Getting started with geospatial data?
RiderOfGiraffes: Close enough is good enough.Given a (lat,lng) of (A,B) and another of (X,Y), each in degrees, the distance between them is (roughly)  sqrt( (A-X)^2 + (B-Y)^2 )

in degrees, and to convert to miles you use 360 degrees is 25000 miles.OK, now that's not quite right, because as your latitude increases the distance between the lines of longitudes decreases.  You need to multiply by cos(lat).  So your answer in miles is:  sqrt( (cos(A)*(A-X))^2 + (B-Y)^2 )/360*25000

Again, close enough is good enough because if you're more than a mile or three then the errors are really small.There is a more exact formula using dot products of 3-space vectors, and then another that takes the non-spherical nature of the Earth into consideration, but the above is good enough to start with.EDIT ...To get the more accurate version, for a (lat/lng) you convert to a unit vector:    X = cos(lat)*cos(lng)
    Y = cos(lat)*sin(lng)
    Z = sin(lat)

Given two lat/lngs, convert to (a,b,c) and (x,y,z).  Now you can take the dot product:    dot = (a,b,c).(x,y,z) = a*x+b*y+c*z

Compute the angle between the vectors:    angle_in_radians = arccos(dot)

Finally, the distance is    distance = radius * angle_in_radians
             = 25000 * angle_in_radians (miles)
             = 40000 * angle_in_radians (kilometers)

As a comment about accuracy and appropriate formulae,... if the points really are close together then the arccos comes out as highly error-prone because of limited accuracy.  The dot product is very, very close to one, so the arccos is unreliable.  This is because when theta is really, really small then the value of cos(theta) is very, very close to 1.  If you subtract it from 1, you get very small rubbish.  Multiply by a large number and you get big rubbish.