"Wlodzimierz ABX Skiba" wrote:
> sorry for the mistake but ...
> you wrote "I find points on the plane perpendicular to
> the input normal, then deform those points, and then find
> the deformed normal which is the vector perpendicular to
> the deformed points on the plane."
> I just don't understand this meaning sentence.
> How you find normal perpendicular to 4 (you wrote that
> you best number is 4) deformed points ? and what plane ?
> 4 points not describe plane.

I have the input normal vector iN. Then I find 4 vectors perpendicular to
it. Those are tangents. Let's call them iT1, iT2, iT3 and iT4. These all lie
in the plane perpendicular to the input normal vector. Then I apply the
deformation to the 4 tangents, and I get the deformed tangents dT1, dT2,
dT3, dT4. These "describe" the deformed plane. To get the deformed normal I
find the vector that is perpendicular to the 4 deformed tangent vectors. And
yes, I average the tangent vectors 2 and 2 first. But they are tangents, not
normals, so I don't average any normals. And the tangent vectors are simply
added together, so it is automatically a weighted average (because a longer
tangent vector will have greater influence).

> every two vectors joined to one point describe triangle
> I just "virtualize" situation :-)

That may make sense for your method, but not for mine. I don't pair tangent
vectors next to each other, I pair opposite-lying tangent vectors...

Rune
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