Physics of falling objects
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A Squared:
So to your knowledge there has been no scientific mathematical model applied on that animation?
As I said, and gave some examples of, nothing I have come up with even remotely resembles that kind of trajectory. In all calculations I have done, even when just considering the B747 as an intact plane just shortened fuselage by the length of the nose, the CoG inevitable goes backwards and causes the plane to stall after a very short period of time. Assuming all engines work completely fine* I do get some upwards movement, but not even in the range of thousands of feet.
But, as I said. These are all ballistic calculations and as we know that is a huge amount of drag that are not taken into consideration when calculating ballistic scenarios.
I have not made any drag (non-ballistic) calculations because honestly they are SOO much work!
*at the point of explosion and in some scenarios for 2s afterwards. I chose 2s completely arbitrary.
The raison d'etre for the "zoom upward" theory was to explain why so may people reported seeing something "streak" upward. It serves no other purpose, other than to provide an explanation for the witness accounts of a fire streak going up.
As I said, and gave some examples of, nothing I have come up with even remotely resembles that kind of trajectory. In all calculations I have done, even when just considering the B747 as an intact plane just shortened fuselage by the length of the nose, the CoG inevitable goes backwards and causes the plane to stall after a very short period of time. Assuming all engines work completely fine* I do get some upwards movement, but not even in the range of thousands of feet.
But, as I said. These are all ballistic calculations and as we know that is a huge amount of drag that are not taken into consideration when calculating ballistic scenarios.
I have not made any drag (non-ballistic) calculations because honestly they are SOO much work!
*at the point of explosion and in some scenarios for 2s afterwards. I chose 2s completely arbitrary.
Last edited by MrSnuggles; 21st Mar 2014 at 14:56. Reason: adding explanation of calculations
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High-flying aircraft that are approaching you "streak upwards"; those that are receding from you "dive downwards". Things come up over the horizon, and they go down below it, even if they are traveling along horizontally.
The "streak" in TWA800 was from the fire initiated by the initial blast in the center tank. It connected to a "fireball" from the bulk of the rest of the fuel burning.
For a fine example of journalistic integrity and inquisitiveness, after a slow start, and some associated conspiracist buffoonery, google "LA missile launch catalina helicopter". Mystery Missile Launch Seen off Calif. Coast - CBS News
The "streak" in TWA800 was from the fire initiated by the initial blast in the center tank. It connected to a "fireball" from the bulk of the rest of the fuel burning.
For a fine example of journalistic integrity and inquisitiveness, after a slow start, and some associated conspiracist buffoonery, google "LA missile launch catalina helicopter". Mystery Missile Launch Seen off Calif. Coast - CBS News
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Originally Posted by MrSnuggles
Do anyone here know if that animation of the trajectory after (whatever happened) is based on scientific calculations or if it is just a conjecture based on eyewitnesses statements?
"Only the FAA radar facility in North Truro, Massachusetts, using specialized processing software from the United States Air Force 84th Radar Evaluation Squadron, was capable of estimating the altitude of TWA 800 after it lost power due to the CWT explosion.[99] However, because of accuracy limitations, this radar data could not be used to determine whether the aircraft climbed after the nose separated.[99] Instead, the NTSB conducted a series of computer simulations to examine the flightpath of the main portion of the fuselage.[100] Hundreds of simulations were run using various combinations of possible times the nose of TWA 800 separated (the exact time was unknown), different models of the behavior of the crippled aircraft (the aerodynamic properties of the aircraft without its nose could only be estimated), and longitudinal radar data (the recorded radar tracks of the east/west position of TWA 800 from various sites differed).[101] These simulations indicated that after the loss of the forward fuselage the remainder of the aircraft continued on in crippled flight, then pitched up while rolling to the left (north),[98] climbing to a maximum altitude between 15,537 feet (4,736 m) and 16,678 feet (5,083 m)[102] from its last recorded altitude, 13,760 feet (4,190 m).[21]"
P.S. A quick sum shows that an object travelling at 400 kts can gain about 7000 ft of height ballistically.
Last edited by HazelNuts39; 21st Mar 2014 at 17:53. Reason: P.S.
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Tourist
I think you'll find that circular orbits produce perfectly fine tides.
Discussion of moon tides is obfuscation. These are due to an eliptic orbit, ie the moon does not face the exact same side to the earth at all times and the tides due to the sun.
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HN39,
You could turn speed into height - at about that rate,
Delta-h proportional to Delta-v-squared.
Lop off the front, but keep the wings intact on the rest, the angle of attack would rise sharply, boosting both lift and drag, and pointing whatever thrust was left upwards, so slowing and climbing until a stall makes sense. I can see 2000 feet gained being reasonable.
That's probably not what the witnesses would have seen as the "streak" though, that's all about how a short fiery track on the sky that ends in a bigger fire appears. Unless you're directly abeam of the track, a horizontal track would still appear to rise or fall above the horizon.
You could turn speed into height - at about that rate,
Delta-h proportional to Delta-v-squared.
Lop off the front, but keep the wings intact on the rest, the angle of attack would rise sharply, boosting both lift and drag, and pointing whatever thrust was left upwards, so slowing and climbing until a stall makes sense. I can see 2000 feet gained being reasonable.
That's probably not what the witnesses would have seen as the "streak" though, that's all about how a short fiery track on the sky that ends in a bigger fire appears. Unless you're directly abeam of the track, a horizontal track would still appear to rise or fall above the horizon.
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HazelNut39
Thankyou for that Wikipedia citation. I have read the NTSB report but apparently failed to notice that part of it.
Regarding the quote.. What object are you calculating? I based my calculations on an airplane (first approximated with a cone, then with a real airplane) albeit in some weird configuration(s). Is your object a sphere (= bullet)? Larger than a bullet?
I'll have to do some equations tomorrow methinks...
awblain:
I must do something wrong here because my calculations are inaccurate with an order of ten if this climb trajectory is factual. Mind you, I do all calculations in the SI units.
P.S. A quick sum shows that an object travelling at 400 kts can gain about 7000 ft of height ballistically.
Regarding the quote.. What object are you calculating? I based my calculations on an airplane (first approximated with a cone, then with a real airplane) albeit in some weird configuration(s). Is your object a sphere (= bullet)? Larger than a bullet?
I'll have to do some equations tomorrow methinks...
awblain:
I must do something wrong here because my calculations are inaccurate with an order of ten if this climb trajectory is factual. Mind you, I do all calculations in the SI units.
Last edited by MrSnuggles; 21st Mar 2014 at 23:27. Reason: adding info to awblain
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Mr S.,
Be careful with my Delta there - it's a "change all the kinetic energy to potential energy" Delta, so gives the maximum possible height gain, and not a normal small-change small-d delta.
(Delta-v)(Delta-v) = 2g (Delta-h).
As 1kt is very close to about 0.5m/s, for 400kt, Delta-h ~ 200x200/20 ~ 2000m, as HN39 said.
It's for a dragless/liftless object, so more like a bullet than an airliner.
For a normal small change delta-v: v delta-v = g delta-h.
There's also the remaining thrust at work in TWA800, assuming the engines kept running.
Be careful with my Delta there - it's a "change all the kinetic energy to potential energy" Delta, so gives the maximum possible height gain, and not a normal small-change small-d delta.
(Delta-v)(Delta-v) = 2g (Delta-h).
As 1kt is very close to about 0.5m/s, for 400kt, Delta-h ~ 200x200/20 ~ 2000m, as HN39 said.
It's for a dragless/liftless object, so more like a bullet than an airliner.
For a normal small change delta-v: v delta-v = g delta-h.
There's also the remaining thrust at work in TWA800, assuming the engines kept running.
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awblain,
You are of course correct in your observation that my 'simple sum' is based on nothing else than the conversion of kinetic into potential energy. That is no more than a first approximation (an upper bound) of what can happen in this sort of scenario.
I probably misunderstood MrSnuggles when he wrote:
You are of course correct in your observation that my 'simple sum' is based on nothing else than the conversion of kinetic into potential energy. That is no more than a first approximation (an upper bound) of what can happen in this sort of scenario.
I probably misunderstood MrSnuggles when he wrote:
These are all ballistic calculations and as we know that is a huge amount of drag that are not taken into consideration when calculating ballistic scenarios.
I have not made any drag (non-ballistic) calculations ...
I have not made any drag (non-ballistic) calculations ...
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HN, It's a good upper bound, and probably just as much use as the NTSB's simulations.
The thing that is striking about the TWA breakup, which I hadn't appreciated, was that the wing box remained intact after the explosion, and all the initial damage was to a collar forward of the wing root, causing the front to separate. From the reconstructed wreckage put back together on the scaffold, I'd assumed it had all come apart together.
This pattern of damage seems rather similar to Lockerbie, which in hindsight isn't too surprising from an explosion in the lower fuselage.
The thing that is striking about the TWA breakup, which I hadn't appreciated, was that the wing box remained intact after the explosion, and all the initial damage was to a collar forward of the wing root, causing the front to separate. From the reconstructed wreckage put back together on the scaffold, I'd assumed it had all come apart together.
This pattern of damage seems rather similar to Lockerbie, which in hindsight isn't too surprising from an explosion in the lower fuselage.
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awblain
You've contradicted yourself:
(Delta-v)(Delta-v) = 2g (Delta-h).
As 1kt is very close to about 0.5m/s, for 400kt, Delta-h ~ 200x200/20 ~ 2000m, as HN39 said.
It's for a dragless/liftless object, so more like a bullet than an airliner.
As 1kt is very close to about 0.5m/s, for 400kt, Delta-h ~ 200x200/20 ~ 2000m, as HN39 said.
It's for a dragless/liftless object, so more like a bullet than an airliner.
saying anything numerical "without drag" is like focussing on a flea while missing a bear it is sitting on.
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Oggers,
You may wish to avoid selectively quoting part of one statement, instead of including the relevant bit that "2000 feet gain being reasonable", and not adding in a separate completely disconnected comment about the bears and fleas, which pertained to a quite separate discussion.
While you're at it, you can also ponder the statement about HN's number that "It's a good upper bound, and probably just as much use as the NTSB's simulations."
You may wish to avoid selectively quoting part of one statement, instead of including the relevant bit that "2000 feet gain being reasonable", and not adding in a separate completely disconnected comment about the bears and fleas, which pertained to a quite separate discussion.
While you're at it, you can also ponder the statement about HN's number that "It's a good upper bound, and probably just as much use as the NTSB's simulations."
In sailing forums, people have amazing arguments about the use of water ballast in yachts - 'cos any fool knows water is no heavier than water so it obviously can't exert a downforce, right?
In motorcycling forums people will argue vehemently that raising your body above the bike by standing up on the footpegs will power the centre of gravity of the bike - 'cos it's obvious to any fool since your weight is then supported by the foot pegs rather than the seat, right?
But I never thought I'd see professional pilots arguing that the density of otherwise identical objects would not affect their terminal velocity!
I guess its a sort of second order "any fool" thing.
1) 'Cos "any fool" knows that heavy objects fall faster, right?
2) Except of course that "any fool" remembers Galileo and the cannon balls and knows that is wrong and that gravity acts at (~)9.8 N/kg on light and heavy objects alike, right?
1 N = 1 kg m s^-2 so 9.8 N/kg = 9.8 m/s^2.
Gravity is a force per unit mass, or an acceleration, while forces due to drag on identical shaped objects of different densities (and hence different masses) are simply forces which result in an additional acceleration which is dependent upon the mass.
Thanks for an entertaining read.
In motorcycling forums people will argue vehemently that raising your body above the bike by standing up on the footpegs will power the centre of gravity of the bike - 'cos it's obvious to any fool since your weight is then supported by the foot pegs rather than the seat, right?
But I never thought I'd see professional pilots arguing that the density of otherwise identical objects would not affect their terminal velocity!
I guess its a sort of second order "any fool" thing.
1) 'Cos "any fool" knows that heavy objects fall faster, right?
2) Except of course that "any fool" remembers Galileo and the cannon balls and knows that is wrong and that gravity acts at (~)9.8 N/kg on light and heavy objects alike, right?
1 N = 1 kg m s^-2 so 9.8 N/kg = 9.8 m/s^2.
Gravity is a force per unit mass, or an acceleration, while forces due to drag on identical shaped objects of different densities (and hence different masses) are simply forces which result in an additional acceleration which is dependent upon the mass.
Thanks for an entertaining read.
Last edited by nonsense; 30th Mar 2014 at 08:57.
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Dunno, it's kind of a difficult post to , but I *think* his point is that the he agrees that the density of a real object falling though a real atmosphere *will* affect the velocity. I *think* it's just a case of him being a bad writer.
I could be wrong though.
I could be wrong though.
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The poor guy's written a decidedly mediocre conference proceeding that advances Mr Stokes' work from the 19th century not one jot.
Video: Skydiver almost gets hit by meteorite
Apart from being an interesting read in itself (a skydiver managed to film a small falling meteor passing him as he descended), this is an interesting reminder that terminal velocity is not the fastest an object can fall, it is the stable falling speed an object will tend towards regardless of whether it starts out slower or faster. It is the speed at which wind resistance up is equal to the mass times the acceleration due to gravity down, resulting in no net acceleration.
Apart from being an interesting read in itself (a skydiver managed to film a small falling meteor passing him as he descended), this is an interesting reminder that terminal velocity is not the fastest an object can fall, it is the stable falling speed an object will tend towards regardless of whether it starts out slower or faster. It is the speed at which wind resistance up is equal to the mass times the acceleration due to gravity down, resulting in no net acceleration.
