How can i get x,y points of squares center ?

In this picture

we have 4 squares. I need 4 x and 4 y points(coordinates) of each square center. Can somebody do it? Because I'm a beginner ...

[Adam Danz]

I'd like to help you but it's very unclear what you need. You should know that no one will write code for you but lots of people will help you to write code by answering more specific questions.

I see 4 black quadrilaterals in the image that could be squares in reality but because the paper in the image is tilted and distorted, they do not appear as squares on my monitor. I think you want to know the (x,y) coordinates of their centers but relative to what? In what coordinate system? Where is (0,0)? Is (0,0) in the center of the paper? Do you need to calculate the centers based on this image (I hope not)?

[mplz]

yes paper is distorted and that 4 squares do not likes square,i know. but how can i solve this problem ,i don't know. i need coordinates of each square center, for example 1 square coordinates is (x,y)>>(192,90),i hope you understand me.

[Adam Danz]

In your example, what are (192, 90)? Pixels? Millimeters? And is 192 and 90 a distance from the bottom corner of the paper on the wall, the center of the paper? The center of the photograph? It's unclear what you need to do.

[mplz]

1_ (192,90)-->it's just an example.

2_ yes, its Pixels.

3_ no,it's not distance.it's x,y coordinates of square center(1 square) and in fact i want 4x and 4y coordinates for those 4 square center.

4_ after(if) solved this problem ,i tell ,what i was looking for.

thank s

[Adam Danz]

Replies to your points...

(3) I understand they are (x,y) coordinates but where is the origin? Where is (0,0)? For example, the coordinate (5,12) is 5 pixels to the right of x=0 and 12 pixels above y=0. In your example, (192, 90) these are also distances from (0,0) but I don't know where (0,0) is in your image.

(4) feel free to follow up with more specific questions but I think the problem is not well defined and that will make it difficult to find a solution.

Answer 1

[Image Analyst]

First of all, use a better exposure. Having a brighter image will give you more contrast and less noise. Then you can just use a bottomhat filter, imbothat() to find the locally dark things. Then use bwareafilt() to extract the 4 largest dark things. Then use regionprops to get their centers and average them. Try the attached script. It makes this:

I'm going out for a bit. See if you can finish that while I'm gone. And let me know how it goes and if you're able to start with a properly lit image.

[mplz]

thanks you very much,you looking great in image analyst.

i think,I have to explain my project now.

in this project we have one paper shooting target and somebody shot with laser gun on shooting target and then matlab calculate his score .but problem is camera location change,so we need to camera calibration because when camera took picture ,this image is distorted and awry and then distance and score are wrong.

calculating score method is by distance between black circle in center of paper and laser bright circle,and when distance is x then score is 10 and when distance is y score is 9 and ....if distance is out of range score is 0.

scores is 10,9,8,7,6,5,4,3,2,1,0 .

those 4 squares was for calibration i thought ,now i think those squares is not valid and right,i don't know ,maybe somebody bring better offer.

[Image Analyst]

See the camera calibration capabilities in the Computer Vision System Toolbox. https://www.mathworks.com/products/computer-vision/features.html#camera-calibration

They're rather extensive and well thought out so it's nothing I can replicate in one or two hundred lines of code.

If I answered your question(s), can you "Accept this answer"? Thanks.