they have determined the trajectory based on huge assumptions and BFO&BTO values having huge margin of error(only 1 Hz equates to ~100 km), in reality you can probably make the trajectory to fit these values for more or less any point on the arc

example(the post from duncansteel.com) :

Byan C.

2014/07/03 at 08:22

I woke up wondering if Duncan thinks I'm some kind of crackpot, so I thought I should send you a clearer explanation. If R represents satellite to plane distance, s, p, r reresents the satellite, plane, and reference satellite respectively. Numbers 1 and 2 represents two points in time and x, y, z are the coordinates. Then,

(x-xs)^2 + (y-ys)^2 + (z-zs)^2 = Rs^2

(x-xr)^2 + (y-yr)^2 + (z-zr)^2 = Rr^2

x^2 + y^2 + z^2 = r^2 where r = earth radius and recognizing that

xs^2 + ys^2 + zs^2 = (r+H)^2 where H is the satellite height above the earth. The same applies to the reference satellite.

Converting BTO data to velocities yield:

V-Vsp-Vps = Vpr

Assuming a constant satellite velocity the Vsp + Vps = Rs2 – Rs1

Similarly, Vpr = Rr2-Rr1 or Rr2 = Rr1 + Vpr

This is where I got into trouble. Since Rr2 depends on Rr1 if I don't check the methodology using the course defined by the BTO data, I don't get a good fit to the BTO data at 19:40. After solving, I plug the plane velocities derived from (x2-x1)/(t2-t1) to a program the solves based on the distance traveled between 2 ping rings. This routine calculate the BTO values based on the actual Vsp at time t. Comparisions indicate that my solution using average Vsp values is yielding reasonable results. To calculate values using the actual Vsp value at time t involves solving 10 equations with 10 unknowns. The good news is that the solution would reduce to a quadratic equation. It is interesting that the solution for the equations based on distance traveled yields 2 roots with different latitudes. The solution to the equations based on velocities yields 2 roots with different longitudes. 1 to the west of the satellite and 1 to the east, effectively resulting in 1 solution. As a check, I plugged in Ra and Rs values for the northern and southern paths, it duplicates those paths.

Imagine my shock after I got this debugged (I keep reversing signs in my vector math) and I get a result that indicates the plane took a large circular path to the south heading eventually back to KLIA. I almost fell off my chair. I was not expecting this. I have to admit that I have not been that excited about anything in a long time, hence my need to hurriedly share this with someone.

I'm sharing this at this time because I believe the actual solution will be difficult to achieve. I could sove the 10 equations, but developing reasonable flight paths is beyond my ability.

I indicated that I thought we should keep this private because if there is a 3rd flight path, this will cause hugh controversy and we need to be certain before presenting such a result. Perhaps, my results are just a fluke of the data, but since I achieved these results with absolutely no manipulation of the raw data, I feel like it needs to be pursued further.

Cheers,

Bryan