Marching Cubes

A High Resolution 3D Surface Construction Algorithm

Slice Data to Volumetric Data(1)

Slice Data to Volumetric Data(2)

Marching Cube Create cells (cubes) Classify each vertex Build an index Get edge list

Based on table look-up Interpolate triangle vertices Obtain polygon list and do shading in

image space

Cube

Consider a cube defined by 8 data values, 4 from slice k, and another 4 from slice k+1

Classify each vertex

Label 1 or 0 as to whether it lies inside or outside the surface

Match!!!

0

1

0

1

1

00

0

Build an indexCreate an index of 8 bits from the binary labeling of each vertex.

Get edge list

Give an index, store a list of edges.

Because symmetry ： 256/2=128rotation ： 128/8=16

256 cases are reduced to 14 cases.

Interpolate triangle vertices

X = i +

i i+1Xi i+1X

= 20

= 10

iso_value=18 iso_value=14

(iso_value - D(i))

(D(i+1) - D(i))

Problems about MC

Empty cells 30-70% of isosurface generation time wa

s spent in examining empty cells. Speed

Ambiguity

The Asymptotic Decider

Resolving the Ambiguity in Marching Cubes

Ambiguity Problem (1) Ambiguous Face : a face that has two diagonally o

ppsed points with the same sign

+

+

Ambiguity Problem (2) Certain Marching Cubes cases have more than

one possible triangulation.

Case 6 Case 3

Mismatch!!!

+

+

+

+

Hole!

Ambiguity Problem (3) To fix it …

Case 6 Case 3 B

Match!!!

+

+

+

+

The goal is to come up with a consistent triangulation

Asymptotic Decider (1) Based on bilinear interpolation over

faces

B01

B00 B10

B11

(s,t)

B(s,t) = (1-s, s) B00 B01B10 B11

1-t t

The contour curves of B:

{(s,t) | B(s,t) = } are hyperbolas

= B00(1- s)(1- t) + B10(s)(1- t) +

B01(1- s)(t) + B11(s)(t)

Asymptotic Decider (2)

(0,0)

(1,1)

Asymptote

(ST

If B(ST >=

(ST

Not Separated

Asymptotic Decider (3)

(1,1)

Asymptote

(ST

(0,0)

If B(ST <

(ST

Separated

Asymptotic Decider (4)

(S1 , 1)

(ST

(S0 , 0)

S B00 - B01 B00 + B11 – B01 – B10

T B00 – B10 B00 + B11 – B01 – B10

B(ST B00 B11 + B10 B01 B00 + B11 – B01 – B10

(0 , T0)

(1 , T1)

B( S) = B( S , 1)B( 0, T) = B( 1 , T)

Asymptotic Decider (5) case 3, 6, 12, 10, 7, 13

(These are the cases with at least one ambiguious faces)