Question

This question is given from 'Numerical Techniques and Statistical Analysis' paper which based on using Matlab program.

However, I don't know how can I solve this tough question.

Could anyone please help me?

Thank you so much.

6. Even today, if you are attempting to sail to Fiji, you need to pass a test by Maritime New Zealand to show that you can navigate without using a GPS! One of the things you need to do is find your (, y) position given a number of circles with known centers, and radii, as shown in Fig. 2. 0 .5 -10 -5 5 15 20 Figure 2: The intersection of 5 different circles of different radii is in this case at the point Write a MATLAB routine to find the intersection of a bunch of circles of different centers and radii. You should expect around 5 to 10 circles, and also anticipate that the intersec- tion may not be perfect, i.e. in marine navigational terms, you should expect a 'cocked hat As a test-case, use the data for the 5 circles centers and radii given in Table 2. This data is similar to, but not exactly the same as, the case presented above in Fig. 2. Also assume for this section, that we are assume the (x, y) data lies on a perfectly flat plane, (not the surface of a sphere for example.) Table 2: Circle centers and radii on a flat plane. radius -7.37637.6512 7.8 3.51991.0563 5.6 5.26464.8721 10.0 2.17830.5596 4.9 -5.4952 2.1826 3.5 For the general case, you should anticipate the following issues: 1. You will probably have more circles than you strictly need. (With just two cir- cles, you possibly have two intersecting points, with three you might just have one common point.) 2. Due to imprecision in the distance measurements, (perhaps using an electronic range finder), you may not have a perfect intersection. In nautical navigation, this was classically known as a cocked hat3 In this case, you will need to find the best position, perhaps the one that minimises the squared errors. Figure 3: This is an actual chart from the bridge of the USS Midway aircraft carrier in San Diego harbour. You can see the small error in position as a crooked hat' penciled on the chart by the navigator 3. You may wish to think of ways to obtain a reasonable starting guess for the center coordinates, and also reasonable upper and lower bounds for these variables. For example, you should be able to easily construct rectangular bounds by looking at the spread of circle center points and their radii. 4. It is entirely possible that one of your 'sights' is completely wrong, so much so, that there is no reasonable solution. You should try to identify such cases, and flag appropriately Hint: Since you are solving a nonlinear system of algebraic equations, you should use either fsolve or lsqnonlin. See www.working-the-sails.com/fixing_a position.html

Answer 1

% Center of the circles

C1=[0,0];

C2=[5,0];

% Radius of the circles

R1=3;

R2=5;

% x1(x,y) on the first circle satisfy the following equation:

%(x1(1)-C1(1))^2+(x1(2)-C1(2))^2-R1^2=0

% x2(x,y) on the second circle satisfy the following equation:

%(x2(1)-C2(1))^2+(x2(2)-C2(2))^2-R2^2=0;

% once intersecting we have:

% x1(1)=x2(1);

% x1(2)=x2(2);

% then

F=@(x) ([(x(1)-C1(1))^2+(x(2)-C1(2))^2-R1^2; ...

(x(1)-C2(1))^2+(x(2)-C2(2))^2-R2^2]);

opt=optimoptions(@fsolve);

opt.Algorithm='levenberg-marquardt';

opt.Display='off';

x=fsolve(F,[C1(1),C1(1)+R1],opt);

fprintf('First intersection point: (%f,%f)\n',x(1),x(2));

x=fsolve(F,[C1(1),C1(1)-R1],opt);

fprintf('Second intersection point: (%f,%f)\n',x(1),x(2));

when you run it you get:

First intersection point: (0.900000,2.861818)

Second intersection point: (0.900000,-2.861818)

The main difference here is that the initial guess or starting points is located on one of the circles (so it satisfies one of the equations and not the other. (0,0) that you start doesn't satisfies any of the equations. The convergence rate would be much worse there. setting starting guess as (x,y)=(C(1), C(2)+R) helps it, and that is always the point right on top of one of the circles. Not necessarily the intersection. And Setting the initial guess like this is by no mean any restriction, and in fact choosing proper initial guess is always encouraged.

Alternatively you could have posed this as a constrained optimization problem. Then you had more flexibility for the initial guess.