Error using atan2 Inputs must be real.
12 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
abouda tebba
el 14 de En. de 2021
Respondida: Steven Lord
el 17 de En. de 2021
this programme and function keep saying :
*Error using atan2 Inputs must be real. line 4 'function file'
*Error line 14 'main programme
i made 'px,py,pz' as coments cause i dont know its supose to help
ps :didn't with it either
Main programme :
clear
clc
clf
a2=6.5;a3=0;d3=1.9;d4=6;
% px=[-0.7;-2.52;-1.13;-4.60;1.00;1.68];
% py=[11.70;10.10;1.54;7.10;5.00;1.18];
% pz=[-0.90;-1.63;-6.93;-1.38;-3.00;-6.94];
t=0:23;
px=30*t'.*cos(t');
py=1200-(30*t'.^2)/(50*pi);
pz=30*t'.*sin(t');
[theta,t1,t2,t3,t4,t5,t6]=part2(a2,a3,d3,d4,px,py,pz);
pxx=cos(t1).*(sin(t2+t3)*d4+a3*cos(t2+t3)+a2*cos(t2))-d3*sin(t1);
pyy=sin(t1).*(sin(t2+t3).*cos(t5)+sin(t2+t3)*d4+a3*cos(t2+t3)+a2*cos(t2));
pzz=cos(t2+t3)*d4-a3*sin(t2+t3)-a2*sin(t2);
figure(1),plot(t,t6,'r')
hold on
plot(t,t1,'k')
plot(t,t2,'b')
plot(t,t3,'m')
plot(t,t4,'g')
plot(t,t5,'c')
grid on
hod off
figure(2)
plot3(0.001*px,0.005*py,0.001*pz,'r',pxx,pyy,pzz,'b')
grid on
hold off
The function file :
function [theta,t1,t2,t3,t4,t5,t6]=part2(a2,a3,d3,d4,px,py,pz)
t1=atan2(px(1),py(1))-atan2(d3,sqrt(px(1)^2+py(1)^2-d3^2));
k=(px(1)^2+py(1)^2+pz(1)^2-a2^2-a3^2-d3^2-d4^2)/2*a2;
t3=atan2(a3,d4)-atan2(k,sqrt(a3^2+d4^2-k(1)^2));
t23=antan2((-a3-a2*cos(t3))*pz(1)-(cos(t1)*px(1)+sin(t1)*py(1))*(d4-a2*sin(t3)),(a2*sin(t3)-d4)*pz(1)+(a3+a2*cos(t3))*(cos(t)*px(1)+sin(t1)*px(1)+sin(t1)*py(1)));
t2=t23-t3;t4=0;t5=0;t6=0;
c1=cos(t1);s1=sin(t1);c2=cos(t2);s2=sin(t2);c3=cos(t3);s3=sin(t3);c23=cos(t23);s23=sin(t23);c4=cos(t4);c5=cos(t5);s5=sin(t5);c6=cos(t6);s6=sin(t6);
r13=-c1*(c23*c4*s5+s23*c5)-s1*s4*s5;
r23=-s1*c(23*c4*s5+s23*c5)+c1*s4*s5;
r33=s23*c4*s5-c23*c5;
r11=c1*(c23*(c4*c5*c6-s4*s6)-s23*s5*c6)+s1*(s4*c5*c6+c4*s6);
r21=s1*(c23*(c4*c5*c6-s4*s6)-s23*s5*c6)-c1*(s4*c5*c6+c4*s6);
r31=-s23*(c4*c5*c6-s4*s6)-c23*s5*c6;
for i=2:24
t1(i,1)=atan2(px(i,1),py(i,1))-atan2(d3,sqrt(px(i,1)^2+py(i,1)^2-d3^2));
k(i,1)=(px(i,1)^2+p(i,1)^2+pz(i,1)^2-a2^2-a3^2-d3^2-d4^2)/2*a2;
t3(i,1)=atan2(a3,d4)-atan2(k(i,1),sqrt(a3^2+d4^2-k(i,1)^2));
t23(i,1)=atan2((-a3-a2*cos(t3(i-1,1)))*pz(i,1)-cos(t1(i-1,1))*px(i,1)+sin(t1(i-1,1))*py(i,1)*(d4-a2*sin(t3(i-1,1))),(a2*sin(t3(i-1,1))-d4)*pz(i,1)+(a3+a2*cos(t3(i-1,1)))*(cos(t1(i-1,1))*px(i,1)+sin(t1(i-1,1))*px(i,1)+sin(t1(i-1,1))*py(i,1)));
t2(i,1)=t23(i,1-t3(i,1));
c1=cos(t1(i-1,1));s1=sin(t1(i-1,1));c2=cos(t2(i-1,1));s2=sin(t2(i-1,1));c3=cos(t3(i-1,1));
s3=sin(t3(i-1,1));c23=cos(t23(i-1,1));s23=sin(t23(i-1,1));c4=cos(t4(i-1,1));s4=sin(t4(i-1,1));
c5=cos(t5(i-1,1));s5=sin(t5(i-1,1));c6=cos(t6(i-1,1));s6=cos(t6(i-1,1));
r13(i-1,1)=-c1*(c23*c4*s5+s23*c5)-s1*s4*s5;
r23(i-1,1)=-s1*(c23*c4*s5+s23*c5)+c1*s4*s5;
r33(i-1,1)=s23*c4*s5-c23*c5;
t4(i,1)=atan2(-r13(i-1,1)*s1+r23(i-1,1)*c1,-r13(i-1,1)*c1*c23-r23(i-1,1)*s1*c23+r33(i-1,1)*s23);
s5=-r13(i-1,1)*(c1*c23*c4+s1*s4)+r23(i-1,1)*(s1*c23*c4-c1*s4)-r33(i-1,1)*s23*c4;
c5=r11(i-1,1)*(-c3*s23)+r23(i-1,1)*(-s1*s23)+r33(i-1,1)*(-c23);
t5(i,1)=atan2(s5,c5);
r11(i-1,1)=c1*(c23*(c4*c5*c6-s4*s6)-s23*s5*c6)+s1*(s4*c5*c6+c4*s6);
s6=-r11(i-1,1)*(c1*c23*s4-s1*c4)-r21(i-1,1)*(s1*c23*s4+c1*c4)+r31(i-1,1)*s23*s4;
c6=r11(i-1,1)*((c1*c23*c4+s1*s4)*c5-c1*s23*s5)+r21(i-1,1)*((s1*c23*c4-c1*s4)*c5-c1*s23*s5)-r31(i-1,1)*(s23*c4*c5+c23*s5);
t6(i,1)=atan2(s6,c6);
end
theta=[t1,t2,t3,t4,t5,t6]
0 comentarios
Respuesta aceptada
Más respuestas (1)
Steven Lord
el 17 de En. de 2021
Set an error breakpoint and run your code. When MATLAB stops on that line, look at the values you're passing into atan2 in the call on that line. My guess is that while you expect the value whose square root you're passing into atan2 to be either positive or zero it's actually very slightly negative due to floating point issues.
x = 0.3-0.2-0.1
In exact arithmetic, x would be 0. But none of one-tenth, two-tenths, or three-tenths can be exactly represented in double precision so it's not. So sqrt(x) is complex not 0.
sqrt(x)
0 comentarios
Ver también
Categorías
Más información sobre Logical en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!