I want to calculate the angle between 2 planes, Reference plane and Plane1. When I feed the X,Y,Z co-ordinates of pointCloud to the function plane_fit.m (by Kevin Mattheus Moerman), I get the coefficients:
reference_plane_coeff: [-0.13766204 -0.070385590 130.69409]
Plane1_coeff: [0.0044337390 -0.0013548643 95.890228]
Next, I find the intersection of both planes, separately on the XZ plane and get a line equation; ref_line_XZ and plane1_line_XZ respectively. For this, I make the second coefficient 0. (Is this right?)
Aref = reference_plane_coeff(1);
Cref = reference_plane_coeff(3);
ref_line_XZ = [Aref Cref];
Arun = Plane1_coeff(1);
Crun = Plane1_coeff(3);
plane1_line_XZ = [Arun Crun];
angle_XZ = acos( dot(ref_line_XZ,plane1_line_XZ ) / (norm(ref_line_XZ) * norm(plane1_line_XZ )) )
I get the angle_XZ value as 0.0012 rad. i.e. 0.0685 degrees
When I plot these planes on a graph and view it, the angle seems to be much more than 0.0012 degrees. I'm talking about the angle made by the two lines after intersection of both planes with the XZ plane.
What am I doing wrong?
Also, when I tried to find angle between its normals, using:
angle_beta_deg = acosd( dot(reference_plane_coeff,Plane1_coeff) / (norm(reference_plane_coeff) * norm(Plane1_coeff)) )
I got the angle as 0.0713.
On visual inspection of both planes' plots and manually calculating from the plot, angle_XZ should be around 9 degrees.
plane_fit.m (by Kevin Mattheus Moerman)
Related
I have two samples of 3D point cloud of human face. The blue point cloud denote target face and the red point cloud indicates the template. The image below shows that the target and template face are aligned in different directions (target face roughly along x-axis, template face roughly along y-axis).
Figure 1:
The region around the nose is displayed in Figure 1.
I want to rotate my target face (blue face) with nasal tip as the center of rotation (I translated the target to the template prior to Figure1 so that the tip of nose, i.e., the centerpt, for both faces are superimposed) to grossly align with the template face (red face). I rotated the target face with the following MATLAB code:
% PCA for the target face
targetFaceptfmt = pointCloud(targetFace); % Convert to point cloud format
point = [templateFace(3522, 1), templateFace(3522, 2), templateFace(3522, 3)]; % The 3522th point in the templateFace is the nasal tip point used as center of rotation later on
radius = 20; % 20mm
[NNTarIndex, NNTarDist] = findNeighborsInRadius(Locationptfmt, point, radius); % Find all vertices within 20 of the nasal tip point on the target face
NNTar = select(Locationptfmt, NNTarIndex); % Select the identified points for PCA
[TarVec,TarSCORE,TarVal] = pca(NNTar.Location); % Do PCA for target face using vertices close to the nasal tip
% PCA for the template face
templateFaceptfmt = pointCloud(templateFace); % Convert to point cloud format
[NNTemIndex, NNTemDist] = findNeighborsInRadius( templateFaceptfmt, point, radius); % Find all vertices within 20 of the nasal tip point on the template
NNTem = select(templateFaceptfmt, NNTemIndex); % Select the identified points for PCA
[TemVec,TemSCORE,TemVal] = pca(NNTem.Location); % Do PCA for template face using vertices close to the nasal tip
% Rotate target face with nasal tip point as the center of rotation
targetFace_r = R * (targetFace-cenertpt)' + centerpt';
targetFace_new = targetFace_r';
where targetFace and templateFace contains coordinates for the unrotated target face and the template face, respectively. The targetFace_r contains coordinates for the target face after rotation around nasal tip, R is the rotation matrix calculated through PCA (See here for source of formula for rotation), and centerpt is the nasal tip point which is used as the center of rotation. I then plotted the transposed targetFace_r, i.e., the targetFace_new, with normals added to each vertex:
Figure 2:
Before rotation, the normals for the target face and template face are generally pointing toward similar directions (Figure 1). After rotation, the target and template face are both aligned along the y-axis (which is what I want), however, normals for the target face and template face point toward opposite directions. Bearing in mind that no changes were made to the template face, I realized that normals of the target face calculated after rotation are flipped. But I do not know why. I used the checkFaceOrientation function of the Rvcg package in R to check if expansion along normals increases centroid size. I was returned TRUE for the template face but FALSE for the target face, which confirms that vertex normals for the target face are flipped.
Vertex normals were calculated in MATLAB as follows:
TR = triangulation(Faces, Vertices); % Triangulation based on face and vertex information
VN = vertexNormal(TR); % Calculate vertext normal
where Faces contains face information, i.e., the connectivity list, and Vertices contains coordiantes for vertices. For target face before rotation, target face after rotation, and template face, vertex normals were calcuated separately. I used the same Faces data for calculation of vertex normal before and after rotating the target face.
The flipped vertex normals resulted in errors for some further analyses. As a result, I have to manually flip the normals to make them pointing similarly to normals of the template face.
Figure 3:
Figure 3 shows that after manually flip the normals, normals of the target and template face are generally pointing similarly in direction.
My question is why does the normals of the target face calculated after rotation flipped? In what case does rotation of 3D point cloud result in flipping of vertex normals?
Some further informaiton that may be useful: the rotation matrix R I obtained is as follows for your reference:
0.0473096146726546 0.867593376108813 -0.495018720950670
0.987013081649028 0.0355601323276586 0.156654567895508
-0.153515396665006 0.496001220483328 0.854643675613313
Since trace(R) = 1 + 2cos(alpha), I calcualted alpha through acos((trace(R)-1)/2)*180/pi, which yielded an angle of rotation of 91.7904, relative to the nasal tip point.
If I'm understanding everything correctly, it looks like your rotation matrix is actually encoding a rotation plus a reflection. If your matrix is approximately:
0.04 0.86 -0.49
0.98 0.03 0.15
-0.15 0.49 0.85
Then the image of each unit vector pointing along the positive axes are:
x = [ 0.04 0.98 -0.15]
y = [ 0.86 0.03 0.49]
z = [-0.49 0.15 0.85]
However, if you take the cross-product of x and y (cross(x, y)), you get approximately [0.49 -0.15 -0.85], which is the negation of z, which implies that the matrix is encoding both a rotation and a reflection. Naturally, multiplying a mesh's vertices by a reflection matrix will reverse the winding order of its polygons, yielding inverted normals.
In the slides that you referenced, it states that the PCA method of generating a rotation matrix should only be considering four different combinations of axes in the 3D case, to ensure that the output matrix obeys the right-hand rule. If all combinations of axes were checked, that would allow PCA to consider both rotated and reflected spaces when searching for a best match. If that were the case, and if there some noise in the data such that the left half of the template is a slightly better match to the right half of the target and vice versa, then the PCA method might generate a reflection matrix like the one you observe. Perhaps you might want to reexamine the logic of how R is generated from the PCA results?
As alluded to in the comments, the direction of your vertex normals will depend on how you've ordered the triangular facets in your Faces matrix. This will follow a right-hand rule, where your fingers follow the vertex order around the triangle and your thumb indicates the surface normal direction. Here's a simple example to help illustrate:
Vertices = [0 0; 0 1; 1 1; 1 0]; % Points clockwise around a unit square in x-y plane
Faces = [1 2 3; 1 3 4]; % Two triangular facets, clockwise vertex ordering
TR = triangulation(Faces, Vertices);
VN = vertexNormal(TR)
VN =
0 0 -1
0 0 -1
0 0 -1
0 0 -1
In this example, Vertices contains the 4 vertices of a unit square in the x-y plane, ordered clockwise if you're looking down from positive z. Two triangular facets are defined in Faces, and the order of the indices in each row traces along the vertices in a clockwise fashion as well. This results in a surface normal for each face that points in the negative z direction. When the vertex normals are computed, they are pointing in the negative z direction as well.
What happens when we flip the order of one triangle so that its points are counter-clockwise?...
Faces = [1 2 3; 1 4 3]; % Second facet is 1 4 3 instead of 1 3 4
TR = triangulation(Faces, Vertices);
VN = vertexNormal(TR)
VN =
0 0 0
0 0 -1
0 0 0
0 0 1
The surface normal of the second triangle will now point in the positive z direction. The vertices that are only used by one triangle (rows 2 and 4) will have vertex normals that match the surface normals, while the vertices shared by each (rows 1 and 3) will have vertex normals of 0 (the two surface normals cancel).
How will this help you with your problem? Well, it's hard to say since I don't know exactly how you are defining Faces and Vertices. However, if you know for certain that every vertex normal in your mesh is pointing in the wrong direction, you can easily flip them all by swapping two columns in your Faces matrix before computing the normals:
Faces = [1 2 3; 1 3 4]; % Clockwise-ordered vertices
TR = triangulation(Faces(:, [1 3 2]), Vertices); % Change to counter-clockwise
VN = vertexNormal(TR)
VN =
0 0 1 % Normals are now pointing in positive z
0 0 1
0 0 1
0 0 1
Having this coordinate system:
And this dominant vertical vanishing point:
I would like to rotate the image around x axis so the vanishing point is at infinity. That means that all vertical lines are parallel.
I am using matlab. I find the line segmentes using LSD and the vanishing point using homogeneous coordinates. I would like to use angle-axis representation, then convert it to a rotation matrix and pass this to imwarp and get the rotated image. Also would be good to know how to rotate the segments. The segments are as (x1,y1,x2,y2).
Image above example:
Vanishin point in homogenous coordinates:
(x,y,z) = 1.0e+05 * [0.4992 -2.2012 0.0026]
Vanishin point in cartesian coordinates (what you see in the image):
(x,y) = [190.1335 -838.3577]
Question: With this vanishing point how do I compute the rotation matrix in the world x axis as explained above?
If all you're doing is rotating the image so that the vector from the origin to the vanishing point, is instead pointing directly vertical, here's an example.
I = imread('cameraman.tif');
figure;imagesc(I);set(gcf,'colormap',gray);
vp=-[190.1335 -838.3577,0]; %3d version,just for cross-product use,-ve ?
y=[0,1,0]; %The vertical axis on the plot
u = cross(vp,y); %you know it's going to be the z-axis
theta = -acos(dot(vp/norm(vp),y)); %-ve ?
rotMat = vrrotvec2mat([u, theta]);
J=imwarp(I,affine2d (rotMat));
figure;imagesc(J);set(gcf,'colormap',gray); %tilted image
You can play with the negatives, and plotting, since I'm not sure about those parts applying to your situation. The negatives may come from plotting upside down, or from rotation of the world vs. camera coordinate system, but I don't have time to think about it right now.
EDIT
If you want to rotation about the X-axis, this might work (adapted from https://www.mathworks.com/matlabcentral/answers/113074-how-to-rotate-an-image-along-y-axis), or check out: Rotate image over X, Y and Z axis in Matlab
[rows, columns, numberOfColorChannels] = size(I);
newRows = rows * cos(theta);
rotatedImage = imresize(I, [newRows, columns]);
I want to simulate depth in a 2D space, If I have a point P1 I suppose that I need to project that given point P1 into a plane x axis rotated "theta" rads clockwise, to get P1'
It seems that P1'.x coord has to be the same as the P1.x and the P1'.y has to b shorter than P1.y. In a 3D world:
cosa = cos(theta)
sina = sin(theta)
P1'.x = P1.x
P1'.y = P1.y * cosa - P1.z * sina
P1'.z = P1.y * sina + P1.z * cosa
Is my P1.z = 0? I tried it and P1'.y = P1.y * cosa doesn't result as expected
Any response would be appreciated, Thanks!
EDIT: What I want, now I rotate camera and translate matrix
EDIT 2: an example of a single line with a start1 point and a end1 point (it's an horizontal line, result expected is a falling line to the "floor" as long as tilt angle increases)
I think it's a sign error or an offset needed (java canvas drawing (0,0) is at top-left), because my new line with a tilt of 0 is the one below of all and with a value of 90ยบ the new line and the original one match
The calculation you are performing is correct if you would like to perform a rotation around the x axis clockwise. If you think of your line as a sheet of paper, a rotation of 0 degrees is you looking directly at the line.
For the example you have given the line is horizontal to the x axis. This will not change on rotation around the x axis (the line and the axis around which it is rotating are parallel to one another). As you rotate between 0 and 90 degrees the y co-ordinates of the line will decrease with P1.y*cos(theta) down to 0 at 90 degrees (think about the piece of paper we have been rotating around it's bottom edge, the x axis, at 90 degrees the paper is flat, and the y axis is perpendicular to the page, thus both edges of the page have the same y co-ordinate, both the side that is the "x-axis" and the opposite parallel side will have y=0).
Thus as you can see for your example this has worked correctly.
EDIT: The reason that multiplying by 90 degrees does not give an exactly zero answer is simply floating point rounding
I want to plot a latitude and longitude using matlab. Using that latitude and longitude as center of the circle, I want to plot a circle of radius 5 Nm.
r = 5/60;
nseg = 100;
x = 25.01;
y = 55.01;
theta = 0 : (2 * pi / nseg) : (2 * pi);
pline_x = r * cos(theta) + x;
pline_y = r * sin(theta) + y;
hold all
geoshow(pline_x, pline_y)
geoshow(x, y)
The circle does not look of what I expected.
Drawing a circle on earth is more complex that it looks like.
Drawing a line or a poly line is simple, because the vertices are defined.
Not so on circle.
a circle is defined by all points having the same distance from center (in meters! not in degrees!!!)
Unfortuantley lat and lon coordinates have not the same scale.
(The distance between two degrees of latidtude is always approx. 111.3 km, while for longitude this is only true at the equator. At the poles the distance between two longitudes approach zero. In Europe the factor is about 0.6. (cos(48deg))
There are two solution, the first is more universal, usefull for nearly all problems.
convert spherical coordinate (of circle center) to cartesian plane with unit = 1m, using a transformation (e.g equidistant transformation, also called equirectangular transf., this transformation works with the cos(centerLat) compensation factor)
calculate points (e.g circle points) in x,y plane using school mathematics.
transform all (x,y) points back to spherical (lat, lon) coordinates, using the inverse transformation of point 1.
Other solution
1. write a function which draws an ellipse in defined rectangle (all cartesian x,y)
2. define bounding of the circle to draw:
2a: calculate north-south diameter of circle/ in degrees: this a bit tricky: the distance is define in meters, you need a transformation to get the latitudeSpan: one degrees of lat is approx 111.3 km (eart circumence / 360.0): With this meters_per_degree value calc the N-S disatcne in degrees.
2b: calculate E-W span in degrees: now more tricky: calculate like 2a, but now divide by cos(centerLatitude) to compensate that E-W distances need more degrees when moving north to have the same meters.
Now draw ellipseInRectangle using N-S and E_W span for heigh and width.
But a circle on a sphere looks on the projected monitor display (or paper) only like a circle in the center of the projection. This shows:
Tissot's Error Ellipse
Imagine a dome with its centre in the +z direction. What I want to do is to move that dome's centre to a different axis (e.g. 20 degrees x axis, 20 degrees y axis, 20 degrees z axis). How can I do that ? Any hint/tip helps.
Add more info:
I've been dabbling with rotation matrices in wiki for a while. The problem is, it is not a commutative operation. RxRyRz is not same as RzRyRx. So based on the way I multiple it I get a different final results. For example, I want my final projection to have 20 degrees from the original X axis, 20 degrees from original Y axis and 20 degrees from original Z axis. Based on the matrix, giving alpha, beta, gamma values 20 (or its corresponding radian) does NOT result the intended rotation. Am I missing something? Is there a matrix that I can just put the intended angles and get it at the end ?
Using a rotation matrix is an easy way to rotate a collection of (x,y,z) points. You can calculate a rotation matrix for your case using the equations in the general rotation section. Note that figuring out the angle values to plug into those equations can be tricky. Think of it as rotating about one axis at a time and remember that the order of your rotations (order of multiplications) does matter.
An alternative to the general rotation equations is to calculate a rotation matrix from axis and angle. It may be easier for you to define correct parameters with this method.
Update: After perusing Wikipedia, I found a simple way to calculate rotation axis and angle between two vectors. Just fill in your starting and ending vectors for a and b here:
a = [0.0 0.0 1.0];
b = [0.5 0.5 0.0];
vectorMag = #(x) sqrt(sum(x.^2));
rotAngle = acos(dot(a,b) / (vectorMag(a) * vectorMag(b)))
rotAxis = cross(a,b)
rotAxis =
-0.5 0.5 0
rotAngle =
1.5708