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

我想从旋转角度和到对象的距离开始计算透视变换(warpPerspective 函数的矩阵).

I want to calculate perspective transform (a matrix for warpPerspective function) starting from angles of rotation and distance to the object.

怎么做?

我在 OE 的某个地方找到了代码.示例程序如下:

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);
}

但不幸的是,这个转换做了一些奇怪的事情

But unfortunately, this transform does something strange

为什么?当 alpha>0 时,上面图像的另一半是什么?以及如何绕其他轴旋转?为什么 dist 这么奇怪?

Why? What is another half of image above when alpha>0? And how to rotate around other axes? Why dist works so strange?

推荐答案

我有足够的时间来思考数学和代码.一两年前我就这样做了.我什至用漂亮的 LaTeX 排版.

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.

我特意设计了我的解决方案,以便无论提供什么旋转角度,整个输入图像都包含在输出帧内,居中,否则为黑色.

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.

我的 warpImage 函数的参数是所有 3 个轴的旋转角度、比例因子和垂直视野角.该函数输出扭曲矩阵、输出图像和输出图像中源图像的角点.

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.

LaTeX 源代码位于此处.

The LaTeX source code is here.

这是一个扭曲相机的测试应用程序

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){
    int c = 0;
    Mat m, disp, warp;
    vector<Point2f> corners;
    VideoCapture cap(0);

    while(c != 033 && cap.isOpened()){
        cap >> m;
        warpImage(m, 5, 50, 0, 1, 30, disp, warp, corners);
        imshow("Disp", disp);
        c = waitKey(1);
    }
}

这篇关于如何从旋转角度计算 OpenCV 的透视变换?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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