How to calculate sway area rate for center of pressure

Hello, i want to calculate sway area (mm2 /s) which is defined as the area enclosed by the center-of-pressure trajectory per unit time. Some useful paper (find below). Please find attached the CoPx and CoPy data.

Hufschmidt A, Dichgans J, Mauritz H, Hufschmidt M (1980) Some methods and parameters of body sway quantifcation and their neurological applications. Arch Psyc Nervenkr 228:135–150

Prieto TE, Myklebust JB, Hofman RG, Lovett EG, Myklebust BM (1996) Measures of postural steadiness: differences between healthy young and elderly adults. IEEE Trans Biomed Eng 43:956–966

Answer 1

[William Rose]

@chris pamp,

Since you already have the data in Excel. there is no need for Matlab. Do the calculation in Excel, using equation 19 of Prieto et al., 1996.

When I opened the file you attached, I received a warning message: "External links have beeen disabled", and the data do not look like COP data. There are no headers and the first row is at time=0.006. Therefore I have not analyzed the data in the spreadsheet you uploaded.

I am attaching an Excel file that demonstrates the calculation of Sway Area using equation 19 of Prieto et al., 1996.

Good luck.

[chris pamp]

Thank you! Could you write it and for matlab as well?

[William Rose]

@chris pamp, please email.

[William Rose]

@chris pamp,

As I mentoined earlier, the data in the file you posted does not look like actual COP data, since it has many values of zero in the columns. THereeefore I will demonstrate the Sway Area calciulaiton with other COP data which I collected.

%Read data from file and remove mean values
data=readmatrix('Test_Data.txt');
t=data(:,2);                    %time
x=data(:,3)-mean(data(:,3));    %COPx (mm)
y=data(:,4)-mean(data(:,4));    %COPy (mm)
N=length(x);
T=t(end)-t(1);

%compute sway area rate
a=abs(x(2:end).*y(1:end-1)-x(1:end-1).*y(2:end));
swayAreaRate=sum(a)/(2*T);

%display results
fprintf('Time=%.2f, Sway Area Rate=%.2f mm^2/s.\n',T,swayAreaRate)

Time=20.61, Sway Area Rate=649.14 mm^2/s.

%plot data
plot(x,y,'-b.');
grid on; xlabel('X (mm)'); ylabel('Y (mm)'); title('C.O.P.')

Try the code above.

[chris pamp]

Thank you very much!

[Mario]

Hello, I'm also working with the COP and it's really usefull your answer, thanks. Can i askyou something? it's possible to calculate the pathlenght? how will do it?

Thanks for the help

[William Rose]

@Mario Yes it is possible, and not difficult. I am traveling now and have only a cell phone. More later.

[William Rose]

@Mario,

This uses a two-column array data which has the x,y coordinates of the center of pressure. In this example, the COP path consists of 100 points around the unit circle.

data=[cos(2*pi*(0:100)/100);sin(2*pi*(0:100)/100)]';
%plot(data(:,1),data(:,2),'-r.') %view the path
n=length(data);
pL1=0;
for i=2:n
    pL1=pL1+sqrt((data(i,1)-data(i-1,1))^2+(data(i,2)-data(i-1,2))^2);
end
fprintf('Path length 1=%.3f.\n',pL1)

Path length 1=6.282.

The length is 2*pi, as expected for this path.*

In Matlab it is a point of pride to eliminate for loops. You can compute the path length without a for loop:

pL2=sum(((diff(data(:,1)).^2+diff(data(:,2)).^2).^0.5));
fprintf('Path length 2=%.3f.\n',pL2)

Path length 2=6.282.

The pL2 code does not use n, so you can eliminate the line n=length(data); also.

Try it. Good luck.

*Not exactly, since the path is a 100-gon, not a perfect circle.

Answer 2

[William Rose]

@chris pamp, You are welcome. Best wishes for your research.