I have solved a system of two second-order differential equations using an implementation of Euler's method in Julia. The below code shows how Euler's method has been called to solve the system in question.
θ1 = 1.1900518004210798; θ2 = 0.3807445263738167
f(t, y) = [y[2], -2(y[1] - θ1) - 4y[2] + 0.5sin(3pi*t),
y[4], -2(y[3] - θ2) - 4(y[4] + abs(y[2])) + 0.5sin(3pi*t)]
y0 = [pi/2, 0, pi/6, 0]; t0 = 0; tFinal = 50; h = 0.001
res = euler(f, y0, t0, tFinal, h)
The result, res, is a vector of four numbers
1.18798735437173
-0.0458294959470722
0.31530569612003573
-0.049213402534541074
The first entry is the angle that the bottom line segment forms with the x-axis while the third entry is the angle that the two line segments form with one another (see below figure).
To create this plot I called plot_robotarm([res[1], res[3]]) which is implemented according to the below code.
function plot_robotarm(thetav)
# Plots a robotarm with angles thetav
R = 1;
xv=zeros(length(thetav)+1)
yv=zeros(length(thetav)+1)
for i in 1:length(thetav)
xv[i+1]=xv[i]+R*cos(thetav[i])
yv[i+1]=yv[i]+R*sin(thetav[i])
end
# Plot with colors
opts = (:circle, 10, 1., :blue, stroke(7, 1., :red))
plt = plot(xv, yv,
marker = opts,
c = :red,
w = 5,
legend = false,
xlims = (0, 2.0),
ylims = (0, 2.0))
display(plt)
end
How can I create an animation that visualizes how consecutive iterations of Euler's method make the robot arm (i.e. the two line segments) move toward the final point at t = 50? (I do not need to plot every iteration, just enough so that it makes for an animation that captures the movement of the robot arm.)
You can use ffmpeg and Luxor.jl's animation features to make an animated GIF. The frame function needs to be modified to reflect graphical display of each step in your program. See the docs for Luxor for more.
using Luxor
using Colors
using BoundaryValueDiffEq
# constants for differential equations and movie
const g = 9.81
const L = 1.0 # pendulum length in meters
const bobd = 0.10 # pendulum bob diameter in meters
const framerate = 50.0 # intended frame rate/sec
const t0 = 0.0 # start time (s)
const tf = 2.3 # end simulation time (s)
const dtframe = 1.0/framerate # time increment per frame
const tspan = LinRange(t0, tf, Int(floor(tf*framerate))) # array of time points in animation
const bgcolor = "black" # gif background
const leaderhue = (0.80, 0.70, 0.20) # gif swing arm hue light gold
const hslcolors = [HSL(col) for col in (distinguishable_colors(
Int(floor(tf*framerate)+3),[RGB(1,1,1)])[2:end])]
const giffilename = "pendulum.gif" # output file
# differential equations copied from docs of DifferentialEquations.jl
simplependulum!(du, u, p, t) = (θ=u[1]; dθ=u[2]; du[1]=dθ; du[2]=-(g/L)*sin(θ))
bc1!(residual, u, p, t) = (residual[1] = u[div(end,2)][1] + pi/2; residual[2] = u[end][1] - pi/2)
bvp1 = TwoPointBVProblem(simplependulum!, bc1!, [pi/2,pi/2], (tspan[1],tspan[end]))
sol2 = solve(bvp1, GeneralMIRK4(), dt=dtframe)
# movie making background
backdrop(scene, framenumber) = background(bgcolor)
function frame(scene, framenumber)
u1, u2 = sol2.u[framenumber]
y, x = L*cos(u1), L*sin(u1)
sethue(leaderhue)
poly([Point(-4.0, 0.0), Point(4.0, 0.0),
Point(160.0x,160.0y)], :fill)
sethue(Colors.HSV(framenumber*4.0, 1, 1))
circle(Point(160.0x,160.0y), 160bobd, :fill)
text(string("frame $framenumber of $(scene.framerange.stop)"),
Point(0.0, -190.0),
halign=:center)
end
muv = Movie(400, 400, "Pendulum Demo", 1:length(tspan))
animate(muv, [Scene(muv, backdrop),
Scene(muv, frame, easingfunction=easeinoutcubic)],
creategif=true, pathname=giffilename)
Related
I have 2 greyscale images that i am trying to align using scalar scaling 1 , rotation matrix [2,2] and translation vector [2,1]. I can calculate image1's transformed coordinates as
y = s*R*x + t;
Below the resulting images are shown.
The first image is image1 before transformation,
the second image is image1 (red) with attempted interpolation using interp2 shown on top of image2 (green)
The third image is when i manually insert the pixel values from image1 into an empty array (that has the same size as image2) using the transformed coordinates.
From this we can see that the coordinate transformation must have been successful, as the images are aligned although not perfectly (which is to be expected since only 2 coordinates were used in calculating s, R and t) .
How come interp2 is not producing a result more similar to when i manually insert pixel values?
Below the code for doing this is included:
Interpolation code
function [transformed_image] = interpolate_image(im_r,im_t,s,R,t)
[m,n] = size(im_t);
% doesn't help if i use get_grid that the other function is using here
[~, grid_xr, grid_yr] = get_ipgrid(im_r);
[x_t, grid_xt, grid_yt] = get_ipgrid(im_t);
y = s*R*x_t + t;
yx = reshape(y(1,:), m,n);
yy = reshape(y(2,:), m,n);
transformed_image = interp2(grid_xr, grid_yr, im_r, yx, yy, 'nearest');
end
function [x, grid_x, grid_y] = get_ipgrid(image)
[m,n] = size(image);
[grid_x,grid_y] = meshgrid(1:n,1:m);
x = [reshape(grid_x, 1, []); reshape(grid_y, 1, [])]; % X is [2xM*N] coordinate pairs
end
The manual code
function [transformed_image] = transform_image(im_r,im_t,s,R,t)
[m,n] = size(im_t);
[x_t, grid_xt, grid_yt] = get_grid(im_t);
y = s*R*x_t + t;
ymat = reshape(y',m,n,2);
yx = ymat(:,:,1);
yy = ymat(:,:,2);
transformed_image = zeros(m,n);
for i = 1:m
for j = 1:n
% make sure coordinates are inside
if (yx(i,j) < m & yy(i,j) < n & yx(i,j) > 0.5 & yy(i,j) > 0.5)
transformed_image(round(yx(i,j)),round(yy(i,j))) = im_r(i,j);
end
end
end
end
function [x, grid_x, grid_y] = get_grid(image)
[m,n] = size(image);
[grid_y,grid_x] = meshgrid(1:n,1:m);
x = [grid_x(:) grid_y(:)]'; % X is [2xM*N] coordinate pairs
end
Can anyone see what i'm doing wrong with interp2? I feel like i have tried everything
Turns out i got interpolation all wrong.
In my question i calculate the coordinates of im1 in im2.
However the way interpolation works is that i need to calculate the coordinates of im2 in im1 such that i can map the image as shown below.
This means that i also calculated the wrong s,R and t since they were used to transform im1 -> im2, where as i needed im2 -> im1. (this is also called the inverse transform). Below is the manual code, that is basically the same as interp2 with nearest neighbour interpolation
function [transformed_image] = transform_image(im_r,im_t,s,R,t)
[m,n] = size(im_t);
[x_t, grid_xt, grid_yt] = get_grid(im_t);
y = s*R*x_t + t;
ymat = reshape(y',m,n,2);
yx = ymat(:,:,1);
yy = ymat(:,:,2);
transformed_image = zeros(m,n);
for i = 1:m
for j = 1:n
% make sure coordinates are inside
if (yx(i,j) < m & yy(i,j) < n & yx(i,j) > 0.5 & yy(i,j) > 0.5)
transformed_image(i,j) = im_r(round(yx(i,j)),round(yy(i,j)));
end
end
end
end
I have an image and I would like to blur it in one specific direction and distance using Matlab.
I found out there is a filter called fspecial('motion',len,theta).
Here there is an example:
I = imread('cameraman.tif');
imshow(I);
H = fspecial('motion',20,45);
MotionBlur = imfilter(I,H,'replicate');
imshow(MotionBlur);
However the blurred picture is blurred in 2 directions! In this case 225 and 45 degrees.
What should it do in order to blur it just in a specific direction (e.g. 45) and not both?
I think you want what's called a "comet" kernel. I'm not sure what kernel is used for the "motion" blur, but I'd guess that it's symmetrical based on the image you provided.
Here is some code to play with that applies the comet kernel in one direction. You'll have to change things around if you want an arbitrary angle. You can see from the output that it's smearing in one direction, since there is a black band on only one side (due to the lack of pixels there).
L = 5; % kernel width
sigma=0.2; % kernel smoothness
I = imread('cameraman.tif');
x = -L:1.0:L;
[X,Y] = meshgrid(x,x);
H1 = exp((-sigma.*X.^2)+(-sigma.*Y.^2));
kernel = H1/sum((H1(:)));
Hflag = double((X>0));
comet_kernel = Hflag.*H1;
comet_kernel=comet_kernel/sum(comet_kernel(:));
smearedImage = conv2(double(I),comet_kernel,'same');
imshow(smearedImage,[]);
Updated code: This will apply an arbitrary rotation to the comet kernel. Note also the difference between sigma in the previous example and sx and sy here, which control the length and width parameters of the kernel, as suggested by Andras in the comments.
L = 5; % kernel width
sx=3;
sy=10;
theta=0;
I = imread('cameraman.tif');
x = -L:1.0:L;
[X,Y] = meshgrid(x,x);
rX = X.*cos(theta)-Y.*sin(theta);
rY = X.*sin(theta)+Y.*cos(theta);
H1 = exp(-((rX./sx).^2)-((rY./sy).^2));
Hflag = double((0.*rX+rY)>0);
H1 = H1.*Hflag;
comet_kernel = H1/sum((H1(:)))
smearedImage = conv2(double(I),comet_kernel,'same');
imshow(smearedImage,[]);
Based on Anger Density's answer I wrote this code that solves my problem completely:
L = 10; % kernel width
sx=0.1;
sy=100;
THETA = ([0,45,90,135,180,225,270,320,360])*pi/180;
for i=1:length(THETA)
theta=(THETA(i)+pi)*-1;
I = imread('cameraman.tif');
x = -L:1.0:L;
[X,Y] = meshgrid(x,x);
rX = X.*cos(theta)-Y.*sin(theta);
rY = X.*sin(theta)+Y.*cos(theta);
H1 = exp(-((rX./sx).^2)-((rY./sy).^2));
Hflag = double((0.*rX+rY)>0);
H1 = H1.*Hflag;
comet_kernel = H1/sum((H1(:)));
smearedImage = conv2(double(I),comet_kernel,'same');
% Fix edges
smearedImage(:,[1:L, end-L:end]) = I(:,[1:L, end-L:end]); % Left/Right edge
smearedImage([1:L, end-L:end], :) = I([1:L, end-L:end], :); % Top/bottom edge
% Keep only inner blur
smearedImage(L:end-L,L:end-L) = min(smearedImage(L:end-L,L:end-L),double(I(L:end-L,L:end-L)));
figure
imshow(smearedImage,[]);
title(num2str(THETA(i)*180/pi))
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);
end
I am trying to implement FRST on python to detect centroids of elliptical objects (e.g. cells in microscopy images), but my implementation does not find seed points (more or less center points) of elliptical objects. This effort comes from duplicating FRST from Segmentation of Overlapping Elliptical Objects in Silhouette Images (https://ieeexplore.ieee.org/document/7300433). I don't know why I have these artifacts. An interesting thing is that I see these patterns (crosses) all in the same direction per object. Any point in the right direction to generate the same result as in the paper (just to find the seed points) will be most welcome.
Original Paper: A Fast Radial Symmetry Transform for Detecting Points of Interest by Loy and Zelinsky (ECCV 2002)
I have also tried the pre-existing python package for FRST: https://pypi.org/project/frst/. This somehow results in the same artifacts. Weird.
First image: Original Image
Second image: Sobel-operated Image
Third image: Magnitude Projection Image
Fourth image: Magnitude Projection Image with positively affected pixels only
Fifth image: FRST'd image: end-product with original image overlaid (shadowed)
Sixth image: FRST'd image by the pre-existing python package with original image overlaid (shadowed).
from scipy.ndimage import gaussian_filter
import numpy as np
from scipy.signal import convolve
# Get orientation projection image
def get_proj_img(image, radius):
workingDims = tuple((e + 2*radius) for e in image.shape)
h,w = image.shape
ori_img = np.zeros(workingDims) # Orientation Projection Image
mag_img = np.zeros(workingDims) # Magnitutde Projection Image
# Kenels for the sobel operator
a1 = np.matrix([1, 2, 1])
a2 = np.matrix([-1, 0, 1])
Kx = a1.T * a2
Ky = a2.T * a1
# Apply the Sobel operator
sobel_x = convolve(image, Kx)
sobel_y = convolve(image, Ky)
sobel_norms = np.hypot(sobel_x, sobel_y)
# Distances to afpx, afpy (affected pixels)
dist_afpx = np.multiply(np.divide(sobel_x, sobel_norms, out = np.zeros(sobel_x.shape), where = sobel_norms!=0), radius)
dist_afpx = np.round(dist_afpx).astype(int)
dist_afpy = np.multiply(np.divide(sobel_y, sobel_norms, out = np.zeros(sobel_y.shape), where = sobel_norms!=0), radius)
dist_afpy = np.round(dist_afpy).astype(int)
for cords, sobel_norm in np.ndenumerate(sobel_norms):
i, j = cords
pos_aff_pix = (i+dist_afpx[i,j], j+dist_afpy[i,j])
neg_aff_pix = (i-dist_afpx[i,j], j-dist_afpy[i,j])
ori_img[pos_aff_pix] += 1
ori_img[neg_aff_pix] -= 1
mag_img[pos_aff_pix] += sobel_norm
mag_img[neg_aff_pix] -= sobel_norm
ori_img = ori_img[:h, :w]
mag_img = mag_img[:h, :w]
print ("Did it go back to the original image size? ")
print (ori_img.shape == image.shape)
# try normalizing ori and mag img
return ori_img, mag_img
def get_sn(ori_img, mag_img, radius, kn, alpha):
ori_img_limited = np.minimum(ori_img, kn)
fn = np.multiply(np.divide(mag_img,kn), np.power((np.absolute(ori_img_limited)/kn), alpha))
# convolute fn with gaussian filter.
sn = gaussian_filter(fn, 0.25*radius)
return sn
def do_frst(image, radius, kn, alpha, ksize = 3):
ori_img, mag_img = get_proj_img(image, radius)
sn = get_sn(ori_img, mag_img, radius, kn, alpha)
return sn
Parameters:
radius = 50
kn = 10
alpha = 2
beta = 0
stdfactor = 0.25
I'm trying to implement Fabian Timm's eye-tracking algorithm [http://www.inb.uni-luebeck.de/publikationen/pdfs/TiBa11b.pdf] (found here: [http://thume.ca/projects/2012/11/04/simple-accurate-eye-center-tracking-in-opencv/]) in numpy and OpenCV and I've hit a snag. I think I've vectorized my implementation decently enough, but it's still not fast enough to run in real time and it doesn't detect pupils with as much accuracy as I had hoped. This is my first time using numpy, so I'm not sure what I've done wrong.
def find_pupil(eye):
eye_len = np.arange(eye.shape[0])
xx,yy = np.meshgrid(eye_len,eye_len) #coordinates
XX,YY = np.meshgrid(xx.ravel(),yy.ravel()) #all distance vectors
Dx,Dy = [YY-XX, YY-XX] #y2-y1, x2-x1 -- simpler this way because YY = XXT
Dlen = np.sqrt(Dx**2+Dy**2)
Dx,Dy = [Dx/Dlen, Dy/Dlen] #normalized
Gx,Gy = np.gradient(eye)
Gmagn = np.sqrt(Gx**2+Gy**2)
Gx,Gy = [Gx/Gmagn,Gy/Gmagn] #normalized
GX,GY = np.meshgrid(Gx.ravel(),Gy.ravel())
X = (GX*Dx+GY*Dy)**2
eye = cv2.bitwise_not(cv2.GaussianBlur(eye,(5,5),0.005*eye.shape[1])) #inverting and blurring eye for use as w
eyem = np.repeat(eye.ravel()[np.newaxis,:],eye.size,0)
C = (np.nansum(eyem*X, axis=0)/eye.size).reshape(eye.shape)
return np.unravel_index(C.argmax(), C.shape)
and the rest of the code:
def find_eyes(face):
left_x, left_y = [int(floor(0.5 * face.shape[0])), int(floor(0.2 * face.shape[1]))]
right_x, right_y = [int(floor(0.1 * face.shape[0])), int(floor(0.2 * face.shape[1]))]
area = int(floor(0.2 * face.shape[0]))
left_eye = (left_x, left_y, area, area)
right_eye = (right_x, right_y, area, area)
return [left_eye,right_eye]
faceCascade = cv2.CascadeClassifier("haarcascade_frontalface_default.xml")
video_capture = cv2.VideoCapture(0)
while True:
# Capture frame-by-frame
ret, frame = video_capture.read()
gray = cv2.cvtColor(frame, cv2.COLOR_BGR2GRAY)
faces = faceCascade.detectMultiScale(
gray,
scaleFactor=1.1,
minNeighbors=5,
minSize=(30, 30),
flags=cv2.CASCADE_SCALE_IMAGE
)
# Draw a rectangle around the faces
for (x, y, w, h) in faces:
cv2.rectangle(frame, (x, y), (x+w, y+h), (0, 255, 0), 2)
roi_gray = gray[y:y+h, x:x+w]
roi_color = frame[y:y+h, x:x+w]
eyes = find_eyes(roi_gray)
for (ex,ey,ew,eh) in eyes:
eye_gray = roi_gray[ey:ey+eh,ex:ex+ew]
eye_color = roi_color[ey:ey+eh,ex:ex+ew]
cv2.rectangle(roi_color,(ex,ey),(ex+ew,ey+eh),(255,0,0),2)
px,py = find_pupil(eye_gray)
cv2.rectangle(eye_color,(px,py),(px+1,py+1),(255,0,0),2)
# Display the resulting frame
cv2.imshow('Video', frame)
if cv2.waitKey(1) & 0xFF == ord('q'):
break
# When everything is done, release the capture
video_capture.release()
cv2.destroyAllWindows()
You can perform many of those operations that save replicated elements and then perform some mathematical opertaions by directly performing the mathematical operatrions after creating singleton dimensions that would allow NumPy broadcasting. Thus, there would be two benefits - On the fly operations to save workspace memory and performance boost. Also, at the end, we can replace the nansum calculation with a simplified version. Thus, with all of that philosophy in mind, here's one modified approach -
def find_pupil_v2(face, x, y, w, h):
eye = face[x:x+w,y:y+h]
eye_len = np.arange(eye.shape[0])
N = eye_len.size**2
eye_len_diff = eye_len[:,None] - eye_len
Dlen = np.sqrt(2*((eye_len_diff)**2))
Dxy0 = eye_len_diff/Dlen
Gx0,Gy0 = np.gradient(eye)
Gmagn = np.sqrt(Gx0**2+Gy0**2)
Gx,Gy = [Gx0/Gmagn,Gy0/Gmagn] #normalized
B0 = Gy[:,:,None]*Dxy0[:,None,:]
C0 = Gx[:,None,:]*Dxy0
X = ((C0.transpose(1,0,2)[:,None,:,:]+B0[:,:,None,:]).reshape(N,N))**2
eye1 = cv2.bitwise_not(cv2.GaussianBlur(eye,(5,5),0.005*eye.shape[1]))
C = (np.nansum(X,0)*eye1.ravel()/eye1.size).reshape(eye1.shape)
return np.unravel_index(C.argmax(), C.shape)
There's one repeat still left in it at Dxy. It might be possible to avoid that step and Dxy0 could be fed directly into the step that uses Dxy to give us X, but I haven't worked through it. Everything's converted to broadcasting based!
Runtime test and output verification -
In [539]: # Inputs with random elements
...: face = np.random.randint(0,10,(256,256)).astype('uint8')
...: x = 40
...: y = 60
...: w = 64
...: h = 64
...:
In [540]: find_pupil(face,x,y,w,h)
Out[540]: (32, 63)
In [541]: find_pupil_v2(face,x,y,w,h)
Out[541]: (32, 63)
In [542]: %timeit find_pupil(face,x,y,w,h)
1 loops, best of 3: 4.15 s per loop
In [543]: %timeit find_pupil_v2(face,x,y,w,h)
1 loops, best of 3: 529 ms per loop
It seems we are getting close to 8x speedup!
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.