如何从旋转角度计算OpenCV的透视变换?

Suz*_*ioc 15 c++ math opencv transformation perspectivecamera

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

怎么做?

我在OE上找到了代码.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;

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);
}
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但不幸的是,这种转变确实很奇怪

在此输入图像描述

为什么?什么是上面的另一半图像alpha>0?以及如何围绕其他轴旋转?为什么dist这么奇怪?

Iwi*_*ist 40

我有时间思考数学和代码.我是在一两年前做过的.我甚至在漂亮的乳胶中排版了这个.

我有意设计了我的解决方案,这样无论提供什么样的旋转角度,整个输入图像都包含在输出框内,居中,否则是黑色.

我的warpImage函数的参数是所有3个轴的旋转角度,比例因子和垂直视场角度.该函数在输出图像内输出扭曲矩阵,输出图像和源图像的角.

数学(代码见下)

第1页 在此输入图像描述

LaTeX源代码在这里.

守则(数学,见上)

这是一个扭曲相机的测试应用程序

#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);
    }
}
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  • @IwillnotexistIdonotexist 再次感谢。这是我在 python 中编写的版本 https://nbviewer.jupyter.org/github/manisoftwartist/perspectiveproj/blob/master/perspective.ipynb 我很想知道您对此有何看法。其次,我想知道如何结合相机内在/外在矩阵来模拟手机相机看场景(在这种情况下是源图像)。 (5认同)
  • 非常感谢。一个小问题。为什么用斜边而不是 d/(2 * tg(fov/2)) 来平移 z (2认同)
  • @IwillnotexistIdonotexist 感谢您的快速回复。我正在尝试在 python 中实现相同的功能。我发现了这个错字。除此之外,我知道`ptsIn`是对象的坐标,参考系原点以(0,0,0)为中心。但是,我不太明白什么是 ```ptsInPt2f``` 和 ```ptOutPt2f```。你能帮我解决这个问题吗? (2认同)
  • @hAcKnRoCk 酷!关于`ptsInPt2f`和`ptsOutPt2f`:`ptsInPt2f`是原图`(TL, TR, BR, BL)`中角点的四个2D坐标,是一个与原图大小相同的矩形原始图片。`ptsOutPt2f` 是输出图像中扭曲目标各个角的四个 2D 坐标。它们用于计算输出透视变换“M”(以及“warpPerspective()”的输入)以扭曲源图像。 (2认同)