给定点到给定椭圆的距离

本文介绍了给定点到给定椭圆的距离的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我有一个椭圆，由中心点"，radiusX和radiusY定义，并且有一个点.我想在椭圆上找到最接近给定点的点.在下图中，该名称为S1.

I have an ellipse, defined by Center Point, radiusX and radiusY, and I have a Point. I want to find the point on the ellipse that is closest to the given point. In the illustration below, that would be S1.

现在我已经有了代码，但是其中有些地方存在逻辑错误，而且我似乎找不到它.我将问题分解为以下代码示例:

Now I already have code, but there is a logical error somewhere in it, and I seem to be unable to find it. I broke the problem down to the following code example:

#include <vector>
#include <opencv2/core/core.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <math.h>

using namespace std;

void dostuff();

int main()
{
    dostuff();
    return 0;
}

typedef std::vector<cv::Point> vectorOfCvPoints;

void dostuff()
{

    const double ellipseCenterX = 250;
    const double ellipseCenterY = 250;
    const double ellipseRadiusX = 150;
    const double ellipseRadiusY = 100;

    vectorOfCvPoints datapoints;

    for (int i = 0; i < 360; i+=5)
    {
        double angle = i / 180.0 * CV_PI;
        double x = ellipseRadiusX * cos(angle);
        double y = ellipseRadiusY * sin(angle);
        x *= 1.4;
        y *= 1.4;
        x += ellipseCenterX;
        y += ellipseCenterY;
        datapoints.push_back(cv::Point(x,y));
    }

    cv::Mat drawing = cv::Mat::zeros( 500, 500, CV_8UC1 );

    for (int i = 0; i < datapoints.size(); i++)
    {
        const cv::Point & curPoint = datapoints[i];
        const double curPointX = curPoint.x;
        const double curPointY = curPoint.y * -1; //transform from image coordinates to geometric coordinates

        double angleToEllipseCenter = atan2(curPointY - ellipseCenterY * -1, curPointX - ellipseCenterX); //ellipseCenterY * -1 for transformation to geometric coords (from image coords)

        double nearestEllipseX = ellipseCenterX + ellipseRadiusX * cos(angleToEllipseCenter);
        double nearestEllipseY = ellipseCenterY * -1 + ellipseRadiusY * sin(angleToEllipseCenter); //ellipseCenterY * -1 for transformation to geometric coords (from image coords)


        cv::Point center(ellipseCenterX, ellipseCenterY);
        cv::Size axes(ellipseRadiusX, ellipseRadiusY);
        cv::ellipse(drawing, center, axes, 0, 0, 360, cv::Scalar(255));
        cv::line(drawing, curPoint, cv::Point(nearestEllipseX,nearestEllipseY*-1), cv::Scalar(180));

    }
    cv::namedWindow( "ellipse", CV_WINDOW_AUTOSIZE );
    cv::imshow( "ellipse", drawing );
    cv::waitKey(0);
}

它产生以下图像:

您可以看到它实际上在椭圆上找到近"点，但不是最近"点.我故意想要的是:(对不起，我的绘画不好)

You can see that it actually finds "near" points on the ellipse, but it are not the "nearest" points. What I intentionally want is this: (excuse my poor drawing)

您可以扩展上一张图像中的线条，它们会越过椭圆的中心，但是上一张图像中的线条却不是这种情况.
希望你能明白.谁能告诉我我在做什么错?

would you extent the lines in the last image, they would cross the center of the ellipse, but this is not the case for the lines in the previous image.
I hope you get the picture. Can anyone tell me what I am doing wrong?

diagram

给定点到给定椭圆的距离

问题描述

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