How does it feel like to be outside a flying airplane? Riding a motorcycle on a freeway at over 200 kph you surely feel the rush of air around you. A typical commercial jet airliner travels at approximately 3-5 times that speed at cruise altitude. Assuming you could open the cabin door closest to the wing fuselage and step out on to the wing, does the air rush feel more 3-5 times more intense than if you were on your bike? Does it mean there is more air resistance to blow you off if your bike could travel that speed? If so, then what would be required to secure yourself to the aircraft in order not to be blown off from the air velocity around you, again assuming that air temperature and thickness of air to support human consciousness isn't a factor?
Then there is a issue of cabin depressurizing, but again assuming this is ignored or resolved through pressure equalization with the outside pressure (oxygen masks will need to be worn by the rest of passengers inside for survival)

It won't be pleasant.  The main result of being exposed outside at altitude, besides the obvious hypothermia, frostbite and hypoxia, will be bruising from the 280-ish knot slipstream (it's the indicated airspeed that matters as far was what you feel, not the true airspeed), and injuries from being flung around by any turbulent flow you are in. Most of your skin that was exposed directly to the slipstream will have bruising.
When I was skydiving I tried tucking in and diving straight down instead of the normal 120 mph flat freefall position, just to see what it was like.  You accelerate to about 150-160 mph.  It got to be a bit uncomfortable at that speed.
A friend's dad once ejected from an F-104 while going just below transonic speed, and I got to ask him about it back in the 70s.  He said he was purple from head to toe except where his face was protected by his helmet visor and oxygen mask (except for the purple mask outline bruise on his face).
Even better, you can read about this British Airways BAC One-Eleven driver, the one who got sucked halfway out the windshield when it blew out back in 1990.  He had broken bones in addition to the other stuff, from being beaten against the fuselage like a rag doll the whole time.   The crew were able to hang on to him because his legs were still hooked around the windshield frame, providing some mechanical advantage for the crew members trying to hold him in over the airstream trying to pull him free.


How are registration numbers assigned? I'm surprised this hasn't been asked before but I could find any question on here for it.

Who picks the tail number for a new aircraft?

Is it automatically assigned by the authorities you are registering with or can the new owner pick any number as long as it's not already I use?

If it's auto generated by the authorities, can you purchase the equivalent of a custom car registration plate if you want a specific number?

I've tagged CAA but would be interested to know if the FAA is the same.

In the US, aircraft numbers are requested or reserved from an available list by the aircraft owner: N-Number Availability Query.  Then the owner, the manufacturer in the case of new aircraft, submits an application for registration with the reserved or available number.

This order of events (to reserve then request assignment) is important as registration numbers are applied to the aircraft, recorded in logbooks, programmed in computers, etc. before the aircraft certification is complete.

In the US, aircraft manufacturers reserve blocks of registration "numbers" long in advance of manufacture.  A registration request is made to the FAA on the appropriate form at which time a registration examiner at the FAA determines the legitimacy of the request: whether the number is assigned or available to be assigned to the aircraft as requested.

The Canadian process is the same.  Available Marks


How do we get Required RVR or Actual RVR for an instrument approach? When executing an instrument approach, when the pilot should start reference the RVR reported. Also, How do we get the required RVR for the current approach in execution and actual RVR. I think actual RVR can be part of METAR data. But I am not sure on the required RVR.

Current RVR is obtained from the METAR, yes, which in turn is obtained from transmissometer(s) on the airport site.

Minimum allowable RVR is obtained from charts (chart legend from Jeppesen, PDF), more specifically the instrument approach plate for approaching an airport. And on the SID chart for departing.

Example:

R is RVR in hundreds of feet, V is Visibility in miles, in this example. Units can change and will be denoted, check full PDF for all examples.


Are the CAA and FAA Class 1 medical certificates equivalent? Is the CAA medical class 1 certificate equivalent to the FAA class 1 medical certificate? How so?

Or does one have to convert to the FAA license in order to achieve the medical certificate?

The CAA medical (UK or any other country with a CAA) is not equivalent to an FAA medical, you have to get an FAA medical to get a FAA license. The classes are similar but they cannot be used in place of the other. In the UK you can get an FAA medical from several AMAs.


How does the F-35B not flip over during transition? As the engine nozzle rotates, the lift-fan cannot. Wouldn't this imbalance of moments around the center of gravity result in the F-35 pitching up onto its back?

As can be seen at 5:14 in this video:

             the F-35B is capable of transitioning from vertical hover to horizontal flight.

It accomplishes this in part by gradually rotating the engine nozzle from the vertical position to the horizontal.
Let us assume that the throttle input stays constant in the following scenario:
As the nozzle rotates, the vertical component of its thrust is reduced (as the horizontal component increases) as

$$thrust_{horizontal} = thrust \cdot \cos(\theta)$$

$\theta$ being the angle between the jet exhaust and the vertical axis.
The problem is, the lift fan does not rotate.  And since it is attached by shaft to the engine, we assume it always rotates at the same RPM as the engine itself.
Therefore, the vertical component of lift of the lift fan does not change for a given throttle input, while the vertical component of the jet engine nozzle does!

So as the nozzle transition occurs, the thrust of the lift fan (ahead of the center of gravity) stays constant, but the vertical thrust behind the center of gravity (rear engine nozzle) reduces: shouldn’t this cause the aircraft to pitch upwards and flip onto its back?
Yet in the video the aircraft seems to pitch up only ten or so degrees (I assume due to the increased coefficient of lift on the wings) but otherwise stays flat in pitch.

The thrust of the lift fan must be somehow modulated to accomplish this, right?  If so how is this done if its shaft speed is tied to that of the engine?

Possible solutions:
1. As the aircraft’s forward velocity increases, the lift of the wings compensates for the reduced vertical lift of the rear engine nozzle.
2. The aircraft pitches up so as to deflect the angle of the lift fan from the vertical to match the angle between the engine nozzle and vertical.

To try and change lift by changing fan blade speed would be too slow to modify the lift for "tiny" and "delicate" changes in how the plane hovers.
Like a helicopter, the F-35 is able to maneuver in very small movements and change very fast (in fact likely better than a helicopter).
So the pilot can "nod" to you, and move around the front nose with tiny, delicate moves. So tiny movements by the F-35 and the flight stick are rapidly accepted by the flight computers.
Now a helicopter uses variable pitch blades to make rapid changes, which allows helicopters to achieve tiny, delicate, and fast "little" moves. So tiny fast changes in lift are required for these delicate moves.
Change of engine throttle to change fan lift would be far too slow (for the F-35, or for a helicopter).
To save weight, and complexity, the F-35 lift fans are counter rotating, and they are light weight carbon fiber.
So, to rapid change lift and have rapid tiny changes in lift, they simply change the area/size of the exit exhaust area of the lift fan.
Think of those "variable" nozzles much like a window shutter to let more or less light into a room.
The shutter looks like this:

So, you can open a bit more, or close a bit more.
However, when you close the shutter more, then where does the air go? It has to go somewhere! (remember, up to 20,000 lbs is created by that fan).
Well, there is a second set of doors that open right behind the one large fan lid.
That second set of doors can be seen here:

Those 2nd set of doors behind the big lift fan lid are to allow venting of the lift fan.
So you need a fast and delicate means to change lift fan output. So amazing and so tiny are these changes, that the pilot can "nod" the nose down a bit. Very tiny changes are achieved.
So to change the fan downforce you thus close or open the lift fan exit (exhaust) area doors. This can be closed or opened rather fast, and also by tiny amounts. This gives delicate control.
However, when you squeeze close this exit area, then the lift fan air has to go some place else, so it is vented out those doors right behind the big lift fan lid door.
So, if you close down the exit hole on the bottom of the F-35 fan, then that air takes another path – right out those 2nd set of doors.
9 out of 10 diagrams on the internet call that second set of doors an auxiliary inlet. However, that is NOT 100% correct. It should be called an auxiliary exhaust.
Now, to be fair, because the cold inlet turbine compressor is also exposed to that vent area, then no doubt that area likely remains "negative" pressure most of the time.
However, this is quite a brilliant design, since that lift fan exit area in general has some amount of "close" and thus "some" amounts of fan exit air is going to be vented (by-passed) right into that "box" behind the lift fan.
And that box just so happens to be right where the hungry for cold air turbine inlets of the massive F135 engine is. (What a nice design!)
So likely that area remains negative due to the compressor inlets. However, that area will go positive with most of the lift fan exhaust closed.
In this case, then those two doors are in fact a lift fan exhaust vent – not an intake vent!
I should also point out that due to vast amounts of cold air hitting the deck with this lift fan system, Lockheed engineers state that the overall deck temperatures are less than a Harrier when landing on the deck of a ship.
So while both Harrier and F-35 tend to heat up the decks on a vertical ship landing, the F-35 actually is less of a heating problem then is the Harrier for such vertical landings.
References to my statements:
From Lockheed:
https://www.lockheedmartin.com/content/dam/lockheed-martin/eo/documents/webt/F-35_Air_Vehicle_Technology_Overview.pdf
page 25:
quote:

With this system, nozzle thrust may be directed in an arc of 41.75-104
degrees (fore/aft aircraft coordinate system), at a rate of 40 degrees
per second.

Ok, so far, above ONLY talks about directing the down thrust.
But, we have this:

Independent control of the three VAVBN actuators provides the
capability to vary the nozzle throat area independent of the vector
angle

To save reading, we have:
VIGV = top intake part (variable )
VAVBN = bottom exit part (also variable).
So above states the VAVBN actuators (the EXIT PART OF THE FAN) can vary the nozzle throat area.
My claims:
Lift fan blades are NOT variable pitch.
There no evidence, no diagram, or ANY information of ANY kind in existence that CHANGES this claim of mine. I stand by this claim.
Lift exit fan can open/close or "vary" the fan output.
Again, the above document 100% supports and verifies my claim.
NOT verified:
My claim that when closing down the 20,000 lbs of fan vent flow, that flow HAS to go some other place, and I am claiming that flow goes to the air box behind. This is a SPECULATION on my part. I do believe this to be what occurs but I do NOT have a link or article that outright states this (some diagrams DO suggest this, but that would not be proof). So this claim of mine that fan air vents to the air box behind is NOT verified by me, and until I can provide solid evidence of this, then that claim should be taken with a grain of salt.
Edit 2
As for the speculation of air venting out of the fan box, this here is a photo.

It is "difficult" to tell if that is a vent door, but does seem to look like one. As noted - I am open to opinions on this - this right now thus is still a speculation on my part.


What are these suffixes used after the call sign? I often play a flight simulator game Infinite Flight (http://www.infinite-flight.com) on my iPhone. The game lets you choose the following suffixes with your call sign.

Heavy
Super
Flight of 2
Flight of 4
Flight of 5
Flight of 6
Flight of 7
Flight of 8
Flight of 9
Flight of 10

I am aware what "Heavy" means but now aware about the rest. Can someone let me know?

Thanks

Heavy - all aircraft with maximum take-off weight greater than 300,000lbs (whether or not they're operating at that weight) It's to highlight to the controller that wake turbulence could be a problem for a following aircraft.

Super - Airbus A380: As for 'Heavy', but more so.

Flight of X - All the rest relate to groups or formations containing that number of aircraft.

These originate in the FAA regulations and are summarised on Wikipedia

There are a number of other unusual suffixes including 'Concorde' (used only by British Airways for the now retired Concorde), Lifeguard (for medical flights), and others


Propeller effect on the L/D of an aircraft (Pusher Propeller configuration) I recently performed an aerodynamic analysis of a fixed-wing UAV with a pusher propeller configuration. I found out that the L/D reduced by about 15% when compared to the body alone without propeller. I am investigating methods to mitigate this problem.

The following are some details about the aircraft,
Wing Span = 2.5 m

Fuselage length = 2 m

MTOW = 25 kgs

Tandem wing configuration. The following image is a representative image.

Courtesy: Raytheon Coyote

Any suggestions will really be helpful.

You need to fair in the tail with a carbon fibre fairing/spinner such that the prop folds back and nests into it.  Like below.


Do helicopters fly holding patterns under IFR? If a helicopter is flying an IFR flight plan, and the airspace around the destination is busy, will controllers tell them to fly holding patterns just like planes do? Or will they hover at the holding fix? Or something else entirely?

In short, yes, helicopters fly procedure turns, hold, etc. just like airplanes.

Hovering at a fix is not practical, as they do not really have the nav capability to perform that, and it is contrary to normal ATC procedures.


What's the exact moment were the spoilers are supposed to be deployed on a A320? Are they deployed when the levers are pulled back to 'IDLE' or only when pulled back to 'Reverse'?

I still can't figure out, 12 years after, why flight TAM JJ3054 crashed into a gas station and a building, in São Paulo, Brazil. The investigation concluded that the captain pulled the left engine lever to reverse, but "by mistake", left the right one on 'CL'.

For those who don't remember the tragedy: 1- right reverse was pinned due to a malfunction; 2 - It was raining at the Congonhas Intl. Airport at the moment of the crash; 3 - the a320 landed on the touchdown mark on 35L runway ( about 1290 meters long); 4- Left reverse deployed but no spoilers nor hydraulic breaks engaged; 4 - Death toll: 199 souls.
To this day, they can't fully determine whether the captain left the right lever on 'CL' or the FDR displayed, erroneously, 80% of thrust on right engine, when it was supposedly on 'Reverse' position. The pedestal was found among the wreckage, twisted and melted beyond recognition.

PL at idle is one of several conditions (weight on wheels, N1, wheel speed etc) that have to be satisfied to get ground lift dumpers to come out on airliners.  Reverse isn't required.  Once on the ground, just moving the PL levers back toward TO thrust makes them come back down.


How much thrust and fuel are required to get a B767 moving from a complete stop? How much thrust is involved, and how much fuel is burned, to get an aircraft like a Boeing 767 - or something similar - moving from a complete stop?

Recently, I was on a 767 that took the "scenic route" around the DFW airport trying to get to a gate.  We were on the tarmac for an unusually long time, and had multiple stops and starts.  I was curious how that might have affected fuel costs as opposed to a shorter time on the tarmac and a more direct path to our gate.

This research paper may be of interest for you. According to their data the 777 burns about 250Kg of fuel on taxi on average with a few data points showing burns well over 700Kg of fuel. This paper which quotes the previously linked paper talks a bit about different fuel saving methods and their impacts, like single engine taxi and what not. In short taxi distance will always burn a bit more fuel, the primary concern is taxing for takeoff (since you don't want to cut into your estimates). Once the plane is back on the ground it becomes less of a safety concern as running out of fuel becomes an inconvenience (that a tug can solve) and less of flight risk.

Here is an interesting paper on fuel burn rates if you were to extend a runway by 500Ft. They provide data/estimates for many air frames.


Why are these steps performed on the A330 emergency electrical checklist? For an Airbus A330 in emergency electrical configuration with both generators and APU FAIL (both engines working), why does the checklist ask you to select RAT MAN ON before slat extension and then LAND REC ON?

If we look at the Airbus A330 manual we will see that the RAT (Ram air turbine) can only power the GREEN hydraulic system and when the EDP (Engine driven pump) of the green system is available, the emergency generator is run by the EDP and not the RAT.
The A330 has two levels of emergency electrical configurations. They are:

A fault in the main electrical buses of the aircraft (AC BUS 1 and AC BUS 2). In this case, even if the engines and their generators are running the aircraft electrical system cannot be supplied. In this situation, the green hydraulic system's EDP powers the emergency electrical generator. This is the first level.

The aircraft falls into level two when there is a dual engine failure situation. So, there are no working generators to supply the aircraft electrics. Here, the RAT extends automatically. The RAT then runs the green hydraulic system which in turn runs the emergency generator.

The level one emergency configuration is always better because it provides the aircraft with more power.
When you are coming into land, while the aircraft is in level two emergency electrical configuration (RAT only), once the slats are extended, the emergency generator is automatically disconnected, and the aircraft is purely powered by the batteries. The reason for this is to allocate RAT only for the operation of the hydraulic systems so that the flight control power is prioritized. When you turn on the land recovery push button, the LAND RECOVERY AC BUS and the LAND RECOVERY DC BUS are automatically powered giving the aircraft the necessary electrical needs for the landing. This includes powering of the SFCC 1 (Slat flap control computer). But in the case of RAT powering the aircraft, only the slat channel is recovered. The landing has to be conducted flapless.

When the aircraft is in level one emergency electrical configuration (emergency generator powered by the green EDP), there is no disconnection of the emergency generator even when slats are extended. The reason why it is still required to extend the RAT in this situation is to allow the RAT to assist the green hydraulic system. The green system managing both the emergency generator and the load from the flight control demands can cause the emergency generator to shut off. With the RAT on, the green system thus gets that extra help to prevent an over demand situation. Here also, pushing the land recovery gives one of the SFCC computers. But with the EDP running the show, both the flaps and slats are available for the landing.


Why is the landing area of an aircraft carrier angled to port side, not starboard? On a modern aircraft carrier, the landing area is not aligned with the axis of the ship.

Given this part of the ship needs to be asymmetric, why do most (if not all) angled flight decks point to the left side of the ship (and not the right)? Is there any advantage to favor this side?

Farhan's answer covered why the deck is angled in the first place. As far as why the landing area of the deck points left instead of right, this is for several reasons:

Angling to port makes the most use of the deck area behind the catapults. Angling starboard would require the catapults to be shifted left off the centerline of the carrier (yes, the carrier does have two additional catapults on the port side, but these can't be used while the carrier is retrieving aircraft, which is more often than you'd think as the carrier maintains regular patrols).
Angling the landing strip toward the "island" (conning tower) of the carrier just doesn't sound like a good idea when you say it. That goes double when you realize the carrier is underway while you're trying to land on it, and therefore you have to lead the landing strip a little. If you don't, you'll end up "behind" the ship and the landing strip. With the landing strip angled to port, a plane on a "lagging" approach (lined up "behind" where the carrier's going to be at touchdown) will have a relatively clear path off to port for a wave-off. Angled to starboard, a plane that's "lagging" the carrier would be pointed right at the tower.
It would be possible to mirror the entire deck layout; island to port, landing area angled starboard. The reason the island is on the starboard side is primarily convention; all U.S. Navy aircraft carriers ever built have been either flush-decked or have had the island to the right. That's thus a familiar sight picture for naval pilots across all eras, and easy to train to. A new carrier design doesn't get the luxury of a whole new crew trained an entirely different way; the Navy's been training pilots since 1910, they've been landing planes on boats since 1927, and they'll do it they way they've been doing it for 88 years, thank you very much.

The Imperial Japanese Navy commissioned two carriers, Akagi and Hiryu, with island-left designs, fielded with the idea that the island-left carrier would sail in formation with an island-right carrier in the same flag group, with opposing traffic patterns that didn't conflict. This procedure never materialized in practice; most carrier groups in both Japanese and U.S. fleets ended up with a single carrier due to their high cost and lower total number, a pattern that continues today, and even when two (or more) carriers were close enough to be called part of the same naval group, they typically maintained a wide enough spacing to permit a left-hand pattern for both ships.


Is there a difference between how commercial jets and GA aircraft use airways? In the answer to this question about why people use airways, one of the commentors pointed out that GA aircraft often use airways in a different fashion than commercial aircraft.  I couldn't tell from context why this is true though.  Is it laziness, spite, lack of knowledge?  Or is it simply that GA aircraft are instructed to use them differently?  Or, even, is it that GA aircraft are typically at a lower altitude so they simply have different airways?

Why do GA aircraft use airways different than large commercial planes?

In Europe, some airways are only available in certain flight level ranges, so a commercial airliner and a GA turbo-prop travelling from the same departure airport to the same destination airport could be travelling on different routes due to airway restrictions in place and the inability of one aircraft to climb to the same level as the other. Below the routing for PC12 operating IFR at FL150 and an A320 operating at FL250.

(FPL-DEXAM-IG
-PC12/M-SDE1FGHIJ1RWXY/LB1
-EDDL1000
-N0160F150 KUMIK L603 KOSEK Z74 HAREM T104 ANORA
-EDDM0150 EDQT
-PBN/A1)

The above flightplan from EDDL to EDDM validates for FL240, but anything above FL245 will not pass, it will error out with:

PROF195: KUMIK L603 KOSEK DOES NOT EXIST IN FL RANGE F245..F999
PROF195: KOSEK Z74 HAREM DOES NOT EXIST IN FL RANGE F245..F999
PROF197: RS: EDUUFFM14 EDUUFUL13 EDUUNTM13 EDUUTGO13 EDUUSLN13 EDUUWUR14:F245..F255 IS CLOSED FOR CRUISING REF:[EUROED41A] ED AIP(ENR3.3 1 NOTE 9)

To fly the same flight in higher levels, you would need another routing:

(FPL-DAXAM-IG
-A320/M-SDE1FGHIJ1RWXY/LB1
-EDDL1000
-N0260F270 DODEN Y853 BOMBI T104 ANORA
-EDDM0150 EDQT
-PBN/A1)

This routing works fine for anything above FL245, below it would again error out with:

PROF195: BOMBI T104 HAREM DOES NOT EXIST IN FL RANGE F000..F245
PROF204: RS: TRAFFIC VIA DODEN Y853 BOMBI IS ON FORBIDDEN ROUTE REF:[ED2318A] Y853 DODEN BOMBI
PROF205: RS: TRAFFIC VIA DODEN Y853 BOMBI:F000..F245 IS OFF MANDATORY ROUTE REF:[ED2318B] Y853 DODEN BOMBI
ROUTE135: THE SID LIMIT IS EXCEEDED FOR AERODROME EDDL [EDDL50120A] CONNECTING TO DODEN


What is the purpose for two nitrogen charged chambers in an oleo strut? What is the purpose of a twin nitrogen charged chambers oleo, such as the one illustrated below? Why is it used and what are the advantages compared to a conventional mono-nitrogen chamber?
Training manual describes its operation as follow:

"Compression of the strut forces hydraulic fluid through the restrictor, thus pressurizing the nitrogen in the upper chambers and thereby absorbing landing shock loads. A flutter plate in the restrictor assembly permits an unrestricted flow through its controlling orifices during compression, however, flow is restricted during subsequent extension, thus eliminating any tendency to bounce during recoil."

While I understand the operation, it is not clear why this design solution was used.

Two chambers at different pressures creates a variable rate spring. Initial landing shock is absorbed by the low pressure chamber until its nitrogen is compressed to that of the higher pressure chamber, which then has sufficient volume to allow the desired amount of further compression at the higher pressure. The lower pressure chamber would have already consumed virtually all its volume.

The illustration is somewhat misleading in that the floating piston in the high pressure chamber should appear bottomed out when the strut is extended. There is no way for a fluid to be in equilibrium at two different pressures.


Am I Required to Use the Runway Advised by a Unicom? Let's say that an uncontrolled airport's ASOS reports wind 060 @ 6 knots.  The airport's unicom advisory states that Runway 22 is in use.  Am I required to use Runway 22, or can I use Runway 4?

You are not required, but it is advisable if there are other aircraft in the pattern.  My home airport often does not have anyone in the pattern and pilots that are based there will sometimes land opposite of the advised runway because their hangars are located at the other end of the runway from the FBO.  This means that they can either land long or do a long rollout after landing and then turn off at their hangars.  If they landed on the advised runway, they would having to taxi all the way back down from the FBO end of the airport or applying brakes to try to make one of the turnoffs to the taxiway.  You just listen on the CTAF and announce your intentions if you don't hear anyone.  It is probably more used when a pilot is flying back and wants to just make a long straight in approach while also ending up at his hangar with minimal wait time.


Why angle of attack is always shown against the relative wind parallel to horizon? I have a question about Stall, which I have difficulty to understand.
According to the theory, stall happens when :
1-  Speed is slow then a certain limit.
2-  Angle of attack is greater than a certain limit.
In the case of angle of attack, there is being spoken about relative wind along the wings.
But here is my confusion.
Relative wind is always shown parallel to the horizon, which strikes the wings.
While in my understanding, in the sky there is equal air and wind everywhere. We create our own hard wind for our wings, by moving fast through the air. Right?
So it should not make any difference in which direction we move, along the horizon, or at a steep angle upwards or downwards, relative to the horizon. The relative hard wind will be created straight in opposite direction where we move fast. Because we are also moving in the same direction as our wings angle of attack.
As I have shown with blue line in my illustration above.
If so, then there is no question about angle of attack at all. Because we always create our own wind by moving fast in any direction in the sky.
It would be different if we are moving parallel to the horizon, but only our wings have a greater angle of attack relative to the fuselage and the horizon. (As it is often explained in the example of, hand out of a car. While car is moving horizontally, but only the angle of our hand changes relative to the car). Then this stall theory is understandable.
If we look at the fighter jets and acrobatic airplanes, then we see that they can climb vertically up against the horizon, at 90 degree bank angle and fly upside down too. And they do no stall, because they create their own relative straight opposite wind by moving fast in any direction in the sky.
So why in normal planes we do have to think about straight only horizontally coming relative wind and accordingly the angle of attack against it?

I think this has been at least hinted at in other answers, but, just to put it succinctly, the diagrams aren't actually intended to show the airflow as always being parallel to the horizon. The airstream is just shown flowing along the horizontal axis of the graph because it's convenient to illustrate it that way. There is no implication intended in the diagram that the horizontal axis of the diagram is actually parallel to the Earth's surface.
The behavior of the airfoil is exactly the same regardless of the actual orientation of the wing and airflow relative to the horizon. All that matters is how the airfoil is oriented relative to the airflow. You can rotate those diagrams any way you want relative to the horizon and what they depict will remain true.
The blue airflow lines you've drawn on the diagrams seem to be assuming that the airplane is always climbing at an angle roughly equal to the angle between the horizon and the wing's chord line, but this is not true. You can fly with a high pitch-up angle while maintaining level flight if you fly slowly enough, for example. You will practice this (called "slow flight") when training for a private pilot certificate. You can (and often do) even descend with a nose-up attitude. And, if you put the nose up too far, you can descend very rapidly with a nose-up attitude after your wings stall. For example, Air France 447 was falling 10,000 feet per minute with a nose-up attitude as one of the pilots held the stick back.


What is this plane seen in the Falcon Crest TV series? Does anyone recognize the make and model of this small plane?  It was used in episode 154 <6.27> "Chain Reaction" of the 1980's TV show "Falcon Crest".  I'm working on a database for the show's picture planes and I am missing the identification of this aircraft and two more planes. Thank you so much for your help.

That's a Cessna 210 Centurion II
.
The Cessna 210 models K, L, M, N and R have the windows shaped like the one shown in the picture from Falcon Crest. If it has a two-bladed propeller, it is a model K, the others have three-bladed props.
The picture below is a Cessna 210M.

source: www.airhistory.net, photo copyright: Phil Vabre


Does a long time to solo hurt your chances of becoming an airline pilot? I've heard that on job applications for airline pilots that they ask the question: "How many hours to Solo?".  If it took someone 30 hours to solo but they had more logged flight time than someone who took 10 hours to solo, would the person who soloed in 10 hours have an advantage?

It depends on the final product presented to the airline.  If an airline had two candidates, one of whom soloed in 10 hours and did well in a Sim Eval but was self absorbed, didn't like to follow procedures if he/she knew better, and didn't work well with people, and a candidate who took 30 hours to solo but was a team player, can get along with anyone, was an eager learner and was emotionally mature, and did a passable job in the Sim Eval, they'd take the "slow learner" in a heartbeat if they were confident the candidate could get through the type course.

In the airline world, raw flying skills are an important baseline, but big picture personality related issues are just as important and personality problems with crews are a bigger long term problem than how long it took someone to catch on at the start.


How are aircraft filmed in sky? We have several spectacular and amazing videos of aircraft, flying high in the sky.  How are these videos filmed, at great altitudes?

Two words: chase planes.

A few more words: The easiest way to film an airplane in flight is with another airplane in flight. Here's one article on how that's done, including images captured from both sides; the chase plane (a Learjet 25B) getting the shots of the 787 Dreamliner, and crew aboard the Dreamliner getting a few snapshots of the chase plane.

Early shots taken from chase planes often used simple handheld cameras; a two-seater would normally be used with the pilot in front and the photographer in back. This is still a totally viable way to do it, but more modern professional shoots utilize aircraft specifically modified for the task by adding internal or external camera mounts, optical-quality viewports, etc. Here's a modified T-33 used by Boeing specifically for air-to-air photography, including some of the initial publicity shots of the 787 in flight; notice the camera mount between the seats that can be used by the GIB (he's using a handheld in this photo:

For lower-speed shots, you can also use a helicopter. Many of these such as news choppers have remote-control cameras on gyro-stabilized mounts, so the vibration of the airframe is minimized in the shot even at high zoom:

These same camera systems can be used for air-to-air shots as well, however the maximum airspeed of the helicopter becomes a limiting factor. Still great for publicity shots of small prop planes, or takeoffs/landings.


Is a winglet better than an equal span extension? Is there indisputable evidence that a winglet improves performance over an equal span extension? Please note: I am only interested in L/D improvements.

Winglets do improve roll performance, that is not what I am looking for. Also, if span is restricted, winglets improve L/D over straight wings. Again, that is not what I am looking for.

Boeing and Airbus use fancy wingtip designs to demonstrate technological sophistication and make incredible-sounding claims about them. This is not the kind of proof I ask for. Is there theoretical or practical evidence comparing winglets with span extensions of equal wetted surface which show that the winglet produces better L/D values at any point of the polar?

Bonus points for a comparison of net lift to drag comparison, so the structural impact of both wing extension and winglet get subtracted from the generated lift. This should be the fairest way to compare both, but it seems such research is not published at all.

Here is what I think you need to come to your own conclusion. First I will give a very general overview over lift creation, and then I will look at three wings:

An unmodified wing
This wing plus a winglet
This wing plus the winglet, but this time folded down into the plane of the wing.

For each I will plot the lift and bending moment distribution. I will assume an elliptic circulation, fully knowing that this is not what most aircraft use. But I have to pick a distribution to make all three cases comparable, and the elliptic one makes things easier. The conclusions can be generalized for other distributions.

This will be a lengthy post (you should know me by now), so thanks to all who persevere through all of it.

Lift creation and induced drag

This topic had been covered before, and I mention it again to show a very simple and elegant way to explain induced drag that does not need vortices. I want to dispel the myth that induced drag is caused by air flowing around the wingtip, and winglets somehow magically can suppress this flow.

Consider a wing with elliptic circulation over span (think of circulation as the product of the local lift coefficient $c_l$ and local chord; it is basically the lift per spanwise increment). The wing bends the air through which it flows slightly downwards, and creates an opposite upwards force, namely lift (Newton's second law). I choose an elliptic distribution because then the downwash is constant over span, which makes the following calculations easier.

The sheet of air coming off behind the wing looks trough-shaped and moves downwards, thereby pressing other air below out of the way and allowing air above to flow inwards and to fill up the vacated volume. That is how the free vortex is created, and air flowing around the wingtips has only a small part in this.

Induced drag is the consequence of the wing bending the airflow downwards. To simplify things, let's assume the wing is just acting on the air with the density $\rho$ flowing with the speed $v$ through a circle with a diameter equal to the span $b$ of the wing. If we just look at this stream tube, the mass flow is $$\frac{dm}{dt} = \frac{b^2}{4}\cdot\pi\cdot\rho\cdot v$$

Lift $L$ is then the impulse change which is caused by the wing. With the downward air speed $v_z$ imparted by the wing, lift is: $$L = \frac{b^2}{4}\cdot\pi\cdot\rho\cdot v\cdot v_z = S\cdot c_L\cdot\frac{v^2}{2}\cdot\rho$$

$S$ is the wing area and $c_L$ the overall lift coefficient. If we now solve for the vertical air speed, we get $$v_z = \frac{S\cdot c_L\cdot\frac{v^2}{2}\cdot\rho}{\frac{b^2}{4}\cdot\pi\cdot\rho\cdot v} = \frac{2\cdot c_L\cdot v}{\pi\cdot AR}$$
with $AR = \frac{b^2}{S}$ the aspect ratio of the wing. Now we can divide the vertical speed by the air speed to calculate the angle by which the air has been deflected by the wing. Let's call it $\alpha_w$: $$\alpha_w = arctan\left(\frac{v_z}{v}\right) = arctan \left(\frac{2\cdot c_L}{\pi\cdot AR}\right)$$

The deflection happens gradually along the wing chord, so the mean local flow angle along the chord is just $\alpha_w / 2$. Lift acts perpendicularly to this local flow, thus is tilted backwards by $\alpha_w / 2$. In coefficients, lift is $c_L$, and the backwards component is $\alpha_w / 2 \cdot c_L$. Let's call this component $c_{Di}$: $$c_{Di} = arctan \left(\frac{c_L}{\pi\cdot AR}\right)\cdot c_L$$

For small $\alpha_w$s the arcus tangens can be neglected, and we get this familiar looking equation for the backwards-pointing component of the reaction force: $$c_{Di} = \frac{c_L^2}{\pi\cdot AR}$$

If the circulation over span has an elliptic distribution, the local change in circulation times the local amount of circulation is constant, and the induced drag $c_{Di}$ is at its minimum. If this would be different, a higher local $v_z$ causes a quadratic increase in local induced drag, so the whole wing will create its lift less efficiently.

Now we know we can calculate induced drag and we understand why the vortex sheet behind the wing rolls up, producing two counter-rotating vortices, all without looking at the details of the wingtip. What counts is that the wing is of finite span, so the stream tube which is influenced by the wing is of finite diameter as well. Of course, in reality there is no clear boundary between air which is affected by the wing and other air which is not. There is a diffuse transition the more one moves away from the wing.

Comparison of wingtips

First the geometries: Here are three wingtips in top and front views for comparison:

Now let's look at the circulation distribution of the simple wing tip:

Again, I choose the elliptic distribution for simplicity. The corresponding bending moment looks like that:

No surprises so far. Now we add a winglet and make it work as best as possible. This means we have to give it an angle of attack where it carries the circulation from the wing over on the winglet and completes the elliptic tapering of circulation down to 0 at the tip:

The grey dashed line is the circulation of the original wing. I adjusted the circulation such that both wings produce the same lift. $b_{WL}$ is the span at the winglet tip, and for the bending moment plot I have folded the spanwise coordinate down on the y axis:

Now the bending moment starts at the wingtip with a nonzero value. Since the sideways force of the winglet is parallel to the wing spar, this bending moment contribution is constant over span. But there is more: Now also circulation at the old wingtip location is nonzero, and we get a substantial lift increase at the outer wing stations. This effect is what causes the additional lift and gives the better aileron response that winglets make possible. But it also increases the root bending moment, because this additional lift acts with the lever arm of the outer wing.

How can we compare the induced drag of the wing with winglets to the original wing? The circulation gradient is lower, that helps. Also the diameter of that stream tube is larger, but it is hard to say by how much. The sideways force on the winglet is created by pushing the vortex sheet aft of the winglet sideways out, so the trough-shaped area should become wider. Empirical evidence hints at an increase in diameter of 45% of the winglet span (see chapter 6 for a discussion of several papers on the topic).

Just for the heck of it, let's assume that the diameter really increases in line with winglet span. Then let's compare that to the straight wing extension, where the same diameter can be assumed with much more certainty:

Now also the lift on the folded-down winglet acts upwards, so the circulation at the center of the wing can be reduced even further. However, now it adds a linearly increasing part to the bending moment, and the outer wing section creates more lift, as before with the wing with winglet:

Here, the root bending moment is higher than in the winglet case. This is a second advantage of winglets: They allow to increase maximum lift with less bending moment increase than a wing extension. But the wing extension puts all parts towards the creation of lift, and not some to the useless creation of side force. Both the extended and the winglet-wing have the same surface friction and (when we assume the same diameter of the hypothetical stream tube) the same induced drag. But since the winglet creates some side force, the remaining wing needs to fly at a higher lift coefficient. Also, the intersection of wing and winglet might be as well rounded as possible, this is where early separation starts at higher angles of attack. None of this affects the straight wing extension.

Most evidence shows that winglets improve L/D over the original wing, but folding the winglet down will more than double its effectiveness in lowering drag. Even if we assume that the winglet is just as good as an equal span extension, still the span extension comes out ahead in L/D improvement because all its lift contributes to overall lift, whereas the winglet produces a side force instead. If no separation occurs at the wing-winglet intersection, both will create the same induced and profile drag (pressure and friction), because both have the same wetted surface and the same local circulation. Again, this gives winglets the benefit of equally low induced drag, which is not supported by most measurements.

The extended wingtip in the example above has interesting characteristics. It is a sweptback (raked) wingtip, which causes the local lift curve slope to be lower than that of the straight wing. This increases its maximum angle of attack and  - assuming that the local area is bigger than what an elliptic wing shape would dictate - makes it possible to keep a nearly elliptic circulation distribution over a wider angle of attack range. The bigger local area is a sensible precaution against the wing tip stalling first, so a raked wing tip will combine benign stall characteristics and very low induced drag.

Compare this to the winglet, which has to be tailored for one polar point: Since changes in wing angle of attack will not change the incidence of the winglet, it cannot adapt as well to different flow conditions as can the extended wing. In sideslip the winglet will mess up the circulation distribution on the wingtip and will act like a deflected spoiler.

Conclusion

Comparing equal winglets and wing extensions gives these basic characteristics:

Both have the same viscous drag at low angle of attack.
Both can create more maximum lift, and both lower induced drag.
The wing extension can create most lift for the given increase in wetted surface.
The wing extension is more than twice as effective in lowering induced drag.
The wing extension gives a better circulation distribution at off-design angle of attack.
The wing extension produces the highest root bending moment for a given amount of lift.

How much the bending moment increase will drive up structural mass depends on the aspect ratio of the original wing. Low aspect ratio wings will not suffer much, but stretching high aspect ratio wings will drive up spar mass considerably. But please note that the winglet also causes higher root bending moments, and it creates less bending moment than the wing extension because it creates some side force instead of pure, useful lift.


Why was the "Dispatch Reliability" of the A320 "low" during its first years? I found the following Statement on the German Wikipedia page about the A320 family of aircrafts.

  Nach erheblichen Schwierigkeiten in den Anfangsjahren erreicht die A320-Familie mittlerweile eine Abflugzuverlässigkeit von 99,7 Prozent.
  (https://de.wikipedia.org/wiki/Airbus-A320-Familie#Wirtschaftliche_Aspekte)

This roughly translates to:

  After significant difficulties in its early years, the A320-family reaches a Dispatch Reliability of 99.7% nowadays.

I could not find more information about this statement, so my question would be why the Dispatch Reliability of said aircraft family was supposedly significantly lower in its early years? Is it normal for completely new types of aircraft to have such a low DR that it's worth mentioning?

The A320 saw entry into service on 18 April 1988 with Air France.
About a year later Flight International covered the dispatch topic. For British Airways – whose initial acquisition of the A320 was through the British Caledonian takeover, and not an order – the problems weren't with the plane, but with the Boeing culture that had to adapt.
Otherwise there was nothing notable.

(...) The "culture difference", says Bracken, materialised in apparently trifling (but actually problematic) ways, such as line mechanics having trouble with Airbus literature. There was nothing wrong with the literature, it was just different.
November 1 came. "It wasn't a very good start," says Bracken, "but the reasons were outside the aircraft." By January, BA's A320 technical dispatch rate had not climbed out of the 96 per cent range, reveals Barry Gosnold, chief engineer technical services.
That is a revealing figure. BA had expected much more, even though the performance is not unusual for a completely new aircraft.The A320's performance with Air Inter at the same time was between 2 and 3 percentage points better, although Air Inter had received its first A320 months later than BA.

Source: AIRBUS A320 AT WORK, Flight International, 17 June 1989


Did John Stringfellow ever build an aeroplane, powered by a steam engine, that really flew? After doing some research, I found two apparently conflicting articles. The first, dated Sep. 23, 1848, describes John Stringfellow as marching through the air in a tent. The second, published about 20 years later (July 10, 1868) quotes Stringfellow as declaring that the world still waited to see him flying.
Question: Did John Stringfellow ever build an aeroplane, powered by a steam engine, that really flew? Does it exist written contemporary evidence (letters, articles, patents, etc.) proving that this inventor designed, built and flew a plane of any kind (manned or just a model)?
1848-09-23, Scientific American, September 23, 1848, col. 3, p. 4.

An Air Navigator.
A series of experiments have lately been made beneath an immense tent
in Cremorne Gardens, London, by a Mr. Stringfellow — a fine name for
suspension. The inventor marches through the air by a machine which
sustains and propels itself through the cicumambient fluid. The
machine excited considerable attention and surprised all the
spectators by its wonderful performance. The next expedition that is
fitted out by the British government to explore the Niger and the
country through which it winds its sluggish and pestilential way,
should employ this Mr. Stringfellow with a number of his machines to
make a flying exploration, untramelled with their heels in mud or
water.

1868-07-10, "Aerial Navigation", English Mechanic and Mirror of Science, July 10, 1868, col. 3, p. 341.

Aerial Navigation ...
Mr. Stringfellow, one of the exhibitors at the Crystal Palace,
enthusiastically declared that the world should yet see that he could
fly with his own machine, and this opinion seemed to be universally
held. ...

I was not able to discover articles of 1848, or even up to 5 years after, that give details about the 1848 flight of Stringfellow's model. However, I found 3 first hand articles, published on July 3, 24 and Oct. 9, 1868, which say that, at an exhibition which took place starting with June 25, 1868, at the Crystal Palace in London, John Stringfellow exhibited two engines. For the larger one, that generated, when started, a little more than 1 HP, he took a prize of 100 pounds. The smaller motor was installed on a working plane which apparently flew but the description is vague being unclear whether the author talks about an event that happened in 1868 or he refers to previous flights, like the one of 1848. Anyway, the conclusion was that despite taking the prize (for the larger standalone engine) it was not clear whether that motor would serve for any practical purposes in aeronautics.
From the numerical values furnished by the commission that awarded the prize, it results that Stringfellow's motor needed:
$150 ft * 3 in^2 = 88.5 liters$ of steam at $80 lbf/in^2 = 5.4 atm$ per minute to generate 1.09 HP continuous power.
Nothing is said about the interval of time during which the boiler was able to maintain this considerable flow of steam. Another remark would be that the steam was expelled from the cylinder at 5.4 atm which means the engine had a very poor efficiency.
"The Aeronautical Society of Great Britain", The Mechanics' Magazine, July 3, 1868, col. 2-3, p. 3.

THE AERONAUTICAL SOCIETY OF GREAT BRITAIN. ...
We have said that there are a few exhibits which are a step in advance
of the rest at the Crystal Palace. Amongst the rest may be noticed a
working model of an aerial steam carriage, by Mr. J. Stringfellow, the
whole, including engine, boiler, water and fuel, weighing about 12
lb.; cylinder, 1 3-16ths inch diameter; 2-inch stroke; works two
propellers, 21 in. diameter, about 600 revolutions per minute; gets up
steam of 100 lb. pressure in five minutes. ...

"The Aeronautical Society Late Exhibition", English Mechanic and Mirror of Science, July 24, 1868, col. 3, p. 378.

THE AERONAUTICAL SOCIETY'S LATE EXHIBITION.
THOSE who predicted that the exhibition of the Aeronautical Society at
the Crystal Palace would prove a failure may, in part be said to have
been true prophets; but the exhibition was not all a failure — far
from it: enough, we think, was given in evidence to cause at least one
section of the members to be of good heart. True, the expected
demonstration of the possibility of flying was not made, and the
rewards offered for flying have not been gained. The flying man put in
no appearance with bis wings, and the only model of this class which
actually worked was the ingenious steam-propelled machine of Mr.
Stringfellow, which, running along a a wire, has enabled our facetious
contemporary, "Punch," to change the inventor's name to Wirefellow,
and which elicited the remark from sundry spectators of a certain
class that "he was wiring into it."

"The Aeronautical Exhibition", The Mechanics Magazine, October 9, 1868, pp. 282-284.

THE AERONAUTICAL EXHIBITION.
IT will be fresh in the memory of our readers that the Aeronautical
Society of Great Britain held their first exhibition at the Crystal
Palace on the 25th of June last and ten following days. ...
The only engines, therefore, which competed for the prize were three,
— one by M. Camille Vert and two by Mr. J. Stringfellow. M. Vert's
engine was a small model of a steam engine, weighing 2 lb., for
propelling a hydrogen gas balloon. Mr. Stringfellow's engine No. 1 was
a light engine and boiler for aerial purposes; it was of 1-horse
power, and weighed 13 lb. No. 2 was a similar engine on a smaller
scale. M. Vert's engine was found to be too small to enable any
accurate test to be applied for ascertaining its power. The jurors
state that it appeared to keep up steam well in the repeated
demonstrations in propelling a model balloon in various directions
within the Crystal Palace building.
Mr. Stringfellow's engine No. 1, from its size and power, may be
considered something more than a mere model. The cylinder was 2 in. in
diameter, stroke 3 in., and it worked with a boiler pressure of 100
lb. per square inch, the engine making 300 revolutions per minute. The
time of getting up steam was noted; in three minutes after lighting
the fire, the pressure was 30 lb., in five minutes 50 lb., and in
seven minutes there was the full working pressure of 100 lb. When
started, the engine had a fair amount of duty to perform in driving
two four-bladed screw-propellers, 3 ft. in diameter, at 300
revolutions per minute.
In the jurors' report, the data for estimating the power are taken as
follows: — Area of piston, 3 in., pressure in cylinder 80 lb. per
square inch, length of stroke 3 in., velocity of piston 150 ft. per
minute, 3 x 80 x 150 = 36,000 foot pounds; this makes rather more than
1-horse power (which is reckoned as 33,000 foot pounds). The weight of
the engine and boiler was only 13 lb., and is probably the lightest
steam engine that has ever been constructed. The engine, boiler, car,
and propellers together, were afterwards weighed, but without water
and fuel, and were found to be 16 lb. The council of the Aeronautical
Society agreed that this engine, as a complete working machine, met
the condition of the Society's award for "the lightest engine in
proportion to its power from whatever source the power may be
derived." The prize of £100 was accordingly allotted to Mr.
Stringfellow. ...
The exhibition, however, proved that the advocates for aerial
locomotion by mechanical means have greatly increased in number, and
their designs for accomplishing this end were as varied as they were
ingenious. But still we have no practical result by which we can
measure the progress of aeronautics in this direction; we cannot say
we have even attained the first step in actual practice. The nearest
approach that was made to this long-desired condition of things was
the demonstration that by means of vertical screws a weight of 100 lb.
might be supported by a constant force of about 90,000 foot-pounds or
three-horse power. Then we have Mr. Stringfellow's engine of one-horse
power and weighing only 13 lb., which, taken in conjunction with the
vertical screws, would appear to indicate the direction in which
mechanical flight may be realized; and this is as far as we have got —
just on the borders of possibility. But we must bear in mind the
enormous expenditure of power the system of vertical screws requires,
and the very flimsy character of an engine and boiler sufficiently
powerful to work them, and sufficiently light to allow of their
working result, if, and when attained, being made available for
transit.
It practically comes to this, that Mr. Stringfellow carried off the
£100 prize for the lightest engine in proportion to its power, but of
what real service that engine will prove to the science of aeronautics
remains still to be seen.


What, if any, would be the most correct term for the aerodynamic flight control surfaces of SpaceX's Starship? SpaceX's Starship uses a unique flight control scheme during descent that I have not seen anywhere except with skydivers: it falls straight down belly-first using four aerodynamic control surfaces at the nose and the bottom to induce drag and thus provide pitch, roll, and yaw control, somewhat like a skydiver uses their four limbs.
 [Time lapse by @asy on Twitter, original source footage by LabPadre.]
This is a pre-flight actuation test of the aerodynamic control surfaces. The Starship is in its launch attitude, i.e. tail-down / engines down. On descent, it would be falling belly-down, where the belly is the side facing away from the camera, i.e. if the vehicle were falling right now, we would be looking at it from the top.
These aerodynamic control surfaces have been called lots of names by the spaceflight community: fins, flaps, wings, flings, wing-dings, flippety-flappety-bits (yes, really). Tim Dodd, the Everyday Astronaut, sometimes jokingly calls them Elonerons. SpaceX themselves only name them once on the Starship website, where they use the term flaps. Elon Musk uses both flaps and body flaps in tweets.
They are, however, different from many of the concepts mentioned above, in that they act perpendicular to the airflow. The closest analogy I can think of, other than a skydiver's limbs, is how the Northrop Grumman B-2 Spirit uses differential speed brakes for yaw control instead of a rudder.
Given that we have so many names for so many different types of control surfaces (e.g. flap, slat, spoiler, aileron, elevator, rudder, speed brake, horizontal and vertical stabilizer) as well as combinations of them (spoileron, flaperon, stabilator (all-moving tail), "ruddervator" (V-tail)), etc., there must be some term we can apply to these?
What I am looking for is a term that concisely conveys what these control surfaces do, in much the same way as everybody knows what an "aileron" or an "elevator" does. Or, alternatively, an answer could be that such a term does not exist.

As far as I know, the only other example of this flight profile in aviation is post-stall maneuvering, especially for flat spin recovery. For planes, control in these flight regimes is mainly accomplished by thrust vectoring and canard/stabilator actuation. The Starship flaps don’t match these.
Another close sibling of this flight profile is the Shuttle reentry. The Shuttle used a body flap as well as conventional Delta-wing control surfaces in combination with RCS to maintain a specific, high AoA during hypersonic reentry, then transitioned to gliding for approach and landing. The Shuttle body flap (under the engines) has the closest resemblance to the Starship flaps out of any space vehicle I know. It has a much smaller actuation range, though.
Of course, the control surface naming depends on how you define the vehicle frame; with thrust direction-forward, the flaps are variable-dihedral. With velocity direction-forward, the flaps are variable-AoA with a wide actuation range.
The noun I would go with would be variable-drag flaps. Spoilers aren’t correct; there isn’t another airfoil to spoil. Airbrakes usually don’t contribute to attitude control.


How to calculate minimum flying speed in distributed propulsion? How to calculate minimum takeoff speed due to distributed propulsion?

Is there a quick way or rule of thumb way to calculate effects of distributed propulsion?

I understand NASA is doing research in this area and is determined to prove an increase in efficiency of 500% in real world tests, although this may include the benefits of electric distributed propulsion.

I understand they swapped the wings out on their Dornier, from 17psf to a 45psf wing loading.  That's an increase of almost 300% in just the aerodynamics!!

So....

I understand the bigger the prop disc, the more efficient the prop.

if I have a small ultralight with 25 hp, and now use two 12.5 hp motors instead with newly optimized props....

What is the approximate % increase in efficiency/thrust as the prop disc is now twice as big? ( assume only a small weight increase)

Now let's say I have a 10' span, with 2 motors with 6' props. The entire wing and tail is in the prop wash.

What is the formula for propeller exit speed(v2)?

Is this my new indicated airspeed(v+v2)?

In other words, if my stall speed is 40mph and my prop exit speed is 20 mph at zero ground speed distributed over the entire wing, can I take off at a 20mph ground speed? i.e IAS= 20mph ground speed + 20 mph prop exit speed= 40mph need for takeoff.

Is this correct, at least conceptually, after all, it's just like a headwind, no?

The better way to look at predicting the performance of a distributed propulsion wing (specifically the NASA X-57 wing concept) is to think of the propellers along the span as a high-lift device that augments the lift of the wing through interaction with the trailing-edge flaps, rather than as headwind-generators. In other words, the propellers increase wing C_L, allowing you to have a smaller wing for the same landing and takeoff distances.

With many small propellers, you get performance that starts to resemble a jet flap or externally blown flap, but without the complexity of internal ducting or non-uniformity due to a small number of engines.

Optimistic rule of thumb from thin airfoil theory is that, for a 2D section
$c_\ell = \frac{\partial c_\ell}{\partial\alpha}\alpha + \frac{\partial c_\ell}{\partial\delta_F}\delta_F$

$\frac{\partial c_\ell}{\partial\alpha} = 2\pi(1+0.151\sqrt{c_J}+0.219c_J)$

$\frac{\partial c_\ell}{\partial\delta_F} = 2\sqrt{\pi c_J}\left(1+0.151\sqrt{c_J}+0.139c_J\right)^{\frac{1}{2}}$

where $c_J = 2\frac{h_J}{c}\frac{V_J}{V_\infty}$

$V_J$ can be calculated from propeller momentum theory based on disk area, input power, and assumed efficiency.

Bear in mind this is for section $c_L$, and will need 3D corrections for the whole vehicle based on percentage of span with propellers, induced flow, etc. It is also reasonable only in the limit of small flap deflections and angles of attack. Wind tunnel data or flight test reports might be a better place to get a first-pass estimate.

https://www.researchgate.net/publication/299641308_Comparison_of_CFD_and_Experimental_Results_of_the_LEAPTech_Distributed_Electric_Propulsion_Blown_Wing

https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19740003719.pdf

https://arc.aiaa.org/doi/abs/10.2514/6.2019-3170


What happens to the pressure differential of airflow under and over the wing as the critical angle of attack is reached? I've been doing a research project for mathematics which involves calculating the relationship between lift and angle of attack. While this may seem pretty easy, it's not for a 12th-grade student like me.

We all know that the pressure differential is 1/2 ρ〖(v2-v1)〗^2, where v2 is the airflow velocity over the wing and v1 the airflow velocity under the wing.

It is also fairly commonly known that as AoA is increased, so does (v2-v1) leading to a higher lift coefficient.

But, my question is: Whenever the critical AoA (AoA at which the aircraft stalls) is reached, how do we calculate pressure differential? What is the new figure for (v2-v1)? How do these airflow speeds change as the critical AoA is reached?

While this may seem pretty easy, it's not for a 12th-grade student like me.

It definitely does not seem easy. It takes solid background in numeric integration, good working knowledge of a numeric library and some serious programming time.

There is a good reason everybody—and I mean aircraft engineers and researchers—just uses XFoil. And that's just the basic analysis—3D calculations require even more complex software packages—that cost hefty sums of money (XFoil is free).

  We all know that the pressure differential is 1/2 ρ〖(v2-v1)〗^2, where v2 is the airflow velocity over the wing and v1 the airflow velocity under the wing.

There is no single $v_1$ and single $v_2$. There is just $\vec{v}(x, y)$ (limiting to 2D analysis like XFoil does for simplicity) that is different at each point. You can't just take $\alpha$ and divine two speeds from it, because the situation is more complex than that.

Remember that:

The air flowing over the wing reaches the trailing edge in significantly shorter time than the air flowing below it!
A thin flat plate is not a particularly good wing, but it does generate some lift.
At the Reynolds numbers involved for full-scale aircraft the flow is rarely laminar beyond 20–30% chord. Turbulence increases the velocity and therefore decreases pressure and increases lift.
In viscous fluid, the velocity must be continuous in all coordinates and that includes the boundary, so the layer directly touching the surface is not moving relative to it. The velocity quickly increases in the boundary layer. The properties of this boundary layer is what governs whether the stream will remain attached or not.
There is no lift in inviscid fluid (= superfluid) and neither would there be in massless fluid. That is, you need to take both inertia and viscosity into account to get any result.

Basically the only way to calculate this is by evaluating the Navier–Stokes equations, which has to be done numerically and in fairly fine grid to achieve any useful precision.

Before numeric integration, there were some simpler analytical methods like the thin airfoil theory, but those didn't work from scratch—some coefficients must be measured experimentally.

  Whenever the critical AoA (AoA at which the aircraft stalls) is reached, how do we calculate pressure differential?

Just like before—by integrating the Navier–Stokes equations.


Why not have Pitot Heat always on? Most (all?) aircraft have a Pitot Heat Switch which must be turned on in certain conditions.

Seems Pitot Heat is one of those things you don't want to forget to switch on when it's needed, or worse, not recognize you need it on and take action based on false readings.

The pre-flight checklists I've read always say something like:

Pitot Heat - As Required

Why not always have it on? Or not have a switch and it just comes on with the aircraft's other power systems?

Is there a downside to it being always on?

I can generally only speak for commercial aviation. The reasoning is probably the same or similar for General Aviation, but seeing as that is such a wide and varied field, with a tremendous number of different aircraft, probe types, etc., someone somewhere could probably find some niche corner-case to dispute the reasoning. My guess would be that, generally speaking, in General Aviation, certain probes are most likely only meant to run for a certain period of time. After all, it's not like a Piper Cub is going on an 18-hour non-stop flight to China like a Boeing 777 might. I'm not saying that a probe will fail after 18 hours of being on, just that repetitive use in that way might necessitate a different type of probe with a more robust design.

In commercial aviation, many modern airliners use certain logic to turn pitot heat on or off. With the exception of a manual override switch equipped on some aircraft, to force it off, the pitot heat is generally on during most flight phases and works as such:

On the ground, with engines off, the pitot heat is turned off. On the ground, with engines running, pitot heat is on low. Finally, once the plane takes off, the pitot probes are put into a "high heat" condition and remain that way for the remainder of the flight (unless manually turned off).

The pitot heat is generally monitored with current sensing relays such that if the pitot heat draws excessive current or not enough current, the crew will be alerted to a pitot heat failure.

Why not keep them full hot all the time? For one, that would most likely cause premature failure of the heating element. Another reason is that, on the ground, pitot covers are normally applied to prevent bugs from building homes inside the tubes as well as preventing other debris from entering. The pitot covers will melt and destroy the pitot probes if the covers are left on while heat is applied. This can also be problematic and dangerous if the pitot cover partially melted, wasn't noticed by ground crews, and the drain holes or entry holes were blocked by the melted cover. Finally, and probably most importantly, the pitot probes get extraordinarily hot. Speaking from experience, and from having touched one that had failed in the "full hot" phase, it takes only a brief second to receive excessive burns on one's hand. Most of us that have been around them in our lifetimes have been burned by them, and a savvy mechanic will always approach them with caution!

So, it makes sense to have the probes off, on the ground, with engines not running. Otherwise, commercial aviation has more or less decided that "full hot" in the air, and low heat on the ground with engines running is a design characteristic.


How are aircraft transported to, and then placed, in an aircraft boneyard? I was looking at the boneyard at Davis Monthan Air Force Base and was wondering:
How are aircraft brought there, and then how are they placed so tightly together?

Most are just flown in to Davis-Monthan AFB and then towed over to the boneyard where they're 'pickled'.  There are a couple different levels of preservation depending on it's potential future (returnable to flight, parts, etc.).  More information is available at the AMARC web site.


Where does the Garmin G1000 Nav ID come from? If you tune in a NAV frequency into your G1000 the ID of the Nav Aid will be shown next to the frequency after some time. In this example SEA:

Source: http://krepelka.com

How does this work? Because we know that there is the possibility that two VORs could have the same frequency, so the frequency is not uniquely bound to an identifier.

I have two assumptions:

The Garmin is listening to the morse code, transfers it into the letters and shows it in the display.
The Garmin is checking the Nav database and searches for the frequency and takes the identifier of the closest, based on the GPS position.

Which one is right, or is there even another way Garmin does it?

Garmin doesn't explicitly state the method in the documentation I can find, but I believe the answer is #1. It decodes the Morse code and displays that value. I say that for two reasons:

First; Garmin, like most successful aviation companies, makes it a habit to reuse technology where it can. Volumes are low and certification costs are high so using the same NAV receiver in more than one product saves costs and increases profit. From their web site, the GNC 225 decodes the ident.

  Automatically displays tuned frequency's navaid or airport identifier
  Automatically decodes Morse signals so you don't have to

Second; the fact that it takes several seconds for the ID to display indicates it's being decoded as it has to be received before it's displayed. If it were pulled from the database, it would be available almost instantly.


Is there any benefit to sweep back on an ultralight? Is there any benefit to sweep back on an ultralight of:

15 deg
30 deg
45 deg.
other?

There are two primary benefits of wing sweep at ultralight speeds. The first is that it acts as variable dihedral -- the higher your angle of attack, the more dihedral effect you get from sweep.

Secondly, when you see ultralights (or hang gliders) with a lot of sweep, it's generally for a tailless design that needs to get the tip fins/rudders and twisted tips and/or elevons back behind the CG location so they have a useful moment arm.

The benefits of sweep in transonic flight (delay of Mach effects onset) obviously don't apply at ultralight aircraft speeds.


Does skin friction change with AoA? Does skin friction change with AoA or stay constant during change in angle of attack?
I think that local airflow velocity change with AoA so this must affect skin friction ..
But what definiton tells?

Let's start by skin friction drag formula:
$$F_s=\overline{C_f}\frac{\rho V^2}{2}A$$
where $\overline{C_f}$ is the average skin friction coefficient, $\rho$ is the density, $V$ is the True Airspeed and $A$ is the surface area of the wing.
By varying the Angle of attack, the term $\frac{\rho}{2}A$ stays constant assuming incompressible flow. As the Angle of attack increases, the velocity of the fluid increases over the top surface of the airfoil due to the lower static pressure being created. Therefore, as the Angle of attack increases, the skin friction drag also increases.
As the Angle of attack increases, the skin friction coefficient will also change. Let's assume two scenarios, one in which the angle of attack is close to zero, and one in which the angle of attack is very high, but the airfoil is not stalled, for instance, $12^{\circ}$. In the first scenario where $\alpha\approx0$, the flow over the airfoil is mostly laminar, that is, the velocity gradient inside the boundary layer is smaller close to the surface $\left(\nabla V=\frac{\partial{V}}{\partial{y}}|_{y=0}\right)$ which exhibits lower skin friction coefficient. Therefore, in scenario #$1$, the drag due to skin friction would be lower.
If the angle of attack is suddenly increased to $12^{\circ}$, scenario #$2$, the transition point on the upper surface would move upstream thus increasing the portion of the turbulent layer over the airfoil. And since the turbulent boundary layer has a higher skin friction coefficient, as a result, the skin friction drag would increase. On the lower surface, things are reversed. On the lower surface, the transition point would move downstream, increasing the laminar portion and lowering the skin friction coefficient. Since the fluid becomes more turbulent around the airfoil with an increasing angle of attack, the net change would be an increase in the skin friction coefficient resulting in the higher skin friction drag.
If we were to account for the compressibility effect on the skin friction coefficient, the skin friction coefficient would decrease. That is, the skin friction coefficient in the compressible flow would be smaller compared to the skin friction coefficient determined by incompressible flow:
$$\overline{C_f}^\prime=\frac{\overline{C_f}}{(1+0.144Ma^2)^{0.65}}$$
where $\overline{C_f}^\prime$ is the skin friction coefficient of the compressible flow, and $Ma$ the Mach number.
To summarize, with an increase in the angle of attack there would be an increase in the skin friction coefficient due to the presence of turbulent flow along with an increase in the fluid velocity over the airfoil, which would result in higher skin friction drag.


How do intake deflectors work? Some turboprop aircraft, for example DHC 6 Twin Otter have “intake deflectors” which should be engaged when taking off from or landing on unpaved surface to prevent foreign object damage. But how does it actually work?

This document describes it nicely. They are also known as inertial seperators:

  The nacelle inlet of many aircraft models includes an inertial separator
  provided by the aircraft manufacturer to prevent heavy particles from
  entering the engine inlet. Most installations incorporate two moveable
  vanes, one upstream of the engine inlet and the other blocking the bypass
  duct. For bypass operation, the inlet vane is lowered and the bypass
  duct vane is opened permitting maximum separating efficiency. In some
  installations, the vanes are fixed in the bypass mode.

  As shown in the sketch, air entering the engine inlet must turn sharply past the inlet vane (shown in the lowered position). Particles heavier than air are carried straight through, by their own inertia, into the bypass duct and dumped overboard.

  The aft radial inlet design provides many advantages such as superior
  anti-icing utilizing the principle of inertial separation, low noise levels and unequalled protection from FOD.


What is the best way to navigate a storm? Let me say I had no choice other than fly through a storm I'm on a one way mission to airdrop a vital cure to an isolated area. I can't turn around, land or go around, I have to go through. Going around it means I run out of fuel and they all die.

How does a pilot navigate a thunderstorm? What altitude has the least lightning, wind and hail?

If you're flying through a thunderstorm, lightning is the least of your concerns. Airplanes are struck by lightning all the time, and while it may mess with your electronics (navigation, radio, etc.), it very rarely causes any more damage than that. But to answer the second part of your question, there's no one place in a thunderstorm that has any more or less chance of lightning than any other.

The big issue with flying through a thunderstorm is with the wind. There's a lot of vertical air movement in thunderstorms, and the wind shear can easily cause structural damage to your aircraft if you're not very careful (and sometimes the wind can damage or destroy aircraft even if you are careful. And that's leaving out other dangers, such as hail. Thunderstorms are not good places to fly).

The accepted method for flying in a thunderstorm is to fly straight through (as most thunderstorms are small enough that going straight gets you out of it faster than turning around) at less than your maneuvering speed (to minimize the chance of structural damage to your wings), accepting any changes of altitude (again, to minimize stress on your wings).


How is rudder used in a aileron roll? Rudder is usually used to maintain coordinated flight. When banking to one side to begin a turn, the rudder is usually pushed towards that side. I guess this is still valid in inverted flight: in normal flight, if the pilot bank the plane to the right, it also push the right pedal (to conteract adverse yaw); in inverted flight, if the pilot bank to the right, the plane will turn to the right seen from the ground, i.e. the left seen from the pilot. The pilot should apply left rudder to keep coordinated flight.

If the pilot want to make a aileron roll, what will be the general action on rudder given the flight goes quickly from normal to inverted and back to normal, and the pilot don't want to initiate a turn?

The key in flying a good roll is to keep the aircraft's nose up in anticipation of the lift requirement when flying in knife-edge or inverted flight. Since in knife-edge the elevator and rudder will assume each others function, you will need rudder as well as elevator for directional and pitch control. Also, depending on adverse yaw, some rudder is required to hold yaw to zero when applying ailerons. The procedure is as follows:

Speed up, so you are fast enough for roll maneuvering and inverted flight.
Pitch up a few degrees. How much depends on wing incidence and zero-lift angle of attack. Look out for some feature on the horizon: This will help you to keep your direction constant.
Stop the pitching motion, then apply full ailerons. Correct the aileron-induced yawing motion by applying gentle rudder (trailing edge in the direction of the up-going aileron). Your goal is to keep the fuselage pointing at the chosen feature throughout the maneuver.
With increasing roll angle, apply rudder to keep the nose up. At the same time, push the stick gently to avoid course deviation. At 90° roll angle, both elevator and rudder have changed their function: The elevator will now control direction, and the rudder will control pitch.
Keep the rotation rate and push the stick more when changing from 90° to 180° roll angle. In inverted flight you need to keep the aircraft trimmed, and depending on the static stability and speed this can require considerable negative elevator deflections. At the same time reduce rudder deflection such that the fuselage will still point into the initial direction. At 180° roll angle both elevator and rudder have returned to their old function, but pulling will now start a dive.
On the way to 270° you will again need to add some rudder, now in the opposite direction, to keep the nose up. Again, elevator and rudder will exchange their functions.
Once you return back to normal flight attitude, stop the rolling motion and reduce rudder deflection back to zero.


Why don't airliners have a navigation GPS as back up in case of a pitot tube malfunction? Why don't airliners have a navigation GPS as "back up" in case of a pitot tube malfunction? Aeropero 603 and Air France 447 plus many other im sure, could have been saved if these aircrafts would have had a simple Navigation GPS as a plan B. The GPS could have given the pilot the basic information needed to keep flying instead of leaving them confused about what the instruments were showing due to the blocked pitot tubes. having reliable information of ground speed, heading and altitude would have helped the pilots recognize the situation they were in, which was caused by Pitot tubes blockage. Most private pilots now use GPS why not the big airliners?
There is many blogs about why airliners do not have GPS tracking so the planes could be found after a crash, but none talking about why GPS is not use to help prevent some of those crashes.

Airliners do generally have GPS receivers, but GPS doesn't give you the information that's needed in this case.

In AF 447, the primary consequence of the blocked pitot tubes was that the flight envelope protection system couldn't do its job properly and therefore automatically disengaged together with the autopilot. The subsequent crash was caused by the pilots' confusion and failure to realize which consequences this had for the aircraft's reaction to their control inputs.

However, what this system needs to function is not something a GPS system can possibly provide. GPS knows how fast the aircraft is moving relative to the earth, but for flight envelope protection you need to know how fast the aircraft is moving relative to the surrounding air, which GPS doesn't have any way of knowing. The differences between these values is caused by wind and can be pretty large. It would be a recipe for disaster to let the computer limit what the pilot can do based on an arbitrary assumption that there's no wind. (And, as Terry comments, there's an additional difference caused by air pressure).

If I remember correctly, the AF 447 investigation concluded that the pitot tubes had thawed and began functioning properly well before impact; it was not the loss of airspeed indication that prevented the pilots from salvaging the situation. By this time the pilots were confused and thought the airspeed indication was still malfunctioning because it showed a much slower speed than they thought they ought to have -- whereas in fact it was working perfectly well.


Why aren't the CVR and FDR required to store data for a longer period? I'll be honest I don't know a lot about the brand new airplanes, but one think I do recall a lot of from reading accident reports is that they typically only recorded the last 30 minutes of data.

When an aircraft costs over 30 million USD, with some costing in excess or $300 million, surely being able to increase the amount of data recorded 10x or 20x pales in comparison to the costs?

CVRs typically store two hours of voice recordings (source).

It simply wouldn't serve any purpose to store longer recordings.
It is extremely rare for an emergency to last for such an extended time.
Qantas 32 was one of the longest emergencies in recent memory, and it was only 1:45.

The MH370 flight was longer, of course, but it still hasn't been recovered.

While I'm sure it is technologically possible to record more voice data, that would primarily record innocuous and routine conversations among the pilots.

I suspect that the pilots, and their unions, are reluctant to have their mundane and irrelevant conversations recorded and inspected.


What does "composite panels on a frame" mean? Reading about composite materials in the 787 on Wikipedia, I came across this:

  Carbon fiber, unlike metal, does not visibly show cracks and fatigue, prompting concerns about the safety risks of widespread use of the material; the rival Airbus A350 uses composite panels on a frame, a more conventional approach, which its contractors regarded as less risky.

I bolded the curious part. What does "composite panels on a frame" mean exactly?

Traditionally, aircraft are built of frames (analogous to a person's skeleton), as shown below, and the smooth outer skin of the aircraft consists of panels fixed to these frames.  There are many frames and, for example, the small size of the windows is because they have to fit between the frames.

Image source: Wikipedia

The A350 uses this traditional construction, except the frames and panels are built of composite materials (carbon fibre reinforced plastic), rather than aluminium. However, the fuselage of the Boeing 787 seems to consist of a composite "tube", with reinforcing ribs running the length of the plane. There seem to be many fewer "ribs" running around the fuselage in this design than in traditional ones.

Image source: Wikipedia


How do pilots cope with pressurization changes? I have tried

drinking a glass of water
yawning
plugging my nostrils and pushing air

but none of these seem to work - I still have a distinct pain in my upper sinus cavities when changing pressure in an airplane. I have the same pain underwater.

Do some pilots have this problem, and do they have a remedy?

There is a chance you suffer from aerosinusitis but you should consult a physician to obtain a proper diagnosis (Disclaimer: I am NOT a doctor). For what its worth commercial planes are pressurized generally to about the pressure you would see at 8000 ft. this equates to about .75 Atm. In the reverse case 33ft underwater already has you at 2 Atm.

When I fly (generally unpressurized) I usually chew gum which i find helps.


How does engine geometry affect flight stability? News items around the recent aviation accidents involving Boing 737 MAX 8 aircraft suggest, that bigger engine diameter demanded for a different (longer) nacelle strut for proper ground clearance which in the end had a negative effect on flight characteristics, necessitating an additional control system MCAS. This reasoning may be correct but it is not self evident IMHO.

One would think that a different engine mount point primarily shifts center of gravity, which can be statically counteracted by shifting other masses. The attack point of thrust, which also generates angular momentum if not coincident with the center of aerodynamic drag, should not be significantly affected by this geometry change.

So how does this geometry change adversely affect margins of inherent static stability and impede return from increased angle of attack to equilibrium?

The problem appears to be mostly the effect of a larger cowling, moved more forward, that generates significant lift when operating at higher AOA and has an effect similar to moving the entire wing forward, or moving the center of gravity aft, partially unloading the horizontal tail in the flaps up low speed regime and making the pitch stability a bit neutralish.

Seaplanes suffer from a bit of a similar effect in yaw from adding floats. The floats add quite a lot of side area forward which reduces the fin's weathervaning tendency.  This is why you usually see additional fins added to the tails of float planes to restore the original yaw stability.

In the case of the MAX it appears the pitch destabilizing effect of moving the cowls forward really caught them by surprise because the problems were only found during flight test.  To restore the original relationship of the effective center of lift to CG would have required some pretty radical redesign, say like adding a plug to the forward fuse to move the operating CG forward, which would also require making the tail bigger... and off we go on the one-thing-effects-another design change merry-go-round without end.  What a mess.  They were really between a rock and a hard place, and an artificial stability system using the stab trim system to work around what was a fairly narrowly focused problem provided a relatively easy escape route.


Can gusts be reported as lower than wind speed? KADS 072350Z 31030G20KT 13SM SKC 24/02 A2980

I’ve never seen a negative gust factor before. Is that a real thing (and if so, what does it mean), or did someone just make a typo?

WMO standard defines a gust as the maximum wind speed exceeding the "mean speed" by 5 m/s (10 knots) during the 10-minute interval. With this definition, I would never expect to see a negative gust factor.
However, the FAA defines it differently (FMH-1):

5.4.3 Wind Speed.
The wind speed shall be determined by averaging the speed over a 2-minute period.
5.4.4 Wind Gust and Squall.
The wind speed data for the most recent 10 minutes shall be examined to evaluate the
occurrence of gusts or squalls. Gusts are indicated by rapid fluctuations in wind speed with a variation of 10 knots or more between peaks and lulls.

Thus, by my interpretation, a U.S. airport METAR can report gust speeds below the mean reported wind speed since it only requires a variation of 10 knots, not an exceedance of 10 knots.
However, reading further in the same document:

5.6 Summary of Wind Observing and Reporting Standards
Wind speed: 2-minute average speed in knots is reported.
Wind gust: The maximum instantaneous speed in knots in the past 10 minutes is reported.

It seems impossible to have a 2-minute average speed which is higher than the maximum instantaneous speed over the past 10 minutes. So following that, it would seem to indicate an error in the measuring or reporting equipment.


Is it legal to take off with an iced windscreen? Looking at this video on Youtube, I'm wondering if it is actually legal to take off with a windscreen iced like that?

For VFR this is a no go, but what about IFR and under what conditions?

When operating under Part 91, taking off with an iced windshield is not permitted. It does not differentiate between IFR and VFR operations.

  § 91.527 Operating in icing conditions.

  (a) No pilot may take off an airplane that has frost, ice, or snow
  adhering to any propeller, windshield, stabilizing or control surface;
  to a powerplant installation; or to an airspeed, altimeter, rate of
  climb, or flight attitude instrument system or wing, except that
  takeoffs may be made with frost under the wing in the area of the fuel
  tanks if authorized by the FAA.

Now in the case that the winshield is simply covered in condensation (not ice), I'm not so sure about that, though it'd be pretty stupid to takeoff without heating the windshield and pretty much zero visibility for no particular reason. In that particular video though, that's definitely ice.


Can asymmetric thrust occur for Single engine airplane at idle power during stall practice? As stated in the title: why can asymmetric thrust occur for Single-propeller-engine airplane at idle-power during stall practice?

The phenomenon you speak of is called P-factor.  It is the result of a non uniform angle of attack for the propeller blades throughout the propeller disk.

A typical piston powered single engine aircraft has a propeller which turns in a clockwise direction as viewed from the cockpit.  In this fashion the propeller blades are ascending to the left of the pilot and descending to the right of the pilot.  During straight and level cruise (high speed) flight where the airplane’s angle of attack (AoA) is minimal the propeller blades on the ascending and descending sides of the propeller disk have approximately the same angle of attack as they rotate causing the thrust force to be uniform over the entire area of the propeller disk.  But as the airplane enters a higher AoA either when maneuvering or maintaining altitude during slow flight, the angle of attack for the propeller blades on the descending side of the propeller disk have a greater angle of attack than do the blades on the ascending side of the disk.  This causes the blades on the descending side to generate more thrust for a given engine speed which results on an uneven thrust loading over the entire propeller disk, resulting in a counter clockwise torque about the airplane’s vertical axis.  P-factor is most pronounced at high engine power settings at the critical AoA for the airplane where the thrust asymmetry is most acute.

P-factor plays an important part during multiengine operations as well.  When an airplane had two or more wing mounted engines all turning in the same direction, there is always one engine such that, if it should fail, the yawing moment generated by the P-factor of the remaining running engine will be the greatest as compared to if the engine which is currently running had failed instead.  This engine is called the critical engine and creates the most difficulty in maintaining directional control should it fail.


What protection did the Concorde have against an uncontrolled decompression? The Concorde's cruising altitude was 60,000 ft. I believe if the cabin ruptured at this altitude, everyone onboard would be almost instantly killed, if not from the Armstrong limit then from hypoxia (as oxygen masks cannot supply sufficient oxygen at that altitude).

Therefore, there should have been extra protection against cabin decompression in the Concorde. This type of accident would have probably caused an immediate stop to the Concorde program.

What safeguards did the Concorde have against an uncontrolled decompression?

This is the main reason why Concorde had such small windows. It was calculated to reduce the rate of decompression to a survivable level in case one window fails. (The prototype, by the way, had much larger windows).

Other than that, I don't know of any other 'special' decompression mitigation measures that would be unique to Concorde.

There were, of course, things that had to be done in a very special manner on Concorde in case of decompression. For example, rapid emergency descent: there was a special program that transfered fuel forward during emergency descent. It could be activated by a single button, and the flight engineer had to keep the program up to date for the current flight conditions. (Generally, keeping the CG/fuel balance was pretty much a full time job for the Concorde flight engineer). But this would hardly help in case of 'uncontrolled' (explosive) decompression.


Why can’t electric motors be used in aviation Why can’t electric motors be used in aviation? Is it a problem with batteries, torque, or what? From what I’m aware motors at high torque are expensive and need a massive footprint, than a high rpm low torque motor. Is the issue the weight and size of a motor to lift 500,000 pounds off the ground? We have motors right now that can spin a turbo fan at a high speed and compute a comparable thrust, but is that thrust under load a different story? Is the energy required way too much for a realistic battery size?

Electric motors can be and are used in aviation - there's hundreds of them on any airliner you fly on.

As for using them for main propulsion, let's first look at the power/weight ratios. The highest number for Tesla's car motors is 8.5 kW/kg. The electric-aircraft-specific Emrax 268 delivers about 12 kW/kg.

In comparison, the Trent XWB delivers 430 kN of thrust at 300m/s flow rate, which equates to 64.5 MW of power, in a 7,550 kg package - a power/weight ratio of 8.5 kW/kg. However this isn't apples to oranges: this ratio is for the whole package, engine and fan, and measures useful output, like wheel power for a car.

In short, turbine engines are still lighter than electric motors, but the difference is not dramatic.
Where all-electric powerplants fail to stack up is range. I've elaborated on it in response to another question - Are there any hybrid electric planes?. The short version is that the maximum possible range of an electric aircraft is 10 nmi for every % of its weight dedicated to the battery. This limits the range of electric aircraft to 300-450 nmi, if sticking with the fuel fractions of known airliners.

But there are aviation applications where this is enough. The most important non-renewable resource consumed by modern aviation - the supply of 1960s Cessnas and Pipers, without which no one could afford to become a pilot - will not last forever. Should authorities permit it, mass-produced Tesla powerplants could power trainers and GA planes at a fraction of the ownership cost of a certified avgas engine.


Is it possible to do a barrel roll in a large aircraft like a B737? Can the bigger planes like B737 do barrel rolling kind of stunts like fighter jets do? I'm sure if it's full of passengers or cargo it would be very difficult and dangerous, but is it possible with an empty aircraft?

I remember an All Nippon Airways incident where the co-pilot did a 131 deg roll by mistake but I don't know if a complete barrel roll is actually possible in large planes?

Yes, it has been done before in an even larger aircraft, the Boeing 367-80, very famously by "Tex" Johnston on August 7, 1955. This aircraft was the prototype for the Model 707.

When a barrel roll maneuver is properly performed the aircraft remains in a positive-G state, so no "inverted flight" is experienced by the aircraft. According to the International Aerobatic Club,

  The Barrel roll is a combination between a loop and a roll. You complete one loop while completing one roll at the same time. The flight path during a barrel roll has the shape of a horizontal cork screw. Imagine a big barrel, with the airplanes wheels rolling along the inside of the barrel in a cork screw path. During a barrel roll, the pilot always experiences positive Gs. The maximum is about 2.5 to 3 G. The minimum about 0.5 G.