If the earth is flat, how come you need to use spherical geometry in order to accurately calculate the distance between two objects? For instance, many "distance by zip code" applications are based on using the coordinates of the center of a zip code area and plugging them into something like the Haversine formula:

R = earth’s radius (mean radius = 6,371km)

Δlat = lat2− lat1

Δlong = long2− long1

a = sin²(Δlat/2) + cos(lat1).cos(lat2).sin²(Δlong/2)

c = 2.atan2(√a, √(1−a))

d = R.c

Since GPS data has been opened up to civilians and GPS data sources (by their very nature) overlap in area, it would be impossible to fake this data in a way that would produce the illusion of a sphere where there wasn't one. This is compounded by the fact that you can verify the above data simply by walking between two coordinates and doing the math.

So you know, how does this fit into the FE way of life?

Your formula is incorrect. On a RE, the distance between two points with geodisic coordinates (latitude, longitude) (λ

_{1}, ϕ

_{1}) and (λ

_{2}, ϕ

_{2}) is:

*d* = *R* * cos^{-1}(sin*λ*_{1}*sin*λ*_{2} + cos*λ*_{1}*cos*λ*_{2}*cos(*ϕ*_{2}-*ϕ*_{1}))

The problem with FE is that we only know that points with constant lattitude correspond to concentric circles with center at the North Pole, but we cannot exactly determine their radius. In any case, the radius depends on the latitude only

*r* = *f*(*λ*).The only thing we know for sure is that

*f*(+90°) = 0, because this point corresponds to the North Pole and that is in the center of the circle. Then, we can find the shortest distnace between two points as:

*d* = √(*r*_{1}^{2}+ *r*_{2}^{2} - 2**r*_{1}**r*_{2}*cos(*ϕ*_{2} - *ϕ*_{1}))

However, one can prove that no choice of the function

*f*(

*λ*) can give the same result for all possible points. The proof is simple. First, notice that both formulas depend only on the difference in longitudes

*ϕ*_{2} -

*ϕ*_{1}, as it should be, because we are free to choose the meridian of zero longitude (which, by convention today is through Greenwich). That's why we can choose in our formula

*ϕ*_{1} = 0 and

*ϕ*_{2} =

*ϕ* in general. Furthermore, let us choose

*λ*_{1} = 90°, because we know the value of the function

*f* for this latitude, and

*λ*_{2} =

*λ* in general. Then, the two formulas give:

*d* = *R**(*π*/2 - *λ*) - on RE

*d* = *f*(*λ*) - on FE

from which we conclude that

*f*(*λ*) = *R**(*π*/2 - *λ*)

where the angle is measured in radians now.

Now, for points on the Equator (

*λ*_{1} =

*λ*_{2} = 0), the two equations give:

*d* = *R**|*ϕ*_{2} - *ϕ*_{1}| - in RE

*d* = *R***π**|sin(*ϕ*_{2} - *ϕ*_{1})/2)| - in FE

These two estimates are obviously not the same. If they were, we would have the function

|sin(*x*)/*x*| = 2/*π*, 0 ≤ *x* ≤ 2*π*

which is certainly not correct. Hence, we cannot find

*f*(

*λ*) making the two distance estimates equal for any two points on the Earth.

However, this does not mean that FE model is not true. One has to give an empirical definition of longitude and latitude before one can use the above formulas. You provided the GPS data as a source for longitude and latitude. Until you provide us with the method with which this system gets the latitude and longitude of any point on Earth, without inverting the RE distance formula I stated above, you don't have a proof of anything.