Plotting an Archimedean Spiral

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Rajbir Singh
Rajbir Singh am 15 Okt. 2019
Bearbeitet: Leroy Tyrone am 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 Kommentare
Rajbir Singh
Rajbir Singh am 15 Okt. 2019
Bearbeitet: Rajbir Singh am 15 Okt. 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 am 16 Okt. 2019
How can i change the rotation (clockwise or anti-cloclwise) of Archimedean Spiral?

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Jos (10584)
Jos (10584) am 15 Okt. 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 Kommentare
Rajbir Singh
Rajbir Singh am 17 Okt. 2019
It works, thanks once again. :)
Leroy Tyrone
Leroy Tyrone am 8 Feb. 2023
Bearbeitet: Leroy Tyrone am 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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