ReadyACARS:
A couple of inconsistencies spring to mind.
1. If the aircraft flew faster, and ran out of fuel sooner, why is the search area so close to Australia? The aircraft left Malaysian radar at 0215 local and the last Inmarsat contact was 0811, so if it was moving faster in the same given time, its distance traveled would be longer, but its time aloft would be slightly shorter.
As I pointed out in post 8741 - for the aircraft to end up more to the northeast it needs to fly slower, not faster ... (shorter distance flown, in the same time). The confusion is probably that the aircraft was flying lower, so it had a FASTER IAS but a SLOWER TAS.
ie:
39,000' 260 kts IAS = 463 kts TAS. (slower speed = faster TAS)
10,000' 290 kts IAS = 335 kts TAS. (faster speed = slower TAS)
http://www.pprune.org/rumours-news/5...ml#post8408255
If your calculation for a required 360 kts groundspeed is correct, to end up in that search location, then we have a range of possible speed-altitudes to consider (assuming no wind and TAS = groundspeed).
312 kts at 10,000' gives 360 kt TAS
285 kts at 15,000' gives 360 kt TAS
262 kts at 20,000' gives 360 kt TAS
240 kts at 25,000' gives 360 kt TAS
Since they said the aircraft was going 'faster', one suspects that the 10,000 ft or 15,000' speed/altitudes are the more correct.
However, if someone here has the fuel-burn speed/altitude tables for a B777, they could easily work out from this information exactly what altitude MH370 was flying at (plus or minus the wind). We have the distance, time and fuel-load, and the inly other variable is the speed-altitude. And i imagine only one speed-altitude configuration will comfortably fit the data.
Here is a copy of the image i posted previously (not to any exact speed-scale).
Yellow = high TAS flight
Scarlet = medium TAS flight
Purple = slow TAS flight
