RE: Captain Bob (#3536)
Quote:
It doesn't matter. Everyone does it. Parallel Track. It has nothing to do with
what happened.
/Unquote
I thought so too, but after reading about SLOP (
Strategic Lateral Offset Procedure - Wikipedia, the free encyclopedia) I felt that maybe 3 NM left of track is possibly significant. With as little evidence as we have, it would seem justified to examine even this detail.
Quote from BEA Interim Rept:
Note: A position report message (AOC type) was received at 2 h 10 min 34 s, between two maintenance
messages. This can be explained by the fact that AOC messages take priority over maintenance messages
/Unquote
Considering that the first WRN message transmitted by ACARS was received at 02:10:10Z and the last position report at 02:10:34Z - I wanted to have a feel for how much time it would take to get 3 NM off track, if one assumes that the airplane is initially on track at 468 kt TAS. For example, in a relatively shallow turn (10 degrees of bank) it would take approx. 80 seconds, or 10 NM along track, and a change of heading of 33 degrees. Similar data also for steeper turns are shown in the following table:
Track deviation [NM] . 3
TAS .[kt] ............ 468
Bankangle .[deg] ...... 10 ..... 15 .... 20 ... 25 .... 30
Loadfactor .......... = 1,015 . 1,035 . 1,064 . 1,103 . 1,155
TurnRadius [NM] ......= 18,1 . 11,9 ... 8,8 ... 6,8 ... 5,5
TurnRate [deg/sec] ...= 0,412 . 0,625 . 0,850 . 1,088 . 1,348
Heading Chg [deg] ... = 33,5 .. 41,6 .. 48,9 .. 55,8 .. 62,8
Time [sec] .......... = 81,3 .. 66,5 .. 57,5 .. 51,3 .. 46,6
Dist along track [NM] = 10,0 .. 7,9 ... 6,6 ... 5,7 ... 4,9
P.S. As you might guess, I'm an engineer, not a pilot, and a newcomer to this forum. If it's irrelevant, don't get angry with me, just tell me and I'll be happy to hit the EDIT button to delete it.
regards,
HN39