CODEWITHSUNDEEP
matlabhelix
Wazir
Wazir

Reputation: 110

Helix in Matlab

I am working to create a helix in Matlab.

Going by the below code:

t = 0:pi/50:20*pi;

(Can you please explain me this syntax or we have to follow this everytime when creating a helix?) 

st = sin(t);
ct = cos(t);
plot3(st,ct,t)

As maximum efficiency in a helix angle is between 40 and 45 degrees, if I want to enter the angle as 42, how is it possible in the code?

It would be very helpful if anyone can share their opinion on this

TIA

Upvotes: 0

Views: 337

Answers (1)

Florian
Florian

Reputation: 1824

What happens in the code is merely an execution of the parametric mathematical description of a helix, which you can read up on wikipedia as

x(t) = cos(t)

y(t) = sin(t)

z(t) = t

The first line of your code generates a vector for values of t from 0 to 20pi in steps of pi/50 (i.e., 1000 steps). Since every 2pi means one full rotation (cos and sin are 2pi-periodic), it coincides to 10 turns of the helix (if you want to change this, let t run up to 2*pi*NumberOfRotations). The other two lines generate corresponding vectors for x and y. plot3 plots a line in 3-D where x and y are passed and as argument for z we pass t since z=t.

To change the slope of the helix, use the more general description given by

x(t) = a*cos(t)

y(t) = a*sin(t)

z(t) = b*t

where a is the radius and b/a is the slope. To get 42° use b = a*atand(42). To make sure the aspect ratio is correct in display, use axis equal; after the plot and maybe axis vis3d; if you want to turn it around.

Upvotes: 1

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