Transformation of coordinates, Concept - matlab

Question

i want to convert x,y,z coordinates to polar coordinates. I am getting (-) in y coordiantes. Can someone explain me why I am getting it. It would be great help.
I am reading these values (xyz , az_elev_r) from a software and can't be changed.I am just not sure of the order of angles( az & elev). Using my code I get -y instead of y. It means there is 180 rotation.My code is:
xyz=[-0.564 3.689 -0.735;
2.011 5.067 -1.031;
-1.181 3.943 -1.825; % Reference values
];
%% az_elev_r-->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

Your az_elev_r matrix is not consistent with your xyz reference.
>> [az, el, r] = cart2sph(xyz(:,1), xyz(:,2), xyz(:,3));
>> rad2deg(az)
ans =
98.6924675475501
68.3527736950233
106.673911589314
Your answers are consistent with the values returned by the sph2cart function. (Example starts with your original input, before the dec2rad replacement.
>> [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
Incidentally, you're code will be more readable if you just use the sph2cart function, and work in radians, unless you are trying to understand the conversions for their own sake.

Answer 2

OpenCV has the code for conversion to polar coordinates and back. This conversion is useful for finding object rotation through correlation or otherwise creating object-centred 'rotation-independent' representation of objects. It is useful to visualize each of the polar coordinates as well as their joint image. The images below should be self_explanatory. The polar plot has angle as a horizontal axis and Radius as a vertical axis, so that 4 peaks correspond to the 4 corners of the input image. The code (C++ with OpenCV) is attached.
//================================
// 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;
}