如何将欧拉角转换为前，上，右向量 [英] How to convert Euler Angles to Front, Up, Right vectors

问题描述

我需要一个给定偏航角，俯仰角和横滚角的函数，可以在世界坐标"中产生前"(或望")，右"和上"向量.

I need a function that given Yaw, Pitch, and Roll, can produce the Front (or Looking At), Right, and Up vectors in "world coordinates".

在我的特定世界空间中，从原点(0,0,0)开始，X是向左的正数，Z是远离观察者/原点的正数，Y是向上的正数.

In my particular world space, starting from the origin (0,0,0), X is positive to the left, Z is positive going away from the viewer/origin, and Y is positive going up.

例如，给定...(角度)

For for example, given... (angles in degrees)

• 偏航= 0，俯仰= 0，侧倾= 0， 预期的输出是:

• yaw=0, pitch=0, roll=0, the expected output is:

• front =(0.0,0.0,1.0)
• right =(-1.0,0.0,0.0)
• up =(0.0,1.0,0.0)

偏航= 90，俯仰= 0，侧倾= 0， 预期的输出是:

yaw=90, pitch=0, roll=0, the expected output is:

• front =(1.0,0.0,0.0)
• right =(0,0,0.0,1.0)
• up =(0.0,1.0,0.0)

偏航= 0，俯仰= 90，侧倾= 0， 预期的输出是:

yaw=0, pitch=90, roll=0, the expected output is:

• front =(0.0,1.0,0.0)
• right =(-1.0,0.0,0.0)
• 向上=(0.0,0.0，-1.0)

偏航= 0，俯仰= 0，侧倾= 90， 预期的输出是:

yaw=0, pitch=0, roll=90, the expected output is:

• front =(0.0,0.0,1.0)
• right =(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，我将很乐意使用glm解决此问题.如果我能通过四元数到达那里，那我也很满意该解决方案，因为我找到了其他教程来描述如何从欧拉角获得四元数.

The language I'm working in is C++, and I will gladly use glm to solve this problem if that makes the most sense. If I can get there through quaternion's I'm fine with that solution as well, since I've found other tutorials that describe how to get a quaternion from euler angles.

推荐答案

下面是一个完整的示例.它不是很像C ++.您可能希望使用一个真实的矩阵类，但是出于演示目的应该没问题.您的问题尚不清楚的一件事是轮换顺序，但可以轻松更改.

Here is a full working example. It isn't very C++-like. You would probably want to use a real matrix class, but it should be ok for demonstration purposes. One thing that isn't clear from your question is the rotation order, but that can easily be changed.

#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;
}

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