目的:练习类的定义和实现,构造函数的使用。 功能:输入地面点坐标,在给定内、外方位元素的情况下,输出对应像点坐标。 要求:地面点坐标的输入利用构造函数。用c++实现
时间: 2024-05-02 09:16:39 浏览: 148
利用C#编写一个坐标方位角计算程序
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以下是一个简单的类定义和实现,用于输入地面点坐标并输出对应的像点坐标:
```
#include <iostream>
#include <cmath>
using namespace std;
class GroundToImage {
private:
double xg, yg, zg, x0, y0, z0, phi, omega, kappa;
double f, Xs, Ys;
public:
GroundToImage(double xg, double yg, double zg, double x0, double y0, double z0, double phi, double omega, double kappa, double f, double Xs, double Ys) {
this->xg = xg;
this->yg = yg;
this->zg = zg;
this->x0 = x0;
this->y0 = y0;
this->z0 = z0;
this->phi = phi;
this->omega = omega;
this->kappa = kappa;
this->f = f;
this->Xs = Xs;
this->Ys = Ys;
}
void groundToImage() {
double X = xg - x0;
double Y = yg - y0;
double Z = zg - z0;
double cosPhi = cos(phi);
double sinPhi = sin(phi);
double cosOmega = cos(omega);
double sinOmega = sin(omega);
double cosKappa = cos(kappa);
double sinKappa = sin(kappa);
double X1 = cosOmega * X + sinOmega * sinPhi * Y + sinOmega * cosPhi * Z;
double Y1 = -sinOmega * cosKappa * X + cosOmega * cosKappa * sinPhi * Y + cosOmega * cosKappa * cosPhi * Z + sinKappa * Xs;
double Z1 = sinKappa * X - cosKappa * sinPhi * Y - cosKappa * cosPhi * Z + f;
double u = f * X1 / Z1 + Xs;
double v = f * Y1 / Z1 + Ys;
cout << "像点坐标为:(" << u << ", " << v << ")" << endl;
}
};
int main() {
double xg, yg, zg, x0, y0, z0, phi, omega, kappa, f, Xs, Ys;
cout << "请输入地面点坐标(xg, yg, zg): ";
cin >> xg >> yg >> zg;
cout << "请输入内方位元素(x0, y0, z0, phi, omega, kappa): ";
cin >> x0 >> y0 >> z0 >> phi >> omega >> kappa;
cout << "请输入相机定位元素(f, Xs, Ys): ";
cin >> f >> Xs >> Ys;
GroundToImage gt(xg, yg, zg, x0, y0, z0, phi, omega, kappa, f, Xs, Ys);
gt.groundToImage();
return 0;
}
```
在上面的代码中,`GroundToImage` 类有一个构造函数用于输入地面点坐标和相机定位元素。然后,`groundToImage()` 函数可以计算出对应的像点坐标。在 `main()` 函数中,我们首先输入地面点坐标、内方位元素和相机定位元素,然后创建 `GroundToImage` 类的实例,并调用 `groundToImage()` 函数输出像点坐标。
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