how to calculate this equation?

2 ビュー (過去 30 日間)
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?
  6 件のコメント
sebastiano della gatta
sebastiano della gatta 2022 年 1 月 5 日
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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laurent jalabert
laurent jalabert 2022 年 1 月 5 日
Hello, maybe I did not understood your problem cause l4=P4-P3; and you want to solve:
(P4-P3)⋅(P4-P3) - l4^2=0, which means (P4-P3)⋅(P4-P3) - (P4-P3)^2 = 0
Do you mean (P4-P3).*(P4-P3) - (P4-P3).^2 = 0 ? which is 0=0
  1 件のコメント
sebastiano della gatta
sebastiano della gatta 2022 年 1 月 5 日
it is not equal to zero because i don't have to use .* because is another operation.

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