這是一個完整的工作示例。這不是非常類似C++的。你可能會想要使用一個真正的矩陣類,但它應該可以用於演示目的。有一點你不清楚你的問題是旋轉順序,但可以很容易地改變。
#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;
}
什麼是沒有旋轉的向量? – 2013-04-18 04:20:48
我想你是問,如果我給它一個偏航= 0,音高= 0,滾動= 0,預期的結果是什麼。 在這種情況下的預期結果是: front =(0.0,0.0,1。0) right =( - 1.0,0.0,0.0) up =(0.0,1.0,0.0) 謝謝。 – 2013-04-18 04:43:12
這與偏航結果相同= 90 – 2013-04-18 04:44:50