how to calculate this equation?

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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 个评论
Star Strider
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
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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采纳的回答

laurent jalabert
laurent jalabert 2022-1-5
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 个评论
sebastiano della gatta
it is not equal to zero because i don't have to use .* because is another operation.

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