Draw arc in Matlab?

In Matlab is there any special function to draw arc with user defined radius, points and angle. If it is not there how is it possible to draw a curve in a figure using user defined radius,angles, points etc. Thank you

Answer 1

[Star Strider]

There are no such functions in core MATLAB, but they’re easy enough to write.

This code plots an arc of stars:

circr = @(radius,rad_ang)  [radius*cos(rad_ang);  radius*sin(rad_ang)];         % Circle Function For Angles In Radians
circd = @(radius,deg_ang)  [radius*cosd(deg_ang);  radius*sind(deg_ang)];       % Circle Function For Angles In Degrees

N = 25;                                                         % Number Of Points In Complete Circle
r_angl = linspace(pi/4, 3*pi/4, N);                             % Angle Defining Arc Segment (radians)
radius = 1.5;                                                   % Arc Radius
xy_r = circr(radius,r_angl);                                    % Matrix (2xN) Of (x,y) Coordinates

figure(1)
plot(xy_r(1,:), xy_r(2,:), 'bp')                                % Draw An Arc Of Blue Stars
axis([-1.25*radius  1.25*radius    0  1.25*radius])             % Set Axis Limits
axis equal                                                      % No Distortion With ‘axis equal’

[rupam baruah]

Thanks a Lot. Can u calculate the length of the arc in Pixel unit?

[Star Strider]

My pleasure.

Calculating the length of the arc mathematically is straightforward.

Calculating it in pixels depends on a number of characteristics of the image. Perhaps the regionprops function will do what you want. Experiment with it.

[kei mcqueen]

Hello thank you for your awesome code, it has helped me with a problem I was having. Is it possible to modify the origin point of the arc? When I try modifying it myself it keeps the origin at 0,0. Thank You.

[Star Strider]

My pleasure.

Add the origin as a third argument:

circr = @(radius,rad_ang,p0)  [radius*cos(rad_ang)+p0(1);  radius*sin(rad_ang)+p0(2)];         % Circle Function For Angles In Radians

Here, ‘p0(1)’ is the x-coordinate of the origin, ‘p0(2)’ is the y-coordinate. Change ‘circd’ similarly.

A vote for my Answer would be appreciated!

[Ram W]

sorry, I have a qestion regarding your last comment, how can I modiy the origin with 3 coordinates (x,y,z) ?

[Amal Fennich]

r_angl = linspace(pi/4, 3*pi/4, N);

what does this meanis beacause I have an angle of 3.433 degree and I do not know haow to implement it here ! any help ?

[Sean Brennan]

In this line, linspace produces a vector of N points, equally spaced (linearly - hence the name), starting from pi/4 to 3pi/4.

If you have an angle of 3.433 degrees to start, and say 90 degrees to end, and you wanted 100 points between these start and end angles, then your command would be

r_angl = linspace(3.433*pi/180, 90*pi/180,100)

Hope this helps!

[Audrey Fernandes]

How can this arc be repeated into a series of arcs? Like the one this the image below?

[Matheus Sousa]

Adding the center points of each circle to xy_r should do it, in the example abova nothing was added so the center is in [0,0], but if you add 3 on xy_r(1,:) and 0 to xy_r(2,:) the position of the arc will change as well.

Answer 2

[Ade Ade]

%Equation of a circle with centre (a,b) is (x-a)^2+ (y-b)^2 = r^2

%Circle Centre (1,1), radius = 10

k=1; %counter

c =1 ; % value of x at the centre of the circle

while c <=11

x(k) = c ;

vv = (c-1)^2 ;

y (k) = 1 + real (sqrt (100 - vv) );

c= c + 0.02;

k=k+1;

end

plot (x, y, 'r')

axis equal

Answer 3

[Sudhir S]

Hello everyone, I'm currently working on a project that requires MATLAB code. This is a 2D blade profile which was design using CATIA V5. I was able to code up till this point, but I got stuck on the arc creation. The radius of the arc is given, and the beginning point is fixed for that arc. The inlet angle is used to determine the finishing point. Could someone assist me in locating a solution?