my graph is not showing....please help

Hi there, im designing lobes for a gerotor and i inputted a code on matlab but the graph is not showing up.
This is tyhe first part of my code only:
%% Geometry Of Pump
%% Inizialisation
N_o = 5;
N_i = 4;
v = 0.6;
D =1;
E = D/(2*(N_o + 2*v));
re = v*E;
rh = E-re;
rp_o = D/2 - (2*re);
rp_i = rp_o*(N_i/N_o);
theta_e = (-2*pi/N_o)*(re/E);
theta_h = 0;
xe = -(rp_o + re)*sin(theta_e) + re*sin(theta_e*((rp_o/re) +1));
ye = (rp_o + re)*cos(theta_e) - re*cos(theta_e*((rp_o/re) +1));
ye = xe +1;
plot(xe, ye)
for theta_e = -(2*pi/N_o)*(re/E):0
[theta_e,rp_o] = pol2cart(xe,ye);
xe = -(rp_o + re)*sin(theta_e) + re*sin(theta_e*((rp_o/re) +1));
ye = (rp_o + re)*cos(theta_e) - re*cos(theta_e*((rp_o/re) +1));
theta_e = theta_e +1;
end
plot(theta_e,rp_o)

4 comentarios

It's because you are only plotting one point. If you add 'o'(see down below) you will be able to see where the point is on the graph.
N_o = 5;
N_i = 4;
v = 0.6;
D =1;
E = D/(2*(N_o + 2*v));
re = v*E;
rh = E-re;
rp_o = D/2 - (2*re);
rp_i = rp_o*(N_i/N_o);
theta_e = (-2*pi/N_o)*(re/E);
theta_h = 0;
xe = -(rp_o + re)*sin(theta_e) + re*sin(theta_e*((rp_o/re) +1));
ye = (rp_o + re)*cos(theta_e) - re*cos(theta_e*((rp_o/re) +1));
ye = xe +1;
plot(xe, ye,'o')
for theta_e = -(2*pi/N_o)*(re/E):0
[theta_e,rp_o] = pol2cart(xe,ye);
xe = -(rp_o + re)*sin(theta_e) + re*sin(theta_e*((rp_o/re) +1));
ye = (rp_o + re)*cos(theta_e) - re*cos(theta_e*((rp_o/re) +1));
theta_e = theta_e +1;
end
plot(theta_e,rp_o,'o')
Mohammad Jawad
Mohammad Jawad el 28 de Oct. de 2020
So how do I plot the whole thing? Please bare with me I'm not very strong in matlab
I'm not very strong in MATLAB either, but I know that you can use "hold on, off" to plot both points onto the same graph.
hold on
plot(theta_e,rp_o,'o')
hold off
If you need more than those points, you will probaly have to rewrite your code, or mabye someone with more experience will answer you.
ye = (rp_o + re)*cos(theta_e) - re*cos(theta_e*((rp_o/re) +1));
ye = xe +1;
Which one of these lines of code do you need, because the 2nd one writes a new value for ye.

Iniciar sesión para comentar.

Respuestas (1)

Elijah McNeil
Elijah McNeil el 28 de Oct. de 2020
Editada: Elijah McNeil el 28 de Oct. de 2020
E = 1/(2*(5 + 2*0.6));
re = 0.6*E;
rh = E-re;
rp_o = 1/2 - (2*re);
rp_i = rp_o*(4/5);
theta_e = (-2*pi/5)*(re/E);
theta_h = 0;
xe = -(rp_o + re)*sin(theta_e) + re*sin(theta_e*((rp_o/re) +1));
ye = (rp_o + re)*cos(theta_e) - re*cos(theta_e*((rp_o/re) +1));
plot(xe, ye,'o')
for theta_e = -(2*pi/N_o)*(re/E):0
[theta_e,rp_o] = pol2cart(xe,ye);
xe = -(rp_o + re)*sin(theta_e) + re*sin(theta_e*((rp_o/re) +1));
ye = (rp_o + re)*cos(theta_e) - re*cos(theta_e*((rp_o/re) +1));
theta_e = theta_e +1;
end
hold on
plot(theta_e,rp_o,'o')
hold off
This is assuming that you don't need this line of code:
ye = xe +1;
If you needed that line of code, then just sub it in for:
ye = (rp_o + re)*cos(theta_e) - re*cos(theta_e*((rp_o/re) +1));
Also, if you have variables that are scalars, you don't really need to include that variable, as it just takes more time to write.

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