Now how about you address the OP and provide some math for a FE?
Read the FAQ
So that's a no. You have no math for FE, likely because you know it shows a FE to be garbage.
Here is an example of math for a FE, using the common FE map: Note: 0 degrees is an example, due to rotational symmetry the same is true for any longitude.
On the equinox, at roughly 6 am (solar time) for people at 0 degrees, east/west, the sun is directly overhead the equator at 90 degrees east.
This would mean the apparent direction to the sun (ignoring elevation) would be NE (i.e. a bearing of 45 degrees) for a person on the equator at 0 degrees east/west.
As you move further north, it would drift towards the east, only reaching ~due east just before the north pole.
As you move further south, the sun would drift further north.
For a given distance d away from the north pole (along 0 degrees east) the sun would appear at a bearing given by:
tan(a)=10000 km/d
So at the southern tip of Chile, some 16 000 km away from the north pole, it should be tan(a)=10/16, and thus a=32 degrees. So it should be north of north east.
Instead, the sun appears roughly due east.
We can also do this the other way to determine how far away the sun is from the north pole (x). Assuming a 1 degree error (to be nice, and meaning we get an underestimate) the sun is off at a bearing of 89 degrees.
Thus, tan(89 deg)=x/16000
Thus x~=915 000 km.
This puts it well outside the range of a FE.
This simple math shows the common FE model to be impossible.