Josh English <eng### [at] spiritonecom> wrote:
: I'm pretty sure I read something about this in Calc 3... I'll have to check
: tonight when I get home.

  The area of the parallelogram is |uxv| (ie. the length of the cross-product
of the two vectors u and v).

-- 
main(i,_){for(_?--i,main(i+2,"FhhQHFIJD|FQTITFN]zRFHhhTBFHhhTBFysdB"[i]
):_;i&&_>1;printf("%s",_-70?_&1?"[]":" ":(_=0,"\n")),_/=2);} /*- Warp -*/

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On Wed, 06 Dec 2000 09:09:28 -0800, Josh English wrote:
>I thought that the area of the parallelogram was equal to the dot product. I
>seem to remember that we were able to find the area with some simple vector
>calculation. Too bad my book is at home.

Nope, the dot product is more closely related to the projection of one
vector onto another [the length of the projection of b onto a is 
vdot(a,b)/vlength(b)].  The length of the cross product is related to 
the orthogonal component [the length of the component of b orthogonal 
to a is vlength(vcross(a,b))/vlength(a)].

Since the area of a parallelogram is base (length of a) times height
(length of the component of b orthogonal to a) it turns out that the 
area of the parallelogram is exactly the length of the cross product.  
In my experience, this fact is rarely useful, but the two above often
are.

-- 
Ron Parker   http://www2.fwi.com/~parkerr/traces.html
My opinions.  Mine.  Not anyone else's.

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Ron Parker <ron### [at] povrayorg> wrote:
: In my experience, this fact is rarely useful

  Perhaps it could be used, as proposed earlier, to calculate the weigthed
average of the normal vectors instead of just the average.
  It makes sense that the normal vector of a larger triangle has more
effect in the normal of the vertex than a smaller triangle.

  Of course doing it this way can result in degenerate triangles, which
is a problem.

-- 
main(i,_){for(_?--i,main(i+2,"FhhQHFIJD|FQTITFN]zRFHhhTBFHhhTBFysdB"[i]
):_;i&&_>1;printf("%s",_-70?_&1?"[]":" ":(_=0,"\n")),_/=2);} /*- Warp -*/

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Thanks, that's what I was able to look up. The scary part was I couldn't find it
on the 'net anywhere, but I wasn't checking anything about the cross product.

I can't wait to teach this stuff, I'll really know it well, then

Josh

Warp wrote:

> Josh English <eng### [at] spiritonecom> wrote:
> : I'm pretty sure I read something about this in Calc 3... I'll have to check
> : tonight when I get home.
>
>   The area of the parallelogram is |uxv| (ie. the length of the cross-product
> of the two vectors u and v).
>
> --
> main(i,_){for(_?--i,main(i+2,"FhhQHFIJD|FQTITFN]zRFHhhTBFHhhTBFysdB"[i]
> ):_;i&&_>1;printf("%s",_-70?_&1?"[]":" ":(_=0,"\n")),_/=2);} /*- Warp -*/

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Better Nate than Lever... I have a small explination that hopefully will
help out here:

http://www.spiritone.com/~english/cyclopedia/smooth2.html

As always, send feedback and WTF's via email.

Josh

Josh English wrote:

> Using Warps read me file for his mesh smoother, I find that normalizing
> the normal vectors of each triangle, then averaging them, works the
> best. Of course, if you are using the vcross command to find your
> normals, remember that the length of the resultant vector has a length
> of the dot product, which will be at it's maximum when the vectors are
> orthogonal (ie, 90 degrees apart).
>
> Since this doens't make a lot of sence, I'll try to have a few pictures
> for you soon.
>
> Josh
>
> Wlodzimierz ABX Skiba wrote:
>
> > I have point O and points P1, P2, ... PN
> > connected as traingles P1OP2, P2OP3, ..., PN-1OPN, PNOP1
> > I want calculate best normal at point O for all this triangles.
> > Should it be simple normalized summary of all normals
> > or waighted one depend of areas of each triangle ?
> > Maybe other way ?
> >
> > ABX
>
> --
> Josh English -- Lexiphanic Lethomaniac
> eng### [at] spiritonecom
> The POV-Ray Cyclopedia http://www.spiritone.com/~english/cyclopedia/

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Josh English <eng### [at] spiritonecom> wrote:
: Better Nate than Lever... I have a small explination that hopefully will
: help out here:

: http://www.spiritone.com/~english/cyclopedia/smooth2.html

  Note that if you use a weighted average to smooth the triangles, you can
get degenerate triangles. This is a bad problem.

  For example, suppose that you have a triangle with a normal vector
pointing at <0,1,0> and an adjacent triangle with its normal vector pointing
at <1,-1,0>. Let's say that the area of the first triangle is 10 square
units and the area of the second triangle is 1 square unit.
  If we calculate the normal vector of a common vertex using the weighted
average of the triangle normal vectors, it will be:

  <0,1,0>*10 + <1,-1,0>*1 = <1,9,0>

  If you apply that <1,9,0> as the normal vector to a vertex of the second
triangle, it will be degenerate. That's because the angle between the
normal vector of the second triangle and that <1,9,0> vector is larger
than 90 degrees.
  (It can be checked from the dot-product; if the dot-product of the
two vectors is negative, then the triangle is degenerate:
  <1,-1,0>.<1,9,0> = 1*1 + (-1)*9 + 0*0 = -8  )


  By the way: It's interesting to note that this same problem can appear
even if we use the average of the normalized normal vectors of the triangles.
  I have never thoight about that...

  However, I would say that the problem is less probable in the latter case.

-- 
main(i,_){for(_?--i,main(i+2,"FhhQHFIJD|FQTITFN]zRFHhhTBFHhhTBFysdB"[i]
):_;i&&_>1;printf("%s",_-70?_&1?"[]":" ":(_=0,"\n")),_/=2);} /*- Warp -*/

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By the way, the link at the beginning of the page is wrong.
  It's not "PovUntils" :)

-- 
main(i,_){for(_?--i,main(i+2,"FhhQHFIJD|FQTITFN]zRFHhhTBFHhhTBFysdB"[i]
):_;i&&_>1;printf("%s",_-70?_&1?"[]":" ":(_=0,"\n")),_/=2);} /*- Warp -*/

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Warp wrote:

> Josh English <eng### [at] spiritonecom> wrote:
> : Better Nate than Lever... I have a small explination that hopefully will
> : help out here:
>
> : http://www.spiritone.com/~english/cyclopedia/smooth2.html
>
>   Note that if you use a weighted average to smooth the triangles, you can
> get degenerate triangles. This is a bad problem.
>
>   For example, suppose that you have a triangle with a normal vector
> pointing at <0,1,0> and an adjacent triangle with its normal vector pointing
> at <1,-1,0>. Let's say that the area of the first triangle is 10 square
> units and the area of the second triangle is 1 square unit.
>   If we calculate the normal vector of a common vertex using the weighted
> average of the triangle normal vectors, it will be:
>
>   <0,1,0>*10 + <1,-1,0>*1 = <1,9,0>
>
>   If you apply that <1,9,0> as the normal vector to a vertex of the second
> triangle, it will be degenerate. That's because the angle between the
> normal vector of the second triangle and that <1,9,0> vector is larger
> than 90 degrees.
>   (It can be checked from the dot-product; if the dot-product of the
> two vectors is negative, then the triangle is degenerate:
>   <1,-1,0>.<1,9,0> = 1*1 + (-1)*9 + 0*0 = -8  )

Thanks for the tip on that. I didn't associate negative dot products with
degenerate triangles. So is it always true that a negative dot product reflects
an obtuse angle?  A cursory check would make it seem that way.

>   By the way: It's interesting to note that this same problem can appear
> even if we use the average of the normalized normal vectors of the triangles.
>   I have never thoight about that...
>
>   However, I would say that the problem is less probable in the latter case.

No, sadly, it's not. I got several degenerate triangles, then I changed n to be
5 or 6, it worked, set it back to 4, and it worked. I have no idea why it did
that, though. This was when the normal at the apex of the triangles was along
<0,1,0>.

Josh

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OOPS. I was looking at my bookmarks screen one one machine and typing on the
other....

It's fixed.

By the way, thank you for that read me file, it made my life very easy on
several occasions.

Josh

Warp wrote:

>   By the way, the link at the beginning of the page is wrong.
>   It's not "PovUntils" :)
>
> --
> main(i,_){for(_?--i,main(i+2,"FhhQHFIJD|FQTITFN]zRFHhhTBFHhhTBFysdB"[i]
> ):_;i&&_>1;printf("%s",_-70?_&1?"[]":" ":(_=0,"\n")),_/=2);} /*- Warp -*/

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Warp wrote in message <3a2fe6ad@news.povray.org>...
>  By the way: It's interesting to note that this same problem can
appear
>even if we use the average of the normalized normal vectors of the
triangles.
>  I have never thoight about that...


than if this appear for both method (weighted and not)
than consider comparison for proper triangles
(picture sended at p.b.i)

ABX

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