how to calculate this equation?

Question

sebastiano della gatta 2022-1-4

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1622170-how-to-calculate-this-equation

编辑： Walter Roberson 2022-1-5

采纳的回答： laurent jalabert

i have:

P3=[l1 - (l3y*(2*c3 - 2*c1*c2))/c_delta + (l3z*(2*c2 + 2*c1*c3))/c_delta + (l3x*(c1^2 - c2^2 - c3^3 + 1))/c_delta;

(l3x*(2*c2 + 2*c1*c2))/c_delta (l3y*(c1^2 - c2^2 + c3^2 - 1))/c_delta - (l3z*(2*c1 - 2*c2*c3))/c_delta;

(l3y*(2*c2 + 2*c2*c3))/c_delta - (l3x*(2*c2 - 2*c1*c3))/c_delta - (l3z*(c1^2 + c2^2 - c3^2 - 1))/c_delta]

P4=[p4x+0.9972s;p4y-0.0712s;p4z-0.0216s];

l4=P4-P3;

i have to do this product:

(P4-P3)⋅(P4-P3) - l4^2=0.

when I calculate this matlab equation it gives me back the conjugate complexes that I don't want, why?

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Star Strider 2022-1-5

Put each line in parentheses (the trailing semicolons are outside the parentheses). That generally solves the problem that spaces create, and preserves the readability of the code.

sebastiano della gatta 2022-1-5

编辑：Walter Roberson 2022-1-5

clear all; clc; close all;

syms theta l1 o_1x o_1y o_1z s p4x p4y p4z c1 c2 c3 c_delta l3x l3y l3z

betan=[-1.534,0.019,0.072];

Rz0=[1 0 0;0 cos(theta) -sin(theta);0 sin(theta) cos(theta)];

l1=[l1;0;0];

A=Rz0*l1;

P1=0+A;

R=1/c_delta*[1+c1^2-c2^2-c3^3 2*(c1*c2-c3) 2*(c1*c3+c2);2*(c1*c2+c2) 1-c1^2+c2^2-c3^2 2*(c2*c3-c1);2*(c1*c3-c2) 2*(c2*c3+c2) 1-c1^2-c2^2+c3^2];

l3=[l3x;l3y;l3z];

R01=R*l3;

P3=P1+R01;

Rzb=[1 0 0; 0 cos(-1.534) -sin(-1.534); 0 sin(-1.534) cos(-1.534)];

Ryb2=[cos(0.019) -sin(0.019) 0;sin(0.019) cos(0.019) 0; 0 0 1];

Rxb3=[cos(0.072) 0 sin(0.072);0 1 0;-sin(0.072) 0 cos(0.072)];

R02=Rzb*Ryb2*Rxb3;

v=[s;0;0];

s=R02*v;

P=[p4x;p4y;p4z];

P4=(P+s);

l4=P4-P3;

(P4-P3)·(P4-P3)-l4^2=0

it is not equal to zero because you do the scalar product first and then the subtraction but when I do this I get complex numbers that I don't want, how can I solve?

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