Activity 13: Photometric Stereo

By capturing multiple images of a surface with light sources at different locations. In this activity, the images used are the following below.

To do this, N sources in 3D space are defined in matrix form

Since I = Vg, we can solve for the surface normal. I is the intensity matrix and V is matrix form for the sources. To get surface normal, we use the equations

With the surface normals determined, the elevation is the computed and plotted in 3D space.

loadmatfile ('C:\Users\nez\Documents\tresloco\ap186_2\activity13\Photos');

browsevar();

I(1,:) = (I1(:))';

I(2,:) = (I2(:))';

I(3,:) = (I3(:))';

I(4,:) = (I4(:))';
V(1,:) = [0.085832 0.17365 0.98106];

V(2,:) = [0.085832 -0.17365 0.98106];

V(3,:) = [0.17365 0 0.98481];

V(4,:) = [0.16318 -0.34202 0.92542];

g = (inv(V'*V)*(V'))*I;

N = size(g);gmag = [];

for i = 1:N(2)

gmag(i) = sqrt(g(1,i)**2 + g(2,i)**2 + g(3,i)**3)+0.0000000001;

end

n(1,:) = g(1,:)./gmag(1,:);

n(2,:) = g(2,:)./gmag(1,:);

n(3,:) = g(3,:)./gmag(1,:);

dfx = -n(1,:)./(n(3,:)+0.0000000001);

dfy = -n(2,:)./(n(3,:)+0.0000000001);

F = cumsum(dfx,1)+cumsum(dfy,2);

new = matrix(F,[128,128]);

plot3d(1:128,1:128,new)

Using this code, a 3D plot of the suface was created.

Grade: 8/10 - I have done the activity well.

Activity 11: Camera Calibration

The activity was done using the image below.

Letting the origin be at the lowest part of the board along the middle, coordinates of 20 corner points were measured. Using Microsoft Paint, the coordinates for each picked corner in the image was determined.

x0 = [0 1 3 0 2 0 0 0 8 3 0 0 0 0 3 8 5 1 6 0];
y0 = [0 0 0 3 0 6 7 3 0 0 8 1 6 8 0 0 0 0 0 2];
z0 = [0 1 4 5 9 11 5 7 0 12 7 10 9 0 6 8 8 10 11 1];
yi = [100 117 154 70 138 37 24 70 243 157 13 92 36 13 154 248 192 120 212 80];
zi = [234 218 163 145 62 26 156 105 251 3 115 41 68 265 123 89 86 41 26 222];for i = 1:20
Q((2*i)+1,:) = [x0(i) y0(i) z0(i) 1 0 0 0 0 -(yi(i)*x0(i)) -(yi(i)*y0(i)) -(yi(i)*z0(i))];
Q((2*i)+2,:) = [0 0 0 0 x0(i) y0(i) z0(i) 1 -(zi(i)*x0(i)) -(zi(i)*y0(i)) -(zi(i)*z0(i))];
d((2*i)+1,:) = [yi(i)];
d((2*i)+2,:) = [zi(i)];
end
Qt = Q'
a = inv(Qt*Q)*Qt*d;

//check
for j = 1:20
y2(j,:) = ((a(1))*x0(j)+(a(2))*y0(j)+(a(3))*z0(j)+a(4))/((a(9))*x0(j)+(a(10))*y0(j)+(a(11))*z0(j)+1);
z2(j,:) = ((a(5))*x0(j)+(a(6))*y0(j)+(a(7))*z0(j)+a(8))/((a(9))*x0(j)+(a(10))*y0(j)+(a(11))*z0(j)+1);
end
The code above used the equation from the manual.

The result was having an average % error of 0.7947295 and 1.245960579 in y and z respectively.

//verification for other points

x0 = [3 7 0];

y0 = [0 0 2];

z0 = [5 2 7];

for k = 1:3

y3(k,:) = ((a(1))*x0(k)+(a(2))*y0(k)+(a(3))*z0(k)+a(4))/((a(9))*x0(k)+(a(10))*y0(k)+(a(11))*z0(k)+1);

z3(k,:) = ((a(5))*x0(k)+(a(6))*y0(k)+(a(7))*z0(k)+a(8))/((a(9))*x0(k)+(a(10))*y0(k)+(a(11))*z0(k)+1);

end

The results show that the method was accurate.

Grade: 7/10 - The activity was passed late.

Acknowledgement: Jorge Presto

Activity 10: Preprocessing Handwritten Text

After cropping a portion from the image above, the new image's Fourier Transform was determined.

im=imread('C:\Users\nez\Documents\tresloco\ap186_2\activity 10\a.jpg');
img = im2gray(im);
imgfft=fft2(img);
scf (1);
imshow(imgfft);
imwrite (imgfft, 'C:\Users\nez\Documents\tresloco\ap186_2\activity 10\imgray.jpg')

With the use of techniques learned from previous activities, the lines were removed using a filter just like the one below.

filter=imread('C:\Users\nez\Documents\tresloco\ap186_2\activity 10\f.jpg');

imf = im2gray(filter);

imffft=fftshift(imf);

imffft=ff1.*ff2; The image is then convolved with the filter, converted to binary, and cleaned using the closing/opening operator.

//filtering the image

a=real(fft2(imffft));

scf(2)imshow(a,[]);

imwrite (a, 'C:\Users\nez\Documents\tresloco\ap186_2\activity 10\imfilter.jpg')

//binarize the image

b=im2bw(a, 127/255);

c=1-b;

scf(3);imshow(c,[]);

imwrite (c, 'C:\Users\nez\Documents\tresloco\ap186_2\activity 10\imbinary.jpg')

//opening operation

SE = ones(2,2);

d=erode(c,SE);

e=dilate(d,SE)scf(4);

imshow(e,[]);

imwrite (e, 'C:\Users\nez\Documents\tresloco\ap186_2\activity 10\imcloseopen.jpg')

After this is done, the image is then labelled.
//labelling the image
imfinal=bwlabel(e);
scf(5);
imshow(imfinal,[]);
//imwrite (imfinal, 'C:\Users\nez\Documents\tresloco\ap186_2\activity 10\imlabel.jpg')

Grade: 6/10 - I passed it too late.

Acknowledgement: Jorge Presto

Activity 9 :Binary Operations

The activity's objective is to find the best estimate of the cell area. Using GIMP, the image above was cropped into 9 parts measuring 256 x 256 pixels. The subimages were then converted to black & white by finding the threshold of the images.
With the code below, the closing and opening operators were used to clean the image of isolated pixels andcells very near each other.
//img = imread("G:\to be posted\186_09\C1_01_bw.jpg");
//img = imread("G:\to be posted\186_09\C1_02_thresh205.jpg");
//img = imread ("G:\to be posted\186_09\C1_03_thresh223.jpg");
//img = imread ("G:\to be posted\186_09\C1_04_thresh194.jpg");
//img = imread ("G:\to be posted\186_09\C1_05_thresh190.jpg");
//img = imread ("G:\to be posted\186_09\C1_06_thresh215.jpg");
//img = imread ("G:\to be posted\186_09\C1_07_thresh200.jpg");
//img = imread ("G:\to be posted\186_09\C1_08_thresh194.jpg");
img = imread ("G:\to be posted\186_09\C1_09_thresh213.jpg");
imgray = im2gray(img);
mat = size(img);
pic = imgray;for i = 1:mat(1) //rows
for j = 1:mat(2) //columns
if pic(i,j) <>
pic(i,j) = 0;
else
pic(i,j) = 1;
end
end
end

SE = [];for i = 1:4 for j =1:4 SE(i,j) = 1; end end
img_co = erode(dilate(pic,SE),SE);
//img_co = dilate(erode(pic,SE),SE);
scf(1);
imshow (img_co4);
imwrite (img_co4, 'G:\to be posted\186_09\C1_09_co.jpg');

After cleaning the images, each blob was then labelled and the area of each was measured.

//labellingimg = imread ("G:\to be posted\186_09\C1_09_co.jpg");
imgray = im2gray (img);pic = imgray;
mat = size(pic);
for x = 1:mat(1);
for y = 1:mat(2);
if pic (x, y) <>
pic (x, y) = 0;
else
pic (x, y) = 1;
end
end
end[L, n] = bwlabel(pic);
scf(3);
imshow(pic, []);
scf(4);
imshow (L+1,rand(n+1,3));

//area
for i = 1: max(bwlabel(pic))
v = find(L == i);
w = size(v);
area(i) = w(2);
end

The following images show the blobs binarized, cleaned, and labelled.

With the use of Microsoft Excel, the histogram was plotted. The values used were in the range of 400 to 700. mean = 498.2 ; standard deviation = 55.19