Plotting an Archimedean Spiral
32 views (last 30 days)
Show older comments
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
Accepted Answer
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
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?
More Answers (0)
See Also
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!