Plotting an Archimedean Spiral

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Rajbir Singh
Rajbir Singh le 15 Oct 2019
Modifié(e) : Leroy Tyrone le 8 Fév 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 commentaires
Rajbir Singh
Rajbir Singh le 15 Oct 2019
Modifié(e) : Rajbir Singh le 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 le 16 Oct 2019
How can i change the rotation (clockwise or anti-cloclwise) of Archimedean Spiral?

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Réponse acceptée

Jos (10584)
Jos (10584) le 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 commentaires
Rajbir Singh
Rajbir Singh le 17 Oct 2019
It works, thanks once again. :)
Leroy Tyrone
Leroy Tyrone le 8 Fév 2023
Modifié(e) : Leroy Tyrone le 8 Fév 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?

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