I am trying to detect each small circle (it is the bead part of the radial tires from the cross-sectional image)located as shown in the image and get their information(optional). To improve the detection process I have performed a few image processing steps including median blurring and binary thresholding (the general binary thresholding and inverse binary thresholding). I am using HoughCicle transform to detect the circles however I stucked and couldn't be able to detect it yet.
Please, any suggestions? Thank you very much.
Cropped image (it is the area where the circle I want to detect appear)
This is the binary image output and cropped it to remove the unnecessary part
As such, I'm trying to detect each circle from the binary image shown in the image below like marked in red. Final preprocessed image
I used the following code
import cv2
import numpy as np
import os
import matplotlib.pyplot as plt
############# preprocessing ##################
img = cv2.imread('BD-2021.png')
median_5 = cv2.medianBlur(img, 5) # median filtering
image_masked = cv2.cvtColor(median_5, cv2.COLOR_BGR2GRAY) # converting to grayscael
res,thresh_img=cv2.threshold(image_masked,230,255,cv2.THRESH_BINARY_INV) # inverse binary
# res,thresh_img_b=cv2.threshold(image_masked,200,255,cv2.THRESH_BINARY) # global binary
height, width = thresh_img.shape
img_crop = thresh_img[int(0.7*height):height,:width]
# reverse_thresh = cv2.cvtColor(thresh_img,cv2.COLOR_GRAY2BGR)
############# Hough circle detection ##################
c = cv2.HoughCircles(img_crop, cv2.HOUGH_GRADIENT,
minDist=2, dp=1, param1=70,
param2=12, minRadius=0,maxRadius=5)
c = np.uint16(np.around(c))
for i in c[0,:]:
# draw the outer circle
cv2.circle(img,(i[0],i[1]),i[2],(0,255,0),2)
# cv2.circle(reverse_thresh,(i[0],i[1]),i[2],(0,255,0),2)
# draw the center of the circle
cv2.circle(img,(i[0],i[1]),2,(0,0,255),3)
# cv2.circle(reverse_thresh,(i[0],i[1]),2,(0,0,255),3)
cv2.imshow('circle detected',img)
cv2.waitKey(0)
cv2.destroyAllWindows()
I would appreciate for any recommendation ? Thank you once again.