Plotting an Archimedean Spiral
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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 个评论
KALYAN ACHARJYA
2019-10-15
Is your plot not an Archimedean Spiral? or ??
Rajbir Singh
2019-10-15
KALYAN ACHARJYA
2019-10-15
Then what is the problem?
Rajbir Singh
2019-10-15
编辑：Rajbir Singh
2019-10-15
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
2019-10-16
采纳的回答
Jos (10584)
2019-10-15
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 个评论
Rajbir Singh
2019-10-16
Rajbir Singh
2019-10-16
Jos (10584)
2019-10-16
hint: use a minus sign :-)
Rajbir Singh
2019-10-17
Leroy Tyrone
2023-2-8
编辑：Leroy Tyrone
2023-2-8
@ 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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