Extended Gaussian Images
Berthold K. P. Horn
Outline
 Discrete Case: Convex Polyhedra
 Continuous Case: Smoothly Curved Objects
 Discrete Approximation: Needle Maps
 Tessellation of the Gaussian sphere:
Orientation Histograms
 Solids of Revolution
Outline
 Discrete Case: Convex Polyhedra
 Continuous Case: Smoothly Curved Objects
 Discrete Approximation: Needle Maps
 Tessellation of the Gaussian sphere:
Orientation Histograms
 Solids of Revolution
Gaussian Image
 What is a Gaussian Image?
Gaussian Image
 What is a Gaussian Image?
 A mapping of surface
normals of an object onto
the unit sphere (Gaussian
sphere.)


Tail lies at the center of the
Gaussian sphere
Head lies on the surface of
the Gaussian sphere.
Example
Example
Extended Gaussian Image
 Extended to include area of each face.


Place point mass on the sphere surface equal to
area of face at the head of each normal.
Alternate representation: Scale the normal
proportional to area.
Properties
 Face position information is lost.


Which faces are connected?
As long as the polyhedron is convex, its
extended Gaussian image is unique.

Algorithms exist that extract a polyhedron from an
extended Gaussian Image.
Properties
 Unaffected by translation
 Rotation of the object causes an equal
rotation of the extended Gaussian image.
 Center of mass lies at the origin of the
sphere.
Center of mass property proof
 Imagine viewing a convex polyhedron from a great
distance:

 v is a vector from the object in the direction of the
viewer.

 s i is the unit normal at each face.
 Oi is the surface area of each face.
 
 A face is visible if si  v  0
 The apparent size due to foreshortening of a face is
 
si  v Oi
Center of mass property proof
 The apparent area of the visible surface is

A(v ) 
 
 si  v Oi
 
{i|si v 0}
 When viewed from the opposite direction:

A(v ) 
 
 si  v Oi
 
{i|si v 0}
Center of mass property proof
 Since the above two equations are equal, we have
   
 
si  v Oi    siOi   v  0

{all i}
{all i}

 Since this is true for all v

 siOi  0
{all i }
 That is, the center of mass of the extended Gaussian
image is at the origin.
Reconstructing a Tetrahedron
 Goal: Find the offset
from the center of
mass for each face.
(a, b, c, d)
 Given: The
surface


 
normals a b c d
and face areas A B C
D.
C
B
A
D
Reconstructing a Tetrahedron
 Tetrahedron property: the distance from the
center of mass to a face is ¼ the distance
from that face to its opposite vertex.
 Start by finding a formula for distance from
face of area D to opposite vertex d. The
desired distance, d, with be ¼ that.
 Repeat for each vertex after doing a cyclical
permutation of the variables.
Reconstructing a Tetrahedron
 Assume D is at the origin, we have:
BC 
C A 
A B

a
,b  
,c  
BC
C A
BC
C
 ( A  C )  ( B  A)
d
( A  C )  ( B  A)
A B  B  C  C  A

A B  B  C  C  A
B
A
D
Reconstructing a Tetrahedron
 Find distance from originto D face by dot product
of A, B, or C vector and d :



[ ABC ]
4d  d  A  d  B  d  C 
A B  B  C  C  A
 Area of the Face:
1
1
D  ( A  C )  ( B  A)  A  B  B  C  C  A
2
2
Reconstructing a Tetrahedron
 Consider the four distinct triple products of
the normals:
 
[ab c ]  
[ ABC ]2
A B B  C C  A

[ab d ]  
[ ABC ]2
B  C C  A A B  B  C  C  A
...
Reconstructing a Tetrahedron
 Multiply these formulae:
      
[ab d ][ ac d ][c ad ]  
[ ABC ]6
( A  B) 2 ( B  C ) 2 (C  A) 2 A  B  B  C  C  A
      
[ab d ][ ac d ][c ad ]
[ ABC ]2

 ( 4d ) 2 ( 2 D )
  2
[ab c ]
A B  B  C  C  A
      
(2 D)[ ab d ][ ac d ][c ad ]
4d 
 
 [ab c ]
Reconstructing a Tetrahedron
 Compute cyclic permutations of each face
and use the previous equation to find the 4
distances of each face from the center of
mass.
 With the normals and distances from the
center of mass, create 4 planes.

The intersection of the planes creates the
tetrahedron.
Outline
 Discrete Case: Convex Polyhedra
 Continuous Case: Smoothly Curved Objects
 Discrete Approximation: Needle Maps
 Tessellation of the Gaussian sphere:
Orientation Histograms
 Solids of Revolution
Gaussian Image
 Create the same mapping between object
normal and Gaussian sphere.


Consider a small patch O on the object.
Each point in patch maps to a point on Gaussian
sphere.
 Creates a patch on the sphere S
Gaussian Curvature
 Gaussian Curvature:



K < 1 curves in (like a crater)
K = 1 plane
K > 1 curves out (like a sphere)
 Integrate K over the object:
 Integrate 1/K over the sphere:
 Use 1/K in definition of extended Gaussian
Image.
Extended Gaussian Image
 Associate 1/K of a point on the surface with
corresponded point on Gaussian sphere.


Let u, v be parameters to identify points on the
object.
Let ξ, η be parameters to identify points on the
sphere.
 Extended Gaussian Image:
Properties
 Center of mass at origin.

Similar proof to discrete case
 Total mass of extended Gaussian Image
equals surface area of object.
 Unaffected by translation
 Rotation of the object causes an equal
rotation of the extended Gaussian image.
Outline
 Discrete Case: Convex Polyhedra
 Continuous Case: Smoothly Curved Objects
 Discrete Approximation: Needle Maps
 Tessellation of the Gaussian sphere:
Orientation Histograms
 Solids of Revolution
Discrete Approximation
 Break the surface into small patches.
 Create a surface normal for each patch.
 Consider the polyhedron formed by the
intersection of the planes tangent to each of
these normals.

Approximation of the original surface.
Discrete Approximation
 How to break into small patches?


Parameterize the model: u, v
Find the normal:
 ru
and rv are found by differentiating the parametric
form of the equation for the surface.

The area of a patch:

No need to compute K
Needle Map
Outline
 Discrete Case: Convex Polyhedra
 Continuous Case: Smoothly Curved Objects
 Discrete Approximation: Needle Maps
 Tessellation of the Gaussian sphere:
Orientation Histograms
 Solids of Revolution
Tessellation
 Need a way to represent in a computer


Tessellate the Gaussian sphere
Create a histogram of orientations
Tessellation Properties
 All cells should have the same area.
 All cells should have the same shape.
 The cells should have regular shapes that are
compact.
 The division should be fine enough to provide good
angular resolution.
 For some rotation, the cells should be brought into
coincidence with themselves.
Simple Case
 Latitude and
Longitude



Easy to Compute
which cell a normal
belongs to
No uniform shape or
area
Alignments only
through rotations about
its axis.
Regular Polyhedra based
Tessellation
 Project a regular
polyhedron onto the
unit sphere.
 Cells have the same
shape and area.
 Sampling is too coarse.

Icosahedron is only 20
regular cells.
Semi-Regular Polyhedra
 Regular polygons for
faces, but not the same
polygon.
 Higher polygon count
 Area difference
between the differing
polygons could be
significant.
Finer Subdivision
 Each polygon can be sub-divided into
triangles.
 How fine do we need?

Lower bound for angular spread with n cells:

Example from paper: n=240, θ=11.5
Geodesic Domes
 Divide Triangular cells into four smaller triangles.




All faces not same area or shape.
Allows for arbitrary fineness.
Start with pentakis dodecahedron or truncated
icosahedron.
Each triangle edge is divided into f sections to create f2
triangles.

Desire that f is a power of 2.
Geodesic Domes
Geodesic Domes
 Create Histogram by assigning each normal to a
cell on the tessellated Gaussian sphere.
 Which cell does a normal belong to?

Find dot product between given vector and normals from
the original regular polyhedron the dome is based on.


Determine which triangle within the face


Normal belongs to the face giving the greatest result.
Use the second greatest normal from above.
Proceed as above hierarchically.


Binary search through the triangles
In practice a lookup table is used
Orientation Histogram
 Compute the number of normals in each cell.


Count them
Create the histogram
 Store in computer as simple array.
 Display graphically with a normal for each
cell or as a grayscale image where the
brightness of each cell is determined by the
count.
Outline
 Discrete Case: Convex Polyhedra
 Continuous Case: Smoothly Curved Objects
 Discrete Approximation: Needle Maps
 Tessellation of the Gaussian sphere:
Orientation Histograms
 Solids of Revolution
Solids of Revolution
 Rotate a curve about an axis
to create a surface.
 Compute the EGI




r is the distance from axis to
arc.
s is distance along arc.
θ is the angle around the axis.
ξ, η are the longitude and
latitude on the sphere.

ξ corresponds to θ
Gaussian Curvature
 Consider a small patch on the solid and the
corresponding patch on the sphere, their
areas are:


Sphere
Solid
 The Gaussian curvature is the limit of the
ratio of the areas as they approach zero:
Gaussian Curvature
 The curvature of the generating curve is the
rate of change of direction with arc length.
Giving:
 Plug that in to the previous equation:
Gaussian Curvature
 From the figure:
 Differentiate with
respect to s:
 Substitute:
Gaussian Curvature
 Sometimes we want to express r as a function
of z.
 From the figure:
 Differentiate with respect to s:
 From the figure:
Gaussian Curvature
 Substituting:
 More Substituting:
Conclusion
 How does it do as a shape descriptor?









Concise to store
Quick to compute
Efficient to match
Discriminating
Invariant to transformations
Invariant to deformations
Insensitive to noise
Insensitive to topology
Robust to degeneracies
Conclusion
 How does it do as a shape descriptor?









J Concise to store
J Quick to compute
J Efficient to match
L Discriminating
L Invariant to transformations
L Invariant to deformations
L Insensitive to noise
J Insensitive to topology
L Robust to degeneracies