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Question about Airspeed vs Groundspeed

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posted on Sep, 14 2015 @ 09:08 AM
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a reply to: F4guy




This SFDR (Solid State Flight Data Recorder) captures indicated airspeed, but since it also records pressure altitude, true airspeed can be calculated.


So when investigators are refering to "computed airspeed",they are refering to true airspeed, which they calculated from those two data sets?




BTW, I assume you're interested because that serial number is the Malaysian 777 shot down over the Ukraine.


Correct.




posted on Sep, 14 2015 @ 05:28 PM
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originally posted by: RogueWave
a reply to: F4guy




This SFDR (Solid State Flight Data Recorder) captures indicated airspeed, but since it also records pressure altitude, true airspeed can be calculated.


So when investigators are refering to "computed airspeed",they are refering to true airspeed, which they calculated from those two data sets?







Not necessarily. It might have been computed from using the radar track by dividing radar return distances by radar sweep interval. "Computed airspeed" is not a technical term normally used in accident investigations. And remember, the report has gone through translations since the report was done by the Dutch Onderzoeksraad Voor Veiligheid (Safety Board). It may be a mistranslation of "calibrated airspeed."



posted on Sep, 14 2015 @ 06:18 PM
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a reply to: F4guy




And remember, the report has gone through translations since the report was done by the Dutch Onderzoeksraad Voor Veiligheid (Safety Board). It may be a mistranslation of "calibrated airspeed."


Nope the Dutch version says "berekend" which means calculated, which is the same as computed, right?

Furthermore, the Dutch version says something like in case of discrepancy between the two versions that the English version is the leading one.




Not necessarily. It might have been computed from using the radar track by dividing radar return distances by radar sweep interval.


But they were refering to data found on the FDR.
edit on 14-9-2015 by RogueWave because: (no reason given)



posted on Sep, 15 2015 @ 08:16 AM
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a reply to: RogueWave

Have you reached any conclusions from reading that report, or are you still working on it?



posted on Sep, 15 2015 @ 09:19 AM
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a reply to: Salander

As long as I don't know what they mean with computed airspeed I can't really draw a conclusion.

I take it that you know what the issue is?



posted on Sep, 15 2015 @ 09:23 AM
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a reply to: RogueWave




Not necessarily. It might have been computed from using the radar track by dividing radar return distances by radar sweep interval.


If this was the case, then we can assume that it is true airspeed.



posted on Sep, 15 2015 @ 02:11 PM
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a reply to: RogueWave

The only thing that I can think of is that actual airspeed must be calculated from FDR raw output since the output is digital. There is a linear engineering units conversion euation from which airspeed can be calculated. The equation transforms a binary value ranging from 0 to 4095 into an airspeed. That does require a calculation. The raw digital data goes from 0 to 4095 with the calculated real value going from 0 to 1024 knots. The old analog FDRs gave real readings using a stylus scratching a metal foil strip. That was back when FDRs only had to withstand 100gs. They now have to withstand 3400 gs, a 2000 degree fire for 30 minutes, and the water pressure at 20,000 feet. Second generation units used magnetic tape. Now, solid-state memory chips are used in most units. The French BEA has put out a really good technical paper on FDRs. www.bea.aero...



posted on Sep, 15 2015 @ 02:13 PM
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originally posted by: RogueWave
a reply to: RogueWave




Not necessarily. It might have been computed from using the radar track by dividing radar return distances by radar sweep interval.


If this was the case, then we can assume that it is true airspeed.

No, that would be groundspeed.



posted on Sep, 15 2015 @ 03:40 PM
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a reply to: F4guy

But we were talking about computed airspeed supposedly being computed from radar data.



posted on Sep, 15 2015 @ 05:09 PM
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originally posted by: RogueWave
a reply to: F4guy

But we were talking about computed airspeed supposedly being computed from radar data.


Right. You would take the distance traveled as shown by the radar and divide by the time between positions. So it would br Dd/Dt which is the definition of groundspeed.



posted on Sep, 15 2015 @ 05:23 PM
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a reply to: F4guy

But you offered this explanation in reference to computed airspeed. Anyways, with 13 kt winds at FL330 at the time, there would hardly be a difference anyway.



posted on Sep, 15 2015 @ 07:18 PM
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originally posted by: RogueWave
a reply to: F4guy

But you offered this explanation in reference to computed airspeed. Anyways, with 13 kt winds at FL330 at the time, there would hardly be a difference anyway.


Depending on direction, there could be as much as a 26 knot variation, or as little as 0.



posted on Sep, 15 2015 @ 07:29 PM
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a reply to: F4guy

Seems to me there could only be a max 13 kt deviation from groundspeed if winds are 13 kt.



posted on Sep, 16 2015 @ 04:37 AM
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originally posted by: RogueWave
a reply to: F4guy

Seems to me there could only be a max 13 kt deviation from groundspeed if winds are 13 kt.

If TAS is 100, and you have a direct headwind f 13, GS is 87. If it's a direct tailwind, GS is 113. The difference between the 2 is 26.



posted on Sep, 16 2015 @ 05:05 AM
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a reply to: Zaphod58




so according to their ground speed they were flying at about mach 1.12 or something like that.


Ok . So is a sonic boom relative to the ground , or relative to the air around the plane . Honest question .
edit on 16-9-2015 by hutch622 because: (no reason given)



posted on Sep, 16 2015 @ 05:16 AM
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a reply to: hutch622

Air. Ground speed is for aspects of that question, a false reading since it can be affected by wind speed.



posted on Sep, 16 2015 @ 05:40 AM
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a reply to: Zaphod58

Having heard of planes going faster than sound relative to the ground i just wondered if you were on the ground if their would be a boom .



posted on Sep, 16 2015 @ 06:09 AM
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a reply to: hutch622

Not in this case, no. The actual speed of the aircraft was subsonic so there was no boom. Their actual airspeed was about 0.9 mach.



posted on Sep, 16 2015 @ 08:33 AM
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a reply to: RogueWave

I'm not sure. The issue with MH17, if that's what you mean, is that there is nothing to support the BUK theory, but some evidence to support cannon fire theory.

As for "computed", I would agree with the other poster that "calibrated" is probably what they meant. Probably "lost in translation", but either way I doubt the numbers change much.



posted on Sep, 16 2015 @ 12:44 PM
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originally posted by: F4guy

originally posted by: RogueWave
a reply to: F4guy

Seems to me there could only be a max 13 kt deviation from groundspeed if winds are 13 kt.

If TAS is 100, and you have a direct headwind f 13, GS is 87. If it's a direct tailwind, GS is 113. The difference between the 2 is 26.


Yes between those two numbers.

But the deviation from groundspeed can never be more than 13 knots, it is either plus or minus 13 kt. Just like I said....




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