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Thanks a lot for the assistance to everybody.
> "Penelope20k" <pen### [at] caramail fr> wrote:
> >> the area of triangle is
> >> Area = abs( (xB*yA - xA*yB) + (xC*yB -x B*yC)+( xA*yC - xC*yA ) ) /2
> >>>> "Le Forgeron" <jgr### [at] freefr>
> > 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?
>> That formula works (???) only in the plane of the triangle. A
>> triangle is always 2D.....
>> the previous formula of penelope2k seems a bit wrong about
>> the computation...
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
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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.
>
If I rotate axis the coordinates are different, so I am really confused : ))
May be I could also use Eron's formula, Area= sqrt (s(s-a)(s-b)(s-c)).
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Le 03.11.2007 10:23, Francesco nous fit lire :
> "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.
>>
>
> If I rotate axis the coordinates are different, so I am really confused : ))
Illumination is on its way... continue.
>
> May be I could also use Eron's formula, Area= sqrt (s(s-a)(s-b)(s-c)).
>
It's Heron... and why do you want to compute the length of the side,
when all you need from the 3 vertices are their coordinaates.
Heron's formula is fine when you only have the lengths.
Actually, from a mesh's triangle:
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.
Of course, you can use Area = 1/4*sqrt( (a+b+c)(a+b-c)(b+c-a)(c+b-a) )
this will avoid the intermediate variable, but the costy part is
computing the length a, b and c anyway!
--
The superior man understands what is right;
the inferior man understands what will sell.
-- Confucius
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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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