Mr Optimistic

As this is a technical forum I was hoping you would flesh this out by showing the calculations. I really don't think it is fair comment to label the guy as unintelligent and expect forum users to simply take your word for it. Has he done something to your sister or do you have another reason for the ad-homs approach?

As neither yourself or awblain have shown any inclination to do the sums I thought I'd do them myself:

According to his paper g is adjusted by the following equation:


gₐ = gₑ x (rₑ ÷ rₐ)²

It is obvious that at low altitudes this is going to make no difference. But what is the upper limit where a tool such as this would be used? A case study would be the Columbia disaster.

The altitude of that break-up was ≈ 64km. The Earth radius is ≈ 6380km at the equator (WGS84). So, rₑ is 6380km and rₐ is 6444km. Using 9.81 for gₑ in the above equation and evaluating for gₐ results in 9.62ms⁻².

By using the equation:

R = V₀ x √(2h ÷ g)

..we can get a feel for how much difference this would make to horizontal range, using the velocity at break-up ≈ 6000ms⁻¹ and evaluating R when g = 9.81 and then again when g = 9.62

This would not include drag. The answers are 685.4km and 692.1km

So, up to about 7km would be the maximum difference this technique would make, before drag. Clearly, drag would reduce this.

Whether or not you think that is trivial depends on one's point of view. I think that if his model allows him to make such a correction he may as well have it. Even if it isn't necessary.

There we have a problem, since the natural frame to choose is the one in which the center of mass is fixed and both bodies mutually orbit around the center of mass. That way you see the motion of the tide-susceptible parts of the Earth clearly.

Only because the Earth is rotating at a different rate to the Earth-Moon system. On the Moon this is not the case, since the Moon is tidally locked, yet there are still tides raised. That discussion of in-out motion also jumps right past where the tides arise.

L1 does not have zero gravity. It has a zero gradient of gravity. If it had zero gravity, things would not be able to be kept orbiting there, but would instead drift off in a straight line.

Oggers,

The Columbia situation is the same point point again there's wood to focus on before worrying about the trees. You don't need to do irrelevant sums, you can just say that different sums need to be done. You don't ask for sums at a PhD exam, you ask for insight, intuition and an understanding of the issues.

Here's one reason why - and one is sufficient to drop the point and think more about it.

Vertical changes in gravity by 1% are a minor detail, since the break up was not at a point - it took tens of seconds, and spread over tens of kilometers. Thus your uncertainty in initial position makes a potential 1% shift in a 700km track neither here nor there.

Remember that the drag for the whole shuttle brings it down in Florida about 4000km after substantial contact with the atmosphere, and the actually distance from break up to debris hitting the ground was only 100-200km, so saying anything numerical "without drag" is like focussing on a flea while missing a bear it is sitting on.

Awblain's address is Pasadena. Judging by his accurate responses I guess he works in an engineering capacity for a tech company perhaps associated with JPL. I studied for a couple of years at a place down the road there on California Blvd. Not sure why anyone would expect to learn much physics on pprune though when there's a world of introductory texts out there.

Swing and a miss.

Y'know, it isn't all that difficult ...

warmer ...

Well sometimes you just have to hazard a guess :)

awblain

No such things as a "natural frame"

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.

Neither of these take away from the explanation.

I'll give you that about the L1 point. I was being simplistic.
In an inertial frame the zero g point is a little further out than L1.

Incidentally, if you are "the" A W Blain, I hope your books are less irritating to read.

Natural frame just means the one that makes the formulation easiest, nothing of cosmological significance implied.

ANCPER,
I miss your entertaining style.

How did you go with conjuring up another mathematical proof that the mass of different objects has no effect on the time taken for those objects, (aircraft pieces in this discussion) to reach the ground?

Quote:
Not a pilot yourself then? And if you are a pilot, then I apologise. You have proved yourself correct with that put-down.

And why on earth would you disagree with A squared?
The post he/she delivered was clear, humorous, accurate and well written and based on "basic physics".

Eagerly awaiting more of your entertaining posts.

Yes, Where is ANCPER? He's gone oddly quiet. I was really looking forward to his next iteration of why I'm wrong.

I do hope nothing has happened to him.

A Squared
 

Glad to see your concerned for my health!

No, I hadn't slunk off anywhere. I was waiting for response higher up the food chain, so had to wait out the w/end to get a response and yes I concede. Damn! :O Yes, it's too dumb to say anything further!

;) Thanks for being a good sport about it.

All is not lost ANCPER.:ok:

If you are a pilot, which doubtful in my opinion, then there may be truth to your assertion that some pilots don't know basic physics.

You being one of them.


But back to the thread...

It is turning out like we need to discuss the physics of disappearing objects rather than falling ones.

If it's any consolation I admit to being confused too.

Good thread this one.

Nice to get away from the MAS hysteria on R&N.

Unfortunately I have made some posts which only contributed to the noise so would be hypocritical to comment. Strange that so many people lost their lives in a manner which may have been awful (hope not) and we are indulging our technical curiosity...and I am guilty of it too.

I just read the first 10 posts of this thread. Man, this internet thingy, it is really something, isn't it?

Any high school physics student could set you all straight; good thing high school physics isn't needed to fly airliners...

Ok, this is on the matter of actual falling objects; airplanes in fact.

To start with: I am NOT entering the weird world of tin foil hats. I could not care less about what happened. The return of Elvis. Evil gnomes. Alien abductions. AlQaeda, IRA and PLO joint operation or any government cover up. I really really DO NOT CARE.

I am only interested in the physics, ok?

So, with this in mind, I would like to know if anyone here knows of ANY kind of calculation that would make, say TWA800, drop its nose and then continue to climb a few hundred metres (thousands of feet). I have tried different ballistic calculations but can only get the plane up to some tens of metres (about hundred feet actually). Some scenarios I tried:

- Approximating a B747 fuselage shortened with the length of the nose but CoG intact from official documents of a whole 747.

- As above but with different CoG.

- Approximating a B747 without a nose, instead a metal skin at the point of breakup, CoG intact from official documents of a whole 747.

- As above but with different CoG.

In all scenarios I assumed that all four engines after-the-fact were providing equal amounts of climb thrust and constant climb ratio per official documents, although I know that may not be the case depending on what wires may or may not have been severed.

Ballistic calculations are not the best way to approach this case because you have to take into account the various drag forces that would make a noseless plane do strange things, but statistically analyzing my calculations I still can't get the plane to make such an extraordinary trajectory even with significance 5%.

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?

Please, no endless tirades about your preferred theory about what happened. CFT explosion sounds good enough for me. I am just baffled by the trajectory - that is my only concern.

Well, for what it's worth, I wondered the same thing while reading the official explanation of the airplane "streaking" upward after the CWT explosion.

Without going thru the numbers again, what I recall was that if you assumed that the airplanes velocity was suddenly directed upward, without significant loss of energy the altitude gain suggested was at leas semi-plausible.

The problems with that of course was that the reason the airplane suddenly turned up was the CWT explosion blew the nose off.

This introduces two problems:

1) The huge amount of drag from having a gaping hole in the fuselage instead of the forward half would not be a low drag feature.

2) The post explosion airplane was assumed to achieve a stable zoom climb following the loss of of the forward fuselage. This is inconsistent with the part of the theory which says that the loss of the forward fuselage caused it to depart stable flight upward. The loss of the forward fuselage either caused it to become unstable, or it didn't. It can't really do both, which is what the theory requires.

You can only figure that it climbed thousands of feet if you completely neglect drad and aerodynamics and focus only on how high the kinetic energy would take it in a vacuum.

The "zooms upward" scenario fails on an entirely different level. 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. The theory being that the post-explosion airplane was what they saw "streak" upward. The problem with that was the distance, altitude and speeds involved. From memory, the airplane was at 14,000 ft (or was it 15K?) when the explosion occurred, and (again from memory) it was something like 40 miles away from the nearest witness who reported something 'streaking upward"

We have all watched airplanes fly. an airliner at 14,000 ft altitude, and even just 10 miles away does not "streak" across the sky. It crawls. So if from the observers perspective it appears to be crawling across the sky, if that same airspeed is suddenly directed perfectly upward without loss of speed, it's not going to appear to "streak" upward any more than it was "streaking" across the sky horizontally.

Now, I'm not trying to start a conspiracy theory discussion either. I have no idea what happened to TWA 400, There are all kinds of reasons why a shoot-down is really implausible, and there are all kinds of reason why a CWT explosion seems the most plausible explanation.

I'm just saying that the post-explosion climb as an explanation for witness reports of fiery streaks going up to the airplane is a complete farce. And that is the only reason it exists.

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.

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

This is what Wikipedia says about your question :

"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.

Tourist

I think you'll find that circular orbits produce perfectly fine tides.

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.

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.

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.

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:

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.

awblain

You've contradicted yourself:

:ok:

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."

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.

Ah, the theatre for scientific illiterates has re-opened.

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 don't see a problem that needs an answer.

Here is a clue. Ballstic coefficient.

Awblain; this is a model for approximating motion you are talking about, a tool for accident investigation, not a tilt at a Nobel prize. If it is subjective discourse which floats your boat, take it to jetblast. That's all I have to say on the matter.

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.