%=========================================s4.m===========================================
%估算水管直徑
I = imread('pipe_out.jpg');                     %讀入已經標記出邊界的圖像
x = 1:L(2)+2;
d = [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0];  %記錄直徑的數組

%求出直徑的斜率
k1 = -cos(th1)/sin(th1);
k2 = -cos(th2)/sin(th2);
k3 = 1/2*(k1+k2);
k = -1/k3;

%繪制出20條直徑，用于估算水管的直徑長度
for i = 1:20,
    y = round(1000-1000*i + k*x);               %直徑的直線方程，相鄰的直線截距相差1000個像素
    
    %求出該直徑的長度，存放到數組里
    x1 = (r1/sin(th1) - 1000*(1-i))/(k + cos(th1)/sin(th1));
    x2 = (r2/sin(th2) - 1000*(1-i))/(k + cos(th2)/sin(th2));
    d(i) = sqrt((x1-x2)^2 + ((1000*(1-i)+k*x1) - (1000*(1-i)+k*x2))^2);
    
    %繪制這條直徑在水管邊緣內的部分，為了顯示清楚，使用紅色并對x方向和y方向各畫3個像素
    for j = 2:L(2)+1,
        if (y(j)>((r2 - j*cos(th2))/sin(th2))) && (y(j)<((r1 - j*cos(th1))/sin(th1))),
            I(y(j)-1,j-1,1) = 255;
            I(y(j)-1,j-1,2) = 0;
            I(y(j)-1,j-1,3) = 0;
            I(y(j)-1,j,1) = 255;
            I(y(j)-1,j,2) = 0;
            I(y(j)-1,j,3) = 0;
            I(y(j)-1,j+1,1) = 255;
            I(y(j)-1,j+1,2) = 0;
            I(y(j)-1,j+1,3) = 0;
            I(y(j),j-1,1) = 255;
            I(y(j),j-1,2) = 0;
            I(y(j),j-1,3) = 0;
            I(y(j),j,1) = 255;
            I(y(j),j,2) = 0;
            I(y(j),j,3) = 0;
            I(y(j),j+1,1) = 255;
            I(y(j),j+1,2) = 0;
            I(y(j),j+1,3) = 0;
            I(y(j)+1,j-1,1) = 255;
            I(y(j)+1,j-1,2) = 0;
            I(y(j)+1,j-1,3) = 0;
            I(y(j)+1,j,1) = 255;
            I(y(j)+1,j,2) = 0;
            I(y(j)+1,j,3) = 0;
            I(y(j)+1,j+1,1) = 255;
            I(y(j)+1,j+1,2) = 0;
            I(y(j)+1,j+1,3) = 0;
        end
    end
end
di = sum(d) / length(d)                     %求出直徑數組的均值，即估算出來的水管直徑
imwrite(I,'result.jpg','jpg');                   %將處理后的圖像輸出到result.jpg
%=======================================s4.m(end)=========================================