What techniques professional pilots use to keep their piloting skills sharp when not flying for an extended period? I don't get to fly as often as I'd like. I know that simulators are good for brushing up on technique, learning new systems, etc.

Are there techniques that pros use (commercial or military) to stay sharp about things when they aren't able to actually fly? Specifically things that I would do not in a simulator, but say I have a couple months where I can't fly, but have an hour commute, or 30 minutes per day that I could do something flying-related to stay sharp.

I realize there is no substitute for actual flying, but I always feel a little rusty (mostly I just feel a little behind the airplane) when I haven't flown for a month or two. Are there things that other pilots do that I could learn to do in my down time that would mitigate this?

Reading a lot is key; staying on top of your checklists too. I noticed I'm behind the airplane more depending on the complexity of the aircraft in question. I keep photocopies of my checklists at home to refresh my mind prior to a flight.

It should also be mentioned the value of sitting in your aircraft on the ground; it costs nothing, and you can take your time working through checklists and scenarios in your mind.


When do the lift and drag vectors contribute a force component along a glider's path of travel as seen from the ground? It is often said that the lift vector helps to propel a glider forwards.  The lift vector has no component acting parallel to the glider's trajectory through the airmass, but in many cases the lift vector does have a component acting parallel to the glider's trajectory as seen from the ground.  Note that for a given airspeed, the direction of the glider's trajectory through the airmass-- i.e. the direction that is exactly opposite to the "relative wind" that is "felt" by the glider-- is not affected by sustained updrafts, downdrafts, headwind, or tailwinds.  But this is not true of the glider's trajectory as seen from the ground.  A glider in a powerful mountain wave updraft may be rising straight up as seen from the ground, while the trajectory through the airmass will still be the normal glide path that would be obtained at whatever airspeed the glider is flying at.

Under what circumstances does the lift vector of a glider contribute a force component along the glider's path of travel (trajectory) as seen from the ground?

Under what circumstances does the drag vector of a glider contribute a force component along the glider's path of travel (trajectory) as seen from the ground?

Likewise for a powered airplane.

Note -- to avoid any possible ambiguity, please be aware that this question is using the word "lift" exclusively to mean an aerodynamic force generated by the aircraft, not to mean rising air.  These are two completely different things.

Note -- the scope of this question is meant to be confined to linear (but not necessarily horizontal) straight-line flight with constant airspeed and groundspeed, at least over the short term.  We're assuming the glider (or airplane) is flying in an airmass that is locally uniform. In other words, if the glider has entered a thermal updraft or penetrated through an abrupt wind shear or flown from sinking air into a ridge updraft or wave updraft, we're assuming that it has been in the new airmass long enough to come into equilibrium, so that net force is zero for the time being.  This is not a question about "dynamic soaring" as practiced by the albatross over the open ocean, r.c. glider pilots flying in loops on the lee side of the hill, someone trying to put into practice Taras Kiceniuk's ideas about exploiting the boundaries between still air and downdrafts as a source of energy, etc.  Those subjects are worthy of an ASE question, or many questions, but this is not that question.

In thermals.

More precisely: Every time when the updraft strength increases. The glider will continue on its original path due to inertia while the airflow will have a positive angle when referenced to the path of travel. Now the aerodynamic force will point slightly forward and accelerate the glider.

Experienced pilots use this by pulling more than 1 g when flying into a thermal and less when updraft strength decreases. This helps them to gain some more energy from the rising air.

Drag is per definition the part of aerodynamic forces parallel to the flow direction, so for drag the answer is: Always. With changes in vertical airspeed, the drag component parallel to the direction of movement will merely change with the cosine of the angle between the aerodynamic and the kinetic coordinate systems.

If you insist on excluding changes in vertical air speed, your question will have no answer regarding lift.


What is the mechanism by which condensation forms above wings? The planes landing at Sydney Airport have what looks to be sheets of steam pouring off the wings. Larger planes have multiple contrails while smaller regional planes only have these sheets.

This is the first sunny day after a large storm. There are patches of dark cloud.

From the link posted by fooot, I've learned that it's condensation, but what is the mechanism by which it forms?  The video shows planes in poor weather, but visibility here is excellent, with 20% cloud coverage.

We know that the pressure decreases over the wing. We can also assume that a particle of air travelling above a wing does it  so fast that does not have the time to exchange energy with the sorrounding ones (called adiabatic process)

We can now look at a pressure/volume diagram for an adiabatic process that goes from "high" pressure (the free undisturbed air in front of the aircraft) to low pressure (the air on top of the wing):

image source (currently unreachable)

As you see from the diagram, if you move along the Adiabat line from the top to the bottom, the temperature drops. If the air has already a temperature near the water condensation one and a high enough humidity, it will condense.
From the Wikipedia article linked before:

  Adiabatic cooling occurs when the pressure on an adiabatically isolated system is decreased, allowing it to expand, thus causing it to do work on its surroundings. When the pressure applied on a parcel of air is reduced, the air in the parcel is allowed to expand; as the volume increases, the temperature falls as internal energy decreases.


Can a Kline-Fogleman airfoil be used for engine turbines? I am currently working on the Kline-Fogleman airfoil for NACA 23015 variant at Reynolds number 208800 with the configuration of the backward facing step of 50% thickness at 50% chord on the bottom surface of the airfoil. The KF airfoil gives higher L/D ratios for all pre-stall AOA and it also stalls at a slighty higher angle than conventional variant of the same series. The highest L/D value from Ansys Fluent was 22.45 at 9deg AOA as compared to conventional NACA 23015 at 9deg AOA being 19.74. I want to know if the KF variant of this airfoil profile can be used in turbine blades of a gas turbine engine and if it will produce higher power output or efficiency. Will it cause any problems due to turbulence and the number of blades and stages?

Also, are there any other applications of KF airfoils other than RC planes?

KF airfoils result from folding paper airplanes. In a way, most paper airplanes employ an airfoil with a stepped contour. At the scale of paper airplanes, this step possibly forces a laminar-turbulent boundary transition which might allow the airfoil to stall at a higher angle of attack.

A regular NACA 23015 has an optimum L/D of close to 50 at a Reynolds number of 200,000 and 9° angle of attack. Airfoiltools gives it as 48.5.

If you doubt the XFOIL data from Airfoiltools, NACA TR 824 gives essentially the same values using wind tunnel data, however, you need to pick Airfoiltool data at a higher Reynolds number to compare the two.

Since the L/D of the conventional airfoil is two an a half times better than what you calculated with ANSYS, I recommend that you review your results. And no, I would not recommend their use in turbines. The exposed corner at the step will face extreme heat loads and film cooling will probably fail there.


What are the specifications of the General Electric J85 engine compressor? On the internet I can't seem to find all the specifications I need.

Mostly I'm looking for the tip and hub radius and the angular velocity of the blades.

Thanks.

That's probably going to be beyond the scope of most on this board.  If you really want an answer to this, I'd contact Thornton Aircraft Company in Van Nuys, CA (http://www.thorntonaircraft.com).  They deal in sales and maintenance on old T-38s and F-5s as well as do type ratings in the jets.  They are intimately familiar with the J-85 power plant and could probably supply you with that info.


How high can an airplane be spotted with eyesight? How high can an airplane be spotted with eyesight? Suppose a plane has a wingspan of 15 m then at what distance it will appear as a faint dot?
What is the correct way to calculate this?

Note: There is some confusion in the comments about this answer. What I am answering here is the question:

  Suppose a plane has a wingspan of 15 m then at what distance it will appear as a faint dot? What is the correct way to calculate this?

This does not imply that you cannot see the airplane at a greater distance any more. It just means you cannot resolve it any more, i.e. it will appear as a faint dot.

This can be estimated with the Rayleigh criterion. The angular resolution of the human eye is given by

$$ \theta = 1.22 \frac{\lambda}{D} $$

where $ \lambda $ is the wavelength of the light and $ D $ is the pupil diameter. Using typical values of $ \lambda = 550 \, \mathrm{nm} $ (highest sensitivity of the human eye) and $ D = 3 \, \mathrm{mm} $ during the day, we get a resolution of

$$ \theta \approx 2.24 \times 10^{-4} \, \mathrm{rad} \approx 0.013^\circ $$

If the aircraft is $ w = 15 \, \mathrm{m} $ wide, the distance at which this corresponds to this angle is given by

$$ d = \frac{w}{\tan \theta} \approx \frac{w}{\theta} \approx 67.1 \, \mathrm{km} $$

Most planes only fly up to FL410, which is about 12.5 km, so you could always distinguish them from a faint point with the naked eye. As Zeiss Ikon pointed out in the comments, the actual resolution of your eyes may be worse because your eyesight is probably not perfect. This is however an upper limit on the resolution.


What is this plane from WWII (probably French or Allied forces)? Picture should have been taken after or during the end of World War II, probably close to Marrakech French base (but that is not sure). Can anyone tell us what these planes are? My grand father could be one of the pilots and I would like to find information on him.

It's an RAF Coastal Command Vickers Wellington, a maritime patrol variant, with a chin radome for its search radar and the nose turret removed and just greenhouse for observing.  The dark projection on top just above the cowling is a fairing for a Direction Finding loop antenna on the fuselage.  The dark shapes farther back are just dark painted areas.  Here's a shot of a similar one:

If you're sure the picture was taken near Morocco, it was probably doing sub hunting in the Mediterranean and nearby Atlantic.  Likely with 179 Squadron, based at Gibraltar.


Do these red stop signs apply to aircraft or only to vehicles? I mean; do planes also see these signs on their way to runway? Or just cars and other vehicles encounter this stop sign on their service road or interconnection roads that are assigned to them?

No, those signs do not apply to aircraft. They are for vehicles and marks entry to the maneuvering area.


What are the differences between porpoise; bounced; and nose-pointed-down landings? I have read some articles, and to tell the truth, I could not clearly see the differences. I would like to know if there are differences or not.

If the nose landing gear (NLG) makes contact first, but the plane settles down on the runway, that's a nose-down landing.
If the landing is normal, i.e. the main landing gear (MLG) touches down first, but the plane bounces, only to settle down on the MLG, that's a bounced landing.
If the plane bounces, after scenarios 1 or 2, but proceeds to land on the NLG, leading to multiple bounces on the NLG, that's a porpoise landing. Risking structural failure.

Source: boldmethod.com


Can two captains fly an airliner? If the First Officer cannot take his role for some reason, and the company only has another Captain available, can two captains fly the airliner, assuming they routinely fly as captains for that company in the aircraft of that type? Can one of them simply take a FO role for that flight?

In other words, is the First Officer a separate certification that must be current?

Yes.

What allows you as a pilot to fly large aircraft (legally under the FAA) is a type rating. As long as the two Captains hold proper type ratings for the aircraft they suffice for the required two crew members needed to fly the plane. However one of the two of them will be performing the duties of the first officer and that should be decided prior to the flight.

First Officer and Captain are rankings given by the airline not any governing body. Both the FO and Captain cary the type cert required to fly the aircraft (as far as I know).

The limited (and quickly dwindling) exception to this is aircraft that require a crew of 3 where the third is a flight engineer. The Flight Engineer position does hold its own certification.


Why does the elliptical wing have the lowest drag? I read that the elliptical wing has the lowest induced drag. Why is that so? I welcome a mathematical and intuitive explanation.

Induced drag is caused by the backward inclination of the lift vector. Lift is defined as the aerodynamic force perpendicular to the flow direction, and since lift is created by deflecting this flow downwards, the resulting force is slightly inclined backwards.

The formula for the deflection angle (called downwash angle) $\alpha_w$: $$\alpha_w = arctan \left(\frac{2\cdot c_L}{\pi\cdot AR}\right)$$

To arrive at the drag, you now need to multiply the local lift with this angle. For small $\alpha_w$s the arcus tangens can be neglected, and we get this familiar looking equation for the backwards-pointing component of the aerodynamic force: $$c_{Di} = \frac{c_L^2}{\pi\cdot AR} = \frac{\alpha_w^2\cdot\pi\cdot AR}{4}$$

Now consider a wing which will do this unevenly over the wingspan. Some parts create a lot of lift and, therefore, a lot of downward deflection. Others produce less lift and less deflection. Due to the quadratic dependency of induced drag with downwash angle, the sum of all parts will always be higher when there are variations in the local downwash angle over span. Only the wing with a constant downwash angle will have the minimum induced drag for a given lift.

This is not necessarily an elliptic wing. By selecting the right twist (incidence variation over wingspan), any planform can be made to produce a constant downwash angle, albeit at only one angle of attack. Only the elliptic wing will combine the constant downwash angle over span with a constant lift coefficient over span such that variations in angle of attack still result in a constant downwash angle.

This is strictly true only in inviscid flow and when elastic deformation and the weight of the wing (which contributes to the need for lift and hence induced drag) are not considered. Also, an elliptic wing will produce nasty stall characteristics. Adding handling, friction and structural weight will shift the optimum to wings which produce more lift near the center. The exact optimum also will depend on the absolute size of the wing because the structural mass fraction goes up with size due to scaling laws.

Nomenclature:
    $AR \:\:$  aspect ratio of the wing
    $c_{Di} \:\:$  Induced drag coefficient
    $c_L \:\:\:$ lift coefficient
    $\pi \:\:\:\;$  3.14159$\dots$


Is it possible to hear a sonic boom when the aircraft is exactly at Mach one? Will an observer on the ground hear a sonic boom if a plane passes overhead at exactly the speed of sound? That is, the plane does not cross the sound barrier. The plane just hits Mach one - and continues to travel at that exact speed.

This is a bit of a problem because the speed of sound changes with altitude, so we assume it remains at the exact same altitude. I imagine there are other caveats also!

I would guess that there is a sonic boom, but I'm not sure.

No, if standard atmospheric conditions apply.

Since speed of sound is proportional to temperature and temperature normally decreases with altitude, the speed of the aircraft at Mach 1 is subsonic in warmer, lower air. This means also that the Mach shock will diffuse and be audible on the ground either as a protracted rumble or not at all, depending on the aircraft's altitude.

Note that I assumed that the listener is located below the flight path of the aircraft. If that would not be the case, the noise of the aircraft diving below the listener altitude would be substantial.


What does VIB stand for in EICAS? From what I know it’s measuring engine vibrations? What do the numbers actually mean? For example, 1.9 VIB:

These are engine vibration measurements. The image you show is the Compact Engine Display:

  3 Vibration (VIB) Indications

  Displayed as digital readout only.

(Boeing 737 NG FCOMv2 - 7.20.1 Engines, APU - Over/Under - Displays)

The value comes either from the EEC (Electronic Engine Control) or the sensor directly:

  Oil pressure, oil temperature, oil quantity, and engine vibration are the secondary
  engine indications. [...] Engine vibration indications are displayed with a round dial/moving pointer.

  The EEC must receive electrical power to supply engine operating data to the
  flight deck engine indications. When the EEC is not powered, N1, N2, oil quantity
  and engine vibration are displayed directly from the engine sensors. Positioning
  the engine start switch to GRD supplies electrical power to the EEC and displays
  pointers/digits for all engine parameters.

(Boeing 737 NG FCOMv2 - 7.20.1 Engines, APU - System Description)

On the lower DU they look like this:

The amber warning line (2) shows the limit of 4.0 units. The QRH defines High Engine Vibration as:

  The vibration level is more than 4.0 units
  Airframe vibrations.

(Boeing 737 NG QRH 7.24)

As to what the actual number really means, the FCOM does not say. I could find this quote on pprune.org, but I cannot verify it:

  Two types of vibration pickups have been used on Boeing airplanes, the electomechanical velocity pickup and the piezoelectric accelerometer. Signals from velocity pickups are converted into vibration displacement, measured in 1/1000 of an inch, peak to peak (mils double amplitude). These units represent the physical displacement of the vibration sensor. Signals from accelerometers are converted into velocity, measured in inches per second (ips). These units represent the peak amplitude of the velovity of the vibration sensor. Boeing airplanes typically use a 0-5 scale for cockpit display of vibration. On earlier applications, the scale provides an indication of absolute vibration units (ips or mils da) as output from the airborne vibration monitoring (AVM) signal conditioner. This approach is not optimum in that a display of displacement units will provide relative insensitivity to low speed rotor vibration. AVM signal conditioners used on 737-300 and latter, 747, 757 and 767 provide output data in scalar units. These scalar units are uniquely tailored for each engine type to match the operating speeds and vibration characteristics of the engine rotors. The scaling used on the 737/CFM56 installation provides display sensitivity for the high speed rotor proportional to the vibration velocity. The low speed rotor vibration is displayed in units proportional to displacement, except near idle conditions where velocity scaling is used.

(Boeing Airliner magazine, OCT-DEC 1987 issue, emphasis mine)

So the number was originally based on acceleration in inches per second, but it is a scalar value today without a direct physical meaning. The only thing the pilots need to know is: around 1.0 is normal, 4.0 is the limit.

This page confirms, that scalar units are used on the CFM56-7B, which is used on the 737 NG series:

  The algorithm detects the highest vibration (in scalar units) of all four sections of the engine during the whole flight and takes a snapshot of the remaining parameters. The vibration data are transfered to the next flight leg upon engine start.

  The Vibro Meter Advanced AVM Signal Conditioner (AAVM) P/N 241-298-002-011/-015 is able to detect #3 or #4 bearing vibrations based on the CFM56-7B engine vibration signature.

(AviaDeCo GmbH)


How does the in flight entertainment system works? I board an aircraft, wait for it to takeoff, then find the headphones somewhere near my seat plug them into the two-pin point, and it works! But, how does it works for hundreds of other passengers simultaneously? Where is all the content stored, how are the screens connected to it, and how powerful is the computer controlling so many screens? Also, why is IFE required to be connected to aircraft's main computing system, which may eventually lead to the hacking of an aircraft?

Side Question : On an average, what is the size of content stored in a typical In-flight entertainment system?

I found a resource specifically for Emirates, so I'll try to explain it with respect to Emirates, though various other Airlines would use more or less the same principle.

  Emirates uses Panasonic Avionics IFE (In Flight Entertainment) system.
  They have the 3000i which is older generation of IFE system on older
  aircrafts and the latest version ex2 on new aircrafts. Media licensing
  is managed, I believe, by the IFE provider via links to the major
  studios. In this tie up, the media is encrypted and stored in portable
  media loaders for the airline to update on the aircrafts. As the media
  is encrypted so the studio doesn't have much objection in giving away
  the rights to air the movies on board rather quickly.

  The Portable media loaders use optical connection for faster data transfer to
  minimize the loading time.

  The Main Aircraft Interface (AI) Server is nothing but a small
  computer which is generally called a LRU (Line Replaceable Unit) a
  small box which controls the entire IFE system of the aircraft the
  back end software can either be windows 95 or Linux depending on the
  type of system. 3000i uses windows. The movies are stored on other
  LRUs which are generally an array of Digital Data Servers or Media
  Servers (MS) or in lay mans terms Hard Drives! These are connected in
  a LAN network and ethernet IP addressing is used in the entire
  aircraft system.

  The data is distributed over Optical fiber in first phase and over
  normal ethernet network to the specific seats. The interface is done
  via Ethernet Switches which intelligently routes the data from the
  Media server to the Area Distribution box to the Seat Electronic box
  and finally to the Smart Display.

Using a central dedicated computer instead of individual computers on different seats is preferred as it is less costly. Also, the IFE is connected to Aircraft's Main IFE cpomputer, and not the Aircraft's main computer, so by hacking it all you get is the files (IFE files), not the aircraft data, though the moving map system data might be first dumped into the IFE system from the Airplane's main computer


Has there ever been a fully automated takeoff, cruise, and landing of a large aircraft similar to commercial airliners? Has there ever been a fully automated takeoff, cruise, and landing of a large aircraft similar to commercial airliners?

In fact, has there ever been a fully automatic flight of any pure aircraft whatsoever? This is gonna sound crazy, but the only thing I know that comes close is the Soviet Space Shuttle Buran. It launched unmanned and returned successfully under computer control, including its glide and landing.

So I'm interested in automatic aircraft similar to large commercial airlines. BTW, remote control toy planes wouldn't count because that is not automated (a person is manually controlling it).

The book The Glass Cage describes a fully autopilot-controlled military test flight in 1947. The plane was a C-54 Skymaster with seven men aboard. According to this source, the pilot aligned the plane on the runway for takeoff, but the takeoff, course control, and landing were performed by autopilot with no human engagement. See also this report from the Chicago Tribune at the time.


Why does An-225 have anhedral wings though it is a cargo aircraft? What are the rationale behind this anhedral high mounted wing configuration? Is that usual for very large carge aircraft?

Anhedral wings will induce roll instability and improve roll maneuverability.

In a large/heavy airplane with a high-wing configuration there is usually excess roll stability, so this type of wings can be pretty common.

Both the high wing configuration and wing sweep contribute a negative sideslip-induced rolling moment, and anhedral is necessary to limit this moment. If sideslip causes strong rolling, the aircraft will have a tendency to dutch roll.


Are Final Tests always required for LRUs? Is an aviation component workshop allowed to release a line-replaceable unit (LRU) without going through a Final Test, if the LRU passes the pre-testing procedure and is found to be a No Fault unit?

I'm assuming by Pre-Testing procedure you are referring to a physical inspection and downloading the fault history.

If there are no logged faults in the fault history, it means that the built-in test (BIT) of the LRU did not detect any faults. BIT is not capable of 100% test coverage. It is also possible that BIT results are inaccurate. A lack of recorded faults is not sufficient to declare a unit airworthy.

Once an LRU is tagged and sent to the service center/shop, it can only be returned to service (RTS) after it has passed the RTS tests specified in the component maintenance manual. After passing the RTS tests, a 'maintenance release document' can be prepared and the unit returned to the spares pool.

In the US, the maintenance release document may be an FAA Form 8130-3, Form 337, or other document containing the information required by 14 CFR Part 43, sections 43.9 and 43.11. The maintenance release document must be signed by an authorized person who is certificated under 14 CFR Part 65. Though not legally required many shops will affix a "yellow tag" to the unit to indicate it's serviceable status.

EASA has similar rules, the primary difference being the required content of the documentation.


Can a military pilot make his airplane invisible for the ATCs? I'm writing a story about the pilot of the Austrian Air Force, who uses his Eurofighter Typhoon aircraft for unintended purposes.

I've read stories in the internets that civil pilots allegedly can make their planes invisible for the flight controllers by turning off some device (transponder?). I'm not sure if it is true, but I saw with my own eyes a pilot of a small aircraft (Cessna Skyhawk), who took off without a transponder (the antenna was broken), was told by the ATC to fix the transponder, landed, and then took off again -- now with a temporary fix that made the antenna work for a while. In other words: Pilots of small aircraft do have the option to fly with or without a transponder (at least on small Austrian airfields). I'm not sure if it's 100 % legal, but sometimes pilots do it.

But a military Eurofighter is different from a little Cessna plane.

Is it technically possible for a pilot of Eurofighter Typhoon to become invisible for the air traffic controllers by switching off the transponder?

Radar systems have 2 aspects, the primary and secondary systems, or the Radar and the interrogator. The radar sends out a RF signal and receives everything that is returned on that same frequency, and performs calculations to determine the azimuth and range of everything it receives. The interrogator sends out radio pulses which transponders in aircraft receive and decode. The transponders then transmit data on a different frequency than what was received. The interrogator then receives that signal and decodes it and correlates it with the primary radar data to tell the operator that 'yes, that is definitely a valid target' and gives the info on the flight to the operator.

You can turn off the transponder, but all that will do is make it so that the ATC won't be able to see a correlated flight, (one that is definitely an aircraft and not just noise coming through the primary radar). The screen would still show a blip where the radar picks up a signature though. If the plane has radar absorbent material or is close enough to the ground, that would get filtered out as noise.

An ATC, who is used to only working with aircraft who want to be seen, and thus keep their transponders on, might be fooled by this. But a military radar operator might not be so easily fooled by this.


How do airplanes regulate the output frequency of AC generators? AC electrical buses are common in airplanes from light piston twins all the way up to the largest airliners. I usually see both a 115v and 36v AC bus with a 400 Hz output frequency. How do most airplanes regulate the output frequency coming from the AC generator?

It's done using mechanical device called constant speed drive. Works almost the same as CVT transmission in a cars. The more complex thing is phase adjustment as airplane usually has multiple engines and generators. But that is not done as different generators always power different buses!


Did Boeing need MCAS because the 737 MAX didn’t meet FAA longitudinal stability requirements, or was it to make the MAX feel like the NG? I’m confused on this. I’ve read in multiple different places that

  MCAS was required due to the easier stick forces at a high angle of attack and a high G force, and because of this, the Max didn’t meet FAA longitudinal stability

Others have said that

  The Max has MCAS because they wanted it to feel like the NG

Confirming the easier forces, but excluding the FAA longitudinal stability possibility. Almost every source contradicts another one, and I’m unable to locate any info through the FAA site.

The Boeing 737 MAX MCAS system is there ONLY to meet the FAA longitudinal stability requirements as specified in FAR Section 25.173, and in particular part (c) which mandates "stick force vs speed curve:, and also FAR Section 25.203 — "Stall characteristics".
FEDERAL AVIATION REGULATIONS Sec. 25.173 — Static longitudinal stability.

Sec. 25.173 — Static longitudinal stability.
Under the conditions specified in §25.175, the characteristics of the elevator control forces (including friction) must be as follows:
(a) A pull must be required to obtain and maintain speeds below the specified trim speed, and a push must be required to obtain and maintain speeds above the specified trim speed. This must be shown at any speed that can be obtained except speeds higher than the landing gear or wing flap operating limit speeds or V FC /M FC,whichever is appropriate, or lower than the minimum speed for steady unstalled flight.
(b) The airspeed must return to within 10 percent of the original trim speed for the climb, approach, and landing conditions specified in §25.175 (a), (c), and (d), and must return to within 7.5 percent of the original trim speed for the cruising condition specified in §25.175(b), when the control force is slowly released from any speed within the range specified in paragraph (a) of this section.
(c) The average gradient of the stable slope of the stick force versus speed curve may not be less than 1 pound for each 6 knots.
(d) Within the free return speed range specified in paragraph (b) of this section, it is permissible for the airplane, without control forces, to stabilize on speeds above or below the desired trim speeds if exceptional attention on the part of the pilot is not required to return to and maintain the desired trim speed and altitude.

FEDERAL AVIATION REGULATIONS Sec. 25.203 — Stall characteristics.

Sec. 25.203 — Stall characteristics.
(a) It must be possible to produce and to correct roll and yaw by unreversed use of the aileron and rudder controls, up to the time the airplane is stalled. No abnormal nose-up pitching may occur. The longitudinal control force must be positive up to and throughout the stall. In addition, it must be possible to promptly prevent stalling and to recover from a stall by normal use of the controls.
(b) For level wing stalls, the roll occurring between the stall and the completion of the recovery may not exceed approximately 20 degrees.
(c) For turning flight stalls, the action of the airplane after the stall may not be so violent or extreme as to make it difficult, with normal piloting skill, to effect a prompt recovery and to regain control of the airplane. The maximum bank angle that occurs during the recovery may not exceed—
(1) Approximately 60 degrees in the original direction of the turn, or 30 degrees in the opposite direction, for deceleration rates up to 1 knot per second; and
(2) Approximately 90 degrees in the original direction of the turn, or 60 degrees in the opposite direction, for deceleration rates in excess of 1 knot per second.

Here is some additional information: http://www.b737.org.uk/mcas.htm

MCAS is a longitudinal stability enhancement. It is not for stall prevention (although indirectly it helps) or to make the MAX handle like the NG (although it does); it was introduced to counteract the non-linear lift generated by the LEAP-1B engine nacelles at high AoA and give a steady increase in stick force as the stall is approached as required by regulation.
The LEAP engine nacelles are larger and had to be mounted slightly higher and further forward from the previous NG CFM56-7 engines to give the necessary ground clearance. This new location and larger size of nacelle cause the vortex flow off the nacelle body to produce lift at high AoA. As the nacelle is ahead of the C of G, this lift causes a slight pitch-up effect (ie a reducing stick force) which could lead the pilot to inadvertently pull the yoke further aft than intended bringing the aircraft closer towards the stall. This abnormal nose-up pitching is not allowable under 14CFR §25.203(a) "Stall characteristics". Several aerodynamic solutions were introduced such as revising the leading edge stall strip and modifying the leading edge vortilons but they were insufficient to pass regulation. MCAS was therefore introduced to give an automatic nose down stabilizer input during elevated AoA when flaps are up.


How much time does an landing/take-off add to an otherwise direct flight? I'm interested to know how much time a descent and ascent to crusing altitude takes as part of a longer journey.

A standard jet-aircraft (for example a A340 or 777) flys for eight hours from City A to City C. While doing so it flies over City B.

Now the same jet flys from City A to City C but as it approaches City B it descends and lands at B. The plane taxis to the gate. As soon as it's there it pushes back, requests take off and returns to the cruise flying to City C.

Approximately how much time is added to the journey by virtue of having descended, landed, taken-off and ascended to cruise ? Assume no ATC delays etc.

Let's take the best case scenario to start with, and assume that the cruise ground speed of the airliner is around 400kts.

For our best case, we also assume that the runway is lined up with the flight path and that the airspace is quiet so the plane is cleared "straight in" with no vectoring for separation or holding patterns.

Descent:

A typical descent profile would start 100nm out, during which the plane would slow pretty uniformly from around 400kts to around 140kt. This gives an average groundspeed of 270kt. So this 100nm is covered in around 25 minutes rather the 15 minutes at full cruise.

The descent has cost us only 10 minutes.

Taxi and turnaround:

At a smallish (and well designed) airport the rollout and taxiing to the stand may only take 5 minutes, plus another 5 to attach the air bridge/stairs and open the door. Passenger and baggage turnaround would typically take around 30 minutes, but you have specifically excluded this in your question so we can ignore it. Then getting departure clearances, startup and taxiing to the active: say another 15 minutes.

So this phase has cost more or less 25 minutes.

Take off and climb out:

Again assuming we're cleared for immediate take off, it takes around 20 minutes to climb to cruise altitude, at an average ground speed of (say) 200 knots. This means we've covered about 70 miles, which the cruising aircraft would have covered in just over 10 minutes.

So this final phase has cost us another 10 minutes.

Total for the best case scenario is 45 minutes.

Of course the best case is not that likely. If the airspace is busy, holding for another 10 minutes before final approach, and 10 minutes further wait for departure and takeoff clearances, would not be surprising. And if the active runway is in the "wrong" direction or vectoring is needed for traffic or noise reasons, that can easily add another 10 minutes or so.

So in the non-best case scenario, a total loss of 1h15 to 1h30 would not be unreasonable.


Use of a METAR when $ is present I fly out of an uncontrolled airport that has an unmanned, automated weather reporting site with a precipitation sensor (AO2).  The METAR often has a \$ symbol at the end of the report, indicating that maintenance is needed on the equipment.  Can I use this METAR, or any METAR with a "\$" as a reliable indication of the current weather?

Do the indications match up and make sense compared with indications of a nearby airfield and other weather sources?

Legally the FARs require that pilots obtain "all available information" (FAR 91.103). $ could mean a number of things and since Metars are still available, you should be able to trust them, but I recommend to check more then one source regardless.


Is it possible to load (and not activate) the approach procedure on a Garmin 1000 or 430? I noticed sometimes when I fly with a G1000, or Garmin 430 and am doing multiple practice approaches I can't always load my next approach. It forces allows me to activate the approach, which would sequence me to the IAF of the next approach, and sometimes I may not be cleared to or ready for that.

Why is activate the only option?

How do you get around this?

If you take a look at the manual for the 430 you will see that you can only load one approach at a time.

Turn the large right knob to highlight ‘Load?’ or ‘Activate?’
(approaches only) and press the ENT Key. (‘Load?’ adds the procedure
to the flight plan without immediately using it for navigation
guidance. This allows the pilot to continue navigating the original
flight plan, but keeps the procedure available on the Active Flight
Plan Page for quick activation when needed.)

The 430 only allows one approach to be loaded then subsequently activate because this is the flow of a general IFR flight and the approach is not technically in the flight plan (its just "primed"). You should know the most likely approach based on winds and weather before even departing. Once en-route you may get assigned a different one but either way you can load the likely approach en-route. If you know your transition point you can even select that as your approach start. If you are unsure, VTF (Vectors To Final) is generally safe with a change when needed. Since the approach is somewhat side loaded and not in the actual flight plan, activating it will defeat all other navigation and take you to the chosen IAF (or nearest leg for VTF).
The other reason that the unit is built this way is approaches rarely follow each other in quick succession. There is often a hold in between or possibly more en-route flight, which is why the suspend button is there (and important for missed approaches).
You are likely missing some of the procedural work that occurs on a real IFR flight. Sims are fun to just hop from field to field flying the ILS but that is not the reality of an IFR flight plan so the hardware is not designed for that.

If you are flying X-Plane or the like and using the mouse to tune the G430 then the apparent programming time is much slower than reality as well. I would advise checking out something like this and you will see how much quicker you can really program a 430 when you have tactile knobs.


Who can perform different maintenance checks? The different types of check (A-check to D-check) require different types of competence and certification.

If I understand correctly, D-checks require special equipment and certification and some airlines developed whole subsidiaries to take care of this task (LHT, AFI,...) in which they send their aircraft to perform a D-check.

Given the existence of those subsidiaries, I assume that a D-check cannot be performed by the airline's employee and must be performed by a company specialized in aircraft maintenance.

A-checks and daily maintenance do not require the same amount of qualifications, certification and equipment and may thus be performed by the airline without sending the aircraft to a specialized place.

For each check (A-check, B-check, C-check and D-check), who is qualified to perform those checks? Can these checks be performed by the airline itself or must a specialized company take responsibility for the aircraft?

Airlines are free to conduct the tests wherever they prefer as long as they meet the required FAA standards.

An extract from American Airlines' Procedures:

"PS" Daily Checks

Each aircraft is checked every 2-3 days in its "PS" (Periodic Service) check. The aircraft is visually inspected and its maintenance log book is checked for entries and maintenance needs. The "PS" check can be performed overnight or during downtime during the flight day. It averages approximately two manhours.

"A" Checks

The "A" check is more detailed than the "PS" check. "A" checks are performed every 7- 9 days (approximately 80 - 100 flight hours). The "A" check is performed at one of 40 stations around American’s system. It averages 10 - 20 man-hours.

"B" Checks

The "B" check is an even more thorough maintenance check. The "B" check is accomplished approximately every two months (roughly 500 - 600 flight hours). Besides specific service performed on the aircraft, a detailed series of systems and operational checks are performed. American always performs "B" checks inside one of its hangars at seven different cities around its system. A "B" check requires approximately 100 man-hours on narrowbody aircraft (those with only one aisle) and approximately 200 - 300 man-hours on widebody aircraft (those with two aisles).

"C" Checks

The "C" check is the most thorough type of maintenance work performed by American. The airframe - virtually the entire aircraft - goes through an exhaustive series of checks, inspections and overhaul work. It is performed at either of American’s heavy maintenance and engineering centers in Tulsa, Oklahoma or the Alliance Maintenance Facility in Fort Worth, Texas. There are different levels of "C" checks depending on the type of aircraft.

Jet Engine Overhauls

Modern jet engines are among the most reliable devices in aviation. American does not replace and overhaul jet engines at a specific number of hours. Instead, American uses a 24- hour-a-day "condition monitoring" process that scientifically tracks the condition of every engine on every aircraft. Besides visual inspection, technicians monitor the internal condition of every engine, using such procedures as boroscope inspections and oil sample spectographs. The goal is to replace and overhaul an engine before a problem can occur. Engine overhauls are performed at the Tulsa and Alliance-Fort Worth Maintenance and Engineering facilities. The engine replacement is usually performed at one of the six "B" check hangar locations around the country.

To answer your question on the qualifications required to perform these checks, we'd have to go to FARs Part 65 which lists the privileges and limitations of a Mechanic Certificate.

In addition to the basic Mechanic Certificate, FAA allows two different ratings.

The first one is the Airframe Rating which, under §65.85, states:

  A mechanic who holds an airframe rating may approve and return to
  service an airframe, an appliance, or any related part after he or she
  has performed, supervised, or inspected minor repairs or alterations.
  He or she may also perform the maintenance actions required for a
  major repair or alteration, and should initiate the appropriate form
  (FAA Form 337, Major Repair and Alteration) associated with that work.
  However, the return to service action must be accomplished by a
  certificated A & P technician holding an Inspection Authorization
  (IA). (Refer to 14 CFR §65.95.)

After that comes the Powerplant Rating under §65.87 which states:

  Similarly, a mechanic holding a powerplant rating has the same
  limitations imposed regarding the powerplant and propeller as the
  airframe technician has on the airframe rating. He or she may perform
  and return to service minor repairs or alterations. He or she may also
  accomplish the work activities required for a major repair or
  alteration, but the work must be signed off for return to service by
  an IA.

That said, after all the checks have been performed, a person with Inspection Authorization (under §65.91) is responsible to approve the aircraft for return to service.

Source: American Airlines Aircraft Maintenance Procedures Fact Sheet


Why does stall speed decreases with increasing aspect ratio? As I have found Cl(max) doesn't change with changing AR. Is it true?
Therefore stall speed which corresponds to Cl(max) and crit AOA must remain constant regardless of AR.
Where am I mistaken?

$c_{L_{max}}$ doesn't change with changing AR. Is it true?

No, it isn't. Since the area near the wingtips cannot produce a pressure difference as large as that on the center wing due to flow around the wingtips, the lift there is reduced. The larger the wingtip part is relative  to the whole wing, the lower the maximum lift will be.

Given the same wing area and mass, a wing with a higher aspect ratio will produce a lower stall speed. Of course, the assumption of an identical mass means that the lower aspect ratio wing has a poorer structural design.


Why use plywood to hold the bearings for the airbrakes on a fibreglass glider? The ASK-21 is mainly a fibreglass construction.

However, the bearings for the airbrakes and their control rods are fitted into small pieces (some 10 cm wide) of what seems to be plywood molded into the fibreglass.

The following picture shows the plate holding the bearing for the airbrake, seen from the rear side in the space in the fuselage where the wing spars are joined. The bolt that sticks out holds the bearing of the linkage that transfers the brake movement to the rods in the wing.

The last picture shows the bearings in the wing where the rod to lift the actual air brake is fastened.

What is the reason for this choice of material - could the bearings not be fitted into the fibreglass structure without extra pieces of plywood?

This use of plywood is not ideal, but it works. The wood soaks up the epoxy and becomes an integral part of the fiberglass construction, and is easy to mold, so laying up the same geometry from glass fibers alone would be much more laborious.

Also, plywood is reasonably isotropic, at least in the two directions of the plywood layers. If you want to achieve similar isotropy with glass fibers, you need to interweave them - just wrapping one roving around the bearing will not do.

You could use other materials - aluminum would do, but doesn't bond as well to the epoxy resin. That Schleicher uses plywood has historic reasons. They used to build gliders with wood, some had steel tube fuselages, but wood was their primary construction material. When they started to build glass fiber gliders (the first was the ASW-12, which was a commercial version of the D-36, on which Gerhard Waibel cut his teeth as a glider designer and won the German gliding championship in 1964) they used what they knew best. And it worked.


When starting a turbine helicopter, why should one slowly increase RPM (throttle)? After the ignition switch is turned ON, my Captain always increases the RPM (throttle) very slowly.
I don't know why, because our helicopter has an ECU.
One day, my Captain asked me "Do you know why we increase RPM slowly?"
so I answered "umm, to prevent a hot start?"
Then my captain said "Nope. It is related to ENG OIL PRESSURE."

Is there any other reason about this slow operation (related to ENG oil pressure)?

Maybe compressor stall or surge?

he probably wanted to make sure the oil pump keeps up with the engine and bearings won't run hot, or that the oil circuit gets primed. turboshaft engines dont like it when running at idle, they tend to overheat; so it better be written somewhere that the engine rpm should be brought up slowly.

is important that whatever procedure, it comes out of the book. I would have expected your captain to direct you to startup procedure then mention the reason, so you could look it up rather than learn it by hearsay.


Why did this Cirrus deploy the parachute while ditching over Pacific Ocean? In this incident near Hawaii, a Cirrus SR22 had fuel problems and deployed its airframe parachute.

My question is why did the pilot deploy the parachute rather than perform a "normal" ditching procedure in the ocean?

For reference, here's the video.

Once your engine fails the airplane is the property of the insurance company, your only goal at that point should be to survive. Using a whole-airframe chute increases your chances of survival whether over land or sea.

The things that kill people in forced landings of any kind are high-g deceleration and post crash fire. A parachute decreases the chances of both as you have very low lateral speed and much less chance of fuel tank leaks. A parachute also will increase the chance the airplane will remain upright, so escaping the aircraft will be much easier - very important in a water landing!

The only reason for a "normal" ditching procedure is that you have no other choice, using a parachute if you have one will almost always be the right option. The only time I would consider otherwise is if I had enough altitude to glide to a safer emergency landing spot. If pulling the chute would mean landing in water or dangerous terrain but gliding would mean reaching land where a forced landing was reasonable I'd probably opt for gliding.


Close Runway 16/34 from one direction Let's say Runway 16/34 used direction 16 for arrival and 34 for departure. Can the runway close 16 (arrival direction) and still working in 34 for departure?

I think your premises are wrong. If an airport has a runway 16/34 then either both landings and departures take place in direction 16 (runway 16 is open) or both take place in direction 34.

In most cases both directions are not used at the same time. Which direction is chosen usually depends on the current wind direction so planes depart and arrive with headwind (not tailwind). Which one is open (16 or 34) can change from day to day, even during a day. Usually one of the directions is used most of the time because the wind usually comes from a certain direction. For example, London Heathrow has two parallel runways and in approx. 70 % of the time the wind comes from the west (westerly operation). Hence, they use runways 27L and 27R most of the time for both arrivals and departures.

There are, however, runways that are restricted to either take-off or landings, like runway 18 in Frankfurt, for example (only for take off), or 07L/25R (only for landing). But again, 07L/25R means: either all planes land in direction (on runway) 07 (towards the east) or all land in direction (on runway) 25 (towards the west).

See also:

What determines in which direction a runway is used?
Why is tailwind during final approach and landing so dangerous?


What is the working principle of trim tabs on the elevators? Trim tabs are used on the tail's horizontal elevators to adjust its neutral position. What is the working principle behind this? How does this affect the sensitivity (the feel) of the pitch to the pilot in a plane with no fly by wire?

The trim tab is like a control surface for the control surface. By changing camber at the rear end of the control surface, it creates a local force which changes the balance of forces such that the force-free deflection angle (sometimes called auswehwinkel after the German word for this angle) is shifted in a direction opposite to the deflection of the tab.

You could do the same with springs along the control rods, or a mass somewhere which pulls the control surface to the same force-free angle. But when speed changes, so do the aerodynamic forces, and your force-free angle will change. The forces caused by the tab change with airspeed, so the force-free angle set by a tab is unaffected by speed.

Don't confuse a trim tab with a Flettner tab: The trim tab stays at the same angle with varying control deflection, while the Flettner tab will move in opposite direction to reduce the stick forces caused by the deflection of the control surface. By setting the zero-deflection position of a Flettner tab, it can in parallel work like a trim tab and shift the zero-force deflection angle to the desired point.

Small aircraft like the Diamond Katana even have an anti-Flettner, where the tab is moving in the same direction as the control surface in order to increase stick forces. This is needed to fulfill certification requirements.

In practice, a combination of both springs and tabs is used to set the desired control forces and their gradient over speed.


Identification of this control panel for a four-engine plane While cataloguing a printer's copper plates I came across a control panel image, and I have attached it here in the hope someone can identify what plane it's from. The Lorenz blind-landing instrument gives an indication of the period.

This is almost definitely the cockpit of a Short Stirling of unknown mark, a 4-engine British heavy bomber from World War II.

The RAF Museum's website has an (admittedly low-resolution) photo that matches quite closely, as does this history website. Here's my thought process:

the placards (not just the photo's labels) are in english, so it's likely either Commonwealth or American-built
the Stirling power quadrant is a match (2 levels of 4 levers each; it's a 4-engine bomber)
engine instruments directly above throttles
the engines have boost gauges; the Bristol Hercules engine of the Stirling was supercharged
odd-shaped cutout on the right side (just above the marked Air Speed Indicator in your photo)
my second link has a photo with a matching CAUTION JETTISON CONTAINERS BEFORE BOMBS placard on the right side
the Lorenz system was used in British (and German) aircraft during WW2, but I can't seem to find any US bombers that used it.
it has a retractable undercarriage; Short's Sunderland flying boat has a lot of the panel in common, but it was wider and didn't have the goofy notch on the right side, and it obviously had no landing gear.

In short, I'm pretty confident this is a Stirling.

Other aircraft I looked up and discarded as not a match:

Avro Lancaster
Avro Shackleton
Boeing B-17
Consolidated B-24
Handley-Page Halifax
Short Sunderland
Avro Vulcan (similar throttle quadrant, otherwise way off)


How does changing the altimeter setting in the US and elsewhere compare? I have been trying to track down an answer to this question and I stumbled upon this website. Apologies if this answer is already out there. I have seen the FAR regulation, 14 CFR 91.121 - Altimeter settings. Is it possible for an ATC controller in the US to give you "Descend and maintain 18,000' (Not FL180)" and thus you would need to change your altimeter setting to local upon reaching?

I also fly ICAO quite often and the same question would apply to the transition level. Could a controller give you, as example, FL150 is transition lvl and 13,000 is transition altitude, "descend and maintain 15,000'." Would you set the local altimeter once level at 15 or wait until cleared to descend below? The source I was looking for ICAO procedures was the ICAO 8168, PANS-OPS, Volume I.

My understanding is that you do not change your altimeter setting until you are cleared below the transition level. I hope my questions aren't too confusing. I appreciate any clarification.

Indeed the US and the rest of the world handle this differently.

In the US, if you are cleared to a flight level or altitude, you change the altimeter setting as you are about to cross the transition level (18,000 @ 29.92).

  [When] descending, even if cleared to an altitude at the time the cruising level is vacated, the altimeters will remain on standard pressure until just prior to the transition level.

Unless cleared in a climb to FL180, then the change is done a bit earlier.

In Europe and elsewhere, the change of the altimeter setting is done as soon as you start climbing or descending to the cleared flight level or altitude.

  climbing and cleared to a FL - set Standard Pressure Setting (QNE/1013 mb);
  descending and cleared to an altitude - set QNH.

Between the TA and TL no altitude/level can be assigned. If the local pressure is too low, what floats (rises) is the TL.

The same for the US:

  If the altimeter setting is lower, the lowest useable FL becomes FL190 or even FL200.  This restriction ensures that a minimum of 1000' vertical separation is maintained between the aircraft at 17000' on QNH and the one at the lowest useable level on Standard Pressure Setting (QNE).

If the TL is FL150 as in your example, you will not be cleared to 15,000 feet. You will be cleared to FL150. Based on the above, you will maintain 1013 (QNE) until cleared to an altitude. The same for the US, you won't be given an altitude of 18,000, it will be FL180 whether going up or down.

(Image source)

Source: SKYbrary.aero


What does the Boeing 777 autopilot do after reaching the last programmed waypoint? As I understand it, a series of waypoints can be programmed, by a pilot, into one of the autopilots on a Boeing 777. If the autopilot is engaged, the autopilot will cause the aircraft to fly to each waypoint in turn, automatically steering the aircraft as necessary.

It is not clear to me whether the autopilot automatically disengages after reaching the final waypoint or whether it maintains (for example) a magnetic heading and altitude.

What happens after reaching the last programmed waypoint if the pilot does not intervene?

The first point I would stress on is that you don't program the autopilot. You program the FMS (Flight Management System), which looks like a very large calculator. In this picture it is the device under then first officer's hand.

On top of the PFD (Primary Flight Display) you see the MCP (Mode Control Panel). This is what controls the modes that the autopilot system will follow, here is another image of it:

Typically the entire flight plan is programmed into the FMC and then during various stages of flight, the autopilot is engaged to help with navigation.

This flight plan includes the departure, enroute waypoints or tracks, and then generally the arrival at the destination; but that is it - it does not include the runway information, etc (as this may be subject to change).

The FMC takes this information and other data (weight, fuel, temperature, weather, etc.) to calculate a flight plan which is then displayed; and the pilot will then select and activate a flight plan.

In most commercial flights however, the flight plan is already loaded in the FMC from the flight operations department and the pilots simply select the predefined flight plan. Punching in a large flight plan takes a significant amount of time so many times this is already downloaded from the flight ops center.

At any stage of flight, the autopilot may be armed and engaged to do one of many things - either maintain thurst, or heading, or control the climb/descent; or follow the flight plan. It may also be disarmed during any stage of flight.

If the autopilot is armed and it set to follow the FMC's flight plan; then after the last waypoint it will follow the waypoint heading.

So, if the last way point is a holding pattern, it will continue on that pattern. If the last waypoint was a marker it will reach that marker, and then continue at that heading and speed; even though there are no more commands coming to it from the FMC (since its at the end of the programmed flight plan).

Note that reaching the end of the flight plan from the FMC does not disengage the autopilot; there are many other things that might do that though.

All the above comes from my time in a 737 and 777 simulator.


What is an air traffic controller's role in handoffs? I know they tell the pilot to tune into the new controller's frequency, but do they also have to do technical stuff? I'm trying to get a sense of how automated hand-offs are. For example, do controllers have to send flight progress strips to the next controller or is that done automatically?

In most modern ATC units, almost all transfers are conducted as “silent transfers”. A silent transfer is a transfer where all necessary information about a flight is sent in advance to the next ATC sector by an automated system. In this case, the controller does not have to manually talk to the next sector, but will simply instruct the pilot to switch to the next frequency prior to the area of responsibility border.
Where manual coordination is required, it is most likely done by the planning controller (PLC). As opposed to the executive controller (EC), who talks to the planes, the PLC (who will be seated right next to the EC) handles coordination with adjacent sectors via dedicated telephone lines.

There are two main parts of a transfer: the transfer of communication and the transfer of control. The transfer of communication is the actual frequency switch – the instruction for the pilot to switch from one frequency to another. The transfer of control is the transfer of responsibility of the flight from one controller to the next.
Before answering your question directly, let us establish the premises of the question.
Note: this answer refers primarily to European procedures, and is largely based on this EUROCONTROL document
In respect to your question, the transfer of control is the interesting part.
Firstly, we must realise that there are two fundamentally different premises for a transfer of control. Either the transfer takes place from one position/sector to another position/sector within the same ATC unit (for example between two area control sectors in the same area control centre), or the transfer takes place from one ATC unit to another (for example when a flight crosses a state border).
In the first case, the adjacent sector will typically already have access to information about the flight, since both sectors rely directly on the same data source. In the second case, this might not be the case.
Additionally, in the first case, controllers will often be physically seated right next to each other. For units where paper flight progress strips are still used (mainly some approach and tower units), this enables controllers to physically pass the flight progress strip to the next controller.
In both cases, there are four required stages which lead up to a transfer of control (refer to ICAO document 4444, section 10.1):

a) Notification of the flight in order to prepare for coordination, as necessary;
b) Coordination of conditions of transfer of control by the transferring ATC unit;
c) Coordination, if necessary, and acceptance of conditions of transfer of control by the accepting ATC unit; and
d) The transfer of control to the accepting ATC unit or control sector.

Doc 4444 also states that:

ATC units shall forward from unit to unit, as the flight progresses, necessary flight plan and control information.

Now, your question is, to what extend is this data transfer process automated. ICAO document 7030 gives us a clue to the background of the answer:

When so agreed between adjacent ATC units, a computer-assisted coordination process shall be introduced to eliminate the need for verbal coordination of boundary estimates and to reduce the amount of manual data input into ATC computers.

In most modern ATC units, almost all transfers are conducted as “silent transfers”. A silent transfer is a transfer where all necessary information about a flight is sent in advance to the next ATC sector by an automated system. In this case, the controller does not have to manually talk to the next sector, but will simply instruct the pilot to switch to the next frequency prior to the area of responsibility border.

Silent transfer of control procedures require that traffic is predictable and sufficient system support is provided to ATCOs to ensure:
-They are in possession of all elements related to the aircraft profile and
-Should the agreed conditions of the transfer change, they have the means to speak to each other such that new conditions are agreed expeditiously and safely.

Silent transfers are based on the exchange of flight data, which happens through connected Flight Dara Processing Systems, which make use of the OLDI protocol.

The operational use of connections between Flight Data Processing Systems (FDPSs) at ACCs for the purpose of replacing verbal "estimates" is referred to as On-Line Data Interchange (OLDI). The implementation of basic OLDI messages is mandatory in accordance with Regulation (EC) No 1032/2006. These messages are defined by the EUROCONTROL Specifications

What this regulation means in practise is that, within Europe, silent transfers are used more often than not. However, you will also see units where a manual coordination prior to a transfer is required, for example flights crossing the border from certain European countries to certain non-European countries. Where manual coordination is required, it is most likely done by the planning controller (PLC). As opposed to the executive controller (EC), who talks to the planes, the PLC (who will be seated right next to the EC) handles coordination with adjacent sectors via dedicated telephone lines.


When is runway slope most important? Based on this other answer, a new question arose.
When is runway slope most important, at take-off or at landing? Or is it equally important in both flight phases?

I would argue that it is most important on takeoff. The reason is that most planes require more distance for takeoff than for landing (partially due to higher weight). Also, it is generally easier to abort a takeoff than a landing (when on the ground at least).

Both landing and takeoff distance calculations are performed (which should take slope into account) to ensure that there is sufficient distance for the aircraft. However, in addition to aborting a takeoff being easier, more can go wrong in a landing. Windshear or improper approach settings can cause an airplane to touch down later or faster, which risks an overrun. During takeoff, the plane has until V1 (takeoff decision speed) to abort if any issues arise.

However, this certainly depends on conditions. Aircraft condition, wind and runway surface condition can increase the distance needed for landing (or aborting a takeoff). If reverse thrust is not available, there is a tailwind, and the runway is wet, that could change the situation.


How does radar detect turbulence and wind speed? In an answer to a questions about turbulence this answer mentioned that aircraft use radar to help detect the most turbulent areas of a storm so that pilots can avoid them.

How does radar do this?  Does it have to do with rain inside the storm moving up and down with the wind?  And if so, can radar detect wind speeds outside of a storm at all?

The answer you are referring to explains how radar is used to fly through a squall line whilst avoiding the most turbulent areas.

Conventional weather radar cannot detect wind speed or turbulence directly, it only detects solid and liquid objects above a threshold size.

The most turbulent areas of thunderstorms are also those areas that contain the most water and ice. This precipitation reflects radar waves very well so the most turbulent areas have the strongest reflections and shows up as red / purple on the scope. The area downwind of the core is usually very turbulent as well.

So the weather radar is merely detecting the presence of water. It is the pilot interpreting the picture who is really detecting the turbulence.

Clear air turbulence  (CAT) does not have condensated water in it and is therefore undetectable by conventional radar. There are systems such as LIDAR that can detect CAT but they generally not installed on aircraft.

As @reirab noted in a comment, a Doppler radar can be used to detect the relative speed of the particles from an aircraft. This is not used to reliably detect wind speed, but when there are large variations in wind speed (a.k.a. turbulence) the Doppler shift in the reflections from the particles causes a spectrum spread. This is an indication for turbulence. The traditional S-band airborne weather radars do not have this capability but the newer X-band radar (e.g. this Honeywell radar (PDF)) can detect wet turbulence to distance of up to 40-60 nautical miles using this technique.

LIDAR (a portmanteau of Light and RADAR) is a technology that uses light instead of radio waves. LIDAR can measure position and relative velocity of aerosols and therefore it can be used to detect turbulence. It has been used successfully to detect wind shears and map wake vortices on final approach from the ground (paperPDF).

Currently investigation (paperPDF, presentationPDF)is going on into airborne use of LIDAR to detect CAT. One of the difficulty is to detect vertical components of the turbulence. This causes very little Doppler shift but at the same time have the most effect on the aircraft (since the vertical component of turbulence causes changes in angle of attack).

As far as I know there are currently no commercial applications of LIDAR for airborne turbulence detection.


How can I know if the Garmin GTX 345 transponder supports modes A and C? How can I know if the GTX 345 supports mode A/C?
I did not find a mention in the installation manual the capability of mode A/C.

It's on the spec page: Garmin GTX 345:

Physical dimensions (unit, rack, connectors); for panel mounts, depth is measured from face of instrument panel): 1.65” x 6.30” x
10.07” (42 x 160 x 256 mm)
Display type: Digital
Weight (unit, rack, connectors): 3.1/3.2 lbs (1.41/1.45 kg)
Voltage range: 14/28 VDC (18/20 W Max)
Transmit power: 200 W minimum
Temperature: -40° C to +70° C
Operating altitude: To 55,000 ft (16,800 m)
Cooling input: Not required
Environmental compliance (TSO Approved): DO-160G
Software compliance (TSO Approved): DO-178 Levels B, C, D, E
Hardware compliance (TSO Approved): DO-254 Level C
TSO compliance (Approved): TSO-C88b (w/opt. alt. encoder; TSO-C112e (Class 1, Level 2els; TSO-C154c (Class A1S); TSO-C157a (Class 1);
TSO-C166b (Class A1S); TSO-C195a (Class C1,C2,C3,C4)
Mount type: Panel
Transponder type: Mode A/C, S and ES
Squawk code selection: Push-button
Traffic Information Services (TIS) alerts compatible: Yes

I have one of these units, and I can confirm that it supports A/C, S, ES, and ADS-B.


How can we design a wing for cruising at very low speeds? I was thinking about a simple super-light aircraft powered with a little low-power piston engine. Is there any way that we can design a wing so the aircraft flies very slow (like 25 km/h as minimum speed) without stalling?
I mean a suitable wing design for this kind of aircraft. (slow one) Other questions:

Which parameters in designing wings must be considered for lowering stall speed?
My thrust power is low so I accept lower cruise speeds.whats the effects of this on my design?
Can we choose thicker airfoils for wing and ignore the drag issues due to low speed?

The first thing you need is a large wingspan $b$. This will help to spread the lift over more air and reduce induced drag, which grows inversely with the square of flight speed $v$:
$$D_i \propto  \frac{1}{v^2} \text{ and } D_i \propto \frac{1}{b^2}$$

Large wings have a high root bending moment, which requires a strong and heavy structure. Next you should add bracing, at least on the lower side of the wing. For that to be effective, you need a high fuselage so the bracing will become efficient. Yes, those wires will add drag, but at low speed this will be a lot less than the added lift-induced drag from lifting a heavier structure.

The single bracing wire can hardly be seen on this picture of the Dedalus human powered aircraft, but it is there and saves a lot of drag at low speed (picture source)

Next, pick the right airfoils. Depending on the local Reynolds number, use thin and highly cambered ones but make sure that the pressure rise on the rear upper surface is slow enough to prevent early separation. One example is the DAE31 outer wing airfoil of the Dedalus aircraft (the inner and mid-span ones are DAE11 and DAE21). At Reynolds numbers of 500,000 to one million, turbulators positioned 2-3% ahead of the separation line will trip the flow and make the airfoil tolerate steeper pressure gradients.

Pick wing chord such that the wing area gives a lift coefficient around 1, details depending on the exact Reynolds number. Lower values go with lower Reynolds numbers.


Can I bring a person with me while getting landing current if they're a licensed, current pilot? It's been over 90 days since I've been flying, but I have a biennial flight review coming up pretty soon and would like to prep for it.

The regulations regarding this seem to be extremely vague, but is it legal for me to fly around, do my landings, log PIC time, and such with another licensed, current pilot in the plane with me?

As long as  you are not the legal pilot in command you can do what you want.  This means the other pilot has to be willing to act as the legal pilot in command on this flight.

  61.57(a) General experience.

  (1) Except as provided in paragraph (e) of this section, no person may
  act as a pilot in command of an aircraft carrying passengers or of an
  aircraft certificated for more than one pilot flight crewmember unless
  that person has made at least three takeoffs and three landings within
  the preceding 90 days, and—

  (i) The person acted as the sole manipulator of the flight controls;

The regulation is clearly required of the legal pilot in command.  If your current passenger is willing to act and take responsibility for your flying as the legal pilot in command then it is allowable.

Unfortunately, the other pilot is taking all the responsibility and gets none of the credit as they, under an airplane certificated for single pilot operations, cannot log any PIC time while you are getting current.

From LOGGING PILOT IN COMMAND TIME (FAA):

  However, two pilots may not simultaneously log PIC when one pilot is
  sole manipulator of the controls and the other is acting as
  pilot-in-command if the regulations governing the flight do not
  require more than one pilot.

If the other pilot is a CFI and it is a training flight then no one has to be landing current as there are no passengers.  See the Kortokrax interpretation.

Relevant excerpt:

  We agree that, for purposes of section 61.57(b), an authorized
  instructor providing instruction in an aircraft is not considered a
  passenger with respect to the person receiving instruction, even where
  the person receiving the instruction is acting as PIC. (The instructor
  must be current, qualified to instruct, and hold a category, class and
  type rating in the aircraft, if a class and type rating is required.)
  The instructor is not a passenger because he is present specifically
  to train the person receiving instruction. Neither is the person
  receiving instruction a passenger with respect to the instructor. This
  training may take place, even though neither pilot has met the
  61.57(b) requirements.


How does one calculate true altitude? "An aircraft is cruising at FL170. The altitude shown with QNH set is 17,500 feet. The OAT is -10 deg C. What is the true altitude?"

The ISA temperature deviation is 9°C.

My calculation is:

17,500'
 + 9 °C x 4' x 17,500/1000
= 18,130'

The answer provided is 18,100'. I don't understand where my mistake is.

This can be calculated by knowing a few formulas, or temperature relationships related to altitude. Note that all these calculations are an estimate based on standard ISA temperature and altitude relationships. See this page for other examples of this type of question.

Note the following from the question:

FL170 represents a pressure altitude of 17,000 ft
We know that the OAT is -10°C
Indicated altitude is 17,500 ft

Now, to solve for true altitude, we need to know a few additional items. We need to know that:

ISA at the surface is 15°C
Standard lapse rate is -2°C per thousand feet
To find true altitude, the difference from indicated altitude is 4 ft per 1°C deviation from ISA for every 1,000 ft

Knowing all this, we can calculate the following:

ISA at 17,000 ft (see 4 and 5 above)
Deviation from ISA (see 2 and 7 above)
True altitude (see 6 and 8 above)

This all might sound complicated, so lets run through an example problem. Let's assume a flight at FL190, with an OAT of -15°C, on a standard pressure day.

Since ISA at the surface is 15°C, using the standard lapse rate of -2°C per thousand feet, we know that ISA at 19,000 ft will be -23°C.

Given an ISA of -23°C, we see that our OAT of -15°C shows a difference of +8°C above ISA. Follow that step carefully, because the signs can be confusing.

Given that the temperature is ISA+8°C, and knowing 6 above, we can calculate our true altitude relative to pressure altitude:

$A_T$ is our estimated true altitude, in feet
$\Delta_{ISA}$ is the difference above ISA, in °C
$A_I$ is our indicated altitude, in feet
$A_{Ik}$ is our indicated altitude, in thousands of feet

$A_T=(4 \times A_{Ik}  \times \Delta_{ISA}) + A_I$
$A_T=(4 \times 19  \times 8) + 19000$
$A_T=608 + 19000$
$A_T=19608$

In some cases, a question might ask for true altitude based on a pressure altitude, instead of an indicated altitude. In such cases, substitute pressure altitude for indicated altitude in the above formula. Remember, this is all an exercise in estimation which is helpful in reinforcing the relationships between temperature, pressure, and altitude.


What would be the Sref i.e. Reference Wing Area for Drag Build Up Calculation in case of a box wing design? Would Sref be sum of areas of both, the front and the back wing? or just one wing? Assuming both the wings have equal areas (image below)
Details - So, I am calculating drag using the component drag build up method and using the below equation for the following parts of my box wing design.

Front Wing
Aft Wing
Fuselage
Tail

as you can see in the equation below, Swet for all different parts are divided by "Sref". For example assuming that both wings have area of 10m^2 and therefore wetted area of roughly about 20m^2 each.
for Cd0 calculation what should use as my Sref? total lift generating area of the aircraft, as in 10+10=20? or just the area of any one wing, 10?
If I use 20, for example, and apply the below formula for the front wing, my (Swet/Sref = 1) whereas it's = 2 if I use, Sref as 10m^2?
Basically, my total CD0 values changes drastically as I change my Sref according to the below equation, why is that?

You could argue for one or for both, but ultimately, it doesn't matter - as long as you are consistent and use the same reference area throughout your calculation/analysis.
If I personally were to choose, I would go for both, on the basis of total lifting area.
EDIT for details ------
Well yet again, it doesn't matter. You will get some value for your $C_{Dmin}$. Then later for other calculations you will use the same $S_{ref}$ and you will get comparable drag coefficients. Finally, when you use $D_{total} \text{ [N]} = C_{D} \cdot 0.5 \cdot \rho \cdot v^2 \cdot S_{ref}$ you will input same $S_{ref}$ and you will get the same Newtons regardless of chosen $S_{ref}$.
What matters is that you're consistent, i.e. if you use $S_{ref} = 10 \text{ m}^2$, use that for everything. If you use $S_{ref} = 20 \text{ m}^2$, use that for everything.
Additionally, do recheck the wetted area argument. $S_{wet} \neq S_{ref}$, because $S_{wet}$ includes both upper and lower surface of each wing, whereas $S_{ref}$ includes only the planform surface.


What material was/is used on the skin of the Blackbird SR71? What material was/is used on the skin of the Blackbird SR71 and how was it cooled.  How was it manufactured to be stronger with age

Titanium. Only the hottest portions of the fuselage (the chines) were cooled, using the fuel as a coolant. As far as I'm aware, there was no treatment to make the material stronger with age.


On which point(s) in a jet engine does the reaction force act? Turbojet and turbofan engines are part of a family of engines referred to as reaction engines to emphasize the fact the aircraft is pushed forward mostly by Newton's 3rd law of motion when gases are accelerated.

Diagram of a turbofan jet engine, source
On which point(s) does the reaction force act?

The thrust produced by the engine is due to the net resultant of various forces acting on various surfaces of the engine. The thrust produced is a function of mass flow rate and the change in velocity ($T = \dot{m} (V_{e}-V_{\inf})$). So, both of these have to be taken into account, rather than acceleration alone. The location of maximum thrust generated varies with the type of engine.

For high bypass jet engines like those used by the modern commercial airliners, most of the thrust is produced by the bypass fan. Though the acceleration is not the greatest here, mass x acceleration is- so this is where the thrust is maximum. The maximum reaction forces are applied there
In case of pure turbojets, the (almost all of the) thrust is produced by the core. For low bypass turbofans, its somewhere in the middle, with the lion's share of thrust produced by the core.

Note that most of the reactive force is applied on the diffuser and the compressor due to the high pressure and the forward facing area (due to cross sectional variation) in this region. This high pressure also acts on the combustion chamber, increasing the reaction force.

In the turbine and nozzle, the variation in the cross section creates a surface area facing the rearward direction, where the gas pressure acts, resulting in an force applied in direction opposite to the compressor. The net result of all these forces gives the thrust.

Image from Aircraft Performance and Design by John Anderson; taken from quora.com


Could TCAS be used as a backup source for position, attitude or air data? Given the number of incidents related to freezing pitot tubes, is there any current/planned implementation to infer aircraft position/speed/attitude/bearing based on TCAS data as a last resort data point? If not, is this at least doable?

Even if TCAS gives only relative data, one may correlate the data at the last moment before pitot data are lost to assist in dead reckoning.

Edit: I am not sure if my question has been misunderstood, or that I am misinterpreting the answers. The question is more like inferring current aircraft's data from transponders of surrounding aircrafts. Let's suppose at time $t_1$ I have planes A, B, and C within my TCAS range. I will possess information regarding their bearing, distance, speed, and altitude. At time $t_2$, even without my own altimeter, I can at least infer my own vertical speed using trigonometry. A, B, C are like ocean currents, a background moving together with me. If I know enough information about them, depending on what TCAS's underlying data stream provides, I can probably infer my own position, altitude, speed, and heading. From my limited understanding, TCAS bearing, distance, and speed are relative to me. However if I have not mistaken, despite what is shown on the screen, altitude is absolute measurement from the other planes' altimeter. One side fixed on a similar triangle, you'll get the rest.

Can TCAS assist in dead reckoning?

First, understand that TCAS runs off secondary (transponder) radar data. Essentially your aircraft with TCAS is like ATC radar, interrogating other transponders, getting their pressure altitude and getting their bearing from your aircraft.  That data combined is processed to create threat advisories.

If displayed on a moving map, the moving map is typically fed positional data for your aircraft from GPS, and the TCAS alert data is overlayed onto the map.

So to address your question, TCAS only provides data about transponders relative to your aircraft.  TCAS does not determine where you are, or where you are going. (Depending upon your display it might appear that way, but that is not part of the TCAS system.)

So the answer to your question is no, TCAS does not effectively create a secondary navigation capability.

Addendum #1
TCAS can only sense altitude through the host aircraft's static port, so if it is iced over, TCAS has invalid altitude information.

TCAS does not sense other air data, such as pitot pressure information (airspeed).

TCAS does not know the host aircraft's position.  Any positional information in TCAS is derived from pressure altitude information the host and all interrogated transponders.  If the host static port is obscured, the information on threats will be invalid, because TCAS displays relative to the host pressure altitude.

In other words, there is no absolute positional information, and all TCAS information is relative to the current host aircraft, and derived from the secondary beacons of interrogated aircraft transponders.


Was the safe landing of Southwest flight 1380 extraordinary or expected? I've been reading with great interest and concern about the recent Southwest flight that resulted in the first fatality in that airline's history.

There's one question I haven't seen answered, though: Was the plane really in serious danger of crashing?

I have long understood that two-engine commercial airliners are designed to be flown safely with one engine. However, are they designed to be flown safely with only one engine as well as a depressurized cabin and broken window? Do these three things in combination make a plane significantly harder to fly or land than just an engine failure?

The passengers aboard this plane had plenty of reason to be terrified, what with oxygen masks dropping from the ceiling and a woman nearly being sucked out of the plane (and ultimately dying from the head trauma she suffered), but the accounts I've read seem to suggest that many thought crashing was the most likely outcome and were preparing for their death.

I don't want to take anything away from the pilot, Tammie Jo Shults, who clearly performed superbly under stress and exhibited the highest level of professionalism, but was her safe landing anything other than the expected outcome?

Based upon the damage done to the aircraft, the fact that it retained a functioning engine, and the location of the engine failure and divert, no there was not a real immediate danger of crashing.

To be fair this incident was an emergency which caused structural damage to the aircraft and breach of the cabin structure, resulting in the loss of life of a single passenger.  It was also a freak accident and I don’t see any evidence that the aircraft or engines used have some kind of inherentt design or engineering fault the prophesizes more failures or future risk.

It’s possible that we did dodge a bullet and that, by pure chance, the failure of the fan blade and where it was ejected from the fan casing could have resulted in far more damage to the aircraft and critical systems like flight controls, fuel, hydraulics, landing gear, major structural members, etc that COULD have resulted in the very realistic probability of a much more devastating accident when the aircraft either attempted to land or could not maintain level flight further.


Why are the MSA direction needles not oriented to north? Why is it when you look at an MSA the arrows are pointed opposite of where they should be?

The arrows are referencing the nav cues, not the VORTAC radials, which define the MSA zones.  It’s cut up that way to define the sectors relative to the reference point that you will be flying in e.g. for the second from left MSA symbol, if you’re flying southeast of the HQM VORTAC between a course heading of 230° TO and 350° TO the VORTAC, this is an acceptable MSA for that segment.  I suspect that choice was made because aircrews are going to be approaching the airport from these areas as opposed to departing and their HSIs, OBSs, RMIs, etc are going to be set to track TO the NAVAID at that point.


How to differentiate two flights with the same flight number that arrive on the same day? I'm currently working on a piece of software that includes flight information. I have access to flight schedules and I'm making each days flight unique based on flight number and the date that it arrives.
I've hit a problem trying to differentiate flights that arrive on the same day though.
Example:

On Sunday, Flight LS1494 leaves at 19:50 and arrives the next day at 00:25.
The next day, on Monday, LS1494 leaves at 16:40 and arrives at 21:15, making it arrive twice in one day.

Is there a standard way to differentiate these? I have one source of data that includes an "Operational Suffix" of "B" for the flight that originally left on Sunday but I'm not sure if that is actually relevant.
I could include the actually arrival time to identify them however sometimes that scheduled time changes and doesn't get updated immediately which can cause problems.

I'm making each days flight unique based on flight number and the date that it arrives.<

A standard used by some (most?) airlines is to refer to the flight by the Flight No and by the departure date (you've already discovered the problem with using arrival dates).

So your example flights would be LS1494/<Sunday date> and LS1494/<Monday date>.


Can you identify this plane? Can you identify this plane?

Unknown airplane

I have just found the photo on a CD together with many other planes and helicopters. I have no clue regarding the place where the picture was taken, the time and the type of the plane.

@Roddy is correct. It is an Romanian IAR-80 all right.

"IAR80" by user:Planck -  licensing policy.. Licensed under Public Domain via Commons.

IAR-80 was produced by IAR (Intreprinderea Aeronautică Română), an airplane manufacturer that was the property of the Romanian State until the end of the WWII. The aircraft first flew in 1938 and was in service till 1952.

The model appears to be the IAR 81. The same image can be found here. It was probably taken in the eastern front.


Is V(L/D)max equivalent to best glide speed regardless of weight, engine, height, etc.? I've read Carson's paper on V(L/D)max, and while theoretically interesting, it does not seem to me to be useful in actually finding the V(L/D)max of an arbitrary aircraft, as it assumes knowledge of total parasite area (f) which can only really be calculated experimentally (and is very hard to calculate at that).

Another paper, by Russ Erb, claims that optimal glide speed, in the emergency section of the Pilot's Handbook for an aircraft, is equal to V(L/D)max for that aircraft.  This sounds wrong intuitively as it does not seem to take into account weight, air pressure, height, or other such variables.  As a math major, I am certainly suspicious of people who say "I'll spare you the derivations."

For an aircraft with an engine, such as a Cessna or a Bonanza, is best glide speed roughly equal to V(L/D)max?

I was told by an experienced pilot that

  Vldmax at weight G = Vldmax at weight L * ( weight G / weight L )^0.5

This would seem to contradict the claim that V(L/D)max = best glide speed, as best glide speed (according to the Bonanza Operating Handbook, as an example), is constant.

According to the FAA's "Best Glide Speed and Distance" handout provided by the General Aviation Joint Steering Committee ("GAJSC"), on most airplanes, the best glide speed is halfway between Vx and Vy. See Exhibit 1


What is the maximum theoretical lift of a plane as a function of engine power and surface of the wings? Maximum theoretical lift of a plane as a function of engine power and surface of the wings?

I know that the maximum static thrust of a propeller turned by an engine of power $P$ can not be greater than:
$$Thrust_{max}=\sqrt[3]{2\pi(\frac{D}{2})^2 \rho P^2}$$
where $\rho$ is the density of the air and $D$ the diameter of the propeller.

I am looking for a formula that gives the maximum theoretical lift of a plane which flies horizontally. After half a page of demonstrations I have found a relation:

$$Lift_{max}=\sqrt[3]{2S \rho P^2}$$
but I am not sure if it is correct ($S$ is the total surface of the wings).

Update 1

Two examples that show the formula for max lift gives lifts smaller than the loaded weight of the plane. I do not know why.

1) Cessna 172: Gross weight - $1,111kg$, Powerplant - $160 hp$, Wing area - $16.2 m^2$

$Lift_{max}=(2 * 16.2 m^2 * 1.2 kg/m^3 * (160 hp)^2)^{1/3}=837 kgf<1,111kgf$

2) de Havilland DHC-6 100 Twin Otter: Gross weight - $5,246kg$, Powerplant - $2*579 eshp$, Wing area - $39 m^2$

$Lift_{max}=(2 * 39 m^2 * 1.2 kg/m^3 * (2 * 579 hp)^2)^{1/3} = 4,198 kgf<5,246kgf$

Update 2

I have tried a different demonstration that stars from the known relations:
$$Lift={0.5 \rho S C_L V^2}$$
$$Drag={0.5 \rho S C_D V^2}$$
$$Power=Drag * V$$
By eliminating the speed of the plane, $V$, the lift result as being:
$$Lift = \sqrt[3]{2S \rho P^2 \frac{C_L^3}{4 C_D^2}}$$
It appears that this new lift can be greater than $\sqrt[3]{2S \rho P^2}$ as long as $\frac{C_L^3}{4 C_D^2}$ does not seem to be restricted to always remain less than 1.

Update 3

Using the formulas for $C_D$ an $L_{max}$, suggested to me by Peter Kampf and mnunos, I have made some evaluations.

$$C_D = C_{D0} + \frac{C_L^2}{\pi\cdot AR\cdot\epsilon}$$

$$L_{max}=\frac{1}{2}\cdot \sqrt{3\pi AR\epsilon}\cdot \sqrt[3]{\frac{P^2\eta^2\rho S}{4 \sqrt{C_{D0}}}}$$

For the assumed values: $\eta=0.7$, $\epsilon=0.8$, $C_{D0}=0.03$, $\rho=1.2\: kg/m^3$ and the data taken from the Cessna 172 general characteristics: $P = 160\:hp$, $S=16.2\:m^2$ and $AR=7.32$, I get:
$$L_{max,\:\eta=0.7}=2199\:kgf$$
However, only $57.451\:hp$ are necessary for lifting $1111\:kg$ (horizontal flight), the rest up to $160\:hp$ can be used for climbing.

From the relation:

$$m_{plane}gV_{climb} = \eta(160\:hp - 57.451\:hp)$$

I get $V_{climb}= 4.913\:m/s$ which is significantly greater than the rate of climb 3.66 m/s the Cessna 172 has.

It seems that working with $C_{D0}$ and $\epsilon$ gives only a quite rough estimate of the max lift. Without knowing the actual $C_L=C_L(C_D)$ little can be said about the real maximum possible lift of a certain plane.

Update 4

The maximum theoretical thrust of an ideal helicopter propeller having the diameter identical to the wingspan, $11\:m$,  of a Cessna 172:

$$T_{max,Diameter\:prop=11\:m}=(2 * pi * (11 m / 2)^2 * 1.2 kg/m^3 * (160 hp)^2)^{1/3} = 1 509.95\:kgf$$

is considerably smaller than the max lift of the plane,  $L_{max,\:\eta=0.7}=2199\:kgf$, I calculated in Update 3. Is such a result logical?  Does it have an explanation?

Your thrust equation is only valid for the static case (initial speed at infinite distance from the propeller is zero), and neglects the propeller's efficiency. For the flight case, I recommend to make a bold assumption for propeller efficiency and to assume that most power will be converted to thrust:

$$T = \frac{P}{v}\cdot\eta_{Prop}$$

A large, slowly turning propeller will have an efficiency approaching 90%, while a small, fast-spinning one will only reach 70%. If operated outside of its design speed, efficiency might even become negative. If you look for an upper boundary of thrust, the equation above sans efficiency will do nicely.

For the maximum lift answer you should expand your drag coefficient a little, as recommended already in this answer. $$c_D = c_{D0} + \frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$$

Now you can equate drag and thrust to get: $$P\cdot\eta_{Prop} = \left(c_{D0} + \frac{c_L^2}{\pi\cdot AR\cdot\epsilon}\right)\cdot S\cdot\frac{\rho}{2}\cdot v^3$$

and solve for lift: $$L = c_L\cdot S\cdot\frac{\rho}{2}\cdot v^2 = \sqrt{\pi\cdot AR\cdot\epsilon\cdot\left(\frac{P}{v}\cdot\eta_{Prop} - c_{D0}\cdot S\cdot\frac{\rho}{2}\cdot v^2\right)}\cdot\sqrt{S\cdot\frac{\rho}{2}\cdot v^2}$$

The minimum power is consumed when the ratio $\frac{c_L^3}{c_D^2}$ reaches its maximum. Expressed differently, this means that $$c_L = \sqrt{3\cdot\pi\cdot AR\cdot\epsilon\cdot c_{D0}}$$

Plugging that back into the lift equation I leave as an exercise to the reader.

As you can see, the answer really depends on the aerodynamic quality of your airfoil. The best you can reach with regular airfoils is a $c_L$ between 1.5 and 1.8, depending on Reynolds number. With slotted flaps this can be doubled, but $c_{D0}$ will go up substantially, so the best ratio $\frac{c_L^3}{c_D^2}$ can be gained with a clean airfoil.

Nomenclature:
    $c_L \:\:\:$ lift coefficient (normally between 0 and 1.5)
    $S \:\:\:\:\:$ Wing surface area
    $\pi \:\:\:\:\:$  3.14159$\dots$
    $AR \:\:$  aspect ratio of the wing (ratio of span to mean chord)
    $\epsilon \:\:\:\:\:$  the wing's Oswald factor (use 0.8 when in doubt)
    $c_{D0} \:$  zero-lift drag coefficient (use 0.02 when in doubt and 0.03 with a fixed gear)