calculating the center of mass

Hi everyone,

I got different datasets, now I wanted to calculate the center of mass the different datasets.

close all, clc

load('dataset1.mat');
plot(x,y, '-'), hold on

load('dataset2.mat');
plot(x,y, '-'), hold on

So I can interpolerate those datasets by just using a "-" inside my plot function but I have no idea how to write a center of mass code, my programming skills in Matlab are quite bad since I programmed for years only with c+, I also tried to find a proper decent code to calculate a center of mass of a 2D-dataset but I just found a thread about a 3D-array....

Answer 1

[Walter Roberson]

As usual, the center of mass of something of uniform density is at the centroid.

For finite approximating points pretty much evenly distributed, the centroid is at the mean.

[Casio]

see my second comment beyond

Answer 2

[Image Analyst]

If you want, you can use poly2mask() to create an image and then use regionprops() to compute the centroid and weightedcentroid.

[Casio]

Ok I wanted to try it with the simpsons method, but how can I calculate the integral when I dont have a function given instad of that I just have a lot of points

[Walter Roberson]

Recall that one of the definitions of integration is the limit of discrete sums. You have discrete locations so although you cannot calculate the integral you can calculate an approximation.

[Casio]

sorry yeah I understood now how the compiler works here, there was some code behind sorry it works now thanks !

[Image Analyst]

Is that the only code in the m-file, or is there more code that you did not show? If it's the only code I think it looks okay. How are you calling it? If there is more code above it, then that could be a problem.

[Casio]

What I am doing wrong I got this function:

my code looks like this with the simpsons method:

why I dont get any result whatever I write in my disp() method it gives me those weird numbers, maybe because I have to use the absolute for the negative values?

[Image Analyst]

I don't know where those numbers are coming from. If you want us to run it you'll have to attach simpsons.m and dataset2.mat.

[Walter Roberson]

The data set is attached above.

[Image Analyst]

You're taking a string and adding the badly-named o (lower case letter O I presume) to it, which has the value 1. So it converts the string to a vector of the ASCII values and adds 1 to those ASCII values. I don't know why you do that. If you want to display a string and a number in the command window, use fprintf():

fprintf('Ergebnis: %f\n', o); % Print directly to command window.

Or construct a string and use disp():

o = 3;
myString = sprintf('Ergebnis: %f\n', o);
disp(myString);

[Casio]

so this is my simpsons method:

    function I = simpsons(f,a,b,n)
    n=numel(f)-1; 
    h=(b-a)/n;
    I= h/3*(f(1)+2*sum(f(3:2:end-2))+4*sum(f(2:2:end))+f(end));
    end

obviously I want to print the result in my command window but that didnt work

[Walter Roberson]

disp(result)

[Image Analyst]

Instead of your lines 19 and 20, do this:

fprintf('Casio, your result = %f.\n', result);

[Casio]

I did that and I still get 0 for my area, so something is wrong in my method?

[Image Analyst]

Evidently so. Please attach the entire m-file(s) with the paper clip icon.

[Casio]

here you got the entire m-files, the datasets are already above

[Walter Roberson]

You have

for j =1:2000
I=simpsons(fi,x(1),x(n),[]);
result=result+I;
end

Why are you doing the same integration 2000 times? Your call to simpsons is independent of your loop control variable, j, so you are doing the same thing every time.

Your x(1) and x(n) are the same, so when you get into your simpsons formula, the difference between them is 0 so your h is 0, which causes the calculated area to be 0.

Your calculation with h in your simpsons routine assumes that the x values only progress. That is not true in your situation: your x reach a maximum and turn back.

simpson's rule is for the area under a curve relative to a baseline of y = 0. That is not the situation you have.

What you need to do is pair off all of the x in one direction compared to the other, take the y at the "upper" curve minus the y at the "lower" curve, and that is the value you should be putting through simpsons.

Though perhaps if you just break the range into two pieces, the upper and lower curve, simpson's the two and subtract the lower from the upper.

[Casio]

so how should I change the code so that the x goes turns around and goes back?

I understood kinda what you propose but I have problems to convert that into my code

[Walter Roberson]

mask = x(1:end-1) <= x(2:end);
xupper = x(mask);
fiupper = fi(mask);
area_upper = simpsons(fiupper, xupper(1), xupper(end), []);
xlower = x(~mask);
filower = fi(~mask);
area_lower = simpsons(filower, xlower(1), xlower(end), []);

result = area_upper - area_lower;

...

Though I am not at all certain how you intend to use this area to calculate the centre of mass.

[Casio]

theres a formular for it right?

but I get a -10.2802 so the area of the data is 10.2802 because an area cant be negative right?

[Image Analyst]

Maybe your xupper and xlower are swapped.

[Walter Roberson]

Your

      h=(b-a)/n;

should have an abs(). Otherwise in your formula the sign of the area changes depending on whether the points were presented in one direction or the other.