no puedo plotear en mi codigo
2 次查看(过去 30 天)
显示 更早的评论
%R1=3;% radio de las bandas
%R2=2;
bi=1; %espaciado;
%h=100;
%descenso
tr=linspace(pi,0,100);
tr1=linspace(2*pi,pi,100);
tr12=linspace(pi,0,100);
%elipse externa
a=[3*cos(tr) 3*cos(tr1)];
b=[3*sin(tr) 3*sin(tr1)-6];
%interna
a1=[2*cos(tr) 2*cos(tr1)];
b1=[2*sin(tr) 2*sin(tr1)-6];
% la separacion para cada banda es = b
c=[(bi*tr12)/2*pi];
c=[c+14 c+13 c+12 c+11 c+10 c+9 c+8 c+7 c+6 c+5 c+4 c+3 c+2 c+1];
%ascenso
tr2= linspace(0,pi,100); tr22=linspace(pi,2*pi,100); tr23=linspace(0,pi,100);
a2=[3*cos(tr2) 3*cos(tr22)];
b2=[3*sin(tr2) 3*sin(tr22)-4];
%interna
a3=[2*cos(tr2) 2*cos(tr22)];
b3=[2*sin(tr2) 2*sin(tr22)-6];
%separacion es de 2
c1=[(bi*tr23)/(2*pi)];
c1=[c1+0 c1+2 c1+4 c1+6 c1+8 c1+10 c1+12 c1+14];
for i=1:3
a=[a a];
b=[b b];
a1=[a1 a1];
b1=[b1 b1];
a2=[a2 a2];
b2=[b2 b2];
a3=[a3 a3];
b3=[b3 b3];
end
%linea cruzada
tz=linspace (0,pi/4,100);
tz1=linspace(pi/4,pi/2,100);
a4=[3*cos(tz) cos(3*pi/2)-tz1];
b4=[3*sin(tz)-6 sin((3*pi/2)-tz1)];
a5=[cos(tz) 3*cos((3*pi/2)-tz1)];
b5=[sin(tz)-6 3*sin((3*pi/2)-tz1)];
c2=c1(end)*ones(1,200);
%graficas
plot3(a2,b2,c1,'r');
hold on
plot3(a3,b3,c1,'m');
plot3(a,b,c,'r');
hold on
plot3(a1,b1,c,'b');
plot3(a4,b4,c2,'g');
plot3(a5,b5,c2,'c');
xlim([-3 3]);
xlabel('axis x')
ylabel('axis y')
zlabel('axis z')
title('banda transportadora')
hold off
1 个评论
Jan
2022-10-30
I've included the error message in the question. It is easier to solve a problem than to guess, what the problem is.
回答(1 个)
Gokul Nath S J
2023-5-26
Hi Ricardo,
It seems that you are receving an error mentioning the vectors are of different length. The variable c1 is of length 800 while the other variables a2 and b2 are 1600. Please note that this difference is attributed to the fact that you are concatinating two matrixes together in the following lines.
a2=[3*cos(tr2) 3*cos(tr22)];
b2=[3*sin(tr2) 3*sin(tr22)-4];
Since the objective of the function is not mentioned in the question, as I workaround, I would suggest you check the dimension of the a2, b2 and c1 variable and understand whether there is a requirement for the concatenation or you might have missed the concatination of the variable c1.
with regards,
Gokul Nath S J
0 个评论
另请参阅
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!