Can you re-register a de-registered aircraft with the same N-number? If I'm selling a plane that has been de-registered, can the new owner re-register the aircraft with the same number (or do they have to get a new one)? Does the airworthiness certificate transfer as well? How long does this process typically take?

The FAA has a page that sort-of takes you through this process.  Basically it boils down to this:

If your registration has expired you still have ~90 days to register the aircraft

  If a renewal application was first received by the Registry after the expiration date, the renewal process no longer applies. An aircraft owner may apply for registration of an unregistered aircraft under 14 C.F.R. §4731(a), by filing an Aircraft Registration Application, AC Form 8050-1, the $5.00 registration fee, and evidence of ownership (if it is not already on file at the Registry).

So it's a "new registration", but you can keep your N number. (Yes, you have to use an official, FAA-Provided 8050-1 form - you can't download them anymore. You're also going to want to talk to someone at the Aircraft Registration Branch or your local FSDO before you start filling out forms because you don't want your form to show up after the 90 day window or you wind up in the other category...)

After about 90 days your N number is Cancelled & unavailable for 5 years

  When an aircraft registration is not renewed, the cancellation of the N-number assigned to that aircraft will take place approximately 90 days after the expiration of an aircraft's registration. Upon cancellation, an N-number will be unavailable for re-assignment or reservation for a period of five years.

At this point you have to talk the aircraft registration branch, but my understanding is you are out of luck -- That N number is void and a new number will be issued when you register the aircraft (at which point you must apply the new N number to the aircraft in accordance with FAR 45, which typically means repainting at least part of the tail).
You must get a new airworthiness certificate bearing the new registration number from your local FSDO.

In both of these cases your airworthiness certificate is invalid* (because without a valid registration you have no authority to operate the aircraft).

In the first case you will not need a replacement airworthiness certificate (because your N number hasn't changed), but if your N number has been cancelled you will  need a new airworthiness certificate issued with the new N number on it.

Total time for all the paperwork to be processed depends on which bucket you fall into -- the aircraft registration branch usually has about a 30-day backlog (check here), plus time for the post office to move the paper from point A to point B, so a lag of a little over 30 days is what you should expect.
If a replacement airworthiness certificate is required that will probably add a few days to the process as well.

you can register the aircraft in a different jurisdiction and use the same last part of the registration and change only the first 1 or 2 codes


What are the specs for runways at international airports? 1) What are the materials that international airport runways are made of? If runways consists of several layers, what are they made of, and what is their depth and other specifications?

2) What are the surface specifications/properties required of such runways?

3) What kind of routine maintenance drills do such runways require?

Runways for international airports are usually man made, which means that they are made up of asphalt or concrete and sometimes are a mixture of both. The depth for the runways are 29cm to 33cm.
Runways are usually around 150ft wide. For an international airport will have a runway around 6,000ft long, but depending on the size of the aircraft that land there a runway will need to be 8,000ft or more.
I found this link which should give you information on runway maintenance.


Is there an official FAA source for time zone of an airport? Looking at this answer, from TypeIA, to the question How to get timezone from airport code (IATA/FAA) .

This person is recommending the NASR database, which is good because it's official FAA stuff, but like other users I'm not able to find any information about time zone for an airport in NASR.

Is this information available via NASR or from any other official FAA source?

I do know that I can derive the information from lat/long etc., but what I'm looking for is the official FAA time zone designation for each airport, if such a thing exists.

EDIT: Sorry, to be clear, I'm looking for a digital version of this data suitable for use in a database or app.

Each airports timezone is listed in the Airport/Facility Directory. I think the FAA is starting to call these "Chart Supplements" now though. You can search for specific airports here but it looks like the results are in .pdf format. The value you want is circled in red here:

EDIT: As far as digital sources, explicit time zone information seems missing from the NASR airport (APT) data here. The next best thing is Latitude/Longitude it seems.

Timezone is always the same for each state, correct? It seems that all you need is the state. Probably that is why their is no official data with a time zone for each airport. From eNASR I just entered "*" for "Airport ID" and (after a while) got over 19,000 airports. They at least have a state code. I assume it would be relatively easy to save and parse that data.

Also have a look at OurAirports. Click on "Site Info" at the top then "Open data downloads". You can download a CSV with over 50,000 airports in the world with a state code for each one in the US.


Why do (some) aircraft shake at low speeds with flaps, slats extended? I landed a couple of times as a passenger with the Embraer 190 (I like Embraer very much) on runways with a length below 2 km.

I noticed that the plane started to shake or vibrate quite strongly once reducing its speed prior to landing. The flaps during this time were fully extended. How can one explain these vibrations? I have to note though, that I often sit in the back of the fuselage, where vibrations tend to be more pronounced.

There are many phenomena which can cause vibrations. Without knowing further details like the frequency, starting moment or the weather conditions on landing it is difficult say. Common causes are:

Stall buffet/ self induced turbulence Normally when a plane lands it flies with low velocity and a high lift coefficient, near to stall. The flow detaches much earlier and there is more turbulence behind the main wing. The tailplane has to fly through this turbulent zone and hence you feel the vibrations. Sometimes deploying the spoilers, even if you're not flying at stall speed is enough to create turbulence which will hit the tailplane. T-tail airplanes are less affected because they're further away from the air zone influenced by the main wing.

Atmospheric turbulence Can be created due to thermals (in that case it should be sunny) or wind hitting ground structures. If you have landed a couple of times and always felt the vibration it probably isn't the answer.

Wake turbulence Shouldn't be your case either, separation minima exist to avoid this problem and it would probably feel different every time you land.

There are many reasons why the plane is shaking.

The plane may be experiencing turbulence, wind causing plane to not balance.
The plane may stall because the angle of attack is too great.
The engine may have problems running at slow slow speeds


Why local(section) lift coefficient(CLL) is higher near the wing tips compared to the root on untwisted tapered wings and untwisted swept(aft) wings? Why local(section) lift coefficient(CLL) is higher near the wing tips compared to the root on untwisted tapered wings and untwisted swept(aft) wings?

I know that in an elliptical wing(untwisted) the absolute angle of attack is the same in all the wing span so, the reduction in lift is 'proportional' to the reduction in area/chord of the section and then the CLL is equal in all the wing span.
I also know that in a retangular wing(untwisted), the effective angle of attack(due to the tip vortex) causes a reduction in the CLL in the tips compared to the wing root.

Even that I understand the flow pattern over swept wings(Attachment line/Spanwise airflow and it's effects on lift and in the boundary layer, adverse pressure gradient and tip stall)I just can't figure out, intuitively, the relation between this flow pattern and the increase in the CLL in the direction of the wing tips.

[ CL = L / S.q ]  Seems to be too simplistic to analise this particular case.

For the swept wing, as you go out along the wing, each section of the wing is effectively flying in the upwash of the tip vortex of the section immediately next to, and inboard of it. So, each section has its angle of attack increased by all of the sections inboard of it. The root section, of course, does not have anything inboard of it, and the tip section is affected (to at least some extent) by all the sections inboard (and ahead) of it.

Birds fly in V-formations to take advantage of this effect. Each bird is flying in the upwash of the tip vortex of the bird ahead and toward the center of the V, and is thus "flying downhill" aerodynamically.

But, that upwash increases the angle of attack, and the effects add up as you move outboard, until they reach a peak effect near the tip.

You can really see this on hang glider wings: if you stand next to one as it's about to take off, you can see the tips appear to be at a negative incidence relative to the wind, and yet they are lifting, because the wing sweep increases their local AoA.

It’s because, on a tapered or a swept wing, the tips are so small that they cannot hold the vortices that are being formed, because of which the vortices are formed inboard. Therefore there is no reduction in the effective angle of attack at the tips whereas there is a reduction in the effective angle of attack inboards and hence, the local lift coefficient is higher closer to the tips as compared to the root.


Can I use a 150mw green laser to scare pigeons and not interfere with air traffic? Pigeons are annoying since they make aggressive cooing noises near my house. Considering many options for their dispersal, it seems lasers are one options since for some reason the lasers drive the pigeons crazy. In one video pigeons are completely dispersed by a 150mw green laser, making it appear to be an attractive option for dispersing my pigeon problem.

However I am concerned that maybe the lasers will continue into the atmosphere past the pigeons, and will affect air traffic....Obviously this is not worth the benefit of dispersing pigeons.

Is this totally out of the question? Are there any rules of thumb one should follow here to be on the safe side to not affect aircraft?

There are cases where you can use a laser pointer safely, such as pointing out constellations in a dark field far from any air traffic, but as a general rule of thumb lasers should not be pointed skyward - not even to get rid of irritating sky rats pigeons.

Should you wind up illuminating an aircraft with your laser you can cause serious problems for the crew - inadvertent or not it could get you a visit from some very unhappy law enforcement officials.

There are many other ways to get rid of pigeons, some of which are used at airports and many of which are more effective (and safer) than your laser idea.
Your local Animal Control group may be able to offer some additional ideas.

(Avoid the "predator bird statue" idea - in my experience pigeons realize it's fake and use it as a toilet.)

This kind of laser is class 3B laser. You may not be allowed to possess it.
Even if you are, this laser come with serious technical restriction such as deactivation if you point it upward.


Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind? We had a discussion with friend and we were talking about if the airplane is on approach and the one is landing with tailwind and the other one is landing with headwind, so both have the same value of IAS but the different value of ground speed. Will the airplanes produce the same value of lift in both cases or will be the value of lift be different? He said that value will be still the same and I was argueing that if you have tailwind you will have to increase angle of attack which means that parasite drag will increase so the value of lift will have to increase.

EDIT: Question is: If two same aircraft are on same glidepath (descending on ils glide slope) with same IAS but different ground speed, because of headwind acting on one aircraft and the tailwind acting on the other aircraft, will the amount of lift produced by one of the airplanes be different or will be both two values be the same?

EDIT: This question is really not about lift vectors. It is asking about the total amount or total value of lift needed with varying headwinds.

Everyone seems to be over complicating this very basic question which asks about two identical aircraft flying down an identical ILS glideslope with the exact same indicated airspeed. The only difference is the amount of headwind.

It is so simple. Rate of descent is a direct result of the amount of lift being produced.

It is a FACT that the aircraft with less headwind will need to produce LESS lift than the aircraft with more headwind, in order to maintain the same indicated airspeed, and follow the glide path.

The rate of descent with a headwind could be -500ft/min while the aircraft with a tailwind could be -1000ft/min.

The aircraft maintaining a -1000ft/min descent because of a tailwind will need to produce LESS lift than an aircraft maintaining a -500ft/min with a headwind.

If both aircraft maintain a constant rate of descent, both aircraft will be generating the same amount of lift, as in both cases the motion is steady state ie not accelerating and therefore lift will be equal to gravity.  The aircraft descending with a tailwind will be doing so at a greater rate of descent, due to the higher ground speed and will require less thrust to maintain this state than the airplane landing in a headwind or still air, as more of the the potential energy can be converted into kinetic energy in the descent than in the case of the airplane descending with a headwind down an identical glideslope.


Why would helicopters not have black boxes? Is there a legitimate argument against having a black box on helicopters? I'm assuming there must be reasons beyond financial ones?

Airline and commuter type aircraft are required to have cockpit voice recorders and flight data recorders. There are no requirements for any other type of aircraft to have them.

Most helicopters are actually small or light aircraft. The helicopter in which Kobe Bryant and others died would be considered a medium sized helicopter, but a small aircraft.

You will find VERY few non-airliner or non-commuter aircraft have any type of “black boxes” as they’re commonly (and erroneously) called. They are not required to, either. Expense, weight, complexity, and maintenance are all reasons. The smaller number of lives at risk might be another. After all, they carry the same number of people as between a sports car and a passenger van. School and passenger buses carry more people.

Saying that, technology has become cheaper and more compact in order to provide de facto recorders. Some modern avionics have that function built in. With the proper app, iPhones and iPads have that function. Some pilots will go the extra step and buy their own sensors to provide more accurate data on Attitude, Heading, Position, Altitude, Ground Speed, Carbon Monoxide presence, and Traffic Data. Add-on apps will even record voice transmissions picked up by the headsets and audio system. This is not mandated by the FAA. It is more for the pilots Situational Awareness and review for training purposes. ForeFlight, CloudAhoy, Stratus, Sentry, and Lightspeed are some common devices and apps used to record data.

Most of the arguments against flight data recorders are moot - because of technology developed in the past decade.  The same technology in a "smart-watch" that allows it to be a telephone, a camera, a computer, and more....is what now makes it possible to have a FDR and CVR that weighs about 6 oz.......one that also has computing power sufficient to contain the entire digital terrain database.....and an algorithm that tells the pilot when he flies too close to the ground.....even when that pilot has no idea where the ground is.  As for wiring every instrument into a data recorder, the Army in 2008-9 worked with a small avionic company to build a camera that took photos of the cockpit instruments -- it cost only $3900 and weighed less than 1 lb.  It's standard equipment on some small helicopters today....and has been used by NTSB in several crashes to determine causes.  Such devices are FAA/EASA approved.  Most arguments against helicopter FDRs and CVRs are based on unsupported claims of "too heavy, too big, too expensive".  Not today, folks, not today!  Wake up to 21st century technology!


Can you be a monocular airline pilot in Europe? I know this question has been asked before but it was for the US and I would like to know about Europe.

I was born with no vision in my left eye, and I have always wanted to become a commercial airline pilot, I have very good vision in my right eye. I just want to know if I could do it in Europe and if I would have any restrictions on my medical like not being able to fly internationally or fly big planes like the a380, and if it’s possible to become a captain. Or if it’s possible to get it without a soda, and would there be any differences. Or can you get a soda in Europe like the us.

UK CAA says

  EASA MED.B.070 (e) states that “Applicants for a Class 1 medical certificate shall be required to have normal fields of vision and normal binocular function”.

Therefore, you will unfortunately have to move out of the EU to become an airline pilot.

The person that speaks about two eyes being important for 'depth perception & peripheral vision' is talking rubbish. My ophthalmologist who teaches at Moorfields says one eye compensates for the lack of the second eye. CAA have understood this and wanted to change the rules but EASA stopped them. In Australia - hardly a country with a poor safety record - monocular commercial pilots are allowed to fly and have been for years. They have been sensible and understood this for years. I think in the US (but am not certain), it may be allowed too. But in Europe we are in the stone age, sadly because of nonsense like this based on pseudo science.


What fraction of air intake becomes bleed air in airliner engines? Gas turbines typically bleed air from the late compressor stages. Some of this air bypasses the combustion chamber and is expanded through the turbine to prevent the blades from becoming too hot. I believe bleed air is also used to provide hydraulic control of the control systems and also for utilities such as air conditioning and electricity generation.

I am interested to know just how much of the air intake to the gas turbine compressor becomes bleed air. I'm also interested in knowing how much of this bleed air goes to each application e.g. Utilities vs. turbine cooling vs. hydraulic control.

I suspect it varies a fair bit so I am mostly interested in how this occurs on larger airliners that would typically fly long international flights.

Indeed about 1-2% of compressed air is bled off for cooling turbine disks and blade roots. When turbine inlet temperatures are higher than 1350K, the turbine blades must be internally cooled as well, and up to 15% of compressor air may be required as bleed air for cooling. Source

Bleed air is also used for aircraft systems. If we take a typical airliner such as the B737NG, compressed air can be supplied by either the engines, the APU or an external air chart, and is used for the following purposes:

Air conditioning & cabin pressurisation, which are both subject to regulation quality standards.
Wing & engine thermal anti-icing
Engine starting
Hydraulic reservoir pressurisation. The reservoirs are pressurised to ensure that hydraulic fuel reaches the hydraulic pumps.
Water tank pressurisation

Compressed air is tapped from the engine 5th and/or 9th compressor stage. Air supply is regulated according to demand and varies according to flight stage - during flight stages with relatively high engine power (take-off, climb, most cruise), 5th stage compressed air only is used. During low engine power conditions such as approach and landing, 9th stage compressed air is used for the pneumatic systems as well. This air is under higher pressure, and hotter, than the 5th stage air.

Exactly how much bleed air is used for the systems is hard to get a reference for, but here is an estimate for the CFM56-7B18:

Bypass ratio = 5.5:1, Air mass flow = 307 kg/s => hot flow = 307 / 5.5 = 55.8 kg/s
Cabin air systems refresh the whole cabin volume every 2-3 mins. Let's take 2 minutes, without recirculation.
Internal volume of a 737NG is about 400 $m^3$ (from fuselage width and length of an -800).
Air density = 0.94 kg/$m^3$ (we'll take cruise conditions, cabin air density @ 2,000 m)
Aircon flow = 400 * 0.94 / 120 = 3.1 kg/s

There are two engines, so the percentage of flow used for the aircon and pressurisation would be 3.1 / (2 * 55.8) = 2.8% maximum. Please again note that this is a ROM value which I will gladly give up for an official reference.

As a magnitude check: the APU brochure mentions 154 pounds of air per minute = 1.16 kg/s for environmental control and engine start. The APU can deliver air conditioning on the ground without the engines running so that would be the right order of magnitude.

Final estimate: bleed air for air conditioning is between 1 - 3 kg/s is about 1 - 3% per engine.

Bleed air is not used for:

Hydraulic power generation. The hydraulic pumps are engine driven pumps (via a gearbox) and electrical pumps, powered by the AC system
Electrical power generation. This is provided by Integrated Drive Generators, from a generator in the APU, and from the batteries through an inverter.

Source: own work from open source on-line pilot training courses.

Internal cooling air might be better termed bypass air as it is not actually bled out of the engine so the energy of compressing this bypass air remains within the overall cycle rather than draining substantial energy from the system, and because turbine combustion is normally oxygen rich this bypassed air does not reduce the total fuel that can be burned.
The pilots can shutoff all accessory bleed air for improved engine performance on takeoff and low altitude climbs. So 0% is the lower limit for bleed air.
Accessory bleed air is used mainly for cabin pressure, engine starting, and deicing. Pressurizing reservoirs is an extremely minor use and only needed at higher altitudes to reduce the chance of pump cavitation and foaming from dissolved gasses.


Are magnetic compass acceleration errors due to the southern side of the compass being weighted? When I was in college, I had an instructor who taught me that the acceleration errors in a magnetic compass were due to a counter-weight installed on the south-facing side of the compass that was intended to counter-balance the magnetic dip on the northern side.

It makes sense that this would reduce or prevent north/south-facing turning errors, and would also create acceleration errors while facing east or west.

For example, flying east the weight would be on the right, south-facing side, of the compass, which is the side labeled as 'N'.  Now imagine a straight line acceleration...

The heavier, right side, would move rearward in relation to the aircraft due to its inertial lag, bringing the, weighted, southern side of the compass (which reads north) closer to the compass window causing the viewed indication be a number closer to north, making the "accelerate north" error, true.

The thing is, I have never seen this explanation on paper, just that it is "magnetic dip" error.  Is it true, and can you find a source that says so, or was my professor wrong?

Some compasses are counterweighted to decrease magnetic dip. Older and/or cheaper compasses for land navigation are this way. Many to most are not weighted, especially in aviation compasses. Weighted compasses would require different compasses for the North and South hemispheres. Instead, aviation wet compasses are built with the low friction mounting point above the CG and magnetic bar. This causes the CG to move to the South polarity end of the compass assembly when the magnetic bar dips downward to the North Pole and to the North polarity end when dipping towards the South Pole. I like this article that explains this phenomenon.

Several decades ago, my unit had the opportunity to train with the Australian 3rd Para Regiment in Australia. Many of us had been using personally owned GI lensatic compasses because the tritium markers worked better than the newer, issued lensatic compasses. We were warned that the older models may not be accurate down under due to them being Northern Hemisphere compasses instead of the universal ones. I did not understand the reasoning for this until many years later.

If that were true you could only use your compass for one specific latitude and only for the northern hemisphere, which means there would have to be label on the compass warning pilots only to navigate between latitude x and y. And international flights would need dozens of compasses in their cockpit, one for each band of latitude where the compass needle is balanced enough. So no, there is no counter weight that balances your compass needle to be precisely horizontal.

The reason why your compass is showing compass errors is that your compass card pivot point is above the needle and mass center of the compass card; and that the earth magnetic lines are not parallel to the horizon but hit the surface at an angle.

As you accelerate your compass card swings back, then your needle is trying to align itself again with earth's magnetic field lines (that are at an angle to the horizon), which causes the indication error. This happens whenever the compass is tilted, e.g. acceleration or deceleration, climbing, descending, banking.

As soon as you imagine the earth magnetic field lines you can understand why it is rotating either left or right.


Do contra-rotating propellers have the same RPM? Image source

Here is a contra-rotating propeller. The front propeller rotates counterclockwise while the back propeller rotates clockwise. My questions are:

Do they rotate at the same speed (RPM)?
If not, which one is rotate faster, and why?

Most if not all contra-rotating propeller systems are geared together so that both sections turn at the same speed.  This is done so that as well as allowing effective conversion of very high power into thrust, torque and P-factor are cancelled, making the aircraft easier to fly (especially in a single-engine or two-into-one installation).

This is why you see the same diameter propeller in front as in the rear -- if they were turning at different rates, the slower one could be larger diameter (a limitation on diameter is transonic conditions at the tips) -- but you always see the same diameter on front and rear.

From efficiency point of view, counter-rotating propellers are worse than a single propeller. Exception to that is counter-rotating rotors on some helicopters where there is enough vertical separation so the upper rotor stream contracts and some fresh air is feed to the bottom rotor. As airplane propellers would work most of the time at high speed, the influx of fresh air is lost, the second propeller is operating in the stream of increased velocity, and so the efficiency is less. To answer your question I'll need more information, as what you want to achieve. For example if the goal is to eliminate torque, the rear propeller needs to rotate faster. If the goal is to reduce fuel burn then one propeller needs to be stopped and feathered.


Why did this plane fly in a zigzag pattern? I was tracking flight NKS739 from LAX to SEA this morning and saw that it made some some odd turns. If you look at the track from FlightRadar24.com you can see that it was heading toward SEA then it took a large turn to the right then to the left then returned to its original course.

This does not look like the usual technical glitch from Flightradar24, and FlightAware shows the same track. There was no weather anywhere in the area. There were many aircraft in the area at the time and none of them made any unusual turns. There are no MOA's in the area (most of it is over a national forest) and none of the other aircraft seem to be avoiding anything in the area. I tried to see if LiveATC.net had a recording but that's Seattle center's area and they don't appear to have a feed for ZSE.

I can't come up with any reason for such an odd deviation. The aircraft arrived in SEA and there didn't seem to be any problem. It immediately turned around and made another flight which was uneventful.

Is there anything that might explain this deviation? Is it just a glitch?

Alrighty then, that took some detective work.
Here's another flight around the same time, SkyWest 4458, which arrived 2 minutes later. Notice it's one deviation, compared to two deviations in your example.

Weather seemed fine, but KSEA has TBFM (Time Based Flow Management) since Aug 2013. If the ATC asks a plane to delay its arrival time, the way it's done is to deviate from the route, then get back on it again—that's if slowing down alone won't meet the time target.
Pilots then use the RTA (Required Time of Arrival) function of the FMC. Initially the function was rarely used, but now with NextGen in the U.S., SESAR in Europe, and similar systems elsewhere, it is frequently used.
It can be arranged way ahead of the airport, i.e., it can be ordered by the approach controller, yet the center controller of another ARTCC delivers the message.
It looks exactly like that. Deviate, if not enough, deviate the other way.
Behind the Scenes
The system is called ERAM (En-Route Automation Modernization).
ERAM detects future conflicts, and based on that, traffic is asked to speed up, slow down, or deviate to slow down further. It reduces/eliminates hold time.
Europe—the first to use it in 2012—refers to it as initial 4D trajectory management. Planes arrive at waypoints at agreed times.
For whatever reason—most probably sudden traffic influx—they were asked to slow down.

The difference between this, and the regular vectoring/doglegging in the terminal area, is the way it is predicted, coordinated, and then relayed to airplanes far from the airport.
Below you can see as the jet turned, it slowed down as well. The initial upward spike is just the tailwind when the plane turned.

Source: flightradar24.com

DeltaLima hits the mark. I used to work there. Undoubtedly traffic management delays going on for some reason, either too much traffic or reduced runway capacity. TBFM issued directly to the aircraft is still in the early stages. I believe most delays are managed by the controllers vectoring, holding or issuing speeds. BTW- speeds are only effective for absorbing very small delays because it takes too much time to work. Old ATC rule; use vectors to get your spacing, speed to keep it.

The "dog leg" vectoring going on looks like it's occurring in the second tier sector, sector 46 (two sectors away from the approach control boundary). The first tier sector may have been at capacity forcing the delays to be pushed out to the high altitude sectors.

FYI - we used to figure about 5 minutes for one turn in hold. Holding needs to be handled at lower altitudes just to keep the plane in the box. I saw someone issue a hold to a 747 at FL390 once and it took the better part of the state to complete one turn!


What is "detonation" in a piston engine? Recently, I saw this question, which included a rather alarming quote:

  Full throttle operation with carburetor heat on can cause detonation.

Woah! That doesn't sound good. In fact, I would have figured aviation engine manufacturers would have worked hard to prevent their engines from exploding, and possibly destroying the rest of the aircraft and/or killing the pilot in the process.

I also found this question, with a rather calm quote:

  What if I have a noise cancelling headset on, and the engine undergoes detonation?

It seems like there should be other cues that your engine just blew itself to bits...

I'm figuring that perhaps "detonation" in this context doesn't actually mean that your engine block turns into shrapnel. I also figure that it doesn't refer to the normal process of piston engines, where fuel is injected and mixed with fuel before detonating. So what does it mean?

Detonation, as the name suggests, is an explosion of the fuel-air mixture inside the cylinder. Under normal operation, the spark plugs each ignite a point in the fuel/air charge, which then propagate through the cylinder and provides a consistent, regulated burn in a process called deflagration. This extends the time that the burning fuel pushes on the cylinder, providing a gentle power stroke.

During detonation, the fuel/air charge (or pockets within the charge) explodes rather than burning smoothly. Because of this explosion, the charge exerts a much higher force on the piston and cylinder, leading to increased noise, vibration, and cylinder head temperatures. The violence of detonation also causes a reduction in power. Mild detonation may increase engine wear, though some engines can operate with mild detonation regularly. However, severe detonation can cause engine failure in minutes. Because of the noise that it makes, detonation is called "engine knock" or "pinging" in cars.

(Source: FAA PHAK, page 6-18)

Detonation can be caused by a number of factors, but generally they relate to a situation outside of the engine's design. Using the wrong fuel can cause detonation: the octane rating of a fuel is basically the amount of pressure it can take before it detonates. Similarly, high cylinder temperatures can cause detonation(1). This is why mixture is set very rich when at high power levels: the excess fuel cools the cylinders, preventing detonation. As noted above, full power with carburetor heat on can cause detonation, as the increased temperature mimics the effect of high cylinder temperatures. Full carburetor heat is required to provide at least a 90 degree F increase in intake air temperature.

Further reading:

Page 6-19 of the Pilot's Handbook of Aeronautical Knowledge.
An article at Savvy Analysis

1) Detonation is distinct from pre-ignition, where the fuel/air charge is ignited by something other than the spark plug. Notably, detonation and pre-ignition often cause each other, as a detonating engine will have a very high CHT (potentially enough to cause pre-ignition), and if the charge pre-ignites as the cylinder is still rising, the continued compression may cause the charge to detonate.

I'd say, sorry, can't retrieve the reference now, that: 'detonation', is not the same as: 'knocking', even if you consider: 'knocking', as the audible expression of a detonation, or preignition or other forms of undesirable combustion inside an Internal Combustion Piston Engine, be the piston alternative or a rotating rotor.

Toyota tested successfully, 7% fuel economy improvement, a cleaner exhaust, in a Wankel Rotary Engine, substituting a glow-plug in the leading site for the ordinary spark plug. Another concept cited here is the presence of two plugs in aviation engines, my idea was that the purpose of a double plug and double ignition system, e.g. two magnetos, is just a redundancy for safety reasons, having a replacement ignition system in the case of failure of one, but the rate and pace of combustion, or flame front speed inside the working chamber of an ICE is also influenced by having one or more plugs, by the presence of hot spots, swirl or quenching inside the combustion chamber, and by the shape of combustion chamber and valve mechanism, if there's any, but the issue is complex.

The books by Harry Ricardo: 'The high-speed internal combustion engine', a classic of ICEs, deal with this and other combustion issues.                  *link to download the 1931 edition: http://www.mediafire.com/download/gn1o0vyvzaz/The+High+Speed+Internal+Combustion+Engine+Ricardo+1931.pdf

Also available was: 'Aircraft Propulsion', by C. Fayette Taylor, Smithsonian Annals of Flight, 1971.
Hope it's useful for you. Regards. Salut †


What is it called when you roll without yawing or pitching? My instructor made me do an exercise that consists of banking the sailplane left and right around 30 degrees without moving from the axis. I need to aim for a specific point, and start with the exercise. This is a coordination exercise but does it have a name? It's like a dutch roll but I need to stay in the roll axis without moving.

In the aerobatic community we would call that the beginning of an Aileron Roll.

It's harder than it looks.

I've heard this maneuver called a "Dutch roll", but that phrase has other meanings too.   It's arguable whether or not this maneuver is even truly possible, or is based on somewhat of an illusion.  If the plane is banked, and you haven't "unloaded" the wing to zero G's, why would the flight path not start to curve to the side, pulling the nose off the point?  To really do this perfectly it looks like you would have to use the rudder in a way that caused significant sideslip, but that was probably not your instructor's intention. (If it was, disregard all but the first sentence of this answer.)


Why do the spoilers move in this video? In this video when the aircraft is turning left(2:25) the spoilers on the wing move. What causes them to move?

Two reasons that are really one reason. In short: Because lift.

In this case though, it's "to get rid of lift". The spoiler's job is to reduce the amount of lift the wing experiences. This can be done symmetrically, on both wings at the same time, to increase the rate of descent or keep an aircraft from lifting back off the ground when in ground effect, or in this case, it can be done asymmetrically, or only on one wing.

In order to turn, an aircraft has to create an imbalance of force, more on the side outside the turn, less on the inside of the turn. The wing to the inside of the turn needs to drop, but it doesn't really want to...it has lots of air pushing it up (when it doesn't, you will have a very bad day)!

The spoiler in this case is just acting as an extension of the aileron. By moving upward with the aileron itself, it spoils the airflow over the top of the wing a bit, reducing the lift, and causing the wing to drop a bit, raising the other wing. It also acts as a larger surface area for the incident airflow to impart force on, which also pushes the wing down.

On the other side, the aileron is doing the opposite, increasing the wing's effective camber, at least near the tip, and increasing lift. Flaps do the same, but also are usually used symmetrically to increase lift on both wings. In some airplanes, you can have a 'flaperon', which can also be used as an additional control surface while also doing it's flap-thing.

That is controlled by the pilot (indirectly via the computer system) who uses the spoiler to control the turn of the airplane instead of ailerons. Notice that the airplane turns when the spoilers move. It does so by killing the lift on one side by "spoiling" the airflow while the lift on the other wing creates the roll and bank moment with the "unspoiled" lift.


What is the definition of load factor and how do you apply it? I'm a little confused about the definition of the load factor, n, as the definitions I found seem to contradict its application in force and moment balance analysis of an aircraft structure.

The definition given in my university script states that the load factor equals the ratio of all external forces acting on an aircraft minus its weight, to the magnitude of the aircraft's weight; this implies that the load factor is the ratio of all aerodynamic forces acting on an aircraft to the magnitude of its weight.

The Wikipedia definition of the load factor states that the load factor is the ratio of the sum total of all aerodynamic forces acting on an aircraft to its weight.

My script definition therefore tallies with the Wikipedia definition. All good so far.

Now, however, there is also a sentence in my script that states that the load factor is a means of including all inertial loads and gravity loads in observing the loading of an aircraft; this implies that weight is included in observing the loading of an aircraft: this makes better sense to me, since, when expressing the loading of an aircraft, it would be easier to say that the aircraft is now being loaded "this many times its weight", compared to this many newtons; this however, contradicts the definitions above.

And now to the force and moment balance analysis issue, I noticed that all components of an aircraft are expressed in terms of its load factor, which is fine, but then in performing the balance, the weight of the components are not considered. How can you ignore the weight force which is always acting on the aircraft when the load factor (per definition) excludes it? An added confusion is in that lift is considered in the moment balance, with all component's loading expressed in terms of load factor; how can you include lift, when it is already incorporated into the load factor of the components; this implies that the part is further loaded above its current loading.

I would appreciate if someone can straighten up what the load factor means, and how should it be applied in force and moment balance calculations.

The load factor is the total acceleration you feel, pointing downwards. In straight and level flight, the load factor is 1: you only feel the acceleration of gravity. So a load factor 1 equates to 9.81 m/s$^2$ (If gravity was higher, let's say 15 m/s$^2$. load factor 1 would equate to 15 m/s$^2$. But that's another story.)

Image source

Bank the aircraft 60° and fly a co-ordinated turn, and you'll experience a downward acceleration of 2g. This is a case that is easy to understand since it is a static situation with constant velocities. If we have a dynamic sine wave for instance, the load factor would be a function of where we are in the cycle, if the aircraft is accelerating up or down. The actual acceleration is added to the gravity vector.

Load factor does not ignore the weight of the aircraft:

In steady horizontal flight, load factor is 1 because it is the same weight that would show up on a scale on the surface of the earth.
All dynamic accelerations are added to 1. If the aircraft accelerates downward with -1g, the load factor is zero and everyone on board is weightless.
Forces are vector entities. For the load vector, we're only counting the resulting accelerations in the z-axis of the aircraft. After summing the forces in 3 degrees of freedom, the z-axis force F drives the airplane up/downwards with an acceleration of a = F/m.

Another case where definitions and language make for very interesting discussion. So we will start with the universal and unchanging force vector, gravity, and how it loads the plane at rest (parked on the tarmac).

Strap in and let's go up!  G "felt" by pilot=1. Aerodynamic load on wing=0. Gravity load on wing= 1  Total load= 1

Go to steady straight level cruise flight.  G "felt"=1
Aerodynamic load=-1 (it is a flying wing as this is easier for now) Gravity load=1 Total load = 0.
Lift and Gravity cancel out.

Top of 1G loop.  G "felt"=1.  Aerodynamic load=-2 (to create 1G inverted for pilot) Gravity load=1
Total load = -1G (+1G WRT inverted pilot and plane).

Now, at top of loop, go Immelmann and level out inverted. G "felt"= -1 (pilot and plane inverted)
Aerodynamic load=-1.
Gravity load=1.  Total load=0.

Roll to level and land.

The structural stress of the aircraft is the sum of gravity and aerodynamic loads.  This is the important concept as the orientation of the aircraft in the gravitational field can change.

The books can be confusing, approaching from reality can make understanding a bit easier.

Answer has been modified to use positive gravity vector as written by @Guha.Gubin and comments from @quiet flyer.

Invite comments from aerobatic flyers!


Are passive or active headsets preferable in an open cockpit? My first headset used passive noise reduction (i.e. padding) and when I - very quickly - upgraded to an ANR headset, my instructor recommended keeping the passive one "in case you ever fly something with an open cockpit". I can't remember if he gave me an explanation, but he definitely said that passive headsets "worked better" in an open cockpit.

I haven't flown anything open so I've never had the chance to put this to the test, but is it (or was it ever) true?

First we should see how active noise reduction works. The simplest definition would be:

Active Noise Cancelling uses electric signals to reduce unwanted sounds.

The following picture on Active noise control depicts it in a simple manner:

Then we can see what unwanted sounds ANR can actually reduce. As mentioned in A Guide To Buying An Aviation Headset:

  ANR headsets work best at frequencies below about 400-450 Hertz. This
  represents the normal frequency range for speech, and also much
  propeller and exhaust noise is in this region. ANR headsets are
  therefore a significant advantage where intelligibility of
  transmissions is affected by engine and propeller noise.

  In general, ANR Headsets offer greater hearing protection in
  high-noise environments but will do little to reduce noise at higher
  frequencies such as wind or airflow noise.

In a flying airplane (even in a slow flight), wind flow would be quite high. This PDF (page 1, paragraph 3) states:

  The wind noise spectrum is dominated by the lower frequencies (< 500
  Hz), although at 27 mph when saturation is present, the wind noise
  level can be greater than 60 dB SPL at 8 kHz.

Since passive noise control is sound reduction by noise-isolating materials such as insulation, sound-absorbing tiles, or a muffler rather than a power source, in an open cockpit, passive NR headsets work better than active NR headsets.

I would never use active cancelling for two reasons:

Simplicity. Doing anything that involves more complexity, batteries, wires, etc, is a bad idea. Keep things simple.
You want to hear everything. The more you hear, the better. If something is wrong with the engine, you want to know that immediately, not find it out later when the plane turns into a brick. The point of the headphones is just to turn down the noise to tolerable levels, not eliminate it. When flying, hearing noises is good.


Is it possible to recover from a flat spin? In Top Gun, Maverick's F-14 enters a flat spin after flying through another F-14's jetwash, forcing him to eject and causing the death of Goose.

Is this scenario possible or was it only created for theatrical purposes?

If not, how does it happen, and how can it be avoided, especially for other types of aircraft such as the B-2?

The F-14 is special; to stop the flat spin you need to pull on the stick. Then the spin can be recovered from.

This is different from what you learn about regular spins, where pushing the stick would be a better recovery method. In a flat spin, the forward fuselage dominates the aerodynamic forces, and we all know there are no control surfaces which could influence this flow. Wings and horizontal surfaces are fully separated, and elevator deflection is mostly useless. As others have pointed out before, a flat spin can only be stopped by shifting the center of gravity forward or deploying a spin chute. The F-14 was an exception.

Spinning the F-14

Early on, the Navy lost several F-14s due to flat spins. They then studied the phenomenon at Pax River, and in the end Bill Bihrle found out that the elevator shields airflow from the two vertical tails of the F-14 when the stick is pushed, but moves out of the way when the stick is pulled full aft. You have to know that the elevator of the F-14 is a full-flying surface, and the movement range is from -20° to +70°. At +70° it is almost in line with the airflow in a flat spin, and now the vertical tails are no longer in the wake of the elevator. They now can reduce the high yaw rate, which in turn reduces the high pitch-up moment of the rotating fuselage. With the lower inertial pitch-up moment, the elevator then has to be moved back to neutral, and the drag from wing and elevator is enough to pitch the aircraft fully down and out of the spin.

Naturally stable flying wings never enter a flat spin; their spin modes are all fairly steep due to the lack of a strong inertial moment from the lengthwise distribution of masses.

Influence of mass distribution

I guess I now need to explain the inertial moment which causes the high pitch attitude in some aircraft. The axis of the spin rotation is close to the aircraft's nose, and the tail has the biggest distance from this axis. In a flat spin the axis of rotation is even more back, close to the center of gravity. All parts of the aircraft rotate with the same yaw rate, and the centrifugal force from this yawing motion grows linearly with distance from the spin axis. This difference in centrifugal force along the lengthwise coordinate of the aircraft pulls the aircraft in a near-horizontal attitude.

Watch this movie of the spinning XB-70 to get an idea how it looks. Depending on elevon settings, the spin axis is somewhere between the cockpit and the forward tip of the wing triangle. With the damaged configuration (watch without sound for no distraction) you get a typical flat spin with the spin axis close to the center of gravity since there are few suction forces acting on most surfaces - only drag and the nose vortex (more on that below) remain.

How to avoid flat spins

If you have a configuration which is prone to flat spins, you need to change the shape of the fuselage tip. A rotating fuselage tip at high angle of attack will produce vortices at its side, and the yawing moment of these vortices will grow stronger with increasing yawing motion. This is why a flat spin is self-stabilizing. One way to suppress these vortices is the placement of small spoiler strips along the nose (which fixes the vortex position and reduces this self-stabilizing effect), and another is to make the nose shape flatter.

Unfortunately, this can only be done long before take-off.

Answer to your question

Regarding the "Top Gun" scenario: If the F-14 is at high angle of attack and the jet wash hits it asymmetrically, the airplane could enter into a spin. But from there it is still some way to the fully developed flat spin, and I would expect that a skilled pilot could recover from this momentary upset.

Flat spin is worst case scenario of pilot mistake and fatal for all types of aircrafts, except fighters, and with thrust to weight ratio of no less, than 0.72-0.75. Thus even military A10 will be lost if it happens, let alone any passenger aircraft. Modern aircrafts have CG moved to nose with massive engines though, and have a chance to move nose down, if at early stage of flat spin. If developed, and plane goes as falling leaf, or rotating stbilised,theres no chance to recover.


Why does this calculation show Gustave Whitehead's propellers were more than 100% efficient? I have tried to calculate the efficiency of Gustave Whitehead's propellers using the information he provided to the American Inventor Magazine in 1902 (see the citation below).
Number of propellers = 2
Engine power per propeller, $P$ = 20 hp
Propeller diameter, $D$ = 6 ft
Static thrust per propeller, $T$ = 254 lbf
Propeller efficiency, $\eta_{prop}$
Gear efficiency, $\eta_{gears}$ < 100%
Air density, $\rho=1.2 \frac{kg}{m^3}$
Doing the calculations: $\eta_{prop} > \sqrt{\frac{T^3}{(P^2 \cdot \pi \cdot \eta_{gears} \cdot \frac{D^2}{2} \cdot \rho )}} = 101.4\%$ which is impossible.
Is there a mistake? If yes, where?
Note: I used the formula that relates static thrust to efficiency, power and propeller diameter.

The Whitehead Flying Machine
Has the End been Finally Attained, and is the Dirigible Balloon to Go?
Editor, American Inventor
Dear Sir: Replying to your recent letter, I take pleasure in sending you the following description of my flying machine No. 22, the latest that I have constructed:
This machine was built in four months with the aid of 14 skilled mechanics and cost about $1,700 to build. It is run by a 40 horse-power kerosene motor of my own design, especially constructed for strength, power and lightness, weighing but 120 pounds complete. It will run for a week at a time if required, without running hot, stopping, or in any possible manner troubling the operator. No electrical apparatus is required for ignition purposes. Ignition is accomplished by its own heat and compression; it runs about 800 revolutions per minute, has five cylinders and no fly-wheel is used. It requires a space 10 inches wide, 4 feet long and 10 inches high. ...
The propellers are 6 feet in diameter and have a projecting blade-surface of 4 square feet each. They are made of wood and are covered with very thin aluminum sheeting. The propellers run about 600 revolutions per minute under full power and turn in opposite directions. When running at full speed they will exert a thrust of 508 pounds. I measured this thrust by attaching the machine to a post by means of a dynamometer and running the engines at full speed. ...
I have no photographs taken yet of No. 22, but send you some of No. 21, as these machines are exactly alike, except the details mentioned. No. 21 has made four trips, the longest one and a half miles, on August 14, 1901.   The wings of both machines measure 30 feet from tip to tip, and the length of the entire machine is 32 feet.   It will run on the ground 50 miles an hour, and in air travel at about 70 miles.  I believe that if wanted it would fly 100 miles an hour. The power carried is considerably more than necessary. ...
Trusting this will interest your readers, I remain, Very truly yours,
GUSTAVE WHITEHEAD
Bridgeport, Conn.

Source: The American Inventor Magazine
, 1 April 1902
.

The equation you referenced shows prop diameter squared, divided by 2, not 4.

Even then, the calculation shows that marketing is to experimentation as about 2:1 :)

I used the formulas from my answer to What is the correct formula to calculate propeller efficiency? and arrived at the following:
we have:

$\beta$=0,
$P$= 20 hp=14,914 W,
$D$=6 ft=1.8±0.3 m,
$T$= 254 pounds=1130 N,
$\rho=1.2 \rm\ kg/m^3$, yielding

$$T_0=\left[\pi\times 1.2\times 14,914^2\times 1.8^2\right]^{1/3}\rm N=1,395±232\ N,$$ with a propeller efficiency of $\eta_{\rm P}$=(1,130/1,395)=81±13%, which seems to me to be a reasonable efficiency.


How frequently do civilian aircraft have mechanical issues to which passengers are oblivious? I know that aircraft are pretty well built and maintained, but I am certain that while flying these aircraft also suffer from some or the other mechanical issues. I believe redundancy plays a huge part in keeping these aircraft in air, but how many times are the passengers really oblivious to the problems? What are some of the most common problems that pilots are used to?

I don't think you'll find a definitive answer because most will be just considered routine items, but if you're including non-safety-critical items then I imagine the answer is "lots". Passenger aircraft are huge, complex, mechanical devices with many many parts. It is accepted and expected that some of these things will occasionally fail.
Most of these can fail gracefully without impacting the flight in any way beyond a note in the tech log. Not only that, but lots of things can be known to have failed but still allow the flight to go ahead as normal.
You may wish to explore the idea of Minimum Equipment Lists. These provide exhaustive lists of items which can fail and flights can still be dispatched as normal. Note this isn't the same as an in-flight failure occurring and continuing to the destination - items in these lists can be failed before you even take off.
For example, here's one for the Airbus A320: https://fsims.faa.gov/wdocs/mmel/a-320%20r21.pdf
Everything listed in that document can be failed in some way, but a flight still operate as normal (with exceptions, as per the remarks column)
For example, and to highlight the level of detail, an Airbus A380 does not need to be grounded because the "PA In Use" notification light doesn't work.

It used to be normal for the airliners of the day to have multiple non-critical fault alarms active at takeoff. Airlines tended to take the view that if an alarm is non-critical then there is no problem, no need to hold the aircraft back. Pilots complained that there were often so many that spotting a critical warning among them could become a problem. Pilots also often disabled them (especially audible bleepers, buzzers and such) to reduce the level of distractions. Keeping the level to manageable proportions is an important part of modern flight deck instrumentation design; the fault indicattions are still there, but not intrusive. Different airlines tolerate different levels of fault indication.
I was once a passenger in a venerable Dakota, still being operated by a commercial airline. Several rows of rivets were missing from the trailing edge of one elevator. Most had been replaced by twisted wire, but some holes were just left. It had evidently passed inspection nonetheless. Most passengers never noticed, and the trip was uneventful.
The general approach to passenger information is often, as they say, "what the eye don't see, the heart don't grieve over".


Are there countries where commercial flying can be conducted with a Mexican DGAC private pilot license? I have a Mexican DGAC licence. Can I work in other countries, or do I need to get another licence? (for example an FAA one if I want to work in the US)

No. If you want to make commercial activities, you need a commercial license.

I remind you the definition of commercial flight


Is there a technical name for a several mile stretch of flight before landing? A friend of mine lives exactly 5 miles from a small but busy airport that accepts commercial traffic (small commercial jets). Yesterday I noticed that many airplanes were flying low right over his house on the approach to the airport. They seemed to be following the same route. The main runway points in the direction of this house. Is there a technical name for the several mile 'air road' that leads to the airport runway?
Edit: Airport is Teterboro Airport (TEB).

The best term to describe it would be the 'final approach segment', which is defined by ICAO as:

  This is the segment in which alignment and descent for landing are made. Final approach may be made to a runway for a straight-in landing, or to an aerodrome for a visual manoeuvre.

The FAA defines the same as:

  The FAS begins at the PFAF and ends at the landing threshold point (LTP).  The FAS is typically aligned with the runway centerline extended. The segment OEA is divided into the FAS OCS and the visual segment obstacle identification surface (OIS).

Note: Both Eurocontrol (which refers to the ICAO document) and Jeppesen define it only for instrument approaches.

A word I hear often is "approach corridor", which kind of describes what you're talking about.


Can the removal of a duty-free sales trolley result in a measurable reduction in emissions? Recently, SAS announced that:

  Withdrawing tax-free sales will reduce the overall weight of the aircraft, which in turn will reduce fuel consumption and emissions.

I could buy that this might be a polite way to say "we don't make any sales, consequently, the trolley with the items is just ballast; throw it away". It's not the first time that I hear it though: some time ago I remember an article in Aegean's Blue magazine about saving 15 tons of CO2 in one year just by replacing the pilots' flight bags with the electronic equivalent (EFBs). Unfortunately I don't remember in which issue was that.

Now assuming that a trolley weights approximately 25kg empty so let's say 30kg with the merchandise. The L/D ratio of a an Airbus A320 is 16,3 and thus a 30kg reduction in the weight will result in approximately 18,05N reduction in drag and and an equal reduction in the thrust required to maintain the same speed.

Compared to the ~150kN that 2 CFM56-5B4 produce in cruise, The 30 Kg reduction in weight will result in 0,00012% a 0,038% to 0,06% reduction in required thrust (depending on the aircraft gross weight), which seems like a drop in the ocean. Can this make a difference measurable by an airline company at the end of the month/semester/year/century?

Yes, the removal of payload will result in a measurable reduction in emissions.

First, I have to correct some math: the engines don't run at 150 kN in cruise - that would indicate a L/D ratio of only 4-5, since a typical A320 weighs around 60 tons mid-flight (give or take 10). Fuel consumption is roughly proportional to weight, so 30 kg out of roughly 60,000 kg is a 0.05% reduction, not a 0.00012% one.

As to whether one would be able to move the throttle down exactly 0.05% to benefit from the reduced weight and leave everything else equal, no, not even the autothrottle is so precise.

But that doesn't mean fuel consumption is unaffected: the balance of an aircraft's weight, lift, drag and thrust always affects its energy. At the exact same throttle setting, the aircraft will climb a fraction of a percent faster, and so finish its climb and go to lower fuel burn at cruise a few seconds sooner. In cruise, maintaining the same flight level, it can spend a minute or two at a bit lower throttle, or trimmed a bit down, which will make it arrive a few seconds sooner.

This is fundamental conservation of energy: if less of it is expended in drag, then either the engines have to work less, or the aircraft will climb or accelerate. This happens all the time - an airliner in level flight doesn't stay in perfect equilibrium throughout, but corrects for energy changes every once in a while. Even if it doesn't change how much fuel is pumped into the tanks that particular day, the effect is cumulative across all sources and persistent through random variations.

Overall the effect of weight on fuel burn is proportionate. Depending on the distance, everything aboard an aircraft can consume 10% to 50% of its weight in fuel each flight. A very rough ballpark is that, over the life of an A320, each pound of operating empty weight will cost an extra ton of fuel.

To be more specific, given the A320's fuel burn of roughly ~2,500 kg/hour, adding the trolley requires an extra 1.25 kg of fuel per hour it's in the air. For a 4-hour flight, that's an extra 5 kg of fuel. Over the 60,000 flight-hours an A320 will serve in total, that's 72,000 kg of fuel. So behind that trolley is 3 large tank trucks' worth of fuel to keep it in the air, and ~225,000 kg of CO2 emissions.

As an aside, in-flight sales of random junk really serve the airline, not the passengers. The profit margin on these items can be up to 100 times the tiny profit margin on economy tickets themselves. With airports packed full of duty-free shops, there's no shortage of shopping opportunities on an air trip. Cramped and burning fuel for every item on board, whether it sells or not, aircraft don't make for a very efficient or practical storefront.

Airlines only make up for this inefficiency by exploiting their cabin crews as free salesmen. Of course, airlines' environmental consciousness tends to correlate strongly with the price of jet fuel, so once sales profit no longer covers the extra burn, it's time to go green.

Based on the other answers, I think the actual answer is No, the difference to emission is not measurable. None of the instruments concerned are sufficiently accurate.

I don't think the engines even have instruments to measure CO2 emissions. The fuel gauges may have enough digits to display a 0.05% change, but they probably aren't calibrated that accurately. Even if you did fit super-precise instruments, you couldn't control for all the other factors that effect fuel consumption.

However, the difference is still possible to estimate, as proved by the people who have done so in this question. When multiplied over a large fleet and a year of operations, it's probably quite accurate.


Would a lighter aircraft require the same amount of power than a heavier one to fly at the same speed? I had a question come up:

  If aircraft A is identical to aircraft B, except that aircraft B weighs 200 kg more. Which of the following statements are true?

The answer I selected was wrong:

  Aircraft A would need less power to fly at the same speed as aircraft B

I pictured that with less weight, a reduced angle of attack would result in less drag and hence less power? Would anyone be able to “lighten” me with this?

Thanks for the feedback, it may have just been a typo..
The 4 multiple choices where:
1) Aircraft A would need less power to fly at the same speed as aircraft B
2) Aircraft A would need the same power to fly at the same speed as aircraft B
3) Aircraft B would have poorer climb performance
4) Aircraft B would require a longer runway for landing

You are right. The heavier aircraft needs more power. With one exception.

The power needed to keep an aircraft in the air is the product of flight speed and drag. When flying at the same speed, the drag of both aircraft is crucial to answer the question. Since both share the same geometry, it is safe to assume that both have about the same friction and parasitic drag. But the heavier aircraft needs to fly at a higher lift coefficient in order to carry its higher mass, so it will create more induced drag.

The one exception is when both aircraft have a highly laminar airfoil with a pronounced drag bucket. If the flight speed is picked such that the heavier aircraft will still fly within this drag bucket but the lighter aircraft will fly at a lift coefficient below the lower boundary of that drag bucket, the difference in friction drag could be higher than the difference in induced drag (which is small at high speed). But that exception would need to be made clear in the question.

To fully judge the correctness of the test, please provide all answers. As @David K suggests, your error might have been not to select all correct answers.

At constant speed the forces of thrust and drag are equal. Since F=m*a, the ratio between accelerations due to these forces is the same regardless of mass. Therefore, they equalize at the same velocity (drag is dependent on velocity) - no more thrust is needed to counteract decceleration due to drag


Is the Meredith effect utilized in the cowling design of air-cooled engines? The P-51 is famous for its advanced-for-its-time aerodynamics, the prime example of which is the radiator design under the belly which utilizes the Meredith effect to, in effect, turn it into a mini jet engine. Now, the cylinder walls and cooling fins of air-cooled radials are much hotter than whatever cooling liquid that's used in the P-51's water cooler, so is it possible to produce a similar cowling for the air-cooled radials? Is there any example of such a cowling on WWII airplanes?

The Meredith effect is to recover radiator drag by using engine heat for thrust, just like a jet engine.  It worked very well in the P-51 Mustang, actually adding a few knots of airspeed.

It also would work well with any source of heat, including a nuclear reactor, so yes it would work with aircooled engines.

The issue is drag from the radiator.  Aircooled radials have much greater frontal drag, so it worked, just not as well.  Efforts to reduce drag in the B-36 bomber resulted in roasted engines, particularly the rearward banks of the "corncob" R-4360 radial.  Overheating is a major concern, as is drag.  Engineers try to find a good balance.

I had to go look up what the "Meredith Effect" was (and edited your question to include a link to Wikipedia).

As I was reading through the Wiki article, I found the answer to your question:

  Many engineers did not understand the operating principles of the effect. A common mistake was the idea that the air-cooled radial engine would benefit most, because its fins ran hotter than the radiator of a liquid-cooled engine, with the mistake persisting even as late as 1949.
  (Source)

So no, the Meredith Effect will not work on an air-cooled engine.


What are the consequences of hail damage to a small GA plane? I recently found a Cessna 172 that had been totaled by the insurance company because of hail damage.

The PDF from that page has detailed pictures which show hail-divots all over the the wings, body and empennage.

Can the plane still fly like this?

What effect is this likely to have on the flight characteristic of the plane?

Like anything in aviation, it depends...

There are a lot of discussions on the internet about it you can find

Here (mooney)

Here (cessna)

And Here

It is a case by case situation and depends on the extent of the hail damage and what/where the damage occurred. In many cases the issue crops up that you can not simply bang the dents out you would need to reskin the panel. This can, depending on the aircraft be a costly process.

There are planes out there flying with light hail damage and it causes no problems, in many cases its purely cosmetic. A lot of planes flying out there have other similar damage (tool drops, bird strikes, impacts from rocks etc) and fly just fine.

An good A&P would most likely be able to tell you if the particular Cessna in question is still airworthy.

Airflow over the wing would be so disrupted that it'd be unsafe to fly, assuming it got off the ground to begin with. It might even be able to take off, but the critical speeds and handling qualities would be so far out of design conditions that you'd have no way of knowing what the airplane would do next.


In the ARINC specification, why is baro altitude (label 204) measured in Knots? I am currently designing a system that reads the baro altitude (on label 204) from an LRU. Ultimately in my software I would like to display the baro altitude reading in real-time using a graphical interface. In doing this, I'd like to have the units of measurement next to the reading. However I am having a bit of trouble in deciding how to display the units of measurement.

In the ARINC specification (attachment 2 "data standards"), label 204 (baro altitude) is listed as having units of knots. Is this an error, or can altitude be in Knots?

I am referencing ARINC specification 429 part 1-17, mark 33 digital information transfer system (DITS) published in May of 2004.

It's an error. Attachment 2B does show that label 204 from the GNLU (Equip ID 056) and GLU (Equip ID 060) as being in knots (highlighted below). That is not correct, it is a BNR in feet as shown for Air Data System (ID 006) and ADIRU (ID 038). Knots only applies to Label 204 from the FMS (ID 002).

It's still incorrect in A429 P1-18, March 2012.

ARINC 429 contact at ARINC Industry Activities, ARINC 429 Maintenance is Jose Godoy. email:jose.godoy@sae-itc.org

Please read all the explanation to understand how the same label might refer to different parameters, therefore to different units.

As a start the label uses only 8 bits which gives a space for only 256 different labels ( 377 labels in octal base), this is obviously insufficient to cover a huge number of parameters of a modern aircraft.

To increase the number of possible parameters, and to cover the complexity of modern systems, the same label according to the emitting computer might refer to a different parameter; more precisely, different equipments might use the same label to send different parameters even though using the same label.

Please refer to the following Website

http://read.pudn.com/downloads111/ebook/462196/429P1-17_Errata1.pdf

You will find that depending on the emitting equipment the same label might refer to : utility airspeed, or baro corrected altitude, or duct temperature, etc etc therefore the units will depend on the equipment that transmits the label.
Label 204 could be send by the flight management, or by the air data computer, or by the fuel and quantity computer, etc, etc.

As a conclusion the same octal label do not refer always to the same parameter.

Is this confusing?

No because a sink computer knows on which bus he is reading and therefore knows the equipment that is sending the data.


Why do some aircraft have multiple switches for the landing lights? A question just popped into my mind: Why does an A320 overhead panel have 2 landing lights switches? What's the logic behind that? Why isn’t there only a single switch to turn on both landing lights?

On a dark morning preflight check, I can verify that the lights all work by turning each one on, one at a time, without leaving the flight deck. Less lifespan on the bulbs & less disruption to the ramp to have several quick flashes of light within a few seconds, than turning everything on, walking outside to observe everything, and then going back inside.
Plus, it's probably a hold-over from days of minimal automation (think, 727 Flight Engineer panel - now entirely replaced by automation & a few switches on the overhead panel) when each {whatever} had its own switch.
I could get creative & invent another scenario or two when I want "this" but not "that" light on, but the preflight is probably the best example of that.

Most of the jets that I fly have their landing lights wired through the down locked switches on the main gear.  This means the landing light automatically turns off as the gear is raised.
It also means this is one of the ways the crew can determine if a main gear is down and locked if they have conflicting information in the cockpit.  It is easier to turn off all lights except the one landing light in question and ask tower if they see any lights than to have both landing lights on and ask the question if they see one landing light.  There is no ambiguity in the first question.  There could be in the second question.


Is it possible to takeoff without flaps extended? related questions:

landing wihtout flaps
takeoff with full flaps

My question focus on passengers jets (goes from Embraer E-jet family to  Airbus A380). I don't know why pilots would attempt to takeoff without flaps but lets say they will.

Considering all the other parameters are good enough (no crosswind, long runway, aircraft light enough,...) and the pilots are aware of this unusual takeoff configuration and adapt their action accordingly (including increasing $V_r$ and $V_2$), is it possible for any passenger jet to takeoff and stay airborn (not like the flight spanair 5022) without extending flaps?

Yes take-off without flaps is possible. The Airbus A300 and Boeing 767 are approved for such take-offs and it is being done regularly. It results in a better climb gradient, especially with one engine out. The one engine out climb gradient is an important and sometimes limiting factor in take-off calculations.

One important constraint is the length of the runway.

Another limitation, perhaps less obvious, is the maximum rolling speed of the wheels. The tires are rated to a limited speed which may be exceeded by a flap-less take-off.

You may be able to take off but as soon as you reach a height of 40-50 feet you lose wing in ground effect and will crash unless you have reached actual free air flying speed. Assuming a completely unobstructed route ahead of the runway the plane will eventually get enough airspeed to fly normally without flaps.


What is the highest possible thrust generated by 15cm ducted fan? I recently developed a sudden interest in flying. I'm wondering whether a pair of small ducted fans could lift a person off the ground.

Lets say the ducted fan is 15cm in diameter. What would be the most thrust a fan like that could produce?

A 15 cm diameter fan could maybe lift 4 kg of mass with a tip speed of 0.6M.
By increasing the power and the blade chord length (in other words, increasing the solidity of the rotor), this may be increased to maybe 6 kg or more, but a single fan would probably never lift more than 10 kg mass.
The thrust of a rotor is $C_T \cdot \rho \cdot \text{Area} \cdot \text{tip speed}^2$.

$\rho$ = air density ($1.225~\mathrm{kg~m^{-3}}$ at sea level).
$\text{Area}$ = area of rotor disc ($\mathrm{m^2}$)
$\text{tip speed}$ = the linear speed at the tip of the rotor = radial speed * radius.
$C_T$ = coefficient of thrust. (is usually in the order of 0.02 to 0.05 for small rotors)

Assuming $C_T$ of $0.03$, the calculation shows $40~\mathrm{N}$ ($=4~\mathrm{kg}$) thrust.
Disc area is one of the most important parameters, even a 30 cm disc would generate 4 times the thrust. 60 cm would make almost 65 kg.
Apparently, a 50 cm diameter disc can possibly generate 45 kg thrust on its own. So two of them could carry and maybe even lift-off a person equipped with this backpack (the person + the system must be less than 90 kg).

Thrust (in lbs) = 9.35(horsepower x diameter of ducted fan in feet)2/3 [power of 2/3]

This is the formula I recall from the book.

Assuming you design your fan for high static thrust:

for a 6 inch diameter to produce say 300lbs of thrust (lifting man and 100lb machine) you will need a 360hp powerplant.


Why do gusts change the airspeed during landing, but turning (which changes where the wind is coming from) does not? As we know it, aircraft are extremely sensitive to gust during its final landing stage. They can lose airspeed if the wind changes to tail, ad thus loose lift or need to adjust their thrust in order o regain the lost speed. On the other side, the opposite happens when encountering a headwind.

On the contrary, we don't take that into account when we consider a fighter jet. When a jet fly's at 20,000 with headwind of 50 knots, turns 90 degrees to the left and has the wind from the right- is it considered to be a change in wind direction or because the whole system is moving along with the plane and the change is more gradual- the aircraft feels no change in actual lift which is derived from the true speed it feels?

Before I get into the question about the turning fighter jet, a few words about airplanes' being extremely sensitive to gusts during final stages of landing. If this was actually true, I'd be dead by now. Seriously. Airplanes are sensitive to gusts during landing, I'll give you that, but I think it's exaggerating beyond reason to say it is extreme.

Now, back to one of the most persistent myths in aviation:

The myth of the downwind turn

The jet in your question does not make a full turn from headwind to downwind, but it does not matter, the physics are the same as in the myth of the downwind turn.

As you stated the plane is moving along with the airmass. So, for the plane, aerodynamically there is no head-, side-, or downwind. Only us on the ground can feel it, as we are locked into to the coordinate system of the earth with our feet.

The airplane is locked into the coordinate system of the airmass, and as long as the airmass itself does not experience acceleration, the airplane flying in it can turn in whichever way, and it will not notice any change in the wind direction.

The direction of the wind is locked into the coordinate system of the earth, so in your example there is no change in wind direction. Not for the jet aerodynamically speaking, and not in the sense of navigation (relative wind changes, but you use the same formulas for the calculations anyway). What does change is the groundspeed of the aircraft, and this is the reason wind (and gusts especially) causes some concerns for landing planes. Takeoff is also affected, but not quite as much as landing.

Ok, I was wrong. In order to better understand what is happening, other than saying that the plane flies embedded in air mass, it is easier to understand that the turn, no matter how fast it is- maintains equilibrium. meaning- it is not abrupt in no time, but a collection and a gradual turn into the tailwind or vice versa. This is easier understood by looking at each particle of air in which the jet fighter is flying. every second millisecond etc.- that the plane turns, the particles travel with it by the same velocity in relation to the ground, making it move altogether. The situation would be different if the aircraft turn 90 degrees in no time, making the air around it change direction in no time- like happens in a windshear or inversion.


Why do privater jets often fly at or above FL 400, but large commercial aircraft are usually in the range of FL 300 to FL 400? With rare exceptions (like the Embraer business jet, which is simply their commercial jet with a different interior), almost all private jets fly in the more rarified air above FL 400.
But, if its more efficient, why don't airliner fly at the same sorts of altitudes on long flights?

FL360 (plus or minus a bit) is the optimal altitude for fuel efficiency, which is the top priority for airlines. And airliners all fly at roughly the same speed (again, for fuel efficiency), so ATC can just line them up in a row and let them follow each other.
The problem is that level is severely congested with all those airliners. You can’t really get out of line or pass someone, which is another reason they’re all flying the same speed—and that has become a vicious cycle because Boeing and Airbus now design their planes to be most efficient at that speed.
Business jets passengers (and thus their operators) value speed over efficiency; if they didn’t, they’d use much cheaper airline flights. However, to avoid the airliner congestion around FL360, they have to climb to FL400 or higher, where they can fly faster and likely get a more direct route.

The most interesting (and most likely the most right) is the safety issue.  An airliner with engines under the wings needs to drop to FL200 in case there is an uncontained engine failure that punctures the fuselage, requiring a descent rate of 10.000 fpm or more.
A private jet, however, has engines in the back. If an engine had an uncontained failure it would be behind the pressure bulkhead, and would not require a rapid descent due to loss of cabin pressure.
Having that - as well as turbojet engines that are low bypass and thus perform better at higher altitudes than a fanjet - makes it possible for the users to fly regularly above FL 400, where there is coincidentally less traffic as well.  And, the relatively large wing aids as well.


What are the symptoms of vortex ring state? What are the signs and symptoms a pilot can recognize, that the helicopter is approaching vortex ring state (VRS) and measures should be taken?

A related question discusses how to prevent or recover from VRS, but it does not address the recognition technique. Similar to stalls (which happen unintentionally), being able to identify a worsening situation before the stall fully develops is better than stall recovery.

Vortex ring state is characterized by rapid descent and reduced effectiveness of control inputs. At the onset, it is characterized by increased vibration and buffet of the airframe, with uncommanded changes in helicopter attitude. Also, any increase in the collective is ineffective and counterproductive.

From skybrary.aero:

  Incipient vortex ring conditions are typically:

  increased vibration and buffet,
  the onset of small amplitude ‘twitches’ in pitch and evidence of longitudinal, lateral and directional instability.

  Established vortex ring conditions are characterised by:

  a very rapid increase in rate of descent towards and beyond 3000fpm,
  reduced effectiveness of cyclic inputs in roll or pitch
  the application of collective pitch failing to arrest the rate of descent and usually increasing it.

From FAA Helicopter Flying Handbook- Helicopter Emergencies and Hazards:

  A fully developed vortex ring state is characterized by an unstable condition in which the helicopter experiences uncommanded pitch and roll oscillations, has little or no collective authority, and achieves a descent rate that may approach 6,000 feet per minute (fpm) if allowed to develop.

The best measure against vortex ring state is quite simple- go forward i.e. increase airspeed. Basically, you enter into forward flight so that the rate of descent doesn't oppose the induced flow. You need 'undisturbed' airflow over the rotor disc. Power is to be applied after gaining sufficient airspeed. From skybrary:

  Incipient Stage ... Keep the collective position unchanged and apply forward cyclic to achieve an accelerative (nose down) attitude so as to increase forward airspeed quickly. As soon as a steady increase in airspeed is indicated, and above 30 KIAS, more power can be applied if necessary without waiting until the best rate of climb speed is reached.

  If this action does not resolve the situation rapidly then it is best to treat the condition as established and take the actions below.
  Established Condition. Recovery can only be effected by changing the airflow around the rotor and will inevitably lead to significant loss of height, which makes recovery from a low level occurrence impossible.

  There are two theoretically possible actions: Moving the cyclic forward and lowering the collective. Combining these actions is likely to produce the quickest recovery with the least height loss.

  Application of forward cyclic should increase airspeed but a large input held for several seconds may be required before significant pitch attitude and consequent speed change is achieved, with a significant nose down attitude resulting. Lowering the collective to reduce power towards auto-rotation, so unstalling some of the inboard portion of the blades, may also be effective but forward airspeed must be gained before power is re-applied during recovery.

Autorotation has also been suggested as a measure for overcoming the vortex ring state (by the FAA handbook, for example), but it may lead to further loss of height, which has to be taken into account. Another issue is that there are some variations between the different helicopter types- in case of tilt-rotor aircraft like V-22, the only way is to go into forward flight.

VRS is settling of a helicopter in its own downwash. Settling rates can be as high as 6000fpm.
To develop VRS you must have all three of the following conditions
1. Speed < ETL.
2. Vertical speed > 300fpm
3. Power on between 20-100%
If any one of these conditions is missing, NO VRS settling.

The best technique to recover from VRS is the Vuichard Recovery. Which is a power ON Side slip to the Right (Counter rotating rotors) or left for Clockwise rotating rotors.


Do aircraft manufacturers like Airbus or Boeing charge the airline, when they assist in aircraft accident investigations? Most of the time, when an aircraft crashes or has an accident, the manufacturer also assists in the investigation. Is this a pro-bono kind of effort on part of the manufacturers, or do they charge the airline for it?

Investigating agencies are in charge of most air accident investigations not involving criminal acts. In the US this is the NTSB, which is required to investigate air accidents per US Code Title 49, Chapter 11. These agencies exist to provide independent investigation of accidents and provide recommendations to regulators.

The investigating agents will do their best to understand the circumstances and causes of an accident. Agents of the NTSB have authority under §1134 of the above mentioned law to gather information about an accident.

  (a) Entry and Inspection.—An officer or employee of the National Transportation Safety Board—
  (1) on display of appropriate credentials and written notice of inspection authority, may enter property where a transportation accident has occurred or wreckage from the accident is located and do anything necessary to conduct an investigation; and
  (2) during reasonable hours, may inspect any record, process, control, or facility related to an accident investigation under this chapter.

Investigations can also benefit from working with people that are already familiar with the critical elements. As part of the investigation, the agency may invite various parties. This could include airlines, manufacturers, air traffic control, workers unions, other agencies, or anyone else with information or expertise that could be helpful. If these parties do not abide by terms for participation, this status can be revoked. The resources spent by parties during an investigation can in some cases be fairly intensive. For example, while supporting the investigation of the crash of British Airways Flight 38, Rolls-Royce and Boeing ran full-scale tests of the engine and the fuel system designs.

Investigating agencies are funded by their governments. The Aviation Safety department of the NTSB has an annual budget of about $50 million. Parties to the investigation represent their respective parties and are paid by their party to do so. Part of the party agreement acknowledges this:

  Participation as a party to an NTSB investigation is a privilege and confers no rights or benefits.

They "pay" that parties get is good press for assisting in the investigation, and helping to both prove and improve the safety of their products and services.

Manufacturers are not necessarily involved in the investigations of an airplane crash while they may lend a hand they have no jurisdiction to cary out an investigation (generally speaking).

This depends on where the accident occurs as well as the aircraft involved. Aircraft Accident investigation is generally the responsibility of the nation and national organization that belongs to it, in which the accident occurs. However in practice not all nations have the facilities or resources to carry out an investigation completely so there is a lot of coordination. The NTSB will generally lend a hand if an American aircraft built or registered is involved or help is requested for something like a blackbox read.

  If an accident or serious incident occurs in a foreign state involving
  a civil aircraft of U.S. Registry, a U.S. operator, or an aircraft of
  U.S. design or U.S. manufacture, where the foreign state is a signator
  to the ICAO Convention, that state is responsible for the
  investigation. In accord with the ICAO Annex 13 SARPS, upon receipt of
  a formal notification of the accident or serious incident that may
  involve significant issues, the NTSB may designate a U.S. Accredited
  Representative and appoint advisors to carry out the Obligations,
  receive the Entitlements, provide Consultation, and receive Safety
  Recommendations from the state of occurrence.

These agencies are government run/funded. The agencies can call on the manufacturer for advice/help such as requesting information on the aircraft or design. Although much of this data is submitted during the type certification process and already on hand.

Often times the manufacturer will offer to aid if at all possible. One thing to consider here is that his can be a conflict of interest. Most countries treat accident investigations as an open crime scene. Since the cause could potentially be a design flaw it is in the best interest of the investigating body to keep makers out of the investigation so as not to sway it.

Boeing does how ever have a team of investigators (that it presumably pays internally) that they send to accidents involving Boeing aircraft.

  International protocol, defined by the International Civil Aviation
  Organization treaty, establishes that a government with jurisdiction
  over an accident site leads an investigation. Accredited
  representatives from the countries of the airplane’s manufacturer,
  designer, operator and registry are invited to join the investigative
  team. For international accidents involving Boeing airplanes, the
  accredited representative is the U.S. National Transportation Safety
  Board. The NTSB leads all domestic investigations. Boeing’s role is to
  support the NTSB as technical adviser.

In a legal sense the NTSB is actually Boeings seat at the international table when it comes to Investigations. While the NTSB can (and often might) call on Boeing directly they are ultimately responsible for the investigations as well as their cost.


As a plane gets slower, why does a certain bank angle make you turn faster? In the video "Maneuvering during Slow Flight" the narrator states that while flying slow the airplane will be less responsive to aileron and other control inputs. He also says that the plane will turn quicker at a certain bank angle than it would if the plane was flying at normal speed. My question is: why does the plane turn quicker when flying at a lower speed?

So basically what I am asking is, as a plane gets slower, why would it start turning faster then it was when in fast flight?

The part of the video I am confused about is at 1:16

Rate of turn is dependent on the following two items:

The horizontal component of lift (centripetal force)
The tangential velocity of the aircraft (true airspeed)

The rate or turn is directly proportional to the horizontal component of lift and inversely proportional to the tangential velocity of the aircraft.

For a given angle of bank, the vertical and horizontal components of lift will be the same, regardless of airspeed in level flight.

Consequently the airplane will experience the same centripetal acceleration, regardless of airspeed.

Since the tangential velocity is slower, any kind of centripetal force will produce a greater rate of turn for a slower flying aircraft as opposed to a faster moving aircraft and this can be shown by the centripetal acceleration equation

$$a_c = \frac{v^2}{r}$$

so both slow flying airplane with a true airspeed $v_s = 100$ knots and fast flying airplane with a true airspeed $v_f = 200$ knots experience the same centripetal acceleration.

$$\dfrac{v_s^2}{r_s} = \dfrac{v_f^2}{r_f} = 4\ \dfrac{v_s^2}{r_f}$$

or, $$\dfrac{1}{r_s} = \dfrac{4}{r_f}$$

Consequently  $r_s < r_f$; in this case $r_f = 4\ r_s$

Since the angular velocity is equal to the tangential speed divided by the radius.

$$\omega = v/r$$

the angular speed of the slower aircraft will be greater than the faster aircraft.

$$\omega_s = v_s/r_s$$

and

$$\omega_f = \dfrac{v_f}{r_f} = \dfrac{2 \  v_s}{4 \ r_s} = \frac{1}{2}w_s$$

So our twice as slow airplane turns twice as fast as the faster one does under these conditions.

A slower plane at a constant bank angle will have to fly at a higher AOA to maintain the same
amount of lift than the faster one.  At a constant bank angle vertical and horizontal lift components are distributed in the same ratio fast or slow.

Why does the slower one turn quicker?  Because of increased drag at higher AOA, there is more resistance to forward motion, while the side profile is identical to the faster one.

The slower one has equal lateral acceleration to the faster one (horizontal lift component) but less forward motion.  The resultant vector (direction of flight) is more in the direction of the turn.  Notice lateral G forces are the same.  In order to do this the faster plane must turn slower!

Also skewing the result in favor of the slower, increased elevator deflection, turned sideways in the bank, helps turn the aircraft faster.

Finally, the prop blast over the tail surfaces (thanks John K) also creates greater turning forces.

The wing pulls the plane sideways, the tail turns it.  Add power in a slow turn.