## matlab 计算经纬度两点间距离

1，简单估计：例子如下 第一个点经纬度 120 24 第二个点经纬度 121 24
D=distance(24,120,24,121) % distance看matlab help
pi=3.1415926;
dx=D*6371*1000*2*pi/360;

2 利用一下程序函数
function r = geodistance( ci , cf , m )

% Calculates the distance in meters between two points on earth surface.
%
% SYNTAX: r = geodistance( coordinates1 , coordinates2 , method ) ;
%
% Where coordinates1 = [longitude1,latitude1] defines the
% initial position and coordinates2 = [longitude2,latitude2]
% defines the final position.
% Coordinates values should be specified in decimal degrees.
% Method can be an integer between 1 and 23, default is m = 6.
% Methods 1 and 2 are based on spherical trigonometry and a
% spheroidal model for the earth, respectively.
% Methods 3 to 23 use Vincenty's formulae, based on ellipsoid
% parameters.
% Here it follows the correspondence between m and the type of
% ellipsoid:
%
% m = 3 -> ANS , m = 4 -> GRS80, m = 5 -> WGS72,
% m = 6 -> WGS84, m = 7 -> NSWC-9Z2,
% m = 8 -> Clarke 1866, m = 9 -> Clarke 1880,
% m = 10 -> Airy 1830,
% m = 11 -> Bessel 1841 (Ethiopia,Indonesia,Japan,Korea),
% m = 12 -> Bessel 1841 (Namibia),
% m = 13 -> Sabah and Sarawak (Everest,Brunei,E.Malaysia),
% m = 14 -> India 1830, m = 15 -> India 1956,
% m = 16 -> W. Malaysia and Singapore 1948,
% m = 17 -> W. Malaysia 1969,
% m = 18 -> Helmert 1906,m = 19 -> Helmert 1960,
% m = 20 -> Hayford International 1924,
% m = 21 -> Hough 1960, m = 22 -> Krassovsky 1940,
% m = 23 -> Modified Fischer 1960,
% m = 24 -> South American 1969.
%
% Important notes:
%
% 1)South latitudes are negative.
% 2)East longitudes are positive.
% 3)Great circle distance is the shortest distance between two points
% on a sphere. This coincides with the circumference of a circle which
% passes through both points and the centre of the sphere.
% 4)Geodesic distance is the shortest distance between two points on a spheroid.
% 5)Normal section distance is formed by a plane on a spheroid containing a
% point at one end of the line and the normal of the point at the other end.
% For all practical purposes, the difference between a normal section and a
% geodesic distance is insignificant.
% 6)The method m=2 assumes a spheroidal model for the earth with an average
% radius of 6364.963 km. It has been derived for use within Australia.
% The formula is estimated to have an accuracy of about 200 metres over 50 km,
% but may deteriorate with longer distances.
% However, it is not symmetric when the points are exchanged.
%
% Examples: A = [150 -30]; B = [150 -31]; L = [151 -80];
% [geodistance(A,B,1) geodistance(A,B,2) geodistance(A,B,3)]
% [geodistance(A,L,1) geodistance(A,L,2) geodistance(A,L,3)]
% geodistance([0 0],[2 3])
% geodistance([2 3],[0 0])
% geodistance([0 0],[2 3],1)
% geodistance([2 3],[0 0],1)
% geodistance([0 0],[2 3],2)
% geodistance([2 3],[0 0],2)
% for m = 1:24
% r(m) = geodistance([150 -30],[151 -80],m);
% end
% plot([1:m],r), box on, grid on

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Second version: 07/11/2007
%
% Contact: [email protected]
%
% Any suggestions to improve the performance of this
% code will be greatly appreciated.
%
% Reference: Geodetic Calculations Methods
% Geoscience Australia
% (Geodesy | Geoscience Australia)
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

r = [ ] ;

if nargin == 2, m = 6; end

longitude1 = pi*ci(1)/180 ;
latitude1 = pi*ci(2)/180 ;

longitude2 = pi*cf(1)/180 ;
latitude2 = pi*cf(2)/180 ;

alla = [0 0 6378160 6378137.0 6378135 6378137.0 6378145 6378206.4 6378249.145,...
6377563.396 6377397.155 6377483.865,...
6377298.556 6377276.345 6377301.243 6377304.063 6377295.664 6378200 6378270 6378388 6378270 6378245,...
6378155 6378160];

allf = [0 0 1/298.25 1/298.257222101 1/298.26 1/298.257223563 1/298.25 1/294.9786982 1/293.465,...
1/299.3249646 1/299.1528128,...
1/299.1528128 1/300.8017 1/300.8017 1/300.8017 1/300.8017 1/300.8017 1/298.3 1/297 1/297 1/297,...
1/298.3 1/298.3 1/298.25];

if ( longitude1 == longitude2)&&( latitude1 == latitude2 )

r = 0;

else

if m == 1 % Great Circle Distance, based on spherical trigonometry

r = 180*1.852*60*acos( ...
sin(latitude1)*sin(latitude2) + cos(latitude1)*cos(latitude2)*cos(longitude2-longitude1) )/pi ;
r = 1000*abs( r );

elseif m == 2 % Spheroidal model for the earth

term1 = 111.08956*( ci(2) - cf(2) + 0.000001 ) ;
term2 = cos( latitude1 + ( (latitude2 - latitude1)/2 ) ) ;
term3 = ( cf(1) - ci(1) + 0.000001 )/( cf(2) - ci(2) + 0.000001 ) ;
r = 1000*abs( term1/cos( atan( term2*term3 ) ) );

else % Apply Vincenty's formulae (as long as the points are not coincident):

a = alla(m);
f = allf(m);

b = a*( 1 - f ) ;

tangens_u1 = ( 1 - f )*tan( latitude1 ) ; u1 = atan( tangens_u1 ) ;
tangens_u2 = ( 1 - f )*tan( latitude2 ) ; u2 = atan( tangens_u2 ) ;
delta_longitude = longitude2 - longitude1 ;
lambda = delta_longitude ;
squared_sin_of_sigma = ( cos(u2)*sin(lambda) )^2 + ( cos(u1)*sin(u2) - sin(u1)*cos(u2)*cos(lambda) )^2 ;
sin_of_sigma = sqrt( squared_sin_of_sigma ) ; % This is zero when the points are coincident...
cos_of_sigma = sin( u1 )*sin( u2 ) + cos( u1 )*cos( u2 )*cos( lambda ) ;
tan_of_sigma = sin_of_sigma/cos_of_sigma ;
sigma = atan( tan_of_sigma ) ;
tangens_of_sigma = sin_of_sigma/cos_of_sigma ;
sin_of_alpha = cos( u1 )*cos( u2 )*sin( lambda )/sin_of_sigma ;
cos_of_alpha = sqrt( 1 - sin_of_alpha^2 ) ;
cos_of_2sigmam = cos_of_sigma - ( 2*sin( u1 )*sin( u2 )/cos_of_alpha^2 ) ;

C = (f/16)*( cos_of_alpha )^2*( 4 + f*( 4 - 3*( cos_of_alpha )^2 ) ) ;

lambda2 = delta_longitude + ( 1 - C )*f*sin_of_alpha*( sigma + C*sin_of_sigma*( ...
cos_of_2sigmam + C*cos_of_sigma*( -1 + 2*( cos_of_2sigmam )^2 ) ) ) ;

while ( abs( lambda - lambda2 ) > 1e-9 )

lambda = lambda2 ;
squared_sin_of_sigma = ( cos(u2)*sin(lambda) )^2 + ( cos(u1)*sin(u2) - sin(u1)*cos(u2)*cos(lambda) )^2 ;
sin_of_sigma = sqrt( squared_sin_of_sigma ) ;
cos_of_sigma = sin( u1 )*sin( u2 ) + cos( u1 )*cos( u2 )*cos( lambda ) ;
tan_of_sigma = sin_of_sigma/cos_of_sigma ;
sigma = atan( tan_of_sigma ) ;
tangens_of_sigma = sin_of_sigma/cos_of_sigma ;
sin_of_alpha = cos( u1 )*cos( u2 )*sin( lambda )/sin_of_sigma ;
cos_of_alpha = sqrt( 1 - sin_of_alpha^2 ) ;
cos_of_2sigmam = cos_of_sigma - ( 2*sin( u1 )*sin( u2 )/cos_of_alpha^2 ) ;

C = (f/16)*( cos_of_alpha )^2*( 4 + f*( 4 - 3*( cos_of_alpha )^2 ) ) ;

lambda2 = delta_longitude + ( 1 - C )*f*sin_of_alpha*( sigma + C*sin_of_sigma*( ...
cos_of_2sigmam + C*cos_of_sigma*( -1 + 2*( cos_of_2sigmam )^2 ) ) ) ;

end % while ( abs(lambda - lambda2 ) > 1e-9 )

u2 = ( cos_of_alpha^2 )*( a^2 - b^2 )/b^2 ;
A = 1 + ( u2/16384 )*( 4096 + u2*( -768 + u2*( 320 - 175*u2 ) ) ) ;
B = ( u2/1024 )*( 256 + u2*( -128 + u2*( 74 - 47*u2 ) ) ) ;
delta_sigma = B*sin_of_sigma*( cos_of_2sigmam + ( B/4 )*( ...
cos_of_sigma*( -1 + 2*cos_of_2sigmam^2 ) - ...
(B/6)*cos_of_2sigmam*( -3 + 4*sin_of_sigma^2 )*( -3 + 4*cos_of_2sigmam^2 ) ) ) ;
r = b*A*( sigma - delta_sigma ) ;

end

end