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I have an unstructured dataset "nm", excel file in the link below.
https://www.dropbox.com/s/k7sdblxg4txjvj8/nm.xls?dl=0
it looks like that in a Mathematica list plot 1.
Could You please help me extracting upper bound envelope of this dataset, similar to the red line marked in picture 2? By extracting an envelope I mean to make another list "nm2" that for a given n produces a maximum m? I need another list "nm2" because then i need to work with it in excel.
Best Regards,
BG
Using a technique found here : obtain all boundary points on a convex hull
data = Import["nm.xls"];
chr = ConvexHullRegion[data];
sr = SignedRegionDistance[chr];
dist = sr /# data;
points = Extract[data, Position[dist, 0.]];
{maxx, maxy} = Max /# Transpose[data];
minx = First[FirstCase[points, {_, maxy}]];
miny = Last[FirstCase[points, {maxx, _}]];
arc = Select[points, First[#] >= minx && Last[#] >= miny &];
f = Interpolation[arc];
Show[ListPlot[data], Plot[f[x], {x, minx, maxx},
PlotStyle -> {Thick, Red}]]
I have this problem I can't figure out and need help.
The problem is about calculating how many balloons are hit by a pellet gun. Balloons positions are described by 3D coordinates (X,Y,Z) and radius R. The gunshot is defined by 3D location of the end of the barrel "p" (Px,Py,Pz) and vector "v" (Vx, Vy, Vz) describing the direction barrel is pointing to.
I've tried to implement the solution suggested here: https://math.stackexchange.com/questions/1939423/calculate-if-vector-intersects-sphere
// C = center of sphere
// r = radius of sphere
// P = point on line
// U = unit vector in direction of line
Q = P - C;
a = U*U; // should be = 1
b = 2*U*Q
c = Q*Q - r*r;
d = b*b - 4*a*c; // discriminant of quadratic
if d < 0 then solutions are complex, so no intersections
if d >= 0 then solutions are real, so there are intersections
But the problem with this is that I get intersection with balloons that are positioned behind the gun. How can I modify this algorithm in order to produce the correct result? Or is my approach maybe wrong?
You need to actually solve the quadratic equation defined by your variables a, b and c.
Often, there are math libraries to do this, something like:
(t1,t2) = QuadraticSolve(a, b, c);
You can also do it manually for each parameter:
t1 = (-b + sqrt(b*b - 4*a*c)) / (2*a)
t2 = (-b - sqrt(b*b - 4*a*c)) / (2*a)
If t1 or t2 is positive then that intersection is in front of your gun.
im trying to find the center of a circle. the only information I have is:
Two random points in the circle and the circle bulge. So far i've manage to calculate the radius of the circle (at least i think i did). Ill post bellow the equasions ive used so far.
these are just random values and will change on user input)
PointA(x = 10, y = 15)
PointB(x = 6, y = 12)
circle_bulge = 0.41
distance = PointB - PointA
radius = (distance / 4) * (circle_bulge + (1 / circle_bulge ))
if this math is incorrect, please let me know, but keep in mind that i need to find the X and Y coordinates of the center of the circle
Here is a picture of the problem:
By definition the bulge is b = tg(Alpha/4)
From the trigonometric formula: tg(2 angle) = 2tg(angle)/(1-tg2(angle))
applied to angle = Alpha/4 and using the definition of bulge:
tg(Alpha/2) = 2 b/(1-b2)
On the other hand
tg(Alpha/2) = s/d
Then
s/d = 2 b/(1-b2) and
d = s(1-b2)/(2 b)
which allows us to calculate d because b is known and s = ||B - A||/2, where ||B - A|| denotes the norm of the vector B - A.
Now, let's calculate
(u,v) = (B - A)/||B - A||
Then ||(u,v)|| = 1, (v,-u) is orthogonal to B - A, and we have
C = (v,-u)d + (A+B)/2
UPDATE
Pseudo code to compute the center
Inputs:
A = (a1, a2), B = (b1, b2) "two points"; b "bulge"
Calculations:
"lengths"
norm := sqrt(square(b1-a1) + square(b2-a2)).
s := norm/2.
d := s * (1-square(b))/(2*b)
"direction"
u := (b1-a1)/ norm.
v := (b2-a2)/ norm.
"center"
c1 := -v*d + (a1+b1)/2.
c2 := u*d + (a2+b2)/2.
Return C := (c1, c2)
Note: There are two solutions for the Center, the other one being
c1 := v*d + (a1+b1)/2.
c2 := -u*d + (a2+b2)/2.
Return C := (c1, c2)
I believe that you have all the required math in this reference: http://autocad.wikia.com/wiki/Arc_relationships
The center of the circunference has to be in the straight line that passes trough the middle point of the segment with perpendicular vector AB. You know the angle of the triangle formed by the three points because you know the bulge so you have to solve a simple equation. Try it out, if you can not I will try to help you.
Ignacio is right pointing out that there will be two solutions.
EDIT
The middle point is given by:
M = (A + B) / 2
The AB vector is gien by:
AB = A-B
The center of the circle C with coordinates (X, Y) has to be in the line given by:
((X, Y) - M) * AB = 0 //where * is the scalar vector product
The isosceles triangle generated by the points A, B and C has an angle opposite of the AB segment:
Angle = 4 arctan(bulge)
now we can compute the distance from the center C to A that we call d1 and half the distance between A and B that we call d2 we know then that
sin (Angle/2) = d2/d1
This will give you the second equation for X and Y that is quadratic (it has two solutions).
Excuse the notation but I do not know how to insert math here :-)
I am trying to determine whether a line segment (i.e. between two points) intersects a sphere. I am not interested in the position of the intersection, just whether or not the segment intersects the sphere surface. Does anyone have any suggestions as to what the most efficient algorithm for this would be? (I'm wondering if there are any algorithms that are simpler than the usual ray-sphere intersection algorithms, since I'm not interested in the intersection position)
If you are only interested if knowing if it intersects or not then your basic algorithm will look like this...
Consider you have the vector of your ray line, A -> B.
You know that the shortest distance between this vector and the centre of the sphere occurs at the intersection of your ray vector and a vector which is at 90 degrees to this which passes through the centre of the sphere.
You hence have two vectors, the equations of which fully completely defined. You can work out the intersection point of the vectors using linear algebra, and hence the length of the line (or more efficiently the square of the length of the line) and test if this is less than the radius (or the square of the radius) of your sphere.
I don't know what the standard way of doing it is, but if you only want to know IF it intersects, here is what I would do.
General rule ... avoid doing sqrt() or other costly operations. When possible, deal with the square of the radius.
Determine if the starting point is inside the radius of the sphere. If you know that this is never the case, then skip this step. If you are inside, your ray will intersect the sphere.
From here on, your starting point is outside the sphere.
Now, imagine the small box that will fit sphere. If you are outside that box, check the x-direction, y-direction and z-direction of the ray to see if it will intersect the side of the box that your ray starts at. This should be a simple sign check, or comparison against zero. If you are outside the and moving away from it, you will never intersect it.
From here on, you are in the more complicated phase. Your starting point is between the imaginary box and the sphere. You can get a simplified expression using calculus and geometry.
The gist of what you want to do is determine if the shortest distance between your ray and the sphere is less than radius of the sphere.
Let your ray be represented by (x0 + it, y0 + jt, z0 + kt), and the centre of your sphere be at (xS, yS, zS). So, we want to find t such that it would give the shortest of (xS - x0 - it, yS - y0 - jt, zS - z0 - kt).
Let x = xS - x0, y = yX - y0, z = zS - z0, D = magnitude of the vector squared
D = x^2 -2*xit + (i*t)^2 + y^2 - 2*yjt + (j*t)^2 + z^2 - 2*zkt + (k*t)^2
D = (i^2 + j^2 + k^2)t^2 - (xi + yj + zk)*2*t + (x^2 + y^2 + z^2)
dD/dt = 0 = 2*t*(i^2 + j^2 + k^2) - 2*(xi + yj + z*k)
t = (xi + yj + z*k) / (i^2 + j^2 + k^2)
Plug t back into the equation for D = .... If the result is less than or equal the square of the sphere's radius, you have an intersection. If it is greater, then there is no intersection.
This page has an exact solution for this problem. Essentially, you are substituting the equation for the line into the equation for the sphere, then computes the discriminant of the resulting quadratic. The values of the discriminant indicate intersection.
Are you still looking for an answer 13 years later? Here is a complete and simple solution
Assume the following:
the line segment is defined by endpoints as 3D vectors v1 and v2
the sphere is centered at vc with radius r
Ne define the three side lengths of a triangle ABC as:
A = v1-vc
B = v2-vc
C = v1-v2
If |A| < r or |B| < r, then we're done; the line segment intersects the sphere
After doing the check above, if the angle between A and B is acute, then we're done; the line segment does not intersect the sphere.
If neither of these conditions are met, then the line segment may or may not intersect the sphere. To find out, we just need to find H, which is the height of the triangle ABC taking C as the base. First we need φ, the angle between A and C:
φ = arccos( dot(A,C) / (|A||C|) )
and then solve for H:
sin(φ) = H/|A|
===> H = |A|sin(φ) = |A| sqrt(1 - (dot(A,C) / (|A||C|))^2)
and we are done. The result is
if H < r, then the line segment intersects the sphere
if H = r, then the line segment is tangent to the sphere
if H > r, then the line segment does not intersect the sphere
Here that is in Python:
import numpy as np
def unit_projection(v1, v2):
'''takes the dot product between v1, v2 after normalization'''
u1 = v1 / np.linalg.norm(v1)
u2 = v2 / np.linalg.norm(v2)
return np.dot(u1, u2)
def angle_between(v1, v2):
'''computes the angle between vectors v1 and v2'''
return np.arccos(np.clip(unit_projection(v1, v2), -1, 1))
def check_intersects_sphere(xa, ya, za, xb, yb, zb, xc, yc, zc, radius):
'''checks if a line segment intersects a sphere'''
v1 = np.array([xa, ya, za])
v2 = np.array([xb, yb, zb])
vc = np.array([xc, yc, zc])
A = v1 - vc
B = v2 - vc
C = v1 - v2
if(np.linalg.norm(A) < radius or np.linalg.norm(B) < radius):
return True
if(angle_between(A, B) < np.pi/2):
return False
H = np.linalg.norm(A) * np.sqrt(1 - unit_projection(A, C)**2)
if(H < radius):
return True
if(H >= radius):
return False
Note that I have written this so that it returns False when either endpoint is on the surface of the sphere, or when the line segment is tangent to the sphere, because it serves my purposes better.
This might be essentially what user Cruachan suggested. A comment there suggests that other answers are "too elaborate". There might be a more elegant way to implement this that uses more compact linear algebra operations and identities, but I suspect that the amount of actual compute required boils down to something like this. If someone sees somewhere to save some effort please do let us know.
Here is a test of the code. The figure below shows several trial line segments originating from a position (-1, 1, 1) , with a unit sphere at (1,1,1). Blue line segments have intersected, red have not.
And here is another figure which verifies that line segments that stop just short of the sphere's surface do not intersect, even if the infinite ray that they belong to does:
Here is the code that generates the image:
import matplotlib.pyplot as plt
radius = 1
xc, yc, zc = 1, 1, 1
xa, ya, za = xc-2, yc, zc
nx, ny, nz = 4, 4, 4
xx = np.linspace(xc-2, xc+2, nx)
yy = np.linspace(yc-2, yc+2, ny)
zz = np.linspace(zc-2, zc+2, nz)
n = nx * ny * nz
XX, YY, ZZ = np.meshgrid(xx, yy, zz)
xb, yb, zb = np.ravel(XX), np.ravel(YY), np.ravel(ZZ)
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
for i in range(n):
if(xb[i] == xa): continue
intersects = check_intersects_sphere(xa, ya, za, xb[i], yb[i], zb[i], xc, yc, zc, radius)
color = ['r', 'b'][int(intersects)]
s = [0.3, 0.7][int(intersects)]
ax.plot([xa, xb[i]], [ya, yb[i]], [za, zb[i]], '-o', color=color, ms=s, lw=s, alpha=s/0.7)
u = np.linspace(0, 2 * np.pi, 100)
v = np.linspace(0, np.pi, 100)
x = np.outer(np.cos(u), np.sin(v)) + xc
y = np.outer(np.sin(u), np.sin(v)) + yc
z = np.outer(np.ones(np.size(u)), np.cos(v)) + zc
ax.plot_surface(x, y, z, rstride=4, cstride=4, color='k', linewidth=0, alpha=0.25, zorder=0)
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('z')
plt.tight_layout()
plt.show()
you sorta have to work that the position anyway if you want accuracy. The only way to improve speed algorithmically is to switch from ray-sphere intersection to ray-bounding-box intersection.
Or you could go deeper and try and improve sqrt and other inner function calls
http://wiki.cgsociety.org/index.php/Ray_Sphere_Intersection
What's the algorithm for computing a least squares plane in (x, y, z) space, given a set of 3D data points? In other words, if I had a bunch of points like (1, 2, 3), (4, 5, 6), (7, 8, 9), etc., how would one go about calculating the best fit plane f(x, y) = ax + by + c? What's the algorithm for getting a, b, and c out of a set of 3D points?
If you have n data points (x[i], y[i], z[i]), compute the 3x3 symmetric matrix A whose entries are:
sum_i x[i]*x[i], sum_i x[i]*y[i], sum_i x[i]
sum_i x[i]*y[i], sum_i y[i]*y[i], sum_i y[i]
sum_i x[i], sum_i y[i], n
Also compute the 3 element vector b:
{sum_i x[i]*z[i], sum_i y[i]*z[i], sum_i z[i]}
Then solve Ax = b for the given A and b. The three components of the solution vector are the coefficients to the least-square fit plane {a,b,c}.
Note that this is the "ordinary least squares" fit, which is appropriate only when z is expected to be a linear function of x and y. If you are looking more generally for a "best fit plane" in 3-space, you may want to learn about "geometric" least squares.
Note also that this will fail if your points are in a line, as your example points are.
The equation for a plane is: ax + by + c = z. So set up matrices like this with all your data:
x_0 y_0 1
A = x_1 y_1 1
...
x_n y_n 1
And
a
x = b
c
And
z_0
B = z_1
...
z_n
In other words: Ax = B. Now solve for x which are your coefficients. But since (I assume) you have more than 3 points, the system is over-determined so you need to use the left pseudo inverse. So the answer is:
a
b = (A^T A)^-1 A^T B
c
And here is some simple Python code with an example:
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
N_POINTS = 10
TARGET_X_SLOPE = 2
TARGET_y_SLOPE = 3
TARGET_OFFSET = 5
EXTENTS = 5
NOISE = 5
# create random data
xs = [np.random.uniform(2*EXTENTS)-EXTENTS for i in range(N_POINTS)]
ys = [np.random.uniform(2*EXTENTS)-EXTENTS for i in range(N_POINTS)]
zs = []
for i in range(N_POINTS):
zs.append(xs[i]*TARGET_X_SLOPE + \
ys[i]*TARGET_y_SLOPE + \
TARGET_OFFSET + np.random.normal(scale=NOISE))
# plot raw data
plt.figure()
ax = plt.subplot(111, projection='3d')
ax.scatter(xs, ys, zs, color='b')
# do fit
tmp_A = []
tmp_b = []
for i in range(len(xs)):
tmp_A.append([xs[i], ys[i], 1])
tmp_b.append(zs[i])
b = np.matrix(tmp_b).T
A = np.matrix(tmp_A)
fit = (A.T * A).I * A.T * b
errors = b - A * fit
residual = np.linalg.norm(errors)
print("solution:")
print("%f x + %f y + %f = z" % (fit[0], fit[1], fit[2]))
print("errors:")
print(errors)
print("residual:")
print(residual)
# plot plane
xlim = ax.get_xlim()
ylim = ax.get_ylim()
X,Y = np.meshgrid(np.arange(xlim[0], xlim[1]),
np.arange(ylim[0], ylim[1]))
Z = np.zeros(X.shape)
for r in range(X.shape[0]):
for c in range(X.shape[1]):
Z[r,c] = fit[0] * X[r,c] + fit[1] * Y[r,c] + fit[2]
ax.plot_wireframe(X,Y,Z, color='k')
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('z')
plt.show()
unless someone tells me how to type equations here, let me just write down the final computations you have to do:
first, given points r_i \n \R, i=1..N, calculate the center of mass of all points:
r_G = \frac{\sum_{i=1}^N r_i}{N}
then, calculate the normal vector n, that together with the base vector r_G defines the plane by calculating the 3x3 matrix A as
A = \sum_{i=1}^N (r_i - r_G)(r_i - r_G)^T
with this matrix, the normal vector n is now given by the eigenvector of A corresponding to the minimal eigenvalue of A.
To find out about the eigenvector/eigenvalue pairs, use any linear algebra library of your choice.
This solution is based on the Rayleight-Ritz Theorem for the Hermitian matrix A.
See 'Least Squares Fitting of Data' by David Eberly for how I came up with this one to minimize the geometric fit (orthogonal distance from points to the plane).
bool Geom_utils::Fit_plane_direct(const arma::mat& pts_in, Plane& plane_out)
{
bool success(false);
int K(pts_in.n_cols);
if(pts_in.n_rows == 3 && K > 2) // check for bad sizing and indeterminate case
{
plane_out._p_3 = (1.0/static_cast<double>(K))*arma::sum(pts_in,1);
arma::mat A(pts_in);
A.each_col() -= plane_out._p_3; //[x1-p, x2-p, ..., xk-p]
arma::mat33 M(A*A.t());
arma::vec3 D;
arma::mat33 V;
if(arma::eig_sym(D,V,M))
{
// diagonalization succeeded
plane_out._n_3 = V.col(0); // in ascending order by default
if(plane_out._n_3(2) < 0)
{
plane_out._n_3 = -plane_out._n_3; // upward pointing
}
success = true;
}
}
return success;
}
Timed at 37 micro seconds fitting a plane to 1000 points (Windows 7, i7, 32bit program)
This reduces to the Total Least Squares problem, that can be solved using SVD decomposition.
C++ code using OpenCV:
float fitPlaneToSetOfPoints(const std::vector<cv::Point3f> &pts, cv::Point3f &p0, cv::Vec3f &nml) {
const int SCALAR_TYPE = CV_32F;
typedef float ScalarType;
// Calculate centroid
p0 = cv::Point3f(0,0,0);
for (int i = 0; i < pts.size(); ++i)
p0 = p0 + conv<cv::Vec3f>(pts[i]);
p0 *= 1.0/pts.size();
// Compose data matrix subtracting the centroid from each point
cv::Mat Q(pts.size(), 3, SCALAR_TYPE);
for (int i = 0; i < pts.size(); ++i) {
Q.at<ScalarType>(i,0) = pts[i].x - p0.x;
Q.at<ScalarType>(i,1) = pts[i].y - p0.y;
Q.at<ScalarType>(i,2) = pts[i].z - p0.z;
}
// Compute SVD decomposition and the Total Least Squares solution, which is the eigenvector corresponding to the least eigenvalue
cv::SVD svd(Q, cv::SVD::MODIFY_A|cv::SVD::FULL_UV);
nml = svd.vt.row(2);
// Calculate the actual RMS error
float err = 0;
for (int i = 0; i < pts.size(); ++i)
err += powf(nml.dot(pts[i] - p0), 2);
err = sqrtf(err / pts.size());
return err;
}
As with any least-squares approach, you proceed like this:
Before you start coding
Write down an equation for a plane in some parameterization, say 0 = ax + by + z + d in thee parameters (a, b, d).
Find an expression D(\vec{v};a, b, d) for the distance from an arbitrary point \vec{v}.
Write down the sum S = \sigma_i=0,n D^2(\vec{x}_i), and simplify until it is expressed in terms of simple sums of the components of v like \sigma v_x, \sigma v_y^2, \sigma v_x*v_z ...
Write down the per parameter minimization expressions dS/dx_0 = 0, dS/dy_0 = 0 ... which gives you a set of three equations in three parameters and the sums from the previous step.
Solve this set of equations for the parameters.
(or for simple cases, just look up the form). Using a symbolic algebra package (like Mathematica) could make you life much easier.
The coding
Write code to form the needed sums and find the parameters from the last set above.
Alternatives
Note that if you actually had only three points, you'd be better just finding the plane that goes through them.
Also, if the analytic solution in unfeasible (not the case for a plane, but possible in general) you can do steps 1 and 2, and use a Monte Carlo minimizer on the sum in step 3.
CGAL::linear_least_squares_fitting_3
Function linear_least_squares_fitting_3 computes the best fitting 3D
line or plane (in the least squares sense) of a set of 3D objects such
as points, segments, triangles, spheres, balls, cuboids or tetrahedra.
http://www.cgal.org/Manual/latest/doc_html/cgal_manual/Principal_component_analysis_ref/Function_linear_least_squares_fitting_3.html
It sounds like all you want to do is linear regression with 2 regressors. The wikipedia page on the subject should tell you all you need to know and then some.
All you'll have to do is to solve the system of equations.
If those are your points:
(1, 2, 3), (4, 5, 6), (7, 8, 9)
That gives you the equations:
3=a*1 + b*2 + c
6=a*4 + b*5 + c
9=a*7 + b*8 + c
So your question actually should be: How do I solve a system of equations?
Therefore I recommend reading this SO question.
If I've misunderstood your question let us know.
EDIT:
Ignore my answer as you probably meant something else.
We first present a linear least-squares plane fitting method that minimizes the residuals between the estimated normal vector and provided points.
Recall that the equation for a plane passing through origin is Ax + By + Cz = 0, where (x, y, z) can be any point on the plane and (A, B, C) is the normal vector perpendicular to this plane.
The equation for a general plane (that may or may not pass through origin) is Ax + By + Cz + D = 0, where the additional coefficient D represents how far the plane is away from the origin, along the direction of the normal vector of the plane. [Note that in this equation (A, B, C) forms a unit normal vector.]
Now, we can apply a trick here and fit the plane using only provided point coordinates. Divide both sides by D and rearrange this term to the right-hand side. This leads to A/D x + B/D y + C/D z = -1. [Note that in this equation (A/D, B/D, C/D) forms a normal vector with length 1/D.]
We can set up a system of linear equations accordingly, and then solve it by an Eigen solver in C++ as follows.
// Example for 5 points
Eigen::Matrix<double, 5, 3> matA; // row: 5 points; column: xyz coordinates
Eigen::Matrix<double, 5, 1> matB = -1 * Eigen::Matrix<double, 5, 1>::Ones();
// Find the plane normal
Eigen::Vector3d normal = matA.colPivHouseholderQr().solve(matB);
// Check if the fitting is healthy
double D = 1 / normal.norm();
normal.normalize(); // normal is a unit vector from now on
bool planeValid = true;
for (int i = 0; i < 5; ++i) { // compare Ax + By + Cz + D with 0.2 (ideally Ax + By + Cz + D = 0)
if ( fabs( normal(0)*matA(i, 0) + normal(1)*matA(i, 1) + normal(2)*matA(i, 2) + D) > 0.2) {
planeValid = false; // 0.2 is an experimental threshold; can be tuned
break;
}
}
We then discuss its equivalence to the typical SVD-based method and their comparison.
The aforementioned linear least-squares (LLS) method fits the general plane equation Ax + By + Cz + D = 0, whereas the SVD-based method replaces D with D = - (Ax0 + By0 + Cz0) and fits the plane equation A(x-x0) + B(y-y0) + C(z-z0) = 0, where (x0, y0, z0) is the mean of all points that serves as the origin of the new local coordinate frame.
Comparison between two methods:
The LLS fitting method is much faster than the SVD-based method, and is suitable for use when points are known to be roughly in a plane shape.
The SVD-based method is more numerically stable when the plane is far away from origin, because the LLS method would require more digits after decimal to be stored and processed in such cases.
The LLS method can detect outliers by checking the dot product residual between each point and the estimated normal vector, whereas the SVD-based method can detect outliers by checking if the smallest eigenvalue of the covariance matrix is significantly smaller than the two larger eigenvalues (i.e. checking the shape of the covariance matrix).
We finally provide a test case in C++ and MATLAB.
// Test case in C++ (using LLS fitting method)
matA(0,0) = 5.4637; matA(0,1) = 10.3354; matA(0,2) = 2.7203;
matA(1,0) = 5.8038; matA(1,1) = 10.2393; matA(1,2) = 2.7354;
matA(2,0) = 5.8565; matA(2,1) = 10.2520; matA(2,2) = 2.3138;
matA(3,0) = 6.0405; matA(3,1) = 10.1836; matA(3,2) = 2.3218;
matA(4,0) = 5.5537; matA(4,1) = 10.3349; matA(4,2) = 1.8796;
// With this sample data, LLS fitting method can produce the following result
// fitted normal vector = (-0.0231143, -0.0838307, -0.00266429)
// unit normal vector = (-0.265682, -0.963574, -0.0306241)
// D = 11.4943
% Test case in MATLAB (using SVD-based method)
points = [5.4637 10.3354 2.7203;
5.8038 10.2393 2.7354;
5.8565 10.2520 2.3138;
6.0405 10.1836 2.3218;
5.5537 10.3349 1.8796]
covariance = cov(points)
[V, D] = eig(covariance)
normal = V(:, 1) % pick the eigenvector that corresponds to the smallest eigenvalue
% normal = (0.2655, 0.9636, 0.0306)