Calculating from Lat/Lng

     The zip code database provides latitude and longitude for the rough center of each zip code. The trick is to calculate distance and direction to any other zip code.


     To start with finding the distance between two points on the globe, we determine the positions of those points in arbitrary coordinates of X, Y, and Z. Here Z is parallel to the axis of the Earth and X and Y are arbitrarily based on the origin of the longitude system. Assuming the Earth is a sphere, the positions are:

Zip Code A is at latitude LatA and longitude LngA
    x_A = R*cos(LatA)*cos(LngA)
    y_A = R*cos(LatA)*sin(LngA)
    z_A = R*sin(LatA)

Zip Code B is at latitude LatB and longitude LngB
    x_B = R*cos(LatB)*cos(LngB)
    y_B = R*cos(LatB)*sin(LngB)
    z_B = R*sin(LatB)

where R is the radius of the Earth.  

     The radius of the Earth is actually not a constant, because the Earth is actually an ellipsoid rather than a sphere. The radius is 0.3% shorter at the poles than at the equator, a difference of some 21.3 kilometers. For distances within the country, you should use the equatorial radius or (if you want to be more precise) the radius at the average latitude in question.

Radius at the equator = 6378137 meters
Radius at the poles =   6356752 meters

Radius at any latitude = 6378137 * (1 - 0.0033493*sin(Lat)^2)

Average Radius between two latitudes = 6378137 * (1 - 0.0033493*sin((LatA+LatB)/2)^2)


     The real distance between the points is the arc along the globe. This is based on finding the angle between the two vectors, which is found from the "dot product". The distance along the arc is equal to the angle (in radians) times the radius (using the average radius defined above). If Ang is the true angle between the points,

R*R*cos(Ang) = x_A*x_B + y_A*y_B + z_A*z_B

cos(Ang) = cos(LatA)*cos(LngA)*cos(LatB)*cos(LngB)
           + cos(LatA)*sin(LngA)*cos(LatB)*sin(LngB)
           + sin(LatA)*sin(LatB)

cos(Ang) = cos(LatA)*cos(LngA)*cos(LatB)*cos(LngB)
           + cos(LatA)*sin(LngA)*cos(LatB)*sin(LngB)
           + sin(LatA)*sin(LatB)

cos(Ang) = cos(LatA)*cos(LatB)*(cos(LngA)*cos(LngB) + sin(LngA)*sin(LngB)) + sin(LatA)*sin(LatB)

distance = radius * Ang

     The above calculation works, and is likely to be used by generalized packages. However, by going through some simple math we can simplify the calculation so that it can be calculated more accurately and in fewer steps.

 [using the general relation cos(a-b)=cos(a)*cos(b)+sin(a)*sin(b)]

cos(Ang) = cos(LatA)*cos(LatB)*(cos(LngA-LngB)) + sin(LatA)*sin(LatB)

cos(Ang) = (cos(LngA-LngB)-1+1)*cos(LatA)*cos(LatB) + sin(LatA)*sin(LatB)

 [use the general relation cos(a-b)=cos(a)*cos(b)+sin(a)*sin(b)]

cos(Ang) = (cos(LngA-LngB)-1)*cos(LatA)*cos(LatB) + cos(LatA)*cos(LatB) + sin(LatA)*sin(LatB)

cos(Ang) = (cos(LngA-LngB)-1)*cos(LatA)*cos(LatB) + cos(LatA-LatB)

Ang = arccos( (cos(LngA-LngB)-1)*cos(LatA)*cos(LatB) + cos(LatA-LatB) )

Computerized Implementation

     The above works fine as a mathematical relation, and it has simplified the number of steps to a fair minimum. There are two concerns for this as a computerized calculation, though:

     The solution for these problems depends on how you are using the calculation. If processsing speed is not important, you can just retain a very large number of places to be sure of accuracy. However, there is also a good approximation which can be made for points which are close together.

For small x, sin(x) ~   x - (x^3/6) + ...
             cos(x) ~ 1 - (x^2/2) + (x^4/24) - ...

So assuming we keep only the x squared term, this means: 

1 - Ang^2/2 = (1-(LngA-LngB)^2/2-1)*cos(LatA)*cos(LatB) + 1 - (LatA-LatB)^2/2

Ang^2 = cos(LatA)*cos(LatB)*((LngA-LngB)^2) + (LatA-LatB)^2

Ang = sqrt(  cos(LatA)*cos(LatB)*((LngA-LngB)^2) + (LatA-LatB)^2  )

     The above relation is good up to x^4 power, which for many purposes is pretty good.

John H. Kim <>
Last modified: Mon Aug 29 09:01:59 2005