I am trying to perform human body animation using translation and orientation data I am given. I have a set of rigid body segments made using patch all centered at (0,0,0) to represent the human body and translated accordingly. I have set up a hierarchy for each of them and performed a transformation matrix for each rigid body segment. The limb segments begin to offset one another and give problems. For example, the rigid body of the arm moves as if it does not have a relative point of origin even though it follows the proper motion. The motion is akin to moving the rigid body from the patch center of gravity? Whereas it is supposed to move with one end being fixed while the other end follows translation data. Can someone let me know what it is that I am doing wrong? Layout of my code is:
% Body segment lengths
xlength = somevalue
ylength = somevalue
zlength = somevalue
% Translation data
Xdata
Ydata
Zdata
% Orientation data
Yaw = rotation about z axis
Pitch = rotation about x axis
Roll = rotation about y axis
Vertices = [xlength*ones(8,1),ylength*ones(8,1),zlength*ones(8,1)]...
.*[-0.5,-0.5,-0.5;
0.5,-0.5,-0.5;
-0.5,0.5,-0.5;
-0.5,-0.5,0.5;
0.5,0.5,-0.5;
-0.5,0.5,0.5;
0.5,-0.5,0.5;
0.5,0.5,0.5];
% Create patches
for i = 1:6
% create faces for patches
end
% create axes
ax = axes(...)
% draw patches
bodysegmentPatch = patch(patchxdata,patchydata,patchzdata)
% create hierarchy using hgtransform
pelvis = hgtransform('Parent',ax);
trunk = hgtransform('Parent',pelvis);
head = hgtransform('Parent',trunk);
leftupperarm = hgtransform('Parent',trunk);
leftforearm = hgtransform('Parent',leftupperarm);
rightupperarm = hgtransform('Parent',trunk);
rightforearm = hgtransform('Parent',rightupperarm);
leftthigh = hgtransform('Parent',pelvis);
leftcalf = hgtransform('Parent',leftthigh);
rightthigh = hgtransform('Parent',pelvis);
rightcalf = hgtransform('Parent',rightthigh);
% set patches to hierarchy
set(pelvisPatch,'Parent',pelvis)
% Animation loop
for i = 1:n
% translation of body segment
bodysegmentT = makehgtform('translate',[x(i) y(i) z(i)]);
% rotation of body segment
bodysegmentR = makehgtform('yrotate',Roll(i),'xrotate',Pitch(i),'zrotate',Yaw(i));
% Create transform matrices
set(pelvis,'Matrix',pelvisR);
set(trunk,'Matrix',trunkR*pelvisR);
set(leftupperarm,'Matrix',leftupperarmT*leftupperarmR*trunkR*pelvisR);
drawnow
end
I'm not sure what your problem is exactly, but without looking at your code too closely my guess is it is likely one of two common mistakes while doing these skeleton transformations.
Your matrix transformations are in the wrong order. Remember that A = A*B is not the same thing as A = B*A. When you do this kind of transformation stack this order is very important.
The objects are rotating around the wrong point. Usually they rotate around the origin. So if you want something to rotate around the center of the object you will have to translate the image to the origin, rotate the image, translate the image back to the original location.
Don't give up! These transformations can be tricky, and often times it may look totally chaotic while in reality the code is really close to being correct.
Related
I want to write a code where an arc stops rotating as soon as it comes in contact with a semi-circle.
I have written a code to do so, my arc does not just touch the circle but it slightly intersects it.
I have put rotation of arc inside a while loop by using linspace() to change theta. I used polyxpoly() for finding intersection. Condition for the while loop is that as long as I have empty array the loop continues, but as soon as I get a value from polyxpoly() my loop stops.
However, at the place of touch the theta value exceeds what I needed, so as a result I get an intersection.
How do I modify the code so that the arc will touch the semi-circle and not intersect it?
Here is the output. Click the link below
Image of intersection but I need touch and not intersection
clc,clear
R = 5; % radius of a circle
r = 10; % radius of arc
aa = 60*pi/180; % arc angle
ap = 0*pi/180; % arc position angle
% defining the semi-circle about the origin
t = linspace(0,pi);
[x,y] = pol2cart(t,R); % circle data
% Shifting circle centre to (3.5,0)
x=x+3.5;
y=y+0;
% defining the arc about the origin
t1 = linspace(0,aa)-aa/2+ap;
[x1,y1] = pol2cart(t1,r); % arc data
% shifting arc-lower-end to (14,0)
delx=14-x1(1); % Finding the x difference between arc-lower-end x-coordinate & 14
dely=0-y1(1); % Finding the y difference between arc-lower-end y-coordinate & 0
x1=x1+delx;
y1=y1+dely;
theta =linspace(0,pi,1000);
i=1;
xc=[];
yc=[];
while isempty(xc)&& isempty(yc)
% create a matrix of these points, which will be useful in future calculations
v = [x1;y1];
% choose a point which will be the center of rotation
x_center = 14;
y_center = 0;
% create a matrix which will be used later in calculations
center = repmat([x_center; y_center], 1, length(x1));
% define a 60 degree counter-clockwise rotation matrix
R = [cos(theta(i)) -sin(theta(i)); sin(theta(i)) cos(theta(i))];
% do the rotation...
s = v - center; % shift points in the plane so that the center of rotation is at the origin
so = R*s; % apply the rotation about the origin
vo = so + center; % shift again so the origin goes back to the desired center of rotation
% this can be done in one line as:
% vo = R*(v - center) + center
% pick out the vectors of rotated x- and y-data
x_rotated = vo(1,:);
y_rotated = vo(2,:);
[xc,yc] = polyxpoly(x_rotated,y_rotated,x,y)
[xc1,yc1] = polyxpoly(x1,y1,x,y)
i=i+1;
end
% make a plot
plot(x,y)
hold on
plot(x1, y1, 'k-', x_rotated, y_rotated, 'r-', x_center, y_center, 'bo');
axis equal
I need to find way it contacts a circle and does not intersect it.
The code is in matlab.
Any suggestions are welcome.
This is the problem of collision detection. Most, if not all, methods I know of in collision detection requires the computer to check for intersections of some sort. It's very difficult (if not impossible) to have two objects "just touch" in simulations, because you'll need the precise (analytically solved) location of the boundaries of those objects.
polyxpoly() is a function to return the intersection of two polygons. So unfortunately, if you insist that the arc cannot touch, then you cannot use polyxpoly(). In that extreme case, you'll have to solve some mathematical equation for when the tip of the arc coincides with a point on the circle perfectly, then simulate up until that point in time.
But realistically, what you need is a finer simulation (although I personally think what you have alone is good enough). So in every simulation step, you calculate a smaller movement, so that when the arc eventually intersects, only a very small amount of the arc intersects.
I have this 3D image generated from the simple code below.
% Input Image size
imageSizeY = 200;
imageSizeX = 120;
imageSizeZ = 100;
%# create coordinates
[rowsInImage, columnsInImage, pagesInImage] = meshgrid(1:imageSizeY, 1:imageSizeX, 1:imageSizeZ);
%# get coordinate array of vertices
vertexCoords = [rowsInImage(:), columnsInImage(:), pagesInImage(:)];
centerY = imageSizeY/2;
centerX = imageSizeX/2;
centerZ = imageSizeZ/2;
radius = 28;
%# calculate distance from center of the cube
sphereVoxels = (rowsInImage - centerY).^2 + (columnsInImage - centerX).^2 + (pagesInImage - centerZ).^2 <= radius.^2;
%# Now, display it using an isosurface and a patch
fv = isosurface(sphereVoxels,0);
patch(fv,'FaceColor',[0 0 .7],'EdgeColor',[0 0 1]); title('Binary volume of a sphere');
view(45,45);
axis equal;
grid on;
xlabel('x-axis [pixels]'); ylabel('y-axis [pixels]'); zlabel('z-axis [pixels]')
I have tried plotting the image with isosurface and some other volume visualization tools, but there remains quite a few surprises for me from the plots.
The code has been written to conform to the image coordinate system (eg. see: vertexCoords) which is a left-handed coordinate system I presume. Nonetheless, the image is displayed in the Cartesian (right-handed) coordinate system. I have tried to see this displayed as the figure below, but that’s simply not happening.
I am wondering if the visualization functions have been written to display the image the way they do.
Image coordinate system:
Going forward, there are other aspects of the code I am to write for example if I have an input image sphereVoxels as in above, in addition to visualizing it, I would want to find north, south east, west, top and bottom locations in the image, as well as number and count the coordinates of the vertices, plus more.
I foresee this would likely become confusing for me if I don’t stick to one coordinate system, and considering that the visualization tools predominantly use the right-hand coordinate system, I would want to stick with that from the onset. However, I really do not know how to go about this.
Right-hand coordinate system:
Any suggestions to get through this?
When you call meshgrid, the dimensions x and y axes are switched (contrary to ndgrid). For example, in your case, it means that rowsInImage is a [120x100x200] = [x,y,z] array and not a [100x120x200] = [y,x,z] array even if meshgrid was called with arguments in the y,x,z order. I would change those two lines to be in the classical x,y,z order :
[columnsInImage, rowsInImage, pagesInImage] = meshgrid(1:imageSizeX, 1:imageSizeY, 1:imageSizeZ);
vertexCoords = [columnsInImage(:), rowsInImage(:), pagesInImage(:)];
Finding the angle between the view vector and the surface normal can be beneficial in getting the visible surfaces since we use it to conduct back face culling techniques and obtaining contours, crease edges of the object.
To obtain the visible surfaces I use the back face culling code below:
N = normals(vertex,faces);
BC = barycenter(vertex,faces);
back_facing = sum(N.*bsxfun(#minus,BC,campos),2)<=0
t.FaceVertexCData = 1*(sum(N.*bsxfun(#minus,BC,campos),2)<=0)
t.FaceVertexCData(sum(N.*bsxfun(#minus,BC,campos),2)>0) = nan;
faces1=faces(t.FaceVertexCData(:)==1,:);
facesv=sort(unique(faces1(:)));
How does one obtain the angle?
r=(sum(N.*bsxfun(#minus,BC,campos),2))
rr=bsxfun(#minus,BC,campos);
V_mag= sqrt(rr(:,1).^2+rr(:,2).^2+rr(:,3).^2);
N_mag= sqrt(N(:,1).^2+N(:,2).^2+N(:,3).^2);
for i = 1:(size(r,1))
A(i)=acosd(r(i)/(N_mag(i).*V_mag(i)));
end
This is what I have done thus far. I am not sure if it is correct and code is slow.
I have an inverted pendulum video here which is 33 second length. The objective is to plot a red point in the center of moving rectangle part of the pendulum and to plot a line along the black stick calculating its angle for every frame.
I have handled the video frame by frame. Then I have used Object Detection In A Cluttered Scene Using Point Feature Matching. It would be good if I had access to the matching point's indexes and then I would easily calculate the angle.
I have thought that I can get the moving rectangle part's region and seek the similar regions in the next frames. But this solution seems too local.
I do not know which techniques to apply.
clear all;
clc;
hVideoFileReader = vision.VideoFileReader;
hVideoPlayer = vision.VideoPlayer;
hVideoFileReader.Filename = 'inverted-pendulum.avi';
hVideoFileReader.VideoOutputDataType = 'single';
while ~isDone(hVideoFileReader)
grayFrame = rgb2gray(step(hVideoFileReader));
frame = step(hVideoFileReader);
if isFirstFrame
part = grayFrame(202:266,202:282); % #moving part's region
isFirstFrame = false;
subplot(1,2,1);
imshow(part);
end
partPoints = detectSURFFeatures(part);
grayFramePoints = detectSURFFeatures(grayFrame);
hold on;
subplot(1,2,1), plot(partPoints .selectStrongest(10));
subplot(1,2,2), imshow(grayFrame);
subplot(1,2,2), plot(grayFramePoints .selectStrongest(20));
frame2 = pointPendulumCenter(frame);
frame3 = plotLineAlongStick(frame2);
step(hVideoPlayer, frame3);
hold off;
end
release(hVideoFileReader);
release(hVideoPlayer);
%% #Function to find the moving part's center point and plot a red dot on it.
function f = pointPendulumCenter(frame)
end
%% #Function to plot a red line along the stick after calculating the angle of it.
function f = plotLineAlongStick(frame)
end
It would make the problem much easier if your camera did not move. If you take your video with a stationary camera (e.g. mounted on a tripod) then you can use vision.ForegroundDetector to segment out moving objects from the static background.
Below is an arbitrary hand-drawn Intensity profile of a line in an image:
The task is to draw the line. The profile can be approximated to an arc of a circle or ellipse.
This I am doing for camera calibration. Since I do not have the actual industrial camera, I am trying to simulate the correction needed for calibration.
The question can be rephrased as I want pixel values which will follow a plot similar to the above. I want to do this using program (Preferably using opencv) and not manually enter these values because I have thousands of pixels in the line.
An algorithm/pseudo code will suffice. Also please note that I do not have any actual Intensity profile, otherwise I would have read those values.
When will you encounter such situation ?
Suppose you take a picture (assuming complete white) from a Camera, your object being placed on table, and camera just above it in vertical direction. The light coming on the center of the picture vertically downward from the camera will be stronger in intensity as compared to the light reflecting at the edges. You measure pixel values across any line in the Image, you will find intensity curve like shown above. Since I dont have camera for the time being, I want to emulate this situation. How to achieve this?
This is not exactly image processing, rather image generation... but anyways.
Since you want an arc, we still need three points on that arc, lets take the first, middle and last point (key characteristics in my opinion):
N = 100; % number of pixels
x1 = 1;
x2 = floor(N/2);
x3 = N;
y1 = 242;
y2 = 255;
y3 = 242;
and now draw a circle arc that contains these points.
This problem is already discussed here for matlab: http://www.mathworks.nl/matlabcentral/newsreader/view_thread/297070
x21 = x2-x1; y21 = y2-y1;
x31 = x3-x1; y31 = y3-y1;
h21 = x21^2+y21^2; h31 = x31^2+y31^2;
d = 2*(x21*y31-x31*y21);
a = x1+(h21*y31-h31*y21)/d; % circle center x
b = y1-(h21*x31-h31*x21)/d; % circle center y
r = sqrt(h21*h31*((x3-x2)^2+(y3-y2)^2))/abs(d); % circle radius
If you assume the middle value is always larger (and thus it's the upper part of the circle you'll have to plot), you can draw this with:
x = x1:x3;
y = b+sqrt(r^2-(x-a).^ 2);
plot(x,y);
you can adjust the visible window with
xlim([1 N]);
ylim([200 260]);
which gives me the following result: