Geodetic to cartesian coordinates
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Benedict Low
am 27 Mär. 2017
Kommentiert: David Goodmanson
am 22 Sep. 2022
Hi,
I work with GPS units for footballers and I want to convert geodetic coordinates to cartesian coordinates. I have tried using Matlab's geodetic2enu function but the coordinates come out funny. As an example, here are the latitude and longitude values of each corner of a football pitch, converted using Matlab's " geodetic2enu" function.
Pitch_dimensions=[50.707025 4.206677;50.70665 4.205923;50.706062 4.207913;50.705676 4.207147];
[xEast, yNorth, zUp] = geodetic2enu(Pitch_dimensions(:,1),Pitch_dimensions(:,2),0,50.70665,4.205923, 0,wgs84Ellipsoid);
Pitch_dimensions=[xEast yNorth];
scatter(Pitch_dimensions(:, 1), Pitch_dimensions(:, 2));
The scatter plot should turn out to be a basic rectangle, but it shows a slanted pitch instead
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/162225/image.png)
.
I just want to display a simple football field with a left-to-right direction of play, and the origin (0,0) to be the bottom-left corner flag of the pitch. Is there a rotation matrix that I need to use? Any help is appreciated.
Thank you.
Best regards,
Ben
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David Goodmanson
am 28 Mär. 2017
Bearbeitet: David Goodmanson
am 29 Mär. 2017
Hello Benedict, (revised)
The four points do form a pretty good rectangle, but it doesn't look like it because the x and y axis scaling on the plot are different. Try 'axis equal' just after the plot command, which should show you a rectangle, hopefully. The following
pd = Pitch_dimensions % in meters
norm(pd(1,:)-pd(2,:)) % width
norm(pd(3,:)-pd(4,:)) % width
norm(pd(1,:)-pd(3,:)) % length
norm(pd(2,:)-pd(4,:)) % length
norm(pd(1,:)-pd(4,:)) % diag
norm(pd(2,:)-pd(3,:)) % diag
shows that there is a bit of inaccuracy in the coordinates, on the order of 1 m, but the diagonals agree within about 2 m so it's a rectangle. (I have to say that I don't have the geodetic2enu function so I had to make my own simplified version, but I think it works acceptably for this purpose).
Pretty long touch lines.
At this point I will assume that you have gone through the same geodetic2enu process for players as for the corners. The result is a 4x2 array for the corners and 11x2 arrays for the teams, plus maybe a 1x2 for the ball, all in meters. Some demo code for that is
% demo code for players and ball, unrotated, in meters
pd = Pitch_dimensions; % in meters
p42 = pd(4,:)-pd(2,:); % touch line
p12 = pd(1,:)-pd(2,:); % goal line
team1 = rand(11,1)*p42+ rand(11,1)*p12;
team2 = rand(11,1)*p42+ rand(11,1)*p12;
ball = rand*p42 + rand*p12;
Here is an example of rotation and plotting. I used plot instead of scatter because I think it is more versatile. Lots of possibilities for a plot. You will see that the boundary is a bit off due to 1 m inaccuracies but you could always make one manually.
% start of real code
pd = Pitch_dimensions;
bound = pd([2 4 3 1 2],:); % boundary points in ccw order for plotting
p42 = pd(4,:) - pd(2,:); % touch line
theta = atan2(p42(2),p42(1)); % theta is in radians
R = [cos(theta) -sin(theta); sin(theta) cos(theta)]; % rotation matrix
team1R = team1*R; team2R = team2*R; % rotation
ballR = ball*R; boundR = bound*R;
figure(1)
plot(boundR(:,1),boundR(:,2),'o-');
hold on
plot(team1R(:,1),team1R(:,2),'sb','MarkerFaceColor','b') % everton
plot(team2R(:,1),team2R(:,2),'sr','MarkerFaceColor','r') % liverpool
plot(ballR(:,1),ballR(:,2),'ok');
axis equal
hold off
I maybe went on too long but it's an interesting topic. Looking at the random player positions reminds me a lot of our intramural soccer team from college.
5 Kommentare
Tania
am 20 Sep. 2022
Hi David,
I've been using the code you provided above to perform a similar rotational task.
In the code above you created a random array for team member data (11x2 arrary), assuming 11 rows of "team members" and one x,y coordinate for each team member.
I'm trying to rotate data in a 6135 x 15 table, where the data is layed out as follows 6135 data points of coordintes seperted into x & y columns for each person.
Time I Person 1 x I Person 1 y I Person 2 x I Person 2 y I Person 3 x I Person 3 y .....etc
When the data is graphed without rotation it looks like the below:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1130440/image.png)
I would like to rotate this date to run length way along the Y axis:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1130445/image.png)
Is there a way to adapt the code written above to rotate a large table of x,y coordinates, instead of single x,y data points?
Thnk you in advance for your help!
David Goodmanson
am 22 Sep. 2022
Hi Tanya,
yes there is. I'll assume that your table is converted to a 6135x15 matrix that I'll call At. Looks like there are 7 players. The 2x2 rotation matrix R works on an mx2 matrix of coordinates. The idea is to temporarily drop the time column to make 6135x14 matrix A , stack up vertically all the player x coordinates into a single 7*6135 column, do the same for y, concatenate those columns side-by-side, do the rotation with your R and then reconstruct a new 6135x14 matrix B. Then put the time back in.
A = At(:,2:end); % temporarily drop the time
n = size(A,1) % times
p = size(A,2)/2 % players
x = A(:,1:2:end); % n x p
y = A(:,2:2:end);
x = x(:); % vertical stack, (n*p) x 1
y = y(:);
xy = [x y]; % (n*p) x 2
xyR = xy*R;
xR = reshape(xyR(:,1),n,p); % n x p
yR = reshape(xyR(:,2),n,p);
B = zeros(size(A)); % n x (2*p)
B(:,1:2:end) = xR;
B(:,2:2:end) = yR;
Bt = [At(:,1) B]; % reinsert time column
Weitere Antworten (1)
Meysam Mahooti
am 1 Nov. 2019
%--------------------------------------------------------------------------
%
% Position: Position vector (r [m]) from geodetic coordinates
% (Longitude [rad], latitude [rad], altitude [m])
%
% Last modified: 2018/01/27 M. Mahooti
%
%--------------------------------------------------------------------------
function r = Position(lon, lat, h)
R_equ = 6378.137e3; % Earth's radius [m]; WGS-84
f = 1/298.257223563; % Flattening; WGS-84
e2 = f*(2-f); % Square of eccentricity
CosLat = cos(lat); % (Co)sine of geodetic latitude
SinLat = sin(lat);
% Position vector
N = R_equ/sqrt(1-e2*SinLat*SinLat);
r(1) = (N+h)*CosLat*cos(lon);
r(2) = (N+h)*CosLat*sin(lon);
r(3) = ((1-e2)*N+h)*SinLat;
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