how can I plot a 3-D helical trajectory

ı need to plot a trajectory like pitcure above but ı could not. Can u help me. the known parameters are that diameter=4 [m], pith (gape) = 2 [m], height = 18 [m]. Also the simulation time is 50 second.

Answer 1

[Matt J]

t=linspace(0,18*pi, 3000);

plot3(2*cos(t), 2*sin(t), t/pi)

[baris]

why did u product the height (18) with pi.

[Matt J]

If the pitch=2, then you need 9 revolutions (2*pi*9=18*pi) to reach a height of 18.

[baris]

but there are more spirals than the pitcure ı attached. What is the reason of that?

[Matt J]

Because the helix in your plot is not to spec. You said you wanted a pitch of 2, but plotted a helix with a pitch of about 5.

[baris]

it is my bad, sir. I thought the pitch is the distance between spirals. I apologize. The known parameters are that if we look over, the helix diameter is 4, the gap beetwen spirals is 2, the heights is 18 and the simulation time is 50 seconds. I also need to get trajectory like down bellow

[Matt J]

OK. But what's the problem? Just increase the pitch.

t=linspace(0,4.5*2*pi, 3000); 
T=t(end);

s=t(end)/50; %rescale to duration 50
t=t/s;

x=@(t) 2*cos(s*t); y=@(t) 2*sin(s*t); z=@(t)s*18*t/T;

plot3(x(t),y(t),z(t))

figure; 
subplot(3,1,1); plot(t,x(t))
subplot(3,1,2); plot(t,y(t))
subplot(3,1,3); plot(t,z(t))