Plotting an Archimedean Spiral

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Rajbir Singh
Rajbir Singh il 15 Ott 2019
Modificato: Leroy Tyrone il 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 Commenti
Rajbir Singh
Rajbir Singh il 15 Ott 2019
Modificato: Rajbir Singh il 15 Ott 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 il 16 Ott 2019
How can i change the rotation (clockwise or anti-cloclwise) of Archimedean Spiral?

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Risposta accettata

Jos (10584)
Jos (10584) il 15 Ott 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 Commenti
Rajbir Singh
Rajbir Singh il 17 Ott 2019
It works, thanks once again. :)
Leroy Tyrone
Leroy Tyrone il 8 Feb 2023
Modificato: Leroy Tyrone il 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?

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