Plotting an Archimedean Spiral

84 views (last 30 days)
Rajbir Singh
Rajbir Singh on 15 Oct 2019
Edited: Leroy Tyrone on 8 Feb 2023
r = 12.5; %outer radius
a = 0; %inner radius
b = 0.01; %incerement per rev
n = (r - a)./(b); %number of revolutions
th = 2*n*pi; %angle
Th = linspace(0,th,1250*720);
x = (a + b.*Th).*cos(Th);
y = (a + b.*Th).*sin(Th);
plot(x,y)
The code executes well r, a, n and b are correct. Th and th both are also correct, but the problem which arises is in the values of x and y.
outer value or last value (desired) should be 12.5, but after execution it gives 78.53 and same corresponds to y.
what can be the solutions of this problem?
  5 Comments
Rajbir Singh
Rajbir Singh on 15 Oct 2019
Edited: Rajbir Singh on 15 Oct 2019
Sir,
The output which i am getting is an Archimedean Spiral, thats fine. But the problem arises with the output values x and y.
According to the software that i am using r, a, n, b, th and Th values are correct.
My desired outer radius for archemedean spiral is 12.5 but it gives 78.53
Rajbir Singh
Rajbir Singh on 16 Oct 2019
How can i change the rotation (clockwise or anti-cloclwise) of Archimedean Spiral?

Sign in to comment.

Accepted Answer

Jos (10584)
Jos (10584) on 15 Oct 2019
In the computation of x and y you wrongly multiply b with Th. You should multipy by Th / (2*pi):
r = 12.5; %outer radius
a = 0; %inner radius
b = 0.5; %incerement per rev % Jos: changed to see the spiral!!
n = (r - a)./(b); %number of revolutions
th = 2*n*pi; %angle
Th = linspace(0,th,1250*720);
x = (a + b.*Th/(2*pi)).*cos(Th);
y = (a + b.*Th/(2*pi)).*sin(Th);
% better:
% i = linspace(0,n,1250*720)
% x = (a+b*i).* cos(2*pi*i)
plot(x,y)
[x(end) y(end)]
  5 Comments
Rajbir Singh
Rajbir Singh on 17 Oct 2019
It works, thanks once again. :)
Leroy Tyrone
Leroy Tyrone on 8 Feb 2023
Edited: Leroy Tyrone on 8 Feb 2023
@ Jos Is it possible to return which revolution in 'n' that each value in 'Th' belongs to? Alnd also to plot the points as equidistant by assigning a variable 's' as arc length?

Sign in to comment.

More Answers (0)

Categories

Find more on MATLAB in Help Center and File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!