我需要一个函数,它给定 Yaw、Pitch 和 Roll,可以在“世界坐标”中生成 Front(或 Looking At)、Right 和 Up vector 。
在我的特定世界空间中,从原点 (0,0,0) 开始,X 向左为正,Z 为正,远离观察者/原点,Y 为正向上。
例如,给定...(以度为单位的角度)
偏航=0,俯仰=0,横滚=0, 预期的输出是:
- 前面 = (0.0,0.0,1.0)
- 右 = (-1.0,0.0,0.0)
- 向上 = (0.0,1.0,0.0)
偏航=90,俯仰=0,横滚=0, 预期的输出是:
- 前面 = (1.0,0.0,0.0)
- 右 = (0,0,0.0,1.0)
- 向上 = (0.0,1.0,0.0)
偏航=0,俯仰=90,横滚=0, 预期的输出是:
- 前面 = (0.0,1.0,0.0)
- 右 = (-1.0,0.0,0.0)
- 向上 = (0.0,0.0,-1.0)
偏航=0,俯仰=0,横滚=90, 预期的输出是:
- 前面 = (0.0,0.0,1.0)
- 右 = (0.0,1.0,0.0)
- 向上 = (1.0,0.0,0.0)
我使用的语言是 C++,如果最有意义的话,我很乐意使用 glm 来解决这个问题。如果我可以通过四元数到达那里,我也可以使用该解决方案,因为我发现其他教程描述了如何从欧拉角获得四元数。
最佳答案
这是一个完整的工作示例。它不是很像 C++。您可能想要使用真正的矩阵类,但出于演示目的应该没问题。您的问题中不清楚的一件事是轮换顺序,但这很容易更改。
#include <iostream>
#include <cmath>
#include <cstdlib>
typedef float Float;
typedef Float Axis[3];
typedef Axis Axes[3];
static void copy(const Axes &from,Axes &to)
{
for (size_t i=0; i!=3; ++i) {
for (size_t j=0; j!=3; ++j) {
to[i][j] = from[i][j];
}
}
}
static void mul(Axes &mat,Axes &b)
{
Axes result;
for (size_t i=0; i!=3; ++i) {
for (size_t j=0; j!=3; ++j) {
Float sum = 0;
for (size_t k=0; k!=3; ++k) {
sum += mat[i][k]*b[k][j];
}
result[i][j] = sum;
}
}
copy(result,mat);
}
static void getAxes(Axes &result,Float yaw,Float pitch,Float roll)
{
Float x = -pitch;
Float y = yaw;
Float z = -roll;
Axes matX = {
{1, 0, 0 },
{0, cos(x),sin(x)},
{0,-sin(x),cos(x)}
};
Axes matY = {
{cos(y),0,-sin(y)},
{ 0,1, 0},
{sin(y),0, cos(y)}
};
Axes matZ = {
{ cos(z),sin(z),0},
{-sin(z),cos(z),0},
{ 0, 0,1}
};
Axes axes = {
{1,0,0},
{0,1,0},
{0,0,1}
};
mul(axes,matX);
mul(axes,matY);
mul(axes,matZ);
copy(axes,result);
}
static void showAxis(const char *desc,const Axis &axis,Float sign)
{
std::cout << " " << desc << " = (";
for (size_t i=0; i!=3; ++i) {
if (i!=0) {
std::cout << ",";
}
std::cout << axis[i]*sign;
}
std::cout << ")\n";
}
static void showAxes(const char *desc,Axes &axes)
{
std::cout << desc << ":\n";
showAxis("front",axes[2],1);
showAxis("right",axes[0],-1);
showAxis("up",axes[1],1);
}
int main(int,char**)
{
Axes axes;
std::cout.setf(std::ios::fixed);
std::cout.precision(1);
getAxes(axes,0,0,0);
showAxes("yaw=0, pitch=0, roll=0",axes);
getAxes(axes,M_PI/2,0,0);
showAxes("yaw=90, pitch=0, roll=0",axes);
getAxes(axes,0,M_PI/2,0);
showAxes("yaw=0, pitch=90, roll=0",axes);
getAxes(axes,0,0,M_PI/2);
showAxes("yaw=0, pitch=0, roll=90",axes);
return 0;
}
关于c++ - 如何将欧拉角转换为前、上、右 vector ,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16074344/