请原谅,因为我对编码完全不熟悉。首先,为了这个项目的目的,我正在使用一个Python绑定到OpenCV的库。
我的相机经过校准,显示鱼眼失真。我分别获得了K和D的以下值,内在相机矩阵和失真矩阵:
K = [[438.76709 0.00000 338.13894]
[0.00000 440.79169 246.80081]
[0.00000 0.00000 1.00000]]
D = [-0.098034379506 0.054022224927 -0.046172648829 -0.009039512970]
Focal length: 2.8mm
Field of view: 145 degrees (from manual)
当我不对图像进行展开并显示它时,我会在拉伸太远(预期)的区域获得带有黑色像素的图像。但是,这并不妨碍物体宽度的计算,因为物体不大,并且填充了20%的图像。
我将把物体放在相机镜头10厘米处。基于我在针孔相机模型上所读到的内容,我将需要控制3D到2D转换的外在参数。但是,我不确定我应该如何推导它。
假设我有2个点的像素坐标(每个点沿着我希望测量距离的边缘),我如何使用这些导出的矩阵找到这两个点之间的真实距离?
另外,如果我的矩形物体与摄影机的主轴不平行,是否有算法计算宽度,即使有这样的条件?
鉴于您的相机和物体之间的距离是固定的,您可以做的是首先找出找到的角之间的像素距离,然后使用像素/毫米比率/比例因子将其转换为毫米对象宽度。
使用的算法是Harris Corner Detection Harris Corner Detection
使用其中的对象捕获帧
cap = cv2.VideoCapture(0)
while(True):
#Capture frame-by-frame
ret, frame = cap.read()
cv2.imshow('LIVE FRAME!', frame)
if cv2.waitKey(1) & 0xFF == ord('q'):
break
#Save it to some location
cv2.imwrite('Your location', frame)
首先使用参考对象校准像素/毫米比率。
#Read Image
image = cv2.imread('Location of your previously saved frame with the object in it.')
object_width = input(int("Enter the width of your object: ")
object_height = input(int("Enter the height of your object: ")
#Find Corners
def find_centroids(dst):
ret, dst = cv2.threshold(dst, 0.01 * dst.max(), 255, 0)
dst = np.uint8(dst)
# find centroids
ret, labels, stats, centroids = cv2.connectedComponentsWithStats(dst)
# define the criteria to stop and refine the corners
criteria = (cv2.TERM_CRITERIA_EPS + cv2.TERM_CRITERIA_MAX_ITER, 100,
0.001)
corners = cv2.cornerSubPix(gray,np.float32(centroids[1:]),(5,5),
(-1,-1),criteria)
return corners
gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
gray = np.float32(gray)
dst = cv2.cornerHarris(gray, 5, 3, 0.04)
dst = cv2.dilate(dst, None)
# Get coordinates of the corners.
corners = find_centroids(dst)
for i in range(0, len(corners)):
print("Pixels found for this object are:",corners[i])
image[dst>0.1*dst.max()] = [0,0,255]
cv2.circle(image, (int(corners[i,0]), int(corners[i,1])), 7, (0,255,0), 2)
for corner in corners:
image[int(corner[1]), int(corner[0])] = [0, 0, 255]
a = len(corners)
print("Number of corners found:",a)
#List to store pixel difference.
distance_pixel = []
#List to store mm distance.
distance_mm = []
P1 = corners[0]
P2 = corners[1]
P3 = corners[2]
P4 = corners[3]
P1P2 = cv2.norm(P2-P1)
P1P3 = cv2.norm(P3-P1)
P2P4 = cv2.norm(P4-P2)
P3P4 = cv2.norm(P4-P3)
pixelsPerMetric_width1 = P1P2 / object_width
pixelsPerMetric_width2 = P3P4 / object_width
pixelsPerMetric_height1 = P1P3 / object_height
pixelsPerMetric_height2 = P2P4 / object_height
#Average of PixelsPerMetric
pixelsPerMetric_avg = pixelsPerMetric_width1 + pixelsPerMetric_width2 + pixelsPerMetric_height1 + pixelsPerMetric_height2
pixelsPerMetric = pixelsPerMetric_avg / 4
print(pixelsPerMetric)
P1P2_mm = P1P2 / pixelsPerMetric
P1P3_mm = P1P3 / pixelsPerMetric
P2P4_mm = P2P4 / pixelsPerMetric
P3P4_mm = P3P4 / pixelsPerMetric
distance_mm.append(P1P2_mm)
distance_mm.append(P1P3_mm)
distance_mm.append(P2P4_mm)
distance_mm.append(P3P4_mm)
distance_pixel.append(P1P2)
distance_pixel.append(P1P3)
distance_pixel.append(P2P4)
distance_pixel.append(P3P4)
打印距离(以像素为单位)和mm(即宽度和高度)
print(distance_pixel)
print(distance_mm)
pixelsPerMetric
是您的比例因子,并给出每毫米的平均像素数。您可以修改此代码以根据您的需要进行操作。
我会使用类似的三角形来确定图像中的宽度与对象宽度成正比,因为(distance of camera to object)/(focal length)
的比例因子在你的情况下是100/2.8
。这将假设物体位于图像的中心(即直接在相机前面)。