使用 Opencv python 进行精确测量

Since I am Looking at very tight tolerance limits, I first did a precision study. I kept the component on the backlight source and took 100 images without moving the part (imagine like a video with 100 frames). I measured the Outer Diameter(OD) of all the 100 images. My mm/pixel ratio is 0.042. I measured the standard deviation of the measurement to find out the precision, which turned out to be around 0.03 mm which is bad. The component nor the setup is touched thus I was expecting a precision of 0.005 mm. But I am off by an order of magnitude. I am using OpenCV's Hough circle to calculate the OD of the component.

代码:

import sys
import pickle
import cv2
import matplotlib.pyplot as plt
import glob
import os
import numpy as np
import pandas as pd

def find_circles(image,dp=1.7,minDist=100,param1=50,param2=50,minRadius=0,maxRadius=0):
    """ finds the center of circular objects in image using hough circle transform

    Keyword arguments
    image -- uint8: numpy ndarray of a single image (no default).
    dp -- Inverse ratio of the accumulator resolution to the image resolution (default 1.7).
    minDist -- Minimum distance in pixel distance between the centers of the detected circles (default 100).
    param1 -- First method-specific parameter (default = 50).
    param2 -- Second method-specific parameter (default = 50).
    minRadius -- Minimum circle radius in pixel distance (default = 0).
    maxRadius -- Maximum circle radius in pixel distance (default = 0).

    Output
    center -- tuple: (x,y).
    radius -- int : radius.
    ERROR if circle is not detected. returns(-1) in this case    
    """

    circles=cv2.HoughCircles(image, 
                             cv2.HOUGH_GRADIENT, 
                             dp = dp, 
                             minDist = minDist, 
                             param1=param1, 
                             param2=param2, 
                             minRadius=minRadius, 
                             maxRadius=maxRadius)
    if circles is not None:
            circles = circles.reshape(circles.shape[1],circles.shape[2])
            return(circles)
    else:
        raise ValueError("ERROR!!!!!! circle not detected try tweaking the parameters or the min and max radius")

def find_od(image_path_list):
    image_path_list.sort()
    print(len(image_path_list))
    result_df = pd.DataFrame(columns=["component_name","measured_dia_pixels","center_in_pixels"])
    for i,name in enumerate(image_path_list):
        img = cv2.imread(name,0) # read the image in grayscale
        ret,thresh_img = cv2.threshold(img, 50, 255, cv2.THRESH_BINARY_INV)
        thresh_img = cv2.bilateralFilter(thresh_img,5,91,91) #smoothing
        edges = cv2.Canny(thresh_img,100,200)
        circles = find_circles(edges,dp=1.7,minDist=100,param1=50,param2=30,minRadius=685,maxRadius=700)
        circles = np.squeeze(circles)
        result_df.loc[i] = os.path.basename(name),circles[2]*2,(circles[0],circles[1])
    result_df.sort_values("component_name",inplace=True)
    result_df.reset_index(drop=True,inplace=True)
    return(result_df)

df = find_od(glob.glob("./images/*"))
mean_d = df.measured_dia_pixels.mean()
std_deviation = np.sqrt(np.mean(np.square([abs(x-mean_d) for x in df.measured_dia_pixels])))

mm_per_pixel = 0.042
print(std_deviation * mm_per_pixel)

输出:0.024

组件的图片:

由于图像是在不干扰设置的情况下拍摄的，因此我预计测量的可重复性约为 0.005 毫米(5 微米)(对于 100 张图像).但事实并非如此.是霍夫圆的问题吗?或者我在这里错过了什么

import diplib as dip
import math

# Load image and set pixel size
img = dip.ImageRead('N6uqw.jpg')
img.SetPixelSize(0.042, "mm")

# Extract object
obj = ~dip.Threshold(dip.Gauss(img))[0]
obj = dip.EdgeObjectsRemove(obj)

# Remove noise
obj = dip.Opening(dip.Closing(obj,9),9)

# Measure object area
lab = dip.Label(obj)
msr = dip.MeasurementTool.Measure(lab,img,['Size'])
objectArea = msr[1]['Size'][0]

# Measure holes
obj = dip.EdgeObjectsRemove(~obj)
lab = dip.Label(obj)
msr = dip.MeasurementTool.Measure(lab,img,['Size'])
sz = msr['Size']
holeAreas = []
for ii in sz.Objects():
   holeAreas.append(sz[ii][0])

# Add hole areas to main object area
objectArea += sum(holeAreas)

print('Object diameter = %f mm' % (2 * math.sqrt(objectArea / math.pi)))
for a in holeAreas:
   print('Hole diameter = %f mm' % (2 * math.sqrt(a / math.pi)))

Object diameter = 57.947768 mm
Hole diameter = 6.540086 mm
Hole diameter = 6.695357 mm
Hole diameter = 15.961935 mm
Hole diameter = 6.511002 mm
Hole diameter = 6.623011 mm