一、问题描述

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

二、推导步骤

轴线n^\bm{\hat{n}}n^在坐标系XYZXYZXYZ下的直线方程为：
{x=x0+nxty=y0+nytz=z0+nzt(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+nxty=y0+nytz=z0+nzt(1)
弧PP′PP'PP′所在平面在坐标系XYZXYZXYZ下的平面方程为：
nx(x−px)+ny(y−py)+nz(z−pz)=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nx2+ny2+nz2=1n_x^2+n_y^2+n_z^2=1nx2+ny2+nz2=1，可以求得：
t0=nx(px−x0)+ny(py−y0)+nz(pz−z0)(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PP′PP'PP′的圆心在坐标系XYZXYZXYZ下的坐标为(xc,yc,zc)(x_c,y_c,z_c)(xc,yc,zc),将t0t_0t0代入式(1)(1)(1)，得圆心坐标：
{xc=x0+nxt0yc=y0+nyt0zc=z0+nzt0(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+nxt0yc=y0+nyt0zc=z0+nzt0(4)
圆半径rrr:
r=(px−xc)2+(py−yc)2+(pz−zc)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OP\bm{OP}OP:
OP=[px−xcpy−ycpz−zc]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′=[nxnynz]T×OP(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XYZXYZXYZ的旋转变换矩阵为：
R3×3=[OPy′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：
{xtemp=rcos(α)ytemp=rsin(α)ztemp=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XYZXYZXYZ的坐标：
[px′py′pz′1]=[R11R12R13xcR21R22R23ycR31R32R33zc0001][xtempytempztemp1](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⎦⎥⎥⎤=⎣⎢⎢⎡R11R21R310R12R22R320R13R23R330xcyczc1⎦⎥⎥⎤⎣⎢⎢⎡xtempytempztemp1⎦⎥⎥⎤(10)

化简式(10)(10)(10)得：
{px′=R11xtemp+R12ytemp+xcpy′=R21xtemp+R22ytemp+ycpz′=R31xtemp+R32ytemp+zc(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′=R11xtemp+R12ytemp+xcpy′=R21xtemp+R22ytemp+ycpz′=R31xtemp+R32ytemp+zc(11)

式(11)(11)(11)可以展开，并写成：
[px′py′pz′1]=T4×4[pxpypz1](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⎣⎢⎢⎡pxpypz1⎦⎥⎥⎤(12)

T4×4=[nx2K+cos(α)nxnyK−nzsin(α)nxnzK+nysin(α)(x0−nxM)K+(nzy0−nyz0)sin(α)nxnyK+nzsin(α)ny2K+cos(α)nynzK−nxsin(α)(y0−nyM)K+(nxz0−nzx0)sin(α)nxnzK−nysin(α)nynzK+nxsin(α)nz2K+cos(α)(z0−nzM)K+(nyx0−nxy0)sin(α)0001]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=⎣⎢⎢⎡nx2K+cos(α)nxnyK+nzsin(α)nxnzK−nysin(α)0nxnyK−nzsin(α)ny2K+cos(α)nynzK+nxsin(α)0nxnzK+nysin(α)nynzK−nxsin(α)nz2K+cos(α)0(x0−nxM)K+(nzy0−nyz0)sin(α)(y0−nyM)K+(nxz0−nzx0)sin(α)(z0−nzM)K+(nyx0−nxy0)sin(α)1⎦⎥⎥⎤
其中，K=1−cos(α)，M=nxx0+nyy0+nzz0K = 1 - cos(\alpha)，M = n_xx_0 + n_yy_0 + n_zz_0K=1−cos(α)，M=nxx0+nyy0+nzz0

三、MATLABMATLABMATLAB代码

clc;
clear;syms t nx ny nz px py pz alpha x0 y0 z0 real
syms tx ty tz axisFlag real% axisFlag = 1 : x轴
% axisFlag = 2 : y轴
% axisFlag = 3 : z轴
% axisFlag = 4 : 过点(0,ty,tz)且平行x轴
% axisFlag = 5 : 过点(tx,0,tz)且平行y轴
% axisFlag = 6 : 过点(tx,ty,0)且平行z轴
% axisFlag = 7 : 过点(0,0,0)且单位方向向量为[nx ny nz]
% axisFlag = 8 : 过点(tx,ty,tz)且单位方向向量为[nx ny nz]axisFlag = 8;
switch axisFlagcase 1x0 = 0;y0 = 0;z0 = 0;nx = 1;ny = 0;nz = 0;case 2x0 = 0;y0 = 0;z0 = 0;nx = 0;ny = 1;nz = 0;case 3x0 = 0;y0 = 0;z0 = 0;nx = 0;ny = 0;nz = 1;case 4x0 = 0;y0 = ty;z0 = tz;nx = 1;ny = 0;nz = 0;case 5x0 = tx;y0 = 0;z0 = tz;nx = 0;ny = 1;nz = 0;case 6x0 = tx;y0 = ty;z0 = 0;nx = 0;ny = 0;nz = 1;case 7x0 = 0;y0 = 0;z0 = 0;case 8x0 = tx;y0 = ty;z0 = tz;otherwisereturn;
end%{
x = nx * t + x0;
y = ny * t + y0;
z = nz * t + z0;
t0 = solve(nx * (x - px) + ny * (y - py) + nz * (z - pz) == 0, t)
%}t0 = nx * (px - x0) + ny * (py - y0) + nz * (pz - z0);
xc = x0 + nx * t0;
yc = y0 + ny * t0;
zc = z0 + nz * t0;
r = sqrt((px - xc)^2 + (py - yc)^2 + (pz - zc)^2);OP = [px - xc; py - yc; pz - zc] / r;
yVector = cross([nx; ny; nz], OP);
R = [OP, yVector, [nx; ny; nz]];xtemp = r * cos(alpha);
ytemp = r * sin(alpha);
p = [R(1,1) * xtemp + R(1,2) * ytemp + xcR(2,1) * xtemp + R(2,2) * ytemp + ycR(3,1) * xtemp + R(3,2) * ytemp + zc];
p = [simplify(p(1)); simplify(p(2)); simplify(p(3))]%% 写成矩阵形式，并验证结果正确性
K = 1 - cos(alpha);
M = nx * x0 + ny * y0 + nz * z0;
T = [nx^2 * K + cos(alpha),  nx * ny * K - nz * sin(alpha),  nx * nz * K + ny * sin(alpha),  (x0 - nx * M) * K + (nz * y0 - ny * z0) * sin(alpha)nx * ny * K + nz * sin(alpha),  ny^2 * K + cos(alpha),  ny * nz * K - nx * sin(alpha),  (y0 - ny * M) * K + (nx * z0 - nz * x0) * sin(alpha)nx * nz * K - ny * sin(alpha),  ny * nz * K + nx * sin(alpha),  nz^2 * K + cos(alpha),  (z0 - nz * M) * K + (ny * x0 - nx * y0) * sin(alpha)0, 0, 0, 1]res = simplify([nx^2 * K + cos(alpha),  nx * ny * K - nz * sin(alpha),  nx * nz * K + ny * sin(alpha),  (x0 - nx * M) * K + (nz * y0 - ny * z0) * sin(alpha)nx * ny * K + nz * sin(alpha),  ny^2 * K + cos(alpha),  ny * nz * K - nx * sin(alpha),  (y0 - ny * M) * K + (nx * z0 - nz * x0) * sin(alpha)nx * nz * K - ny * sin(alpha),  ny * nz * K + nx * sin(alpha),  nz^2 * K + cos(alpha),  (z0 - nz * M) * K + (ny * x0 - nx * y0) * sin(alpha)0, 0, 0, 1] * [px; py; pz; 1] - [p(1); p(2); p(3); 1])

四、总结

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(x0,y0,z0)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xxx轴，则(x0,y0,z0)=(0,0,0),(nx,ny,nz)=(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，
T4×4=[10000cos(α)−sin(α)00sin(α)cos(α)00001](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=⎣⎢⎢⎡10000cos(α)sin(α)00−sin(α)cos(α)00001⎦⎥⎥⎤(13)
2.当旋转轴为yyy轴，则(x0,y0,z0)=(0,0,0),(nx,ny,nz)=(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，
T4×4=[cos(α)0sin(α)00100−sin(α)0cos(α)00001](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)00100sin(α)0cos(α)00001⎦⎥⎥⎤(14)
3.当旋转轴为zzz轴，则(x0,y0,z0)=(0,0,0),(nx,ny,nz)=(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，
T4×4=[cos(α)−sin(α)00sin(α)cos(α)0000100001](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)0000100001⎦⎥⎥⎤(15)
4.当旋转轴过点(0,ty,tz)(0,t_y,t_z)(0,ty,tz)且平行于xxx轴，则(x0,y0,z0)=(0,ty,tz),(nx,ny,nz)=(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，
T4×4=[10000cos(α)−sin(α)tyK+tzsin(α)0sin(α)cos(α)tzK−tysin(α)0001](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=⎣⎢⎢⎡10000cos(α)sin(α)00−sin(α)cos(α)00tyK+tzsin(α)tzK−tysin(α)1⎦⎥⎥⎤(16)
5.当旋转轴过点(tx,0,tz)(t_x,0,t_z)(tx,0,tz)且平行于yyy轴，则(x0,y0,z0)=(tx,0,tz),(nx,ny,nz)=(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，
T4×4=[cos(α)0sin(α)txK−tzsin(α)0100−sin(α)0cos(α)tzK+txsin(α)0001](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)00100sin(α)0cos(α)0txK−tzsin(α)0tzK+txsin(α)1⎦⎥⎥⎤(17)
6.当旋转轴过点(tx,ty,0)(t_x,t_y,0)(tx,ty,0)且平行于zzz轴，则(x0,y0,z0)=(tx,ty,0),(nx,ny,nz)=(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，
T4×4=[cos(α)−sin(α)0txK+tysin(α)sin(α)cos(α)0tyK−txsin(α)00100001](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)000010txK+tysin(α)tyK−txsin(α)01⎦⎥⎥⎤(18)
7.当旋转轴过原点(0,0,0)(0,0,0)(0,0,0)，则(x0,y0,z0)=(0,0,0)(x_0,y_0,z_0)=(0,0,0)(x0,y0,z0)=(0,0,0)时，
T4×4=[nx2K+cos(α)nxnyK−nzsin(α)nxnzK+nysin(α)0nxnyK+nzsin(α)ny2K+cos(α)nynzK−nxsin(α)0nxnzK−nysin(α)nynzK+nxsin(α)nz2K+cos(α)00001](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=⎣⎢⎢⎡nx2K+cos(α)nxnyK+nzsin(α)nxnzK−nysin(α)0nxnyK−nzsin(α)ny2K+cos(α)nynzK+nxsin(α)0nxnzK+nysin(α)nynzK−nxsin(α)nz2K+cos(α)00001⎦⎥⎥⎤(19)
T4×4T_{4\times4}T4×4的前333行333列就是将三维旋转的轴-角表示转化为旋转矩阵表示：
R3×3=[nx2K+cos(α)nxnyK−nzsin(α)nxnzK+nysin(α)nxnyK+nzsin(α)ny2K+cos(α)nynzK−nxsin(α)nxnzK−nysin(α)nynzK+nxsin(α)nz2K+cos(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=⎣⎡nx2K+cos(α)nxnyK+nzsin(α)nxnzK−nysin(α)nxnyK−nzsin(α)ny2K+cos(α)nynzK+nxsin(α)nxnzK+nysin(α)nynzK−nxsin(α)nz2K+cos(α)⎦⎤(20)
其中，K=1−cos(α)K = 1 - cos(\alpha)K=1−cos(α)

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三维空间任意一点绕任意轴线旋转

文章目录

一、问题描述

二、推导步骤

三、MATLABMATLABMATLAB代码

四、总结

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