Activity 17 - Video Processing

In this activity, video processing was done. First, the video is ripped to create images representing frames in the video. The images obtained were then processed.
The video used in this activity is a rolling object on an inclined plane. The objective of this activity is to obtain the acceleration seen from the video and compare it to theoretical values given some parameters.

(I cannot upload the video because of large file size)..

The images obtained from the video were converted to binary images. The centroid of the rolling cylinder for all the images were obtained and plot the distance from the origin versus the frame. The plot is given below:




Conversion of values from pixel to millimeter and frame to time were performed through known values in the video (for example, the size of the rolling cylinder and the frame rate). Curve fitting was then performed to the plot to obtain a polynomial equation of distance that depends on the constant acceleration and velocity which is given by:
d = 331.69 t^2 + 21.002 t^2
from the equation, the acceleration is given by 331.69x2 = 663.38 mm/sec^2 (which is the acceleration along the x - axis since the video was viewed from the top).
Theoretical calculation of the acceleration given by the equations below revealed that for that same parameter, the acceleration is 592.71 mm/sec^2. Large difference from theoretical and experimental values maybe due to the centroid finding of the object.

For this activity, I would give myself a grade of 8. I would like to acknowledge the help of Abraham Latimer Camba for the theoretical computation of the acceleration.

Appendix:

//15 frames per second
I = [];
se = [1 1;1 1];
for i =1:15
I = imread("vid" + string(i) + ".jpeg");
I1 = im2bw(I,0.7);
I2 = erode(dilate(I1,se),se);
I3 = erode(I2,se);
imwrite(I3,"videos" + string(i)+".jpeg");
[x1,y1] = find(I3==1);
x(i)=mean(x1);
y(i)=mean(y1);
end
distancey = y - min(y);
distancex = x - min(x);
distance = sqrt(distancex^2 + distancey^2);
for i = 1:15
velocity(1) = distancey(1);
velocity(i+1) = distance(i+1) - distancey(i);
end
for i = 1:15
acceleration(1) = 0;
acceleration(i+1) = velocity(i+1) - velocity(i);
end
scf(0);plot(distance);
scf(1);plot(velocity);
scf(2);plot(acceleration);

Activity 16 - Color Image Segmentation

In this activity, a sample from the whole image is picked out to get the Region Of Interest (ROI) from the image.
Example of this is the human skin recognition.
There are two basic techniques in segmentation; (1) Probability Distribution Estimation and (2) Histogram Backprojection.

The image and the used in this activity is shown below:
Figure 1. The original image
Figure 2. Sample from the original image

Probability Distribution Estimation

In this technique, a sample from the original image is cropped from the original image. Per pixel, the r value and the g of both the original image and the cropped image are normalized by dividing it to the sum of r, g and b values of that pixel. The probability that the the r value and the g value of the original image belong to the ROI is given by the equation:

Equation 1

The resulting ROI is shown below:Figure 3

Histogram Backprojection

In this technique, the 2D histogram of the portion of the original image was obtained. The obtained histogram was normalized to obtain the probability distribution function(PDF). The r and the g values of the original image per pixel were back-projected (or replaced) by finding the r and the g values from the PDF. The resulting image is shown below:

Figure 4

Comparing the two techniques, for me the Probability Distribution Estimation is a better technique for segmentation because of the expected ROI was obtained. In contrast to the Histogram Back-projection, not all the expected ROI was obtained.


For this activity, I will give ,myself a grade of 8 because of the late submission.
I would like to acknowledge Abraham Latimer Camba for helping me in the Histogram Back-projection part.



Appendix:
Source code


//First: Segmentation via probability
I = imread("F:\AP 186\act16\pic.jpg");
I2 = imread("F:\AP 186\act16\small.jpg");
n1 = size(I);
scf(1);imshow(I);
r1 = I(:,:,1)./(I(:,:,1)+I(:,:,2)+I(:,:,3));
g1 = I(:,:,2)./(I(:,:,1)+I(:,:,2)+I(:,:,3));
b1 = I(:,:,3)./(I(:,:,1)+I(:,:,2)+I(:,:,3));
r2 = I2(:,:,1)./(I2(:,:,1)+I2(:,:,2)+I2(:,:,3));
g2 = I2(:,:,2)./(I2(:,:,1)+I2(:,:,2)+I2(:,:,3));
b2 = I2(:,:,3)./(I2(:,:,1)+I2(:,:,2)+I2(:,:,3));
pr = (1/(stdev(r2)*sqrt(2*%pi)))*exp(-1*((r1-mean(r2)).^2/(2*stdev(r2))));
pg = (1/(stdev(g2)*sqrt(2*%pi)))*exp(-1*((g1-mean(g2)).^2/(2*stdev(g2))));
new = (pr.*pg);
scf(3);imshow((new),[]);

//Second: Segmentation via Histogram


r2_1 = floor(r2*255);
g2_1 = floor(g2*255);
n = size(r2);
Hist = zeros(256,256);
for i = 1:n(1)
for j =1:n(2)
x = r2_1(i,j)+1;
y = g2_1(i,j)+1;
Hist(x,y) = Hist(x,y) +1;
end
end
scf(5);plot3d(0:255,0:255,Hist);
Hist = Hist/max(Hist);

scf(2);imshow(log(Hist+0.0000000001),[]);
r1 = round(r1*255);
g1 = round(g1*255);
for i = 1:n1(1)
for j = 1:n1(2)
T(i,j) = Hist(r1(i,j)+1,g1(i,j)+1);
end
end
scf(4);imshow(T,[]);
imwrite(new/max(new),"F:\AP 186\act16\1_1.jpg");
imwrite(T,"F:\AP 186\act16\2_2.jpg");

Activity 15 - Color Image Processing

In this activity, images having unbalanced colored images were enhanced by;

(1) White Balancing,
(2) Gray World Balancing

White Balancing:

White balancing technique uses a known white object from the image for balancing. The red, green and blue of the RGB values of the known white object in the image is used as "NORMALIZING" or as divider for all the RGB values of all the pixels in the image.

Gray World Balancing:

Gray balancing technique uses the mean of all the red, green and blue values present in the image as the divider for all the RGB values of the pixels present in the image.

Here are some examples of images that were white balanced and gray balanced:

For this activity, I will give myself a grade of 7 out of 10 because of the late submission.
I would like to acknowledge Mark Leo for helping me debug errors on my program.

Appendix:

Source code for White Balancing created in Scilab:

I = imread("C:\Documents and Settings\AP186user15\Desktop\act15\outside - daylight.jpg");
imshow(I);
n = size(I);
RGB = round(locate(1,flag=1));
r = I((RGB(1)),(RGB(2)),1);
g = I((RGB(1)),(RGB(2)),2);
b = I((RGB(1)),(RGB(2)),3);
Ibal(:,:,1) = I(:,:,1)/r;
Ibal(:,:,2) = I(:,:,2)/g;
Ibal(:,:,3) = I(:,:,3)/b;

index=find(Ibal>1.0);
Ibal(index)=1.0;
//Inew(:,:,1) = Ibal(:,:,1)/max(I(:,:,1));
//Inew(:,:,2) = Ibal(:,:,2)/max(I(:,:,2));
//Inew(:,:,3) = Ibal(:,:,3)/max(I(:,:,3));
imwrite(Ibal,"C:\Documents and Settings\AP186user15\Desktop\act15\outside - daylight_bal.jpg");

Source code for Gray World Balancing:
I = imread("C:\Documents and Settings\AP186user15\Desktop\act15\o- incandescent.jpg");
//imshow(I);
//n = size(I);
//RGB = round(locate(1,flag=1));
r = mean(I(:,:,1));
g = mean(I(:,:,2));
b = mean(I(:,:,3));
Ibal(:,:,1) = I(:,:,1)/r;
Ibal(:,:,2) = I(:,:,2)/g;
Ibal(:,:,3) = I(:,:,3)/b;

index=find(Ibal>1.0);
Ibal(index)=1.0;
Ibal = 0.75*Ibal;
//Inew(:,:,1) = Ibal(:,:,1)/max(I(:,:,1));
//Inew(:,:,2) = Ibal(:,:,2)/max(I(:,:,2));
//Inew(:,:,3) = Ibal(:,:,3)/max(I(:,:,3));
imwrite(Ibal,"C:\Documents and Settings\AP186user15\Desktop\act15\o- incandescent_bal_gray.jpg");

Activity 14 - Streometry