One of the many baffling aspects of the QZ8501 story so far is why the plane disappeared from radar screens when it did. Did the plane suffer some kind of catastrophic event that caused the plane’s transponder to cease functioning? Or did something else occur?

I believe that we now have enough information to answer that question.

All we know about the plane’s final moments comes via two images that were apparently leaked from the official inquiry. The first (figure 1, above) is said to be a screen grab from an air traffic control (ATC) screen shortly before the plane disappeared. The second (figure 2, after the jump) is a screen grab taken very shortly afterward, this time from what looks to be some kind of analysis software, showing the plane’s speed, heading, rate of climb, and so forth.

According to Embry-Riddle Aeronautical University professor Martin Lauth, who helped me to understand the symbology of figure 1, the yellow arrow is pointing at the symbol for the plane in question, here designated “AWQ8501.” The number to the right, 353, is the ground speed of the plane in knots. The number below, 363, indicates that the plane was at 36,300 feet, and the white arrow to the right of it shows that the plane was climbing.

Next, let’s talk about the four white lines coming from the QZ8501 symbol, starting with the one heading more or less straight down and connecting it to “AWQ8501.” That line just indicates which symbol the tag corresponds to. Moving clockwise, we next find a much shorter line sticking to the left. This is a visual indicator of how far the plane will move in a certain amount of time — controllers typically set it for anywhere from one to three minutes, and in this case it seems to set for one minute. We already know the speed of the plane, but this line tells us its heading: a little south of due west, on a heading of 265 degrees true.

That, of course, is not the direction the plane was supposed to be heading. It was supposed to be going to RAFIS, marked by the next line, working clockwise. So what we see here apparently is a plane that’s either diverting to the west to avoid weather or is simply out of control.

Finally, we see a line that projects from the QZ8501 symbol almost directly to the north and continues past the edge of the image. This appears to indicate the radar station which is collecting the displayed information, in this case a facility at the town of Pontianak on the island of Borneo. The data on the screen either comes from the plane’s Mode S transponder via secondary surveillance radar, or is routed through the same transponder to a different systems called ADS-B.

Either way, the thing to understand is that these are line-of-sight systems. That means that if an aircraft gets too low, the transmission of information will be blocked by the curvature of the earth. How low? Well, of course, it depends on hour far apart the plane and the antenna are. In this case, the plane was about 200 nautical miles away. A back-of-the-envelope calculation suggests that the altitude below which the plane will be unable to transmit to the station is about 20,000 feet. (By the way, a tip of the hat to Mike Exner and Don Thompson for turning me on to all of this.)

In the final ATC screen grab we see that the plane was at 24,025 feet and descending at 11,518.75 ft/min, which is equivalent to 130 mph straight down.

This is never something that a commercial jet wants to do, but we now understand that we have no reason to believe that the plane’s electronics were compromised: rather, we would expect the plane to vanish from ADS-B and radar screens around this time anyway, for the simple reason that it will henceforth be hidden by the curvature of the earth.

In other words, for all we know the plane’s avionics might well have been in perfect working order at this point.

Oh, and by the way, I’m also attaching a screen grab (figure 3) from Google Earth showing the locations of the things I talk about above, including the 200 nm range ring around Pontianak. I’m also including, for sheer curiosity value, the 0:11 and 0:19 ping rings from MH370. I am in no way advocating that this is the case, but I think it’s worth noting that there is a non-zero probability that QZ8501 (just by chance!) came to rest on top of the wreckage of MH370.

Or, to look at it another way, it’s possible that the searchers looking for 8501 could find 370 first.

UPDATE: Below, I’ve added the

More from ex-RMAF Capt. Rahmat:

“I spoke personally to members of the criminal probe, bg check on all pax cleared by Interpol”

https://twitter.com/nihonmama/status/552274053755830273

Criminal investigation??

Not sure what that means in the context of Malayasia. I have long felt we have had too many number crunchers and not enough buckle draggers on this problem, but I do not have a whole lot of confidence in the people doing the criminal investigation.

I don’t think MH370 is going to be found in the current search area. Not saying it is not there. It is simply a formidable task. It would sure help if some clues from the investigative side were forthcoming.

Gysbreght,

Thanks for your comments. Yes, I meant AP is off (MH370). Relevant scenarios are of my current interest. With the constant thrust of approximately 160 kN (also subject of optimization of BTO/BFO residuals), which would correspond to 80% of the maximum of a single engine thrust (if I am not mistaken) and nearly-constant mass, I am getting pretty good BTO/BFO match, including the second phone call. Altitude varies between ~5 to 7 km in that scenario; the terminal point appears to be not far from the Chinese ping location. There is a gap in Fugro bathy survey nearby – somewhere there. However, nearly-constant mass is not consistent with the other model assumptions, unless I implement some correction on the lift force or thrust itself.

You say “thrust reduces with altitude”. Can you recommend any dependency of the thrust on pressure, air temperature, mach, etc.? I guess this statement is valid for both GE and RR engines.

You say “B777 would maintain constant CAS”. Wouldn’t this be another mode of A/T? What is a likely mode of A/T in case of aborted landing, btw?

Does your answer imply that the angle of attack may automatically be adjusted in A/T mode? If yes, what flight parameters control the attack angle (in other words what is the target of the adjustment of the attack angle)?

Apparently I need some more reading re auto-throttle for better understanding of your answer.

Oleksandr:

As an introduction:

LAL B777 Ops Manual 4.20.8 and 4.20.9

“The autothrottle can be operated without using the flight director or the autopilot.

In this condition, the autothrottle operates in either the THR REF, SPD, HOLD or

IDLE modes.

(…)

Autothrottle Thrust Lever Operation

The autothrottle system moves either or both thrust levers to provide speed or

thrust control, depending on the mode engaged.

The thrust levers can be manually positioned without disengaging the autothrottle.

After manual positioning, the autothrottle system repositions the thrust levers to

comply with the engaged mode. The autothrottle system does not reposition the

thrust levers while in HOLD mode.”

THR REF mode: Thrust set to the reference thrust limit displayed on EICAS

SPD mode: Thrust applied to maintain target airspeed set using the MCP or FMC

HOLD mode: The A/T removes power from the servo motors and will not move the thrust levers

IDLE mode: The A/T moves the thrust levers to the aft stop

In summary, with A/P off, the only A/T ‘constant thrust’ mode between minimum and maximum thrust levels is HOLD mode. In this mode the thrust is defined by the manually set position of the thrust levers and will be nominally constant, but may vary somewhat with changing ambient temperature, pressure and Mach number, as controlled by the engine electronic control system (EEC or FADEC).

Addressing your questions:

“You say “thrust reduces with altitude”. Can you recommend any dependency of the thrust on pressure, air temperature, mach, etc.?”

Thrust at a constant intermediate thrust lever setting and constant air temperature and Mach typically varies proportionally to ambient pressure, but the system designers may have chosen something else. The dependency on temperature and Mach cannot be generalized and will differ between engine types.

“You say “B777 would maintain constant CAS”. Wouldn’t this be another mode of A/T?”

No, the thrust controls the rate of climb or descent.

“What is a likely mode of A/T in case of aborted landing, btw?”

In an aborted landing usually TOGA (Takeoff/Go-Around) thrust is selected, but some systems select a lower thrust level to avoid excessive rate of climb or acceleration.

“Does your answer imply that the angle of attack may automatically be adjusted in A/T mode? If yes, what flight parameters control the attack angle (in other words what is the target of the adjustment of the attack angle)?” Constant CAS means constant angle of attack at a given weight. In the longer term it is not entirely clear how the system responds to the weight reducing as fuel is consumed. If CAS is maintained then AoA will reduce. If AoA is maintained then CAS will reduce as the square root of weight.

Some data before trying to guess what might happen:

1. The aircraft does glide (seems that someone has some doubts about it). The speed target with two engines out is green dot; for the 320, 62000Kg. weight (GW) and FL 300 is 216Kt. At FL225 and below 200 kt.. It can make 2.5 miles per 1000 feet (about 1600 ft/min for the 320).

2. Both engines out is a critical situation. You have to read and perform DUAL ENGINE FAILURE list and this could be very triky. The aircraft goes into emergency electrical configuration because it loses the two main generators. This causes that only the equipments connected to emergency bars will operate. Specifically: VHF1 and ATC1. If the transponder switch was in ATC2, the eco on the ATC radar screen vanished, and if pilots did not use VHF1 no signal was emitted. The RAT automatically extends to recover blue hydraulic system. The APU cannot be started until reaching 25,000 feet and is just under 22500 when can be used for bleeding to pressurize both cabins (pneumatic system is lost). Do not rule out a possible emergency descent. Only the captain (CM1) has an active display (PFD) to fly, and emergency instruments barely can be seen because the lighting in the cabin is missed except just one light and it’s dark outside. By the other hand, the aircraft is immersed in a maelstrom of storms, severe turbulence, strong updrafts and downdrafts and total darkness. Now we could talk about the “perfect storm”.

What could happen?. The aircraft entered the cumulonimbus, a cloud with an incredible high energy. Encountered a strong updraft (can reach above 4000ft / min values.) and was catapulted up (it crossed FL363 at a very high rate of climb outperforming normal parameters). Severe icing, hail …. caused the engines to flame out. The plane disappeared from the radar control screen because the active transponder was ATC2 and pilots did not change to ATC1. They lost communications because they did not use the VHF1 perhaps because they were absorbed in fighting to control the aircraft in an extreme dangerous situation. In the VHF1 121.5 is selected as standard while in VHF2 the ATC frequency is selected. They initiated a descend at green dot or an emergency descent. If in addition they encountered a downdraft , the aircraft was “sucked” at a high rate of descent. The pilots controlled the plane in extremis and attempted a sea landing. There is an emergency list for that. I do not know if they could complete it and therfore if they pressed DITCHING switch in order to close all possible inputs of water in the aircraft after a sea landing.

Finally, the strong swell and maybe an uncontrolled sea landing with two engines out, or a first strong impact with one plane, etc. prevented a pax evacuation because a sudden flooding into passenger cabin (maybe even fractured). Perhaps a depressuration during descent could happen if pilots did not start the APU and due to that the oxygen masks dropped.

It’s just one of so many possible theories.

@DennisW:

Thanks for your Japan comments (in other thread) – just saw them today.

And re the criminal investigation, here’s more of the convo with the ex-RMAF. You’ll note that two of my questions weren’t answered:

https://twitter.com/rahmatomar/status/552284354190520321

Getting interesting. See my previous response to Rand. If true that ALL MH370 crew cleared by Interpol AND considering PM Najib Razak stated publicly his belief that MH370 was hijacked (and IF that turns out to be the case) then who were the culprits – and where were they?

Gysbreght,

Thanks for your explanation.

Do you think the maximum thrust 400 to 417 kN (according to http:||en.wikipedia.org|wiki|Boeing_777 – replace | with /) corresponds to the maximum thrust at the sea level? Would you expect the maximum thrust to fall to around 200 kN at 5.5 km altitude due to twice lower pressure?

If TO/GA was the last human input, what would happen next? Would the aircraft reach a certain altitude and become ‘trapped’ in some layer around that altitude, subject to ambient conditions?

“Constant CAS means constant angle of attack at a given weight. In the longer term it is not entirely clear how the system responds to the weight reducing as fuel is consumed”. This is exactly my problem under “constant thrust” assumption, which appears to be incompatible with the decreasing weight of the aircraft.

8501 tail section said to have been found. CNN showing photos.

Oleksandr,

The thrust values in the Wikipedia table is probably the rated takeoff thrust of the uninstalled engine on a teststand, at sealevel standard temperature and pressure. The thrust at altitude will be different due to altitude, temperature, Mach, installation losses and the rating structure adopted by the engine manufacturer.

If TOGA was the last human input, assuming A/P off and the airplane trimmed for a reasonable speed, and the thrust is increased very slowly in still air, the airplane would initially climb to a certain altitude, and then continue climbing at a very small rate of climb as fuel is consumed, and the thrust required to maintain altitude reduces with reducing weight.

I don’t understand the problem mentioned in your last paragraph.

Gysbreght,

Thanks again. In my last paragraph I meant that vertical momentum can be described by

d(M*W)/dt = -M*g + Lift,

or by more accurate expressions taking into account the vertical component of drag force, vertical component of Coriolis force, minor variations in g, etc. Here M is the aircraft’s mass, Lift is the lift force, g is the gravitational acceleration.

In my model I assumed

Lift = CL*AD*TAS^2,

where CL is the constant lift coefficient and AD is the air density. Indeed, CL depends on the attack angle, so that constant CL implies constant attack angle.

The term dM/dt * W is small, and thus it can be neglected. So, in the simplest form

dW/dt = -g + CL*AD*TAS^2/M.

This expression is very delicate. By saying “…and then continue climbing at a very small rate of climb as fuel is consumed…”, this would imply very small dW/dt. For example, the residual acceleration of as little as 0.001g would result in 1 m/s vertical speed in just 100 s.

Now, if for horizontal speed one assumes

dU/dt = TH/M – CD*AD*TAS^2/M,

where TH is the constant thrust, CD is the drag coefficient, the equation for dW/dt takes the form

dW/dt = -g + CL/CD*(TH/M-dU/dt).

If the horizontal acceleration is also small, then

-g+CL/CD*TH/M = 0 (approximately).

If M varies significantly, such a balance is impossible to achieve, unless considering possibility for CL or TH to vary. CL can vary due to changing angle of attack; TH can vary due to different ambient conditions as you noted.

In summary, to make your description of what would happen in case of TOGA was the last human input feasible, it appears to be required to implement varying coefficient for the lift force and/or thrust. However, varying thrust would imply uneven fuel consumption, and hence explicit modelling of M=M(t).

I am interested in this scenario, because with the constant (or nearly constant) mass, there is a trajectory that satisfies BTO/BFO. So it is not just “what if”. On top of it, I think such a scenario makes more sense than IG’s AP scenario.

(PL = Performance Limit; FMT = Final Major Turn)

Fig.3 of the ATSB’s Jun.26 report shows the March PL (FMT=18:22.

Fig.3 of the ATSB’s Oct.15 report shows the March PL (FMT=18:40).

Overlaying these two charts (rescaling and rotating 6° for comparability), we see how dramatically the PL has retracted since March.

I’m trying to isolate the “available fuel at 18:22” component of this change. I think I need to subtract:

1) approximation error inherent my “rotation method” to adjust PL for change in FMT: wind, satellite drift, FMT in degrees – all (I think) immaterial, as previously discussed.

2) (primary) effect of wind: I believe the ATSB ignored wind for March PL, but includes it now..?

Question for the modeling experts: to what extent do you think wind impacted the Oct.15 PL? A quick sumproduct of (TAS-GS)*time from Richard’s 13.1 model suggests this would have retracted the PL by a mere 8 nmi, i.e. also fairly immaterial.

Anything else to consider?

If I’m not off base, they retracted the PL by over 250nmi for the “increased speed” @Malacca. Is this reasonable?

When they announced in August the search was moving back south, I thought they’d REVERSED this available fuel reduction. But apparently, the “wild altitude swings” were UNRELATED to the fuel reduction, because the latter seems to have been retained, despite abandonment of the former. If “wild altitude swings” wasn’t responsible for the Mar.28 reduction in assumed available fuel, what WAS? The primary radar track has had horizontal speed pinned down since mid-March.

(Clarification: I recognize impact of wind on PL would vary by location, and that, even if computed correctly, the 8 nmi is only relevant to the PL near the IG-modeled path. Just trying to confirm principles.)

Oleksandr,

First a small correction. Your second equation should read: Lift = CL*AD*TAS^2* S, where S is the reference wing area.

Second, may I suggest to use a more conventional notation where:

L = lift

D = drag

T = thrust

m = mass

W = m*g = weight

Greek letter rho = air density

Greek letter gamma = flight path angle

M = Mach

V = airspeed

vz = V*sin(gamma) = vertical speed

Then your last equation may be written as:

T/W = CD/CL, which is the horizontal force equation for unaccelerated level flight, since CD/CL = D/L and L = W.

In a simple model compressibility effects on aerodynamic coefficients may be ignored, then CD and CL vary with angle of attack only, so CD/CL is constant for constant angle of attack.

When weight reduces for constant thrust, T/W becomes greater than CD/CL, and we get a gradient of climb equal to:

Sin(gamma) = T/W – CD/CL

In an isothermal atmosphere (above the tropopause) thrust reduces with altitude proportional to atmospheric pressure, so as a first approximation the airplane could be assumed to follow a path where the weight divided by the atmospheric pressure is constant, and hence T/W is constant.

Strictly speaking the last equation assumes constant TAS, but it can be shown that in this simple model TAS is constant.

Apologies – second equation should read:

Lift = CL*AD*TAS^2*S/2

Hi Gysbreght,

Thanks.

Re reference wing area S: it is implicitly accounted in my model in CL and CD as long as the attack angle remains constant.

Re: “the airplane could be assumed to follow a path where the weight divided by the atmospheric pressure is constant, and hence T/W is constant”. Here is a problem. Thrust as a function of the air pressure implies its dependence on the altitude. But the altitude depends on vz integrated over the time. On top of it add a dependence of W on the thrust integrated over the time (assuming the thrust is proportional to the fuel consumption). If T/W is constant and TAS is constant, then vz is constant, and then it means linearly increasing altitude alt(t)=alt0+a*t, t is the time, a is a coefficient, alt0 is the initial altitude. Up to 10 km pressure approximately linearly decreases with the altitude, meaning that the ambient pressure changes as P(t)=P0 – b*t, where P0 is the pressure at the sea level, b – is the constant. Thus the ratio T/W is constant only if m is proportional to (P0-b*t), or m = m0 – b*m0/P0*t, where m0 is the initial mass. However, it is just a particular case of the initial mass, so that in general case T/W is not a constant.

In case it directs attention to my query, please allow me to demonstrate its pertinence:

The CURRENT MH370 search zone is predicated on the JOINT assumption that a) the signal data is accurate, AND b) fuel exhaustion occurred at 00:19. If not b), then [search elsewhere].

If the 250nm in my prior post cannot be explained, then perhaps the JIT is sandbagging the PL w.r.t. latitude (Jun.Fig.3 v Oct.Fig.3) as WELL as longitude (Oct.Fig.2 vs. Oct.Fig.3, previously discussed).

Oleksandr:

I believe you are making a simple problem very complicated. I would say that if at time t1 the weight is W1 and the airplane is at an altitude where the pressure is p1, and at time t2 the weight is W2, then the pressure at the altitude at time t2 is p2 = p1*W2/W1.

@Brock

The biggest search zone dependency is one you did not even mention – the assumption of a fixed AP mode since the turn South. I have had serious heartburn with this assumption from the very beginning. Without it, the plane could be just about anywhere.

Nihonmama – Since MH370 went missing interpol’s watch lists would have expanded enormously – sobering. I wouldn’t forget the Iranians. A lot of their “asylum seekers” are actually official exports. Large numbers of them here are male, 18-35 with govt/military backgrounds/connections. Check out this paywalled piece from The Australian:

MAN Haron Monis, the gunman behind last month’s deadly Sydney siege, would have been closely connected to Iran’s sprawling intelligence establishment before his arrival in Australia, his former lawyer claims.

Nazir Daawar, who represented Monis after he was charged in 2011 for sending offensive letters to the families of Australian soldiers killed in Afghanistan, said his former client had been a privileged member of the Iranian elite prior to his arrival in Australia.

Mr Daawar said Monis’s job as the director of a travel agency could not have been obtained without high-level connections to Iran’s intelligence services.

He said in Iran such jobs were considered positions of considerable authority, as travel agents were involved in obtaining visas in and out of the country.

“Knowing the Iranian community and their system, a person who did not have these connections would not be put into that position,’’ Mr Daawar told The Australian.

On December 15, Monis took 17 people hostage in the Lindt cafe in Sydney’s Martin Place. The siege ended 16 hours later with Monis and two hostages being shot dead.

The self-styled sheik and religious fanatic is understood to have arrived in Australia in 1996 on a business visa before applying for, and obtaining, refugee status. In letters reportedly written by him after his arrival in Australia, Monis claims to have had inside knowledge of the workings of the Iranian intelligence services. Monis had been a figure of occasional interest to Australia’s police and security services.

Mr Daawar described his former client as an often charming and charismatic man, but one given to outbursts of temper.

He said he refused to represent Monis after the Iranian ignored his advice not to address the media as the charges against him were being heard. “He was a very good talker. He could run a religious talk with power, with charisma. Even the first time he came to my office, he impressed me with the way he behaved,’’ Mr Daawar said.

Monis was investigated by ASIO in 2008 after he sent a series of offensive letters to the families of killed Diggers.

There is little doubt that the final judgment of the security agencies was that Monis was mentally unbalanced and was considered more of a pest than a threat.

The nature of his relationship with security agencies after he arrived in the country in 1996 remains unclear.

Sources familiar with ASIO’s practices say there is little doubt the agency would have examined his claimed relationship with the Iranian intelligence services.

However, it is not clear what conclusions, if any, they formed on Monis in the mid-1990s, or what, if any, bearing those judgments had on Monis’s application for a protection visa.

Link to chart overlay:

https://drive.google.com/file/d/0B-r3yuaF2p72OVpHRDVHT2thaGc/view?usp=sharing

I tested at two points 440kts and 400kts, and came up with the same % reduction in post-radar range (as my earlier work had predicted): 12.6%.

In nmi, this is 215nmi at 400kts, and 267nmi at 440kts. These reductions are BEFORE subtracting the wind component.

Further to it, Iran has demonstrated long term lethargy on the subject of extradition treaties so that people who slip out cannot be easily returned there. Hmm.

@all

So help me out here, please. I am making a presentation to the local (Gualala, CA) Lions Club on MH370. Does anyone know why the AES does not pre-compensate for ROC? I have been unable to find any background on this question.

I am still working on a title, and I need one before Friday of this week.

Choices:

“Where MH370 Really Went Down”

“MH370 – Experts Can be Expensive”

“MH370 – A Lesson in Understanding the Data Limitations”

“MH370 – The Devil is NOT in the Details”

Dennis: AES pre compensation – – probably because it is unnecessary for the originally intended purpose, and therefore an unnecessary complication. The functionality is there to achieve reliable communications, and it can do this without adjustments for ROC.

Thanks for the reply Flitzer. Seems like from a SW perspective it would be trivial to include it. I was thinking maybe it was a legacy from pre-GPS/INS days.

Was it that the old ‘selective availability’ GPS feature (now turned off) introduced (deliberately) large GPS altitude errors that would have caused an ROC correction to increase frequency error, rather than reduce it? So better to not use altitude change at all. The equipment and software seems to have been designed a long time ago, back in that era of GPS.

@ Richard Cole:

“Back in that era of GPS” airplanes had no altitude information?

Gysbreght,

by saying “if at time t1 the weight is W1 and the airplane is at an altitude where the pressure is p1, and at time t2 the weight is W2” you explicitly define dependence of the altitude on the mass, and hence the dependence of vertical velocity component and vertical acceleration component on the time. However, the acceleration is already a function of the thrust and lift coefficient. This results in the inconsistency (unless you assume that the attack angle also changes somehow).

I googled an interesting book, which contains some information I was looking for:

Ahmed F. El-Sayed. 2008. Aircraft Propulsion and Gas Turbine Engines. CRC Press, Taylor&Francis.

Oleksandr,

With vertical speed about 11 fpm and constant I wouldn’t worry too much about vertical acceleration.

Dennis,

Re yours “Without it, the plane could be just about anywhere.”. Not really. Discard IG’s AP hypothesis and suggest a better one, and you will have another priority search area.

Gysbreght,

Zero vertical acceleration requires very accurate balance of vertical forcing, such as in your model T/W – CD/CL = 0, because any systematic residual may have a large impact on these 11 fpm as a result of integration over the time. So, you cannot simultaneously postulate altitude change and achieve balance of vertical forcing.

A gentleman by the name of Isaac Newton taught us that a body moves at constant speed and direction unless a force is acting upon it.

The equation:

Sin(gamma) = T/W – CD/CL

expresses an equilibrium of the forces acting upon the airplane and is valid for climbing, level, or descending flight at constant speed, i.e. zero acceleration.

Gysbreght,

Now I am getting a bit confused. I should have probably said T/W – CD/CL = constant rather than =0, but now you say it “expresses an equilibrium of the forces acting upon the airplane and is valid for climbing, level, or descending”. If W is changing (decreasing due to fuel consumption), while CD and CL are constants, how gamma can be a constant in general case?

oleksandr,

T and W both change as ambient pressure, the ratio T/W remains constant.

Yes, you are confused.

Gysbreght,

Now I’m even more confused. How can the ratio T/W be a constant if W depends on the fuel consumption, while T is assumed to be a function of the air pressure (and, perhaps, air temperature)?

What do you specifically (in the form of functions or so) mean by “T and W both change as ambient pressure”?

Oleksandr,

I have clearly stated that, to keep it simple, temperature is assumed to be constant. At a given setting of the thrust levers, which typically commands an rpm (rotational speed), and constant Mach number, the thrust T of a turbojet engine is then proportional to ambient pressure p, i.e. T/p is constant. If weight W reduces due to fuel consumption and the airplane climbs at such a rate that W/p is constant, then T/W is constant. For the fuel flow of MH370 of 6.1 tons/hour that results in a RoC of approximately 11 feet per minute. At constant TAS it also results in constant CL and hence constant CD, AoA and pitch attitude.

Apologies if my explanation did not make that sufficiently clear.

Selective availability added pseudo-random errors to GPS positions of about 50m horizontally and 100m vertically. Calculating ranges of climb from an altitude jumping by 100m would give poor result. Of course INS data could be used (which would not have this problem), but I got the impression from the Honeywell manual that the system was designed to use an INS or GPS navigation data feed.

http://www.bbc.com/news/world-asia-30706298

“Part of the tail of the Airbus A320-200 was spotted by teams involving divers and unmanned underwater vehicles, search and rescue chief Bambang Soelistyo said in Jakarta.

It is the first significant piece of wreckage from the crash to be identified and was found in an area some 30km (19 miles) from the initial search area.

The part found has the AirAsia mark on it, Mr Soelistyo said. It is buried in mud, in water 30m (98ft)deep, and is believed to be upside down.”

http://www.bbc.com/news/world-asia-30706298

Pictures of said tail at bbc

Gysbreght,

Thanks; I think I start understanding your model.

“If weight W reduces due to fuel consumption and the airplane climbs at such a rate that W/p is constant…”

To make W/p = const let W(t)= k*p(z(t)), where z is the altitude of the aircraft, k is some constant. Hence

g * dm/dt = k*(dp/dz)*(dz/dt) == k*(dp/dz) *vz.

Fuel consumption rate q = -dm/dt. Then

q = -k/g * vz * (dp/dz).

Note dp/dz =: -f(z) is a known function of the altitude z, whatever simplified atmospheric model you want to take. So,

q = k/g * vz * f(z(t)).

vz = constant (climb rate) according to your statement. This yields z(t) = z0 + vz*t, where z0 is the initial altitude. Inserting into the previous:

q = k/g *vz * f(z0 + vz*t).

One of the implications is a varying fuel consumption rate.

The other implication is that we know that T~q at a constant temperature, hence q=k1*p, where k1 is a constant. Then

k1*p(z0+vz*t) = k/g *vz * f(z0 + vz*t).

This gives f(z) = (k1*g/k/vz)*p(z) == k2 * p(z)

Recalling that dp/dz = -f:

p(z) = p0* exp(-k2*z).

This really corresponds to the pressure distribution in the atmosphere with a constant temperature, where k2 = 1.186e-4 m^-1.

The three unknowns k1, k and vz can now be determined from the thrust, fuel consumption, and the relation k1*g/k/vz = 1.186e-4.

Am I right?

Oleksandr,

I didn’t examine your derivation in detail, because it is really quite simple. The weights at the start and at the end of the segment you’re considering define the pressure ratio between the altitudes at those points. Calculation of the corresponding altitude differential is then a matter of applying the hydrostatic equation and the ideal gas law, or looking up in a table for the standard atmosphere above the tropopause.

Gysbreght,

It was not a derivation, but rather a proof that only exponential pressure profile is compatible with your model… I’m sure originally it was derived in the inverse order.

Oleksandr,

I suppose you know that an isothermal atmosphere has an exponential pressure profile.

I don’t understand your second sentence. It was not a derivation, but it was derived in the inverse order?

Gysbreght,

I guess among the original assumptions in you model were the following: (1) isothermal atmosphere; (2) thrust is proportional to the air pressure; (3) fuel consumption is proportional to the thrust. It happens that the aircraft’s mass changes according to the exp law in such a case, and thus T/W = const, which consequently results in the constant vz. I did not initially realize that the fuel consumption varies in your model, also exponentially. Also initially I did not link “constant temperature -> exp pressure profile”, but rather thought that the assumption of the constant temperature is needed to eliminate a dependence of the thrust on it. I came to points (1) and (3) in a reverse order, i.e. considering vz = const in your model.

Oleksandr,

The original assumptions were: (1) constant thrust setting (rpm); (2) constant angle of attack; and (3) constant ambient temperature.

(1) and (3) together result in thrust proportional to air pressure.

(2) results in constant CD/CL

Changing ambient pressure proportionally to aircraft mass then results in constant T/W and hence constant vz. Fuel consumption proportional to thrust, i.e. constant specific fuel consumption, and constant TAS complete the overall picture so that everything is nice and tidy.

The “exponentional pressure profile” of the atmosphere results from the hydrostatic equation and the ideal gas law, and is not limited to constant temperature. The standard atmosphere below the tropopause is based on a constant temperature lapse rate and the pressure also varies exponentially with altitude, but the exponent differs from that at constant temperature.

Assumptions (1), (2) and (3) are not essential for answering your original question, they are only made because they result in simple equations that can be solved analytically. If different assumptions are made the airplane will still climb, but to determine how it climbs requires detail knowledge of engine, airplane, and control system characteristics and numerical calculation of the flight path.