如何从旋转角度计算OpenCV的透视变换？ [英] How to calculate perspective transform for OpenCV from rotation angles?
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问题描述
我想计算透视变换（warpPerspective函数的矩阵），从旋转角度和到对象的距离。
如何做？ >
我在OE的某处找到了代码。示例程序如下：
#include< opencv2 / objdetect / objdetect.hpp>
#include< opencv2 / highgui / highgui.hpp>
#include< opencv2 / imgproc / imgproc.hpp>
#include< iostream>
#include< math.h>
using namespace std;
using namespace cv;
垫框架;
int alpha_int;
int dist_int;
int f_int;
double w;
double h;
double alpha;
double dist;
double f;
void redraw（）{
alpha =（double）alpha_int / 1000 .;
// dist = 1 /（dist_int+1）;
// dist = dist_int + 1;
dist = dist_int-50;
f = f_int + 1;
cout<< alpha =<< α< endl;
cout<< dist =< dist<< endl;
cout<< f =< f<< endl;
//投影2D - > 3D矩阵
Mat A1 =（Mat_< double>（4,3）<<
1,0，-w / 2，
0，1，-h / b $ b 0，0，1，
0，0，1）;
//围绕X轴的旋转矩阵
Mat R =（Mat_< double>（4,4）<<
1，0，0，0，
0，cos（alpha），-sin（alpha），0，
0，sin（alpha），cos（alpha），0，
0，0，0，
// Z轴上的转换矩阵
Mat T =（Mat_< double>（4,4）<<
1，0，0，0，
0，1，0，0，
0，0，1，dist，
0，0，0，1）
// Camera Intrisecs matrix 3D - > 2D
Mat A2 =（Mat_< double>（3,4）<<
f，0，w / 2,0，
0，f，h / b $ b 0，0，1，0）;
Mat m = A2 *（T *（R * A1））;
cout<< R =<< endl<< R < endl;
cout<< A1 =<< endl<< A1<< endl;
cout<< R * A1 =< endl<< （R * A1）< endl;
cout<< T =<< endl<< T < endl;
cout<< T *（R * A1）= endl<< （T *（R * A1））< endl;
cout<< A2 =<< endl<< A2 << endl;
cout<< A2 *（T *（R * A1））=< endl<< （A2 *（T *（R * A1））） endl;
cout<< m =< endl<< m < endl;
Mat frame1;
warpPerspective（frame，frame1，m，frame.size（），INTER_CUBIC | WARP_INVERSE_MAP）;
imshow（Frame，frame）;
imshow（Frame1，frame1）;
}
void callback（int，void *）{
redraw（）;
}
void main（）{
frame = imread（FruitSample_small.png，CV_LOAD_IMAGE_COLOR）;
imshow（Frame，frame）;
w = frame.size（）。width;
h = frame.size（）。height;
createTrackbar（alpha，Frame，& alpha_int，100& callback）;
dist_int = 50;
createTrackbar（dist，Frame，& dist_int，100，& callback）;
createTrackbar（f，Frame，& f_int，100，& callback）;
redraw（）;
waitKey（-1）;
}
但不幸的是，这种转换有些奇怪。
>
为什么？当 alpha> 0 时，上图的另一半是什么？如何围绕其他轴旋转？为什么 dist 这么奇怪？
解决方案
时间来思考数学和代码。我这一年或两年前做过。我甚至排版在美丽的LaTeX。
我故意设计我的解决方案，使得无论提供什么旋转角度，整个输入图像包含，居中，在输出框架，否则为黑色。
我的 warpImage 函数的参数是所有3个轴的旋转角度，比例因子和垂直场 - 的视角。该函数输出经线矩阵，输出图像和输出图像中源图像的角。
数学（代码如下）
LaTeX源代码是在这里。
代码（数学，见上文）
这是一个测试应用程序， p>
#include< opencv2 / core / core.hpp>
#include< opencv2 / imgproc / imgproc.hpp>
#include< opencv2 / highgui / highgui.hpp>
#include< math.h>
使用命名空间cv;
using namespace std;
static double rad2Deg（double rad）{return rad *（180 / M_PI）;} //将弧度转换为度
static double deg2Rad（double deg）{return deg *（M_PI / 180）;} //将度数转换为弧度
void warpMatrix（Size sz，
double theta，
double phi，
double gamma，
double scale，
double fovy，
Mat& M，
vector< Point2f> * corner）{
double st = sin（deg2Rad（theta））;
double ct = cos（deg2Rad（theta））;
double sp = sin（deg2Rad（phi））;
double cp = cos（deg2Rad（phi））;
double sg = sin（deg2Rad（gamma））;
double cg = cos（deg2Rad（gamma））;
double halfFovy = fovy * 0.5;
double d = hypot（sz.width，sz.height）;
double sideLength = scale * d / cos（deg2Rad（halfFovy））;
double h = d /（2.0 * sin（deg2Rad（halfFovy）））;
double n = h-（d / 2.0）;
double f = h +（d / 2.0）;
Mat F = Mat（4,4，CV_64FC1）; //分配4x4转换矩阵F
Mat Rtheta = Mat :: eye（4,4，CV_64FC1）; //分配4x4旋转矩阵围绕Z轴由θ度量
Mat Rphi = Mat :: eye（4,4，CV_64FC1）; //分配围绕X轴的4x4旋转矩阵phi度
Mat Rgamma = Mat： ：eye（4,4，CV_64FC1）; //分配围绕Y轴的4x4旋转矩阵为gamma度
Mat T = Mat :: eye（4,4，CV_64FC1）; //分配4x4沿Z轴的平移矩阵-h单位
Mat P = Mat :: zeros（4,4，CV_64FC1）; //分配4x4投影矩阵
// Rtheta
R double（双面）（0,0）= R双（1,1）= ct;
R itta.at double（0,1）= - st; R itta.at double（1,0）= st;
// Rphi
Rphi.at double（1,1）= Rphi.at double（2,2）= cp;
Rphi.at double（1,2）= - sp; Rphi.at double（2,1）= sp;
// Rgamma
Rgamma.at< double>（0,0）= Rgamma.at double（2,2）= cg;
Rgamma.at< double>（0,2）= sg; Rgamma.at double（2,0）= sg;
// T
T.at< double>（2,3）= - h;
// P
P.at double（0,0）= P.at double（1,1）= 1.0 / tan（deg2Rad（halfFovy））;
P.at double（2,2）= - （f + n）/（f-n）;
P.at double（2,3）= - （2.0 * f * n）/（f-n）;
P.at< double>（3,2）= - 1.0;
//组合转换
F = P * T * Rphi * Rtheta * Rgamma; //矩阵乘以产生主矩阵
//转换4x4分
double ptsIn [4 * 3];
double ptsOut [4 * 3];
double halfW = sz.width / 2，halfH = sz.height / 2;
ptsIn [0] = - halfW; ptsIn [1] = halfH;
ptsIn [3] = halfW; ptsIn [4] = halfH;
ptsIn [6] = halfW; ptsIn [7] = - halfH;
ptsIn [9] = - halfW; ptsIn [10] = - halfH;
ptsIn [2] = ptsIn [5] = ptsIn [8] = ptsIn [11] = 0; //为所有4个组件设置Z分量为零
Mat ptsInMat（1， 4，CV_64FC3，ptsIn）;
Mat ptsOutMat（1,4，CV_64FC3，ptsOut）;
perspectiveTransform（ptsInMat，ptsOutMat，F）; //转换点
//获取3x3变换和变形图像
Point2f ptsInPt2f [4]
Point2f ptsOutPt2f [4];
for（int i = 0; i <4; i ++）{
Point2f ptIn（ptsIn [i * 3 + 0]，ptsIn [i * 3 + 1]）;
Point2f ptOut（ptsOut [i * 3 + 0]，ptsOut [i * 3 + 1]）;
ptsInPt2f [i] = ptIn + Point2f（halfW，halfH）;
ptsOutPt2f [i] =（ptOut + Point2f（1,1））*（sideLength * 0.5）;
}
M = getPerspectiveTransform（ptsInPt2f，ptsOutPt2f）;
//加载角矢量
if（corners）{
corners-> clear（）;
corners-> push_back（ptsOutPt2f [0]）; // Push Top left corner
corners-> push_back（ptsOutPt2f [1]）; // Push Top right corner
corners- > push_back（ptsOutPt2f [2]）; // Push Bottom Right corner
corners-> push_back（ptsOutPt2f [3]）; // Push Bottom Left corner
}
}
void warpImage（const Mat& src，
double theta，
double phi，
double gamma，
double scale，
double fovy，
Mat& dst，
Mat& M，
vector< Point2f>& corners）{
double halfFovy = fovy * 0.5;
double d = hypot（src.cols，src.rows）;
double sideLength = scale * d / cos（deg2Rad（halfFovy））;
warpMatrix（src.size（），theta，phi，gamma，scale，fovy，M，& corner）; //计算warp矩阵
warpPerspective（src，dst， Size（sideLength，sideLength））; //做实际的图像扭曲
}
int main（void）{
Mat m，disp，warp;
vector< Point2f>角落
VideoCapture cap（0）;
while（cap.isOpened（））{
cap>> m;
warpImage（m，5，50，0，1，30，disp，warp，corners）;
imshow（Disp，disp）;
}
}
I want to calculate perspective transform (a matrix for warpPerspective function) starting from angles of rotation and distance to the object.
How to do that?
I found the code somewhere on OE. Sample program is below:
#include <opencv2/objdetect/objdetect.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <opencv2/imgproc/imgproc.hpp>
#include <iostream>
#include <math.h>
using namespace std;
using namespace cv;
Mat frame;
int alpha_int;
int dist_int;
int f_int;
double w;
double h;
double alpha;
double dist;
double f;
void redraw() {
alpha = (double)alpha_int/1000.;
//dist = 1./(dist_int+1);
//dist = dist_int+1;
dist = dist_int-50;
f = f_int+1;
cout << "alpha = " << alpha << endl;
cout << "dist = " << dist << endl;
cout << "f = " << f << endl;
// Projection 2D -> 3D matrix
Mat A1 = (Mat_<double>(4,3) <<
1, 0, -w/2,
0, 1, -h/2,
0, 0, 1,
0, 0, 1);
// Rotation matrices around the X axis
Mat R = (Mat_<double>(4, 4) <<
1, 0, 0, 0,
0, cos(alpha), -sin(alpha), 0,
0, sin(alpha), cos(alpha), 0,
0, 0, 0, 1);
// Translation matrix on the Z axis
Mat T = (Mat_<double>(4, 4) <<
1, 0, 0, 0,
0, 1, 0, 0,
0, 0, 1, dist,
0, 0, 0, 1);
// Camera Intrisecs matrix 3D -> 2D
Mat A2 = (Mat_<double>(3,4) <<
f, 0, w/2, 0,
0, f, h/2, 0,
0, 0, 1, 0);
Mat m = A2 * (T * (R * A1));
cout << "R=" << endl << R << endl;
cout << "A1=" << endl << A1 << endl;
cout << "R*A1=" << endl << (R*A1) << endl;
cout << "T=" << endl << T << endl;
cout << "T * (R * A1)=" << endl << (T * (R * A1)) << endl;
cout << "A2=" << endl << A2 << endl;
cout << "A2 * (T * (R * A1))=" << endl << (A2 * (T * (R * A1))) << endl;
cout << "m=" << endl << m << endl;
Mat frame1;
warpPerspective( frame, frame1, m, frame.size(), INTER_CUBIC | WARP_INVERSE_MAP);
imshow("Frame", frame);
imshow("Frame1", frame1);
}
void callback(int, void* ) {
redraw();
}
void main() {
frame = imread("FruitSample_small.png", CV_LOAD_IMAGE_COLOR);
imshow("Frame", frame);
w = frame.size().width;
h = frame.size().height;
createTrackbar("alpha", "Frame", &alpha_int, 100, &callback);
dist_int = 50;
createTrackbar("dist", "Frame", &dist_int, 100, &callback);
createTrackbar("f", "Frame", &f_int, 100, &callback);
redraw();
waitKey(-1);
}
But unfortunately, this transform does something strange
Why? What is another half of image above when alpha>0? And how to rotate around other axes? Why dist works so strange?
解决方案
I have had the luxury of time to think out both math and code. I did this a year or two ago. I even typeset this in beautiful LaTeX.
I intentionally designed my solution so that no matter what rotation angles are provided, the entire input image is contained, centered, within the output frame, which is otherwise black.
The arguments to my warpImage function are rotation angles in all 3 axes, the scale factor and the vertical field-of-view angle. The function outputs the warp matrix, the output image and the corners of the source image within the output image.
The Math (for code, look below)
The LaTeX source code is here.
The Code (for math, look above)
Here is a test application that warps the camera
#include <opencv2/core/core.hpp>
#include <opencv2/imgproc/imgproc.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <math.h>
using namespace cv;
using namespace std;
static double rad2Deg(double rad){return rad*(180/M_PI);}//Convert radians to degrees
static double deg2Rad(double deg){return deg*(M_PI/180);}//Convert degrees to radians
void warpMatrix(Size sz,
double theta,
double phi,
double gamma,
double scale,
double fovy,
Mat& M,
vector<Point2f>* corners){
double st=sin(deg2Rad(theta));
double ct=cos(deg2Rad(theta));
double sp=sin(deg2Rad(phi));
double cp=cos(deg2Rad(phi));
double sg=sin(deg2Rad(gamma));
double cg=cos(deg2Rad(gamma));
double halfFovy=fovy*0.5;
double d=hypot(sz.width,sz.height);
double sideLength=scale*d/cos(deg2Rad(halfFovy));
double h=d/(2.0*sin(deg2Rad(halfFovy)));
double n=h-(d/2.0);
double f=h+(d/2.0);
Mat F=Mat(4,4,CV_64FC1);//Allocate 4x4 transformation matrix F
Mat Rtheta=Mat::eye(4,4,CV_64FC1);//Allocate 4x4 rotation matrix around Z-axis by theta degrees
Mat Rphi=Mat::eye(4,4,CV_64FC1);//Allocate 4x4 rotation matrix around X-axis by phi degrees
Mat Rgamma=Mat::eye(4,4,CV_64FC1);//Allocate 4x4 rotation matrix around Y-axis by gamma degrees
Mat T=Mat::eye(4,4,CV_64FC1);//Allocate 4x4 translation matrix along Z-axis by -h units
Mat P=Mat::zeros(4,4,CV_64FC1);//Allocate 4x4 projection matrix
//Rtheta
Rtheta.at<double>(0,0)=Rtheta.at<double>(1,1)=ct;
Rtheta.at<double>(0,1)=-st;Rtheta.at<double>(1,0)=st;
//Rphi
Rphi.at<double>(1,1)=Rphi.at<double>(2,2)=cp;
Rphi.at<double>(1,2)=-sp;Rphi.at<double>(2,1)=sp;
//Rgamma
Rgamma.at<double>(0,0)=Rgamma.at<double>(2,2)=cg;
Rgamma.at<double>(0,2)=sg;Rgamma.at<double>(2,0)=sg;
//T
T.at<double>(2,3)=-h;
//P
P.at<double>(0,0)=P.at<double>(1,1)=1.0/tan(deg2Rad(halfFovy));
P.at<double>(2,2)=-(f+n)/(f-n);
P.at<double>(2,3)=-(2.0*f*n)/(f-n);
P.at<double>(3,2)=-1.0;
//Compose transformations
F=P*T*Rphi*Rtheta*Rgamma;//Matrix-multiply to produce master matrix
//Transform 4x4 points
double ptsIn [4*3];
double ptsOut[4*3];
double halfW=sz.width/2, halfH=sz.height/2;
ptsIn[0]=-halfW;ptsIn[ 1]= halfH;
ptsIn[3]= halfW;ptsIn[ 4]= halfH;
ptsIn[6]= halfW;ptsIn[ 7]=-halfH;
ptsIn[9]=-halfW;ptsIn[10]=-halfH;
ptsIn[2]=ptsIn[5]=ptsIn[8]=ptsIn[11]=0;//Set Z component to zero for all 4 components
Mat ptsInMat(1,4,CV_64FC3,ptsIn);
Mat ptsOutMat(1,4,CV_64FC3,ptsOut);
perspectiveTransform(ptsInMat,ptsOutMat,F);//Transform points
//Get 3x3 transform and warp image
Point2f ptsInPt2f[4];
Point2f ptsOutPt2f[4];
for(int i=0;i<4;i++){
Point2f ptIn (ptsIn [i*3+0], ptsIn [i*3+1]);
Point2f ptOut(ptsOut[i*3+0], ptsOut[i*3+1]);
ptsInPt2f[i] = ptIn+Point2f(halfW,halfH);
ptsOutPt2f[i] = (ptOut+Point2f(1,1))*(sideLength*0.5);
}
M=getPerspectiveTransform(ptsInPt2f,ptsOutPt2f);
//Load corners vector
if(corners){
corners->clear();
corners->push_back(ptsOutPt2f[0]);//Push Top Left corner
corners->push_back(ptsOutPt2f[1]);//Push Top Right corner
corners->push_back(ptsOutPt2f[2]);//Push Bottom Right corner
corners->push_back(ptsOutPt2f[3]);//Push Bottom Left corner
}
}
void warpImage(const Mat &src,
double theta,
double phi,
double gamma,
double scale,
double fovy,
Mat& dst,
Mat& M,
vector<Point2f> &corners){
double halfFovy=fovy*0.5;
double d=hypot(src.cols,src.rows);
double sideLength=scale*d/cos(deg2Rad(halfFovy));
warpMatrix(src.size(),theta,phi,gamma, scale,fovy,M,&corners);//Compute warp matrix
warpPerspective(src,dst,M,Size(sideLength,sideLength));//Do actual image warp
}
int main(void){
Mat m, disp, warp;
vector<Point2f> corners;
VideoCapture cap(0);
while(cap.isOpened()){
cap >> m;
warpImage(m, 5, 50, 0, 1, 30, disp, warp, corners);
imshow("Disp", disp);
}
}
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