{"id":638,"date":"2022-08-10T18:43:53","date_gmt":"2022-08-10T18:43:53","guid":{"rendered":"http:\/\/fenstar.co\/wp\/?p=638"},"modified":"2023-01-21T17:56:45","modified_gmt":"2023-01-21T17:56:45","slug":"more-texture-generation-wood-grain","status":"publish","type":"post","link":"https:\/\/fenstar.co\/wp\/?p=638","title":{"rendered":"More texture generation (Wood grain)"},"content":{"rendered":"

I started thinking about the trunk of a tree. Essentially, you have a bunch of concentric growth rings. I visualized this as something looking like the image below.<\/p>\n

 <\/p>\n

\"The<\/a>

The center of the tree trunk would be in the middle of the left side.<\/p><\/div>\n

Obviously, as you get closer to the top the growth rings shrink and disappear, but I figured it was best to start simple.<\/p>\n

This was represented by the\u00a0following code\u00a0(GUI Octave)<\/p>\n

for i = 1:y\r\nfor j = 1:x\r\nT(i,j) = sqrt((i-(y\/2))^2+j^2);\r\nendfor\r\nendfor<\/pre>\n
T=abs(sin(T\/s));<\/pre>\n
Where x, y, and s were 128, 128, and 8 respectively.<\/p>\n
So, if we view this as the trunk, we can “cut” a board out of it by taking a slice.<\/p>\n
\"Visualization<\/a>
Visualization of “cutting” a board from the trunk of a tree.<\/p><\/div>\n
This can be achieved with the code<\/p>\n
for i =1:y\r\nfor j =1:x\r\nIM(i,j) = T(i,(x\/2));\r\nendfor\r\nendfor<\/pre>\n
which returns this super boring image<\/p>\n
\"ZzZzZz...<\/a>
ZzZzZz… oh, sorry, I fell asleep there for a moment.<\/p><\/div>\n
We can also simulate, to some extent, the tree shrinking as we go up the trunk by sloping the “cut”<\/p>\n
for i =1:y\r\n for j =1:x\r\n IM(i,j) = T(i,int8((x\/2)+j\/(x\/4)));\r\n endfor\r\n endfor<\/pre>\n
\"Sort<\/a>
Sort of looks a bit like wood grain.<\/p><\/div>\n
in reality, this is obviously not exactly<\/em> equivalent to what a perfectly straight tree would look like as the growth rings shrunk, but it is pretty close so we are going to use this for now.<\/p>\n
Of course, growth rings are not perfect circles in the real world, but in the same vain as the slope before, we can simulate irregularities\u00a0by applying an offset.<\/p>\n
I randomly generated and smoothed a matrix which I combined with the slice to create the image shown after the code below.<\/p>\n
My code, which I admit is not great, is shown below. (The lack of\u00a0matrix convolution is 100% not because I failed linear algebra and\/or I am too lazy to remember)<\/p>\n
x = 128;\r\ny = x;\r\ns = 3;\r\nq=2;\r\nv=8;\r\n\r\nR = rand(y,x);\r\nS = zeros(y,x);\r\n\r\nfor i =q+1:(y-q)\r\nfor j =q+1:(x-q)\r\nfor k = -q:q\r\nfor l = -q:q\r\nS(i,j) = S(i,j)+R(i+k,j+l);\r\nendfor\r\nendfor\r\nendfor\r\nendfor\r\n\r\nS=S-((2*q+1)^2)\/2;\r\nS=S*s\/v;\r\n\r\nfor i = 1:y\r\nfor j = 1:x\r\nT(i,j) = sqrt((i-(y\/2))^2+j^2);\r\nendfor\r\nendfor\r\n\r\nT=abs(sin(T\/s));\r\n\r\nfor i =2:(y-2)\r\nfor j =2:(x-2)\r\nIM(i,j) = T(i,int8(S(i,j)+j\/(x\/4))+(x\/2));\r\nendfor\r\nendfor\r\n\r\nimshow(IM)\r\n<\/pre>\n
T is the cross cut of the tree, R is a random value matrix, S is a smoothed version of R, and IM is everything combined into the final image shown below.<\/p>\n
Or, equivalently, for those that don’t want to waste time, you can use the following code.<\/p>\n
x = 128;\r\ny = x;\r\ns = 3;\r\nq=2;\r\nv=8;\r\n\r\nR = rand(y,x);\r\nS = ones(2*q+1,2*q+1);\r\n\r\nZ= conv2(R,S,\"same\");\r\nS=Z-((2*q+1)^2)\/2;\r\nS=S*s\/v;\r\n\r\nfor i = 1:y\r\nfor j = 1:x\r\nT(i,j) = sqrt((i-(y\/2))^2+j^2);\r\nendfor\r\nendfor\r\n\r\nT=abs(sin(T\/s));\r\n\r\nfor i =2:(y-2)\r\nfor j =2:(x-2)\r\nIM(i,j) = T(i,int8(S(i,j)+j\/(x\/4))+(x\/2));\r\nendfor\r\nendfor\r\n\r\nimshow(IM)\r\n<\/pre>\n
\"Getting<\/a>
Getting closer<\/p><\/div>\n
Granted, this isn’t exactly a high resolution image and we are missing a lot of things\u00a0including, but not limited to<\/p>\n