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About joohooo

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  1. 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....
  2. [quote name='max343' timestamp='1354207366' post='5005355'] [quote name='joohooo' timestamp='1354205457' post='5005341'] 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 ! [/quote] Why not? For tetrahedron: A*d/3 == Det(v1-v0,v2-v0,v3-v0)/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. [/quote] 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 ?
  3. 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 !
  4. 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
  5. Hum, I'm really sorry but I don't understand your demonstration. If you take vector field F (x,y,z) -> (x-x0,0,0)/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 !
  6. Thanks to your response but I don't see how it is a "simple" result of the divergence theorem because you don't compute with element surface but with tetrahedron where the base is each triangle of the parametrized surface and the common point is the origin of the 3d space (0,0,0). The divergence theorem said that the volume is equal to the outward flux of the surface, isn't it ? Here we don't use surface but volume of tetrahedron composed with origin ? Could you explain me with math if it's possible ?
  7. Hi, I start a new topic from an older because I have a tiebreaker ! the older topic : [url=""]http://www.gamedev.n...ume-for-a-mesh/[/url] My question is: How can we prove the formula V= sum(each tetrahedron volume with the origin) [source lang="cpp"] float volume = 0; int* index = I; for (i = 0; i < T; i++) { Vector3 v0 = V[*index++]; Vector3 v1 = V[*index++]; Vector3 v2 = V[*index++]; volume += Dot(v0,Cross(v1,v2)); } volume /= 6; [/source] I know I must use 3D integrals and Fourier moments but I don't know how I can do that ! Please someone give me some clues about a possible math proof ? Thanks by advance PS: I already try to find this formula with the gauss theorem as said in the older topic but I never find the result but an approximate result! I obtain the formuma here but with some addition in more...