Related
Calibration:
I have calibrated the camera using this vision toolbox in Matlab. I used checkerboard images to do so. After calibration I get the cameraParams
which contains:
Camera Extrinsics
RotationMatrices: [3x3x18 double]
TranslationVectors: [18x3 double]
and
Camera Intrinsics
IntrinsicMatrix: [3x3 double]
FocalLength: [1.0446e+03 1.0428e+03]
PrincipalPoint: [604.1474 359.7477]
Skew: 3.5436
Aim:
I have recorded trajectories of some objects in motion using this camera. Each object corresponds to a single point in a frame. Now, I want to project the points such that I get a top-view.
Note all these points I wish to transform are are the on the same plane.
ex: [xcor_i,ycor_i ]
-101.7000 -77.4040
-102.4200 -77.4040
KEYPOINT: This plane is perpendicular to one of images of checkerboard used for calibration. For that image(below), I know the height of origin of the checkerboard of from ground(193.040 cm). And the plane to project the points on is parallel to the ground and perpendicular to this image.
Code
(Ref:https://stackoverflow.com/a/27260492/3646408 and answer by #Dima below):
function generate_homographic_matrix()
%% Calibrate camera
% Define images to process
path=['.' filesep 'Images' filesep];
list_imgs=dir([path '*.jpg']);
list_imgs_path=strcat(path,{list_imgs.name});
% Detect checkerboards in images
[imagePoints, boardSize, imagesUsed] = detectCheckerboardPoints(list_imgs_path);
imageFileNames = list_imgs_path(imagesUsed);
% Generate world coordinates of the corners of the squares
squareSize = 27; % in units of 'mm'
worldPoints = generateCheckerboardPoints(boardSize, squareSize);
% Calibrate the camera
[cameraParams, imagesUsed, estimationErrors] = estimateCameraParameters(imagePoints, worldPoints, ...
'EstimateSkew', true, 'EstimateTangentialDistortion', true, ...
'NumRadialDistortionCoefficients', 3, 'WorldUnits', 'mm');
%% Compute homography for peripendicular plane to checkerboard
% Detect the checkerboard
im=imread(['.' filesep 'Images' filesep 'exp_19.jpg']); %exp_19.jpg is the checkerboard orthogonal to the floor
[imagePoints, boardSize] = detectCheckerboardPoints(im);
% Compute rotation and translation of the camera.
[Rc, Tc] = extrinsics(imagePoints, worldPoints, cameraParams);
% Rc(rotation of the calibration view w.r.t the camera) = [x y z])
%then the floor has rotation Rf = [z x -y].(Normal vector of the floor goes up.)
Rf=[Rc(:,3),Rc(:,1),Rc(:,2)*-1];
% Translate it to the floor
H=452;%distance btw origin and floor
Fc = Rc * [0; H; 0];
Tc = Tc + Fc';
% Combine rotation and translation into one matrix:
Rf(3, :) = Tc;
% Compute the homography between the checkerboard and the image plane:
H = Rf * cameraParams.IntrinsicMatrix;
save('homographic_matrix.mat','H')
end
%% Transform points
function [x_transf,y_transf] =transform_points(xcor_i,ycor_i)
% creates a projective2D object and then transforms the points forward to
% get a top-view
% xcor_i and ycor_i are 1d vectors comprising of the x-coordinates and
% y-coordinates of trajectories.
data=load('homographic_matrix.mat');
homo_matrix=data.H;
tform=projective2d(inv(homo_matrix));
[x_transf,y_transf] = transformPointsForward(tform,xcor_i,ycor_i);
end
Quoting text from OReilly Learning OpenCV Pg 412:
"Once we have the homography matrix and the height parameter set as we wish, we could
then remove the chessboard and drive the cart around, making a bird’s-eye view video
of the path..."
This what I essentially wish to achieve.
Abhishek,
I don't entirely understand what you are trying to do. Are your points on a plane, and are you trying to create a bird's eye view of that plane?
If so, then you need to know the extrinsics, R and t, describing the relationship between that plane and the camera. One way to get R and t is to place a checkerboard on the plane, and then use the extrinsics function.
After that, you can follow the directions in the question you cited to get the homography. Once you have the homography, you can create a projective2D object, and use its transformPointsForward method to transform your points.
Since you have the size of squares on the grid, then given 2 points that you know are connected by an edge of size E (in real world units), you can calculate their 3D position.
Taking the camera intrinsic matrix K and the 3D position C and the camera orientation matrix R, you can calculate a ray to each of the points p by doing:
D = R^T * K^-1 * p
Each 3D point is defined as:
P = C + t*D
and you have the constraint that ||P1-P2|| = E
then it's a matter of solving for t1,t2 and finding the 3D position of the two points.
In order to create a top view, you can take the 3D points and project them using a camera model for that top view to generate a new image.
If all your points are on a single plane, it's enough to calculate the position of 3 points, and you can extrapolate the rest.
If your points are located on a plane that you know one coordinate of, you can do it simply for each point. For example, if you know that your camera is located at height h=C.z, and you want to find the 3D location of points in the frame, given that they are on the floor (z=0), then all you have to do is calculate the direction D as above, and then:
t=abs( (h-0)/D.z )
The 0 represent the height of the plane. Substitute for any other value for other planes.
Now that you have the value of t, you can calculate the 3D position of each point: P=C+t*D.
Then, to create a top view, create a new camera position and rotation to match your required projection, and you can project each point onto this camera's image plane.
If you want a full image, you can interpolate positions and fill in the blanks where no feature point was present.
For more details, you can always read: http://www.robots.ox.ac.uk/~vgg/hzbook/index.html
I'm trying to estimate a position based on signal strength received from 4 Wi-Fi Access Points. I measure the signal strength from 4 access points located in each corner of a square room with 100 square meters (10x10). I recorded the signal strengths in a known position (x, y) = (9.5, 1.5) using an Android phone. Now I want to check how accurate can a multilateration method be under the circumstances.
Using MATLAB, I applied a formula to calculate distance using the signal strength. The following MATLAB function shows the application of the formula:
function [ d_vect ] = distance( RSS )
% Calculate distance from signal strength
result = (27.55 - (20 * log10(2400)) + abs(RSS)) / 20;
d_vect = power(10, result);
end
The input RSS is a vector with the four signal strengths measured in the test point (x,y) = (9.5, 1.5). The RSS vector looks like this:
RSS =
-57.6000
-60.4000
-44.7000
-54.4000
and the resultant vector with all the estimated distances to each access points looks like this:
d_vect =
7.5386
10.4061
1.7072
5.2154
Now I want to estimate my position based on these distances and the access points position in order to find the error between the estimated position and the known position (9.5, 1.5). I want to find the intersection area (In order to estimate a position) between four circles where each access point is the center of one of the circles and the distance is the radius of the circle.
I want to find the grey area as shown in this image :
http://www.biologycorner.com/resources/venn4.gif
If you want an alternative way of estimating the location without estimating the intersection of circles you can use trilateration. It is a common technique in navigation (e.g. GPS) to estimate a position given a set of distance measurements.
Also, if you wanted the area because you also need an estimate of the uncertainty of the position I would recommend solving the trilateration problem using least squares which will easily give you an estimate of the parameters involved and an error propagation to yield an uncertainty of the location.
I found an answear that solved perfectly the question. It is explained in detail in this link:
https://gis.stackexchange.com/questions/40660/trilateration-algorithm-for-n-amount-of-points
I also developed some MATLAB code for the problem. Here it goes:
Estimate distances from the Access Points:
function [ d_vect ] = distance( RSS )
result = (27.55 - (20 * log10(2400)) + abs(RSS)) / 20;
d_vect = power(10, result);
end
The trilateration function:
function [] = trilat( X, d, real1, real2 )
cla
circles(X(1), X(5), d(1), 'edgecolor', [0 0 0],'facecolor', 'none','linewidth',4); %AP1 - black
circles(X(2), X(6), d(2), 'edgecolor', [0 1 0],'facecolor', 'none','linewidth',4); %AP2 - green
circles(X(3), X(7), d(3), 'edgecolor', [0 1 1],'facecolor', 'none','linewidth',4); %AP3 - cyan
circles(X(4), X(8), d(4), 'edgecolor', [1 1 0],'facecolor', 'none','linewidth',4); %AP4 - yellow
axis([0 10 0 10])
hold on
tbl = table(X, d);
d = d.^2;
weights = d.^(-1);
weights = transpose(weights);
beta0 = [5, 5];
modelfun = #(b,X)(abs(b(1)-X(:,1)).^2+abs(b(2)-X(:,2)).^2).^(1/2);
mdl = fitnlm(tbl,modelfun,beta0, 'Weights', weights);
b = mdl.Coefficients{1:2,{'Estimate'}}
scatter(b(1), b(2), 70, [0 0 1], 'filled')
scatter(real1, real2, 70, [1 0 0], 'filled')
hold off
end
Where,
X: matrix with APs coordinates
d: distance estimation vector
real1: real position x
real2: real position y
If you have three sets of measurements with (x,y) coordinates of location and corresponding signal strength. such as:
m1 = (x1,y1,s1)
m2 = (x2,y2,s2)
m3 = (x3,y3,s3)
Then you can calculate distances between each of the point locations:
d12 = Sqrt((x1 - x2)^2 + (y1 - y2)^2)
d13 = Sqrt((x1 - x3)^2 + (y1 - y3)^2)
d23 = Sqrt((x2 - x3)^2 + (y2 - y3)^2)
Now consider that each signal strength measurement signifies an emitter for that signal, that comes from a location somewhere at a distance. That distance would be a radius from the location where the signal strength was measured, because one would not know at this point the direction from where the signal came from. Also, the weaker the signal... the larger the radius. In other words, the signal strength measurement would be inversely proportional to the radius. The smaller the signal strength the larger the radius, and vice versa. So, calculate the proportional, although not yet accurate, radius's of our three points:
r1 = 1/s1
r2 = 1/s2
r3 = 1/s3
So now, at each point pair, set apart by their distance we can calculate a constant (C) where the radius's from each location will just touch one another. For example, for the point pair 1 & 2:
Ca * r1 + Ca * r2 = d12
... solving for the constant Ca:
Ca = d12 / (r1 + r2)
... and we can do this for the other two pairs, as well.
Cb = d13 / (r1 + r3)
Cc = d23 / (r2 + r3)
All right... select the largest C constant, either Ca, Cb, or Cc. Then, use the parametric equation for a circle to find where the coordinates meet. I will explain.
The parametric equation for a circle is:
x = radius * Cos(theta)
y = radius * Sin(theta)
If Ca was the largest constant found, then you would compare points 1 & 2, such as:
Ca * r1 * Cos(theta1) == Ca * r2 * Cos(theta2) &&
Ca * r1 * Sin(theta1) == Ca * r2 * Sin(theta2)
... iterating theta1 and theta2 from 0 to 360 degrees, for both circles. You might write code like:
for theta1 in 0 ..< 360 {
for theta2 in 0 ..< 360 {
if( abs(Ca*r1*cos(theta1) - Ca*r2*cos(theta2)) < 0.01 && abs(Ca*r1*sin(theta1) - Ca*r2*sin(theta2)) < 0.01 ) {
print("point is: (", Ca*r1*cos(theta1), Ca*r1*sin(theta1),")")
}
}
}
Depending on what your tolerance was for a match, you wouldn't have to do too many iterations around the circumferences of each signal radius to determine an estimate for the location of the signal source.
So basically you need to intersect 4 circles. There can be many approaches to it, and there are two that will generate the exact intersection area.
First approach is to start with one circle, intersect it with the second circle, then intersect the resulting area with the third circle and so on. that is, on each step you know current intersection area, and you intersect it with a new circle. The intersection area will always be a region bounded by circle arcs, so to intersect it with a new circle you walk along the boundary of the area and check whether each bounding arc intersects with a new circle. If it does, then you leave only the part of the arc that lies inside a new circle, remember that you should continue with an arc from a new circle, and continue traversing the boundary until you find the next intersection.
Another approach that seems to result in a worse time complexity, but in your case of 4 circles this will not be important, is to find all the intersection points of two circles and choose only those points that are of interest for you, that is which lie inside all other circles. These points will be the corners of your area, and then it is rather easy to reconstruct the area. After googling a bit, I have even found a live demo of this approach.
I am in need of an idea! I want to model the vascular network on the eye in 3D. I have made statistics on the branching behaviour in relation to vessel diameter, length etc. What I am stuck at right now is the visualization:
The eye is approximated as a sphere E with center in origo C = [0, 0, 0] and a radius r.
What I want to achieve is that based on the following input parameters, it should be able to draw a segment on the surface/perimeter of E:
Input:
Cartesian position of previous segment ending: P_0 = [x_0, y_0, z_0]
Segment length: L
Segment diameter: d
Desired angle relative to the previous segment: a (1)
Output:
Cartesian position of resulting segment ending: P_1 = [x_1, y_1, z_1]
What I do now, is the following:
From P_0, generate a sphere with radius L, representing all the points we could possibly draw to with the correct length. This set is called pool.
Limit pool to only include points with a distance to C between r*0.95 and r, so only the points around the perimeter of the eye are included.
Select only the point that would generate a relative angle (2) closest to the desired angle a.
The problem is, that whatever angle a I desire, is actually not what is measured by the dot product. Say I want an angle at 0 (i.e. that the new segment is following the same direction as the previous`, what I actually get is an angle around 30 degrees because of the curvature of the sphere. I guess what I want is more the 2D angle when looking from an angle orthogonal from the sphere to the branching point. Please take a look at the screenshots below for a visualization.
Any ideas?
(1) The reason for this is, that the child node with the greatest diameter is usually follows the path of the previous segment, whereas smaller child nodes tend to angle differently.
(2) Calculated by acos(dot(v1/norm(v1), v2/norm(v2)))
Screenshots explaining the problem:
Yellow line: previous segment
Red line: "new" segment to one of the points (not neccesarily the correct one)
Blue x'es: Pool (text=angle in radians)
I will restate the problem with my own notation:
Given two points P and Q on the surface of a sphere centered at C with radius r, find a new point T such that the angle of the turn from PQ to QT is A and the length of QT is L.
Because the segments are small in relation to the sphere, we will use a locally-planar approximation of the sphere at the pivot point Q. (If this isn't an okay assumption, you need to be more explicit in your question.)
You can then compute T as follows.
// First compute an aligned orthonormal basis {U,V,W}.
// - {U,V} should be a basis for the plane tangent at Q.
// - W should be normal to the plane tangent at Q.
// - U should be in the direction PQ in the plane tangent at Q
W = normalize(Q - C)
U = normalize(Q - P)
U = normalize(U - W * dotprod(W, U))
V = normalize(crossprod(W, U))
// Next compute the next point S in the plane tangent at Q.
// In a regular plane, the parametric equation of a unit circle
// centered at the origin is:
// f(A) = (cos A, sin A) = (1,0) cos A + (0,1) sin A
// We just do the same thing, but with the {U,V} basis instead
// of the standard basis {(1,0),(0,1)}.
S = Q + L * (U cos A + V sin A)
// Finally project S onto the sphere, obtaining the segment QT.
T = C + r * normalize(S - C)
I need to slide a window over a 3d volume. The sliding is only on one layer of the 3d volume, i.e for each x,y with one specific z.
This is what I want to do in a loop:
for each x,y,z, for example:
px =9; py =9; pz =12;
a = rand(50,50,50);
[x y z] = meshgrid(1:50,1:50,1:50);
r = 3;
%-------------loop starts here:
% creating a shaped window, for example sphere of radius r
inds = find((x-px).^2 + (y-py).^2 + (z-pz).^2 <= r.^2);
% getting the relevant indices, here, it is the sphere around px,py,pz
[i,j,k] = ind2sub(size(a),inds);
% adjust the center of the sphere to be at (0,0,0) instead of (px,py,pz)
adj_inds = bsxfun(#minus,[i,j,k],[px,py,pz]);
% Computing for each sphere some kind of median point
cx = sum(a(inds).*adj_inds(:,1))./sum(a(inds));
cy = sum(a(inds).*adj_inds(:,2))./sum(a(inds));
cz = sum(a(inds).*adj_inds(:,3))./sum(a(inds));
%Saving the result: the distance between the new point and the center of the sphere.
res(yc,xc) = sqrt(sum([cx,cy,cz].^2));
%-------------
Now, all of this should happen many many time, ( ~300000), loop takes ages, convolution returns 3d volume (for each x,y,z) while I need to perform this only for each (x,y) and a list of z's.
Help please...
Thanks
matlabit
I'm trying to create a dataset of raw volumetric data consisting of geometrical shapes. The point is to use volume ray casting to project them in 2D but first I want to create the volume manually.
The geometry is consisting of one cylinder that is in the middle of the volume, along the Z axis and 2 smaller cylinders that are around the first one, deriving from rotations around the axes.
Here is my function so far:
function cyl= createCylinders(a, b, c, rad1, h1, rad2, h2)
% a : data width
% b : data height
% c : data depth
% rad1: radius of the big center cylinder
% rad2: radius of the smaller cylinders
% h1: height of the big center cylinder
% h2: height of the smaller cylinders
[Y X Z] =meshgrid(1:a,1:b,1:c); %matlab saves in a different order so X must be Y
centerX = a/2;
centerY = b/2;
centerZ = c/2;
theta = 0; %around y
fi = pi/4; %around x
% First cylinder
cyl = zeros(a,b,c);
% create for infinite height
R = sqrt((X-centerX).^2 + (Y-centerY).^2);
startZ = ceil(c/2) - floor(h1/2);
endZ = startZ + h1 - 1;
% then trim it to height = h1
temp = zeros(a,b,h1);
temp( R(:,:,startZ:endZ)<rad1 ) = 255;
cyl(:,:,startZ:endZ) = temp;
% Second cylinder
cyl2 = zeros(a,b,c);
A = (X-centerX)*cos(theta) + (Y-centerY)*sin(theta)*sin(fi) + (Z-centerZ)*cos(fi)*sin(theta);
B = (Y-centerY)*cos(fi) - (Z-centerZ)*sin(fi);
% create again for infinite height
R2 = sqrt(A.^2+B.^2);
cyl2(R2<rad2) = 255;
%then use 2 planes to trim outside of the limits
N = [ cos(fi)*sin(theta) -sin(fi) cos(fi)*cos(theta) ];
P = (rad2).*N + [ centerX centerY centerZ];
T = (X-P(1))*N(1) + (Y-P(2))*N(2) + (Z-P(3))*N(3);
cyl2(T<0) = 0;
P = (rad2+h2).*N + [ centerX centerY centerZ];
T = (X-P(1))*N(1) + (Y-P(2))*N(2) + (Z-P(3))*N(3);
cyl2(T>0) = 0;
% Third cylinder
% ...
cyl = cyl + cyl2;
cyl = uint8(round(cyl));
% ...
The concept is that the first cylinder is created and then "cut" according to the z-axis value, to define its height. The other cylinder is created using the relation A2 + B 2 = R2 where A and B are rotated accordingly using the rotation matrices only around x and y axes, using Ry(θ)Rx(φ) as described here.
Until now everything seems to be working, because I have implemented code (tested that it works well) to display the projection and the cylinders seem to have correct rotation when they are not "trimmed" from infinite height.
I calculate N which is the vector [0 0 1] aka z-axis rotated in the same way as the cylinder. Then I find two points P of the same distances that I want the cylinder's edges to be and calculate the plane equations T according to that points and normal vector. Lastly, I trim according to that equality. Or at least that's what I think I'm doing, because after the trimming I usually don't get anything (every value is zero). Or, the best thing I could get when I was experimenting was cylinders trimmed, but the planes of the top and bottom where not oriented well.
I would appreciate any help or corrections at my code, because I've been looking at the geometry equations and I can't find where the mistake is.
Edit:
This is a quick screenshot of the object I'm trying to create. NOTE that the cylinders are opaque in the volume data, all the inside is considered as homogeneous material.
I think instead of:
T = (X-P(1))*N(1) + (Y-P(2))*N(2) + (Z-P(3))*N(3);
you should try the following at both places:
T = (X-P(1)) + (Y-P(2)) + (Z-P(3));
Multiplying by N is to account for the direction of the axis of the 2nd cylinder which you have already done just above that step.