Le 03.11.2007 11:27, Le Forgeron nous fit lire :
> Area = 1/4*sqrt( (a+b+c)(a+b-c)(b+c-a)(c+b-a) )
That's bogus...

Area = 1/4*sqrt((a+b+c)(a+b-c)(b+c-a)(c+a-b))
Of course!


-- 
The superior man understands what is right;
the inferior man understands what will sell.
-- Confucius

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Le Forgeron <jgr### [at] freefr> wrote:

> On one hand, two vectors differences, a cross product, three
> squaring, a sum and a square root ending with a division.
>
> On the other end, three vectors differences, nine squaring, three
> sums, three square root, an intermediate variable (yet a sum and a
> division), then three differences, a product of four element, and a
> square root.
>


You've been convincing :)

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Francesco <nas### [at] gmailcom> wrote:
> > > Mesh...are not in the same plane, share no common edges... )

> Don't they share edges? Ok, they are not in the same plane but they share at
> least one edge, or not?

  You can create meshes with completely separate, non-touching triangles
which don't share anything.

-- 
                                                          - Warp

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"Francesco" <nas### [at] gmailcom> wrote:

> Calculating area triangle per triangle I can always suppose z=0 (translating and
> rotating axis) as triangles are 2D. So I think the formula posted by Penelope20k
> is good.
>
>
> Francesco

YES ! it is not only good but great. it is the shortest route from points to
area I have seen.

just I am not able to mentally make the leap to 3D without a lengthy conversion
to 3 side lengths. If I could, it might be the shortest route in 3D.


also note that 666... is common to all tetra and pyramid volumes?!?

  ; //e $tetrahedron_volume(383 156 0, 645 254 0, 465 289 0,400 200 200) =
893666.666667
  ; //e $pyramid_volume(276 308 0, 383 123 0, 666 324 0, 276 308 0, 400 300 200)
= 2462066.666667


I have finished the volume macros

#macro tetrahedron_volume(A,B,C,D)
  #local result = abs(vdot(D-A,vcross(D-B,D-C))) / 6
  result
#end
#macro pyramid_volume(A,B,C,D,E)
  // perimeter order for 1st 4 points
  #local result =
     tetrahedron_volume(A,B,C,E) + tetrahedron_volume(A,C,D,E);
  result
#end

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"alphaQuad" <alp### [at] earthlinknet> wrote:

> also note that 666... is common to all tetra and pyramid volumes?!?

no, it was coincidental. or was it synchronicity that keeps getting me.


new users: dont forget the missing ; in that macro

still hope someone will make the leap from Pene's 2d to a short,
or shortEST 3D method

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"alphaQuad" <alp### [at] earthlinknet> wrote:
> "Francesco" <nas### [at] gmailcom> wrote:
>
> > Calculating area triangle per triangle I can always suppose z=0 (translating and
> > rotating axis) as triangles are 2D. So I think the formula posted by Penelope20k
> > is good.
> >
> >
> > Francesco
>
> YES ! it is not only good but great. it is the shortest route from points to
> area I have seen.
>
> just I am not able to mentally make the leap to 3D without a lengthy conversion
> to 3 side lengths. If I could, it might be the shortest route in 3D.


apparently I am in similiar company as no else can either.

the task:
Area = abs(   (xB*yA - xA*yB)  + (xC*yB -x B*yC)+( xA*yC - xC*yA ) ) /2

make shortest 3D version

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Le 04.11.2007 17:16, alphaQuad nous fit lire :

> apparently I am in similiar company as no else can either.
> 

Stop whining... it won't help to get any result.

> the task:
> Area = abs(   (xB*yA - xA*yB)  + (xC*yB -x B*yC)+( xA*yC - xC*yA ) ) /2

This formula looks bogus.

We already know that

Area = || AB x AC ||
           -------------
                   2

Let's make a numerical application with A(0,-1) B(1,0) and C(0,1)

 abs ( (-1 - 0) + (0 - 1) + ( 0 +1 ) ) /2 ---> abs ( -1 ) /2 --> 1/2
(wrong)

AB: <1,1,0>
AC: <0,2,0>

AB x AC: <0,0,-2>

Area = length(AB x AC)/2 = 1. (good!)

> 
> make shortest 3D version

The crossproduct is fine, what do you need ?

-- 
The superior man understands what is right;
the inferior man understands what will sell.
-- Confucius

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Le Forgeron <jgr### [at] freefr> wrote:
> Le 04.11.2007 17:16, alphaQuad nous fit lire :
>
> > apparently I am in similiar company as no else can either.
> >
>
> Stop whining... it won't help to get any result.
>
> > the task:
> > Area = abs(   (xB*yA - xA*yB)  + (xC*yB -x B*yC)+( xA*yC - xC*yA ) ) /2
>
> This formula looks bogus.
>
> We already know that
>
> Area = || AB x AC ||
>            -------------
>                    2
>
> Let's make a numerical application with A(0,-1) B(1,0) and C(0,1)
>
>  abs ( (-1 - 0) + (0 - 1) + ( 0 +1 ) ) /2 ---> abs ( -1 ) /2 --> 1/2
> (wrong)
I get ( (-1 - 0) + (0 - 1) + ( 0 + 0 ) ) /2= -1, which is correct.

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Le 05.11.2007 21:30, Grassblade nous fit lire :
> Le Forgeron <jgr### [at] freefr> wrote:
>> Le 04.11.2007 17:16, alphaQuad nous fit lire :
>>> the task:
>>> Area = abs(   (xB*yA - xA*yB)  + (xC*yB -x B*yC)+( xA*yC - xC*yA ) ) /2

>> Let's make a numerical application with A(0,-1) B(1,0) and C(0,1)
>>
>>  abs ( (-1 - 0) + (0 - 1) + ( 0 +1 ) ) /2 ---> abs ( -1 ) /2 --> 1/2
>> (wrong)
> I get ( (-1 - 0) + (0 - 1) + ( 0 + 0 ) ) /2= -1, which is correct.
>
Ok, my fault.

So, I let you with alphaQuad trying to generalize that to 3D.

And I will keep using length(crossproduct(B-A,C-A))/2.

-- 
The superior man understands what is right;
the inferior man understands what will sell.
-- Confucius

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Le Forgeron <jgr### [at] freefr> wrote:
> Le 05.11.2007 21:30, Grassblade nous fit lire :
> > Le Forgeron <jgr### [at] freefr> wrote:
> >> Le 04.11.2007 17:16, alphaQuad nous fit lire :
> >>> the task:
> >>> Area = abs(   (xB*yA - xA*yB)  + (xC*yB -x B*yC)+( xA*yC - xC*yA ) ) /2
>
> >> Let's make a numerical application with A(0,-1) B(1,0) and C(0,1)
> >>
> >>  abs ( (-1 - 0) + (0 - 1) + ( 0 +1 ) ) /2 ---> abs ( -1 ) /2 --> 1/2
> >> (wrong)
> > I get ( (-1 - 0) + (0 - 1) + ( 0 + 0 ) ) /2= -1, which is correct.
> >
> Ok, my fault.
>
> So, I let you with alphaQuad trying to generalize that to 3D.
>
> And I will keep using length(crossproduct(B-A,C-A))/2.
>
> --
> The superior man understands what is right;
> the inferior man understands what will sell.
> -- Confucius

I didn't say anything against cross product, did I? It's a useful tool, but it
doesn't work on four or more dimensions. What if I go nuts and want a mesh in
4D?
But to take you up to the task, let's start with the big formula on wikipedia.
With no loss of generality, we can consider A to be in the origin by simply
taking the difference between B and A and C and A. Consequently two of the
determinants are nil, and the square root cancels out with the square. Sarrus'
rule yields: (xB*yC+yB*zC+zB*xC-xB*zC-yB*xC-zB*yC)/2, or
(xB*(yC-zC)+yB*(zC-xC)+zB*(xC-yC))/2. You can reintroduce the coordinates of A
by substitution if you want. I bet that's exactly the same result as the cross
product. :-p

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