Apache <apa### [at] yahoocom> wrote in message
news:3c868ca3$1@news.povray.org...
> Don't think so. sqrt needs more cpu cycles. But maybe I'm wrong!

PoV's going to use square root anyway to find those lengths.
 -Shay

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>>> universal formula for area of triangle:
>>> S=sqrt(p*(p-a)*(p-b)*(p-c))

>> Is that faster than using half the cross product of two of the sides?
>> Something like:
>> area = vlength(vcross(P1-P2,P3-P2))/2;

> Don't think so. sqrt needs more cpu cycles. But maybe I'm wrong!

Vlength uses sqrt internally. But the major slowdown is Povrays interpreter
and not the cpu.

Regards
Hugo

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Josh Seagoe wrote:
> 

> > universal formula for area of triangle:
> >
> > S=sqrt(p*(p-a)*(p-b)*(p-c))
> >
> > where a,b,c are lengths of edges and p is half of perimeter = (a+b+c)/2
> >
> 
> Is that faster than using half the cross product of two of the sides?
> (cosidering vcross is internal...)
> 
> Something like:
> area = vlength(vcross(P1-P2,P3-P2))/2;

I have not looked at their mesh code yet,
but I assume that they are calculating
the normal vectors for the triangles by
just taking the cross product of a pair
of each triangle's "edge vectors".

And if this is the case, then I don't see
the point in first calculating this cross
product, normalize it and then multiply
it by your area expression.

I.e.:

vnormalize(vcross(P1 - P2, P3 - P2))
*vlength(vcross(P1 - P2, P3 - P2))/2

They could just use the cross product
directly:

vcross(P1 - P2, P3 - P2)/2

And they don't even have to divide it
by 2 prior to the summing. Because
every triangle normal will be "double"
length.

The final normalizing of the sum of
all the "weighted" normal vectors will
eliminate it anyway.


Tor Olav

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> 
> On Tue, 5 Mar 2002 09:12:47 -0600, "Shay" <sah### [at] simcopartscom> wrote:
> > The weighted normals look better. Is this the case with every mesh?
> 
> This is just more accurate method of averaging.
> 
> > This goes against everything I have seen concerning mesh normals.
> 
> Not exactly. Weighting is just longer and result is visible when there are
> differences in triangle area.
> 
> > By a weighted average, do you mean simply not normalizing the normal of each
> > triangle?
> 
> Considering you have N triangles with vertex P:
> - take all N normals as previously
> - normalize them as previously
> - new step: multiply/adjust according to area of triangle
> - add all N normals as previously
> - normalize this sum as previously (I'm not sure it is necessary)
> 
> So only new thing is muliplying. This is done to save larger triangles more
> float.

Wlodek,
if the triangle normals are calculated
with cross products, then step 2 and 3
above are not necessary.

Also see my reply in this thread to Josh
Seagoe some minutes ago:

http://news.povray.org/%3C3C86A730.785EA5C1%40hotmail.com%3E
news://news.povray.org/3C86A730.785EA5C1%40hotmail.com


Tor Olav

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On Thu, 07 Mar 2002 00:56:33 +0100, Tor Olav Kristensen
<tor### [at] hotmailcom> wrote:
> Wlodek,
> if the triangle normals are calculated
> with cross products, then step 2 and 3
> above are not necessary.

I feeled this...

ABX

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On Wed, 6 Mar 2002 12:45:09 -0600, "Shay" <sah### [at] simcopartscom> wrote:
> > #local triangle_area_engine=function{(x+y+z)*(y+z-x)*(x+y-z)*(x+z-y)/16}
> > #local triangle_area=
> >   function(x1,y1,z1,x2,y2,z2,x3,y3,z3)
> >     {sqrt(triangle_area_engine(
> >         f_r(x2-x1,y2-y1,z2-z1),
> >         f_r(x3-x2,y3-y2,z3-z2),
> >         f_r(x3-x1,y3-y1,z3-z1)
> >     ))};
>
> #include "functions.inc"
> #local triangle_area_engine=function{(x+y+z)*(y+z-x)*(x+y-z)*(x+z-y)/16}
> #local Triangle_Area =  triangle_area_engine(Side_1_Length, Side_2_Length,
Side_3_Length);
>
> Seems right to me, but I might be missing something.

You forgot about sqrt applied but since others mentioned that not normalized
cross product is sufficient then I think my method is wasting of pov-time ;-)

ABX

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