I'm working on some deformation macros that deforms the point vectors and
the normal vectors in a mesh.

Deforming the points is easy. Deforming the normals is a bit more
complicated.
To find a deformed normal I try to find the deformed points very close to
the start point of the normal, and see how the direction changes. However, I
can't figure out quite how to do it.

Say the macro Deform(V) will return a deformed version of V.
If I have the point P I can find the deformed point dP:
#declare dP = Deform(P)

If I have the normal N, how can I find the deformed normal dN?

Rune
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In article <3a60d0eb@news.povray.org>, "Rune" <run### [at] inamecom> 
wrote:

> Say the macro Deform(V) will return a deformed version of V.
> If I have the point P I can find the deformed point dP:
> #declare dP = Deform(P)
> 
> If I have the normal N, how can I find the deformed normal dN?

Try something like:
#declare SmallValue = 0.01;
#declare dN = vnormalize(Deform(P+V*SmallValue) - dP);

In other words, deform the point and a second point a short distance 
along the normal from it, and use the direction from dP to the new point 
as your normal. This should work in most circumstances, and you could 
decrease SmallValue for more accuracy.

-- 
Christopher James Huff
Personal: chr### [at] maccom, http://homepage.mac.com/chrishuff/
TAG: chr### [at] tagpovrayorg, http://tag.povray.org/

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"Chris Huff" wrote:
> Try something like:
> #declare SmallValue = 0.01;
> #declare dN = vnormalize(Deform(P+V*SmallValue) - dP);
>
> In other words, deform the point and a second point a short
> distance along the normal from it, and use the direction from
> dP to the new point as your normal. This should work in most
> circumstances, and you could decrease SmallValue for more
> accuracy.

Nope, I've already tried that, and it doesn't work.
Imagine a box deformed by a shearing matrix. The normals on the skewed sides
should change, but they won't using your method. I think I need several
points for reference (two points forming vectors perpendicular to the normal
vector?), but I haven't had any luck so far.

Rune
--
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\ all be found at http://rsj.mobilixnet.dk (updated January 6)
/ Also visit http://www.povrayusers.org

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In article <3a60edd6@news.povray.org>, "Rune" <run### [at] inamecom> 
wrote:

> Nope, I've already tried that, and it doesn't work. Imagine a box 
> deformed by a shearing matrix. The normals on the skewed sides should 
> change, but they won't using your method. I think I need several 
> points for reference (two points forming vectors perpendicular to the 
> normal vector?), but I haven't had any luck so far.

Right, this only approximates, and only "works" for some 
distortions...are you only using matrix transformations? Or were you 
only using shearing matrices as an example?
I'll be thinking about this problem for a while...

-- 
Christopher James Huff
Personal: chr### [at] maccom, http://homepage.mac.com/chrishuff/
TAG: chr### [at] tagpovrayorg, http://tag.povray.org/

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Rune wrote:
> 
> "Chris Huff" wrote:
> > Try something like:
> > #declare SmallValue = 0.01;
> > #declare dN = vnormalize(Deform(P+V*SmallValue) - dP);
> >
> > In other words, deform the point and a second point a short
> > distance along the normal from it, and use the direction from
> > dP to the new point as your normal. This should work in most
> > circumstances, and you could decrease SmallValue for more
> > accuracy.
> 
> Nope, I've already tried that, and it doesn't work.
> Imagine a box deformed by a shearing matrix. The normals on the skewed sides
> should change, but they won't using your method. I think I need several
> points for reference (two points forming vectors perpendicular to the normal
> vector?), but I haven't had any luck so far.

If N is the normal at the point P, then how about trying this ?

#declare dN = vnormalize(Deform(P + 0.01*vnormalize(N)) - Deform(P));


Tor Olav
-- 
mailto:tor### [at] hotmailcom
http://www.crosswinds.net/~tok/tokrays.html

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Rune wrote:
>...
> Imagine a box deformed by a shearing matrix. The normals on the skewed sides
> should change ...

Hmmm ... Are you sure about this ?

If so, then how would you define a 
normal of a surface at a given point ?


Tor Olav
-- 
mailto:tor### [at] hotmailcom
http://www.crosswinds.net/~tok/tokrays.html

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In article <3A61148C.66317755@online.no>, Tor Olav Kristensen 
<tor### [at] onlineno> wrote:

> If N is the normal at the point P, then how about trying this ?
> 
> #declare dN = vnormalize(Deform(P + 0.01*vnormalize(N)) - Deform(P));

That is exactly the same as what I suggested (and which won't work), 
though it is slightly reworded and I assumed the normal was already 
normalized.

-- 
Christopher James Huff
Personal: chr### [at] maccom, http://homepage.mac.com/chrishuff/
TAG: chr### [at] tagpovrayorg, http://tag.povray.org/

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In article <3A61158D.9D409B1B@online.no>, Tor Olav Kristensen 
<tor### [at] onlineno> wrote:

> Hmmm ... Are you sure about this ?

The normal to the -x side of a box is -x. The top of the box is then 
sheared +x so the -x and +x faces are at a 45 degree angle...but points 
still remain in the same xz plane as they were before, the -y and +y 
faces remain perpendicular to the y axis. The normal should be at a 45 
degree angle now, <-sqrt(2)/2, sqrt(2)/2, 0> to be precise, but since 
the two sample points were in the same xz plane, their position relative 
to each other is the same, and the normal is still -x.


> If so, then how would you define a 
> normal of a surface at a given point ?

The vector perpendicular to the surface at that point.

-- 
Christopher James Huff
Personal: chr### [at] maccom, http://homepage.mac.com/chrishuff/
TAG: chr### [at] tagpovrayorg, http://tag.povray.org/

<><

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Chris Huff wrote:
> 
> In article <3A61158D.9D409B1B@online.no>, Tor Olav Kristensen
> <tor### [at] onlineno> wrote:
> 
> > Hmmm ... Are you sure about this ?
> 
> The normal to the -x side of a box is -x. The top of the box is then
> sheared +x so the -x and +x faces are at a 45 degree angle...but points
> still remain in the same xz plane as they were before, the -y and +y
> faces remain perpendicular to the y axis. The normal should be at a 45
> degree angle now, <-sqrt(2)/2, sqrt(2)/2, 0> to be precise, but since
> the two sample points were in the same xz plane, their position relative
> to each other is the same, and the normal is still -x.

I see.

Then would this work ?

#declare dP = Deform(P);
#declare vN = vnormalize(N);

#declare AA = 0.01;
#declare BB = 0.1;

#declare dNpx = vnormalize(Deform(P + AA*(vN + x*BB)) - dP);
#declare dNmx = vnormalize(Deform(P + AA*(vN - x*BB)) - dP);

#declare dNpy = vnormalize(Deform(P + AA*(vN + y*BB)) - dP);
#declare dNmy = vnormalize(Deform(P + AA*(vN - y*BB)) - dP);

#declare dNpz = vnormalize(Deform(P + AA*(vN + z*BB)) - dP);
#declare dNmz = vnormalize(Deform(P + AA*(vN - z*BB)) - dP);

#declare dN = vnormalize(dNpx + dNmx + dNpy + dNmy + dNpz + dNmz);


Regards,

Tor Olav
--
mailto:tor### [at] hotmailcom
http://www.crosswinds.net/~tok/tokrays.html

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Tor Olav Kristensen wrote:
> 
> Chris Huff wrote:
> >
> > In article <3A61158D.9D409B1B@online.no>, Tor Olav Kristensen
> > <tor### [at] onlineno> wrote:
> >
> > > Hmmm ... Are you sure about this ?
> >
> > The normal to the -x side of a box is -x. The top of the box is then
> > sheared +x so the -x and +x faces are at a 45 degree angle...but points
> > still remain in the same xz plane as they were before, the -y and +y
> > faces remain perpendicular to the y axis. The normal should be at a 45
> > degree angle now, <-sqrt(2)/2, sqrt(2)/2, 0> to be precise, but since
> > the two sample points were in the same xz plane, their position relative
> > to each other is the same, and the normal is still -x.
> 
> I see.
> 
> Then would this work ?
> 
> #declare dP = Deform(P);
> #declare vN = vnormalize(N);
> 
> #declare AA = 0.01;
> #declare BB = 0.1;
> 
> #declare dNpx = vnormalize(Deform(P + AA*(vN + x*BB)) - dP);
> #declare dNmx = vnormalize(Deform(P + AA*(vN - x*BB)) - dP);
> 
> #declare dNpy = vnormalize(Deform(P + AA*(vN + y*BB)) - dP);
> #declare dNmy = vnormalize(Deform(P + AA*(vN - y*BB)) - dP);
> 
> #declare dNpz = vnormalize(Deform(P + AA*(vN + z*BB)) - dP);
> #declare dNmz = vnormalize(Deform(P + AA*(vN - z*BB)) - dP);
> 
> #declare dN = vnormalize(dNpx + dNmx + dNpy + dNmy + dNpz + dNmz);

I see now that this will probably not work either.

I'll think of it while I sleep.  :)


Goodnight from Tor Olav
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http://www.crosswinds.net/~tok/tokrays.html

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