Drawing an arc on the end of a straight line

I am trying to find out a way to plot an arc on the end of a line in Matlab.

For instance:

I have an array that runs from -1 to +1, and I wish for the central section (-0.7:0.7) to be 0, and then at each side a quarter circle of radius 0.3 going up into the +y direction. (See image if my poor description of the geometry is doing no good).

I cannot figure out the equation of a line to do the curves, and I can't seem to translate anything on the internet to this problem, where I need to find the equation of that line as a function of x, where the x-axis runs from 1:20, and each curve section is from 1:30 and 70:200 respectively.

Any help is much appreciated, I'm sure this can't be that hard, but I'm out of ideas.

Answer 1

[Star Strider]

Recalling basic geometry, for an angle vector going from 0 to 2*pi, the equation for a complete circle with radius ‘r’ is given as:

angle = linspace(0,2*pi);
r = 1;
x = r*cos(angle);
y = r*sin(angle);

figure(1)
plot(x, y)
grid
axis equal

You can translate the circle (move its centre) by adding appropriate values to the ‘x’ and ‘y’ vectors, and changing its radius involves changing the value assigned to ‘r’. So to move it to be centred at (3,5), add 3 to ‘x’ and 5 to ‘y’. Plotting only a segment of the circle requires you to change the first and second arguments to the linspace call.

Experiment with it! You’ll have your finished plot in a few minutes.

[Image Analyst]

By the way, you can make rounded rectangles with the rectangle() function but they don't give you the coordinates of the outline. That would give you what you want, if they only exported it, but they don't. I've put it in as a request for enhancement.

[Chris [e-mail]]

Thanks Star Strider.

I follow this concept, but I am struggling with mapping it to my problem (I may be being incredibly dense).

I can visualise creating the circle (the section from pi:1.5*pi) and translating it's position, but I don't understand how to get the formula above to represent a y value, where y is a function of x, and my x values range from say 1:30.

In the example given you're plotting y versus x, where x is already a function. I need x to be a linear integer progression (i.e 1:1:30 in my example).

[Star Strider]

It’s a slightly challenging problem, and involves creating two quarter-circle arcs and connecting them with a line. This code is reasonably flexible, so changing the length of the center connecting line will reposition the arcs automatically.

The only part that may not be immediately straightforward is the ‘circ’ function. It calculates the x- and y-coordinates of a circle, so you need give it only the three arguments it asks for: ‘a’, the angle you want to define the circle; ‘r’, the radius, and ‘ofst’, the [x,y] (in that order) offset at which you want the centre. I will let you analyse the code beyond that to see how it works. It could be a bit more efficient (using the more efficient bsxfun rather than repmat), but if it’s a one-off, that likely doesn’t matter.

The code:

R4crc = linspace(-pi/2, 0, 30);             % Angle Defining Right Arc
L4crc = linspace(pi, 3*pi/2, 30);           % Angle Defining Left Arc
CL = [-20:20];                              % Connecting Line
r = 30;                                     % Radius
circ = @(a,r,ofst) r.*[cos(a); sin(a)]+repmat(ofst(:),1,length(a));
Rcxy = circ(R4crc,r,[CL(end),r]);          % Calculate Right Arc
Lcxy = circ(L4crc,r,[CL(1),r]);            % Calculate Left Arc

LnWd = 2;                                   % Line Width
LnCl = 'k';                                 % Line Colour

figure(1)
plot(CL, zeros(size(CL)), 'LineWidth',LnWd, 'Color',LnCl)
hold on
plot(Rcxy(1,:), Rcxy(2,:), 'LineWidth',LnWd, 'Color',LnCl)
plot(Lcxy(1,:), Lcxy(2,:), 'LineWidth',LnWd, 'Color',LnCl)
hold off
grid
axis equal

produces this plot:

[Image Analyst]

Nice answer. I voted for it. Chris might also want to vote for it and then accept it since it seems like this definitely answers the question. By the way, creating an arc is in the FAQ: http://matlab.wikia.com/wiki/FAQ#How_do_I_create_an_arc.3F

[Star Strider]

Thank you, Image Analyst.

[Chris [e-mail]]

Star Strider, that is perfect - that circ function is very neat.

Thank you very much.

[Star Strider]

My pleasure!

If my Answer solved your problem, please Accept it!

[Moby Philip]

Hello @Star Strider can you please elaborate why you choose R4crc = linspace(-pi/2, 0, 30); as from -90. It should be 0 right because you are in X-Axis.

[Star Strider]

I chose it because it works in this instance.