surface normals and principal component analysis (pca)

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Surface normals and principal component analysis (PCA)

3DM slides byMarc van Kreveld

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Normal of a surface

• Defined at points on the surface: normal of the tangent plane to the surface at that point

• Well-defined and unique inside the facets of any polyhedron

• At edges and vertices,the tangent plane is notunique or not defined(convex/reflex edge) normal is undefined

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Normal of a surface

• On a smooth surface without a boundary, the normal is unique and well-defined everywhere (smooth simply means that the derivatives of the surface exist everywhere)

• On a smooth surface (manifold) with boundary, the normal is not defined on the boundary

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Normal of a surface

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Normal of a surface

• The normal at edges or vertices is often definedin some convenient way: some average of normalsof incident triangles

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Normal of a surface

• No matter what choice we make at a vertex, a piecewise linear surface will not have a continuously changing normal visible after computing illumination

not normals!they would be parallel

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Curvature

• The rate of change of the normal is the curvature

higher curvature

lower curvature

infinite curvature

zero curvature

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Curvature

• A circle is a shape that has constant curvature everywhere

• The same is true for a line, whose curvature is zero everywhere

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Curvature

• Curvature can be positive or negative• Intuitively, the magnitude of the curvature is the

curvature of the circle that looks most like the curve, close to the point of interest

negative curvature

positive curvature

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Curvature

• The curvature at any point on a circle is the inverse of its radius

• The (absolute) curvature at any point on a curve is the curvature of the circle through that point that has the same first and second derivative at that point (so it is defined only for points where the curve is C2)

r curvature = 1/r

r

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Curvature

• For a 3D surface, there are curvatures in all directions in the tangent plane

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Curvature

negative

positive

inside

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Properties at a point

• A point on a smooth surface has various properties:– location (“zero-th” derivative)– normal (first derivative) / tangent plane– two/many curvatures (second derivative)

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Normal of a point in a point set?

• Can we estimate the normal for each point in a scanned point cloud? This would help reconstruction (e.g. for RANSAC)

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Normal of a point in a point set

• Main idea of various different methods, to estimate the normal of a point q in a point cloud:– collect some nearest neighbors of q, for instance 12– fit a plane to q and its 12 neighbors– use the normal of this plane as the estimated normal for q

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Normal estimation at a point

• Risk: the 12 nearest neighbors of q are not nicely spread in all directions on the plane the computed normal could even be perpendicular to the real normal!

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Normal estimation at a point

• Also: the quality of normals of points near edges of the scanned shape is often not so good

• We want a way of knowing how good the estimated normal seems to be

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Principal component analysis

• General technique for data analysis• Uses the statistical concept of correlation• Uses the linear algebra concept of eigenvectors• Can be used for normal estimation and tells

something about the quality (clearness, obviousness) of the normal

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Correlation

• Degree of correspondence/changing together of two variables measured from objects– in a population of people, length and weight are correlated– in decathlon, performance on 100 meters and long jump

are correlated (so are shot put and discus throw)

Pearson’s correlation coefficient

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Covariance, correlation

• For two variables x and y, their covariance is defined as

(x,y) = E[ (x – E[x]) (y – E[y]) ]= E[xy] – E[x] E[y]

• E[x] is the expected value of x, equal to the mean x

• Note that the variance 2(x) = (x,x), the covariance of x with itself, where (x) is the standard deviation

• Correlation x,y = (x,y) / ( (x) (y))

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Covariance

• For a data set of pairs (x1, y1), (x2, y2), …, (xn, yn), the covariance can be computed as

where x and y are the mean values of xi and yi

1𝑛∑𝑖=1

𝑛

(𝑥𝑖−𝑥 )(𝑦 𝑖− 𝑦 )

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Data matrix

• Suppose we have weight w, length l, and blood pressure b of seven people

• Let the mean of w, l, and b be w, l, and b

• Assume the measurements have been adjusted by subtracting the appropriate mean

• Then the data matrix is X =

• Note: Each row has zero mean, the data is mean-centered

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Covariance matrix

• The covariance matrix is XXT

• This is in the example:

• The covariance matrix is square and symmetric• The main diagonal contains the variances• Off-diagonal are the covariance values

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• PCA is a linear transformation (3 x 3 in our example) that makes new base vectors such that– the first base vector has a direction that realizes the largest

possible variance (when projected onto a line)– the second base vector is orthogonal to the first and

realizes the largest possible variance among those vectors– the third base vector is orthogonal to the first and second

base vector and …– … and so on …

• Hence, PCA is an orthogonal linear transformation

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• In 2D, after finding the first base vector, the second one is immediately determined because of the requirement of orthogonality

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• In 3D, after the first base vector is found, the data is projected onto a plane with this base vector as its normal, and we find the second base vector in this plane as the direction with largest variance in that plane

(this “removes” the variance captured by the first base vector)

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• After the first two base vectors are found, the data is projected onto a line orthogonal to the first two base vectors and the third base vector is foundon this line (trivial)

it is simply givenby the cross product of the first two base vectors

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• The subsequent variances we find are decreasing in value and give an “importance” to the base vectors

• The mind process explains why principal component analysis can be used for dimension reduction: maybe all the variance in, say, 10 measurement types can be explained using 4 or 3 (new) dimensions

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• In actual computation, all base vectors are found at once using linear algebra techniques

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Eigenvectors of a matrix

• A non-zero vector v is an eigenvector of a matrix X if X v = v for some scalar , and is called an eigenvalue corresponding to v

• Example 1: (1,1) is an eigenvector of because = 3

In words: the matrix keeps the direction of this eigenvector the same, but its length is scaled by the eigenvalue 3

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• A non-zero vector v is an eigenvector of a matrix X if X v = v for some scalar , and is called an eigenvalue corresponding to v

• Example 1: (1, –1) is also an eigenvector of because =

In words: the matrix keeps the direction and length of (1, –1) the same because its eigenvalue is 1

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• Consider the transformation animated

Blue vectors: (1,1)

Pink vectors: (1, –1) and (–1, 1)

Red vectors are not eigenvectors (they change direction)

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• If v is an eigenvector, then any vector parallel to v is also an eigenvector (with the same eigenvalue!)

• If the eigenvalue is –1 (negative in general), then the eigenvector will be reversed in direction by the matrix

• Only square matrices have eigenvectors and eigenvalues

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Eigenvectors, a 2D example

• Find the eigenvectors and eigenvalues of • We need: Av = v by definition, or (A – I) v = 0

(in words: our matrix minus times the identity matrix applied to v is the zero vector)

• This is the case exactly when det(A – I) = 0• det(A – I) = det =

= (–2 – )(2 – ) – (–3)1 = 2 – 1 = 0

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Eigenvectors, a 2D example

• 2 – 1 = 0 gives = 1 or = –1• The corresponding eigenvectors can be obtained by

filling in each and solving a set of equations

• The polynomial in given by det(A – I) is called the characteristic polynomial

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Questions1. Determine the eigenvectors and eigenvalues of

What does the matrix do? Does that explain the eigenvectors and values?

2. Determine the eigenvectors and eigenvalues of What does the matrix do? Does that explain the eigenvectors and values?

3. Determine the eigenvectors and eigenvalues of

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• Recall: PCA is an orthogonal linear transformation• The new base vectors are the eigenvectors of the

covariance matrix!• The eigenvalues are the variances of the data points

when projected onto a line with the direction of the eigenvector

• Geometrically, PCA is a rotation around the multi-dimensional mean (point) so that the base vectors align with the principal components(which is why the data matrix must be mean centered)

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PCA example

• Assume the data pairs (1,1), (1,2), (3,2), (4,2), and (6,3)• X = 15/5 = 3 and Y = 10/5 = 2 • The mean-centered data becomes

(-2,-1), (-2,0), (0,0), (1,0), and (3,1)• The data matrix X = • The covariance matrix XXT = • The characteristic polynomial is det

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PCA example

• The characteristic polynomial is det = ((18 – )(2 – ) – 25) = (2 – 20 + 11)

• When setting it to zero we can omit the factor • We get = , 1 19.43 and 0.57

as the eigenvalues of the covariance matrix

• We always choose the eigenvalues to be in decreasing order: 1 > > …

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PCA example

• The first eigenvalue 1 19.43 corresponds to an eigenvector (1, 0.29) or anything parallel to it

• The second eigenvalue 0.57 corresponds to an eigenvector (–0.29, 1) or anything parallel to it

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PCA example

• The data points and the mean-centered data points

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PCA example

• The first principal component (purple): (1, 0.29)

• Orthogonal projection onto the orange line (direction of first eigenvector) yields the largest possible variance

• The first eigenvalue 1 19.43 is the sum of the squared distances to the mean (variance times 5) for this projection

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PCA example

• Enlarged, and the non-squared distances shown

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PCA example

• The second principal component (green): (–0.29, 1)

• Orthogonal projection onto the dark blue line (direction of second eigenvector) yields the remaining variance

• The second eigenvalue 2 0.57 is the sum of the squared distances to the mean (variance times 5) for this projection

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PCA example

• The fact that the first eigenvalue is much larger than the second means that there is a direction that captures most of the variance of the data a line exists that fits well with the data

• When both eigenvalues are equally large, the data is spread equally in all directions

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PCA, eigenvectors and eigenvalues

• In the pictures, identify the eigenvectors and state how different the eigenvalues appear to be

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PCA observations in 3D

• If the first eigenvalue is large and the other two are small, then the data points lie approximately on a line– through the 3D mean– with orientation parallel to the first eigenvector

• If the first two eigenvalues are large and the third eigenvalue is small, then the points lie approximately on a plane– through the 3D mean– with orientation spanned by the first two eigenvectors /

with normal parallel to the third eigenvector

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PCA and local normal estimation

• Recall that we wanted to estimate the normal at every point in a point cloud

• Recall that we decided to use the 12 nearest neighbors for any point q, and find a fitting plane for q and its 12 nearest neighbors

qAssume we have the 3D coordinates of these points measured in meters

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• Treat the 13 points and their three coordinates as data with three measurements, x, y, and z: we have a 3 x 13 data matrix

• Apply PCA to get three eigenvalues 1 ,2 ,3 , (in decreasing order) and eigenvectors v1 , v2 , and v3

• If the 13 points lie roughly in a plane, then 3 is small and the plane contains directions parallel to v1 , v2

• The estimated normal is the perpendicular to v1 , v2 , so it is the third eigenvector v3

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• How large should the eigenvalues 1 ,2 3 be, to get a reliable normal?

• This depends on – scanning density– point distribution– scanning accuracy– curvature of the surface

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• Example 1: Assume– the surface is a perfect plane (no curvature)– scanning yields uniform distribution– density is 100 pt/m2

– accuracy is 0.03 m

• Then 3 will be less than 13 x 0.032 = 0.0117;we expect the 13 points to roughly lie in a cylinder of radius 0.203 m and thickness 0.06 m, and 1 and 2 should each be about 0.1 (10x as large as 3, but this is just a rough guess)

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• Then 3 will be less than 13 x 0.032 = 0.0117;we expect the 13 points to roughly lie in a cylinder of radius 0.203 m and thickness 0.06 m, and 1 and 2 should each be about 0.1 (10x as large as 3, but this is just a rough guess)

0.06 m

0.203 m

best fit plane is half-way

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• Then 3 will be less than 13 x 0.032 = 0.0117;we expect the 13 points to roughly lie in a circle of radius 0.09 m, and 1 and 2 should each be about 0.026 (maybe 3 times as large as 3)

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• Then 3 will be close to 0.02 ( 13 x 0.042 );we expect the 13 points to roughly lie in a circle of radius 0.09 m, and 1 and 2 should each be about 0.026 (maybe hardly larger than 3)

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• When the density goes up and/or the accuracy goes down, we may need to use more than 12 nearest neighbors to observe a considerable difference in eigenvalues for points on a plane, and estimate the normal correctly

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• More neighbors means more reliable covariance estimations, so a better normal, for flat surfaces

• … but for surfaces with considerable curvature, or close to edges of a flat surface, more neighbors means that the quality of normals goes down(because more often, points across the edges will be included in the nearest neighbors)

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• Example 4: Assume– the surface is a perfect plane (no curvature)– scanning is by LiDAR, with dense lines– density is 100 pt/m2


• We will find one largeeigenvalue and twosmall ones

q

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• Example 5: Assume– point q lies in a tree

(leaf or thin branch)– distribution is uniform– density is 100 pt/m2


• We will find threeeigenvalues that don’tdiffer too much (but itis rather unpredictable)

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Another way for normal estimation

• Compute the Voronoi diagram of the points and choose the direction from each point to the furthest Voronoi vertex bounding its cell

• For unbounded cells,take the middledirection of the unbounded rays

• Need to resolveinside/outside

• Works in any dimension

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Curvature estimation

• Recall: in 3D a curvature at point p exists for any direction in the tangent plane of p

• “The” curvature at p is the maximum occurring curvature

• Basic approach: fit a suitable ball or quadratic surface with p on its boundary that locally is a good fit for the point set. Then compute the curvature of the ball or surface at p

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Curvature estimation

• Or: choose many directions in the tangent plane and fit a circle tangent to p in the plane normal to the tangent plane and a chosen direction

• Choose the circle with smallest radius over all directions

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Summary

• Eigenvectors and eigenvalues are a central concept in linear algebra and they are generally useful

• Local normal estimation can be done by principal component analysis on the 12 nearest neighbors, which essentially comes down to eigenvector and eigenvalue computation of 3x3 matrices

• There are other methods for local normal estimation, but these are generally less reliable and may not indicate how “good” the normal is

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Questions1. Perform principal component analysis on the following four

data points: (1, 3), (2, 2), (4, 2), (5, 5) [mean-center first!]2. Estimate the normal from 5 points in 3D, namely:

(0,1,1), (4,2,0), (8,5,1), (-6,-2,1), (-6,-6,-3)How clear do you think the estimated normal is?

3. Can the estimated normal always be obtained as the third eigenvector? If not, what can you do?

4. Suppose the density and accuracy suggest that you need the 100 nearest neighbors to get a good normal estimate, but you don’t want to manipulate large matrices (3 x 100) for every point, what can you do?

surface normals and principal component analysis (pca)

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