cut image from centre at an angle of 30 degrees

I'd like to cut an image into for example 18 parts,so that anothers script of mine will run for each part of the image, can you help me a bit cause i'm stuck. i am sharing the original image and the image which show how to cut the image in 18 parts

[Jan]

What do you want as output? The corrdinates of cropping area, a binary mask, an RGB image with the masked pixels set to black, white or blue?

[KSSV]

The user asked the same question: https://in.mathworks.com/matlabcentral/answers/448361-how-to-crop-an-image-equally?s_tid=prof_contriblnk

[Ruksana siddique]

sir.. do you see the attachment. i attach an image.. given below.. sir in this image you can see that i take 3 circle (shown by dotted lines) ok .. according to its gray portion. and more from the centre of circle i draw 6 lines (shown by dark white lines) which cut the circle at 60 degree. .. but i dont know how to seprate all these 18 parts. i want each part cut. this image have total 18 outputs (shown by yellow numbers). i watnt to cut all these parts seprately.

[Jan]

Your description does not allow to know, what you exactly do. "I take 3 circles" is not clear. How do you do this? Neither "seprate all these 18 parts" nor "each part cut" explains, what you want as output. Remember, that images are arrays, such that they are rectangular in every case.

You have posted two questions for the same problem and bump the thread by posting a pseudo-answer, but you still did not mention, what you want as output. So how can we help you?

[Ruksana siddique]

I want to crop the 10 and 16 part shown in image by numbering .. is it possible

Answer 1

[Image Analyst]

You should be able to figure it ouit from the FAQ: How_do_I_create_an_arc

If you tried and can't figure it out, let us know.

[Ruksana siddique]

OK. . thank you so much sir. .i will let you know

Answer 2

[Ruksana siddique]

sir do ythisu know how can I do

[Jan]

Please do not bump your question by postuing pseudo-answers. This wastes the time of the readers.

That this is a duplicate question in addition is not an efficient usage of the forum also.