檢測兩個圓形物體之間的角度差

我正在嘗試檢測兩個圓形物體之間的角度差，如下圖 2 所示。

我正在考慮以一些小角度旋轉影像之一。每次旋轉一張影像，將計算旋轉影像和另一張影像之間的 SSIM。SSIM 最大的角度就是角度差。

但是，找到極端從來都不是一個容易的問題。所以我的問題是：在這種情況下，是否可以使用另一種演算法（opencv）？

圖片#1

圖片#2

編輯：

謝謝@Micka，我只是按照他建議的方式進行操作，并像@Yves Daoust 所說的那樣洗掉黑色區域以縮短處理時間。這是我的最終結果：

原始影像 旋轉 移位影像

uj5u.com熱心網友回復：

這是一種方法：

1. 檢測圓（例如我假設圓在影像中心，半徑是影像寬度的 50%）
2. 按極坐標展開圓形影像
3. 確保第二張圖片在第一張圖片中完全可見，沒有“圓端溢位”
4. 簡單的模板匹配

以下代碼的結果：

min: 9.54111e 07
pos: [0, 2470]
angle-right: 317.571
angle-left: -42.4286

我認為這在一般情況下應該作業得很好。

int main()
{
    // load images
    cv::Mat image1 = cv::imread("C:/data/StackOverflow/circleAngle/circleAngle1.jpg");
    cv::Mat image2 = cv::imread("C:/data/StackOverflow/circleAngle/circleAngle2.jpg");

    // generate circle information. Here I assume image center and image is filled by the circles.
    // use houghCircles or a RANSAC based circle detection instead, if necessary
    cv::Point2f center1 = cv::Point2f(image1.cols/2.0f, image1.rows/2.0f);
    cv::Point2f center2 = cv::Point2f(image2.cols / 2.0f, image2.rows / 2.0f);
    float radius1 = image1.cols / 2.0f;
    float radius2 = image2.cols / 2.0f;

    cv::Mat unrolled1, unrolled2;
    // define a size for the unrolling. Best might be to choose the arc-length of the circle. The smaller you choose this, the less resolution is available (the more pixel information of the circle is lost during warping)
    cv::Size unrolledSize(radius1, image1.cols * 2);

    // unroll the circles by warpPolar
    cv::warpPolar(image1, unrolled1, unrolledSize, center1, radius1, cv::WARP_POLAR_LINEAR);
    cv::warpPolar(image2, unrolled2, unrolledSize, center2, radius2, cv::WARP_POLAR_LINEAR);

    // double the first image (720° of the circle), so that the second image is fully included without a "circle end overflow"
    cv::Mat doubleImg1;
    cv::vconcat(unrolled1, unrolled1, doubleImg1);

    // the height of the unrolled image is exactly 360° of the circle
    double degreesPerPixel = 360.0 / unrolledSize.height;

    // template matching. Maybe correlation could be the better matching metric
    cv::Mat matchingResult;
    cv::matchTemplate(doubleImg1, unrolled2, matchingResult, cv::TemplateMatchModes::TM_SQDIFF);

    double minVal; double maxVal; cv::Point minLoc; cv::Point maxLoc;
    cv::Point matchLoc;
    cv::minMaxLoc(matchingResult, &minVal, &maxVal, &minLoc, &maxLoc, cv::Mat());

    std::cout << "min: " << minVal << std::endl;
    std::cout << "pos: " << minLoc << std::endl;

    // angles in clockwise direction:
    std::cout << "angle-right: " << minLoc.y * degreesPerPixel << std::endl;
    std::cout << "angle-left: " << minLoc.y * degreesPerPixel -360.0 << std::endl;
    double foundAngle = minLoc.y * degreesPerPixel;
    
    // visualizations:
    // display the matched position
    cv::Rect pos = cv::Rect(minLoc, cv::Size(unrolled2.cols, unrolled2.rows));
    cv::rectangle(doubleImg1, pos, cv::Scalar(0, 255, 0), 4);

    // resize because the images are too big
    cv::Mat resizedResult;
    cv::resize(doubleImg1, resizedResult, cv::Size(), 0.2, 0.2);
    
    cv::resize(unrolled1, unrolled1, cv::Size(), 0.2, 0.2);
    cv::resize(unrolled2, unrolled2, cv::Size(), 0.2, 0.2);

    double startAngleUpright = 0;
    cv::ellipse(image1, center1, cv::Size(100, 100), 0, startAngleUpright, startAngleUpright   foundAngle, cv::Scalar::all(255), -1, 0);

    cv::resize(image1, image1, cv::Size(), 0.5, 0.5);
    cv::imshow("image1", image1);

    cv::imshow("unrolled1", unrolled1);
    cv::imshow("unrolled2", unrolled2);

    cv::imshow("resized", resizedResult);

    cv::waitKey(0);

    
}

這是中間影像和結果的樣子：

展開影像 1 / 展開 2 / 展開 1 (720°) / 展開 1 (720°) 中展開 2 的最佳匹配：

標籤：Python opencv

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