plotting an N segments circle

Hi all,

I am trying to plot an N segments shape using this line of code:

N =10;

theta = (0: (2*pi)/N : 2 * pi) ;

theta_centers = (pi/2: (2*pi)/N : 4*pi) ; % this is to find the centers of the prevoius line

xb = 0.02*sin(theta) ;

yb = 0.02*cos(theta) ;

xbc = 0.02*sin(theta_centers) ; % to find the centers of xb

ybc = 0.02*cos(theta_centers) ;

plot(xb,yb,'r',xbc,ybc,'o')

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after running this code, the 'o' are not on the main function lines....

WHY?

thanks

Answer 1

[Star Strider]

WHY?

Because you are plotting a decagon rather than a circle. If you want them to be on the line, you need to reduce their radius to times the circle radius, so:

Ns = N;                                            % Number Of Sides Of Polygon
ro = sin(pi*(Ns-2)/(2*Ns));
xbc = ro*0.02*sin(theta_centers) ;  % to find the centers of xb 
ybc = ro*0.02*cos(theta_centers) ;

I will let you ponder that derivation in detail.

However it will help to know that each segment of the circle subtends an angle of , so half of that is . The angle formed by the radius passing through the center of the chord of each segment is a right angle, , so the remaining angle is . Multiply the sine of that angle by the radius of the circle (here 0.02) to put the ‘o’ markers on the lines between the segments.

[Al Alothman]

Thank you so much!

I've noticed for even numbers of N, (xb,yb) & (xbs,ybc) overlaps, as for 20,40,60...

I appriciate your help

[Star Strider]

My pleasure.

‘I've noticed for even numbers of N, (xb,yb) & (xbs,ybc) overlaps, as for 20,40,60...’

It depends on how you define the polygon. As an experiment (and to test my code), I came up with this before I posted my Answer:

N = 60;
theta = linspace(0, 2*pi, N);
theta_centers = theta + pi/(N-1);
xb = 0.02*sin(theta) ; 
yb = 0.02*cos(theta) ;
Ns = N - 1;                                         % Number Of Sides Of Polygon
ro = sin(pi*(Ns-2)/(2*Ns));
xbc = ro*0.02*sin(theta_centers) ;  % to find the centers of xb 
ybc = ro*0.02*cos(theta_centers) ;
figure
plot(xb,yb,'+-r',xbc,ybc,'o')      
axis equal
axis([-0.017  -0.012    -0.017  -0.012])            % Zoom Axis (Delete To See Entire Circle)

There is never an overlap, and the ‘o’ markers are always mid-way between the polygon ‘+’ angle markers. (The ‘Ns’ variable is defined differently here because of the way linspace creates its vectors. The rest of the code is the same.)

Answer 2

[Al Alothman]

Thank you so much!

in my previous code, I started my theta from (pi/2: (2*pi)/N : 4*pi) to get the first edge of the polygon at 0, now it is starts away from the xb = 0,02 and yb = 0

any advice?

Thanks!

[Star Strider]

As always, my pleasure!

My code only does the circumference one time. (The plot looks really strange otherwise.) If you want it to start at pi/2 (or any other angle), add that value to the ‘theta’ vector created by linspace:

theta = linspace(0, 2*pi, N) + pi/2;

The rest of my code remains the same. (The polygon then rotates by the added angle.)