Finding the first point of great circle Intersection

Question

I have a problem I've been trying to solve and I cannot come up with the answer. I have written a function in Matlab which, given two lat/lon points and two bearings, will return the two great circle points of intersection.

However, what I really need is the first great circle point of intersection along the two initial headings. I.e. if two airplanes begin at lat/lon points 1 and 2, with initial bearings of bearing1 and bearing2, which of the two great circle intersection points is the first one they encounter? There are many solutions (using haversine) which will give me the closer of the two points, but I don't actually care about which is closer, I care about which I will encounter first given specific start points and headings. There are many cases where the closer of the two intersections is actually the second intersection encountered.

Now, I realize I could do this with lots of conditional statements for handling the different cases, but I figure there's got to be a way to handle it with regard to the order I take all my cross products (function code given below), but I simply can't come up with the right solution! I should also mention that this function is going to be used in a large computationally intensive model, and so I'm thinking the solution to this problem needs to be rather elegant/speedy. Can anyone help me with this problem?

The following is not my function code (I can't list that here), but it is the pseudo code that my function was based off of:

 %Given inputs of lat1,lon1,Bearing1,lat2,lon2,Bearing2:
%Calculate arbitrary secondary point along same initial bearing from first
%point
dAngle = 45;
lat3 = asind( sind(lat1)*cosd(dAngle) + cosd(lat1)*sind(dAngle)*cosd(Bearing1));
lon3 = lon1 + atan2( sind(Bearing1)*sind(dAngle)*cosd(lat1), cosd(dAngle)-sind(lat1)*sind(lat3) )*180/pi;
lat4 = asind( sind(lat2)*cosd(dAngle) + cosd(lat2)*sind(dAngle)*cosd(Bearing2));
lon4 = lon2 + atan2( sind(Bearing2)*sind(dAngle)*cosd(lat2), cosd(dAngle)-sind(lat2)*sind(lat4) )*180/pi;


%% Calculate unit vectors
% We now have two points defining each of the two great circles.  We need
% to calculate unit vectors from the center of the Earth to each of these
% points
[Uvec1(1),Uvec1(2),Uvec1(3)] = sph2cart(lon1*pi/180,lat1*pi/180,1);
[Uvec2(1),Uvec2(2),Uvec2(3)] = sph2cart(lon2*pi/180,lat2*pi/180,1);
[Uvec3(1),Uvec3(2),Uvec3(3)] = sph2cart(lon3*pi/180,lat3*pi/180,1);
[Uvec4(1),Uvec4(2),Uvec4(3)] = sph2cart(lon4*pi/180,lat4*pi/180,1);


%% Cross product
%Need to calculate the the "plane normals" for each of the two great
%circles
N1 = cross(Uvec1,Uvec3);
N2 = cross(Uvec2,Uvec4);


%% Plane of intersecting line
%With two plane normals, the cross prodcut defines their intersecting line
L = cross(N1,N2);
L = L./norm(L);
L2 = -L;
L2 = L2./norm(L2);


%% Convert normalized intersection line to geodetic coordinates
[lonRad,latRad,~]=cart2sph(L(1),L(2),L(3));
lonDeg = lonRad*180/pi;
latDeg = latRad*180/pi;
[lonRad,latRad,~]=cart2sph(L2(1),L2(2),L2(3));
lonDeg2 = lonRad*180/pi;
latDeg2 = latRad*180/pi;

UPDATE: A user on the Mathworks forums pointed this out:

Actually they might each reach a different point. I might have misunderstood the question, but the way you worded it suggests that both trajectories will converge towards the same point, which is not true. If you image the first point being a little after one intersection and the second point being a little after the other intersection, you have a situation were each "plane" will travel towards the intersection that was the closest to the other plane at the beginning.

I didn't think about this. It is actually very possible that each of the planes intersects a different great circle intersection first. Which just made things much more complicated...

Answer 1

Presumably you have a velocity vector for the direction of the plane (the direction in which you want to look for your first intersection - the "bearing vector on the surface of the earth"). You didn't actually specify this. Let us call it v.

You also have the cartesian coordinates of two points, P1 and P2, as well as the initial position of the plane, P0. Each is assumed to be already on the unit sphere (length = 1).

Now we want to know which is the shorter distance - to P1, or P2 - so we need to know the angle "as seen from the direction of the normal vector". For this we need both the sin and the cos - then we can find the angle using atan2. We obtain sin from the cross product (remember all vectors were normalized), and cos from the dot product. To get the sin "as seen from the right direction", we take dot product with the normal vector.

Here is a piece of code that puts it all together - I am using very simple coordinates for the points and direction vector so I can confirm in my head that this gives the correct answer:

% sample points of P0...P2 and v
P0 = [1 0 0];
P1 = [0 1 0];
P2 = [0 -1 0];
v = [0 1 0];
% compute the "start direction normal":
n0 = cross(P0, v);
n0 = n0 / norm( n0 );  % unit vector 

% compute cross and dot products:
cr01 = cross(P0, P1);
cr02 = cross(P0, P2);
cos01 = dot(P0, P1);
cos02 = dot(P0, P2);

% to get sin with correct sign, take dot product with direction normal:
sin01 = dot(cr01, n0);
sin02 = dot(cr02, n0);
% note - since P0 P1 and P2 are all in the same plane
% cr02 and cr02 are either pointing in the same direction as n0
% or the opposite direction. In the latter case we get a sign flip for the sin
% in the former case this does nothing

% Now get the angle, mapped from 0 to 2 pi:
ang1 = mod(atan2(sin01, cos01), 2*pi);
ang2 = mod(atan2(sin02, cos02), 2*pi);

if( ang1 < ang2 ) 
  fprintf(1,'point 1 is reached first\n');
  % point 1 is the first one reached
else
  fprintf(1,'point 2 is reached first\n');
  % point 2 is the first
end

When you change the direction of the velocity vector (pointing towards P2 instead of P1), the program correctly tells you "point 2 is reached first".

Let me know if this works for you!