The meaning of that formula is really quite simple. Imagine you take two same-sized small areas of an image, the blue one and the red one:
The window function equals 0 outside the red rectangle (for simplicity, we can assume the window is simply constant within the red rectangle). So the window function selects which pixels you want to look at and assigns ...
One important thing to understand is that after extracting the keypoints, you only obtain information about their position, and sometimes their coverage area (usually approximated by a circle or ellipse) in the image. While the information about keypoint position might sometimes be useful, it does not say much about the keypoints themselves.
Depending on ...
The source code of OpenCV is available, so I'd recommend just taking some time going through the code. The relevant file for this particular function is:
I've not looked into it in detail, but it looks like
int cvFindChessboardCorners( const void* arr, CvSize pattern_size,
As suggested above, the Matlab Canny edge detector calculates the gradient using a "derivative of a Gaussian filter" (as stated in the documentation). In other words, Matlab does a Gaussian blur of the image and then finds the gradient of that smoothed image... all using a single fancy filter. [If you want to know the details, just type in edit edge as ...
Are there any other objects that can move beside people? If there aren't, you can just find the blobs (connected components) in your foreground mask, and these are your people.
They can also "collide" one with another, creating one blob instead of two. In this case, you can do a motion tracking and resolve the ambiguity by using the fact that the ...
This is a non-linear filter, the operation being performed is not a convolution and cannot be represented by a filter kernel. It is sometimes called a mode filter, by analogy to the median filter, since the value taken by a pixel is the mode of the distribution of the neighboring values. OpenCV does median filtering (cvSmooth with CV_MEDIAN parameter), I don'...
If you are not worried about the speed or exact contour of hand, below is a simple solution.
The method is like this : You take each contour and find distance to other contours. If distance is less than 50, they are nearby and you put them together. If not, they are put as different.
So checking distance to each contour is a time consuming process. Takes a ...
I think what you are asking is about the feasibility of your pedestrian algorithm.
There are two general strategies for this kind of problems:
(bottom-to-top) Consider it as a pure detection problem, where in
each frame you detect only pedestrians. Once you detect them, a)
counting their number in a frame is fairly easy; and b) tracking any
of them in ...
A first attempt using Matlab:
im = imread('squares.jpg');
im2 = rgb2gray(im);
se = strel('disk', 15);
for i = 1:16;
t = 60+i*5; % try out a range of bw thresholds to see what works best
labelled = bwlabel(im2>t); % label regions in the BW image
closed = imclose(labelled, se); % close small regions
cleared = imclearborder(~closed,4); % ...
You have to take images for calibration from different points of view and angles, with as big difference between angles as possible (all three Euler angles should vary), but so that pattern diameter was still fitting to camera field of view. The more views are you using the better calibration will be. That is needed because during the calibration you detect ...
I can see a number of possible problems with this approach. I speak from my own experience here from improving a pedestrian counting system with a very similar approach, so I don't mean to be discouraging. On the contrary, I'd like to warn you of possible hurdles you may have to overcome in order to build an accurate and robust system.
Firstly, background ...
I had tried something else to improve my result in question. Below solution is on the assumption that first square(orange) is always detected in step 1. And it is practical due to its high contrast color compare to background. Even the result I showed in question has detected it correctly
Step 1 : Find as many squares possible
I split the image to R,G,B,H,...
I have been in the "in order to use well tested methods I would need an extensive database of examples which I don't have" position in a very small company that "couldn't afford it". I regret very much that I didn't simply do whatever was necessary to get as much such data as possible. I think it would have made a world of difference to them in the end.
You can simply specify a ROI for that region and convert it into HSV.
For eg (below is pseudo-code in Python-OpenCV)
# define ROI of RGB image 'img'
roi = img[r1:r2, c1:c2]
# convert it into HSV
hsv = cv2.cvtColor(roi,cv2.COLOR_BGR2HSV)
Now it gives you the hsv values of the region.
But one or two difficulties there:
Your object may comprise some part ...
In general this is not possible, since you do not have full spectral information for each pixel, but only coordinates in some low-dimensional color space (i.e. RGB).
If your image data is in RGB, then the best you can do is to simply take the green color channel (which will be most responsive to 550 nm) and throw away the red and blue.
Depending on the ...
Even if my answer comes too late for you, maybe other people find this useful. I have the codes for an openCV Pose from Homography. I found the method at this really useful website, euclideanspace.
void cameraPoseFromHomography(const Mat& H, Mat& pose)
pose = Mat::eye(3, 4, CV_64FC1); //3x4 matrix
float norm1 = (float)norm(H.col(0));
I am not giving you a complete algorithm here, but since I worked on a similar project, I can give you some hints and tips.
First of all, changing an image taken from one perspective to a different one relatively easy only for planar surfaces. E.g., you have an image of a tall building with a road, and since buildings are usually build vertical in the air ...
Use bilateral filter or anisotropic diffusion first.
The effect of anisotropic diffusion is as the following:
. The MATLAB code can be found here.
Here is its effect on your image:
Finally, non-local means is a also a good way to get rid of the noise. You might also want to take a look into that. I warn you though, it is slow.
In general, this is an image segmentation problem (http://en.wikipedia.org/wiki/Image_segmentation) into which you would be trying to isolate the focused to the non-focused regions of the image.
Optical lenses are equivalent to low pass filters anyway and the effect of a low pass filter on a signal is to smooth it out by limiting the higher frequency ...
You might look at trying Autocorrelation for this. Here is an SO answer describing how to perform autocorrelation with Matlab using FFTs. This could be extended for two dimensions.
I implemented your test case in numpy as follows:
a = np.zeros(300)
a[::30] = 1
This gives the following plot:
As you can see, the peaks pop up at +/- ...
One simple way for quantification of contract that I can think of is through use of image histogram. Following is my suggestion
Compute Histogram of the Image
From the counts compute entropy
If you just want to try it out you can use the matlab inbuilt function http://www.mathworks.ch/ch/help/images/ref/entropy.html
You can use the entropy value of the ...
Use the haartraining methods of opencv according to this tutorial http://note.sonots.com/SciSoftware/haartraining.html -- this should give the best results, but I haven't worked with haartraining myself so far...
I would suggest to use methods of "markerless tracking" of the individual tiles of the board. You can implement this ...
Here are a few ideas:
First convert from color to grayscale. It looks like you have fairly good contrast already. There are various methods to perform this conversion; choose the simplest at first: gray = (red + green + blue)/3. Quite often you don't need anything better than that. For some applications, using just the green color plane is sufficient.
Laplace of Gaussian
The Laplace of Gaussian (LoG) of image $f$ can be written as
\nabla^2 (f * g) = f * \nabla^2 g
with $g$ the Gaussian kernel and $*$ the convolution. That is, the Laplace of the image smoothed by a Gaussian kernel is identical to the image convolved with the Laplace of the Gaussian kernel. This convolution can be further expanded, ...
Parallel lines in the image do intersect at a vanishing point. Therefore simply hypothesizing lines (a gradient direction at a point suffices to describe it) and voting (see Hough voting) would suffice to identify this point. One could then record all the lines that casted votes to this very point and identify them. Care must be taken as it is difficult to ...
Just an idea with no guarantee of success:
isolate the red blobs (e.g. mark them as white, the rest of the image as black)
perform a distance transform for the white blobs (every pixel indicates the distance to the nearest black pixel)
perform a non-maxima-suppression (ideally only the centers of the circles remain)
deal with non ideal conditions (filter ...
You can keep track of the last 10-15 frames and count the number of white pixel of these frames. You can differentiate if it is a vehicle or not, since the graph will be high for 10-15 frames you will be able to say that it is a vehicle otherwise it is noise.
To count multiple cars in multiple side by side lanes you can check the pixels column wise, I mean ...
Note: This method is going to be really slow.
Generate a mask that looks like the contours of a ideal object. Similar to this:
then slide (position,scale,rotation) the mask over the image and match it with the contour of the real image (perhaps blurred a bit to get softer response) to calculate how similar they are, the (position,scale,rotation) with the ...