Given that most of us don't sit in airline boardrooms it's difficult to say for sure why an airline favors one particular aircraft over another. But I'll have a go anyway. The A380 was initially marketed as having unbeatable costs per passenger compared to similar aircraft, i.e. the 747, so if you wanted to carry a lot of people a long way it was a logical choice.

In reality, those unbeatable costs only come if you can actually fill the seats. The introduction of the A380 coincided with rising fuel costs and the Global Financial Crisis. Less passengers and more expensive fuel helped the business case for heavy twins such as the 777, 787 and now the A350, and most airlines have preferred those types to the A380.

Emirates love the A380 because of some fairly unique circumstances:

Emirates operates a hub and spoke model in an excellent geographic location. It has its main base in Dubai and most of its business comes from connecting passengers through this one city - i.e. passengers travelling between Asia, Africa, Australia and Europe.
Emirates carried 58 million passengers in FY2017-18. How do you carry this many people through one airport, without running out of space? The answer is in bigger aircraft (and soon a second, bigger airport). Other airlines with such passenger numbers usually have multiple bases, and more point-to-point operations.
Finally, Emirates is less sensitive to extra operating expenses than other airlines due to its low cost base. This is largely thanks to favorable government conditions (such as labor and tax laws). Some even argue that they receive outright subsidies from their government.


What the weather condition can affect for flight?

It's actually fairly simple when talking about weather related dispatch delays for the airlines.

Number one items:

In summer, thunderstorms, especially those associated with fronts and squall lines.

In winter, snowstorms with heavy accumulations, or severe freezing rain.

And beyond that, in both summer and winter, extreme low visibility, as in fog.


The same reasons that run-flat tires aren't more common on cars or trucks are amplified with aircraft.

They're more expensive and can wear out faster. They will also be much heavier. This has to be traded against the cost of an occasional flat tire.
They typically have a maximum speed when flat. Aircraft tires need to operate at much faster speeds than vehicle tires. Many of the examples you provided were made worse by under-inflated tires.
They typically have very limited range when running flat. Again, many of your examples involved under-inflated tires. If they left a regular tire to burst, the same would happen to a run-flat tire.
Many tire failures are due to the tires bursting, not just a simple puncture that makes them go flat. A run-flat design will not help in this case, and may even make the fragments more dangerous.

All of your airliner examples were on pretty old aircraft models, and many included additional factors. Regulations and procedures change over time in response to these sorts of accidents. When properly maintained and operated, aircraft are designed so that a disintegrating tire won't cause serious hazards.


About how much drag does a non-operating engine create?

An inoperative engine creates much less drag than a flat plate of the same cross section. According to Sighard Hoerner's Fluid Dynamic Drag, the drag coefficient of a flat plate is 1.17. An engine nacelle has rounded intake lips which help the flow to stay attached while flowing around the nacelle. The closest of the generic bodies in the table below would be the sphere (drag coefficient of 0.47).

Figure 33 from Sighard Hoerner's Fluid Dynamic Drag, Chapter 3. Left column: Bodies of rotation; right column: Cross sections of 2D-bodies.

Much depends on the detail of flow separation at the forward corner, and here modern engines are rather good. If the flow stays attached, drag will be much lower than with the massive separation around and behind the flat plate. Air flowing out from the inside and over the corner of the flat plate will need some space to "turn around", effectively increasing the blocked cross section that the outside flow experiences.

Note that the reference area for all values here is the cross section of the body perpendicular to the flow, while the reference area for airplanes is their wing area.

Depending on the type, the diameter of one GE90 engine is 134 or 135 inches. Two of those power a Boeing 777 which has a wing area of 427.8 m². The ratio of engine frontal area to wing area is thus 1/23.335. The (estimated) drag coefficient of the nacelle of 0.5 would shrink to 0.0214 when referenced to the wing area. This would increase the total drag by about 50%, so instead of an L/D of 18 without engines, the airplane would have an L/D of 12 with inoperative engines.


Changing the interception angle in autopilot for OBS or VOR in the G1000 from 30 to something more aggressive

The manual does not state any way to do that. You can adjust the vertical navigation flight path angle, but not the horizontal intercept angle. You'll have to fly that yourself. In addition, in this case you should be telling the G1000 where you want it to go, rather than which heading to go. If you just tell it the heading it will never figure out what it has to do to get you to that nearby airport, so in this case it is working as advertised.


What can cause a watchdog failure in an INDRA MSSR?

1) What I expect is that there is a transmitter (or multiple) on 1090 MHz to the west of the RADAR that does not conform to the ICAO standard. I would start by looking at the decoded replies that the MSSR is receiving (not in ASTERIX format, but raw datagrams format. I believe the INDRA MSSRs can provide this). This may give a hint as to what the cause is(e.g. a certain 24 bit address range, certain downlink formats only, received power level, etc.).

2) Are there any military bases to the west of your RADAR? Or RADAR manufacturers? A transponder manufacturer? It may help to contact them. (Likely they will deny any experiments are going on, but the problem may suddenly disappear. It happened before).

3) Atmospheric tunneling may cause a specific transmission path only to exist under certain weather conditions (e.g. inversion layers) . Can you find any correlation between the interference and weather conditions?

4) Do you have multilateration systems covering the same area. The multilateration system may be able to locate the source.

If nothing else helps, you may have to install an array of 1090 MHz receivers and multilaterate the source.


What does N mean after a runway visual range (in a METAR)?

RVR often has a letter at the end to indicate the visibility trend, so you can tell if it's going to improve (and maybe you should circle overhead for a while) or worsen (and you should consider going to your alternate).

U means an "upward" trend: visibility is improving. D means a "downward" trend: visibility is worsening. N means no change.


Turbo-electric engine vs high by-pass-ratio Turbo fan, which config. would seem more efficient? (Liquid Hydrogen Fuel)

Turbo-electric configurations do not offer any meaningful advantages for commercial airliners.

Aircraft engines have extremely high power-to-weight ratios. These would be greatly degraded by adding motors and generators.
For instance, the RR T406 offers power in excess of 10 kW/kg. Large turbofans are even better. Most electric generators and motors are worse. The best have the same power-to-weight, but since they only transfer power, this increases the weight of the engine system by a factor of 2.5-3.

Wings are already the optimal location for both the generator and the propulsor.
You want your heavy engine where the lift is created, and that's the wings. You want your propulsors away from the fuselage, and that's either the wings or the tail.
Thus, there is no need to move the power any distance.

Hydrogen is a suitable fuel for turbofan engines and requires no other intermediate devices to convert it to thrust.
There is no significant need for hybrid airliners, as there's no stops to recover energy in flight.
Weight means fuel burn. Engines are heavy: in an A320neo, for example, they take up 14% of the weight. Increasing it 2.5x will make the aircraft 20% heavier. After adding payload and unusable fuel, this results in 15% more fuel burn, if the motors and generators were perfectly efficient.
Since they are not perfectly efficient, a naive replacement would result in even worse fuel consumption. This could be offset by perfecting propulsor size, speed, and placement, but the end result is still that a turbo-electric aircraft will be less efficient than a similar turbofan aircraft.
As to hydrogen itself, it is not the best aviation fuel. However, it is usable. The overall energy efficiency will be worse, due to the loss of cabin space to tanks. Whether that can be offset by clean production is a question outside of aviation.
Regardless, a 15-year horizon is so short that it can be safely said that hydrogen or electric or hybrid airliners will not be anything more than concepts and experiments in this timespan. 75 years, that could be a topic for debate, on appropriate sites.


How to calculate Equivalent Airspeed immediately from Calibrated Airspeed?

I wanted to add to the very detailed and helpful answer above to clarify some things as hopefully it might help someone. Most pilot ground studies courses approach this from the viewpoint of how to go from IAS to CAS to EAS to TAS but skip a lot of detail, which I think is what is frustrating the OP. So here's a deep dive. Also, the previous answer ends with "Solving it for EAS = f(CAS) is left to the reader". Since that's sort of what the OP was asking I thought it might be worth a crack at it, although from a different angle.

IAS (Indicated Air Speed)

This is the speed that the air speed indicator in the cockpit displays. The best way to think of the ASI it that it's a pressure gauge that measures differential pressure (pitot minus static) only it's calibrated in knots (or whatever.) As you say in your question, IAS is a function of (p_total - p_static). Is to what that function is, we'll come to that in a moment.

CAS (Calibrated Air Speed)

What the instrument should have read if the instrument and the pitot/static system were perfect. In other words, the IAS after correcting for position errors (e.g. effect of flap setting on pitot/static system) and instrument errors. CAS can be derived from IAS by using look-up tables but these are aicraft-specific. In modern systems, this is done automatically by the Air Data Computer, so the ASI will directly read CAS. In other words, CAS = f(IAS, hardware errors, Mach, AoA, flap setting, ...) To find out this function, you have to do computational fluid dynamics, wind tunnel testing, flight testing or all three. So it's non-trivial. I can't just give you an equation for it. The good news is, in straight and level flight, the errors aren't usually very big.

EAS (Equivalent Air Speed)

This is where things get weird and annoying. Many aviation courses and textbooks treat EAS as a "stepping stone" to TAS but it's almost never used that way in reality and these sources never tell you how to calculate EAS from CAS. It's usually easier to go straight to TAS from CAS, e.g. using a CRP-5 flight computer (a kind of circular slide rule) or by let the aircraft work it out for you via the Air Data Computer. As a pilot, you very rarely need to actually calculate EAS. CAS, TAS and Mach number are usually far more useful.

EAS is supposed to be a version of CAS corrected for "compressibility errors". This might lead you to think EAS is necessary because the ASI uses some simple non-compressible flow assumptions, which need to be corrected at higher Mach numbers. However, at sea level, EAS is the same as CAS, regardless of airspeed, by definition Wikipedia's definition. I.e. you can be in a very compressible flow regime, like Mach 0.8, and EAS is still the same as CAS. This initially makes no sense until you realise that the definition of EAS has to make an assumption about how an air speed indicator actually works.

One might think the airspeed indicator is calibrated by simply reversing the well-known formula (from Bernoulli) for static pressure, namely:

$q = \frac{1}{2}\rho v^2 $,

like this:

$v_{IAS} = \sqrt{\frac{2q}{\rho}} $,

where $\rho$ is the air density and $q$ is the dynamic (pitot minus static) pressure.

This is not so. Here's where we get to "things they don't tell you in ATPL ground exam courses." It seems that the standard definitions of EAS makes the assumption that the airspeed indicator actually works by using isentropic flow relations. This would require it to know things like the static pressure (it only senses differential) and the local speed of sound (which requires knowledge of outside air temperature.) However, it is still possible to use this approach if you "cheat" and put in sea level values for all the things the instrument doesn't know. In other words, the ASI calculates IAS according to:

$v_{IAS} = a_0 * \sqrt{5\left[\left(\frac{q}{p_0}+1\right)^{\frac{2}{7}} - 1\right]} $

Where $a_0$ is the ISA (international standard atmosphere) sea level speed of sound (340 m/s), $q$ the dynamic pressure as before (aka impact pressure) and $p_0$ is the ISA sea level pressure (101325 Pa).

Note that the magic numbers 5 and 2/7 arise because of the value of $\gamma$ (gamma), the ratio of specific heats for air, which equals 1.4. The more general form is:

$v_{IAS} = a_0 * \sqrt{\frac{2}{\gamma-1}\left[\left(\frac{q}{p_0}+1\right)^{\frac{\gamma-1}{\gamma}} - 1\right]} $

This relation is a well known thing but I don't know how to derive it. Yet. (In any case, this answer is already long enough.)

It turns out that if you take a binomial approximation of the 2/7 power and neglect second order terms and higher, you get this:

$q \approx \frac{1}{2} 1.227 v^2$

The value 1.227 being very close to the ISA sea level density of 1.225, so it is equivalent to Bernoulli for low Mach numbers. Anyway, I digress.

Calculating EAS

From here on, we'll asssume IAS=CAS. Now let us assume you want to find the EAS given only the IAS/CAS and the altitude. Normally, as a pilot, perhaps tackling an ATPL exam question, you'd use a CRP-5, but let's assume we are proper nerds and we want to code this (e.g. in python) or get a better undersanding of how an ADC (Air Data Computer) or ADIRU (Air Data Inertial Reference Unit) works.

The EAS is a function of the CAS, the dynamic pressure and the static pressure.

Finding static pressure from altitude

The static pressure can be inferred from the altitude using the International Standard Atmosphere. For the troposphere (i.e. up to 11km or 36,080ft) the pressure is calculated by solving the hydrostatic equation for a constant lapse rate. Dealing with the tropopause and above is left as an exercise for the reader :-P

$p_s = p_0 \left( \frac{T_s}{T_0}\right)^{\frac{g}{L R}} $,

where $p_s$ is the static pressure at altitude, $p_0$ is the ISA sea level pressure (101325 Pa), $T_0$ is the ISA sea level temperature (288.15K, $g$ is the acceleration due to gravity (9.81m/s2), $L$ is the lapse rate (0.0065K/m), $R$ is the specific gas constant for air (287 J/K/Kg) and $T_s$ is the static temperature at altitude:

$T_s = T_0 - L h$,

were $h$ is the altitude in metres.

Finding the dynamic pressure from CAS

Note that the ASI does not tell you (the pilot) the dynamic pressure directly, it only tells you it in terms of the IAS/CAS. So we have to "reverse" the formula above:

$q = \left[\frac{\left(\frac{v_{CAS}}{a_0}\right)^2}{5} + 1\right]^\frac{7}{2} - 1$

Finding the Mach number

Next we need to find the Mach number. Note that this is a very similar equation to how the ASI calculates airspeed, but this time we use the real static pressure, not the sea-level value:

$M = \sqrt{5\left[\left(\frac{q}{p_s}+1\right)^{\frac{2}{7}} - 1\right]} $

The EAS is then defined as:

$v_{EAS} = a_0 M \sqrt{\frac{p_s}{p_0}} $

In other words, EAS uses the real Mach number but the sea level speed of sound to obtain a speed, and then multiplies this by the root of the pressure ratio. Putting this all together:

$v_{EAS} = a_0 \sqrt{5\left[\left(\frac{q}{p_s}+1\right)^{\frac{2}{7}} - 1\right]} \sqrt{\frac{p_s}{p_0}} $

Epic.

TAS

TAS can be obtained from EAS, which in effect gives another formal definition of EAS:

$v_{TAS} = v_{EAS} \sqrt{\frac{\rho_0}{\rho}} $,

where $\rho$ (rho) is the actual air density at altitude (which can be calculated from temperature and pressure using the ideal gas law ($\rho = \frac{p_s}{R T_s}$) and $\rho_0$ is the ISA sea level density (1.225 kg/m3.) In other words, TAS is just the EAS divided by the root of the density ratio.

However, it's actually easier just to calculate TAS directly, which saves knowing about densities:

$v_{TAS} = a M $

where $M$ is the Mach number as calcuated above and $a$ is the local speed of sound (not the sea level value like the in the EAS formula,) which can be calculated from $a = \sqrt{\gamma R T_s}$.

#!/usr/bin/env python3
#
# Calculate EAS and TAS, given CAS and a pressure altitude
# Warning: only works in troposphere, i.e. below 11km
#
# By Halzephron 2020
# THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
# IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
# FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL
# THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
# LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
# OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
# SOFTWARE.

from math import pi, sqrt
import sys

ms_per_kt = 0.51444
feet_per_metre = 3.28
T_0C = 273.15 # 0 degrees C in Kelvin

# Some useful constants

g=9.81          # Acceleration due to gravity
R=287           # Specific gas constant for air
L=0.0065        # Lapse rate in K/m
T0 = 288.15     # ISA sea level temp in K
p0 = 101325     # ISA sea level pressure in Pa
k = 1.4         # k is a shorthand for Gamma, the ratio of specific heats for air
lss0 = sqrt(k*R*T0) # ISA sea level speed sound
rho0 = 1.225    # ISA sea level density in Kg/m3

# Return knots given m/s
def kt(m):
    return m/ms_per_kt

# Return pressure ratio given a Mach number and static pressure,
# assuming compressible flow
def compressible_pitot(M):
    return (M*M*(k-1)/2 + 1) ** (k/(k-1)) - 1

# Return Mach number, given a pressure ratio d=p_d/p_s
def pitot_to_Mach(d):
    return sqrt(((d+1)**((k-1)/k) - 1)*2/(k-1))

# Given an altitude h, return the temperature, assuming we're
# using the International Standard Atmosphere and are flying
# in the troposphere.
def temperature(h):
    return T0 - h*L

# Given an altitude h, return the local spead of sound, assuming
# we're using the International Standard Atmosphere and are flying
# in the troposphere.
def lss(h):
    return sqrt(k*R*temperature(h))

# Given an altitude h, return the pressure, assuming we're
# using the International Standard Atmosphere and are flying
# in the troposphere.
def pressure(h):
    return p0 * (temperature(h) / T0) ** (g / L / R)

# Given an altitude h, return the density, assuming we're
# using the International Standard Atmosphere and are flying
# in the troposphere.
def density(h):
    return pressure(h) / (R * temperature(h))

if len(sys.argv) < 2:
    print("usage: {} CAS ALT".format(sys.argv[0]))
    exit(0)

cas = float(sys.argv[1])*ms_per_kt # Convert kts to m/s
alt = float(sys.argv[2])/feet_per_metre  # Convert ft to m

ps = pressure(alt)
lss = lss(alt)
oat = temperature(alt)
rho = density(alt)

# First we need to "reverse" the air speed indicator to find the dynamic
# or "impact" pressure. It is tempting to assume that an airspeed
# indicator relates airspeed to dynamic pressure using Bernoulli's
# 0.5 * rho * v**2, but it that is not the case here. Instead,
# a sort of modified compressible flow equation is used. A "pseudo"
# mach number is found as a function of a pressure ratio, assuming the
# static pressure is equal to ISA conditions. The airspeed is then found
# by assuming the local speed of sound is that of ISA sea level.
pd = compressible_pitot(cas/lss0) * p0

# Find Mach number
M = pitot_to_Mach(pd / ps)

# Calculate EAS (equivalent air speed)

eas = lss0 * M * sqrt(ps/p0)

# Calculate TAS (true air speed)

tas = lss * M

# Or we could have used tas = eas / sqrt(rho/rho0)

print(f"Pressure Altitude:     {alt*feet_per_metre:5.0f}   ft")
print(f"Static Temperature:      {oat-T_0C:5.1f} C")
print(f"Density Ratio:            {rho/rho0:6.3f}")
print(f"Pressure Ratio:           {ps/p0:6.3f}")
print(f"Static Pressure:        {ps/1e2:6.1f} mb")
print(f"Dynamic Pressure:       {pd/1e2:6.1f} mb")
print(f"Total Pressure:         {(pd+ps)/1e2:6.1f} mb")
print(f"CAS:                    {kt(cas):6.1f} kt")
print(f"EAS:                    {kt(eas):6.1f} kt")
print(f"TAS:                    {kt(tas):6.1f} kt")
print(f"LSS:                    {kt(lss):6.1f} kt")
print(f"Mach:                      {M:4.2f}")


Are crosswind landings easier with larger airplanes?

Generally yes, since a 20kt crosswind requires considerably less crab angle on final at 135kt compared to 70kt.  The CRJ900 has a demonstrated max crosswind component of 32 kt (If I had those conditions in a 172 and had to put it down, I think I'd just land across the runway).
There is also a significant technique difference once you get above, say, 100,000 lbs, or depending on the engine location.  On heavies where you only have a few degrees of banking margin when the gears touch before the nacelles hit, the technique is to fly wings level crabbing into wind into the flare, and give a large boot of rudder just prior to touchdown, and get it down before you start drifting off to the side (you are effectively in a skidding turn, but there is a delay before you start to actually change direction and you have to achieve ground contact in that delay period, so you make it snappy and don't work on teasing out a smooth touchdown).
On smaller aircraft it's more the traditional sideslip technique, wing low with opposite rudder maintained to keep runway alignment through the flare and touchdown. My jet time was on CRJs and I would land in crosswinds slightly wing low with rudder to keep aligned, similar to landing a glider in a crosswind (it took about 7-9 deg of bank to risk tip contact and a crosswind would never need much more than 3-5, although tip strikes do happen in the CRJ fleet from time to time when pilots have an oops on landing).
Transport airplanes generally don't have a maximum crosswind component as a published limitation, but will list a "demonstrated" crosswind component with the statement "not considered limiting".  It means that it's only the maximum component demonstrated in testing, and you are free to land in stronger ones, but the manufacturer doesn't vouch for the ability of the plane to handle it.


What is that instrument with two knobs and what looks like range rings?

WX-10 Stormscope

The WX-10 System consists of a display, remote processor, and an antenna. This Stormscope offers thunderstorm information on a bright green CRT display. The more the dots on the display are clustered together, the stronger the storm activity. Up to 256 dots may be displayed. The WX-10 also has a Forward mode to concentrate the display and processing power on the airspace ahead of you.
— bennettavionics.com

See also: Wikipedia: Lightning detection § Aviation use


Is it possible to revive a pilot who has had a medical condition in a seated position in a small aircraft?

I'm going to preface this by saying I'm not a doctor, however I am trained in first aid and CPR.

If you are asking whether it would be possible to revive someone who is in respiratory or cardiac arrest to the point where they could fly again then the answer is absolutely not.

The movies make it seem like you can give CPR and 2 minutes later the person is running around shooting guns, however that's very far from the truth. CPR and rescue breathing are techniques which can, if done properly, keep someone's blood moving enough that radical emergency medicine may save their life. CPR cracks and/or breaks ribs and can cause internal injuries even when it's done right, rescue breathing doesn't generally cause injury, however anyone who has needed either will require days, weeks or even months of intensive care and recovery.

As to whether it would be possible to administer CPR or rescue breathing to someone in a light airplane then I would say yes, with extreme difficulty and a low chance of success. CPR requires a lot of force, you use the full weight of your upper body concentrated on one single point on the person's chest, unless you could lay the person flat on a hard surface you'd probably be wasting your time, so if the seats reclined a person in the back seat may be able to lean over and do it, although that's extremely chancy of working due to the angle and the cushioning of the seat. Rescue breathing is possible with a seated person, provided you can open the airway, getting the angle on that would be very tricky in the constrained area of a small cockpit. Again, if the seat could recline a person in the back would have a better chance of doing this.

Most importantly to remember is that if the sole pilot dies or becomes incapacitated someone else has to fly the airplane. Struggling to get into position to perform CPR or mouth to mouth on a seated person would be a great way to snag the stick and send the airplane out of control. If there's a back seater by all means let them give it a shot, but your best bet at survival is getting control of the airplane and calling ATC for someone to talk you through things, it's happened before and has a remarkably good survival rate.


Why can't horizontal stabilizers be longer than they are today?

The horizontal stabilizers can be longer, they just don't need to be any longer than they are.

Each additional square inch will add induced drag and parasitic (form / profile) drag which costs fuel so they're not made any bigger than necessary to provide adequate control of the plane.


My question is, when there's no defined cloud layer, what is
  considered "cloud" and what isn't? If the minima require you to stay
  500 feet below clouds, where exactly is that? With metars showing no
  ceiling how do you know where it is safe to fly and where there might
  be an ifr flight descending that can't see you?

The FAA defines clouds in AC-006B

  A cloud is a visible aggregate of minute water droplets and/or ice
  particles in the atmosphere above the Earth’s surface. Fog differs
  from cloud only in that the base of fog is at the Earth’s surface
  while clouds are above the surface. Clouds are like signposts in the
  sky that provide information on air motion, stability, and moisture.
  Clouds help pilots visualize weather conditions and potential weather
  hazards.

However legally and practically speaking you can fly as high as you like if there are no clouds provided you can maintain situational awareness and proper cloud clearances from any adjacent clouds (including those under you). There are a few practical limits to this. First off Class A airspace starts at FL180 so you cant fly above 18,000 Ft. without being on an IFR flight plan since you cant fly VFR into class A. Further more, since you seem to be asking about small aircraft, you will more than likely be limited by your airframes maximum altitude before hitting the Class A limit. You will also need O2 onboard to fly above 12,500 ft.

The risk of a descending aircraft above you is always a concern be it VFR or IFR flying as they generally cant see below them and you cant always see above you when flying. In many cases the IFR flight will have some help from ATC who will see you on radar even if you are flying VFR and squawking 1200 on your transponder.


What is the difference between "x" and "+" configuration in missiles?

Mounting missiles below a wing or stocking them in weapon bays is much easier with X-Configured missiles. Look at these photos:

K-5M Missile Source

AIM-120 Missile Source

"+" mounted missiles would need more complex hard points and larger weapon bays.

Once fired the configuration doesn't matter anymore as all four control surfaces are continuously used to guide and stabilize the missile along its 3d trajectory.

If the missile only moves in the horizontal or vertical plane, the X configuration is better. For a fixed force F the angle of attack of the control surfaces ($\alpha_{cs}$) is smaller and can therefore be set faster, making the missile more agile. Assuming* a linear relationship between the the control surface force and its angle of attack:

$$\alpha_{csx} = \frac{F}{q \cdot S \cdot C_{L\alpha}}\cdot \frac{\sqrt{2}}{4}$$
$$\alpha_{cs+} = \frac{F}{q \cdot S \cdot C_{L\alpha}}\cdot \frac{1}{2}$$

For a quadratic relationship* between the drag ($D_{cs}$) and the angle of attack of the control surfaces there is no difference in the total drag.

$$D_{cs+} = q \cdot S \cdot (4\cdot C_{D0}+2 \cdot C_{D\alpha} \alpha_{cs+}^2) = 4\cdot q \cdot S \cdot C_{D0} + \frac{1}{2} C_{D\alpha} \cdot (\frac{F}{q \cdot S \cdot C_{L\alpha}})^2$$

$$D_{csx} = q \cdot S \cdot (4\cdot C_{D0}+4 \cdot C_{D\alpha} \alpha_{csx}^2) = 4\cdot q \cdot S \cdot C_{D0} + \frac{1}{2} C_{D\alpha} \cdot (\frac{F}{q \cdot S \cdot C_{L\alpha}})^2$$

*If someone has better assumptions for lift and drag please comment or correct the post.


How do I calculate the centre of gravity on the lateral (transverse) y-axis?

Transport airplanes have lateral C of G limits in terms of x distance from the centerline, but this only a certification requirement (related to roll authority in a critical case situation) and is only expressed for the flight crew as a fuel imbalance limit, fuel being the dominant factor, and is not accounted for in weight and balance data that the crew uses to determine if loading is correct.
The datum is the x=0 axis, with the C of G to the left, on, or right of it.  To calculate it, you would need an empty lateral C of G position, determined by weighing the airplane and working out the empty C of G offset to left or right of the x axis based on the weight difference between the left and right main gear weighing points, and the distance of those points from x.
Then you can add fuel and passengers and calculate the change if any to the lateral C of G. Since the passengers are normally randomly distributed on the left and right side of the cabin, the passengers can normally be considered neutral. Then it's only the effect of an unequal fuel load.
But as I said, for practical purposes, the fuel imbalance limits in the airplane's operating limitations account for the requirement to stay within maximum lateral C of G limits, and it's not necessary to calculate it.


Why cycle the magnetos before engine shutdown?

This is a hot-mag check to verify that when you shut the engine down, you are really putting it in a safe(r) state. Think about how the mag-switch works, when you switch to R or L, it grounds the P-Lead on the other side (so switching to R grounds the P-Lead on the Left side). If you don't see the RPM drop, you may have a hot mag on that side.

An un-grounded mag is a dangerous situation in that it can cause the aircraft engine to turn over if you accidentally (or intentionally) move the prop when you are parking the aircraft, or when somebody walks by and runs into your prop.

So during this mag check you are only looking for an RPM drop or a change that tells you that the mag is shut off.


How precise must a pilot be when flying a holding pattern?

Flying holds does not have to be rocket science and the holds do not have to be perfect ovals.  In fact when applying proper wind correction the pattern will not be a perfect oval.  There are two main parameters you need to meet to be safe and keep ATC happy when flying a holding pattern:

Maintain your altitude:  Sometimes an ATC controller will assign multiple aircraft to a published hold but assign different altitudes.  Any deviation of altitude could reduce separation to unsafe levels.
Stay on the cleared side of the hold as much as possible:  While there is some leeway if you cross over to the uncleared side (parallel entries typically do cross over to the uncleared side) you want to stay on the cleared side as much as possible.  This not only helps with separation of traffic and terrain but also helps keep you from drifting away from the fix especially if you are encountering winds aloft.

As far as your question concerning the time it takes to fly a holding pattern that is not that important as long as you are not drifting away from the fix.  Lets say you have a tailwind when flying towards the holding fix which will turn into a headwind when turning outbound.  The goal you want to obtain is 1 minute inbound legs.  But to maintain that you will need to increase you time on the outbound leg since you are flying into a headwind and losing ground speed.  That means your holding pattern will take longer to complete but ATC will not care about that.  ATC wants you to keep you altitude and not drift away from the fix.  Whatever you have to do to safely make that happen is fine with them.

Question about 8 shaped oval:  I assume you are talking about if you accidentally turn into the uncleared side of the holding pattern and fly a figure 8 to get back on course and how to avoid that.  The rule of thumb is apply two or three times the amount of wind correction on the inbound course to the outbound course.  The reason this is done to account for the difference in turning radius when turning into the wind versus turning away from the wind.  The resulting pattern will not be a perfect oval but will look like the pattern below.  (Image borrowed from this math stack exchange post).

If you would like more info on what airspace ATC actually expects you to use for a hold click here and see wbeard52's answer.


What permissions or approval are needed to land in Antarctica?

For this we need to reference the US Antarctic Program Inter-agency Air Operations Manual

First off this will tell us who is in charge down there (at least for US bases) as its not quite an ice frontier free for all,

  Oversight of the on-ice operations of the civil aircraft is
  NSF-delegated to the Dept of the Interior/OAS Office of Aviation
  Services (OAS) in accordance with the SLA between OAS/AIL. Joint use
  of the USAP airspace in Antarctica requires close coordination between
  the DoD and NSF (and its delegations).

It looks like if you want to fly down there you need to file form DD 1801

  2.3.4. Intercontinental flights between New Zealand and Antarctica will require USAP aircrew to file a DD Form 1801 (International Flight
  Plan) in Christchurch Base Operations to Christchurch Clearance
  Delivery. Flights utilizing the Macquarrie route will require a FIR
  estimate for Melbourne and Auckland boundaries.

The document does not list any approach procedures for NZSP but it does list them for McMurdo Station and they are presumably similar,

  2.7.2.1. IFR. McMurdo ATC retains responsibility for all IFR aircraft. Aircraft will be  USAP Inter-agency Air Operations Manual, 1 August
  2012 40 instructed to contact Tower prior to entering Class E
  airspace. Mac

  2.7.2.2. VFR. Aircraft are normally instructed to contact tower no closer than 10 miles from the airport for landing instructions.

  2.7.2.3. Aircraft operating in Class D airspace will be in radio contact with the tower and should remain at 1,000 feet AGL or higher
  (500 feet for helicopters) until commencing final descent.

They even have reg's for uncontrolled fields

  2.8.1. For operations at uncontrolled McMurdo area airfields, fixed/rotary wing aircraft pilots will follow uncontrolled field
  procedures.

  2.8.1.1. Flight plans will be filed with Mac Center and advisory service provided.
  2.8.1.2. Fifteen (15) minutes prior to departure from an uncontrolled airfield, the pilot will notify Mac Center/Tower of planned activity
  via landline or radio.

  2.8.1.3. Mac Center/Tower will notify Aircraft Rescue and Fire Fighting (ARFF) (who notifies the USAP Airfield Manager) on FM Channel
  2 of intended uncontrolled airfield operations.

  2.8.1.4. Pilots will announce their intentions on the Common Traffic Advisory Frequency (CTAF) VHF 129.7 MHz.

Interestingly it looks like McMurdo is the primary landing strip however based on the letter of 1.2.1 the South Pole Station (which Im pretty sure is NZSP) only accepts ski equipped aircraft so you would at least need skis.

  1.2.1. Primary airfields for the USAP are located in the McMurdo Station area and at the South Pole Station (ski-equipped aircraft
  only). The primary heliport is located at McMurdo Station. Additional
  ice runways, skiways, and helicopter landing sites are established
  throughout the continent supporting the USAP science activities as
  planned by the NSF Prime Contractor’s Fixed/ Rotary Wing Coordinators.

According to the doc a lot of the info is published in the Antarctic Flight Information Manual which looks like you may need to set up some kind of account here to download or "contact [your] affiliate National Antarctic Program".

All this in mind, I have not yet found any documentation or out right publications that seem to forbid flying there as long as procedures are followed. With that in mind you clearly need to file some flight plans and related info, I would think you would need a good reason to be going there to get approval. The reality is there may be more info on this buried in the antarctic treaty but in practice I dont know how that sort of stuff gets enforced.


Why would maximum fly-time (endurance) not coincide with max L/D operating point?

Maximum dwell time or maximum endurance occurs when the power required is minimum. Hence, in this case, the maximum endurance speed is one where the power required is minimum, while in case of maximum range speed, the thrust required is minimum.

For maximum endurance, we must minimize the fuel consumed per unit time i.e. the fuel flow. For maximum range, we must minimize the fuel used per unit distance traveled.

In case of propeller aircraft, the fuel flow rate is proportional to the power produced. Hence, the maximum endurance occurs at a point where the power is minimum. For (turbo)jets, the minimum fuel flow occurs when the thrust is minimum. Hence the maximum endurance occurs when the L/D is maximum. For turbofans, it is somewhere in-between.

Consider a propeller aircraft in a steady, level flight. For determining the condition where the energy expenditure is minimum, we have,

$P = W (\frac{C_{D}}{C_{L}})V$

is minimum. For steady flight, we have,

$V = \sqrt{\frac{W}{\frac{1}{2} \rho S C_{L}}}$

This gives,

$P = \sqrt{\frac{W}{\frac{1}{2} \rho S}}(\frac{C_{D}}{C_{L}^{\frac{3}{2}}})$

Thus, for propeller aircraft, the minimum power and maximum endurance occurs when $\frac{C_{L}^{\frac{3}{2}}}{C_{D}}$, rather than $\frac{C_{L}}{C_{D}}$ is maximum. Due to this, the minimum power (maximum endurance) condition occurs at a speed which is 76% of the minimum drag (maximum range) condition.

Image from eaa1000.av.org

Also, see here and here

Thrust is a force that moves the aircraft. In steady, level flight, this is equal to the drag (if it is more/less, the aircraft will accelerate/decelerate). Power is the rate of doing work i.e. the energy consumed per unit time or rate of energy expenditure (by the a/c powerplant). This is why we are considering minimum power i.e. rate of energy expenditure for determining endurance.

Power is the product of force (thrust) and velocity. Think of it in this way- as speed increases, the drag decreases, reaches a minimum and then increases. However, as the power is product of drag (i.e. thrust) and velocity, it too follows the a similar path; however, the minimum is reached before the minimum drag. That speed gives the maximum endurance.

For jet engined aircraft, the speeds are different. In this case, the speed corresponding to minimum $\frac{C_{L}}{C_{D}}$ gives maximum endurance, while the speed corresponding to $\frac{C_{L}^{\frac{1}{2}}}{C_{D}}$ gives maximum range. Also, see here


How hot do internal parts of jet engine get?

This parameter is called Turbine Inlet Temperature (TIT)

  Turbine Inlet Temperature (TIT) is the temperature of the combustion
  chamber exhaust gases as they enter the turbine unit. The gas
  temperature is measured by a number of thermocouples mounted in the
  exhaust stream and is presented on a flight deck gauge in either
  degrees Fahrenheit or degrees Celcius.

Essentially it's the temperature going out of the combustion chamber, representing the highest temperature the turbine would face.

Here's a figure showing how we are doing recently. So we are at around 1800K now and may reach 2000K soon.

For the temperature of the actual part under cooling, here is the current status of materials: we are at around the 1350K mark and may approach 1400K soon.


Can I become a Commercial Pilot if I had OCD?

Maybe. It depends on how severe the OCD was, the root causes of the disorder, the length of time since the last episode and the likelihood of future recurrence. The aviation medical system is set up to be conservative for good reasons, the cockpit is a bad place to become fixated.

If your condition is/was stress related then I would expect it to be a no, flying can be very stressful at times, you have to be able to handle fluid, difficult scenarios.

Your best answer will be from an Aviation Medical Examiner, I suggest you speak to one, it's what they're for.


What does it mean "two plane thrust vectoring nozzle"?

This statement is using "plane" in the mathematical-geometric sense, which is of course very confusing when we talk about a part of a "plane" in the "flying machine" sense.

So, a "one-plane" thrust vectoring nozzle can only move up and down or left and right (typically up and down). So, this gives you pitch control using the thrust, without requiring airflow over the elevators. This is sometimes called "1D" thrust-vectoring, but it is also (confusingly) sometimes called "2D" thrust-vectoring (see the last paragraph).

A "two-plane" thrust vectoring nozzle can move up and down and left and right. So, this gives you yaw control without using the rudder in addition to pitch control. This is sometimes called "2D" thrust-vectoring, but it is also (confusingly) sometimes called "3D" thrust-vectoring (see the last paragraph).

In a two-engine aircraft, in both cases, you can add another "virtual" axis by having independent vectoring of the two nozzles, e.g. one up and one down, which gives you roll control. That is why, for two-engined aircraft, we sometimes talk about "2D" and "3D" even if the individual nozzles only can move on one or two dimensions.


Is there a taxi turning radius requirement for normal category aircraft?

There is no legal requirement. In fact there is no stipulation that the aircraft be maneuverable on the ground.

Good examples of this are early aircraft (often biplanes) with tail skids. In the middle part of the last century there where gliders that had no wheels - just a center skid. Many current gliders have just one wheel and it is not steerable.

"The nose gear of a few aircraft with tricycle-type landing
gear is not controllable." FAA Airframe Handbook pg 13-4

These type of aircraft are usually steered using differential braking. Examples are: Grumman (AA1, AA5, AG5), Diamond (DA1, DA4), Rallye except the 235C, early "straight 35" Bonanza, Columbias, T-28, T-34C, Nanchang CJ-6A, HU-16 Albatross, IAR 823, etc.

The HU-16 Albatross does not have steerable gear

Aircraft without wheels

Glider without steerable gear


Controllability vs. Stability in PHAK. Is it an error?

Your understanding of controllability and stability is correct, for the following definitions:

Controllability: the response of an aircraft in steady flight, on pilot control inputs. For instance deflecting the ailerons: a high resulting roll rate means a fast response.
Stability: the tendency of an aircraft air frame to return to trimmed position after a disturbance in air stream. Note that in this definition of stability, the airframe itself does all the work in getting back to neutral position, no pilot input required.

So the above is for steady, trimmed flight. The paragraph that you cite is about how to provide flight control inputs to recover from a spin. Now we need both control deflections and aerodynamic stability to recover. If the CoG is too far aft, there is not enough stability and the airframe may not recover even if the correct control inputs are given.

Use of the word Controllability is correct here, since it refers to controllability of recovery from a spin.


How frequently do aircraft get struck by lightning?

It is estimated that on average, each airplane in the U.S. commercial fleet is struck lightly by lightning more than once each year.

Edward J. Rupke, via Scientific American.

  The Federal Aviation Administration estimates that on average, lightning hits each airliner in U.S. service once a year.

Washington Post

It looks like commercial airliners get struck about once a year. I was also able to find an FAA source, but it doesn't look like it's the original source.


Why don't horizontal stabilizers have winglets?

Horizontal stabilizers don't generate as much pressure difference as the wings. Generally the stab deflections are very small in flight, and there's so much other drag during landing -- when the stab gets the most use -- that wing vortices from the tail are probably the least of your problems. Stabilizer winglets on their own, when there's no pressure difference, would simply result in drag, whereas the main wing winglets are constantly effective while flying.

Furthermore, the weight and trouble of adding little winglets to the horizontal stabilizers with larger actuators and hinges, for example, would probably outweigh the very slight aerodynamic benefit. I could also imagine that since the airflow over latter part of the wing is complex and varied, find a good design solution would be difficult.

As @Federico points out, the DA42 has them, but that's a composite airframe rather than the metal construction you see in most Cessnas, for instance, where you have to be more conservative in construction.


What techniques, tools, or instruments are used while performing zero-g parabolic dives?

I'm an aerobatic pilot, with exactly 0 hours of turbine time, so this is a guess:

Climb to altitude and establish a predetermined airspeed.
Gently push to an 'A' degree nose down attitude, and maintain throttles at 'B'(accelerate to the highest airspeed permitted)
When airspeed indicator shows 'C', gently pull up to a nose-up attitude of 'D' (use the G meter to ensure you do not exceed 'E' Gs)
When airspeed indicator indicates 'F', reference the G meter and slowly push over to 0 Gs.
Monitor the changing pitch attitude and airspeed.  Warn the passengers when certain limits are passed.  (i.e., 10 seconds to recovery)
When nose-down pitch of 'G' is achieved, stop pushing and begin pulling back to level flight, taking extreme care not to exceed Vne (or other appropriate limit) and the G limit of the airplane.

Instruments required: Altimeter, Airspeed Indicator, G-meter, Attitude Indicator


What is the difference between a barrel roll and an aileron roll?

The difference between an aileron roll and a barrel roll is that an aileron roll's centre of rotation is very close to or on the aircraft. A barrel roll has its centre of rotation around a point further away from the aircraft itself.

The difference can be appreciated in this image:

Image Source You can find out about the difference in feeling that these two manoeuvres have.

Having stated these things, the manoeuvre shown in the GIF is certainly an aileron roll.


How close does a plane have to come to the ground before triggering the ground collision alert?

There are 7 modes of warning

Mode 1. Large descent rate near terrain. Gear and flap position not a factor in the warning.

Mode 2. Large closure rate with terrain.

Mode 2A. Occurs for a large closure rate if the flaps are not in the
landing configuration and glideslope deviation is more than 2
dots.

Mode 2B. Alerts for a large closure rate if the flaps are in
the landing configuration.

Mode 3. Alerts for a large altitude loss during takeoff or
during a missed approach with the flaps not in the landing
configuration or landing gear not down.

Mode 4. Alerts when the airplane is too close to the
terrain and the landing gear or flaps are not in the landing
configuration

Mode 4A alert when the landing gear is not
down and gives the aural messages TOO LOW GEAR at
low airspeeds, or TOO LOW TERRAIN at high airspeeds.

Mode 4B alerts when the landing gear is down and the flaps are not in landing configuration and gives the aural messages TOO LOW FLAPS if the airplane is at low airspeed, or TOO LOW TERRAIN at high airspeeds.

Mode 5 alerts when the airplane goes below
the glideslope during approach if the landing gear is down.

Mode 6 gives aural callouts when the airplane descends through set altitudes with the landing gear down. It gives callouts like MINIMUMS or DECISION HEIGHT.

Mode 7 gives windshear warnings.

Every one of these modes is altitude, speed and aircraft configuration dependent.

In the Enhanced GPWS, these triggers are also changed based on the position of the aircraft.

Regulatory requirement 14 CFR 121.354

Technical requirements TSO-151C


Where does the term 'steam cockpit' come from?

Well, to answer your second question first: no.  No aircraft instruments actually run on steam.  There are lots of safety issues related to generating and using the steam on board an aircraft that are prohibitive.

As far as why they are called steam gauges, take a look at one and compare it to a traditional flight instrument:

They are both round, both have a calibrated scale printed on  them, and both have a needle which points at the current value.  They actually look pretty similar!


What's exact offset degree dividing localizer and LDA?

The maximum angular offset for a LOC is 3° for FAA and 5° for ICAO. Everything else is named either LDA (FAA) or IGS (ICAO).

In ICAO wording, any ILS which doesn't meet all ILS requirements is an instrument guidance system (IGS).
Localizer type directional aid (LDA) is a designation used by FAA.

Guidance /aid refers to the fact that this system is not a precision approach, only an assistance for for the approach.

The most infamous IGS (at least for sim users) is probably the one used at Kai Tak RWY 13 until the airport was decommissioned. Current ones include Sion LSGS RWY 25 which terminates before the runway (see also Washington KDCA RWY 19 as commented by @Gerry), Innsbruck LOWI with a 5° offset localizer or Lugano LSGS RWY 01 for its 6.6° glideslope.

Many of those non-standard approaches are required to access an aerodrome at the bottom of a valley. All require special training and certification before a pilot is authorized to use them.

ICAO defines approach characteristics in Doc 8168 Procedures for Air Navigation Services (PANS-OPS), volume II: Construction of Visual and Instrument Flight Procedures. According to SKYbrary (Eurocontrol) "PANS-OPS procedures are the international standard and are used throughout Europe and in many other countries world-wide".
FAA equivalent are the United States Standard for Terminal Instrument Procedures (TERPS, Order 8260.3). According to the same source: TERPS are used by "USA and certain other countries. These include Canada, Korea, Saudi Arabia and Taiwan".

ICAO PANS-OPS

Final approach segment:

  (5.2.1) When runway aligned track guidance is not possible it may be
  offset up to 5 degrees without OCA/H penalty (see 5.4.3.1, “Aligned
  straight-in approach”). Beyond these limits (or where other
  requirements cannot be met) a circling approach shall be used.

ILS Cat I:

  (2.1.1) The localizer course line shall intersect the runway extended
  centre line:

  a) at an angle not exceeding 5°; and b) at a point where the nominal
  glide path reaches a height of at least 55 m (180 ft) above threshold.
  This is called intercept height.

  (2.1.2) The procedure shall be annotated: “localizer offset ...
  degrees” (tenth of degrees). The general arrangement is shown in
  Figure II-1-2-1.

FAA TERPS

Localizer without glide path. Final Segment alignment (8-1-2):

  Localizers aligned within three degrees of the RCL are identified as
  localizers. If the alignment exceeds 3 degrees, they will be
  identified as LDA facilities.

Localizer with glide path (ILS). Final segment alignment (10-2-2 a):

  Where a unique operational requirement indicates a need to offset the
  course  from RCL, the offset must not exceed three degrees. The offset
  course must intersect the runway centerline at a point 1100 to 1200
  feet inside the DA point (see figure 10-2-2 ). For offset courses the
  minimum HAT is 250 feet and the minimum RVR is 2400.

For LDA with GS (10-2-2 b):

  The final course maximum offset from RCL extended is 15 degrees. The
  final course must cross the RCL extended at least 3000 feet from LTP,
  but no more than 5200 feet from LTP.


Why is horizontal stabilizer set to 4° up in airbus aircraft for take off?

Both landing and take-off are critical phases of flight, and it is standard procedure to configure the aircraft for those phases to reduce the workload on the crew.

On landing, aircraft are configured for a stabilized approach, which means that the aircraft can be flown to the runway with a minimum of configuration changes to flaps/slats, trim, power, and drag devices.

On take-off, the aircraft is configured for take-off, and trim is set so as to provide a slight nose-up. If the trim is set nose low, it requires a lot of pull to "unstick" the aircraft from the runway. If the trim is set too nose high, then the aircraft will want to nose up before the aircraft may be ready to fly, and then to continue a nose-up attitude after rotation, requiring more than a minimal effort to keep the a proper departure attitude.

Nose-up trim can induce what is called a trim stall, which is when the trim is set so far back that the aircraft will exceed the critical angle on take-off:

  ELEVATOR TRIM STALL
  The elevator trim stall maneuver shows what can happen
  when full power is applied for a go-around and
  positive control of the airplane is not maintained.
  [Figure 4-8] Such a situation may occur during a
  go-around procedure from a normal landing approach
  or a simulated forced landing approach, or
  immediately after a takeoff. The objective of the
  demonstration is to show the importance of making
  smooth power applications, overcoming strong trim
  forces and maintaining positive control of the airplane
  to hold safe flight attitudes, and using proper and
  timely trim techniques.

  Source: Airplane Flying Handbook p. 4-10


Identify canard, low-wing, single pusher-prop plane over Upstate NY

There are a number of near-copies of the Long EZ, with nearly identical wing and canard shape; they differ in number of seats, performance, range, and, yes, several of them have retractable gear.  I think there's at least one that's pressurized.

Without a high quality photo (possibly even with one, if shot from the ground), it may be impossible to determine what you actually saw -- but a single pusher propeller, low mounted cranked/swept wing with tip fins, and swept canard originated with the Long EZ (before that, there was the Vari-EZ with unswept canard, and the Vari-Viggen with a different wing sweep profile).


Why is aircraft marshalling still necessary?

There are automated systems that provide the same function as a marshal. They are in place at some airports but not all since upgrades cost money and not all airports may see a use for it.

At smaller fields marshals often serve multiple roles aside from just guiding aircraft they may also handle baggage, pump fuel, drive tugs and all the other various ramp jobs. In this case it may be more cost effective to keep a marshal around than instal a costly automated system that only replaces one of their jobs.


There are some bases that are joint-use anyway, and in that case, it would be almost exactly like an emergency landing at a civilian field. Then most bases will do little if you are truly an emergency. The times this would come into play are sensitive bases such as Nellis AFB or Groom Lake, or bases that have significant national security interests, such as Andrews AFB, or potentially a base with nuclear bombers.

In these cases, you would probably get a fighter escort if you reported intentions soon enough or entered restricted airspace. Most bases will not provide a fighter escort, as there is little they can do to help, and you should not be a threat to anything from the air.

If you declared an emergency and nothing was actually wrong, you would almost certainly be arrested. If there was an actual emergency, you would probably be detained at a sensitive base, but at most bases they would just keep you confined to unsensitive visitor's areas. Here is an incident where the pilot was forced to land at Travis AFB (which only has C-5, C-17, and KC-10 aircraft). The pilot followed emergency procedures and everything turned out well. But the report notes:

  "Had he not made contact with RAPCON, security forces would have to respond as if the aircraft was hostile," Wilson said. "His proper communication allowed us to clear the area for him and notify the appropriate agencies."

As your aircraft is not flyable at the moment, it would have to stay there until it was repaired, but in most cases you should be able to fly it out after it was fixed. In a sensitive area, I have heard reports of airplanes being disassembled and shipped off the base, but nothing concrete about it.

However, at most of the most prohibited areas the prohibited airfield is rarely going to be you best option (e.g. it would be unlikely for Groom Lake to be your only option, unless you were snooping around the edges, which they would not take kindly to). But just remember - in an emergency, you are in control until your plane is on the ground. So make
the conservative decision and land safely. If it is an emergency, and you weren't pushing limits before the emergency, the most you should get is a stern talking-to and a short detainment.


Would a wingtip stall still be bad if there was no aileron there?

Short answer: yes.
Control effectiveness: We do care about ailerons
The fact that there is an aileron at the wingtip implies that flow separation (stall) at the wingtip results in significant decrease or complete loss of aileron effectiveness during a tip stall.  As a result, as you imply by your question, designing aircraft to begin their stall at the wing root (or, otherwise, to implement some device such that control effectiveness is maintained at high angles of attack) is a typical practice.  Peter Kämpf has a very detailed answer about this (and other topics related to stalls in swept-wing aircraft) here, and there's another answer from AEhere Supports Monica and another one JohnK that gives a summary as well.
But what if the aileron wasn't there?
However, if the aileron did not exist, loss of lift (and increase in drag) at the wingtips is still of great concern.  Insofar as the stall itself goes, when I think of the dangers of losing lift at my wingtips, I think of spins, which look like the below.  This "classic" spin entry is caused by a sideslip (in this case, to the left wing) with the stick pulling back to put the aircraft into a stall.  In this case, the left wing is operating at a higher angle of attack than the right and stalls first, creating the conditions for a spin.

When I did spin training in a Grob 103 sailplane (with your typical control setup of elevators, rudder, and ailerons), the mantra was put the stick neutral, push the rudder opposite the spin, then recover in pitch.  The ailerons were not a part of the recovery (until, perhaps the very very end once the airplane has started flying again and you find yourself not quite level).  This is not the case for all stalls: forward stalls (stalls with near-zero sideslip and coordinated controls) are relatively simple to recover from, and I've held Grob 103s in a forward stall for extended periods of time (though they also have pretty benign stall characteristics).
Turning stalls or stalls in uncoordinated flight are where you get get in trouble with spins, and the more your airplane wants to stall at the wingtip first, the more dramatic (and interesting) will be the aircraft's departure from controlled flight.  This is, in part, because of the yawing and rolling moment caused by the lift and drag differentials between wings -- and, if those differentials are between wingtips, the moment arm for those forces is that much more.  Note: you are not usually using ailerons during this maneuver--in fact, if ailerons are deflected counter to the direction of the spin (per intuition), it can make the spin worse.  That said, Peter Kämpf pointed out that using the ailerons to point the aircraft in the direction of the spin can make it easier to recover from.
However, more to the point: regardless of if ailerons are present or not, rolling an airplane usually produces a restoring roll moment.  As I move a wing upward, per the diagram below, the wing "sees" air coming from above the wing, which decreases the angle of attack (relative wind coming from above the chord line is a negative angle of attack).  The converse occurs for a wing that is descending.  A negative angle of attack corresponds to a decreasing lift coefficient and vice versa: hence, the lift of the wing with change with the wing's motion and, in some way, "damp" the wing's heaving motion.

However, keep in mind that we are talking about something a little more nuanced.
If we go back to the picture of an airplane entering a spin (above--1st picture), the right wing is close to stall and is operating at a very high angle of attack and a relatively large lift coefficient.  The left wing is stalled (operating at an even higher angle of attack than the right wing): it's lift coefficient is much lower than that of the right wing: the airplane wants to roll.  However, if we reference the chart of angle of attack vs. lift coefficient below (stall is denoted by the first peak of lift coefficient around 10 or 12 degrees angle of attack), the stalled wing is operating "post-stall" or, potentially, somewhere in the 15-20 degree range where an increase in angle of of attack no longer corresponds to an increase in lift coefficient--but, rather, a decrease.  Even if the stalled wing if past this range, the lift coefficient increases relatively slowly, and drag increases much faster, slowing the wing down and rapidly decreasing the wing's lift production (lift is directly proportional to the lift coefficient, but changes with the square of velocity).

In short, the aircraft already wants to roll towards the stalled wing and the lift coefficient of the stalled wing is, potentially, decreasing as the wing drops (which increases the local angle of attack).  Now, the opposite wing is moving upward and, as such, its angle of attack is decreasing and it's lift coefficient (which is already higher than the stalled wing) is also decreasing.  However, drag is also decreasing as well, so the wing can travel a little faster - and, thus, produce even more lift.  If the lift production of the stalled wing remains less than that of the other wing, the aircraft will continue to roll.  Which, as you might imagine, is a $Bad \hspace{1ex}Day^{TM}$ (unless you can get both wings flying again).
Peter does mention some issues with tip stalls in the answer I linked earlier as well, specifically in swept-wing aircraft.  In swept-wing designs, lack of aerodynamic or geometric washout resulted in substantial nose-up moments at stall (when the wingtips stalled first) due to the physical geometry of the aircraft.  The tip of the wing, which was no longer producing significant amounts of lift, is probably located behind the aircraft's center of gravity.  As a result, loss of lift at that location would decrease the nose-down moment generated by the wings, which is exactly the opposite of what you want to be produced when in a stall.  A stall at the root would obviate this.  In short, how bad the results of a wingtip stall will be are tied to the design of the aircraft -- and swept wings, in most cases, are a design choice tied to lateral stability or design for high speed flight.  A design choice for one characteristic influences others.
But what if we didn't use ailerons at all?
The most complete answer would speak to how tailerons or elevons (or a horizontal tail whose surfaces can be actuated independently of each other) change spin recovery technique.  For example, I was reading this document on F/A-18C/D spin recovery technique, and it notes that proper spin recovery in the F/A-18 is accomplished by a lateral stick movement.  The Hornet, apparently (can't say I've flown one, but I know there are folks on here who have), has a very violent stall/spin departure mode in which the stick should be kept neutral, but, once the spin is established, the stick is moved laterally to recover.  Examples I've seen go into the direction of the spin, which would match with the prior note about turning into the spin with ailerons.  I anticipate that this leverages the sheer size of those horizontal surfaces and that they have some degree of effectiveness (as opposed to the rudders, whose effectiveness may be reduced due to the wake from the other vertical tail).
Regardless, with reference to the question of "would a tip stall be bad if I was using tailerons anyway", the earlier points about how the severity of the stall is tied to aircraft design still stand.  How bad it is depends on the design of the aircraft, how the CG is located relative to the wing, how pitching moments are produced and balanced in the aircraft's design, and so on -- all factors that are not necessarily tied to the selection of roll control surface.


Can an airplane fly with on-off control surfaces?

Theoretically it could work that way. A maneuver that needs a certain control surface deflection for a certain amount of time could be done with full 30 degree deflection in a shorter time.

Fine control and trimming would be more difficult. Small adjustments would require a very short time at full deflection. Control laws would have to take into account the step response of the system to determine how quickly the surface can reach the full deflection. By pulsing a deflection command like with PWM, you could determine a modulation that would keep the surface near some nominal deflection.

For an RC plane's systems this may be fine. With larger aircraft you would need to consider the effect of these large inputs into the control system, and structural effects, including whether 30 degree deflection is possible in the current flight conditions.


How do I interpret these values in NASA's C-MAPSS dataset files?

difference between nominal and fault file

It is found inside the .tar.gz file. From the README file contained in the page you linked:

  the Simulation_Info.txt contains a summary of the number of flights,
  the fault type that was simulated, if it is nominal or not, and if not
  what flight and time sample the fault was introduced.

  There are 27 parameteres used, among these which parameters will affect fan, HPC, HPT, LPC, LPT

I assume you are asking what are those 27 parameters.
From the README file that is found in the page you linked:

  what is the threshold value for these parameters or when do we say that the parameter has reached its max value and above which it would lead to fault condition.

A detailed answer is out of scope for this site, it heavily depends on the failure simulated. The documentation of the various tests should have enough details to understand.

Additionally, hard absolute values are definitely of no use here, different engines will have different characteristics: the numbers valid for the engine simulated here will be applicable to basically no real engine.


Does the 300 hp IO550 conversion require an auxiliary tank on the Cessna 182?

This article gives a slightly dated (2002) overview of the engine upgrade options for a 182. There's no mention of auxiliary tanks being required, but it does note that for at least one option they're typically a good idea:

  The IO-550 engine is a guzzler and if you don't have long-range tanks,
  you'll need them. Indeed, you'll probably want to consider Flint tip
  tanks for your Skylane.

It's possible that you've seen conversions where, for whatever reason, a cabin tank was added instead of - or even in addition to - tip tanks.

If you're looking into an upgrade then if you aren't already a member you should consider joining the Cessna Pilots Association and search through their forums (members only) for whatever you can find. I found one thread on upgrading the C182 to an IO-550 but it didn't mention the fuel tanks at all, although I didn't search thoroughly.


Why do you need to subtract when using a climb/descent performance chart for an altitude other than sea level?

If you look closely at the example, it's describing a situation where you're cruising at 5500 feet, and landing at an airport with a "pressure altitude" (elevation) of 680 feet. The chart, however, is indexed to zero-altitude. You can't just reduce your cruise altitude by the elevation of the airport, because performance varies with altitude (indicated by the curvature of the arcs for fuel, time, and distance).

To find what you'd need to descend to the airport, you must run the numbers twice - once for descending to sea level from cruise (3 gallons, 9 minutes, 30 NM), and once for descending from the airport to sea level (1 gallon, 2 minutes, 4 NM). Note the two sets of dashed arrows on the chart, providing the two example calculations. You then subtract the numbers for the airport from the numbers for the cruise altitude, as you would hopefully not descend below the altitude of the airport!


What device in western countries allows ATC to get clearances or search flight plans automatically?

The European equivalent would be the Flight Data Processing System (FDPS). The FDPS receives flight plans over the Online data interchange (OLDI) / ATS Message Handling System (AMHS) from the Integrated Initial Flight Plan Processing System (IFPS) which is operated by EUROCONTROL. In addition local flight plan can be added to the FDPS.

Aircraft operators file their flight plan with the IFPS, which checks the flight plan for correctness (syntax errors, route errors, etc) and determines whether the plan is acceptable to the air traffic service providers. If necessary, plans are rejected or modified. Once the plan is acceptable, it is distributed to all Air Traffic Control units (ATCU) along the route.

In the ATCU, the FDPS will receive the flight plan and store it in its flight plan database.

It must be noted that the FDPS is much more than just a flight plan database. It is at the heart of the air traffic control unit. It correlates flight plan data with actual radar tracks, updates flight plan with inputs from air traffic controllers and it supports local flow management functions, conflict detection systems and much more.


How does a Machmeter behave if the static port is blocked during a constant CAS climb?

A Machmeter doesn't use the true airspeed (TAS) or the local speed of sound for its readings since the TAS measurement can't be obtained without getting the Mach number first.
The calibrated airspeed (CAS) shown to the pilot is also used by the Machmeter, it measures the speed based on the pressure differential between the pitot tube and the static port.

If $P(pitot)$ $>$ $P(static)$, then the speed reading will be positive.

The bigger the difference, the faster the speed.

If $P(pitot)$ $<$ $P(static)$, then the speed will read zero since the speed indicator doesn't allow negative speeds.

A blocked static in a climb will cause (2.), and it would require an extremely fast flying to get the gauge to show any speed reading above zero, probably even faster than the plane can handle.
The blocked static will also freeze the altimeter and show zero vertical speed. Glossing over how such a climb can be maintained, let's assume there are different pitot-static sources, one for flying and one for the Machmeter.
Answer:
After the static-port is blocked in a climb, the Machmeter will within minutes (or seconds if it's a fast climb) drop to zero, just like its CAS reading.
Do note:

Aircraft flight instruments use pressure differential, not temperature, to compute the Mach number.
— Wikipedia

and

The true airspeed is calculated by the Air Data Computer from the Mach number and the outside air temperature.
— Is the Mach number shown by an Air Data Computer considered a "true" Mach number?

As you can see, it is the Mach number that is used to know the TAS, not the other way around.

Pre climb and blocking.

Blocked, then climbed at same speed.

(Source)


What is the meaning of “American 755, turn right, heading 320"?

American 755, turn right, heading 320.

Aviation phraseology is direct, specific, and concise. The three items here:

The controller is instructing the aircraft identified as American 755
to make a right turn
such that they maintain a heading (direction) of 320 degrees (5 degrees north of north-west).

Though you may not be interested right now, at some level you might have to get technical, as communication involves defining terms :)

In case you are interested, for references and further reading see this page from the FAA. Especially of note:

Pilot/Controller Glossary - contains definitions for radio communications.
Aeronautical Information Manual - contains instructions for pilots.
JO 7110.65 - (more technical) the handbook for controllers. Everything they say should be in there.


What happened to the Boeing 737.X design?

Boeing was considering designing an all-new airplane to replace the 737 around 2020. Having been in production for over 40 years and through 3 major versions, it seemed like it was finally time to leave the 737 type behind.

However, Airbus decided to upgrade the main competition to the 737 to create the A320neo. The A320neo is more efficient than the 737NG and will be entering service in 2016. Boeing watched orders accumulate for the A320neo and felt that they would lose too much market share in the single aisle market if they waited until 2020 to introduce a new plane to compete with the A320neo.

The high number of orders for the A320neo convinced Boeing to scrap the new plane concept, and focus on just new engines for the 737, in order to get a plane on the market to compete with the A320neo. This allowed the 737MAX to be scheduled to enter service in 2017, less than 2 years after the A320neo. While the MAX does not have as many improvements as a new design would have, it has helped narrow the gap in orders to the A320 series. A new plane still 5 or 6 years away while the A320neo enters service would not have nearly as many orders, meaning the Airbus would be quickly increasing their market share with the A320neo.

A similar situation has happened with the 777, although it is still a much more recent design. The 777X is a more significant redesign than the 737MAX, but the update is needed to compete with the new A350 that is already in service.


Have there been any cases of collision/airprox of small drones with manned aircraft?

CASA (Australia's aviation authorities) has a report on an airprox between a crop duster and a surveying mini-drone:

ATSB incident report


Do laser Christmas lights have the same dangers as laser pointers aimed at planes?

A fine beam is needed do draw patterns. A regular led would require a moving focusing mirror, something expensive.

There is a non-standard warning: "Laser aperture, Max output < 1 mW" near the leds:

As the leds are in the visible spectrum, this is equivalent in power to a Class 2 laser.

  A Class 2 laser is considered to be safe because the blink reflex
  (glare aversion response to bright lights) will limit the exposure to
  no more than 0.25 seconds. (Wikipedia).

This image shows the effect of a 5 mW laser light. You may use it for a 1 mW laser light by dividing the distances by 5:

Adapted from original on Wikipedia

or from the original site:

  A 0.99 mW Class 2 laser beam can temporarily flashblind a pilot or driver, causing afterimages, within 240 ft (73 m) of the laser.
  It can cause glare, blocking a pilot or driver's vision, within 1050 ft (320 m) of the laser.
  It can cause distraction, being brighter than surrounding lights, within 2 miles (3.2 km) of the laser.

The above distances are for a tightly focused beam, which is probably not the case of an inexpensive laser device. The effects of a more diverged beam are weaker.

More info: This hazard chart.

To sum up:

Most countries have regulations for dazzling a pilot, regardless of the mean and power used, it's the effect that is taken into account.
A class 2 laser led couldn't blind someone unless the person is at a few meters (15 m for temporary blindness and 3 m for eye damages, if the eye is forced to stay open).
As the class is not mentioned on the label, I would be careful when seeing this item.
There are actually 2 laser leds, a red and a green, is it 2x1 mW or 2x0.5 mW? In practical the lasers probably can't illuminate the same eye, due to the divergence of their beams.
To draw a pattern a laser light must move, so the time of illumination of a point is reduced when the device is working correctly.


What is the difference between IMU (Inertial Measurement Unit) and AHRS (Attitude and Heading Reference System)?

An INS calculates and updates the vehicle's position (latitude and longitude), alongside the orientation. It needs to be initialized on ground, with the aircraft completely still.

An AHRS does not record/update the position. It outputs real time orientation (attitude and heading) only. It thus can be initialized in flight.

(Image source)

Image shows the INS panel of a Boeing 747-400, notice the ATT (Attitude) option, that's an AHRS-only mode. NAV includes both AHRS and navigation.

ATT can be initiated in-flight, it requires a manual entry for the heading while flying straight and level. So in other words, an INS includes AHRS-functionality.

  Sidenote on terminology: INS nowadays is predominantly used to describe an Inertial Platform, Inertial Reference System (IRS), and Inertial Reference Unit (IRU/IMU); and often used to describe all three of these units (as one system) typically installed on a single aircraft.


How long does flight planning take / when is it done / how is it done?

The pilots don't do the flight planning. It is done 1-2 hours before the flight by dispatchers using computer software.

On a quick turnaround, one pilot goes to a printer to retrieve the flight plan. The other stays with the aircraft and prepares for the next flight.

Sometimes someone will bring the flight plan directly to the aircraft. Some airlines have stopped using paper flight plans and instead download to an iPad.

The pilots review the flight plan to ensure it meets all requirements, and can request a reroute, or additional fuel it they feel it is needed.


Why do modern fighter jets use 20mm guns?

20mm is the smallest caliber that is practical to load with a bursting charge. Some fighter jets use 23mm, 27mm, or 30mm guns.

The reason 20mm has become one of the most common aircraft gun calibers is that it produces the lightest gun that meets the minimum requirements to be useful in air combat: enough rate of fire to hit a maneuvering jet, good enough ballistics for meaningful range, enough bursting charge capacity for meaningful damage.

The majority of modern combat jets do not carry armor, but they are built with considerable redundancy and internal subdivision. A bursting charge (explosive payload) is needed to do meaningful damage to a modern combat aircraft.

Smaller calibers can cause some damage that will require repairs, but not enough to rely on them to bring down a jet in combat. With missiles prioritized, most air forces want the lightest gun that's still effective.

Larger calibers have a cost in design. A large caliber rapid-firing gun will be much heavier, its propellant gases (muzzle exhaust) can cause flameouts in the engine, and firing will throw the aircraft's aim off center. A slower-firing one can be suitably light, but still has the same recoil force. Another issue with large calibers is collateral damage - 30mm+ rounds send fragments a couple hundred meters, far enough to damage friendly aircraft in air combat maneuvering.

The 20mm rounds used by combat aircraft have limited armor-piercing capability anyway (comparable to .50 BMG). Armored aircraft are rare, and some exceptions include CAS aircraft such as the A10 and to a lesser extent the Su-25 and the Su-34. This armor is mostly designed to protect them against ground fire, not enemy fighters.


Why does camouflage make some planes fly slower?

Same reason gliders keep their wings waxed.  It wasn't the camo per se, it was the dull matte field-applied paint finish that included all sorts of imperfections, and to a small degree, the weight of the paint vs natural finish.  At a microscopic level, the surface of the matte finish was much rougher than a gloss or unpainted metal surface.  The microscopically rougher surface would have a stronger ability to inhibit air molecules flowing right against the surface (skin friction).  Add to that the surface imperfections introduced by the typical wartime spray-and-get-it-out-the-door process, vs smooth bare aluminum.

Think of the inhibiting effect on wind velocity close to the ground of a forested area vs a grassy plain, and shrink it down to the micro level.  The effect is most pronounced where the flow is relatively smooth, or laminar against the surface, which is most areas there the cross section is getting fatter (leading edges of wings and control surfaces, the nose, etc).

The Mustang would suffer a bit more than others from this with flat paint because of its (somewhat) laminar wing that had a maximum thickness well aft, which means it would have more surface area that is sensitive to skin friction, than a non-laminar wing aircraft like a Spitfire or Typhoon.  So if the Mustang was penalized 5kt with flat paint, a Spitfire might have only been penalized 3kt because it had less surface area that was (relatively) laminar in the first place.


Can a pilot take off at his own discretion?

Generally speaking a pilot can (physically) do whatever they want: ATC doesn't have a remote control. If you push the throttles forward and pull back on the yoke you will take off, no matter what mean things the folks in the tower are shouting at you over the radio.

Practically speaking doing so can get you into trouble: You may be met in the air by planes with more guns than yours, or projectiles from the ground may be fired in such a way as to intersect your flight path ("You might get intercepted or shot down by very annoyed military types.") -- even if that doesn't happen you may face administrative or civil action for ignoring ATC instructions in the jurisdiction where you did this.

I'm not sure what the ATC policies are in Turkey, but the US FAA has a phrase specifically for this: UNABLE TO ISSUE DEPARTURE CLEARANCE. DEPARTURE WILL BE AT YOUR OWN RISK. (you'll find it in JO 7110.65W, on page 3-3-1).

The emergency authority of a pilot is also recognized worldwide (at least in ICAO states - you'll find it in Annex 2, 2.3.1, conveniently quoted on Wikipedia), so if the pilot in command deems that remaining on the ground constitutes an emergency they can violate regulations to the extent necessary to meet the emergency (i.e. forego getting a clearance and take off even if ATC is saying not to).


Does an airfoil in supersonic flow have zero circulation?

Maybe I'm missing something obvious to you, but why do you think there's zero circulation?

Take the simplest case of an infinitely thin 2D flat plate in linearized supersonic flow. The solution in the perturbed velocity potential ($\hat{\phi}$) is:

$$\hat{\phi}=
    \begin{cases}
    0 & x\pm\beta z<0 \\
    \frac{1}{\beta}\alpha(x-\beta z) & 0\le x-\beta z\le c, & z>0 \text{ (upper surface)}\\
    -\frac{1}{\beta}\alpha(x+\beta z) & 0\le x+\beta z\le c, & z<0 \text{ (lower surface)}\\
    \pm\frac{1}{\beta}(\alpha c) & x\pm\beta z>0
    \end{cases}
$$

where $c$ is the chord length of the flat plate, $x$ and $z$ are the horizontal and vertical coordinates, $\alpha$ is the angle of attack and $\beta=\sqrt{M_\infty^2-1}$ is the supersonic Prandtl-Glauert factor.

The velocity field ($\vec{V}$) can  be obtained as:

$$\vec{V} = \begin{bmatrix}V_\infty \\ 0\end{bmatrix}+V_\infty\nabla{\hat{\phi}} =
\begin{bmatrix}V_\infty\pm\frac{1}{\beta}\alpha V_\infty \\ -\alpha V_\infty\end{bmatrix}$$

where $V_\infty$ is the free-stream airspeed. The first equation is plus for upper surface and minus for lower surface.

Take an infinitely thin closed path ($C$) that wrap around the upper and lower surfaces, and calculate the circulation ($\Gamma$):

$$\Gamma = \oint_C\vec{V} \cdot \hat{n} ds=\frac{2\alpha c V_\infty}{\beta}$$

This is obviously non-zero. In fact, this can be readily seen by the potential jump in the wake.

And when we calculate the lift coefficient ($C_l$) of this flat airfoil using the Kutta-Joukowski Theorem ($L'=\rho_\infty V_\infty \Gamma$), we retrieve the classic supersonic lift relation:

$$C_l=\frac{L'}{1/2\rho_\infty V_\infty^2 c}=\frac{4\alpha}{\beta}$$


Why are rotary wing aircraft less efficient than fixed wing aircraft?

Quick answer:

The rotary wing is moving at different speeds over the radius, and on the left and right side in forward flight, and since there can be only one optimum speed, most of the rotor is working at less than optimum speeds, lowering the lift-to-drag ratio overall.
The slim rotor blade needs a reflex airfoil, an airfoil with low pitching moment, which is less efficient at creating high lift coefficients. If it would use a regular wing airfoil, it would twist horribly.


Is a Phenom 300 single pilot able to fly non-commercially in Europe?

The EASA type certificate says yes granted the cockpit is approved and the required additional equipment is met (refer to EASA AFM).

Minimum Flight Crew:
One pilot (in the left pilot seat) plus additional
equipment as specified in the Limitations Section of the
EASA Approved Airplane Flight Manual

The related note:

Approval for operation with a minimum crew of one pilot (in the left pilot seat) is
based upon the cockpit equipment installation and arrangement evaluated during ANAC
certification testing. No significant changes may be made to the installed cockpit equipment or
arrangement (EFIS, autopilot, avionics, etc.), except as permitted by the approved MMEL,
without prior approval from the responsible Aircraft Certification Office.


How to join a traffic pattern when passing overhead?

What is the correct way?
Is there a difference between controlled and uncontrolled airports? What about the procedure when approaching from other angles?

Well I'll address your second quesition first: There is ABSOLUTELY a difference between controlled and uncontrolled fields. At a controlled field you do what ATC tells you to do (and if there's any ambiguity you ask them to clarify what they want you to do).

At uncontrolled fields in the USA the standard way to enter the VFR traffic pattern is a 45-degree intercept to the downwind leg. The FAA's pronouncements on the issue can be found in AC 90-66A.
A brief and relevant excerpt:

a. Prior to entering the traffic pattern at an airport without an operating control tower, aircraft should avoid the flow of traffic until established on the entry leg. For example, wind and landing direction indicators can be checked while at an altitude above the traffic pattern. When the proper traffic pattern direction has been determined, the pilot should then proceed to a point well clear of the pattern before descending to the pattern altitude.
b, Arriving aircraft should be at the appropriate traffic pattern altitude before entering the traffic pattern. Entry to the downwind leg should be at a 45-degree angle abeam the midpoint of the runway.

(Emphasis Added)
You will find a rich variety of opinions (a polite way of saying "hidebound arguments") among pilots on the correctness of various other entries, but the 45-to-downwind is what's expected by most pilots in the US, and it is generally the safest way to enter the pattern (you will be where everyone else expects you to be, and you will have plenty of time to look at the pattern and figure out how you're going to sequence into it).
The 45-degree intercept assumes that you are on the side of the runway where you will be flying your downwind leg. If you are on the opposite side you need to position yourself to maneuver for a 45-degree intercept. I am personally a proponent of the technique illustrated below:

Cross midfield 500-1000 feet above pattern altitude.

At airports where there are multiple pattern altitudes specified, such as a "jet pattern" above the light aircraft pattern, cross at least 500 feet above the highest pattern altitude to ensure traffic separation.

Proceed away from the airport and clear of the traffic pattern before descending.
Make a descending turn to intercept the 45-degree entry

Ideally time your turn and descent such that you're at pattern altitude when you intercept the 45-degree track. If you expect to need more time to descend plan for a longer entry.

Enter the traffic pattern on a 45-degree-to-downwind.

Note that this is subtly different from your "Option B" (which has the descending turn leading directly into the downwind) in order to comply with the advisory circular's instruction that you should be "well clear of the pattern before descending".
I cribbed the pattern entry graphic from Alliance Airways, with some minor modifications, as it was the best illustration of the technique I could find.


Small planes tend to have keys, but not all.  The bigger the plane the less likely a key is needed and once you get into jets, I can't think of any that need keys for engine starts.  The starting and securing of engines with no keys comes down to switches and knobs.

Likewise with exterior doors, on smaller planes including private end executive jets, you can find keyed exterior doors and luggage compartments.  With transport jets at airlines you won't find keys. To secure those doors the jet bridge is pulled away from the airplane so there is no physical access to the boarding door.


Do any airports have runways or taxiways that cross active rail lines at grade?

There is at least one:

In Gisborne, New Zealand, there is a freight rail line which crosses the south end of runway 14/32 at Gisborne Airport (GIS). It appears to still be an active line for freight (logging) haulage, as well as occasional passenger excursions.

Image source: Google Maps, plus own annotation.

This article has some additional pictures and info as well.


How does envelope protection work in Airbus vs. Boeing aircraft?

There are two sorts of "autopilots", and it is important to make a distinction between the two. One is for the behaviour of the aircraft around its Centre of Gravity (CoG), the other one is for defining the path of the CoG.

The Inner Loop autopilot: behaviour around CoG, or the aircraft attitude control. This autopilot should not be called autopilot since its characteristics are actually shielded from the pilot: it does its job without showing what it is doing. It only moves the control surfaces (elevator, ailerons) or helicopter swash plate, and not the flying controls (stick, wheel). To the pilot, the system is invisible and just results in a stable platform, in the case of the A320 and B777 a platform with envelope protection that does not allow for the aeroplane to enter a situation that would be dangerous. In helicopters this system is referred to as Stability Augmentation System (SAS).
The Outer Loop autopilot: defining the path of the CoG, which is ultimately what pilots are paid for :). They are in control of the flight path and administer this control by deflecting the flying controls (stick, pedals). The pilot can delegate this control to a feedback system that provides inputs to the same flying controls, via an actuator with a slip clutch. With an old fashioned mechanical linkage, the actuator moves the whole circuit from stick to surfaces (or swash plate in a helicopter); the pilot can override this input by applying a force greater than the override force of the slip clutch.

So normally, by design, the Outer Loop autopilot shows the pilots what it is doing by moving the flight path input devices for them, in a way that is transparent and can be intuitively overridden by applying a high force. It is this bit where the A320 has a different philosophy: the input stick never moves other than by the hand of the pilot. This main distinguishing feature between the two types of autopilot does not exist in an A320. It does in a B777, which also has an inner loop autopilot plus envelope protection, and an outer loop autopilot which actually does move the column/yoke. So the B777 is functionally identical to the A320 in every meaningful way, plus retains the feature of moving flying controls. It has no mechanical linkage, and it uses two separate sets of actuators for this function: one set for the surfaces, and one for the flying controls.

This is the situation on both Airbus > A320 and Boeing > 777, when all systems are functioning. These aircraft are all dynamically stable: aerodynamics take care of bringing back the aircraft attitude to a neutral position. The F16 for instance is aerodynamically unstable in order to provide high manoeuvrability: it always wants to pitch and roll already, and the flight control computers provide constant rapid inputs to elevators and ailerons to maintain attitude by active control. This is not the case with passenger aircraft, which do not need to be as manoeuvrable as a fighter jet and just need to bring the passengers home safely.

Therefore, both on the Airbuses & Boeings, the aircraft can fly without the Inner Loop autopilots as well, without any problems. The aircraft just lose their envelope protection and now it is possible to command them into potentially dangerous situations such as fully developed stalls. In both types, it is always possible to disconnect the outer loop autopilot. For the inner loop ones:

The B737 does not have one.
In the A320, it is not possible to disconnect the inner loop autopilots if all is functioning correctly, except by pulling circuit breakers for resetting the ELACs as this site indicates;
In the B777 there is a guarded switch that allows disconnection of the envelope protection as @Cpt Reynolds pointed out.

In the latter two types, the system disconnects features by itself if failures are detected, to allow control in degraded mode. Both manufacturers have implemented a direct mode as the lowest mode, where surface deflection is a direct function of stick deflection. There is however no direct mechanical cable connection to the surfaces, even the direct mode is still an electrical input, so technically still Fly By Wire.

Above picture is from Aircraft Systems Mechanical, electrical, and avionics subsystems integration Third Edition Ian Moir Allan Seabridge, and shows the top level comparison of the implementation of the flight control loops. The redundancy features are different, functionality is much the same: both implementations prevent excessive AoA that brings the aircraft into a stall, and excessive bank angle.

So although the functionality of modern Airbuses and Boeings is very similar, we do often see the myth that Airbus ranks the automated systems higher than the pilot. There is a difference in user interface implementation, in that Airbus has sticks that are not coupled with each other, and have no means to be driven other than by hand. At the time of implementation, this was a new feature that pilots were not accustomed to, as was the envelope protection system. New features can be poorly understood by inexperienced users, and the aviation world is conservative: when the DC3 introduced a 3/4 control wheel, pilots were protesting that they might be gripping the bit of the wheel that was missing and that they were used to.

Ultimately, the most relevant fact is that both Airbus and Boeing have excellent and very similar safety numbers, and that both manufacturers make aircraft that enable the pilots to bring the passengers home safely in 99.9999999 % of flights.


Is this paper's description of ATC services over the internet actually practical?

For the scope of the article you should read Internet as IP VPN.In practice you would have at least dual links with some kind of SLA attached. All this is much cheaper than having and maintaining your own radar on site.  At my employer we have a system which gets information from various radars over an IP VPN ( been there , done that ).