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Plotting an Archimedean Spiral

조회 수: 79 (최근 30일)
Rajbir Singh
Rajbir Singh 2019년 10월 15일
편집: Leroy Tyrone 2023년 2월 8일
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?
  댓글 수: 5
Rajbir Singh
Rajbir Singh 2019년 10월 15일
편집: Rajbir Singh 2019년 10월 15일
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 2019년 10월 16일
How can i change the rotation (clockwise or anti-cloclwise) of Archimedean Spiral?

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채택된 답변

Jos (10584)
Jos (10584) 2019년 10월 15일
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)]
  댓글 수: 5
Rajbir Singh
Rajbir Singh 2019년 10월 17일
It works, thanks once again. :)
Leroy Tyrone
Leroy Tyrone 2023년 2월 8일
편집: Leroy Tyrone 2023년 2월 8일
@ 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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