What was the flight route from New York to Paris that Air France took in 1955? I have a photograph of me and my sister in 1955 or 1956 boarding Air France to Paris. I seem to remember the route was:

New York - Gander
Gander - Shannon
Shannon - Orly

Is that correct? I know the return flight (westbound) was that route. Does anybody know how long it took?

The route was IdleWild (renamed JFK in 1963) or LaGuardia -> Gander, Canada -> Shannon, Ireland -> Paris. The trip took a total of about 18 hours going eastbound with tailwinds, and 24 hours going westbound with headwinds.

Gander was opened in 1938 and continues to be the preferred refuelling stop for aircraft that can't make the hop across the north Atlantic in one go, simply because it is the northeasternmost point on the North American continent where such a facility could be built. For a while it was the largest airport in the world, and was used extensively by the military during WW2 as a stopover point for aircraft being ferried to Europe.

Air France used the route from the beginning of their transatlantic service in 1946 until the late 1950's, when the jet age made it possible to fly the route non-stop.

Sometimes they used the southern route.
ORL to Azores to Gander to IDL.
Or ORL to Azores to Bermuda to IDL.

Sometimes instead of Gsnder they stopped at Goose Bay.

AND even with jets, they still had to stop for refueling on westbound flights. They often still do on certain aircraft flying the route.


Are modern aircraft (fully loaded and fueled) able to operate from hot and high airports? Some decades ago, aircraft had really difficulties to take off and land at high altitude airports in hot weather.

I'm not sure whether advanced FADEC and better mastering of aerodynamics might today allow a heavily loaded and fully fueled FedEx MD11 to take off from 5000+ feet elevation airport in hot weather.

I'm curious to know about how much progress there was over the last 20 years or so.

1. Thrust / FADEC
FADEC does not increase the thrust of jet engines. If anything it limits the thrust; the MD-11 has an emergency power overboost bar that disables the FADEC protections.
(YouTube) Instructor holding the aforementioned bar.

Whereas in crisis (for example, imminent terrain contact), a non-FADEC engine can produce significantly more than its rated thrust, a FADEC engine will always operate within its limits.
— Wikipedia

2. Lift
Aerodynamic advances sure help. But they can't make the air denser. Hot and high conditions still limit aircraft performance.
3. Tires
Tires are better now, but even with an extremely long runway, they can't withstand the speed needed to lift off fully fueled and loaded from a hot and high airport. (Higher speed is needed because of the thinner air and high takeoff weight.)
4. Number of engines
The more engines you have, the better performance you get, namely if an engine fails on takeoff, you'll have more power left on a quad-jet versus a twin-jet (3/4 vs. 1/2).

MD-11
In this answer, you see the MD-11F needs to stop for fuel mid-journey because it takes off from a hot and high airport.

The answer to the main question is no—the tires would burst at the required high takeoff roll speed given an adequate runway length and zero (hypothetical) climb limitations.

Boeing 787 (21 years later)
In this other answer, you see a not particularly high airport, but hot, limits the amount of passengers on the 787 headed from Vegas to Europe.

The first linked answer examines the physics behind hot and high, and the 787 answer examines a real life scenario with real numbers.

"High and hot" covers a wide range of conditions.  In addition, total takeoff distances are subject to a number of factors including, but not limited to takeoff weight, field elevation, type and condition of runway, surface winds, obstacle clearance, etc.

So an MD-11 with a low payload and a meaningful headwind can depart the same airport successfully that could not be done if it were loaded at or near MTOW with a tailwind.  You will have to be a little more specific.


Why is the L/D ratio numerically equal to the glide ratio? *L/D is ratio of two forces, lift and drag.
*Glide ratio is ratio of two distances.
Can you explain mathematically how they end up  numerically equal and are they always equal?

The sum of all work done on the airplane must be zero in a glide. That is, the work done by gravity must equal the work done by drag. Note also that lift equals the force of gravity in equilibrium
Work is force times distance. If gravity is X times larger than drag, this means that the distance in vertical direction must be X times smaller than in the horizontal direction.

First a caveat-- please keep in mind that this answer is focussed on steady-state linear gliding flight.  Throughout this answer, when we talk about what happens when we change the angle-of-attack and L/D ratio, we're assuming that we've waited long enough for the glider to settle back into equilibrium at whatever airspeed is appropriate for the new angle-of-attack and L/D ratio.
All three of the vector diagrams below represent the exact same physical situation-- sustained, steady-state gliding flight along a linear flight path in still air. We've represented the aerodynamic forces in black, and the gravitational force-- the weight vector-- in green. The left-hand and right-hand diagrams are easier to understand than the middle diagram, because the force vectors are arranged into a closed polygon (a triangle in this case), which shows us instantly that the net force is zero. On the right-hand diagram, we've also included the airspeed vector, and its horizontal and vertical components.

Lift is defined as acting at right angles to the relative wind, i.e. the undisturbed apparent airflow created by the aircraft's path of travel through the airmass.  I.e. lift is defined as acting at right angles to the flight path through the airmass.  Drag is defined as acting parallel to the relative wind, i.e. parallel to the aircraft's path of travel through the airmass.
Note that K is the glide angle, and note that on the right-hand diagram, the triangle formed by the force vectors and the triangle formed by the velocity vectors are similar triangles.  This is why the still-air glide ratio -- the ratio of horizontal speed to sink rate in still air -- is identical to the ratio of Lift to Drag.
The still-air best glide ratio -- i.e. the best ratio of horizontal distance travelled to altitude lost in still air -- must coincide with the best ratio of Lift to Drag.
Note that in a steady-state glide, Lift is less than Weight by the same proportion that horizontal speed is less than airspeed.  For many calculations, it is a close enough approximation to assume that Lift and Weight are the same, and that horizontal speed and airspeed are the same.  However, our vector triangle is very sensitive to this distinction -- overlooking it forces our vector triangle to collapse into a straight line, which forces the sink rate to be zero and gives us a glide angle of zero, i.e. a glide ratio of infinity.
Also note that the geometry of the triangle is such that the maximum Lift to Drag ratio also coincides with the minimum Drag value, and the maximum Lift value.
Note that the minimum Drag value does not coincide with the minimum value of the Drag coefficient.  The Drag coefficient is minimized at very low angles-of-attack, i.e. at very high airspeeds.  So we need to be careful about what we mean when we say we want to minimize Drag in order to maximize the glide ratio.  This is only true when we are speaking about Drag as an actual force, not as a coefficient.  The same is true for Lift.  While (for reasonably high L/D ratios) Lift varies much less than Drag as we change the L/D ratio, it is still true that in gliding flight, Lift is slightly higher at the point of max L/D ratio than at any other L/D ratio.  But the Lift coefficient is not maximized at the point of max L/D ratio-- rather, it is maximized near the stall angle-of-attack.  Yet in spite of all this, it is an arithmetic fact that when we maximize the L/D ratio, we also maximize the ratio of Lift coefficient / Drag coefficient -- even though we are not individually maximizing either the Lift coefficient or the Drag coefficient.
Warning: do not mistake the right-hand diagram for an indication that to minimize the sink rate, we simply need to minimize the drag vector.  The force vector triangle and the airspeed vector triangle are geometrically similar, but  they are not identical.  As airspeed changes, the entire airspeed vector triangle scales up or down in size, while the force vector triangle is locked in size by the fact that the Weight vector is constant.  If we miss this relationship, we may tend to assume that the minimum sink rate airspeed must coincide with the airspeed for minimum Drag, which is also the airspeed for the maximum L/D ratio.  This is not the case-- as we'll see below, the airspeed for the minimum sink rate is lower than the airspeed for minimum Drag, which is also the airspeed for the maximum L/D ratio.
Every possible angle-of-attack of the wing is associated with a specific lift coefficient, a specific drag coefficient, and a specific Lift / Drag ratio. If we ignore effects due to changes in Reynolds number associated with changes in airspeed, we can make the approximation that the same angle-of-attack of the wing will always be associated with the same lift coefficient, drag coefficient, and Lift / Drag ratio regardless of whether the aircraft is heavy or light-- a change in weight is expressed as a change in the airspeed associated with any given angle-of-attack, but no change in the lift coefficient, drag coefficient, or Lift / Drag ratio associated with any given angle-of-attack.  If we increase weight, any given L/D ratio-- including the maximum L/D ratio-- will now occur at a higher airspeed.  The increase in airspeed will be proportional to the square root of the increase wing loading.
If the still-air best glide ratio occurs at the same angle-of-attack that gives the best L/D ratio, then it clearly follows that we can minimize the sink rate by increasing the angle-of-attack to slow down slightly, even if this degrades the the L/D ratio and the glide ratio a bit.  In other words, if we fly more slowly down a slightly steeper glide path, we can minimize the sink rate.  Obviously there is a limit to this process-- as we continue to increase the angle-of-attack and degrade the L/D ratio, then at some point well before we reach a full-blown stall, we'll reach a point where the increased steepness of the glide path is too much to be offset by the decreased airspeed, and the sink rate will start to increase again.
In fact, the highest L/D ratio occurs at the angle-of-attack that gives the highest ratio of Cl/Cd, while the minimum sink rate occurs at the angle-of-attack that gives the highest ratio of (Cl^3) / (Cd^2), which is also the highest ratio of ((L^3) / (D^2)) / airspeed.  It's clear that when we want to minimize the sink rate, we gain more by maximizing the lift coefficient than by minimizing the drag coefficient, so we should choose an angle-of-attack that is somewhat higher than the angle-of-attack for the best L/D or Cl/Cd ratio.  This means that the airspeed for the minimum sink rate is always slower than the airspeed for the best glide ratio in still air.
Vector diagrams like the ones included in this answer are surprisingly rare in books about soaring flight.  Materials published by the FAA sometimes attempt to include such a diagram, but they are invariably badly botched, with the length of the various force vectors altered so that they could not possibly form a closed vector triangle, thus eliminating any possibility for the reader to understand the relationship between the force vectors and the glide ratio.  (One might optimistically hope that they are trying to make the point that a glider is not always flying in a straight line with no acceleration?)  (See for example figure 3-1 on page 3-2 of the FAA's "Glider Flying Handbook" (2013) -- here they've even included a Thrust vector!)  Even a theory-rich book like "New Soaring Pilot" by Welch, Welch, and Irving (3rd edition, 1977) is entirely lacking in any presentation of the Lift-Drag-Weight diagram for steady-state gliding flight.  However, see page 233 of that book for a "polar curve" clearly showing that the max L/D ratio occurs at a higher airspeed than the minimum sink rate.
A couple of additional notes--

Our vector triangles also show that the maximum value of (horizontal speed / sink rate) and the maximum value of (airspeed / sink rate) both occur exactly at the maximum L/D ratio, so we can also find the airspeed for maximum glide angle by looking at the plot of (airspeed / sink rate) versus airspeed and picking the airspeed where (airspeed / sink rate) is maximized or (sink rate / airspeed) is minimized.  Naturally, this will not be the airspeed for minimum sink rate.  The airspeed for minimum sink rate will clearly always be lower than the airspeed for minimum (sink rate / airspeed).

Occasionally one encounters a suggestion that the best L/D speed or the minimum sink rate speed has something to do with the point where Induced Drag is minimized.  The vector diagrams above demonstrate that as long as Weight is fixed, when we maximize L/D, we also minimize Drag.  So at the best glide-ratio airspeed, where the L/D ratio is maximized, Drag is minimized.  Since Induced Drag decreases with airspeed and Parasitic Drag increases with airspeed, it follows that at the point of max L/D -- i.e. at the point of minimum Drag -- half the drag must be Induced Drag and half the drag must be Parasitic Drag.  This is the only point where Drag can be at a minimum.  Clearly, the point of best L/D is cannot be tied to a minimum in Induced Drag or a maximum in Lift / Induced Drag.  The ratio of Lift / Induced Drag is highest at very low angles-of-attack, i.e. at very high airspeeds.

Can you explain mathematically how they end up numerically equal and
are they always equal?

The L/D ratio and the glide ratio are always equal in steady-state linear unpowered flight in still air, but they are not equal if there is some thrust from a motor.  They also are are not equal if there is horizontal or vertical wind.  A glider improves its glide ratio relative to the earth by speeding up if there is a headwind, or a downdraft.  A glider improves its glide ratio relative to the earth by slowing down to near the minimum sink rate speed if there is a tailwind, or a weak updraft (not strong enough to permit actual climbing).  The exact speed to fly to optimize the glide ratio relative to the earth in any combination of headwind or tailwind and lift or sink can easily be graphically determined from a graph of sink rate versus horizontal speed, which for all practical purposes can be approximated by a graph of sink rate versus airspeed.  See for examples figures 5-13 and 5-14 on page 5-10 of the FAA's "Glider Flying Handbook" (2013).
For more on L/D ratios and gliding flight and powered flight, see this related A.S.E. answer, and other linked answers-- 'Gravitational' power vs. engine power


Could someone really die in the way that Goose did in the movie "Top Gun"? Looking at this question I remembered a question I always wanted to ask:

In the movie "Top Gun" one of the main characters dies when trying to eject from the F-14. Here is the scene where this happens. Did an incident like this ever occur in real life, or is there at least a theoretical possibility that a pilot could be killed this way?

Top Gun was filmed in the mid 80s, which is quite a while ago. If there were incidents, did they do something to fix that issue? And this question goes not only for F-14s but also other aircraft with ejection seats.

Some sources I found claiming the event was real:

  A flat spin creates a low-pressure area and stalls the canopy when
  ejected. Producers wanted a midair crash but based this accident on a
  real-life incident instead.

(source)

  The way Goose died was actually based on a real life incident that
  happened to a 14 crew in the Navy. Don't remember the specifics, just
  found this out recently while reading an article in one of last months
  flying mags (not Flying)

(source)

  I have a photo of the 'real' Goose and it's true that he died of seat
  ejections but not exactly like the movie.

(source, with pictures)

There's also an interview of Steven C. Schallhorn, who apparently was one of top gun instructors interviewed for the movie:

  Dr. Schallhorn said during his interview for the movie, he explained
  how a flat spin could actually happen with an F-14 and that it
  complicates ejection. "The aerodynamics of the F-14 flat spin affect
  the timing of the ejection sequence," Dr. Schallhorn said. "The canopy
  is jettisoned, followed by the ejection of the back seat, followed by
  the front seat. In a flat spin, the canopy, when it ejects, bobbles
  for an extra few hundredths of a second above the aircraft. That
  upsets the carefully engineered sequence because the guy in the back
  could then hit the canopy. That sequence made it into the movie."

(source)

Interesting discussion. But one thing that isn't addressed is the time elapsed between the canopy jettison and the first seat to leave the plane.
This is relevant because of the flat spin.
An f-14 is 63' long. If you put the center axis of the spin roughly in the middle that means the pilots are probably half of that distance to the nose of the aircraft, so say 15 feet from the axis.
If the nose of the aircraft makes a full 360 degree rotation every second, that would mean the pilots are traveling around their circumference of the circle at nearly 100' per second. (The nose cone would be traveling at about 200'/s)
That means that if the canopy fires, and a .5 second time elapses before the first seat goes, the pilots should be half a rotation away (or ~50')  from the canopy.
But of course there is the problem of the momentum of the canopy and air resistance.
The momentum of the plane would have been accelerating the canopy at the same rate while the two were attached. However, once the canopy was jettisoned (and we assume the canopy is jettisoned in a perpendicular vector to the aircraft in that moment), the canopy should immediately begin to slow its rate of rotation as compared to the rest of the system.
Turbulent air above the fuselage considered, the canopy should have been slowing it's rate of rotation, even if only slightly. This would create "z-axis" distance between the pilots and the canopy.
Finally, there is the variable of longditudinal acceleration. The same reason why maverick couldn't reach the ejection handles above his head. The rotation would have been creating an "outward" force on the canopy while it was spinning with the aircraft. So when the canopy was jettisoned, that momentum would have carried through and pushed the canopy off the nose of the aircraft. I know people have mentioned that there was a "backward" jettison path designed into the f-14 which theoretically could have exactly cancelled out the forward momentum of the canopy but this would be a very big coincidence.
Anyway, my theoretical conclusion would be - assuming there was some kind of time elapse between canopy and first seat - is that by the time the first seat fired, the canopy should be "lagging" behind the rotation of the cockpit, as well as having been moving forward off the nose of the aircraft, leaving the pilots to clear the aircraft safely.
But that wouldn't make as good movie. :D


What visibility and cloud clearances are required for airplanes conducting visual approaches? What visibility and cloud clearances are required for airplanes conducting visual approaches during IFR while flying under FAA regulations? Is the visual approach a purely VFR maneuver requiring the usual VFR weather minimums for the airspace being flown? Or does the visual approach maneuver flown under IFR supersede normal VFR weather minimums?

This depends:

Under part 91, you just have to remain clear of clouds and have a minimum of a 1,000 ft. ceiling and 3 SM visibility.

Commercial operations (121 and 135) are further restricted by their Operations Specifications and have to maintain VFR weather minimums appropriate for the airspace that they are in (among other things, see below).

Here is an excerpt from the AIM (there is more details there for different operations, with the differences between controlled and uncontrolled fields, etc.)

  5−4−23(a). Visual Approach

  a. A visual approach is conducted on an IFR flight plan and authorizes a pilot to proceed visually and clear of clouds to the
  airport. The pilot must have either the airport or the preceding
  identified aircraft in sight. This approach must be authorized and
  controlled by the appropriate air traffic control facility. Reported
  weather at the airport must have a ceiling at or above 1,000 feet and
  visibility 3 miles or greater. ATC may authorize this type approach
  when it will be operationally beneficial. Visual approaches are an IFR
  procedure conducted under IFR in visual meteorological conditions.
  Cloud clearance requirements of 14 CFR Section 91.155 are not
  applicable, unless required by operation specification.

The Part 135 Ops Spec for this contains the following requirements for a visual approach:

  b. Terminal arrival IFR - Visual approach or a Charted Visual Flight
  Procedure (CVFP).
  The flightcrew may accept a visual approach or a
  CVFP provided all the following conditions exist. The flightcrew may
  not accept a visual approach or a CVFP unless the limitations and
  provisions of subparagraph f. of this operations specification are
  met.

  (1) The flight is operated and remains in Class B, C, or D
  airspace, within 35 miles of the destination airport in Class E
  airspace, or the airspace beneath the designated transition area.

  (2)
  The flight is under the control of an Air Traffic Control (ATC)
  facility.

  (3) The flightcrew must maintain the basic cloud clearance
  as specified in Section 91.155.

  (4) For a visual approach without a
  CVFP - The flightcrew must be able to establish and maintain visual
  contact with the airport or maintain visual contact with the traffic
  to be followed as directed by ATC. In addition, the following
  provisions and weather conditions at the airport during the approach
  must be met:

  (a) Reported visibility must be as specified in Section
  91.155, but not lower than a visibility of three miles and reported ceiling must be 1,000 feet or greater, or

  (b) When in the terminal
  area with the reported visibility not lower than three miles and
  ceiling not reported, the flightcrew may continue to a landing if the
  runway of intended landing is in sight and the flightcrew can maintain
  visual contact with the runway throughout the approach and landing,
  and

  (c) Ceiling and cloud clearance must be as such to allow the
  flightcrew to maintain the minimum altitudes prescribed in Section
  91.129, 91.130, or 91.131, as applicable for the airspace class in which the flight is operated.

  (5) For a CVFP - The flightcrew must be
  able to establish and maintain visual contact with the airport or the
  charted visual landmark(s) for the CVFP throughout the approach and
  landing. In addition, the weather conditions at the airport at the
  time of the approach must be reported to be at or above the weather
  minima established for the CVFP, but never lower than the VFR landing
  weather minima stated in Section 135.205 in uncontrolled airspace.

  ...

  f. Special Limitations and Provisions for Visual Flight Rules. All VFR operations authorized by this operations specification shall be
  conducted in accordance with the following limitations and provisions.

  (1) The certificate holder must identify obstacles and use airport
  obstacle data which ensures that the performance requirements of Part
  135 are met.

  (2) The weather conditions must allow the flightcrew sufficient
  visibility to identify and avoid obstacles and safely maneuver using
  external visual references and to maintain minimum altitudes.

What visibility and cloud clearances are required?

VFR, as per 91.129, 91.130, or 91.131 for the various classes of airspace


Why can't an Airbus A330 dump fuel in an emergency? It was reported recently that the Swiss flight LX18 from Zurich to New York had detected an engine issue shortly after take-off and decided to land back in Zurich.

Apparently, the Airbus A330-343 cannot just dump the fuel and, fully tanked, was too heavy to land safely, so it had to fly for 5 hours in a loop just to burn the excess kerosene.

Would it have been too complicated to implement a fuel draining mechanism?
What would have happened if they couldn't fly for 5 hours and had to emergency land right away? Wouldn't this inability to drop fuel be a safety hazard?

To supplement Jimmy's answer, if they had to land right away, they could have; it just would've resulted in an overweight landing being recorded, and which on most airliners triggers a special inspection of the landing gear and its attaching structure, and if nothing is permanently bent or cracked or broken, you are good to go.

An overweight landing in itself shouldn't result in damage unless the landing was hard, the overweight condition using up some of the structural safety margin, and if you know you are landing overweight you will take extra care to land smoothly.

Depending on the airplane, an overweight landing may apply a "penalty" so to speak, accelerating some inspection interval, or advancing a life limit on a structural component in the gear, or some other negative impact maintenance cost wise.  So there may be a significant incentive to avoid it from a long term cost perspective even if it means wasting 5 hours of fuel going round and round in a hold.

In the case mentioned, the crew would have consulted with the airline's Maintenance Control organization on the private company radio frequency to discuss whether to land overweight or not, and if there wasn't an urgent need to land, the Capt would have agreed with Maint Control and burn off the fuel to avoid the overweight landing and its ramifications.

Also consider the location. Assuming the a/c can do so, there would be no issue (apart from pollution) in dumping fuel over sea. But you want to spray jet-A all over Zurich? Plenty of ignition sources on the ground. Really not good.


Why have cargo hybrid airships not succeeded in the market? Using a Lighter Than Air (LTA) Hybrid Airship to efficiently transport cargo to remote areas seems like an idea that should flourish.  It allows access to remote places in the world that have no airports, harbors, or roads leading to them by providing a point-to-point delivery system with vertical take-off and landing, all while providing a greener solution.  I see at least three players all trying to enter this market which seems perpetually stuck as coming about in the next 4-5 years and all seem to have benefited from the U.S. Army's Long Endurance Multi-intelligence Vehicle (LEMV) project that was canceled in January 2013.

Hybrid Air Vehicles in the U.K. has partnered with Northrop Grumman owns a prototype that came out of the LEMV project and plans to build a 50-ton capacity airship in about five years.
Aeroscraft in the U.S. has a prototype and plans to build an airship that carries a 66-ton load as shown on the History channel.
Lockheed Martin has a P-791 / SkyTug prototype derived from the LEMV program and hopes to build a SkyFreighter to lift a 70-ton cargo.

It appears none of these companies has any current orders and are all looking for funding to build their proposed cargo airships.  Apparently an investment of on the order of 50 to 250 million USD would allow one of these to be built for commercial use.  While this is a lot of money, it is small compared to other aviation investments. For example, the Boeing 787 Dreamliner required 32 billion USD as of 2011.

Today, all such airships use noble gas helium instead of the very flammable hydrogen used in the Hindenburg accident, so the historical fear of using such ships should not apply to the new technologies.  These proposed ships are planned to travel at between 70 and 115 MPH, so while not as fast as airplanes, they do compete with trucking and shipping speeds.

So the question is, given the relatively small cost to develop one of these cargo lift systems, why hasn't a company like Boeing with deep pockets wanted to invest in such systems and these companies continue to search for funding?  Is there some kind of flaw in the proposal that such a new cargo shipping approach would save money?  Is it that the market is too small to interest big companies? The hangar space required to store and maintain these ships is huge and little such infrastructure currently exists, perhaps that is their downfall?  Alternatively, maybe their maintenance requirements are their Achilles heel?

First of all, there is the question of demand. There aren't enough cases where a very heavy load needs to be moved to a remote location by air. There are probably even fewer where there are funds to actually pay for this option. The An-225 is currently the largest aircraft in operation, and it was only constructed under special circumstances. Although has certainly found use, they still haven't even found the funds to finish a second one which is partially built.

When this option is needed today, helicopters are deployed on cargo aircraft to the area. The helicopters may have to make multiple trips but can do the job. So the subset of cases is even smaller; where heavy cargo needs to be delivered to a remote area and also cannot be split into shipments that helicopters can handle. Deploying a large airship in this manner is also much harder. Either the airship would have to fly itself there, which could take a long time, or it would have to be reduced to a transportable size.

The airship would also need to be protected from the weather somehow. This could require either a very large (and expensive) hangar. Without a hangar, the airship must be tethered, and restraining such a large surface area in the wind would be difficult.

Wind would also affect operation of the airship. Helicopters are of course limited by weather conditions, but wind would have a much greater effect on a large and slow airship.

If the airship could be fairly easily collapsible, it would solve the issues of storage and transportation. However, this is not a simple problem to solve. This would require a large development budget. If you are comparing this to the cost of the 787 development, where are you going to find buyers for 1000 of these airships in order to break even and make them affordable?

The issue is that there needs to be a market for this kind of transportation, which currently competes at least partially with existing helicopters. Why invest in a new type of vehicle when the current option works fine for most cases?

Well actually hybrid airships kind of defeat the main advantage of airships (Does not require fuel to have lift) While using surface area for lift occasionally isn't a bad idea, when 40 percent of the lift is in surface area and 60 percent is the helium it makes it slow and inefficient and it still requires a runway. Lockheed's largest airship has a top speed of 60, while the Hindenburg, which was created using 90 year old technology, had a top speed of 80.
but it is actually be more cost effective to send 8 airships across the pacific multiple times carrying cargo than it would be to send a single enormous, slow innefficient cargo ship across the Atlantic. I did the math a while ago so I do t remember the numbers, but it could be something like 2.3 million dollars more cost effective and take about one month less time to transfer the same amount of cargo a regular sized cargo ship against 8 cargo airships with 200 tons carrying capacity flying at 150 mph. This isn't taking airship manufacturing Vs. cargo ship manufacturing into account, this is simply fuel based. This doesn't count personell either.     If I were to take manufacturing into account, It would be even more in favor of the airships because it is quicker to produce a hydrogen fueled airship and costs a WHOOOOOOLE lot less.


Is it possible to upgrade a Boeing 747-8 and Airbus A380's engines for greater speed? Is it possible to upgrade the Boeing 747-8 and Airbus A380's engines (upgrading to a newer engines) for greater speed, if the engines were more efficient and used less fuel at higher speeds relative to distance?

If so, what are the maximum speeds that the airframes or entire aircraft could handle or are designed to handle?

I just wanted to comment on this question since some of my code is on the A380's GE engine.

Those engines are optimized for fuel efficiency at their cruising altitude and speed.  The tradeoff that you fight against when designing the engine for cruising altitude is that it makes the engine hard to start, and the stall margin (the pressure and fuel combo you need to keep the engine from stalling) is razor thin when they are on the ground.

Just to echo what Kevin said, any improvements to allowing the engine to start without stalling would surely go to the fuel efficiency, not to speed.

The answers here are BONKERS!

The sound barrier is the limit for speed on 747s and 380s.  Therefore stronger engines will not make top speed higher.  Rest in pieces.

For the more common case of cruising speed, new engines might make the airplane cruise faster for a given cost (eg fuel cost), but that would be a hard sell.

ASIDE: I have heard that 747s have the highest cruising speed.


What is ground effect? We've all heard the "acts like a cushion of air" explanation tossed casually around by CFIs.

There's also plenty of books and reference materials that give a detailed, accurate, and complete explanation of ground effect, and which are incomprehensible to most student pilots.

What's the simplest way to accurately explain ground effect?

It's impossible to give a correct AND simple explanation I think. I would explain it like this if I was asked:

The downwash from the wings and the downward component of the wingtip vortices create a higher than normal pressure area below the wings because the air hits the ground and can't "escape", increasing lift.

Induced drag is reduced because wingtip vortices (causing drag) can't fully develop close to the ground.

Near the ground, the ground can push up on the airplane, giving it more lift.


What is a take off roll? Reading about 737-900ER take off roll (3000 meters) from another question, I would like to know what this means.

Is it intended that the aircraft needs 3000 meters before it can pitch up to take off or at 3000 meters is has acquired sufficient height and should be airborne?

The take-off roll, also called ground roll, is the roll distance before the aircraft becomes airborne.

In case of the 737-900ER, it will be in the order of 2000 meter.

The quoted 3000 meter is probably the take-off distance required (TODR) or take-off run required (TORR) at maximum take off weight. That assumes a full throttle acceleration to about half of the runway (1500 m), then an engine failure, continued acceleration on a single engine, rotation and take-off at 3000m. In case of TODR, the climb to a height of 35 ft is included as well (for FAR Part 23 aircraft it is 50 ft).

This means the 737-900ER requires a runway of 3000 meter length to take-off with an engine failure or to stop when the failure happens before the decision speed (V1). Since engine failures are unpredictable this is the minimum runway length required for take-off planning.

This all assumes ICAO standard atmosphere conditions. For runways at higher altitude ( > MSL) / higher temperature (> 15 deg C) the TORR will be even higher.

737's typically take off in 3000 to 4000 feet depending on cargo and configuration.

The TORR and other statistics are runway length guidance figures that are much longer than the actual distance the aircraft rolls before it leaves the ground.


Can hydrogen gas be an alternative fuel source for a gas turbine engine? I want to do research for an alternative fuel source for a commercial aircraft that will reduce the cost for flights.

Will the hydrogen gas be ignited in the combustion chamber and produce enough thrust to power the aircraft for flying? As for the problem that people had stated before for storing the gasses in a high pressure gas tank, can an increase in the aircraft weight be reduced by replacing the tank with usage of a hydrogen generator instead of a whole bulky tank that will weigh more?

By using a generator which converts water into hydrogen gas, we can just use water as our main fuel source which is more cost efficient?

In a flame of pure hydrogen gas, burning in air, the hydrogen (H2) reacts with oxygen (O2) to form water (H2O) and releases energy. If carried out in atmospheric air instead of pure oxygen, as is usually the case, hydrogen combustion may yield small amounts of nitrogen oxides, along with the water vapor.

Yes, and it has been demonstrated 30 years ago on the Tupolev 155. This is/was a hydrogen-powered version of the Russian Tu-154B tri-jet. Only one has been built and has since been retired after demonstrating the use of liquid hydrogen in 5 experimental flights. In total, the Tu-155 performed about 100 flights with several fuels, among them hydrogen and liquefied natural gas.

Tu-155 (top), cut-out (center) and layout (bottom). Source: https://www.aviaru.net/pr/?id=11633

Hydrogen is actually better for a jet engine: It is gaseous at normal temperature, so there is no delay for the evaporation step to happen as there is for liquid fuels before mixing and combustion can start. Also, hydrogen burns in a wide variety of mixing ratios with oxygen, so flame-outs are much less likely, making a smaller combustion chamber possible.

The downsides, of course, are storage and the small molecular size of hydrogen. It is very hard to contain, and it needs big volumes to store a given amount of chemical energy. Pressurized storage at 200 bars holds only 18 kg/m³ or 45 times less than kerosene. With 142 MJ/kg, hydrogen holds three times as much chemical energy than kerosene, but then the volumetric efficiency of kerosene is still a factor of 13.3 better.

Cryogenic storage swaps pressure for low temperature: Below 33 K and above 13 bar, hydrogen becomes a liquid and storage density increases to 30 kg/m³. Still, cryogenic hydrogen storage needs 4 times the volume of the same amount of energy stored as kerosene, plus the isolation and the energy to cool it down and compress it.

I was wondering the same thing. Here’s my idea. Before the flight, use electrolysis to get a lot of hydrogen from water. Then compress the hydrogen into a smaller container because its a gas so you can do that. Use that as your fuel for the engine instead of water, and use the oxygen in the air to combine with the hydrogen instead.

So you would have your engine suck in air, combine it with hydrogen, use a bit of electricity to light it with a spark or something, then shoot it out the back of the engine.

Also I’ve never done anything with jet engines so I have no idea if this would work.


Why do most aerobatic aircraft have conventional landing gear? I was browsing images of aerobatic aircraft on Google and noticed that the vast majority of them have conventional "tail-dragger" landing gear.

Is there any practical reason why this is the case, or rather an aesthetic preference?

The tailwheel is much lighter and smaller than a nose wheel. That means reducing airplane overall weight and drag (because the landing gear is usually fixed). And it's probably cheaper, too.

Fighter jets and related trainers outnumber (and outperform) light aerobatic aircraft, they all use tricycle gear.

Most aircraft were tail draggers until after WWII, so it's really an issue of the aircraft design dates to a period when everything had conventional gear. There are no particular advantages to it, but an airframe cannot be converted from one to the other (the balance would be off) so if you want a classic aerobat you also get the classic landing gear.


How changes in Air masses, Air flow direction and intensity affect lift? I have a question regarding the airflow an aircraft experiences during 3 distinct scenarios:

An aircraft during landing, experiencing a sudden gust of 40 knots from the tail. What would be the immediate consequance- Airspeed, lift?
All this if the aircraft does not correct the situation with pitch/ engine thrust.
An airliner cruising at 270 IAS, suddenly experiencing into an area of tailwind- meaning- the tailwind component of the wind aloft changed and increased by 40 knots. How would the airspeed look- just the following moment, and from there- lift? Regard to the situation without any compensating action by the pilot,
A fighter jet, at 20,000 feet, turning from at 25 degrees per second, meaning doing 90 degrees in under 4 seconds, from a situation of 50 knots headwind to no headwind. If the air mass is stable and blowing at a constant 50 knots, would there be any change in the characteristics of flight to the jet, meaning deteriorating airspeed which he/she would have to compensate for using pitch/thrust adjustment? In this situation, we assume that the pilot doesn't care for maintaining ground speed or track, just keeping the turn at a constant rate of 25 degrees per second.

My question is (and we discussed it in the previous question)- is there any diminishing effect to the fast change of airflow that was just ahead and changed to sideways/behind- when the change is caused by the aircraft itself, just as witnessed (by me recently) in an airliner when suddenly encountering a different air mass (I witnessed increasing Ground speed but decreasing IAS, until the Autothroottles commanded a slight increase in thurst).

For sake of simplicity, lets state that lift is produced by airflow over and under the wings (the rest of the airframe also contributes, but in the scope of this question we can neglect that).

Lift for any given airfoil is basically dependent on airspeed and angle of attack. Up to a certain limit any increase in airspeed or angle of attack will amount to increasing lift and vice versa. So, the question:

Case 1. and 2. The situation is the same. Sudden tailwind of 40kts will lead to a sudden 40kt drop of airspeed and sudden decrease of lift. The groundspeed will start to increase (slowly, because of inertia of the plane). If immediate corrective action is not taken, some altitude will be lost. Near the ground this would be catastrophic.

Case 3. I'm a assuming a level turn here: During or after the turn the wind will not have any effect on the airspeed or lift of the aircraft.

The aircraft is flying embedded in an airmass. If we are considering aerodynamics (lift and stuff), all changes made in the flightpath of the plane are made in relation to the airmass. Not the wind, not the ground. The wind is simply the movement of the airmass in relation to the ground.

If the wind does not change, nothing changes for the plane (overly simplified, of course, there's temp, pressure etc, but that's not on subject here). In cases 1 and 2, the wind changes, as the velocity is momentarily different (gust). In case 3 there is no change in the wind.

Your example of the event you've witnessed, with increasing groundspeed and decreasing airspeed is a simple case of flying into a region of tailwind. Tailwind would simultaneously reduce your airspeed and push you to go faster in relation to the ground, increasing your groundspeed. This event might happen as a result of flying somewhat perpendicularily through top or bottom part of a jetstream.

Since this subject has proven to be somewhat difficult to comprehend (here and in this question too), let us end this answer in a humouresque article on Flying Magazine website, that might help understand the misconception of turning in wind: The Last Word on Downwind Turns, Really.

As a sidenote, in case 3 you have described a rather steep turn without increase in engine power, the aircraft will bleed energy during the turn resulting in loss of airspeed, if airspeed is not maintained with power or glide. I'm also pretty sure that a constant 25°/second rate of turn is just too much to sustain for most fighters, even with full afterburners, and it is definitely too much for any pilot to handle for long. F-16 has a max rate of turn of about 27°/sec at about 380kts and 9g's.

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

Whooa that was interesting!


Calculate g's per radian I'm currently attempting to assign handling qualities to the A-4D, and I would like to know how to find the term:

$ n / \alpha = \overline{q} \times C_{L\alpha} / (W / S) $

where $C_{L\alpha}$ is the lift-curve slope. I think I could solve for the lift-curve slope by using the value $1/T_{\theta2}$ or $1/T_{P2}$, the phugoid frequency, somehow, but I cannot find that equation.

I'm not given a chart of the change in lift-curve slope per Mach number.

The solution that I needed is the following image:


Why are winglets swept back and not front? A winglet is defined by its airfoil shape, profile, sweep angle and height. In case of the sweep angle, why does it always mean that the winglet is swept back. What is the disadvantage of having a front sweep angle

A winglet makes lift like a wing. That's why it's called a wing...let.  It's basically a sail (originally they were called "tip sails") designed to exploit the otherwise wasted circulation of air from bottom to top around the tip, placed so that the local flow curling around the tip strikes it at a positive angle of attack.  It makes lift/thrust and, being a wing, creates a downwash while doing so, which means some of the circulation is being redirected back the other way (an "outwash" in other words).

So the overall effect is the vortex field is weakened, and some its energy is used to make lift and/or thrust.  If the winglet is straight up, its lift component is straight inboard and forward, and is purely thrust and a (wasted) bending moment on the winglet root (which leads to angled winglets).

If it's angled, its lift component is a combination of vertical lift that contributes to holding up the plane, and thrust.   So the dihedral angle is a function of how much the designer wanted the winglet's lift force to be extra lift and how much was to be thrust, the resulting angle being a compromise sweet spot between inboard lift and vertical lift in that particular wing's tip flow pattern (nobody seems to use straight up winglets any more).

The sweep angle is for the same reason the main wing is swept, to reduce high Mach# drag (remember it's not just a fin "blocking" airflow;  it's a wing making lift like the main wing), and also to place the winglet in the optimal location in the circulation for exploiting the energy in the flow.

Simply because it's easier to pull something bendy than to push it.

When you load a forward swept aerofoil, it will bend in a direction that increases the load, so it has to be stronger (and heavier) than a rearward swept aerofoil.

The technical term is 'aeroelasticity'.


How heavy is too heavy to land on a grass strip? This is actually a tricky question, because it's not just about gross weight, but about how much weight each individual tire imparts into the ground.  This, of course, depends on the design of the suspension and the number of wheels.  Thus a really big craft, with a well designed undercarriage, can land on grass just fine (like a C-130 for example).

The point being, "how heavy is too heavy" has more to do with specific weight supported by each wheel during landing, and less to do with total landing weight.

So, kind of a two parter.  Firstly, how much weight can a grass field generally support when an aircraft is landing?  Secondly, how do aircraft designers make landing gear that can land on a grass field without plowing into the ground and ripping off?

Your second question is really easy to answer:
Q: How do aircraft designers make landing gear that can land on a grass field without plowing into the ground and ripping off?
A: Big Fat Low-Pressure Tires.

A wide, low-pressure tire has a larger contact patch with the ground, spreading the aircraft's weight over that larger area helps prevent the tire from digging into the ground, and lets the aircraft roll more easily on rough or soft surfaces.

Your first question - How much weight can a grass strip support? is a much tougher one.  The bearing strength of an unimproved surface depends on a number of factors (some big ones: The composition of the runway bed, its drainage, and what kind of surface cover - turf, rocks, etc. - it has).

If you look at aerial photos of Edwards Air Force Base you'll notice part of the runway complex includes a dry lake bed:

Some pretty heavy stuff has taken off from and landed on those runways - space shuttles have landed there.
Similarly TACA Airlines Flight 110, a Boeing 737, landed on a levee at NASA's Michoud Assembly Facility (and was subsequently repaired and took off again, though they used an access road for the takeoff rather than the grass levee).

Conversely if you tried to land a Cub on a grass strip that's saturated with water after a rainy month there's a chance your tires might sink into the mud.

CBR Technologies has a nice presentation on unpaved runways, which includes some discussion of how their bearing strength (and thus maximum landing weights, tire pressure limits, etc.) can be calculated, but each unimproved runway will be different, and the values will likely change over time.

Any aircraft can land on a grass field. It is actually easier and safer to land on a grass field than a hard surface. The reason that hard surface runways are used is because a dirt runway will develop ruts and require frequent resurfacing, but a hard surface can be landed on repeatedly with no change. Also, if you land on dirt it kicks up a lot of dust and possibly rocks which are bad.
Landing gear will not "rip off" an aircraft unless it hits an immovable object, like a brick wall. The landing gear is the toughest part of a plane, so if you hit a brick wall what happens is that the plane rips off the landing gear, not the other way around. I know of small planes that have hit steel highway dividers and come to a dead stop. You have to have a very severe sort of accident to pull apart an aircraft from the landing gear. Normally wings and stuff will come off way before the landing gear does. The landing gear is heavily integrated with the fuselage, so they basically act like a unit.
Dirt is very good for landing because the aircraft comes to a stop more quickly. With concrete you just keep rolling and rolling, which is much more dangerous. If I had to land off field, I would look for the muddiest place I could find. Shallow water is excellent too for the same reason: it slows you down quickly.

To respond to some of the asphalt-loving comments... In any off field landing you should ALWAYS land on dirt, grass or mud (even better). Landing on streets and highways is incredibly dangerous because there are cars and people on roads and other dangerous obstacles such as railings, signposts, fire hydrants, traffic lights and highway lights. Also, in any settled areas there will be wires and steel cables strung across the roads which are invisible from the air and can cut a plane in half.
People who advise landing off field on asphalt are just power pilots who don't have off field experience, so they don't know what they are talking about. As any glider pilot or bush pilot will tell you grass is way way safer than trying to land on a road. Even at an airport, it is much safer and easier to land on the grass strip at the airport than the runway.


How can flying high minimize drag and increase fuel efficiency? As a plane goes higher, the engines have to work harder to compensate for the air density, therefore it will require more fuel in order to provide the same power at lower altitudes.

But I always hear that flying high means less fuel burnt.

How can that be the case?

Flying higher means less drag because the air is thinner; therefore, you can fly faster at altitude and hence travel farther on less fuel.

However, flying higher also means less oxygen available to burn your fuel, so available horsepower decreases with altitude.

There is a special altitude at which these two effects (drag reduction and available power reduction) achieve a crossover point. This is called the critical altitude and it is there where you achieve optimum cruise and best economy.

If you want to fly higher than critical altitude, you will fly slower because the power loss is greater than the drag reduction and you will spend more time in the air.

This is a good question but first we must clarify something.

There is a major difference between propeller driven aircraft and pass through turbine propelled aircraft. Without getting to technical in this explanation, let’s leave propeller driven aircraft alone for the moment and keep our answer to the confines of high bypass ratio turbine driven engines.

The long definitive answer involves a great deal of physics, another subject we will not get to rhetorical about at this time. We will however consider Bernoulli’s principles concerning fluids and the airfoil or wing. This brings us to our first thought; consider the atmosphere, or air, around us as a fluid. Now that was simple enough.

When an airplane takes flight it must defeat its greatest foe, Newton’s contribution to this question, gravity. In its most simple terms, the ability to defeat gravity requires an opposing force that is greater than the weight of the mass being lifted. In aviation this comes from the lift created, of course, by the wings.

To help make this possible on airplanes the wings are shaped with a leading edge, that is the part that first hits the air, then a highly rounded upper surface, known as camber, and much flatter lower surface which come together at the trailing edge. (Just a moment more, I’m getting to your answer.) Now we apply Bernoulli’s principal.

Looking at air as a fluid, in principle Bernoulli’s theory is this. The air traveling over the top of the wing is moving at a faster speed than the air traveling under the wing. In theory, this would cause greater pressure under the wing (lift) and less pressure over the wing as the air is rushing from the leading edge to the trailing edge. The reasoning here is that when the air molecules are separated by the leading edge of the wing, the molecules passing over the top, which has a longer surface due to the added camber, must move faster in order to reach the trailing edge at the same moment as the air passing below the wing. Now we get to the part concerning thicker (heavier) air mass and thinner (lighter) air mass and the effects on “lift”.

In order to for the wings to produce the lift needed to defeat gravity you must have thrust and lots of it! Now we enter into the power (engines) production that is required to create flight. As we all know, an airplane must be moving into the air mass in order to obtain the effects required to produce flight also known as the lighter than air effect. So in principle, the shortened version, the air passing over and under the wings must produce a lift equivalency greater than the weight of the aircraft. There is a formula but we’ll skip that here and just state this, the airmass flowing around the wings must be equal to or greater than the mass of the airplane. There, we have flight... well sort of.

You see the weight of air at sea level is much greater than the air at say 30,000 feet. Here again, in principle, the amount of air flowing across the wing on takeoff, by mass, is more at mean sea level than it would be at 5,000 feet. Since this is a principle of mass verses mass (in this case in order to defeat the effects of gravity), the fluid principles of Bernoulli’s would apply well here. So what are the effects on engine performance?

Modern power plants today are engineered to direct less airflow through the engine or turbine section of the power plant. The ratios have changed dramatically in the last decade with most engines diverting 80% or more of the air flow around the core and exiting as additional thrust at the tail. These engines, as one could imagine, consume extreme amounts of air as it flows through and around the core, or power unit, and exits the tail as super heated energy and thrust. To maintain this thrust the amount of air mass must remain fairly constant. So, how is this accomplished at 38,000 feet? And what about the thinner air passing over the airfoils?

Well, here is where physics and the dynamics of air mass come into play but in a much simpler way than you might expect. Remember the role that the wing plays in all of this? Air pressures above and below the wings can still be easily maintained at higher altitudes. Why? Because of the much greater amount of air passing over the wing surfaces at speeds often over 500 mph. You see, it’s the air molecules that hold the secret to air mass and density. At speeds like this the air density is super compressed to allow far more lift than needed to maintain flight. Then there is the engine question.

Since air is composed of gasses (mainly nitrogen with about 20% oxygen and other gasses mixed in) at the molecular level, they too can be “squeezed” under tremendous pressure and therefore compressed to create an even mixture through a jet engine. This ability to compress huge amounts of air while flying at high altitudes again makes up for loss of mass and density. Since the engine is moving between 500 and 600 mph, the intake or ingesting of plenty of air through the core and surround is no problem.

The added bonus is this; thin air creates less friction on the aircraft and much cooler air temperatures (normally around -60 degrees) as the air flows through the engine. There is also a slight reduction in fuel cost flying at these higher altitudes as we normally reduce our throttles 10 to 20 percent while at cruising altitude. This of course is dependent on the aircraft type and any head or tail wind!


Do any propellers/propfans use stator vanes? Why is this not a commonplace feature? I understand that axial-flow turbine engines generally include stator vanes behind every(?) compressor and turbine stage. This makes quite intuitively sense, because without them the rotating blades would mostly swirl the air around through the engine, which would not only prevent the stages from properly transporting/compressing/expanding the air in axial direction, but also put useless rotational kinetic energy into that air. The latter is presumably also the reason why turbofans have such guide vanes behind the fan, too.

All of this applies just as well to unducted fans/propellers. Yet I've never seen stator blades behind an open propeller. There are contra-rotating coaxial propellers of course, the idea of which seems to be similar: get rid of the useless solenoidal air motion. But actively contra-rotating both props adds a lot of complexity, and seems somewhat unnecessary if the goal is mostly to remove large part of the swirling motion. Cancelling all torque is not necessary for a plane.

Especially, I would think stators could also be shaped to help retrieve some energy from the tip vortices like a duct does, without however interfering with the inflow that seems to be the reason why ducts make props inefficient at low speed.

When a compressor rotor stage speeds up the flow, the stator:

Slows it back down, turning kinetic energy into pressure (see image here from the book The Jet Engine)
Presents the flow at the correct angle for the next rotor stage.

A nozzle guide vanes stage (what a turbine stator stage likes to be called) comes before the turbine stage, and it does the same but in reverse (turns pressure into velocity) and adjusts the flow direction for the upcoming turbine stage.

The de-swirl function of the stator stage behind a fan in a turbofan (the outlet guide vanes – OGVs) is beneficial because of how much air the fan has acted on.

The number (and size) of blades tell the whole story. It's often said that propellers move air by a little, and jet engines by a lot. To put that into perspective, the typical pressure ratio rise behind a fan is 1.4–1.6, whereas behind a propeller it's less than 1.02.$^1$

This is where a turbofan's OGVs become beneficial. Adding a static stage of blades behind a propeller (with or without a case) that doesn't accelerate the air by that much to recover meaningful pressure (compared to a turbofan's fan, not much kinetic energy to begin with) in the axial direction won't help. On the contrary it will add drag (air hitting a static object), and weight.

    Actively contra-rotating both props adds a lot of complexity, and seems somewhat unnecessary if the goal is mostly to remove large part of the swirling motion. Cancelling all torque is not necessary for a plane.

A propeller is like a wing, it needs to act on (hit) air to create a normal force. If we substitute the second set of moving prop blades with static ones, we've just lost a lot of thrust.

$^1$: https://history.nasa.gov/SP-468/ch10-3.htm

Quick answer is that it boils down to thrust vs drag.  Propellers create thrust as airfoils and mass movers, fans are less efficient mass movers that can move much more mass if sufficient power is available.  This is why fans are teamed with jet engines.  Brute force.

A stator will not benefit the more efficient propeller, but I would not wish to discourage research.

The "reason ducts make props" link provided above is outstanding in that it shows each more efficient propulsion method falling by the wayside as speed increases due to drag,
while the turbojet keeps improving.


How often does an airline pilot use the rudder pedals during flight? How often does an airline pilot use the rudder pedals during flight? Is it only during takeoff and landing?

I am interested to know the rudder pedal usage time in minutes per average short range flight (A320, Domestic flights, ~1-2 hr flight duration). When on autopilot, the rudder control will be automatic. There are even automatic rudder trim features for most of the commercial aircraft.

Transport category aircraft have a yaw damper that will maintain coordinated flight when the aircraft is in the air unless the flying pilot overrides it with the rudder. It operates even if the autopilot is not in use. The usual procedure is that you take your feet off the rudder pedals as soon as you have lifted off and don't put them back on the rudder pedals until on short final.

The pilot will possibly use the rudder shortly before or going into the landing flare to slip into a crosswind and/or to align the aircraft with the runway if they were flying a crabbed final. However, with most if not all transport category there's nothing wrong with touching down in a crab correctly being flown to correct for a crosswind if desired.

So whether a short range light or a long haul flight, the amount of time in the air that the pilot needs their feet on the rudder pedals is minimal and is in seconds, at most a minute or so.

An exception, of course, is if there is an engine failure, in which case the pilot will use the rudder to counteract the adverse yaw from the asymmetrical thrust.

I am not sure about usage during autopilot, but without autopilot, you always keep your feet on the rudder pedals. Depending on the number of engine, it can have an engine configuration where it bias towards one side (as with one engine in front, for say a Cessna 172). To make a coordinated turn, one need the rudder, and to counteract the wind/drift to keep a straight course. So, pretty much always. There are no rudder trim to my knowledge, as the wind is never constant all along a mission, even if following a straight heading the whole time.


Do you need strong muscles as far as applying the yoke is concerned, in order to be able to deal with some emergencies? In some of the plane accidents, the pilot had to use all his might to apply the yoke in certain direction to save the plane from imminent crash; in some cases even crew members had to help with the force. So it seems certain degree of physical strength is needed to be a pilot in view of possible emergencies, though it doesn't seem to be explicitly required?

For the Airbus it is not actually require any strength, all the technology is fly by wire, this means that the actual control (Side stick) is not directly connected to any cable, so been this said, there is no actual force to fly any airbus plane. For the Boeing plane, at least for the 737, that's the one I know, when it loses all the hydraulic systems, the aircraft will become on a manual reversion mode and it will need strength from the pilot to move the aircraft controls that are linked to the surfaces by cables, but this is a remote possibility to happen, and if it is the case both pilots can apply force to the control column

It would be very helpful to narrow this question down as to whether or not "physical strength" is needed due to a failure of a built in mechanical/hydraulic system
(like losing power steering in your car), or simply in a cable driven control surface system, found in recreational aircraft.

Enormous physical strength is NOT required to be a pilot, although good mental and physical condition is a must.  Endurance of concentration and stamina of strength do help as you may be required to "go around" after a long flight.

More strength is needed at higher speeds due to aerodynamic forces, but that is part of Vne.  Please do not imagine three people tugging at the controls to save a Piper Cub.

So, depending on your abilities, careful research and expert instruction will get you going in a plane that suits you.


Why do we say certain things three times (e.g., "Mayday, Mayday, Mayday"), rather than two or four? Repetition is a key characteristic of communication in the control tower, cockpit, and control room.  Some phrases, like "Mayday" get repeated. The speaker says the same thing three times. We know this is for redundancy.

Why exactly three times?

Why not twice or four times? Is there research suggesting three is the most effective number, or is there a historical reason for the convention?

Yep, the critical commands are repeated 3 times.  This ensures there is ABSOLUTELY zero doubt in anyone's mind (especially on a big crew airplane) of what needs to be done in a critical situation.  It also standardizes these criticalities across different aircraft and aircrew cultures.  "Bail out, bail out, bail out"  "Eject, eject, eject"  "Abort abort abort" "Pan-pan, pan-pan, pan-pan."  I was 27 years a USAF pilot, and this is how the training has worked for over 50 years.  I only saw these terms used 2-3 times, but it certainly gets your attention and amps up the sense of urgency.  A little history:  back in the day of very poor radio communications, it was necessary to repeat to "get someone's attention" or in the event a single "mayday" didn't come across when the transmit button was pressed.

the reason for the repetition of mayday mayday mayday is for receiver of the messages can hear the callings, if the first mayday calling is breaking, the second mayday calling maybe be heard, and totally sure the third mayday the messages needs to be convey along with the mayday


Why does the FMC in B737 limit the climb speed to 230 knots on the VNAV mode until the leading edge devices are fully retracted? In the VNAN climb on B737, the FMC changes the target speed from V2+20kt at liftoff, to 230kt for acceleration at the preset accleration height, then to 250kt (below 10,000ft) as the leading edge devices are fully retracted along with the trailing edge flaps.

My question is: Why is the climb speed limited to 230kt on the VNAV mode until the leading edge devices are fully retracted, when there are only operating speed limits to trailing edge flaps as below?

For your information, I've found a discussion on this topic on another aviation website but no one seems to have a definite answer to this.

https://www.pprune.org/tech-log/570676-737ng-led-s-speed-limitations-ops.html

Does anybody have a good answer to this?

The non normal checklist QRH will answer your question.  If the LE flap transit light remains illuminated I.e the leading edge flaps and slats haven’t retracted or disagree with the flap handle position, then the QRH directs you to limit the speed to  230 kits until you have rectified the issue.  VNAV is honouring that restriction.

Hope that answers your question.

Because at flaps 1, the leading edges are deployed until the flaps are up therefore 250K speed restriction


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

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

The incredibly simple answer is that:
You have to be sitting upright in a chair, in a seat belt, for take off and landing.
It's that simple - an "all-bunk" configuration (like a sleeping-bus) is a non-starter.
That's all there is to it.


Who said "A little of all of us goes with every pilot we lose"? The only attribution I’ve ever seen for the quotation below is “Anonymous FAA Inspector.”

  Whenever we talk about a pilot who has been killed in a flying accident, we should all keep one thing in mind. He called upon the sum of all his knowledge and made a judgment. He believed it so strongly that he knowingly bet his life on it. That his judgment was faulty was a tragedy, not stupidity. Every inspector, supervisor, and contemporary who ever spoke to him had an opportunity to influence his judgment, so a little of all of us goes with every pilot we lose.

Who originally said or wrote these words? What were the motivating circumstances?

William Garvey in "Viewpoint"
The authoritative reference is in this article:
Aftermath: Lessons Learned  (Aviation Week)
Excerpt:

The last words of William Garvey's January Viewpoint -- ". . . so a
little of all of us goes with every pilot we lose." -- struck me
("Aftermath," page 7). I had just returned from the small town of
Vesoul, France, where I attended a ceremony marking the 40th
anniversary of the death of a Mirage III pilot. Lt. Patrick de
Barbeyrac Saint-Maurice died while flying an intercept training
mission near Vesoul on Nov. 3, 1966. The cause of the accident was
never fully determined, other than the ...
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Do not have enough reputation to comment, but will add this as a potential answer source:

  — Captain A. G. Lamplugh, British Aviation Insurance Group, London. c. early 1930's. This famous phrase has been reproduced on posters and plaques many times, almost always with the attribution of 'anonymous.' I was told at a book signing that André Priester (one of the first Pan Am employee's) may have said it, and decided to check this with the late R. E. G. Davies, then curator of air transport history at the Smithsonian and author of a book on Pan Am. Ron called me back and told me the phrase pre-dates Priester. His research showed the originator of the phrase was Captain Lamplugh, who was quite well known in British aviation circles after WWI

Source: http://www.skygod.com/quotes/safety.html


Why wasn't the 737 MAX's landing gear lengthened enough to mount the engines under the wings, rather than in front of them? The reason Boeing originally added the MCAS to the 737 MAX was to compensate for a slight pitch-up moment generated by the engine nacelles at very high angles of attack.  This pitch-up moment occurred because the LEAP engine used on the MAX series is larger and longer than the CFM56 used on the 737 Classic/NG, so Boeing mounted the new engines further forward and higher up to fit them on the plane; the larger, forwarder nacelles added more planform area to the front part of the aircraft, causing the MAX's center of lift at high attack angles to be slightly forwarder than on the Classic or NG, causing the MAX to pitch up slightly more at high attack angles.

However, the reason the Classic and NG have their engines mounted in front of the wing is that the 737 originally used the much slimmer JT8D engine, allowing it to use a very short landing gear to make it sit lower on the ground and ease the process of loading baggage and other cargo by hand.  When the Classics came along, Boeing decided to keep the 737-100/-200-length landing gear, which forced them to use a smaller-fan variant of the CFM56, move all the engine accessories from the bottom to the sides, and mount it in front of the wing (instead of directly below it) in order to give the new, larger engine sufficient ground clearance; with the NGs, which use a CFM56 variant that is essentially the same size as that used on the Classics (with a diameter all of one inch greater), they simply kept the same nacelle, pylon, and landing gear designs as those used on the Classics.

The MAX, in contrast, has a longer landing gear than all other 737s (handloading of baggage not really being that much of a thing anymore at most 737-served airports); given that they were giving the MAX a longer landing gear anyway, why not make it long enough to let the engines fit under the wings and finally eliminate the engine-mounting kludge inherited all the way from the 737-300?

The main gear sits between the wings and retracts inwards, such that the wheels will rest in the lower fuselage. Lengthening the main gear would either require to shift the gear attachment points outward or to retract one leg a bit forward and the other a bit backwards, such that both wheels will sit behind each other instead of side-by-side, as they do now - see below (source).

Shifting the gear attachment points out is clearly feasible but results in a major wing redesign. That would cost money so Boeing management decided that a software fix would be the better solution.

Placing the wheels in tandem would equally be a major and costly modification. A lot of systems would need to be relocated and all modifications would require testing. Boeing management cleary did not want to pursue this option.

Another quick fix is shortening the gear upon retraction. Decompressing the oleo strut in order to shorten the gear upon retraction is already standard in many fighter jets. Another solution is to add another hinge such that the gear leg will fold inside the gear bay. Such a solution was used on the Grumman F-8F Bearcat and is pictured below (source).

From what I understand, this particular solution was applied to the longest 737 version to allow enough tail clearance for rotation, but added weight and complexity. Of course it could be applied to the rest of the 737 MAX fleet as well, but again Boeing management tried to save money and weight by implementing a software-only solution.

But the engine location would be fine if a larger horizontal tail would be added, increasing stability at high angle of attack and making a quick software fix unnecessary. Again, this would add cost and weight and make the aircraft less competitive.

The fuselage is not wide enough and therefore there is not enough space in the wheel well to store two main gears, had they been extended further.

MAX 10 employs a new landing gear design that shrinks the legs slightly, but even that is just on the limit of what is possible, so if Boeing increased engine size for all models and used that mechanism for MAX 7, 8 and 9, they would have been unable to develop the MAX 10.


Are taxiing aircraft to hold short of taxiway hold short markings if not instructed by ATC to do so? At some airports, some taxiways have taxiway hold short markings (single dashed yellow lines against a black background). Do taxiing aircraft by default need to hold short of these markings if they're not instructed by ATC to do so?

This is one biggest pet-peeves of mine that even old, experienced pilots have missed after years of flying.

Taxi instructions are just that: Instructions, not clearances. Mind you there are some clearances when on the ground aswell, crossing a runway, entering a runway, backtracking, maybe some others.. mostly involving the runway. But a general text like ‘Taxi to holding point Charlie via Golf’ is an instruction.

Ok, so the difference is just pedantic but there is one big caveat: when you are cleared to land, cross, enter on a runway you are guaranteed the runway is, well, Clear and will continue to be so for the ammount of time you need it to land/take-off.

This is not true for taxi instructions. You are instructed on what path to take on the ground, but the taxiways and route to your runway is most definitelly not cleared of other aircraft (unless we are talking about low visibiliy ops, but just bear with me for now). When taxing you are still supposed to look out for other trafic and try to avoid hitting other airplanes taxing around as much as you can.

You might be instructed to go via one taxi path and then later told to give way to another aircraft comming from your left. See how the path has not been ‘cleared’? It is not a clearance.

All this talk to sum up: entering/crossing a runway is a clearance. Crossing a (non-mandatory) taxiway holding point is not. Is the answer slowly building up ?

Crossing a runway hold short makrking (two dased lines and two solid lines) you need clearance.

Crossing a taxiway holding marking (or intermediate holding position, or cat 2/3 holding when not in low-vis) does not require one.

A taxi clearance from ATC to a specific point on the airport gives the pilot authorization to cross all hold short markings or hold position markings for intersecting taxiways up to the clearance limit of his taxi instructions without stopping. If a pilot is given a hold short clearance at a certain position, he must stop and ask ATC for permission to proceed. If there is no conflicting traffic as the pilot reaches the hold position, ATC will normally give prior authorization to cross that hold point before the aircraft reaches it. Any position markings that require the aircraft to hold will be given in the original taxi clearance and would have been repeated back when receiving the taxi clearance. If not previously cleared beyond a hold point, the pilot must STOP until further clearance is authorized.  Runway Holding Position Markings identify the locations on a taxiway where an aircraft MUST STOP when a clearance has not been issued to proceed onto the runway and no part of the aircraft can extend beyond the holding position marking. This is usually the last position of a taxi clearance limit and normally would be a change of frequencies for take-off authorization. One point to remember: if at any time you are in doubt of of your instructions,  STOP! Don’t be embarrassed to ask questions of the controller. It’s okay to ask for directions or for
progressive taxi instructions, if you are not familiar with the airport or route of taxi that has been assigned.


Can I use PVC pipes for the rib on the wings of my homemade airplane? Can I use a PVC pipe as the rib of the wings of my homemade airplane?A rib is already load tested, and it failed at 40kg of weight and it is supposed to hold just 10kg in flight.

Yes, of course you can.

However, it's probably not a very good idea.

PVC pipes are pretty strong and light; circular-section PVC pipes have an excellent strength-to-weight ratio.

Being strong and light isn't enough though. Construction materials need to have the appropriate properties (like being strong and light) in the right conditions.

In the case of an aeroplane wing, that includes the conditions that aeroplane wings are subjected to: rapidly-changing stresses, vibration, sustained load, torsion and so on. You need to know: will PVC remain strong under such conditions?

Then presumably, you will need to find various ways of joining your PVC sections to each other and to other materials. How will these joins be made? How will they behave under the conditions an aeroplane wing is subjected to? How will the joining methods affect the properties of the PVC pipes?

Aircraft designers tend towards the conservative; even the most radical of aircraft designers would not be considered daring compared to the radical leaps taken by designers in other fields. That's because the chief reward for brave leaps in aviation is death.

All the same, over its history, aviation has tried all kinds of things (usually, quite carefully). If aviation has successfully used PVC pipes in aircraft construction, there'll be a record of it and a tradition of it. The successful pioneers of PVC-pipe-based aviation will not, however, have embarked on their adventures on the basis of an encouraging Stack Exchange reply.

If you don't already know the answers to the questions I mentioned above, and are not already in possession of reliable information about techniques, tools, practices and theory for working safely with PVC load-bearing structures, I'd prefer to use PVC pipes for plumbing only.

Analyze this problem in terms of structure dynamics.  A structural member of circular cross section is isotropic with respect to stiffness; that is to say it resists bending equally from all directions.  This is not the best choice for an airframe.  Because the lifting members of an airframe need to resist force primarily along the vertical axis, the underlying structure of those members should be anisoptropic.  An example of an anisotropic cross section is an 'I-beam'.  That kind of section yields most bending resistance per pound of material.


What is a "runaway supercharger"? During World War 2, my uncle was a navigator for a B-17G bomber based in the UK. On return from a bombing run over Frankfurt, the crew was forced to bail out over Belgium at 3000 feet altitude. My uncle and 3 of his crewmembers were rescued by Belgian resistance members and hidden for 8 months when Allied forces liberated western Belgium.
In the Escape and Evasion report the pilot gave on his return to 8th Army HQ, he wrote that the crew bailed out because "No 2 and 4 engines had runaway superchargers, lost oil. Would not feather with full oil pressure."
Could someone help me out here in understanding what's the likely causal sequence of events here? Why would the plane experience a "runaway supercharger?" Was that the cause of lost oil, or was it the other way around - some sort of oil loss causing loss of control of the supercharger?
Or was it that the propellers for 2 and 4 engines wouldn't feather properly (due to mechanical trouble or maybe flak damage?) causing the engine to overheat and burn oil?

The B-17 had General Electric exhaust-gas-driven turbochargers (or more correctly, turbo-superchargers) that had pilot selectable boost levels. A selector dial in the cockpit controlled the operation of the wastegate that controls whether exhaust goes through the turbine and provide engine boost, or is bypassed directly outside.  The pilot's indication of the state of boost being provided to the engines would come from the engine manifold pressure gauges.
"Runaway superchargers" would suggest that the wastegate control systems on those engines were malfunctioning, such as the waste gate failing closed and directing all the exhaust flow through the turbo, causing the turbos to over boost the engines (the "running away" part), which would cause sky-high manifold pressures, detonation, and send the cylinder head temperatures through the roof.  This would necessitate a precautionary shutdown to keep them from blowing cylinders.
During the shutdown, even though they had oil pressure indication for the engines, the propellers, which operate hydraulically using engine oil, wouldn't feather. There is a feathering pump for each engine, normally electrically operated, that draws oil from somewhere in the system and can drive the prop into feather even without any engine oil pressure, as long as the pump itself has a source of oil to its suction inlet.
The turbosupercharger waste gate uses engine oil pressure for servo power for regulating the waste gate, that is also used to lubricate the turbo's bearings.
So... thinking it through, the turbo waste gate needs engine oil, and the feathering pump needs engine oil.  My guess is that there were oil system problems that impacted the waste gate control systems on those two engines, causing the boost runaway, and the same problem was also starving the engine's feathering pumps so that they were unable to complete the feathering of the propellers.  Maybe flak damage, or just a technical issue with the system (it would be odd for flak shrapnel to cause identical problems in two engines on opposite sides of the aircraft).
A B-17 at low altitude, well lightened, having burned off most of its fuel, dumped its bombs, and expended ammunition, would probably have been able to make it back to the UK on two engines, IF the two dead ones had feathered properly.  Having two unfeathered propellers windmilling away however, is massively draggy, and would have turned it into a two-engine glider, leaving the option of a forced landing or bailing out; and it looks like they chose the latter.

This is speculative, but in diesel engines, if a turbocharger has a failed oil seal on the inlet side, the engine can then run on the oil leaking through the seal.
Since diesel engines are shut off by stopping the fuel supply, the engine then "runs away" and typically runs until it burns all it's oil, and seizes.
If these engines had a sufficiently high compression ratio, it's possible that they could auto-ignite on oil leaking into the intake.
This would support both the oil pressure issues and "runaway" description. I'm not sure if it's applicable to these engines in particular, though. Note that any engine can conceivably "diesel" if it has a sufficient compression ratio, or a fuel that ignites at a low enough temperature.
https://en.wikipedia.org/wiki/Diesel_engine_runaway


How are these Airbus A350s flying in formation without being affected by wake turbulence? The first photo in this article appears to depict three Airbus A350s flying closely in tandem at approximately the same altitude.

I thought that a large jetliner would leave a trail of wake turbulence behind it that would be unsafe for several minutes.  Is there something special about the Airbus A350 that makes this formation possible?  Am I misinterpreting the photo?

The V formation in the second photo seems much more reasonable.

You are right that wake turbulence would be a danger when flying behind another aircraft like that.

The perspective of the photo makes it hard to see the actual position of the trailing aircraft relative to the ones in front of it. Pilots do this sort of thing in airshows when they fly in formation. The trailing aircraft are generally slightly below the ones in front of it. This allows good visibility of the plane they are following. At the higher speeds that these jets are flying at, the wake vortex will not drop enough to impact an aircraft fairly close behind and below.

See this old photo of the Thunderbirds. From this perspective, it is easier to see that the trailing aircraft is also below the flight path of the leading aircraft.

What I'm going to say is my own opinion. I don't claim it's true, that's just my understanding of wake turbulences... I'm eager to learn, so feel free to correct me if I'm wrong :)

I'm assuming wake turbulence is moving air (airflow) caused by an aircraft flying ahead, wether airflow moving down, up, sideways or spiralling - I understand the pressure difference - Without going deep in particles motion (air molecules) I think you just have to consider four things :

your position behind a the leading aircraft : right behind its fuselage, right behind one of its wing tip, beyound its wingtip trail, from the axis of the leading aircraft.
your distance behind the leading aircraft : a few feet just behind (without colliding with) or miles away ?
wake size increases as you go far behind the aircraft producing the wake turbulence, but its strength decreases accordingly.

The point 3 tells you where NOT to position your aircraft behind another. Especially at a (relative) location where the wake is still powerfull, and its size is large enough to affect your controls surfaces and wing envelope, and send your aircraft in an uncomfortable motion, brutal gain or lost of lift, differential lift, etc.

So, right behind another aircraft, you can fly without been affected (much) by the wake turbulence, because the wake size is not large enough to send your aircraft sideways, however, you still encounters slight turbulences, especially if you fly just right behind the wingtips. Worst situation is when you fly several hundreds feets behind the aircraft : large volume of strong spiralling airflow that you can't visualize. Miles aways, the airflow velocity is assumed to have dissipate enough because of its lost of strength due to expansion and air resistance.

Of course, the wake turbulence of a Cessna is not the same of an A380. Weaker and smaller for a Cessna, that dissipates pretty quickly. This leads me to introduce point four :

The size of the plane creating the wake, and the size of the following plane. Surprisingly, you can put an A350 right behind another A350, but you can't put a Cessna right behind an A350. This, because of the three points above : The Cessna will likely be affected by the slightest amount of airflow while the A350 wont much, because of surfaces size (wings/ailerons) The wake (especially the wingtip vortex) hasen't grow enough to hit the following A350 hard enough.

However, that doesn't mean there aren't risks. Main risk in such close formation is vibration. Test pilots are trained to identify their origin and deal with.

And also, I don't have a comparable aircraft in size, so I'll take a DC10, or a B52 : I think you can't put such earlier generation of aircraft right behind an A350 (or 787) because of Wing technology (this related to the question "does the A350 has something special ?") Old aircraft have much less flexible wings, that makes them really sensible to brutal airflow. I'm not a physicist, but more flexible wings better distribute the load factors along said wing, considerably reducing vibrations and fatigue.

That's why wake turbulences are more dangerous at lower speed : the wake can grow enough to compromise the flight enveloppe of the following aircraft even at the tightest separation. That's why wake turbulence are a serious issue at lower altitudes where the moving airflow is really dense and affect surfaces greatly. Logically, wide wakes should dissipate more quickly in dense atmosphere, but the (tiny) spiralling vortex can last several minutes and is still very dangerous especially for smaller aircraft.
Fighter jets in airshow and flight formation :
 a) They fly at high speed => not enough expansion to affect nearby followers
 b) tandem formation requires lining up with leading aircraft => good distribution in differential lift.
 c) in tandem formation, followers are usually below (rarely above) the preceding aircraft.

Finally, a Cessna control surfaces (and flaps) are more sensible than the ones that equip a 777. Never get a Cessna below and behind a 777 in a five to ten minutes timespan unless you know where the wake turbulence path has drifted and/or landed.

Side note : I'm not a pilot. I'm just interrested in how an aircraft flies. There are tons of documents out there about wake turbulences, wake vortices, stall schemas, etc. Some are contradictory, some are complementary. At a molecule level, the magic of an airflow looks like a perfect example of the theory of chaos. But the above is the summary of what I found here and there. That's no science, just opinion.


Does the F-16 have a speed limiter? In an aircraft like F-16, how do the pilot prevent going faster than Vne (velocity, not exceed)?

From the block 50 manual it seems to indicate Vne is 800KIAS/2.05M.

I believe even the earliest F-16s could go mach 2, so the newer with their much more powerful engine should easy exceed 2.05 mach.

Is there an automated system that reduce throttle when exceeding?

Section V of the Block 50 F-16 Flight Manual, is the "operating limitations". It says:

  MAXIMUM AIRSPEED OPERATING LIMITATIONS
  Refer to figure 5 3. Maximum operating airspeed is 800 knots from sea level to 30,000 feet MSL. Above 30,000 feet MSL, the aircraft is limited to 2.05 mach.

The introduction to section V states (my emphasis):

  The aircraft and system limitations that must be observed during normal operations are presented in this section..

What does "observe" mean? One might assume that means the pilot can just sit back and watch them automatically occur, or the pilot must "observe" them by knowing and acting apon them. As evidence it is most likely this second interpretation, the list of contents in this section appears to be many things that the pilot needs to do.

It does not seem like a list of things the aircraft does automatically.

But more signifficantly, the contents of the flight manual also includes a section describing the flight law control system (FLCS) limiters:

and:

There is no mention of maximum speed.

So, it looks strongly like the FLCS limits 4 things (AOA/G, roll, yaw, rudder authority), and warns you about some other things (low speed), but does NOT limit maximum speed.

No. Canopy plastic don't withstand the speed,also the inlet lip temperature must be kept just inside the safety margins. I read about early pilots going in a slope downward and been killed by overspeed and flutter at the moment. 814Knots is maximum safe for me. 1.9Mach is been done at the40. 000feet.


When a hold is distance based, do you still used a timed entry? Let's say you have to do a parallel or tear-drop entry to enter this distance-based holding pattern.  Do you still fly outbound from the fix for one minute and then make your turn inbound to re-intercept the fix or are you taking it to 4 NM?

I haven't found any specific guidance for hand flying entries to RNAV (RNP) holding patterns. But part of the concept of RNAV is that the navigation system can provide the complete desired path including hold entries.

Referring to RTCA DO-236C RNP MASPS and the requirements for holding operations, I find the following: The dimensions of the hold pattern are defined based on altitude, airspeed, bank angle, and an allowance for wind. The length of the inbound/outbound leg is defined by the database or entry. In your case this is 4 NM, and is defined as d1. The width of the pattern (d2) is computed based on the above variables.

Without the details, we can still make an approximation. If you're flying at 120 kts and performing a 2 min turn (1 min for 180 deg), your pattern width (d2) would be about 1.3 NM.

The rules include the following statement:

  When performing an entry into the holding pattern, the navigation
  system is permitted to use an inbound leg length (d1) that is equal to
  the holding pattern width (d2) instead of the length defined in the
  database or from pilot entry.

So the end result is that your entry cannot exceed the 4 NM leg length, and could have an inbound leg length as short as 1.3 NM.

No the hold makes use of a straight line distance from a GPS identified waypoint and is not timed.  This can also be done with holds on a VOR/DME and or VORTAC using a DME or GPS distance from the Navaid.