for a plane; its normal; a

camera

> > position. I guess that's all what it takes for input.

>

> If anything, it's too much. All you really need is the 'd' parameter

> for the plane's normal equation:

>

> plane = { P | P.N = d }

>

> Any of those "x,y,z coordinate" triplets should do. Call it P0,

> the reference for that plane, and the normal equation can be

> rewritten:

>

> (P-P0).N = 0

>

> The test you have to make is on the sign of the left hand side of this

> equation, with the camera position for 'P'. If it's positive, that

> means the point is on that side of the plane the normal points out,

> i.e. the plane is visible. Otherwise you're looking at it's back

> side.

>

> > I assume I'm correct in stating that if the (transformed!) z

> > coordinates are in front of the camera position the plane should be

> > visible (apart from its normal vector).

>

> No. Those (x,y,z) triplets are just examples of points in that plane.

> A random point on the plane can be *anywhere* on the viewer's z axis,

> independently of whether the plane is visible or not. That's because

> a lot of planes will be only *partly* visible. Half of any typical

> plane is behind the viewer. The only exception to this are planes

> orthogonal to the view direction.

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