Yep,
I understand your last topic and that's why I decided to use this theorem (it's a very interesting thing).
But my worry is completely in the proof because we use the origin in the vector field (make a lot of sense to simplify the formula) AND I use the origin for one of my vertice (opposing vertex) in the Det formula to obtain the dot product of a cross product of three points (after simplification)...
Or in the formula we obtain A*d/3 where d == <x,n> because (<x,n><x0,n>)/3==<x,n>/3. To simplify the det formula I use x (opposing vertex)=Origin=xo. And it's here my problem....
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joohooo
Member Since 28 Nov 2012Offline Last Active Dec 04 2012 10:03 AM
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In Topic: Compute volume from mesh Mathematical demonstration
03 December 2012  02:09 AM
In Topic: Compute volume from mesh Mathematical demonstration
02 December 2012  02:19 PM
Yes So it doesn't prove my formula with vertices ??
The aim of this formula is to not use distance but just vertices to gain time !
Why not?
For tetrahedron: A*d/3 == Det(v1v0,v2v0,v3v0)/6, where 'A' is the area of some triangle and 'd' is the distance from the opposing vertex to the plane of the triangle. In the Det formula v1,v2,v3 define the same triangle, while v0 is the opposing vertex.
One more time thanks !
But ... I have one more question?? (yes again)
I agree with you for the Det but in the proof we have already use the origin for x0 ! Or in my formula the origin will be your V0 in the det formula as the opposing vertex for each triangle in the mesh. It's not a wrong thing to use the origin for the X0 in F and for the opposing vertex. In my mind , it's look like a ruse... no ?
In Topic: Compute volume from mesh Mathematical demonstration
29 November 2012  10:10 AM
Yes So it doesn't prove my formula with vertices ??
The aim of this formula is to not use distance but just vertices to gain time !
The aim of this formula is to not use distance but just vertices to gain time !
In Topic: Compute volume from mesh Mathematical demonstration
29 November 2012  09:23 AM
Ok well,
I don't know it was possible to use points in the function F ? I believed it was just vector field ?
I have a last question (in my mind easy but I want to be sure) : I use only vertices in the code above (v0 v1 v2).
Or your variable d is a distance not a point.
How can we say that A*(d<x0,n>)3 == 1/2 (P1^P2).d/3 where d must be a vertice ? (n is not obligatory (1,1,1)?)
Moreover if i use the formule of area it's also distance and not vertices....
Many thanks
I don't know it was possible to use points in the function F ? I believed it was just vector field ?
I have a last question (in my mind easy but I want to be sure) : I use only vertices in the code above (v0 v1 v2).
Or your variable d is a distance not a point.
How can we say that A*(d<x0,n>)3 == 1/2 (P1^P2).d/3 where d must be a vertice ? (n is not obligatory (1,1,1)?)
Moreover if i use the formule of area it's also distance and not vertices....
Many thanks
In Topic: Compute volume from mesh Mathematical demonstration
29 November 2012  04:41 AM
Hum, I'm really sorry but I don't understand your demonstration.
If you take vector field F (x,y,z) > (xx0,0,0)/3 the divergence is 1/3, not 1 ??
Moreover I don't understand how you insert your inner product d_i  <n_i,x0> and directly the area of a triangle after the gauss theorem
=> Int[<F,dS>] = Sum[A_i * (d_i  <n_i,x_0>)]/3 ??
One more time thanks !
If you take vector field F (x,y,z) > (xx0,0,0)/3 the divergence is 1/3, not 1 ??
Moreover I don't understand how you insert your inner product d_i  <n_i,x0> and directly the area of a triangle after the gauss theorem
=> Int[<F,dS>] = Sum[A_i * (d_i  <n_i,x_0>)]/3 ??
One more time thanks !