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Location of orthogonal center and calculation of voronoi area/volume

Location of orthogonal center and calculation of voronoi area/volume. Nitin Bhardwaj Gang Feng 12/9. A quick look at voronoi plane.  ((x-c 2 ) 2 -r 2 2 ).  ((x-c 1 ) 2 -r 1 2 ). Voronoi plane is defined as:  ((x-c 1 ) 2 -r 1 2 ) =  ((x-c 1 ) 2 -r 1 2 ). (x-c 2 ). r 2.

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Location of orthogonal center and calculation of voronoi area/volume

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  1. Location of orthogonal center and calculation of voronoi area/volume Nitin Bhardwaj Gang Feng 12/9

  2. A quick look at voronoi plane ((x-c2)2-r22) ((x-c1)2-r12) • Voronoi plane is defined as: ((x-c1)2-r12) = ((x-c1)2-r12) (x-c2) r2 (x-c1) r1

  3. Voronoi volume for a 2-d case A P1 • Compute the white area by calculating the coordinates for each vertex. P2 Orthogonal Center OC lies inside OC C B

  4. Strategy: • Every time we draw a voronoi plane, we deduct the volume of the total chunk taken away by it… When calculating the voronoi vol for A, deduct the volume VBDE from VABC. We define the volume VBDE as VAC A D B C E

  5. Voronoi volume for a 2-d case A P1 • VA=VABC-cACVAC-cABVAB+ cACABVAC VAB • cAC= 1(B and OC are at the same side of the egde AC)/0 (1(B and OC are at the different side of the egde AC) • cACAB = 1(OC is in the ΔABC)/ 0 (OC is not in the ΔABC) P2 Orthogonal Center (OC) lies inside VAC VAB OC C B VAC VAB

  6. What happens if it lies outside?? A VA=VABC-VAC-VAB since VAC VAB=0. P1 P2 VAB VAC OC • So the strategy for 2-d • Cut the first two planes and locate the orthogonal center. • Apply the relevant formula for two cases (outside or inside)

  7. How do we know where is the orthogonal center?? LAC A • Cut the first two plane and let them meet at, say D. If LAC(B)*LAC(D) > 0 …(1) LAB(C)*LAB(D) > 0 …(2) LBC(A)*LBC(D) > 0 …(3) then D lies inside.. P1 P2 D C B Else it lies outside. If (1), (2) are true but (3) is false it lies in the gray region Its easier to put values in the equation rather than solving the equations (for eg to solve for affine dependency eqns)

  8. Where all can the orthogonal center lie? • The orthogonal center will not lie outside the star. • We can even calculate the area of such a star structure. For example, for an equilateral triangle the area is 135* ΔABC A B C Thicker lies are the main triangle lines. Thinner lines of the same color are perpendicular to the main lines.

  9. What for 3-d…?? • Strategy remains the same.. • Cut a plane and subtract the volume (instead of an area) taken away by it. B For Voronoi vol of A: Cut the first plane, for A and C and subtract VEFGC from VABCD F D A G E C

  10. How do we locate a point with respect to the tetrahedra? B Given point say (E) lies inside if PABC(D)*PABC(E) > 0 …(1) PABD(C)*PADB(E) > 0 …(2) PBCD(A)*PBCD(E) > 0 …(3) PACD(B)*PACD(E) > 0 …(4) else it lies outside. PABC E E D A C Location can further be refined by analysing the above eqns For eg. If 1, 2 and 4 are true, 3 is false then E lies outside the facePBCD.

  11. Location! Location! Location! B • Lies Outside an edge if if exactly 2 are true. • For eg if 1 and 4 are false, it lies outside the edge AC. • Similarly lies outside an vertex if exactly 3 are false. PABC PACD D A C E

  12. Calculation of volume… C Denote V1 as the vol taken away by the first plane and V2 by the second. So volume so far taken away: V1+V2- V1 V2 P2 P1 E A D F Target vertex B

  13. How do we know if the planes intersect inside? • Or, when does V1 V2 disappear? Take the mid-point of following two points: p1: the intersection of the common edge and the plane containing the target vertex and the 2 vertices for which the voronoi planes have been drawn (F) p2: the intersection of the common edge and the plane containing the 4th vertex and the 2 vertices for which the voronoi planes have been drawn (E) If this mid-point lies inside the tetrahedra, the planes intersect inside it else they intersect outside.

  14. Bring in the third plane pls… • 3 possible cases: • It does not touch any of the • other two planes • It touches both but not the edge • It touches the edge In the third case total Area taken away is: V1+V2+V3- V1 V2 -V1 V3 - V2 V3 +V1 V2  V3 E F

  15. Area in the first case. D • VA=VABCD-V1-V2-V3+V1V2 C A B

  16. Area in the second case D VA=VABCD-V1-V2-V3+V1V2 +V1V3 +V2V3 C A B

  17. Area in the second case D VA=VABCD-V1-V2-V3+V1V2 +V1V3 +V2V3 -V2V2V3 C A B

  18. Location of the orthogonal center supplements calculation of the area!! C • We can directly calculate the area from the co-ordinates of E and F that were used to locate the orthogonal center P2 P1 E A D F Target vertex B

  19. Conclusion • We use simple geometrical protocols to calculate the area/volume. • Location of orthogonal center and calculation of area go hand in hand. • A flow-chart can be developed for more systematic depiction.

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