从Tait-Bryan角度的四元数

问题描述

我使用SharpDX编写相机，并在四元数的帮助下旋转相机。

I am writing a camera with SharpDX and rotate it with the help of a quaternion.

• 相机旋转设置为pitch X旋转），偏航（Y旋转）和滚动（Z旋转），因此称为Tiat-Bryan角度（这些不欧拉角，其将具有XYX旋转，而不是XYZ） / li>
  
• 我使用的是左手坐标系：X +向右，Y +向上，Z +向下。


• （俯仰），正值使相机向下看。如果我绕Y旋转（偏航），正值使相机向左看。如果我绕Z旋转（滚动），相机顺时针旋转。

• The camera rotation is set with pitch (X rotation), yaw (Y rotation) and roll (Z rotation), so called "Tiat-Bryan" angles (these are not Euler angles which would have an XYX rotation, and not XYZ).
• I am using a left-handed coordinate system: X+ is rightwards, Y+ is upwards, Z+ is downwards the screen.
• If I rotate around X ("pitching"), positive values make the camera look downwards. If I rotate around Y ("yawing"), positive values make the camera look leftwards. If I rotate around Z ("rolling"), the camera rolls clockwise.

我已经能够获得俯仰，从四元数。以下C＃代码：

I was already able to get pitch, yaw and roll from the quaternion, s. the following C# code:

public static class QuaternionExtensions
{
    public static Vector3 GetPitchYawRoll(this Quaternion q)
    {
        return new Vector3(q.GetPitch(), q.GetYaw(), q.GetRoll());
    }

    public static float GetPitch(this Quaternion q)
    {
        return q.GetK().GetPitch();
    }

    public static float GetYaw(this Quaternion q)
    {
        return q.GetK().GetYaw();
    }

    public static float GetRoll(this Quaternion q)
    {
        // This is M12 * M22 of rotation matrix
        float xx = q.X * q.X;
        float xy = q.X * q.Y;
        float zz = q.Z * q.Z;
        float wz = q.W * q.Z;
        return (float)Math.Atan2(2f * (xy - wz), 1f - 2f * (xx + zz));
    }

    public static Vector3 GetK(this Quaternion q)
    {
        float xz = q.X * q.Z;
        float wy = q.W * q.Y;
        float yz = q.Y * q.Z;
        float wx = q.W * q.X;
        float xx = q.X * q.X;
        float yy = q.Y * q.Y;
        return new Vector3(
            2f * (xz - wy),
            2f * (yz + wx),
            1f - 2f * (xx + yy));
    }

}

public static class Vector3Extensions
{
    public static float GetPitch(this Vector3 v)
    {
        return (float)-Math.Atan2(v.Y, Math.Sqrt(v.X * v.X + v.Z * v.Z));
    }

    public static float GetYaw(this Vector3 v)
    {
        return (float)-Math.Atan2(v.X, v.Z);
    }
}

我不知道其他方法。如何通过指定俯仰，偏航和滚动来获取四元数（例如，从 GetPitchYawRoll 中返回的内容）？

I don't know the other way round. How to get the quaternion by specifying pitch, yaw and roll (e.g. what's returned in the Vector3 from GetPitchYawRoll)?

推荐答案

来自BlitzBasic.com论坛的Kryzon帮助我在这里（s。他的帖子）：

Kryzon from the BlitzBasic.com forum helped me out here (s. his post at http://blitzbasic.com/Community/post.php?topic=102458&post=1224801):

internal static class QuaternionExtensions
{
    internal static Vector3 GetYawPitchRollVector(this Quaternion q)
    {
        return new Vector3(q.GetYaw(), q.GetPitch(), q.GetRoll());
    }

    private static float GetYaw(this Quaternion q)
    {
        float x2 = q.X * q.X;
        float y2 = q.Y * q.Y;
        return (float)Math.Atan2(2f * q.Y * q.W - 2f * q.Z * q.X, 1f - 2f * y2 - 2f * x2);
    }

    private static float GetPitch(this Quaternion q)
    {
        return (float)-Math.Asin(2f * q.Z * q.Y + 2f * q.X * q.W);
    }

    private static float GetRoll(this Quaternion q)
    {
        float x2 = q.X * q.X;
        float z2 = q.Z * q.Z;
        return (float)-Math.Atan2(2f * q.Z * q.W - 2f * q.Y * q.X, 1f - 2f * z2 - 2f * x2);
    }
}

internal static class Vector3Extensions
{
    internal static Quaternion GetYawPitchRollQuaternion(this Vector3 v)
    {
        float c1 = (float)Math.Cos(-v.Z / 2.0);
        float c2 = (float)Math.Cos(-v.Y / 2.0);
        float c3 = (float)Math.Cos( v.X / 2.0);
        float c1c2 = c1 * c2;
        float s1 = (float)Math.Sin(-v.Z / 2.0);
        float s2 = (float)Math.Sin(-v.Y / 2.0);
        float s3 = (float)Math.Sin( v.X / 2.0);
        float s1s2 = s1 * s2;

        return new Quaternion(
            c1 * s2 * c3 - s1 * c2 * s3,
            c1c2 * s3 + s1s2 * c3,
            s1 * c2 * c3 + c1 * s2 * s3,
            c1c2 * c3 - s1s2 * s3);
    }
}

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