r = sqrt(xx_avr-2*x00*x_avr+x00*x00 + yy_avr-2*y00*y_avr+y00*y00 + zz_avr-2*z00*z_avr+z00*z00); %拟合出的球半径r
时间: 2024-03-11 18:42:48 浏览: 16
这段代码是用来计算拟合出的球的半径r的。其中,xx_avr、x00、x_avr、yy_avr、y00、y_avr、zz_avr、z00、z_avr分别表示对应的变量值。
具体计算过程如下:
1. 首先计算平均值:
x_avr = (x1 + x2 + ... + xn) / n
y_avr = (y1 + y2 + ... + yn) / n
z_avr = (z1 + z2 + ... + zn) / n
2. 然后计算平方和:
xx_avr = (x1^2 + x2^2 + ... + xn^2) / n
yy_avr = (y1^2 + y2^2 + ... + yn^2) / n
zz_avr = (z1^2 + z2^2 + ... + zn^2) / n
3. 最后根据公式计算球的半径r:
r = sqrt(xx_avr - 2*x00*x_avr + x00^2 + yy_avr - 2*y00*y_avr + y00^2 + zz_avr - 2*z00*z_avr + z00^2)
相关问题
如何将下边maole中的代码改为matlabfig:=proc(P_x,P_y,P_z,R,mu,varphi,C) with(plots);with(ColorTools); local Rb; local Rf := implicitplot(x^2 + y^2 = R^2, x = -R .. R, y = -R .. 2, color =red, thickness = 2): local s_r:=1/2*(P_x + P_z)*(1 - R^2/r^2) - 1/2*(P_z - P_x)*(1 - 4*R^2/r^2 + 3*R^4/r^4)*cos(2*theta); local s_t:=1/2*(P_x + P_z)*(1 + R^2/r^2) + 1/2*(P_z - P_x)*(1 + 3*R^4/r^4)*cos(2*theta); local s_rt:=1/2*(-P_z + P_x)*(1 + 2*R^2/r^2 - 3*R^4/r^4)*sin(2*theta); local s_y:=P_y - 2*mu*(-P_z + P_x)*R^2*cos(2*theta)/r^2; local s_3:=(s_r + s_t)/2 - sqrt(((s_r - s_t)/2)^2 + s_rt^2); local s_1:=(s_r + s_t)/2 + sqrt(((s_r - s_t)/2)^2 + s_rt^2); local s_2:=s_y; local J2:=(s_1^2+s_2^2+s_3^2-s_1*s_2-s_1*s_3-s_2*s_3)/3; local I1:=s_1+s_2+s_3; local p:=(s_1+s_2+s_3)/3; local J3:=(s_1-p)*(s_2-p)*(s_3-p); local k:=(arcsin((-3*sqrt(3)*J3)/(2*sqrt(J2*J2*J2))))/3; Rb:=implicitplot(I1/3*sin(varphi)+C*cos(varphi)-sqrt(J2)*((1/sqrt(3))*sin(varphi)*sin(k)+cos(k)),r=R..R*20,theta=0..2*Pi,coords=polar,thickness=2); display(Rf,Rb) end proc:
以下是将maole中的代码改为MATLAB的代码,MATLAB中没有ColorTools模块,因此省略相关部分:
```matlab
function matlabfig = proc(P_x, P_y, P_z, R, mu, varphi, C)
syms r theta;
Rf = ezplot(r^2*cos(theta)^2 + r^2*sin(theta)^2 - R^2 == 0, [-R, R, -R, 2]);
set(Rf, 'Color', 'red', 'LineWidth', 2);
s_r = 1/2*(P_x + P_z)*(1 - R^2/r^2) - 1/2*(P_z - P_x)*(1 - 4*R^2/r^2 + 3*R^4/r^4)*cos(2*theta);
s_t = 1/2*(P_x + P_z)*(1 + R^2/r^2) + 1/2*(P_z - P_x)*(1 + 3*R^4/r^4)*cos(2*theta);
s_rt = 1/2*(-P_z + P_x)*(1 + 2*R^2/r^2 - 3*R^4/r^4)*sin(2*theta);
s_y = P_y - 2*mu*(-P_z + P_x)*R^2*cos(2*theta)/r^2;
s_3 = (s_r + s_t)/2 - sqrt(((s_r - s_t)/2)^2 + s_rt^2);
s_1 = (s_r + s_t)/2 + sqrt(((s_r - s_t)/2)^2 + s_rt^2);
s_2 = s_y;
J2 = (s_1^2+s_2^2+s_3^2-s_1*s_2-s_1*s_3-s_2*s_3)/3;
I1 = s_1+s_2+s_3;
p = (s_1+s_2+s_3)/3;
J3 = (s_1-p)*(s_2-p)*(s_3-p);
k = (asin((-3*sqrt(3)*J3)/(2*sqrt(J2^2*J2))))/3;
Rb = ezplot(I1/3*sin(varphi)+C*cos(varphi)-sqrt(J2)*((1/sqrt(3))*sin(varphi)*sin(k)+cos(k)), [R, R*20, 0, 2*pi]);
set(Rb, 'LineWidth', 2);
matlabfig = [Rf, Rb];
end
```
r,x已知,r= (1-X*X-Y*Y)/(X*X-2*X+Y*Y+1),x = 2*Y/ (X*X - 2*X + Y*Y + 1),求X,Y的表达式
我们可以通过将已知条件代入$r$和$x$的表达式,来得到关于$X$和$Y$的方程组。具体来说,我们有:
$$\begin{cases}r = \frac{1-X^2-Y^2}{X^2-2X+Y^2+1}\\ x = \frac{2Y}{X^2-2X+Y^2+1}\end{cases}$$
将$r$代入$x$的表达式中,得到:
$$x = \frac{2Y}{X^2-2X+Y^2+1} = \frac{2Y}{(X-1)^2+Y^2}$$
将$x$代入$r$的表达式中,得到:
$$r = \frac{1-X^2-Y^2}{(X-1)^2+Y^2} \cdot \frac{(X-1)^2+Y^2}{X^2-2X+Y^2+1} = \frac{1-X^2-Y^2}{X^2-2X+Y^2+1} \cdot \frac{1}{(X-1)^2+Y^2}$$
展开和整理可得:
$$(X^2-2X+Y^2+1)r = 1-X^2-Y^2$$
移项得到:
$$X^2(1-r) + 2X(r-1) + Y^2(1-r) + r - 1 = 0$$
这是一个关于$X$和$Y$的二次方程。解方程可以得到$X$和$Y$的表达式:
$$X = \frac{2r-2 \pm 2\sqrt{r^2-2r+5}}{2(1-r)} = r+1 \pm \frac{\sqrt{r^2-2r+5}}{1-r}$$
$$Y = \pm \frac{\sqrt{2r-2-X^2+2X}}{1-X}$$
注意这里的$X$只能取$r+1 + \frac{\sqrt{r^2-2r+5}}{1-r}$,因为另一个解会使得$Y$的分母为零,不符合题意。