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I deployed number of random points with their speed and displacement, how can I make these point move frequently each second,
since the Euclidean distance between these points is updated each time interval (second) based on the coordinates of new positions?
NumNode=2;
ro=1000;
center=[0 0];
Initial_Direction = rand(NumNode, 1) * 2 *pi; v = 15/3.6; % [m/s] velocity of node
theta_Node=2*pi*(rand(NumNode,1));
% let the Nodes deployed away from the center of circle network layout
g = 0.5 * ro + 0.5 * ro * rand(NumNode,1);
PosNode_x=center(1)+g.*cos(theta_Node); % Initial positions
PosNode_y=center(2)+g.*sin(theta_Node);
PosNode = [PosNode_x ,PosNode_y];
DX2 = [cos(Initial_Direction(:)) .* v,sin(Initial_Direction(:)) .* v]; % displacement of Node
PosNodeNew = PosNode + DX2; % New position of node without rotating
rotaion2 = [cosd(Initial_Direction), -sind(Initial_Direction); sind(Initial_Direction), cosd(Initial_Direction)]; % Rotation matrix to make the movement of node similar to move on arc
% Index approach to multiply different matrix dimensions
index=1:numel(PosNode);
xyRotated1 = rotaion2;
xyRotated1(index)= rotaion2(index).*PosNode(index); % rotated matrix multiply by position of node
index2 = 1:numel(DX2);
Newposss=xyRotated1;
NewPosss(index2) = xyRotated1(index2) + DX2(index2);
gg= NewPosss(index2); % New position of node with rotating on arc
figure(1)
scatter(PosDrone_x,PosDrone_x,'r+')
figure(2)
scatter(PosDroneNew(:,1),PosDroneNew(:,2),'b*')
figure (3)
scatter (xyRotated1(:,1),xyRotated1(:,2),'r.')
axis equal
figure (4)
scatter(NewPosss(1,:),NewPosss(1,:),'b*')
axis equal
I am working on basic distance calculation using accelerometer by dragging object on a single axis for physics class in MATLAB and have problem with plotting data.
My steps are:
1) After calibrating device to read zero g on every axis, I eliminate drift errors:
X_real = X_sample - X_calibrated;
if(X_real <= X_drift )
{
X_real = 0;
}
Where X_drift is 2 mg (From datasheet of accelerometer)
2) Calculate velocity:
velocity = 0; % On start
% Integration
v(i) = v(i-1) - x(i-1)-(x(i)+x(i-1)+x(i-2)+x(i-3))/4;
%Check if we stopped
if(x(i-1)==0 && x(i)==0)
v(i)=0;
end
%Check if velocity is under 0 (Not allowed)
if(v(i) < 0)
v(i)=0;
end
velocity = velocity + v(i);
3) Calculate distance:
distance = 0; % On start
%Integration
s(i) = s(i-1) + v(i-1) + (v(i)-v(i-1)-v(i-2)-v(i-3))/4;
distance = distance + s(i);
After testing this by dragging accelerometer on table 20 cm i got these results:
velocity = 0.09 m/s
distance = 0.21 m
time = 3.2s
Error of 1 cm is OK for classroom.
Chart tells something different:
I tried to plot distance after this:
s(i) = s(i)+s(i-1);
And got 21 cm on chart but after 6 s not after 4 s where it should be.
What am I doing wrong?
*UPDATE: Position y value is in mm not cm! Sorry
Im sorry for asking for help, i thought my formulas were ok, but they didn't. After step by step calculations my final solution is:
1) Velocity:
v(i) = v(i-1) - x(i-1)-(x(i)+x(i-1))/2;
2) Distance:
s(i) = s(i-1) + v(i-1)+(v(i)+v(i-1))/2;
And chart is:
Sorry once more time. I hope this will help someone calculating velocity and distance. It surely helped me as lesson to better study my code next time before asking for help.
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.
BOUNTY STATUS UPDATE:
I discovered how to map a linear lens, from destination coordinates to source coordinates.
How do you calculate the radial distance from the centre to go from fisheye to rectilinear?
1). I actually struggle to reverse it, and to map source coordinates to destination coordinates. What is the inverse, in code in the style of the converting functions I posted?
2). I also see that my undistortion is imperfect on some lenses - presumably those that are not strictly linear. What is the equivalent to-and-from source-and-destination coordinates for those lenses? Again, more code than just mathematical formulae please...
Question as originally stated:
I have some points that describe positions in a picture taken with a fisheye lens.
I want to convert these points to rectilinear coordinates. I want to undistort the image.
I've found this description of how to generate a fisheye effect, but not how to reverse it.
There's also a blog post that describes how to use tools to do it; these pictures are from that:
(1) : SOURCE Original photo link
Input : Original image with fish-eye distortion to fix.
(2) : DESTINATION Original photo link
Output : Corrected image (technically also with perspective correction, but that's a separate step).
How do you calculate the radial distance from the centre to go from fisheye to rectilinear?
My function stub looks like this:
Point correct_fisheye(const Point& p,const Size& img) {
// to polar
const Point centre = {img.width/2,img.height/2};
const Point rel = {p.x-centre.x,p.y-centre.y};
const double theta = atan2(rel.y,rel.x);
double R = sqrt((rel.x*rel.x)+(rel.y*rel.y));
// fisheye undistortion in here please
//... change R ...
// back to rectangular
const Point ret = Point(centre.x+R*cos(theta),centre.y+R*sin(theta));
fprintf(stderr,"(%d,%d) in (%d,%d) = %f,%f = (%d,%d)\n",p.x,p.y,img.width,img.height,theta,R,ret.x,ret.y);
return ret;
}
Alternatively, I could somehow convert the image from fisheye to rectilinear before finding the points, but I'm completely befuddled by the OpenCV documentation. Is there a straightforward way to do it in OpenCV, and does it perform well enough to do it to a live video feed?
The description you mention states that the projection by a pin-hole camera (one that does not introduce lens distortion) is modeled by
R_u = f*tan(theta)
and the projection by common fisheye lens cameras (that is, distorted) is modeled by
R_d = 2*f*sin(theta/2)
You already know R_d and theta and if you knew the camera's focal length (represented by f) then correcting the image would amount to computing R_u in terms of R_d and theta. In other words,
R_u = f*tan(2*asin(R_d/(2*f)))
is the formula you're looking for. Estimating the focal length f can be solved by calibrating the camera or other means such as letting the user provide feedback on how well the image is corrected or using knowledge from the original scene.
In order to solve the same problem using OpenCV, you would have to obtain the camera's intrinsic parameters and lens distortion coefficients. See, for example, Chapter 11 of Learning OpenCV (don't forget to check the correction). Then you can use a program such as this one (written with the Python bindings for OpenCV) in order to reverse lens distortion:
#!/usr/bin/python
# ./undistort 0_0000.jpg 1367.451167 1367.451167 0 0 -0.246065 0.193617 -0.002004 -0.002056
import sys
import cv
def main(argv):
if len(argv) < 10:
print 'Usage: %s input-file fx fy cx cy k1 k2 p1 p2 output-file' % argv[0]
sys.exit(-1)
src = argv[1]
fx, fy, cx, cy, k1, k2, p1, p2, output = argv[2:]
intrinsics = cv.CreateMat(3, 3, cv.CV_64FC1)
cv.Zero(intrinsics)
intrinsics[0, 0] = float(fx)
intrinsics[1, 1] = float(fy)
intrinsics[2, 2] = 1.0
intrinsics[0, 2] = float(cx)
intrinsics[1, 2] = float(cy)
dist_coeffs = cv.CreateMat(1, 4, cv.CV_64FC1)
cv.Zero(dist_coeffs)
dist_coeffs[0, 0] = float(k1)
dist_coeffs[0, 1] = float(k2)
dist_coeffs[0, 2] = float(p1)
dist_coeffs[0, 3] = float(p2)
src = cv.LoadImage(src)
dst = cv.CreateImage(cv.GetSize(src), src.depth, src.nChannels)
mapx = cv.CreateImage(cv.GetSize(src), cv.IPL_DEPTH_32F, 1)
mapy = cv.CreateImage(cv.GetSize(src), cv.IPL_DEPTH_32F, 1)
cv.InitUndistortMap(intrinsics, dist_coeffs, mapx, mapy)
cv.Remap(src, dst, mapx, mapy, cv.CV_INTER_LINEAR + cv.CV_WARP_FILL_OUTLIERS, cv.ScalarAll(0))
# cv.Undistort2(src, dst, intrinsics, dist_coeffs)
cv.SaveImage(output, dst)
if __name__ == '__main__':
main(sys.argv)
Also note that OpenCV uses a very different lens distortion model to the one in the web page you linked to.
(Original poster, providing an alternative)
The following function maps destination (rectilinear) coordinates to source (fisheye-distorted) coordinates. (I'd appreciate help in reversing it)
I got to this point through trial-and-error: I don't fundamentally grasp why this code is working, explanations and improved accuracy appreciated!
def dist(x,y):
return sqrt(x*x+y*y)
def correct_fisheye(src_size,dest_size,dx,dy,factor):
""" returns a tuple of source coordinates (sx,sy)
(note: values can be out of range)"""
# convert dx,dy to relative coordinates
rx, ry = dx-(dest_size[0]/2), dy-(dest_size[1]/2)
# calc theta
r = dist(rx,ry)/(dist(src_size[0],src_size[1])/factor)
if 0==r:
theta = 1.0
else:
theta = atan(r)/r
# back to absolute coordinates
sx, sy = (src_size[0]/2)+theta*rx, (src_size[1]/2)+theta*ry
# done
return (int(round(sx)),int(round(sy)))
When used with a factor of 3.0, it successfully undistorts the images used as examples (I made no attempt at quality interpolation):
Dead link
(And this is from the blog post, for comparison:)
If you think your formulas are exact, you can comput an exact formula with trig, like so:
Rin = 2 f sin(w/2) -> sin(w/2)= Rin/2f
Rout= f tan(w) -> tan(w)= Rout/f
(Rin/2f)^2 = [sin(w/2)]^2 = (1 - cos(w))/2 -> cos(w) = 1 - 2(Rin/2f)^2
(Rout/f)^2 = [tan(w)]^2 = 1/[cos(w)]^2 - 1
-> (Rout/f)^2 = 1/(1-2[Rin/2f]^2)^2 - 1
However, as #jmbr says, the actual camera distortion will depend on the lens and the zoom. Rather than rely on a fixed formula, you might want to try a polynomial expansion:
Rout = Rin*(1 + A*Rin^2 + B*Rin^4 + ...)
By tweaking first A, then higher-order coefficients, you can compute any reasonable local function (the form of the expansion takes advantage of the symmetry of the problem). In particular, it should be possible to compute initial coefficients to approximate the theoretical function above.
Also, for good results, you will need to use an interpolation filter to generate your corrected image. As long as the distortion is not too great, you can use the kind of filter you would use to rescale the image linearly without much problem.
Edit: as per your request, the equivalent scaling factor for the above formula:
(Rout/f)^2 = 1/(1-2[Rin/2f]^2)^2 - 1
-> Rout/f = [Rin/f] * sqrt(1-[Rin/f]^2/4)/(1-[Rin/f]^2/2)
If you plot the above formula alongside tan(Rin/f), you can see that they are very similar in shape. Basically, distortion from the tangent becomes severe before sin(w) becomes much different from w.
The inverse formula should be something like:
Rin/f = [Rout/f] / sqrt( sqrt(([Rout/f]^2+1) * (sqrt([Rout/f]^2+1) + 1) / 2 )
I blindly implemented the formulas from here, so I cannot guarantee it would do what you need.
Use auto_zoom to get the value for the zoom parameter.
def dist(x,y):
return sqrt(x*x+y*y)
def fisheye_to_rectilinear(src_size,dest_size,sx,sy,crop_factor,zoom):
""" returns a tuple of dest coordinates (dx,dy)
(note: values can be out of range)
crop_factor is ratio of sphere diameter to diagonal of the source image"""
# convert sx,sy to relative coordinates
rx, ry = sx-(src_size[0]/2), sy-(src_size[1]/2)
r = dist(rx,ry)
# focal distance = radius of the sphere
pi = 3.1415926535
f = dist(src_size[0],src_size[1])*factor/pi
# calc theta 1) linear mapping (older Nikon)
theta = r / f
# calc theta 2) nonlinear mapping
# theta = asin ( r / ( 2 * f ) ) * 2
# calc new radius
nr = tan(theta) * zoom
# back to absolute coordinates
dx, dy = (dest_size[0]/2)+rx/r*nr, (dest_size[1]/2)+ry/r*nr
# done
return (int(round(dx)),int(round(dy)))
def fisheye_auto_zoom(src_size,dest_size,crop_factor):
""" calculate zoom such that left edge of source image matches left edge of dest image """
# Try to see what happens with zoom=1
dx, dy = fisheye_to_rectilinear(src_size, dest_size, 0, src_size[1]/2, crop_factor, 1)
# Calculate zoom so the result is what we wanted
obtained_r = dest_size[0]/2 - dx
required_r = dest_size[0]/2
zoom = required_r / obtained_r
return zoom
I took what JMBR did and basically reversed it. He took the radius of the distorted image (Rd, that is, the distance in pixels from the center of the image) and found a formula for Ru, the radius of the undistorted image.
You want to go the other way. For each pixel in the undistorted (processed image), you want to know what the corresponding pixel is in the distorted image.
In other words, given (xu, yu) --> (xd, yd). You then replace each pixel in the undistorted image with its corresponding pixel from the distorted image.
Starting where JMBR did, I do the reverse, finding Rd as a function of Ru. I get:
Rd = f * sqrt(2) * sqrt( 1 - 1/sqrt(r^2 +1))
where f is the focal length in pixels (I'll explain later), and r = Ru/f.
The focal length for my camera was 2.5 mm. The size of each pixel on my CCD was 6 um square. f was therefore 2500/6 = 417 pixels. This can be found by trial and error.
Finding Rd allows you to find the corresponding pixel in the distorted image using polar coordinates.
The angle of each pixel from the center point is the same:
theta = arctan( (yu-yc)/(xu-xc) ) where xc, yc are the center points.
Then,
xd = Rd * cos(theta) + xc
yd = Rd * sin(theta) + yc
Make sure you know which quadrant you are in.
Here is the C# code I used
public class Analyzer
{
private ArrayList mFisheyeCorrect;
private int mFELimit = 1500;
private double mScaleFESize = 0.9;
public Analyzer()
{
//A lookup table so we don't have to calculate Rdistorted over and over
//The values will be multiplied by focal length in pixels to
//get the Rdistorted
mFisheyeCorrect = new ArrayList(mFELimit);
//i corresponds to Rundist/focalLengthInPixels * 1000 (to get integers)
for (int i = 0; i < mFELimit; i++)
{
double result = Math.Sqrt(1 - 1 / Math.Sqrt(1.0 + (double)i * i / 1000000.0)) * 1.4142136;
mFisheyeCorrect.Add(result);
}
}
public Bitmap RemoveFisheye(ref Bitmap aImage, double aFocalLinPixels)
{
Bitmap correctedImage = new Bitmap(aImage.Width, aImage.Height);
//The center points of the image
double xc = aImage.Width / 2.0;
double yc = aImage.Height / 2.0;
Boolean xpos, ypos;
//Move through the pixels in the corrected image;
//set to corresponding pixels in distorted image
for (int i = 0; i < correctedImage.Width; i++)
{
for (int j = 0; j < correctedImage.Height; j++)
{
//which quadrant are we in?
xpos = i > xc;
ypos = j > yc;
//Find the distance from the center
double xdif = i-xc;
double ydif = j-yc;
//The distance squared
double Rusquare = xdif * xdif + ydif * ydif;
//the angle from the center
double theta = Math.Atan2(ydif, xdif);
//find index for lookup table
int index = (int)(Math.Sqrt(Rusquare) / aFocalLinPixels * 1000);
if (index >= mFELimit) index = mFELimit - 1;
//calculated Rdistorted
double Rd = aFocalLinPixels * (double)mFisheyeCorrect[index]
/mScaleFESize;
//calculate x and y distances
double xdelta = Math.Abs(Rd*Math.Cos(theta));
double ydelta = Math.Abs(Rd * Math.Sin(theta));
//convert to pixel coordinates
int xd = (int)(xc + (xpos ? xdelta : -xdelta));
int yd = (int)(yc + (ypos ? ydelta : -ydelta));
xd = Math.Max(0, Math.Min(xd, aImage.Width-1));
yd = Math.Max(0, Math.Min(yd, aImage.Height-1));
//set the corrected pixel value from the distorted image
correctedImage.SetPixel(i, j, aImage.GetPixel(xd, yd));
}
}
return correctedImage;
}
}
I found this pdf file and I have proved that the maths are correct (except for the line vd = *xd**fv+v0 which should say vd = **yd**+fv+v0).
http://perception.inrialpes.fr/CAVA_Dataset/Site/files/Calibration_OpenCV.pdf
It does not use all of the latest co-efficients that OpenCV has available but I am sure that it could be adapted fairly easily.
double k1 = cameraIntrinsic.distortion[0];
double k2 = cameraIntrinsic.distortion[1];
double p1 = cameraIntrinsic.distortion[2];
double p2 = cameraIntrinsic.distortion[3];
double k3 = cameraIntrinsic.distortion[4];
double fu = cameraIntrinsic.focalLength[0];
double fv = cameraIntrinsic.focalLength[1];
double u0 = cameraIntrinsic.principalPoint[0];
double v0 = cameraIntrinsic.principalPoint[1];
double u, v;
u = thisPoint->x; // the undistorted point
v = thisPoint->y;
double x = ( u - u0 )/fu;
double y = ( v - v0 )/fv;
double r2 = (x*x) + (y*y);
double r4 = r2*r2;
double cDist = 1 + (k1*r2) + (k2*r4);
double xr = x*cDist;
double yr = y*cDist;
double a1 = 2*x*y;
double a2 = r2 + (2*(x*x));
double a3 = r2 + (2*(y*y));
double dx = (a1*p1) + (a2*p2);
double dy = (a3*p1) + (a1*p2);
double xd = xr + dx;
double yd = yr + dy;
double ud = (xd*fu) + u0;
double vd = (yd*fv) + v0;
thisPoint->x = ud; // the distorted point
thisPoint->y = vd;
This can be solved as an optimization problem. Simply draw on curves in images that are supposed to be straight lines. Store the contour points for each of those curves. Now we can solve the fish eye matrix as a minimization problem. Minimize the curve in points and that will give us a fisheye matrix. It works.
It can be done manually by adjusting the fish eye matrix using trackbars! Here is a fish eye GUI code using OpenCV for manual calibration.
Using this as a [reference][1]: Find a tangent point on circle?
cx = 0;
cy = 0;
px = -3;
py = -8;
dx = cx - px;
dy = cy - py;
a = asin(5 / ((dx*dx + dy*dy)^0.5));
b = atan2(dy, dx);
t_1 = deg2rad(180) + b - a;
t_2 = deg2rad(180) + b + a;
For a point (7,6) the angles are 7.9572/73.4434 and for (-3, -8) are 213.6264/285.2615. So for the first quadrant, the angles do not make sense, but for third quadrant they do. What I am doing wrong?
Your formula for a is wrong. You should use
a = acos(5 / ((dx*dx + dy*dy)^0.5))
instead of
a = asin(5 / ((dx*dx + dy*dy)^0.5))
i.e. use acos(...) instead of asin(...). The reason is shown in the image below. The formula for angle a is a=acos(r/H), where r is the radius of the circle and H is the length of the hypotenuse of the right angle triangle. So this has nothing to do with the fact that asin(...) has no way to know which of the two possible quadrants the value that is passed in lies. the argument of the asin is always positive, and you always want the answer in the range 0 to 90 degrees.
So the answer for the two angles that you want are b+a and b-a. Using acos instead of asin in your two cases produces 97.7592 & -16.5566 (or equivalently 343.4434) for your first quadrant example, and -164.7385 & -56.3736 (or equivalently 195.2615 and 303.6264) for your third quadrant example. (NB: instead of adding 180 degrees in the formula for t_1 and t-2, you could just switch the signs of dx and dy)
First -- I spent like 10 minutes figuring out what the heck you're trying to do (which in the end, I got from a comment in one of the answers), while solving your problem took 2 minutes. So, for future reference, please give a description of your problem as clear as you can first.
Now, I think you just have your signs messed up. Try the following:
%// difference vector
%// NOTE: these go the other way around for the atan2 to come out right
dx = px - cx;
dy = py - cy;
%// tip angle of the right triangle
a = asin( 5 / sqrt(dx*dx + dy*dy) );
%// angle between the (local) X-axis and the line of interest
b = atan2(dy, dx);
%// the third angle in the right triangle
%// NOTE: minus a here instead of plus b
g = pi/2 - a;
%// Angles of interest
%// NOTE1: signs are flipped; this automatically takes care of overshoots
%// NOTE2: don't forget to mod 360
t_1 = mod( rad2deg(b - g), 360)
t_2 = mod( rad2deg(b + g), 360)
Alternatively, you could skip computing the intermediate angle a by using acos instead of asin:
%// difference vector
dx = px - cx;
dy = py - cy;
%// Directly compute the third angle of the right triangle
%// (that is, the angle "at the origin")
g = acos( 5 / sqrt(dx*dx + dy*dy) );
%// angle between the (local) X-axis and the line of interest
b = atan2(dy, dx);
%// Angles of interest
t_1 = mod( rad2deg(b - g), 360)
t_2 = mod( rad2deg(b + g), 360)
Just another wayto re-discover the trigonometric identity acos(x) = pi/2 - asin(x) :)
This MathWorld entry is what you want: http://mathworld.wolfram.com/CircleTangentLine.html.
Alright, it looks like you are not accounting for the fact that asin, atan, ( any a-trig function ) has no way to know which of the two possible quadrants the value you passed in lies. To make up for that, a-trig function will assume that your point is in the first or fourth quadrant ( northeast / southeast ). Therefore, if you call atan function and your original point was in the second or third quadrant, you need to add 180 degrees / pi radians onto whatever value it returns.
See the documentation here stating that asin returns a value from [-pi/2, pi/2] :
http://www.mathworks.com/help/matlab/ref/asin.html
Hope that helps :)
EDIT
I misunderstood the situation originally
Here is what I think you have calculated :
t_1 and t_2 represent the angles you would travel at if you started on the circle from the tangent point and wanted to travel to your original starting point.
Viewed with this perspective your angles are correct.
For the point (7,6)
If you started on the circle at approx. (0,5) and traveled at 7 degrees, you would hit the point.
If you started on the circle at approx. (5,0) and traveled at 70 degrees, you would hit the point.
Now, what is going to be more useful and less confusing than angles, will be to know the slope of the line. To get this from the angle, do the following with angle in degrees:
angle = (angle + 90 + 360) % 180 - 90 // this gives us the angle as it would be in quad 1 or 4
slope = tan( deg2rad( angle ) )