Calculate distance from geographics points

Hi, I have two vectors of coordinates, latitude and longitude, of points describing the centerline of a track. I want to create a distance channel to plot some quantities (e.g. velocity) as function of this distance, how can I create a distance channel of the same numerosity of the two vectors of coordinates?

Answer 1

[KSSV]

USe Haversine's formula:

Let (lon1,lat1) and (lon2,lat2) be your coordinates in degrees.

% Haversine formula
dlon = lon2 - lon1 ; 
dlat = lat2 - lat1 ; 
a = sind(dlat/2)^2 + cos(dlat1)*cosd(lat2)*sin(dlon/2)^2 ; 
c = 2*asind(sqrt(a)) ; 
d = 6367*c ;      % distance in km
    

[RAFFAELE MAGLIONE]

I tried, but I had trouble when trying to apply this formula to all points in the vector with a for loop. I also tried to use the 'distance' function from Mapping Toolbox, but the distance vector returned a too small distance.

The distance channel I want is from red points in figure.

I did as follows:

distgc(1) = 0;
for i=2:length(Latitude_cen)-1
    dist(i) = dist(i-1) + distance(Latitude_cen(i),Longitude_cen(i),Latitude_cen(i+1),Longitude_cen(i+1))
end

[KSSV]

Let lon and be your m*1 arrays (say all red points in a column vector).

% Haversine formula
dlon = diff(lon) ; 
dlat = diff(lat) ; 
a = sind(dlat/2).^2 + cos(dlat1).*cosd(lat2).*sin(dlon/2).^2 ; 
c = 2*asind(sqrt(a)) ; 
d = 6367*c ;      % distance in km

[RAFFAELE MAGLIONE]

What are exactly 'dlat1' and 'lat2'?

[KSSV]

Let (lon1,lat1) and (lon2,lat2) be your coordinates in degrees.

[RAFFAELE MAGLIONE]

Perhaps I have not explained myself well, I have two vectors available, the Latitude vector of the red points and the Longitude vector of the red points. I thought of applying the Haversine formula you proposed, but not having only two points, but about 7000, do I have to do a for loop with which I calculate the distance between one point and the next with the Haversine formula? Because I tried and it comes out a very small overall distance

[Bruno Luong]

lon1 = lon(1:end-1);
lon2 = lon(2:end);
lat1 = lat(1:end-1);
lat2 = lat(2:end);

dlon = lon2 - lon1 ; 
dlat = lat2 - lat1 ; 
a = (1-cos(dlat))/2 + cos(lat1).*cos(lat2).*(1-cos(dlon))/2 ; 
c = 2*asin(sqrt(a)) ;
earthradius = 6367;
dd = earthradius*c;
d = cumsum([0 dd])

[RAFFAELE MAGLIONE]

This seems to work better, but return a too long distance, the final valure is about 114km

[Bruno Luong]

Then reduce the earthradius perhaps. Frankly, what is the point to state something (too long, 114 km) that nobody can check.

[RAFFAELE MAGLIONE]

The point is that I tried (for verification) to apply this formula on a set of points of which I already have the distance channel obtained by deriving by integrating the speed acquired through sensors. With Haversine's formula, the distance is two orders of magnitude higher, so I said 114 is too much. For the radius of the earth I don't think I can reduce it to my liking ...

[Bruno Luong]

Just wonder do you have you long / lat in degree or radian?

My code supposes long/lat in radian as you see it use cos function on them.

180/pi % ratio bewteen rad and deg

ans = 57.2958

[RAFFAELE MAGLIONE]

I was working with degrees, now with radian return the exact length, thank you!