Do any airlines publish their policy on CRM / Do Not Pair? A question came up recently on Skeptics.se regarding the implementation of "Do Not Pair" lists in Airline operations. I am skeptical of the answer given (irony?) which ascribes the implementation of the Do Not Pair list to the accident on board Northwest Airlink Flight 5719 - an incident which was in part caused by bad CRM.

However, the NTSB report does not suggest any sort of changes to the pairing of Captain/FO's which is what makes me skeptical that anything such as this actually exists within the industry.

Is there any documented evidence that Captains or FO's can in any way affect who they get paired up with? A single example would suffice.

I worked in the industry for 16 years with multiple airlines.

In two of these we had 'do not pair'. Most airlines use computer systems to do their crew planning. In the systems I have used, you can enter against each crew member, others they should not fly with.

The decision as to the use of this laid with Fleet management or Cabin Crew management and was only used in extremist.

In both airlines however, this was not a documented procedure.
Hope this helps.


What is the "Centre of Pressure"? This question may sound naive but surprisingly it puzzles me.

A wing has a centre of pressure, does it mean that lift acts on point and not on the whole chord of an airfoil?

Center of pressure is not a point "where the lift is most concentrated". Center of pressure has very little relation to how is lift distributed along the wing (you can have exactly same centre of pressure for infinite number of different distributions of lift).

The center of pressure is mathematical sum of all local pressure vectors along whole wing surface. But it is only, more-or-less, thought summing, this point has no special meaning in terms of stress distribution or so (at least no straightforward one).

You could as well continue to put all local pressure points into equation of movement and you would get exactly the same result. But equations with plenty small force vectors would be unnecessary complicated and if you do no need this level of complexity, it is easier to sum all forces in advance and use the summed value instead.

Note that if you would want, say, for example, calculate stresses inside the wing, you can not do like this. You would be forced to put each local pressure point into the analysis. But as long as you could assume whole wing to be solid (which is usually acceptable approximation for lot of simple problems), using single summed lift vector is easier.

If you simply sum all fractional pressure vectors, you get total vector of aerodynamic force. Which is enough to know how big force and in which direction acts on the airplane. But in order to be able to analyze rotational effects, you need to know in which point this total force acts (because same force vector shifted into different point of solid body will cause different rotation). And this is where center of pressure comes into the play, as it is the point where thought summed aerodynamic force should act in order to affect airplane (as a whole) by same translational and rotational effects as actual pressure distribution along the wing does.


What is it called when an airplane has to circle because it can't land? Sometimes, a plane is required to circle around an airport repeatedly because for whatever reason, it is not able or permitted to land just yet.

This state in which a plane is stuck in the air in this way has a name, what is it?


Can someone please tell me all the units for the propeller thrust equations? This answer is excellent:-

Is there any equation to bind velocity, thrust and power?

except that it is (in practice) unusable, since all the units are missing.

I'm guessing we use meters for length (prop diameter), and watts for power.

I'm not sure what units we use for mass (grams or kilograms?) - but either way - neither of those work (grams gives answers 10x too small, and kg gives answers 1000x too large).

I have a 10,000 watt motor
My propeller is 1.25meters diameter
At full pelt, using bathroom scales on the wall, it pushes me with 35kg force

Using 0.8 for the prop efficiency, and 0.9 for the motor efficiency, and 1.225  grams/l for normal air density, the referenced formula gives 367448 grams thrust - 10 times too much.

So yeah... what are all the units supposed to be?

Any good formula (that is, the one without magic numbers baked in) will accept any units as long as they are used consistently.

If you use SI, then use every figure in SI, and you'll get the result in SI.

Force is measured in Newtons, which is different to the unit of mass (kg), specifically $N =  kg⋅m/s^2$. The factor is the gravity acceleration $≈9.81\ m/s^2$ (which comes from the original definition of kg). So your measured thrust is about 343 N.

Because people are so used to treating force (weight) as mass, the 'force' units like kilogram-force (kgf) or pound-force (lbf) are commonly used, often implicitly. But this only leads to traps like you experienced and should never be used scientifically.

Density is $kg/m^3$, which is numerically equivalent to $g/l$ you used.

(In the context of propellers, one needs to be careful about the units of rotation. Strictly speaking, the SI unit of angular velocity is $rad/s$. But in practice coefficients are defined for 'n' in revolutions per second, $1/s$ (or even revolutions per minute, rpm, which is a non-standard unit). Fundamentally, radian is a dimensionless ratio, so it does not affect the resulting units (over 1/s) when used directly in formulas, and the error is not always obvious).


Why can't aircraft be cleared to land before the preceding aircraft turns off the runway? For example, let's say we have an Airbus A330 in front, which has just landed, well past the threshold and looks like it's going to be alright. Next in line to land is a small general aviation aircraft, such as a Cessna 172.

Now it looks like there's little to no risk of the two aircraft colliding and there's already more than enough space to stop the aircraft, so why can't a controller use their judgement to be able to clear the aircraft behind to land? It seems like that it's a waste of time when you know that there's going to be more than ample time for the previous aircraft to turn off, so what's the problem with that?

Actually, the FAA doesn't always require the preceding airplane to clear the runway if they have the required separation unless it is dark (when it is very hard to visually estimate distances).

According to the FAA Air Traffic Control Order Section 10. Arrival Procedures and Separation:

  3-10-6. ANTICIPATING SEPARATION

  a. Landing clearance to succeeding aircraft in a landing sequence need
  not be withheld if you observe the positions of the aircraft and
  determine that prescribed runway separation will exist when the
  aircraft crosses the landing threshold. Issue traffic information to
  the succeeding aircraft if a preceding arrival has not been previously
  reported and when traffic will be departing prior to their arrival.

  EXAMPLE- “American Two Forty-Five, Runway One-Eight, cleared to land,
  number two following a United Seven-Thirty-Seven two mile final.
  Traffic will depart prior to your arrival.”

  “American Two Forty-Five, Runway One-Eight, cleared to land. Traffic
  will depart prior to your arrival.”

  NOTE- Landing sequence number is optional at tower facilities where
  the arrival sequence to the runway is established by the approach
  control.

  b. Anticipating separation must not be applied when conducting LUAW
  operations, except as authorized in paragraph 3-10-5b2. Issue
  applicable traffic information when using this provision.

  EXAMPLE- “American Two Forty-Five, Runway One-Eight, cleared to land.
  Traffic will be a Boeing Seven-Fifty-Seven holding in position.”

And the separation is specified in the same document:

  3-10-3. SAME RUNWAY SEPARATION

  a. Separate an arriving aircraft from another aircraft using the same
  runway by ensuring that the arriving aircraft does not cross the
  landing threshold until one of the following conditions exists or
  unless authorized in para 3-10-10, Altitude Restricted Low Approach.

  The other aircraft has landed and is clear of the runway. (See FIG 3-10-1.) Between sunrise and sunset, if you can determine distances by
  reference to suitable landmarks and the other aircraft has landed, it
  need not be clear of the runway if the following minimum distance from
  the landing threshold exists:

  (a) When a Category I aircraft is landing behind a Category I or II-
  3,000 feet.

  (b) When a Category II aircraft is landing behind a Category I or II-
  4,500 feet.


The relation between Euler-angle rate and body-axis rates? A
paper  about quadcopter dynamics and control
shows the relation between Euler-angle rate and body-axis rates as:

Q1: why are there $R_{v2}^{b}(\dot{\phi})$, $R_{v1}^{v2}(\dot{\theta})$, and $R_{v}^{v1}(\dot{\psi})$?
I understand these are rotational matrices, but how could we use the $\dot{\phi}$, $\dot{\theta}$ and $\dot{\psi}$ to replace $\phi$, $\theta$ and $\psi$?
Q2: is the eqn(7) the approximate representation of the relation between Euler-angle rate and body-axis rates?  if YES, what are accurate representation?

Let me use the following graph to try to clarify your confusion, which is excerpted from Etkins, Dynamics of Flight:

The body axis is rotated from the [flat] Earth axis via a series of Euler angle rotations, $\psi$ (heading), $\theta$ (pitch), $\phi$ (roll), in this sequence. In the nomenclature of the graph, we've rotated from $(x_E, y_E, z_E)$ (Earth axis) to $(x_3, y_3, z_3)$ (body axis).
Now angular velocity is always expressed in the body axis, so the roll rate $p$ is about $x_3$, the pitch rate $q$ is about $y_3$ and yaw rate $r$ is about $z_3$. Our goal is to relate an instantaneous change in Euler angles ($\dot{\phi}$, $\dot{\theta}$, $\dot{\psi}$) to the angular velocity ($p$, $q$, $r$). Since the rotation is instantaneous, it does not add to the existing Euler angles.
We start off backwards, with the last rotation, $\dot{\phi}$. This rotation is about $x_3$, which is the same unit vector as $p$. Therefore, no frame transformation is required. i.e. $\dot{\phi} = r$. That's why your rotation matrix, $R_{v2}^b$, is identity.
Next, let's look at $\dot{\theta}$. This rotation is about the $y_2$ axis, which lags behind the body axis by a roll rotation ($\phi$). So to bring it to the body axis, we need to pre-multiply by $R_{v1}^{v2}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos{\phi} & \sin{\phi} \\ 0 & -\sin{\phi} & \cos{\phi} \end{bmatrix}$.
Last, $\dot{\psi}$ is about $y_1$, which lags behind by two rotations: $R_{v1}^{v2}R_v^{v1}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos{\phi} & \sin{\phi} \\ 0 & -\sin{\phi} & \cos{\phi} \end{bmatrix}\begin{bmatrix} \cos{\theta} & 0 & -\sin{\theta} \\ 0 & 1 & 0 \\ \sin{\theta} & 0 & \cos{\theta} \end{bmatrix}$.
When you add everything together, we have the kinematic relation:
$$\begin{bmatrix}\dot{\phi} \\ \dot{\theta} \\ \dot{\psi}\end{bmatrix} = \begin{bmatrix}1 & 0 & -\sin{\theta} \\ 0 & \cos{\phi} & \sin{\phi}\cos{\theta} \\ 0 & -\sin{\phi} & \cos{\phi}\cos{\theta}\end{bmatrix} \begin{bmatrix}p \\ q \\ r\end{bmatrix}$$
To answer your second question, no, this is not an approximation because the angular velocity is, by definition, an instantaneous change. Magic of calculus. The paper is incorrect to say that the rates are small; the magnitudes of the rates have no bearing on the derivation of the kinematic relation.


What is a lambda wing? I was reading a description of the Boeing X-45 that described it as having "lambda wings." What is a lambda wing and why is it called that?

Indeed, it is due to resemblance with Greek letter 'lambda', though not the capital one ($ \Lambda $), but the small one: $ \lambda $.

I am attaching a pic of X-45 with me drawing a lambda over it. Typically, the wing has two taper ratios. The inboard wing tapers heavily, corresponding to legs of small lambda, and the outboard wing tapers slightly or do not taper at all.

Other examples include B-2 bomber:


How much operating documentation is produced for an airliner by the manufacturer? A long, long time ago, when I was a young child, I read in a book or magazine that the 747 weighed 170 tonnes (or whatever) - but that the manuals it came with weighed even more! (I recall the same article explained that the aircraft contained over 100 miles of wiring; it was that kind of article.)

I can well believe that the total amount of technical documentation (design, production, operation, maintenance, etc) associated with an airliner could weigh more than a hundred tonnes.

But, just how much documentation would actually be delivered to the customer? Could it indeed be 170+ tonnes' worth?

But, just how much documentation would actually be delivered to the customer? Could it indeed be 170+ tonnes' worth?

The days of delivery of air carrier aircraft technical manuals being delivered by paper manuals are over, so it's certain that 170 tonnes of paper aren't now delivered. Whether or not, if you printed one copy of every manual available and then weighed them, that would reach 170 tonnes per airplane, I have no idea. That sounds a little high though to me, but if you included the weight of all of the engineering paper generated back when they actually produced blueprints, that would certainly be very substantial.

Boeing, as an example, delivers manuals as PDFs, perhaps by other means as well.

A few points:

The carrier gets what they need, and that will vary greatly. For example, if the carrier is receiving a passenger airplane, there's no need for them to have documentation concerning pallet tie down information for the aircraft's main deck.
Manuals need to be updated, and now you just get an current PDF rather than a few pages to be inserted in paper copies you already have.
Individual airplanes get manuals that are customized for that airplane, so you're talking about a lot of different manuals for a model of aircraft.
If the airplane goes through a modification program, the company doing the modification will typically supply manuals (also digitally) that deal with the modification. This is especially true for older passenger aircraft being converted to freighters.

Boeing supplies sample manuals for information expected to be common across variants of a model. Here's a 767-300F sample manual link. Of course, the information in an airplane's individual manuals takes precedence over that in sample manuals.

Perhaps if you printed every copy of every manual of every update, maybe you would get to the weight of a single aircraft of that model.


What is the Efficiency of Cessna-150's Propeller? From that Cessna-150's data, it used Continental 0-200-A 100 HP (75 kW) at 2,750 RPM and with Fixed Pitch propeller with its diameter=69 inches. If power is delivered maximum to the propeller (through direct shaft), then what is power of the propeller to push (thrust) the airplane?

Engine (100HP) ==> Propeller ==> Thrust (? %)

Additional Data of Cessna-150:
Weight of loaded: 680 kgs.
Weight of Empty: 437 kgs.
Wing loading: 9.4 psf (45.8 kg/m²)
Power loading: 15 lb/hp (6.81 kg/hp)
Maximum Engine Power: 100HP (75kW).
Wing Span: 10.2 m.

more detail about the plane can be seen here: http://www.skytamer.com/Cessna_150C.html

Probably quite low... After a series of gliding experiments made with a Luscombe 8E, and published by the AIAA, the conclusion was that the prop efficiency was around 62%...

https://engineering.purdue.edu/~andrisan/Courses/AAE490A_S2010/Buffer/AIAA-46372-872.pdf


Why do most commercial solid-state Inertial Measurement Units (IMUs) measure angular rate and not angular acceleration? The vast majority of MEMS-based IMUs are formed by an accelerometer and a gyroscope, providing linear acceleration and angular rate measurements respectively.

However, having available the angular acceleration would be useful for a more precise kinematic model when integrating to find the pose. Judging by this IEEE review, there are sensors for angular acceleration, therefore it doesn't seem to exist a fundamental limitation in the ability to have direct measurements of it.

Why is it that the standard has become to have angular rate as the source of information for rotational motion and not angular acceleration?

My answer assumes that you're asking about having IMUs measure angular acceleration instead of angular rate; if meant they should measure angular acceleration in addition to angular rate, then I apologize for misunderstanding you.

  However, having available the angular acceleration would be useful for a more precise kinematic model when integrating to find the pose.

I'm not sure where you're getting that information. When integrating to find the attitude, the angular rate is much more useful than the angular acceleration.

An attitude indicator essentially just takes the angular rate and integrates it. However, there are two sources of error here: errors from the sensor, and the fact that the direction of gravity changes as the earth rotates and as you move across the surface. But both of these errors can be corrected by mixing proper acceleration into the calculation.

On the other hand, if you only had access to angular acceleration data, you'd need to integrate twice to find the attitude: once to come up with an estimate of the angular rate and again to come up with an estimate of the attitude. (Why would you want to have to integrate to come up with an estimate of the angular rate, instead of just measuring the angular rate?)

This would introduce a third source of error into the attitude indicator: errors that result from accumulated errors in the angular rate estimate. I think these errors would be much more difficult to correct.

I think that the biggest problem is that in a coordinated turn, the angular acceleration is 0 and the lateral proper acceleration is also 0, so without measuring angular rate, an IMU would be unable to distinguish a coordinated turn from straight-and-level flight! That's a critical flaw.

In short, you need a reliable measurement of angular rate, and that can't be obtained from a measurement of angular acceleration.


Why are Leading Edge Flaps (LEFs) create lift while down at slow speeds but decrease lift while down at higher speeds? When taking off or landing, both LEFs and trailing edge flaps (TEFs) for fighter aircraft are in the down position to provide more lift. However, when flying around at higher speeds (let's say 500-600 knots), the left LEF going down and the right LEF going up causes the aircraft to roll left.

My personal theory is that the dominant form of drag (induced) at slow speeds means that the increased camber provides more lift although slightly decreasing the angle of attack (which is good if you are close to the stall angle of the wing). However, at higher airspeeds the slight decrease in angle of attack caused by the downward LEF and also the larger parasitic drag causes the decrease in lift of the wing.

Does this theory hold up? Or is there more to it that I'm not considering?

Leading edge faps also known as slats don't really create lift. The effect on lift coefficient is minor (less than 10% in general) but they help increase the stall angle of attack.
As a summary, flaps creates lift, by increasing surface area and profil camber whereas slats postpone the stall and allow for higher AOA by reducing pressure drop around the leading edge. This increases the theoretical maximal lift by increasing the stall AOA but doesn't change much the lift coefficient.
Slats are ment to work at high AOA by design, at higher speed your AOA is very small, the slat is thus not working properly, disrupting the flow on the suction side of the wind and reducing actual AOA, thus reducing lift of that particular wing while simultaneously increasing drag.


Could airliners use a steam or electric powered catapult for take-off? This would have the obvious benefits:

Less runway needed to take-off
Fuel savings
Faster turn around times for airlines

Could this be implemented in busy airports?

Let's see what the savings are:

A mid-sized airliner carries maybe 20% of its mass in fuel. This fuel has an energy density of 43 MJ per kg. Of that chemical energy at most 40% is converted into useable work. Heck, let's make this 25% so we are really conservative. Thus, the energy for the whole trip is $$E_{\text{trip}} = 0.2 \cdot 0.25 \cdot 43,000,000 \, \frac{\mathrm{J}}{\mathrm{kg}} \cdot \text{mass} = 2,150,000 \, \frac{\mathrm{J}}{\mathrm{kg}} \cdot \text{mass}$$

Now assume that this airliner saves the energy to accelerate from 0 to 150 knots by using a catapult. This energy is $$E_{\text{accel}} = \frac{v_{\text{takeoff}}^2 - v_0^2}{2} \cdot \text{mass} = 2,977.35 \,  \frac{\mathrm{m}^2}{\mathrm{s}^2} \cdot \text{mass}$$

Since I picked metric units, conversion is easy: $1 \, \mathrm{J} = 1 \, \mathrm{Ws} = 1 \, \frac{\mathrm{kg} \cdot \mathrm{m}^2}{\mathrm{s}^2}$. I use $\text{mass}$ for the take-off mass so you don't think it is the unit meter. Now let's put that into proportion:
$$\frac{E_{\text{accel}}}{E_{\text{trip}}} = 0.001385$$

Using the catapult saves 0.1385% of the energy needed to fly a typical airliner trip, assuming the same efficiencies during acceleration as during the flight. If we take into account that jet engines are most efficient during cruise, let's double the fuel need for acceleration and make it 0.277%. Granted, it is more for short-range flights, but still insignificant to what is needed to move the aircraft 10 km up into the sky and then for a couple of hundred miles through the air at Mach 0.8. In terms of fuel mass, these 0.277% are taken from 20% of the take-off mass. So the fuel needed to accelerate to v$_0$ is 0,000554 times take-off mass.

To make a catapult launch feasible, you need to add some strength to the nose gear and the forward fuselage. The typical landing gear fraction of the take-off mass is about 3%, and the nose gear is 10% - 15% of that, so $m_{\text{nosegear}} = 0.00375 \cdot \text{mass}$. Relative to the nose gear mass, the fuel saving from using a catapult launch is $\frac{0.000554}{0.00375} = 0.0148$ or 15% of the nose gear mass. Thus, the reinforcements need to add less than 15% to the mass of the nose gear.

If we assume an acceleration of ½ g = 4.903 m/s², the take-off run to accelerate to 150 kts is 607 m. I expect that even this moderate acceleration (which requires a pulling force of half of the lift at take-off) would translate into much higher mass increases than those 15% of the nose gear mass.


Is there any difference between Airbus and Boeing on yaw damper technology? Modern aircraft have yaw damper systems installed. Is there any difference between the two big manufacturers on yaw damper systems, such as the location, etc.?
I have to do some research on this topic and haven't found any answers yet.

The difference is more between conventional flight controls vs. fly-by-wire rather than Airbus vs. Boeing.
Both Airbus and Boeing used separate yaw dampers in their non fly-by-wire aircraft, which applied their commands with separate actuators to the rudder control servos, without feedback to the mechanically linked rudder pedals in the flight deck. The exact details will differ for every aircraft of course, but the general design is quite similar. Consider e.g. the Airbus A310 and the Boeing 747:
Airbus A310:

The rudder is controlled by 3 actuators, which are commanded by a
single cable run from the rudder pedals. Rudder artificial feel is
provided by a spring-loaded rod.
Additional inputs to the rudder come from:

the rudder trim, and
the 2 yaw dampers.

[...]
The yaw damper ensures the following functions:

dutch roll damping
turn coordination which becomes active if sufficient control wheel deflection is applied [...]
yaw compensation in case of engine failure, provided the AP is engaged in CMD with SRS (takeoff) or GO AROUND mode annunciated on
FMA.

Yaw damper command is transmitted to the rudder but there is no
deflection feedback sent to the rudder pedals.

(Airbus A310 FCOM - Flight Controls - Rudder Control)
Boeing 747-400:

Yaw Dampers
Two independent yaw damper systems operate continously in flight to
improve airplane directional stability and provide turn coordination.
The upper and lower yaw damper actuators are powered by hydraulic
systems 3 and 2 respectively. Yaw damper inputs do not result in
rudder pedal motion.

(Boeing 747-400 FCOMv2 9.20.9 - Flight Controls - System Description)

On the other hand, fly-by-wire aircraft usually integrate the yaw damper logic into the fly-by-wire commands sent to the actuators. Consider e.g. the Airbus A320 and the Boeing 777:
Airbus A320:

Electrical Rudder Control
The yaw damping and turn coordination functions are automatic. The
ELACs compute yaw orders for coordinating turns and damping yaw
oscillations, and transmit them to the FACs.
Rudder Actuation
Three independent hydraulic servojacks, operating in parallel, actuate
the rudder. In automatic operation (yaw damping, turn coordination) a
green servo actuator drives all three servojacks. A yellow servo
actuator remains synchronized and takes over if there is a failure.
There is no feedback to the rudder pedals from the yaw damping and
turn coordination functions.

(Airbus A320 FCOM - Flight Controls - Description)
Boeing 777:

The yaw control law contains the usual functionality employed on other Boeing jetliners, such as the yaw
damper and rudder ratio changer (which compensates a rudder command as a function of airspeed).
However, the 777 FBW rudder control system has no separate actuators, linkages, and wiring for these
functions, as have been used in previous airplane models. Rather, the command for these functions are
calculated in the PFCs and included as part of the normal rudder command to the main rudder actuators.
This reduces weight, complexity, maintenance, and spares required to be stocked.

(Boeing B-777: Fly-By-Wire Flight Controls 11.8.2 - Yaw Control)


What is the difference between the NTSB's "Aviation Accident Database" and their "Aviation Accident Reports" list? I'm trying to understand why there are thousands more records in the "Aviation Accident Database," located here:

https://www.ntsb.gov/_layouts/ntsb.aviation/index.aspx

than there are in the "Aviation Accident Reports" list, located here:

https://www.ntsb.gov/investigations/AccidentReports/Pages/aviation.aspx

For example, in 1991 there are 2,462 records in the accidents database and only 8 records in the accident reports list. What determines whether or not an accident ends up in the list of reports? What is the difference between these two data sets?

The NTSB will investigate every aircraft accident in the US, and any that foreign authorities delegate to them. As you found, there thousands of accidents ever year. Most of these are not fatal. The NTSB will at least gather the factual information and do their best to arrive at a probable cause.

With limited resources, the NTSB must decide which accidents are important enough to do a major investigation. You can see a map and list of the most important current investigations here, along with the following explanation (emphasis mine):

  This listing of investigations includes those open investigations involving accidents or incidents which resulted in significant loss of life, physical damage, involve issues of importance to public safety, or which have generated particular public interest. This represents only a small portion of the investigations conducted by the NTSB, most of which occur at the regional level.

There is no official criteria and not all of the incidents listed will become major investigations that result in full reports. The NTSB will decide based on the situation and factual information whether a full report will be produced. These reports are much more in depth, involve official parties that may submit their own findings and recommendations, and are deliberated on by the board in a public meeting before being officially released along with formal safety recommendations.

The board also periodically releases safety recommendation reports on issues that may not be related to a single major accident, but that are evidenced by multiple accidents that may not have received major reports.


What thrust is used for thrust to weight ratio? Performing some preliminary calculations for an aircraft design I have and was wondering what thrust I should use to calculate thrust to weight? The thrust calculations I have, vary due to speed (since in general force is a change in linear momentum) and so the faster the aircraft is traveling, the lower the thrust I have is. Should I use the initial thrust when the aircraft is initially accelerated, an average thrust of some sort or a different thrust estimate altogether?
These are two different equations I have used to estimate thrust (For propellers)

Well, it depends on what you want to know. If you are interested in thrust-to-weight ratio in certain conditions, you should use thrust (and weight) in those conditions.
In terms of standard performance specifications, the maximum static thrust (zero speed, zero altitude, ISA conditions) is normally used. This is often the only thrust point specified anyway. The weight is often less certain; it would be fair to use MTOW for this figure, but some "practical" weight/configuration is often used instead, because it gives better (and possibly "more relevant") result. Technically, the conditions should be explicitly specified.
If you are interested which thrust formula is more correct, this is a separate question.


What is this strange thing under Tu-95 which looks like a horse-fly's eye? I was seeing this photo and saw an strange thing under the bomber. Too much looks like a compound eye. What could be it?

According to this site it is a MAK-UL IR Threat Warning System, an infrared based missile approach warning system. Here is a closer picture:

(picture by Miroslav Gyűrösi)


How did pilots avoid thunderstorms and related weather before “reliable” airborne weather radar was introduced on airliners? How did airline and or mail pilots avoid hazardous weather especially at night before airborne weather radar systems were in use?

They looked out the window!

If you are talking about a developing thunderstorm, especially during the day, they can be quite easy to see in front of you provided decent visibility. Here is some good footage of what it looks like to fly around a towering cell, you will note its fairly easy to see. At night, lightning, even in the clouds tends to be a prominent effect and is a good indicator of "dont fly there". Here is a good picture of that effect (as well as a little story about flying around it).

Of course, summer haze, the dark of a moonless night, shifty weather, or just plain old mistakes sometimes lands planes right where they should not be sometimes with disastrous results.

Here is an interesting report on thunderstorms and aircrafts from 1948 that may have some interesting info for you.


What is the stall AoA for the space shuttle? Once flying as (fast) glider in order to land, what was the angle of attack it would have stalled? I'm really interested in the angle of attack for such wings and not the speed.

As far as I can tell, NASA has never given a straight-forward answer to this question. The theoretical stall AoA is likely in the 33-40° range (see reason for that guess below). However, at operational airspeeds, the orbiter becomes aerodynamically unstable at a lower angle of attack than the stall angle. Therefore, the stall angle never mattered for shuttle operations.

Just to be clear, this answer is only referring to the very late phases of shuttle flight (TAEM and approach and landing). The 40° of early entry might be beyond the wings' stall angle, but it doesn't really matter since dynamic pressure is low and the shuttle is on more of a ballistic trajectory.

What does NASA say?

From an old NASA answer to this question:

  The question: "What is the Space Shuttle stall speed?" is a bit of a
  trick question because, as a delta wing aircraft, the Orbiter does not
  have a clearly defined stall point.

  ...

  We have determined the minimum flyable speed based on the longitudinal
  stability characteristics.  We start by evaluating the maximum angle of
  attack, alpha, for which the Orbiter is aerodynamically stable in the
  pitch axis.  This is the angle of attack at which the pitching moment
  partial derivative with respect to alpha changes from negative (stable)
  to positive (unstable).  This angle of attack is identified as the
  maximum usable angle of attack and it varies by Mach number.

From The Aerodynamic Challenges of the Design and Development of the Space Shuttle Orbiter document:

  The low-speed longitudinal characteristics shown in figure 29 demonstrate stall-free characteristics over operating flight conditions.

The diagram on the left shows that the lift coefficient continues to increase through at least ≈25° angle of attack (meaning the stall angle must be greater than 25°). δe = elevon deflection.

The only* reference to stalling in the Shuttle Crew Operations Manual is this diagram which depicts a stall speed (not angle). The NASA answer I referenced before gives more context to this graph.

Shuttle Crew Operations Manual - Section 4.9-1

*There are a couple references to elevon stalls in the Operations Manual, but I'm not counting those.

Why does the stall angle not matter?

As we've already seen, the shuttle would exceed aerodynamic/structural limits before reaching the stall angle of attack at most airspeeds. Additionally, there are a couple other reasons for not worrying much about the stall angle.

Guidance always aimed to keep the orbiter on the "front side" of the L/D curve. This means that guidance would never command an angle of attack higher than required for "best glide". If it needed to dump energy, it would do so by deploying the speed brake and/or pitching nose down. It would never dump energy by pitching up (except during the flare). If the orbiter was low on energy, it would pitch for best glide. This means airspeed would always stay high and in the range where other factors are more limiting than the stall angle. Pitch-up maneuvers would be limited by dynamic pressure and Nz (normal acceleration — g's).

The only time the shuttle fell into the "back side" of the L/D curve was during the flare and landing. Incidentally, touchdown speed appears to be roughly the speed where the orbiter is finally slow enough that it might actually stall before becoming unstable for other reasons (see "VA" in the graph above - this is "maneuvering speed", just like in airplanes). However, at touchdown, the limiting angle is the tailscrape angle (15°), which is much lower than the stall angle.

Shuttle Crew Operations Manual - Section 7.4-17

What's the real answer?

Although not operationally significant, I think the specific question you're trying to ask is: what AoA produces the maximum lift coefficient? Generally speaking, that's considered the stall angle for most airfoils, since increasing AoA beyond that point results in decreased lift.

While I don't have a lift coefficient graph for the shuttle which goes beyond 25°, this paper discusses the results from studies on various delta and double-delta wing designs.

None of the studied designs have the shuttle's exact 81/45 sweep angles. However, given that they all have a maximum lift coefficient somewhere in the 33-40° AoA range, I suspect the shuttle was similar. I can't prove that unless someone finds the appropriate data. I'm certainly not an expert on delta wings, so my interpretation could also be wrong.


How much is an airliner's drag coefficient? How much is an airliner's Cd, using the frontal area as reference point. I want to compare values with cars and others objects, tear drops, cube etc.
Is Cd measured at zero lift AoA or when the plane is flying in a straight line?

Is Cd measured at zero lift AoA or when the plane is flying in a straight line?

Both.
$C_D$ is a function, not a number.
The "numbers" are the partial derivatives: $C_{D_\alpha}$ measures how $C_D$ changes w.r.t. the angle of attack $\alpha$, for example.


Is Autopilot NAV setting useful? In flight simulator, using the NAV setting directs the Autopilot to follow the FMS' pre-setup flight plan and will navigate the aircraft through each of the configured waypoints.

In real life, however, it seems almost a guarantee that ATC will provide some heading which is not necessarily the most direct route, ie. what your flight plan had configured. ATC will request you turn right or left somewhere around the waypoint you expected, but not where you or the NAV system would have turned.

Even in the simulator, if you use the built-in ATC, it will give headings that are not what the flight plan had expected... making the NAV button not very useful except when flying VFR without following.

It seems the autopilot HDG feature is used more often. Is this true in real life as well?

There are essentially two scenarios on departure: an initial vector, or flying the departure procedure track immediately "off the ground."

In the former case, the instructions might be along the lines of, "fly heading 120, expect vectors to JEBBB" or "... to join the WHOZT5 departure."  In that case, shortly after takeoff, the pilot would select Heading mode and fly the specified heading.  At some point after that, ATC would issue a clearance to proceed direct to a fix, and then the filed route would be flown from that point on.  So upon receiving this clearance, the pilot would proceed direct to the given waypoint by engaging NAV (or, given modern nomenclature, LNAV).

In the other case, the first leg of the departure corresponds to the "straight ahead" track of the departure runway, and LNAV is activated immediately after departure, and the aircraft will fly the magenta line as published.  So in this case, Heading mode isn't used, unless ATC needs to vector the aircraft off of the published track.  In that case, they can either issue a vector, or they can clear the aircraft to a later point in the procedure or along the route.

In modern ATC, it is very uncommon for an aircraft to be getting vectors enroute, simply because it is more workload for the controller with no benefit.  Once you're away from the terminal area, they clear you on whatever route you've filed (or been assigned), and they don't have to do any more navigating for you.  If they need to turn you off course for traffic, they will, but that's the exception, not the norm.


Why not mount airliner jet engines above the wings? It seems that most airliners put the engines below the wings, i.e. the Boeing 7*7 series and the Airbus A3** series. This requires long (heavy) landing gear, and the engines are close to the ground which may cause damage to the engines (ingest debris) or the ground.

Why not mount the engines above the wings? In addition to the shorter (lighter, stronger) landing gear, the plane would be easier to service and embark / disembark. What are the major disadvantages to having the engines above the wings for civilian airliners?

The question "How does the mounting location of a jet engine affect aircraft performance?" addresses the performance issues affecting engine placement but answers do not cover over-wing placement extensively and focus only on performance and not other operational issues or design advantages.

Maintenance

Having the engines below the wing makes them fairly easy to access, even if some kind of lift is needed. Putting them above the wing would make them much harder to get to. Dropped a tool? Better start on that wing repair too. This includes both general access to the engine and the procedures for removing and installing the engine.

Structure

An engine below the wing is fairly stable, since it is hanging from above. It naturally wants to keep hanging downwards. Mounting the engine above the wing would be less stable. It naturally wants to tip over.

Noise

If an engine is under the wing, it helps to block some of the noise from at least the mid cabin. An engine above the wing would expose much more of the cabin to noise. It could reduce the ground noise while in flight, though.

Aerodynamics

The top of the wing is the most sensitive part aerodynamically. This is why items like fuel tank access panels are placed on the bottom. Having an engine and pylon on top would affect the flow over the top of the wing.

Safety

Engines falling off the wings tends to be a bad thing. But if they were above the wing, they would tend to fall down onto the wing, which would still not be good. It would be harder to have then break away in a crash. Also, in the event of an uncontained failure, the wing would not provide any protection to the fuselage.

Of course there are benefits, such as the reduced noise, less ground clearance required, no interference between the jet blast and the flaps, and less chance of FOD being ingested. As new engines get larger fan diameters, the constraints become more difficult. Some concepts of the blended wing body design put the engines on top of the body. But for the current design of aircraft, the benefits of having the engines below the wing are greater than those of the alternatives.


Who to contact when inbound to adjoining class deltas? Given two class delta airports that share a boundary, say you wanted to transition through one to land at the other.  Who would you call first?  The transitioning controller, or the destination controller?

(skyvector.com)

Example: approaching KOXR from the east for landing through KCMA, VFR w/o flight-following.  Could the KOXR controller clear me through KCMA airspace?  Being conservative, I suppose I could call both but wonder if I'd have time for a frequency change before entering KOXR.

You should contact the tower of the Delta you intend on transitioning through/entering first. Tell them you want to transition through their airspace. If they clear you to enter their airspace you can request a frequency change while still inside the first delta and contact the second. This is the same procedure as say departing a Delta that abuts a Bravo then taking a Bravo transition. You could also ask for a handoff to the second delta from the controller at the first.

For example I fly out of KPNE often a Delta abutting a Bravo, if I want to fly south through the bravo before take off I can call the tower with something along the lines of,

  "Archer 123PB holding short at 33 ready for departure to the south to KWWD.
  Can I get a bravo handoff as well?"

An example in your case might be

  Camarillo tower, Piper 123PB 5 miles to the east looking to transition through your airspace to KOXR

The tower may respond with

  Piper 123PB approved as requested maintain 1500 ft. hold course and overfly the field

Then as you are a bit more east (most likely past the field in this case)

  Camarillo tower requesting frequency change

Tower

  Frequency change approved

You can then call OXR with

  Oxnard tower Piper 123PB 3 miles to the east inbound for a landing at OXR with information {current AITS}

This is going to happen fast and you will need your frequencies in hand for this. You can ask the first tower for a handoff which they may orchestrate but you should be prepared for both options. If you have some kind of a pre-programmable radio this is a great example of where queueing up your frequencies will help.

Your other option (in this specific case) weather permitting you could overfly KCMA above 2000 ft. since the delta ends at 2000 Ft. and simply request entrance to the second delta. In this case you never actually enter the first delta and no clearance is required.

In some cases, there are prior agreements that may not match the charts for adjacent airspaces.


What is this WWII four-engine plane on skis? I've seen this plane but I don't know its name:

Looks like a TB-3 or a variant thereof. According to Wikipedia, this four-engine all steel(!) heavy bomber first flew in 1930 and was the world's first cantilever aircraft in this class. The skis do not appear to have pertained exclusively to any specific variant, given that the first flight already used them.

It was used up to and including the Second World War, and a number of experimental variants were developed from it. One of the most interesting was the Zveno project, an experiment with parasite fighters. Here you can see the Zveno-SPB variant of the TB-3 with two I-16 fighters under the wings:

The Zveno-SPB saw operational usage and even success in raids against the Romanian oilfields during WW2.


Could adding a gearbox to the Trent 900 improve the efficiency of the A380? Could adding a gearbox to the Trent 900 improve the efficiency of the A380 so that it performs better than a smaller jet, i.e with economies of scale? Is the absence of a gearbox in the Trent 900 the main impediment to the efficiency of the A380?

It would not be a simple task to add a gearbox to an existing engine like the Trent 900. The whole engine is designed to operate as a whole, with each of the three shafts rotating at an optimal RPM. The gearbox would have to be designed to work with the existing engine hardware, which would mean compromises on efficiency. The engine would also have to be modified to find room to fit the gearbox in.

A gearbox also does not automatically mean an engine will be more efficient. The geared PW1000G achieves similar efficiency to the CFM LEAP which does not have a gearbox.

It's much more likely that a new engine would be designed for the A380. Airbus has previously said that this was planned for some point in the future. However, a new engine is not cheap for either the engine company to develop or for Airbus to integrate. The A330 and A320 programs have gone this route but they are also designs with high demand.

A new engine would have to generate enough demand to justify the cost, and it does not seem that Airbus sees this demand yet. Airbus has been struggling to find new orders for the A380 and has been slowing down production. It's not clear that better efficiency is enough to make the plane sell. Four engine planes are inherently limited in their efficiency versus twin engine planes. The four engines add more drag and weight versus two engines, and increase maintenance costs.

The A380 also requires special accommodation in the air and on the ground for its large size. The 747 has been around for much longer and does not need as many special accommodations. Boeing went to the expense of redesigning the wing and adding new engines for improved efficiency, yet is still struggling to find orders.

More recently Airbus has announced that they are considering "A380 plus" which will include new winglets, wing changes, and other design tweaks to improve efficiency. This will be a much cheaper way to add efficiency than a new engine, if that efficiency will mean more orders.


How can I progress from tactical aircraft mechanic in the air force to an airline? I am currently 17 years old and next year I will be graduating from  high school. My plan is to go to the Air force and become a Tactical Aircraft Mechanic. Obviously I am going to go through BMT and then Tech school. But after I learn that career I want to work at an Airline. How does that work specifically?

This answer assumes you intend to join the US Air Force and seek certification as a mechanic with the US FAA.

I would suggest you keep a personal log of all your training and all the work that you do while in the Air Force. This, along with your service record, can then be used as evidence to support an application to the FAA to be allowed to take the exams for the Airframe and Powerplant Mechanic (A&P) that will allow you to seek work with airlines. You take three exams -- General, Airframe and Powerplant, then an oral and practical exam with a Designated Examiner. If you pass all of that, you are issued your license as a mechanic and off you go!


Why don't airbus aircraft have raked wingtips? Most airbus aircraft have sharkfin winglets or blended winglets. Why is it that none of them have raked wingtips like the 777-300ER?

Probably because Boeing holds a number of patents related to raked wingtips, like US5039032 A High taper wing tip extension and US6089502 A Blunt-leading-edge raked wingtips, among others.


Is the total lift of a quadcopter the sum of the lift from each propeller? In this diagram each rotor R1,R2,R3,R4 has an individual lift force of 490 N. All combined, does the total force on this aircraft equal to 1960 N?

The aircraft in question is similar to a quadcopter.

Yes.  490 + 490 + 490 + 490 = 1960.  For practical purposes, the thrusts from individual rotors do not interfere constructively or destructively:  they neither cancel each other out, nor do they help each other.  This case is pretty much the same as multiple propellers mounted on a fixed wing.

The situation becomes more complicated only if some rotors overlap, or are coaxial.


When leaving ground effect, why is there a nose-up change in moment? The Airplane Flight Handbook states:

  The aircraft leaving ground effect will

  Require an increase in AOA to maintain the same CL
  Experience an increase in induced drag and thrust required
  Experience a decrease in stability and a nose-up change in moment
  Experience a reduction in static source pressure and increase in indicated airspeed

I'm having a hard time understanding the number 3 here.

What's the aerodynamic reason for the nose-up change in moment after leaving the ground effect?

Is it because the downwash from the wings increases as the aircraft leaves the ground effect, thus placing a greater downward pressure on the horizontal stabilizer?

The aerodynamic centre of a wing shifts when it is in ground effect. Out of ground effect, the wing produces lift by accelerating air downwards. In ground effect the ground prevents the air from being deflected downwards, and more lift is produced due to increased pressure underneath the wing. This pressure has a different distribution than out-of-ground effect pressure, hence the change in pitching moment.

Consider a cross section of a wing: free stream lifting force centre is at the quarter chord point. If the wing was supported by just static pressure only, like a hovercraft, the force centre point would be at 50% chord.

  Actually not sure now if the horizontal tail has much influence but
  I'll leave the original answer underneath:

Ground effect works by increasing static pressure underneath the wing and the horizontal tail. The aircraft flies out of ground effect in a nose-up attitude, and depending on tail configuration it can be possible for the main wing to be out of ground effect while the horizontal tail still experiences increased pressure underneath, therefore creating a nose-down pitching moment. Once the horizontal tail flies out of ground effect as well, the nose-down pitching moment disappears. I would not expect that T-tails have this change in pitching moment but I have never flown one.


How can I learn air traffic control as a high-schooler? I really love aviation and I'm thinking about going for a career in air traffic control, but I can't find anything on the web that will help me learn all about it and how to perform basic operations. Is there anything on the web that I could use?

Join the Vatsim network.  This is a network of simulated pilots flying X-Plane, FSX, etc., and simulated controllers.  The simulated controllers use, as much as they can in a hobbyist network, real phraseology and procedures.  I personally knew several real-world controllers who controlled on the Vatsim network in their off-time, and I know several hobbyist controllers who went on to become real air traffic controllers.

While controlling on a hobby network is not the same as controlling real airplanes, you won't learn a lot you'll have to relearn.  That's because the hobbyist controllers on Vatsim make a serious effort to follow real procedures.  When they discuss air traffic control with each other, you frequently see references to the ".65", the official bible of ATC.


What name is given to an abeam waypoint in the FMS? In a Flight Management System, what name is given to an abeam waypoint in case of DIR TO?
Is this name format different in Thales and Honeywell FMS?

From the Thales Pilot Guide:

  If ABEAM PTS are desired:
  ABEAM PTS [3R] .......... Confirm/Press
  If Option 1 is defaulted, ABEAM PTS is displayed yellow, if not press [3R].

  ND shows direct track (dashed yellow), as entered, superimposed on existing ACTIVE F-PLN.
  Abeam WPTs are named ABxxxxx, xxxxx representing the first five characters of the relative WPTs.
  Ex: ABADAIB, ABBOKNO, ABDEVRO

Honeywell FMS is different:

  Abeam Points (ABEAM PTS)

  Push -

  creates place bearing distance waypoint on the Direct To leg abeam
  the bypassed waypoint if the bypassed waypoint was a database
  airport, navaid, NDB, or waypoint
  creates latitude/longitude waypoint on the Direct To leg abeam the
  bypassed waypoint if the bypassed waypoint was a
  latitude/longitude waypoint

So in the Honeywell FMS the abeam points will use the Place Bearing Distance name or the Lat/Lon name depending on the type of original waypoint.


What must ATC do, following a reported laser sighting by a plane? Supposing a pilot reports having a laser beam shone on them to ATC, what action should they take?

Is the pilot allowed to take evasive action?

Can other planes request clearance to avoid the area?

Here's what the regulations say ATC should do:

Broadcast on all appropriate frequencies a general caution warning every 5 minutes for 20 minutes following the last report;

Include information on the ATIS broadcast for one hour following the last report;

Report the incident through the Domestic Events Network (DEN), or to their overlying TRACON or ARTCC facility.

As for the pilots, lasers are regarded as an emergency, so you are permitted to take evasive action.
Other aircraft are also permitted to request clearance to avoid the area.

Here are the relevant FAA orders:
FAAO JO 7110.65:

10-2-14. UNAUTHORIZED LASER ILLUMINATION OF AIRCRAFT
a. When a laser event is reported to an air traffic facility, broadcast on all appropriate frequencies a general caution warning every five minutes for 20 minutes following the last report.
PHRASEOLOGY- UNAUTHORIZED LASER ILLUMINATION EVENT, (location), (altitude).
b. Terminal facilities must include reported unauthorized laser illumination events on the ATIS broadcast for one hour following the last report. Include the time, location, altitude, color, and direction of the laser as reported by the pilot.
NOTE- All personnel can expect aircrews to regard lasers as an inflight emergency and may take evasive action to avoid laser illumination. Additionally, other aircraft may request clearance to avoid the area.

FAAO JO 7210.3:

2-1-27. REPORTING UNAUTHORIZED LASER ILLUMINATION OF AIRCRAFT
All FAA Air Traffic Control facilities, Federal Contract Towers and Flight Service Stations shall report unauthorized laser illumination incidents through the Domestic Events Network (DEN), providing the following information:
a. UTC date and time of event.
b. Call Sign, or aircraft registration number.
c. Type of aircraft.
d. Nearest major city.
e. Altitude.
f. Location of event (e.g., latitude/longitude and/or Fixed Radial Distance (FRD)).
g. Brief description of the event.
h. Any other pertinent information.
NOTE- Facilities without direct access to the DEN shall forward the information through the overlying TRACON or ARTCC facility


Does mid-air refueling have very hard constraints on weather? I'm guessing mid-air refueling has constraints on windspeed and wind direction, visibility, and precipitation. I may have missed some, but those alone make me feel like mid-air refueling can't be done most of the time (?)

About wind direction though, I guess you could just fly in the "right direction", but then this would also have constraints because the "right" direction could lead away from your ultimate destination.

(also, is there a mid-air refueling tag I couldn't find?)

The receiver has to have visual contact with the tanker to complete the rejoin, so finding a clear block of airspace (i.e. not IMC) is necessary. Beyond that, none of the factors mentioned really matter much.  Turbulence can make refueling more difficult, although it's often possible to find an altitude with a decent ride.

Mostly, AR can be done; the times when it can't are much more the exception.


Was there any kind of preventative collision training for dogfights during WWI and WWII? Watching some old films on YouTube about dogfighting has me curious about this. Definitely seems like most dogfights around this time were extremely close, involving tight formations and turns. Was there any kind of preventative collision training that the pilots went through around this time? Or was the goal just to get them through training ASAP?

No, there was no anti-collision training aside from "dodging the fireball" after enemy aircraft shootdown. The training resources and timeline didn't permit much training beyond safe employment of the aircraft gun.


Which 737 generations doesn't have Alpha Vane (Angle of Attack sensor)? I was wondering if all generations have this angle of attack sensor, especially after recent 737 MAX Lion Air accident.

I have experience on the 737 -100,-200, Classic, Next Gen and the Max8 and even the 707 every one has an AOA sensor.  Different part numbers to be sure but they still have them.

Obviously it senses angle of attack and feeds numerous systems: stall warning and auto slat come to mind immediately.


Which are some planes shaped like an inverted wing? To understand my question better, let's consider the priciple behind the lift generated by the wing - The upper surface of the air-foil is curved to deflect the air downwards using the Coanda effect. According to Newton's 3rd Law, an equal reaction is produced by the downwards deflection of air in the form of lift.

So here's my question - Why aren't some planes designed to look like a massive cross-section of a wing from the side? e.g. A Cessna C172 looks like an inverted wing (with the curved side on the bottom, which might create 'inverse lift' by pulling it down. From that aspect, is the Cirrus SR22 and the Cessna TTx more lift-friendly? Their upper part of the fuselage is relatively more curved than their lower part when viewed from the side. Or am I making a mistake - is it due to high-wing low-wing design philosophies?

Your intuitive feeling that the fuselage could be used to create additional lift is correct. Indeed, some aircraft are specifically shaped such that fuselage could provide substantial lift (and that's excluding any 'flying wing' design where fuselage is completely blended with the wing).

However, these are generally supersonic fighter aircraft. At supersonic speeds, even the 'normal' tubular fuselage will create significant lift, given a decent angle of attack. Moreover, the engine air intakes, if properly designed, will create even more lift.

In GA aircraft, on the other hand, the effect must be negligible, and structural considerations (yes, including high/low wing) are much more important.

The primary aerodynamic consideration for the fuselage is, generally, minimising its drag and reducing interference with other parts, primarily the wing. In this regard, by the way, C172 (or any high-wing aircraft) is more 'lift-friendly'. High-wing design has less negative wing/fuselage interference; in fact, it can be even positive. In simple terms, the part of the wing above the fuselage still creates lift, and probably more of it than the whole fuselage (although I have no numbers at hand).

I also must duly note that the principle of lift generation has little to do with the geometric length of the upper and lower surfaces. It is a popular misconception. The air indeed moves faster over the upper surface, but this is due to entirely different reasons. There are several good explanations of lift on this site.


What kind of problems can a flight have if passenger weight is miscalculated? There was a widely reported news story about a TUI Boeing 737 flight, where a bug in reservations system caused 38 passenger on the flight to be allocated a 38kg standard weight as opposed to the correct figure of 69kg. In total, this caused the report to state that the weight of passengers was 1200kg lighter than it actually was.
My question is - how serious of a problem is that? Would the pilot notice this discrepancies? What kind of problems could appear due to pilots using the wrong passenger weight in calculations and flight configuration?

In general, an incorrect takeoff weight can have serious consequences for an airliner, especially when the actual TOW is close to the RTOW (regulated takeoff weight, see e.g. this question: What is RTOW and how is it different from MTOW?). It is possible that the aircraft is unable to takeoff within the available runway length. This is particularly important in case of an engine failure during the takeoff, since normal performance with both engines operating typically has enough margins. Even on a long runway, the pilots might rotate too early because of an incorrect rotation speed (VR), which could result in a tail strike.
In this particular case, the difference in performance was minimal. The AAIB (Air Accidents Investigation Branch of the UK) investigated the incident and concluded:

Whilst an incorrect takeoff weight was used for aircraft performance planning, the thrust
required for the actual TOW and environmental conditions (88.2% N1
) was marginally less
than the thrust used for the takeoff (88.3% N1
). This meant the safe operation of the aircraft
was not compromised.

The report also shows the difference in the performance calculations with the incorrect (left) and correct (right) TOW:

The difference in the V speeds for takeoff is just one knot, which is not significant enough to have any adverse impact on the takeoff.


Why don't all aircraft use/need nacelle chines? Most modern commercial aircraft (e.g. 737, 767, 777, 787, 747-8, A380, etc.) feature inboard nacelle chines to improve flow over the upper surface of the wing at high angles of attack. However, many other similarly purposed aircraft (e.g. DC-8, L-1011, 757, 747-400) do not have nacelle chines. If these chines are so critical for optimum takeoff and landing performance, how are so many planes able to function just fine without them?

The chines help when the nacelle is very close to the wing. When the engine of Boeing's 737 was changed from the slim JT-8D to the higher bypass ratio CFM-56 on the 737-300 and later, the chines were essential to avoid large flow separation. Generally, aircraft with higher landing gear can afford to have more space between nacelle and wing and do not need fixes like chines. But when a new engine has been fitted to an existing design, chances are that some vortex generators are needed to fix the new flow conditions.

For the nitpickers: The 737-100 and 200 had underslung nacelles which left no space between wing and nacelle, but that is aerodynamically similar to a nacelle at a larger distance, because no narrow flow paths are created which produce steep pressure gradients. Only a closely mounted, but still separate nacelle will have these flow conditions which need to be remedied by energizing the boundary layer.


How can a small ducted fan produce a large amount of thrust? This 15.2 cm ducted fan by Schuebeler can produce 175 Newtons of static thrust. How is this possible in such a small diameter?
If this seems normal, then why and how can this ducted fan produce that much force?

Thrust depends greatly on diameter, so your curiosity is reasonable.
Why:  A first-principles answer says that (static) thrust depends directly on the rate of mass flow through the duct.  Because the duct's area is small, the speed of air through the duct must be large, and thus what must also be large is the fan's (static) pitch speed and its RPM.
See https://www.grc.nasa.gov/www/k-12/airplane/thrsteq.html for the case $p_e = p_0$.
For 175 N thrust, Schuebeler says the fan turns at 20,000 rpm and needs 12 kW.  Those numbers are not vastly different from an unshrouded 15 cm propeller, or a small turbine (a JetCat P220 does 220 N at 725 mL/min, converted via kerosene's MJ/L to roughly 450 kW; there's much more waste heat in a turbine's exhaust than in a ducted fan's gradually warming battery pack).
How:  Tight manufacturing tolerances, so the blades don't vibrate from imbalance, so the blades don't deform or break at high rpm, and so not too much air leaks around the gap between the blade tips and the walls of the duct.  (The same applies to turbines.)


What does it take to turn a 747 around 180 degrees? Assume that a 747 is flying at cruising altitude and speed. If the pilot decided (for a presumably good reason), to turn the plane 180 degrees,

How quickly could the turn be completed?
How wide would the turn radius be?
How much altitude would be lost, if we're not assuming a level turn?

EDIT: I learned something new from both of the great answers below. I accepted the one that was easier to follow. Since both answers assumed a level flight, I have added a #3.

How long, and how wide of a turn depends entirely on three things:

What is the true airspeed.
What is the bank angle of the turn?
Is level flight maintained during the turn.

Since it's easiest to calculate, and presumably what you intended, we'll assume #3 is true (level flight). However, if you really needed to get turned around fast, turning in a descent will give you a tight turn without the extra g-forces.

Cruise Speed for a 747 depends on altitude, generation, company guidelines, etc, but we'll go with Mach 0.85 for this example, which seems to be in the ballpark. If we say it's cruising at 35,000 feet in standard atmospheric conditions, that comes out to 490 knots TAS.

Next, pick a bank angle. A reasonable bank angle for that airplane with passengers on board is 25°, and you could probably get away with 30°. Anything beyond that will cause passengers to complain. Of course, if it's an emergency you might consider something higher. Just remember, the higher the bank angle, the stronger the g-forces which will be felt by the people inside, and the aircraft itself, which has structural limitations. The G-force can be calculated using:

$$
gforce = \frac{1}{\cos(bank)}
$$

For an example, let's select 25° bank since that's the most realistic.

Radius

Calculate the radius of the turn using this formula, slightly modified from Wikipedia to give nmi instead of feet:

$$
Radius\ of\ turn\ in\ nautical\ miles= \frac{velocity^2}{68579 \times \tan(bank)}
$$

Which gives us:

$$
7.51nmi = \frac{490^2}{68579 \times \tan(25°)}
$$

So the turn itself would be about 15 nautical miles wide (≈ 91,000 feet) not accounting for wind.

Distance Traveled

Using basic geometry, the distance traveled for a 180° turn is $d = r\pi$ (half the circumference of a circle), so:

$$
23.59= 7.51\pi
$$

Giving us 23.59 nautical miles traveled.

Duration

$$
time\ in\ minutes = \frac{distance \times 60}{TAS}
$$

Therefore, at 490 knots (nautical miles per hour) and 25° bank, it would take about 2 minutes, 53 seconds to complete the turn.

$$
2.89 = \frac{23.59 \times 60}{490}
$$

You could go through and calculate the results for any speed and bank angle you want. This is true regardless of the type of airplane in question.

Turning in a Descent

Your edited question asks how much altitude would be lost if it were not a level turn. There is also no single answer to this question because it depends entirely on how you perform the maneuver.

One reason you may have to descend in a turn is because the increased lift results un increased induced drag, and the aircraft engines may not have enough power to compensate, resulting in the decreased airspeed, and possibly a stall. In this case, the equations for calculating the turn radius are exactly the same as in level flight. The amount of altitude lost will depend on the descent rate required in order maintain airspeed, which will vary depending on the engine power available, the weight of the aircraft, and the drag curve at the given speed and AoA. Peter Kämpf's answer gives an example of what this might look like for a 747 in a 1.5g turn.

Another version of a turning descent would be an accelerated descent. The advantage in this case it that it would allow turns to be made without additional g-force. The disadvantage is that you'll be accelerating downwards instead of descending at a constant rate. This can get out of hand very quickly and would only be good for very short duration turns.

To illustrate this disastrous option, let's look at what would happen in a 1g turn at the same 25° and 490 knots TAS. Since the math is more complicated, it'll be easier to talk about this portion in metric units. Here's a conversion table:

1 nautical mile = 1852 meters
1 knot = 0.514444 meters per second
1 foot = 0.3048 meters

First, in order to maintain a constant 1g, the lift vector is simply rotated in the turn (instead of rotated and increased in order to maintain altitude). Therefore, the magnitude of our lift vector will be equal to the acceleration due to gravity near the Earth's surface. We'll round that to 9.8 m/s/s.

Inward Acceleration

Our acceleration towards the center of the turn $a_c$ (the portion of our lift vector pointed inwards instead of up) can be determined using the following equation where $\theta$ is the bank angle:

$$
\sin(\theta) = \frac{a_c}{g}
$$

Solving for $a_c$:

$$
a_c = g\sin(\theta)
$$

Therefore

$$
4.142 = 9.8\sin(25°)
$$

Radius

So 4.142 meters per second per second is how fast, at 25° of bank, and 1g, we'll be accelerating towards the center of the turn. With that, information, along with our known velocity, we can calculate the radius of the turn using this equation where $v$ is our velocity in meters per second, and $R$ is the radius in meters:

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

Plug in our numbers:

$$
15341 = \frac{(490 \times 0.514444)^2}{4.142}
$$

This comes out to a radius of 15341 meters (8.28 nautical miles).

Distance Traveled

Now that we have our radius, we can calculate how long it will take us to turn 180°. This part is the same equation as before.

$$
48195 = 15341\pi
$$

Giving us 48195 meters (26 nautical miles).

Duration

Distance divided by velocity gives us duration.

$$
t = \frac{d}{v}
$$

In our case:

$$
191.2\ seconds = \frac{48195}{490 \times 0.514444}
$$

So, it'll take 3 minutes 11 seconds to complete the turn.

Downward Acceleration

The last value we need before we can calculate the altitude lost in the turn is to calculate how fast we're going to be accelerating towards the ground in this turn. Lets first calculate the upward portion of our lift vector:

$$
a_u = g\cos(\theta)
$$

Therefore

$$
8.882 = 9.8\cos(25°)
$$

Our lift vector is accelerating us upwards at 8.882 m/s/s, while gravity is trying to pull us down at -9.8 m/s/s. That comes out to a net vector of -0.918 m/s/s.

Altitude Lost

This part requires a little bit of calculus because our descent rate is accelerating. The integral of acceleration is velocity ($v = at$), and the integral of velocity is distance:

$$
d = \frac{at^2}{2}
$$

So, lets calculate the change in vertical distance (altitude):

$$
-16780 = \frac{-0.918 \times 191.2^2}{2}
$$

So, theoretically, our airplane lost 16,780 meters of altitude (55,052 feet). Of course, since we were only 35,000 feet in the air to begin with, this is rather bad news for us. Besides hitting the ground, you would also come close to risking structural damage to the airframe because in this example, 490 knots is considered the speed over the ground, but the total velocity would be higher (about 596 knots) by the end of the turn due to the sink rate.

You'll also notice that, assuming we had the altitude to lose in the first place, the turn took longer to complete than in level flight. This is because the magnitude of the lift vector was smaller.

You can feel free to experiment with other speeds and bank angles, and you could also experiment with higher g turns (simply replace $g$ in the equations with $2g$ or similar). In some cases higher than 1g, you may actually gain altitude, though it's unlikely a 747 would be able to sustain a high-g maneuver which gains altitude.

As a quick second example, consider a 2g turn with 80° of bank at 400 knots:

$$
a_c = 19.302 = 2g\sin(80°) \\
R = 2193.8 = \frac{400 \times 0.514444}{a_c} \\
d = 6892 = R\pi \\
t = 33.5 seconds \\
a_u = 3.404 = 2g\cos(80°)
$$

Giving a total altitude lost of 3589 meters (11,775 feet) over 33.5 seconds. However, at the end of the turn, you would have a sink rate of 214 meters (703 feet) per second. That's over 42,000 feet per minute. It would likely be possible to recover from that in the altitude remaining, but it would not be pleasant.


Why does manifold pressure increase with power? Whilst writing an answer to this question, I remembered something I don't really understand.

The partial vacuum in the inlet manifold is caused by the piston descending with the inlet valve open during the "suck" phase of the Otto cycle.  It is this vacuum that causes fuel to be pushed into the intake, the greater the vacuum, the greater the fuel added to the inlet charge and the greater the power.

Since the needle valve is on the manifold side of the throttle butterfly along with the input to the MAP gauge, why does the manifold pressure increase with power?  I would have thought more power, more "suck", lower pressure; more pressure, less "suck", less fuel.

I'm obviously wrong since that's not how it works but why?

The answers to this question get close but I'm still missing something (and I'm sure will soon be embarrassed by the obviousness of the answer).

Manifold pressure is actually a measure of the vacuum pressure between the throttle and the cylinders. The more the throttle is opened, the closer that manifold pressure returns to atmospheric pressure.

The amount of fuel provided to the engine depends on the pressure difference between the manifold and the throttle body.  The throttle body pressure will decrease as the throttle is increased, because of greater rate airflow across the venturi (faster air -> lower pressure).  The manifold pressure will increase as throttle is increased, due to the greater flow of fuel/air mix into the manifold (greater mass of air flowing into a fixed volume -> higher pressure).  So the pressure difference will get larger as the throttle is increased.

This link has a pretty good description of what's going on.


Can I generate & transmit ACARS message via software? Is there any software available for generating ACARS message? If yes how to transmit that over RF?.
I want to do that for academic purpose. My settings are GNU radio and HackRF one.

You are unlikely to have a permission for sending technical messages over radio channels used by real aircraft.
But if you just want to make your drone to send you ACARS messages over WiFi, there are 144 open source projects currently on GitHub to deal with these messages. Most of them are parsers, not senders, but anyway selecting the best projects and then understanding the code should give enough insights about the message format. Also, parsers can be used for checking that your sender works properly.


Which are the advantages of low pressure levels during flight? (i.e during cruise flight in a commercial aircraft) I understand that due to health and security reasons, there must be a minimum in the pressure inside the cabin (75 kPa). But in class I have been told that maintaining low pressure levels has advantages, which ones?

You want the lowest pressure that can be tolerated by most of the population for the purpose of maintaining adequate blood oxygen levels, which is generally considered to be 8000 ft above sea level. At 8000 ft, someone with degraded lung function, like a smoker, is effectively at 10000 ft plus and is already at the threshold of having marginal blood oxygen for peak brain function.

At 40000 ft this requires a pressure differential of a little over 8 psi between the interior and exterior and you build the pressure hull to that and no more (it's a certification requirement).  As per min's comment this is partly to keep from having to make the pressure hull any stronger than it needs to be (every pound over the minimum is non-revenue ballast) and also to minimize engine bleed demand, which can be thought of as lost energy being used to keep the balloon of the pressure hull inflated instead of making thrust.  So, bottom line is it would be better for the airplane, both structurally and from an energy use standpoint, to operate with lower pressure differential and higher cabin altitude, but now the crew would have to start using supplemental oxygen and more people in the back would be getting sick.

The 787, because of its carbon fuselage, is able to go with a higher differential structurally, and its electric-pump pressurization system is a bit more efficient than taking bleed air directly, so Boeing had the option to certify to a higher differential pressure, 9 psi, which gives a 6000 ft cabin at service ceiling.  They could have used 8 psi and just shaved off more weight and used slightly lighter air pumps, but instead they increased the differential.  This was done mostly as a cabin comfort sales feature and was thought of as a bigger sales benefit than going with the weight saving route.


Why an external RAT on this early propeller plane at Charlotte? At the Charlotte airport, I caught a glimpse of the Bellanca 14-9L with State Airlines livery they have on display. They call the Bellanca the first plane with retractable landing gear but that's not what struck my attention:

There is a ram air turbine generator visible  and permanently located between the landing gear (circled). I only know of RATs as providing emergency power for airliners -- and they are only deployed (pushed outside) when they are needed. The other use for them is apparently on cropdusters.

Why did this airplane need a RAT permanently installed? Wouldn't it be much more power efficient and aerodynamic to provide DC/AC power through a magneto or alternator on the engine?

Late note: I looked at pictures of other Bellanca 14-9L and they don't show a RAT. Perhaps this means that this plane was not a passenger craft (as its "State Airlines" livery indicates) but was in real life a cropduster?

(Photo by author. Released cc by-sa)

Many older aircraft could be equipped with an optional external air driven generator, including, for example, the Luscombe 8A. Some air driven generator models were granted a Supplemental Type Certificate (STC) allowing them to be added to specific aircraft, such as the venerable J-3 Cub. The reason these were used is that, as you surmise, the engine is not equipped with any mechanism to drive an alternator or generator, i.e., there is no pulley, no gear mechanism, etc. Examples of those engines, still in use today, are the A-65, A-75 and original C-85. Newer designs of air driven generators are still available today, such as the BPE14 TURBO STAND-BY ALTERNATOR. The disadvantages include increased drag and exposure to ground obstacles, debris and of course, birds and bugs!


What is the meaning and difference between CTAF, UNICOM and MULTICOM in the US? What is the meaning and difference between CTAF, UNICOM and MULTICOM in the US? Could you please give examples of radio communications with the correct phraseology to be adopted in each case?

From the Pilot/Controller Glossary

  UNICOM− A nongovernment communication facility which may provide airport information at certain airports. Locations and frequencies of UNICOMs are shown on aeronautical charts and publications.

For example, an FBO may have a unicom frequency that they use to provide information to incoming aircraft and provide traffic advisories (not to be confused with instructions), wind directions, weather, and preferred runways. Unicom frequencies have a ground station at the airport in question.

It is important to note that when talking on a Unicom, the person on the ground that you are talking to in the FBO or airport office is not an air traffic controller.

An example of talking on this would go something like:

Pilot: "KXYZ Unicom, N12345 request traffic advisory"
Unicom: "N12345 No traffic in the pattern, winds XX @ YYY, recommend runway ZZ"
Pilot: "KXYZ Traffic, N12345 will be entering the 45 downwind for runway ZZ"

And so on just like a CTAF if the unicom/CTAF are the same, otherwise switch to the CTAF and make announcements.

  MULTICOM− A mobile service not open to public correspondence used to provide communications essential to conduct the activities being performed by
  or directed from private aircraft.

This is an advisory frequency that does not typically have a ground station associated with it. Pilots are to self-coordinate on the Multicom frequency but there is no ground station at the airport for that frequency. Often many small uncontrolled airports will use the same multicom frequency so you can hear traffic at multiple airports while on a mulitcom.

This is the typical position/intention reports that you normally make at untowered fields, only you are typically just broadcasting your intentions. Sometimes you may have to work with other incoming aircraft such as reporting each other in sight or telling somebody you'll extend downwind to follow them in, etc.

  COMMON TRAFFIC ADVISORY FREQUENCY (CTAF)− A frequency designed for the purpose of carrying out airport advisory practices while operating to or from an airport without an operating control tower. The CTAF may be a UNICOM, Multicom, FSS, or tower frequency and is identified in appropriate aeronautical publications.

Usually CTAF is used as a multicom, but there are cases where it shares the unicom frequency. If the field is towered and in operation, you'll be working with the tower to coordinate your actions. If the field is non-towered or the tower is not in operation, you'll make position reports and work in the frequency just like a multicom.


When must FAA reply to my letter? I have sent letters requesting action and/or clarifications to the FAA medical certifications branch.  How long can FAA legally delay before they are required to answer my questions in writing?

The FAA publication "MEDICAL FACTS FOR PILOTS" Publication #OK-06-002 states the time periods shown below. I don't know of any rule that mandates a response in any specific time period.

We will respond to your request for duplicate medical certificates within 3 working days of receipt.

We will respond to your request for full review of medical records within 30 working days of receipt of all the necessary records and testing.


What official government document should a pilot check to determine what magnetic north a VOR was set to when it was commissioned? Today I learned that VOR stations are all set to magnetic north at the time of commission, and since magnetic north moves, the VOR is not necessarily in-line with magnetic north in the present.

Where can a pilot look to read numerically what the VOR is set to as its magnetic north?  I understand a sectional chart could be measured, but I want to know where I can read the setting in a manual.

For example, is it contained in chart supplements?

The chart supplement does list the magnetic variation of the VORs. Looking under the "radio aids to navigation" section for a nearby airport, it's listed at the end of the line, right after the elevation.

  BATTLE GROUND (H) VORTACW 116.6 BTG Chan 113 N45º44.87´ W122º35.49´
  160º 9.6 NM to fld. 253/21E.

Another source is the radio fix and holding data records from the FAA. Among other information, you can find the magnetic variation in the remarks section, along with the year it was set.

  REMARKS:
  ...
  BTG MAG VAR 1975/E21

It's not official, but the easiest place I've found to get this is to search for the navaid on AirNav.


Is there a reason for a jet liner to raise the nose wheel early during the take-off run? This video shows a 727 apparently rotating too soon.

The take-off run starts at 2:00 minutes and at 2:11, you can see the yoke is pulled back by the elevators rising.

It seems to me that this is no mistake and that the early pull, held until the main wheels leave the ground and accelerating with the nose wheel held off, is very deliberate.

Is there a solid operational reason for doing this?

It's a 727 take-off technique used on unpaved surfaces.

Glimpse of the real deal here.

Possibly an old-timer was showing someone how it was done. Or someone was doing a recurrent for it.

It's to avoid nose gear damage if it hit a rock, or damage to other parts if said rock/debris were to be deflected. Also to avoid digging the nose gear into the soft field—reducing the rolling friction.


For large jets, what is the primary means of slowing down after landing? I understand that a large(r) commercial jet slows down after landing using the following methods:

Wheel brakes
Reverse thrust
Spoilers/flaps/airbrakes

What is the effectiveness of each compared to the others? What if one of the first 2 fails? Can an airliner usually be stopped without using wheel brakes or without using reverse thrust?

And what if the runway is slick because of rain, snow or ice? Does that change the equation?

The brakes are much more effective than anything else, at least for the jet I have experience in (EMB-145).

Our landing performance data typically assumes full braking application, spoilers deployed and no reverse thrust.  The airplane will most certainly stop without using reverse thrust, just not as fast.  I never tried not using the brakes, and if I were to I would want a light airplane and a very long runway.

As Lnafziger pointed out in the comments, the brakes are part of redundant systems so that they are always available.  In the EMB-145, the brakes were serviced by two independent* hydraulic systems. The inboard brakes were on one system while the outboards were on the other for a total of four wheels each with big carbon fiber brakes.

To address the comments on slick runways, assuming braking action allows an attempt to land in the first place, you have to be careful. With reverse thrust alone you may not be able to maintain the runway centerline and directional control may become difficult through uneven deployment of the reversers and other factors.  In this scenario the spoilers are helpful to increase traction and brake effectiveness.  In short, if the runway is slick then the only way I'm attempting a no-wheel-brake landing is after declaring an emergency.

Finally, to address the question in the title, the wheel brakes are always the primary means of slowing the airplane.

*You could transfer fluid between our systems through the parking brake.


Why is the reheat exhaust flame of some fighter aircraft blue while for some it is orange? Why is the afterburner flame of some fighter aircraft blue while for some it is orange?

I agree with Simon's comments. Lighting and ambient lights make all the difference.

The flame is always equally bright, so if the ambient light is lower, the difference between the flame and the rest of the image is larger. When the difference exceeds the dynamic range of the film or sensor, the image of the flame becomes saturated in all channels and ends up as white. Note, that all the images with white flame are night ones.

Also the blue sensor is slightly sensitive to near infrared (being about half wavelength of the blue light), so when high sensitivity/long time are used, this causes the blue glow around the white flames on night images, while on day images with lower sensitivity it is not noticeable.


Can "a drone" damage the wing of "an aeroplane" to the extent it actually threatens its ability to fly? I stumbled across this video by the University of Dayton, showcasing the amount of damage a certain type of drone can do to a wing, and they wrote a whole article about it. While most of said article sounds like marketing for their research institute with very little actual information about the video they just released, the key thing to take away seems to be that:

researchers in UDRI’s Impact Physics group launched a 2.1-pound DJI Phantom 2 quadcopter at the wing of a Mooney M20 aircraft. The drone did not shatter on impact, but tore open the leading edge of the wing as it bore into the structure, damaging its main spar.
[...]
The researchers then fired a similarly weighted gel “bird” into a different part of the wing to compare results. “The bird did more apparent damage to the leading edge of the wing, but the Phantom penetrated deeper into the wing and damaged the main spar, which the bird did not do.”

If we trust that the researchers accounted for the fact that a damaged wing's structure is obviously compromised in places and that they also accounted for bones making an actual bird slightly more rigid than pure gel, as well as repeating the experiments often enough to make above statement with a reasonable degree of confidence -- none of which the article mentions explicitly -- then what conclusions can we draw from this?
It looks dramatic. Sounds dramatic, too, if you know nothing about aviation at all, which I guess was the intention when they wrote the article.
But how significant is the damage really?
The article doesn't say.
One YouTube comment by one "Stephen Mann", appearing to claim piloting experience states:

You can throw, at 200 MPH, anything weighing 2 pounds at a sheet of 2024-O .04" aluminum, typical of aircraft wings, and get this damage.  I've seen worse hail damage.

and notes that:

the Mooney wing uses a thinner aluminum (0.038") than an airliner (0.04 to 0.05").  This made the Mooney lighter and faster, which was a marketing design.
I've had more than a few bird strikes during landing (none during takeoff that I was aware of).  Some were Canadian geese that weigh 10-15 pounds.  There was never any damage to the aircraft beyond cleaning up the bird guts from the wing.

while generally playing down the significance of the test:

Typically, no aircraft below 400 ft., where most drones operate, will be flying that fast.  On final approach most airliners are flying between 130 to 160 knots, and by the time a landing aircraft is below 500 ft. the flaps and slats are deployed and the A/C is probably going 120 knots.  General aviation aircraft, like the Mooney in this video, are typically flying between 70-110 knots on final approach.  Even slower on takeoff.

I'm reluctant to accept information from YouTube comments at face value -- especially since it may be biased. But if true, this may just reduce the usefulness of the test significantly. And although I guess that's quite possible, I'd like to think they put some thoughts into these experiments before conducting them.
So where does that leave us?
Can "a drone" bring down "a plane" by crashing into it? Most likely, if we're talking about these big US military type drones like the RQ-4.
But can a "hobby drone" (for lack of a better word) like the one shown in the video by the University of Dayton seriously threaten the fly-ability of "the average plane" (if such a thing exists)?
Or let's say we start at the Mooney and scale the damage up for larger planes. Would a plane with that kind of damage get into trouble whatsoever or would it just require a probably costly repair?

Mr Mann doesn't have a clue what he's talking about.

Wings aren't made from 2024-O aluminum, which is fully annealed and very soft.  The usual heat treat state is T3 through T6.

A Mooney would NOT have .040" wing skins.  The skins will be .016, .020, .025 and maybe .032 in high stress areas around the wing roots. Leading edges would be .025 or maybe .032 inboard, and likely .020 or .016 near the tips and on the control surfaces.  Skin sizes are for structural efficiency and rigidity for minimum weight, not "marketing"... geez.

His approach speed comments are more or less correct, but a Canada Goose will do significant damage to any aircraft it hits in flight and has a good chance of bringing down a Mooney at any contact point.  A Mooney running into a goose at the wing will, at minimum, completely crunch the leading edge in back to the main spar.  A goose will do significant damage to the slat or leading edge of any airliner.  There was a CRJ900 that hit a goose size bird just below the windshield on the takeoff roll near V1, and it came back into the pilot's instrument panel, distributing guts and feathers all over the place.

Geese are bad news and are a vexing problem in the spring in central North America.  The terrain is all browns and if you are descending with a flock below and in front they are nearly invisible until you are passing through their level.

A goose hitting the windshield in front of the pilot or horizontal tail of a Mooney or similar light aircraft will probably bring it down.  On the wing, maybe.

A 2lb drone will do obvious damage as the test showed but would not bring down a Mooney unless it hit the windshield, incapacitating the pilot.  It would certainly go though a windshield.  1/8" plexiglass has roughly the same tensile strength as .020" 2024T3.   You wouldn't want to hit anything heavier than a few ounces with 1/8" of acrylic in front of your face as the only barrier.

So from that standpoint, drones of any size can be considered a hazard to aircraft.


Why are there no "sleeper" planes? On a recent long-haul flight I sat by the windows and was dreaming away staring into infinity out of the window. Then a thought came up, in Asia sleeping busses are very common. This made me think, why are there no dedicated sleeper planes? Especially for the long-haul 10 to 13 hours this would be ideal.

There certainly are; it's called "Business Class" and you'll find such areas with seats that recline into what are essentially beds on most long-haul aircraft. However, you simply can't fit as many people on board when they use up so much room, so airlines have to charge much more for these to make it economically viable. It's often twice the price of a normal ticket or even more.


Is there any source of live ATC audio with subtitles? I'd like to know if there is a way to hear ATC and pilots' communications, with subtitles.

Not live with subtitles, but there are certainly plenty of ATC/pilot recordings on youtube
https://www.youtube.com/results?search_query=atc+conversations

Plus Live ATC streams. I can't access these (lousy cell phone coverage), give them a try.
https://www.liveatc.net/topfeeds.php


How do uncoordinated turns (skid, slip) affect a helicopter's turn radius? Is a helicopter's flight path (track made good) an arc of constant radius regardless if it's a coordinated or uncoordinated turn? (Assume nil wind conditions and that the pilot maintains the angle of bank, altitude and airspeed constant in both cases.)

I'm confused because texts mention slipping and skidding as sliding downhill and skidding uphill from the turn respectively, which to me reads like the radius of the turn changes.

[EDIT] Updated assumptions to include bank angle and airspeed.

Given your assumptions about bank angle and airspeed and my assumption that coordinated turn is determined by acceleration inside aircraft pointing downward "perpendicular to the floor" (if you mean anything else by coordinated turn, try to be more specific on it), then you have one exact radius for coordinated turn. That one where centrifugal force generated by airspeed and curvature summed with gravity gives vector matching the (fixed) bank angle. If you want local acceleration to point in other direction (making slip or skid), only free variable remaining is turn radius.

So yes, given airspeed and bank angle is constant, turn radius is the parameter determining if turn is coordinated or not.

In a slip the local acceleration in aircraft points toward that side you want to turn (or, in other words, aircraft's floor is banked too much outside the turn). To get this, centrifugal force has to be less compared to coordinated turn (centrifugal force is only quality you allowed to change), so turn radius have to be larger. To get a skid turn radius have to be smaller for the same reason.

Inertial and gravity acceleration as seen and felt inside an aircraft during a turn with bank angle $\alpha$:

where $\vec{g}$ gravity acceleration and $\vec{a_c}$ centrifugal acceleration. Resulting $\vec{a}$ is total acceleration felt by pilot. From basic mechanics $a_c=v^2/r$ where $v$ and $r$ is speed (either TAS or GS, both are the same here) and turn radius respectively. Turn radius for coordinated turn is $\tan\alpha=a_c/g\Rightarrow r_{\rm coord}=v^2/(g\tan\alpha)$. Any smaller radius with the same bank and speed leads to higher $a_c$ and thus skid, larger radius results in slip. Note that this is true for any aircraft, not specific for helicopter.


Does the PFD/ND XFR work when the PFD is on the ND screen on the Airbus A320? So I know the PFD has priority over the ND and that's why if the PFD screen fails its automatically switched to the ND screen. If the ND screens fails it does not get automatically transferred to the PFD screen because the PFD has priority. You can view the ND screen on the PFD screen by pressing the PFD/ND XFR button. So my question is if the PFD is on the ND screen because the PFD screen failed if you press the PFD/ND XFR will it show the ND information on the ND screen, and when you press it again it will go back to the PFD?

From the FCOM A320:

As you can see if you fail the PFDU and the PFD is transferred on the NDU, by pressing XFR PFD/ND you will have the ND appearing on the NDU

Full description: