Question

我想将x，y，z坐标转换为极坐标。我在y coordiantes得到（ - ）。有人可以解释我为什么得到它。这将是很大的帮助。

我正在从软件中读取这些值（xyz，az_elev_r）并且无法更改。我只是不确定角度的顺序（az和amp）。使用我的代码我得到-y而不是y。这意味着有180个旋转。我的代码是：

xyz=[-0.564 3.689 -0.735;
      2.011 5.067 -1.031;
     -1.181 3.943 -1.825; % Reference values
    ];

%% az_elev_r - ＆gt; xyz

 az_elev_r=[ 261.30 -11.24 3.80;
               291.65 -10.692 5.548;
               253.34 -23.897 4.50]; % Also Reference (degree)



az_elev_r(:,1:2)=deg2rad(az_elev_r(:,1:2));

r=az_elev_r(:,3);
az=az_elev_r(:,1);
elev=az_elev_r(:,2);


x=r.*cos(az).*cos(elev)
y=r.*sin(az).*cos(elev)
z=r.*sin(elev)

Answer 1

您的az_elev_r矩阵与您的xyz引用不一致。

>> [az, el, r] = cart2sph(xyz(:,1), xyz(:,2), xyz(:,3));
>> rad2deg(az)
ans =
          98.6924675475501
          68.3527736950233
          106.673911589314

您的答案与sph2cart函数返回的值一致。（示例以dec2rad替换之前的原始输入开头。

>> [x, y, z] = sph2cart(deg2rad(az_elev_r(:,1)), deg2rad(az_elev_r(:,2)), az_elev_r(:,3))
x =
         -0.563766229670505
          2.01131973806906
         -1.17951822049783
y =
         -3.68422880893852
         -5.06709019311118
         -3.94153436658676
z =
         -0.740692730942158
         -1.02931719412937
         -1.82292172199717

顺便提一下，如果您只使用sph2cart函数，并且使用弧度工作，那么您的代码将更具可读性，除非您试图了解转换本身。

Answer 2

OpenCV具有转换为极坐标和返回的代码。此转换对于通过关联查找对象旋转或以其他方式创建以对象为中心的“旋转无关”对象表示非常有用。可视化每个极坐标以及它们的关节图像是有用的。下面的图像应该是自我解释的。极坐标图的角度为水平轴，半径为垂直轴，因此4个峰对应于输入图像的4个角。附加了代码（带有OpenCV的C ++）。 Polar coordinate visualization

//================================
// Name        : PolarCoord.cpp
// Author      : V.Ivanchenko cudassimo@gmail.com
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//======================================

#include <iostream>
#include "opencv.hpp"
using namespace std;
using namespace cv;

#define VALID(x, y, w, h) ((x)>=0 && (y)>=0 && (x)<(w) && (y)<(h)) // validates index

/*
 *   1. Original binary image HxW CV_8U
 *              |
 *              |
 *              V
 *   2. Two coordinate Mats HxW CV_32F
 *              |
 *              |
 *              V
 *   3. Visualization CV_8U
 *      a. gray HxW for a single coordinate image
 *      b. binary Rx360 for two coordinate images
 */

// convert a binary 2D image into two Mats with float coordiantes
void imageToCoord(const Mat& img, Mat& X, Mat& Y, bool centered = true) {
    if (img.empty())
        return;

    int h = img.rows;
    int w = img.cols;
    X.create(h, w, CV_32F);
    Y.create(h, w, CV_32F);
    float Cx = w/2.0f;
    float Cy = h/2.0f;

    for (int i=0; i<h; ++i){
        const uchar* img_row = img.ptr<uchar>(i);
        float* x_row = X.ptr<float>(i);
        float* y_row = Y.ptr<float>(i);

        for (int j=0; j<w; ++j) {
            if (img_row[j]>0) {
                float x = j;
                float y = i;
                if (centered) {
                    x-=Cx;
                    y-=Cy;
                }
                x_row[j] = x;
                y_row[j] = y;
            }
        } // j
    } // i
} //imageToCoord()

// convert a single float ploar coord Mat to a gray image
void polarToImg(const Mat& PolarCoord, Mat& img) {
    if (PolarCoord.empty())
        return;

    int h = PolarCoord.rows;
    int w = PolarCoord.cols;
    img.create(h, w, CV_8U);
    float maxVal = std::numeric_limits<float>::min();

    // find maxVal
    for (int i=0; i<h; ++i){
        const float* x_row = PolarCoord.ptr<float>(i);
        for (int j=0; j<w; ++j) {
            if (maxVal < x_row[j])
                maxVal = x_row[j];
        } // j
    } // i

    // create an image
    if (maxVal>0) {
        float k = 255.0/maxVal;
        for (int i=0; i<h; ++i){
            uchar* img_row = img.ptr<uchar>(i);
            const float* x_row = PolarCoord.ptr<float>(i);
            for (int j=0; j<w; ++j) {
                img_row[j] = saturate_cast<uchar>(k*x_row[j]);
            }// j
        } // i
    } // if
} // plarToImg()

// convert two polar coord Mats to a binary image
void polarToImg(const Mat& radius, const Mat& angle, Mat& img) {
    if (angle.empty() || radius.empty())
        return;

    int h = angle.rows;
    int w = angle.cols;
    assert(radius.cols==w && radius.rows==h);
    const int imgH = sqrt(h*h+w*w)+0.5f; // radius
    const int imgW = 360;                 // angle, deg
    img.create(imgH, imgW, CV_8U);

    // create an image
    for (int i=0; i<h; ++i){
        const float* ang_row = angle.ptr<float>(i);
        const float* r_row = radius.ptr<float>(i);

        for (int j=0; j<w; ++j) {
            int x = ang_row[j] + 0.5f;
            int y = r_row[j] + 0.5f;

            if (x>0) {
                cout<<x<<endl;
            }
            if (VALID(x, y, imgW, imgH))
                img.at<uchar>(y, x) = 255;
            else {
                cout<<"Invalid x, y: "<<x<<", "<<y<<endl;
            }
        }// j
    } // i
} // plarToImg()

int main() {
    cout << "Cartesian to polar" << endl; // prints "Syntax training in openCV"
    const int W=400, H=400;
    Mat Minput(H, W, CV_8U);
    Minput(Rect(W/4, H/4, W/2, H/2)) = 255;
    Mat X, Y, Angle, Radius, Mr, Mang, Mpolar;

    // processing
    imageToCoord(Minput, X, Y);             // extract coordinates
    cartToPolar(X, Y, Radius, Angle, true);// convert coordiantes

    // visualize
    polarToImg(Radius, Mr);
    polarToImg(Angle, Mang);
    polarToImg(Radius, Angle, Mpolar);

    // debug
    //cout<<Mpolar<<endl;

    namedWindow("input", 0);
    namedWindow("angle", 0);
    namedWindow("radius", 0);
    namedWindow("Polar", 0);

    const int winw=200, winh=200;
    resizeWindow("input", winw, winh);
    resizeWindow("angle", winw, winh);
    resizeWindow("radius", winw, winh);
    resizeWindow("Polar", 360, (int)sqrt(H*H + W*W));

    moveWindow("input", 0, 0);
    moveWindow("angle", winw, 0);
    moveWindow("radius", 2*winw, 0);
    moveWindow("Polar", 3*winw, 0);

    imshow("input", Minput);
    imshow("angle", Mang);
    imshow("radius", Mr);
    imshow("Polar", Mpolar);
    waitKey(-1);

    return 0;
}

坐标转换，概念

2 个答案: