三维空间任意一点绕任意轴线旋转

  对于三维空间任意一点 P ( p x , p y , p z ) P(p_x,p_y,p_z) P(px​,py​,pz​),求绕任意轴线旋转角度 α \alpha α得到新的点 P ′ ( p x ′ , p y ′ , p z ′ ) P'(p'_x,p'_y,p'_z) P′(px′​,py′​,pz′​)。轴线的单位方向向量为 n ^ ( n x 2 + n y 2 + n z 2 = 1 ) \bm{\hat{n}}(n_x^2+n_y^2+n_z^2=1) n^(nx2​+ny2​+nz2​=1)，且过点 Q ( x 0 , y 0 , z 0 ) Q(x_0,y_0,z_0) Q(x0​,y0​,z0​)。

  轴线 n ^ \bm{\hat{n}} n^在坐标系 X Y Z XYZ XYZ下的直线方程为： { x = x 0 + n x t y = y 0 + n y t z = z 0 + n z t (1) \left\{ \begin{array}{c} x=x_0+n_xt \\ y=y_0+n_yt \\ \tag 1 z=z_0+n_zt\end{array}\right. ⎩⎨⎧​x=x0​+nx​ty=y0​+ny​tz=z0​+nz​t​(1)   弧 P P ′ PP' PP′所在平面在坐标系 X Y Z XYZ XYZ下的平面方程为： n x ( x − p x ) + n y ( y − p y ) + n z ( z − p z ) = 0 (2) n_x(x-p_x)+ n_y(y-p_y)+ n_z(z-p_z)=0 \tag 2 nx​(x−px​)+ny​(y−py​)+nz​(z−pz​)=0(2)   根据式(1)和式(2)，且 n x 2 + n y 2 + n z 2 = 1 n_x^2+n_y^2+n_z^2=1 nx2​+ny2​+nz2​=1，可以求得： t 0 = n x ( p x − x 0 ) + n y ( p y − y 0 ) + n z ( p z − z 0 ) (3) t_0 = n_x(p_x - x_0) + n_y(p_y - y_0) + n_z(p_z -z_0) \tag 3 t0​=nx​(px​−x0​)+ny​(py​−y0​)+nz​(pz​−z0​)(3)   设弧 P P ′ PP' PP′的圆心在坐标系 X Y Z XYZ XYZ下的坐标为 ( x c , y c , z c ) (x_c,y_c,z_c) (xc​,yc​,zc​),将 t 0 t_0 t0​代入式 ( 1 ) (1) (1)，得圆心坐标： { x c = x 0 + n x t 0 y c = y 0 + n y t 0 z c = z 0 + n z t 0 (4) \left\{ \begin{array}{c} x_c=x_0+n_xt_0 \\ y_c=y_0+n_yt_0 \\ \tag 4 z_c=z_0+n_zt_0\end{array}\right. ⎩⎨⎧​xc​=x0​+nx​t0​yc​=y0​+ny​t0​zc​=z0​+nz​t0​​(4)   圆半径 r r r: r = ( p x − x c ) 2 + ( p y − y c ) 2 + ( p z − z c ) 2 (5) r = \sqrt{(p_x-x_c)^2 + (p_y-y_c)^2 + (p_z-z_c)^2} \tag 5 r=(px​−xc​)2+(py​−yc​)2+(pz​−zc​)2 ​(5)   向量 O P \bm{OP} OP: O P = [ p x − x c    p y − y c    p z − z c ] T / r (6) \bm{OP}=[p_x-x_c \ \ p_y-y_c \ \ p_z-z_c]^T / r \tag 6 OP=[px​−xc​  py​−yc​  pz​−zc​]T/r(6)   如上图，建立坐标系 x ′ y ′ z ′ x'y'z' x′y′z′，根据右手法则： y ′ = [ n x    n y    n z ] T × O P (7) \bm{y'}=[n_x\ \ n_y\ \ n_z]^T\times \bm{OP}\tag 7 y′=[nx​  ny​  nz​]T×OP(7)   坐标系 x ′ y ′ z ′ x'y'z' x′y′z′与坐标系 X Y Z XYZ XYZ的旋转变换矩阵为： R 3 × 3 = [ O P    y ′    n ^ ] (8) R_{3\times3}=[\bm{OP}\ \ \bm{y'}\ \ \bm{\hat{n}} ]\tag 8 R3×3​=[OP  y′  n^](8)   点 P ′ P' P′在坐标系 x ′ y ′ z ′ x'y'z' x′y′z′的坐标为： { x t e m p = r c o s ( α ) y t e m p = r s i n ( α ) z t e m p = 0 (9) \left\{ \begin{array}{c} x_{temp}=rcos(\alpha) \\ y_{temp}=rsin(\alpha) \\ \tag 9 z_{temp}=0 \\ \end{array}\right. ⎩⎨⎧​xtemp​=rcos(α)ytemp​=rsin(α)ztemp​=0​(9)   利用齐次变换，将点 P ′ P' P′在坐标系 x ′ y ′ z ′ x'y'z' x′y′z′的坐标变换到坐标系 X Y Z XYZ XYZ的坐标： [ p x ′ p y ′ p z ′ 1 ] = [ R 11 R 12 R 13 x c R 21 R 22 R 23 y c R 31 R 32 R 33 z c 0 0 0 1 ] [ x t e m p y t e m p z t e m p 1 ] (10) \left[ \begin{matrix} p'_x \\ p'_y \\ p'_z \\ 1 \end{matrix} \right] = \left[ \begin{matrix} R_{11} & R_{12} & R_{13} & x_c \\ R_{21} & R_{22} & R_{23} & y_c \\ R_{31} & R_{32} & R_{33} & z_c \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right] \left[ \begin{matrix} x_{temp} \\ y_{temp} \\ z_{temp} \\ 1 \end{matrix} \right] \tag{10} ⎣⎢⎢⎡​px′​py′​pz′​1​⎦⎥⎥⎤​=⎣⎢⎢⎡​R11​R21​R31​0​R12​R22​R32​0​R13​R23​R33​0​xc​yc​zc​1​⎦⎥⎥⎤​⎣⎢⎢⎡​xtemp​ytemp​ztemp​1​⎦⎥⎥⎤​(10)

  化简式 ( 10 ) (10) (10)得： { p x ′ = R 11 x t e m p + R 12 y t e m p + x c p y ′ = R 21 x t e m p + R 22 y t e m p + y c p z ′ = R 31 x t e m p + R 32 y t e m p + z c (11) \left\{ \begin{array}{c} p'_x=R_{11}x_{temp}+R_{12}y_{temp}+ x_c\\ p'_y=R_{21}x_{temp}+R_{22}y_{temp}+ y_c\\ p'_z=R_{31}x_{temp}+R_{32}y_{temp}+ z_c \tag{11} \end{array}\right. ⎩⎨⎧​px′​=R11​xtemp​+R12​ytemp​+xc​py′​=R21​xtemp​+R22​ytemp​+yc​pz′​=R31​xtemp​+R32​ytemp​+zc​​(11)

  式 ( 11 ) (11) (11)可以展开，并写成： [ p x ′ p y ′ p z ′ 1 ] = T 4 × 4 [ p x p y p z 1 ] (12) \left[ \begin{matrix} p'_x \\ p'_y \\ p'_z \\ 1 \end{matrix} \right] = T_{4\times4} \left[ \begin{matrix} p_x \\ p_y \\ p_z \\ 1 \end{matrix} \right]\tag{12} ⎣⎢⎢⎡​px′​py′​pz′​1​⎦⎥⎥⎤​=T4×4​⎣⎢⎢⎡​px​py​pz​1​⎦⎥⎥⎤​(12)

T 4 × 4 = [ n x 2 K + c o s ( α ) n x n y K − n z s i n ( α ) n x n z K + n y s i n ( α ) ( x 0 − n x M ) K + ( n z y 0 − n y z 0 ) s i n ( α ) n x n y K + n z s i n ( α ) n y 2 K + c o s ( α ) n y n z K − n x s i n ( α ) ( y 0 − n y M ) K + ( n x z 0 − n z x 0 ) s i n ( α ) n x n z K − n y s i n ( α ) n y n z K + n x s i n ( α ) n z 2 K + c o s ( α ) ( z 0 − n z M ) K + ( n y x 0 − n x y 0 ) s i n ( α ) 0 0 0 1 ] T_{4\times4}=\left[ \begin{matrix} n_x^2 K + cos(\alpha) & n_x n_yK - n_z sin(\alpha) & n_xn_z K + n_y sin(\alpha) & (x_0 - n_xM)K + (n_zy_0 - n_yz_0)sin(\alpha) \\ n_xn_yK + n_zsin(\alpha) & n_y^2 K + cos(\alpha) & n_yn_zK - n_xsin(\alpha) & (y_0 - n_yM)K + (n_xz_0 - n_zx_0)sin(\alpha) \\ n_xn_zK - n_ysin(\alpha) & n_yn_zK + n_xsin(\alpha) & n_z^2K + cos(\alpha) & (z_0 - n_zM)K + (n_yx_0 - n_xy_0)sin(\alpha) \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right] \\ T4×4​=⎣⎢⎢⎡​nx2​K+cos(α)nx​ny​K+nz​sin(α)nx​nz​K−ny​sin(α)0​nx​ny​K−nz​sin(α)ny2​K+cos(α)ny​nz​K+nx​sin(α)0​nx​nz​K+ny​sin(α)ny​nz​K−nx​sin(α)nz2​K+cos(α)0​(x0​−nx​M)K+(nz​y0​−ny​z0​)sin(α)(y0​−ny​M)K+(nx​z0​−nz​x0​)sin(α)(z0​−nz​M)K+(ny​x0​−nx​y0​)sin(α)1​⎦⎥⎥⎤​   其中， K = 1 − c o s ( α ) ， M = n x x 0 + n y y 0 + n z z 0 K = 1 - cos(\alpha)，M = n_xx_0 + n_yy_0 + n_zz_0 K=1−cos(α)，M=nx​x0​+ny​y0​+nz​z0​

  本文算法的结论具有普遍性，当轴线的单位方向向量 n ^ \bm{\hat{n}} n^和经过的点 Q ( x 0 , y 0 , z 0 ) Q(x_0,y_0,z_0) Q(x0​,y0​,z0​)取特殊值时，可以得到许多很有用的结论。   1.当旋转轴为 x x x轴，则 ( x 0 , y 0 , z 0 ) = ( 0 , 0 , 0 ) , ( n x , n y , n z ) = ( 1 , 0 , 0 ) (x_0,y_0,z_0)=(0,0,0),(n_x,n_y,n_z)=(1,0,0) (x0​,y0​,z0​)=(0,0,0),(nx​,ny​,nz​)=(1,0,0)时， T 4 × 4 = [ 1 0 0 0 0 c o s ( α ) − s i n ( α ) 0 0 s i n ( α ) c o s ( α ) 0 0 0 0 1 ] (13) T_{4\times4}=\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & cos(\alpha) & -sin(\alpha) & 0 \\ 0 & sin(\alpha) & cos(\alpha) & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right] \tag{13} T4×4​=⎣⎢⎢⎡​1000​0cos(α)sin(α)0​0−sin(α)cos(α)0​0001​⎦⎥⎥⎤​(13)   2.当旋转轴为 y y y轴，则 ( x 0 , y 0 , z 0 ) = ( 0 , 0 , 0 ) , ( n x , n y , n z ) = ( 0 , 1 , 0 ) (x_0,y_0,z_0)=(0,0,0),(n_x,n_y,n_z)=(0,1,0) (x0​,y0​,z0​)=(0,0,0),(nx​,ny​,nz​)=(0,1,0)时， T 4 × 4 = [ c o s ( α ) 0 s i n ( α ) 0 0 1 0 0 − s i n ( α ) 0 c o s ( α ) 0 0 0 0 1 ] (14) T_{4\times4}=\left[ \begin{matrix} cos(\alpha) & 0 & sin(\alpha) & 0 \\ 0 & 1 & 0 & 0 \\ -sin(\alpha) & 0 & cos(\alpha) & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right] \tag{14} T4×4​=⎣⎢⎢⎡​cos(α)0−sin(α)0​0100​sin(α)0cos(α)0​0001​⎦⎥⎥⎤​(14)   3.当旋转轴为 z z z轴，则 ( x 0 , y 0 , z 0 ) = ( 0 , 0 , 0 ) , ( n x , n y , n z ) = ( 0 , 0 , 1 ) (x_0,y_0,z_0)=(0,0,0),(n_x,n_y,n_z)=(0,0,1) (x0​,y0​,z0​)=(0,0,0),(nx​,ny​,nz​)=(0,0,1)时， T 4 × 4 = [ c o s ( α ) − s i n ( α ) 0 0 s i n ( α ) c o s ( α ) 0 0 0 0 1 0 0 0 0 1 ] (15) T_{4\times4}=\left[ \begin{matrix} cos(\alpha) & -sin(\alpha) & 0 & 0 \\ sin(\alpha) & cos(\alpha) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right] \tag{15} T4×4​=⎣⎢⎢⎡​cos(α)sin(α)00​−sin(α)cos(α)00​0010​0001​⎦⎥⎥⎤​(15)   4.当旋转轴过点 ( 0 , t y , t z ) (0,t_y,t_z) (0,ty​,tz​)且平行于 x x x轴，则 ( x 0 , y 0 , z 0 ) = ( 0 , t y , t z ) , ( n x , n y , n z ) = ( 1 , 0 , 0 ) (x_0,y_0,z_0)=(0,t_y,t_z),(n_x,n_y,n_z)=(1,0,0) (x0​,y0​,z0​)=(0,ty​,tz​),(nx​,ny​,nz​)=(1,0,0)时， T 4 × 4 = [ 1 0 0 0 0 c o s ( α ) − s i n ( α ) t y K + t z s i n ( α ) 0 s i n ( α ) c o s ( α ) t z K − t y s i n ( α ) 0 0 0 1 ] (16) T_{4\times4}=\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & cos(\alpha) & -sin(\alpha) & t_yK+t_zsin(\alpha) \\ 0 & sin(\alpha) & cos(\alpha) & t_zK-t_ysin(\alpha) \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right] \tag{16} T4×4​=⎣⎢⎢⎡​1000​0cos(α)sin(α)0​0−sin(α)cos(α)0​0ty​K+tz​sin(α)tz​K−ty​sin(α)1​⎦⎥⎥⎤​(16)   5.当旋转轴过点 ( t x , 0 , t z ) (t_x,0,t_z) (tx​,0,tz​)且平行于 y y y轴，则 ( x 0 , y 0 , z 0 ) = ( t x , 0 , t z ) , ( n x , n y , n z ) = ( 0 , 1 , 0 ) (x_0,y_0,z_0)=(t_x,0,t_z),(n_x,n_y,n_z)=(0,1,0) (x0​,y0​,z0​)=(tx​,0,tz​),(nx​,ny​,nz​)=(0,1,0)时， T 4 × 4 = [ c o s ( α ) 0 s i n ( α ) t x K − t z s i n ( α ) 0 1 0 0 − s i n ( α ) 0 c o s ( α ) t z K + t x s i n ( α ) 0 0 0 1 ] (17) T_{4\times4}=\left[ \begin{matrix} cos(\alpha) & 0 & sin(\alpha) & t_xK- t_zsin(\alpha) \\ 0 & 1 & 0 & 0 \\ -sin(\alpha) & 0 & cos(\alpha) & t_zK + t_xsin(\alpha)\\ 0 & 0 & 0 & 1 \\ \end{matrix} \right] \tag{17} T4×4​=⎣⎢⎢⎡​cos(α)0−sin(α)0​0100​sin(α)0cos(α)0​tx​K−tz​sin(α)0tz​K+tx​sin(α)1​⎦⎥⎥⎤​(17)   6.当旋转轴过点 ( t x , t y , 0 ) (t_x,t_y,0) (tx​,ty​,0)且平行于 z z z轴，则 ( x 0 , y 0 , z 0 ) = ( t x , t y , 0 ) , ( n x , n y , n z ) = ( 0 , 0 , 1 ) (x_0,y_0,z_0)=(t_x,t_y,0),(n_x,n_y,n_z)=(0,0,1) (x0​,y0​,z0​)=(tx​,ty​,0),(nx​,ny​,nz​)=(0,0,1)时， T 4 × 4 = [ c o s ( α ) − s i n ( α ) 0 t x K + t y s i n ( α ) s i n ( α ) c o s ( α ) 0 t y K − t x s i n ( α ) 0 0 1 0 0 0 0 1 ] (18) T_{4\times4}=\left[ \begin{matrix} cos(\alpha) & -sin(\alpha) & 0 & t_xK+t_ysin(\alpha)\\ sin(\alpha) & cos(\alpha) & 0 & t_yK- t_xsin(\alpha)\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right] \tag{18} T4×4​=⎣⎢⎢⎡​cos(α)sin(α)00​−sin(α)cos(α)00​0010​tx​K+ty​sin(α)ty​K−tx​sin(α)01​⎦⎥⎥⎤​(18)   7.当旋转轴过原点 ( 0 , 0 , 0 ) (0,0,0) (0,0,0)，则 ( x 0 , y 0 , z 0 ) = ( 0 , 0 , 0 ) (x_0,y_0,z_0)=(0,0,0) (x0​,y0​,z0​)=(0,0,0)时， T 4 × 4 = [ n x 2 K + c o s ( α ) n x n y K − n z s i n ( α ) n x n z K + n y s i n ( α ) 0 n x n y K + n z s i n ( α ) n y 2 K + c o s ( α ) n y n z K − n x s i n ( α ) 0 n x n z K − n y s i n ( α ) n y n z K + n x s i n ( α ) n z 2 K + c o s ( α ) 0 0 0 0 1 ] (19) T_{4\times4}=\left[ \begin{matrix} n_x^2 K + cos(\alpha) & n_x n_yK - n_z sin(\alpha) & n_xn_z K + n_y sin(\alpha) & 0 \\ n_xn_yK + n_zsin(\alpha) & n_y^2 K + cos(\alpha) & n_yn_zK - n_xsin(\alpha) & 0 \\ n_xn_zK - n_ysin(\alpha) & n_yn_zK + n_xsin(\alpha) & n_z^2K + cos(\alpha) & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right] \tag{19} T4×4​=⎣⎢⎢⎡​nx2​K+cos(α)nx​ny​K+nz​sin(α)nx​nz​K−ny​sin(α)0​nx​ny​K−nz​sin(α)ny2​K+cos(α)ny​nz​K+nx​sin(α)0​nx​nz​K+ny​sin(α)ny​nz​K−nx​sin(α)nz2​K+cos(α)0​0001​⎦⎥⎥⎤​(19)    T 4 × 4 T_{4\times4} T4×4​的前 3 3 3行 3 3 3列就是将三维旋转的轴-角表示转化为旋转矩阵表示： R 3 × 3 = [ n x 2 K + c o s ( α ) n x n y K − n z s i n ( α ) n x n z K + n y s i n ( α ) n x n y K + n z s i n ( α ) n y 2 K + c o s ( α ) n y n z K − n x s i n ( α ) n x n z K − n y s i n ( α ) n y n z K + n x s i n ( α ) n z 2 K + c o s ( α ) ] (20) R_{3\times3}=\left[ \begin{matrix} n_x^2 K + cos(\alpha) & n_x n_yK - n_z sin(\alpha) & n_xn_z K + n_y sin(\alpha)\\ n_xn_yK + n_zsin(\alpha) & n_y^2 K + cos(\alpha) & n_yn_zK - n_xsin(\alpha) \\ n_xn_zK - n_ysin(\alpha) & n_yn_zK + n_xsin(\alpha) & n_z^2K + cos(\alpha) \\ \end{matrix} \right] \tag{20} R3×3​=⎣⎡​nx2​K+cos(α)nx​ny​K+nz​sin(α)nx​nz​K−ny​sin(α)​nx​ny​K−nz​sin(α)ny2​K+cos(α)ny​nz​K+nx​sin(α)​nx​nz​K+ny​sin(α)ny​nz​K−nx​sin(α)nz2​K+cos(α)​⎦⎤​(20)   其中， K = 1 − c o s ( α ) K = 1 - cos(\alpha) K=1−cos(α)