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How to calculate the Euclidean distance beetwen all points of Latitude Longitude pairs?

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I have a 399 cities array with LON LAT coordinates (first column for the Longitudes), like the picture below.
How can I calculate the 399x399 matrix with all distances between this 399 cities?
I used pdist and squareform but the result are small number. Am I correct?
D = pdist(XY);
Z = squareform(D);
For example, the two first points (-50.3125 -23.3005; -48.9918 -24.6617) have a Euclidean distance between them of 216 km (see picture below).
Thank you!
  2 Comments
David Franco
David Franco on 21 Apr 2020
In time: I found that the values in my Z matrix is in fact the 2-norm (or Euclidean distance) between all my 399 points. But how can I convert them to km? Only with haversine formula?

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Accepted Answer

David Franco
David Franco on 21 Apr 2020
function [d1,d2] = pos2dist(point1,point2)
% Distance:
% d1: distance in km based on Haversine formula
% d2: distance in km based on Pythagoras theorem
% Inputs:
% point1: lat lon of origin point [lat lon]
% point2: lat lon of destination point [lat lon]
%
% Outputs:
% d1: distance calculated by Haversine formula
% d2: distance calculated based on Pythagoran theorem
%
% Example 1, short distance:
% point1 = [-43 172];
% point2 = [-44 171];
% [d1 d2] = pos2dist(point1,point2)
% d1 =
% 137.365669065197 (km)
% d2 =
% 137.368179013869 (km)
%
% Example 2, longer distance:
% point1 = [-43 172];
% point2 = [20 -108];
% [d1 d2] = pos2dist(point1,point2)
% d1 =
% 10734.8931427602 (km)
% d2 =
% 31303.4535270825 (km)
radius = 6371; % Earth radius
lat1 = point1(1)*pi/180;
lat2 = point2(1)*pi/180;
lon1 = point1(2)*pi/180;
lon2 = point2(2)*pi/180;
deltaLat = lat2-lat1;
deltaLon = lon2-lon1;
a = sin((deltaLat)/2)^2 + cos(lat1)*cos(lat2) * sin(deltaLon/2)^2;
c = 2*atan2(sqrt(a),sqrt(1-a));
x = deltaLon*cos((lat1+lat2)/2);
y = deltaLat;
d1 = radius*c; % Haversine distance
d2 = radius*sqrt(x*x + y*y); % Pythagoran distance
end
Thanks Geoff!

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