Plane flips over after crash-landing in Somalia

[Dimitri Cherchenko]

A plane has crashed and flipped over on landing at the airport in Somalia's capital but there were no fatalities among the more than 30 people on board.
https://cimg1.ibsrv.net/gimg/pprune.org-vbulletin/976x549/_125951589_whatsappimage2022_07_18at10_39_55_ef950c84a3ad7e0 6737ec7cf901cab93792e53b3.jpg
BBC News (https://www.bbc.com/news/world-africa-62206497)

[DaveReidUK]

1992 former Lufthansa Cityline F50, been around the houses a bit since then.

* Informational post only, no judgement implied. :O

[Lexif]

More pics at avherald: Accident: Jubba F50 in Mogadishu on Jul 18th 2022, flipped over on landing (http://avherald.com/h?article=4fbd0537)
Nose gear is missing a wheel and turned 90°, and left main gear missing. The fuselage looks surprisingly intact, but there is a tear in the skin next to the right prop. Glad there were no serious injuries.

[oceancrosser]

Second F50 crash in the region in two days. If this continues the F50 will be extinct soon.
East Africa has not been kind to the Fokker 50.

[pax britanica]

Extra ordinary pics, F50 fuselage seems to be a very rugged construction being seemingly intact after landing on its back and having both wings torn off. I suppose it is similar if not the same as the F27 /FH227 forebears designed and built in the 1950s where ruggedness for this type of aircraft was essential given the kind of places they would routinely operate to. Mogadishu sounds an incredibly scary place to operate into.

[Doors to Automatic]

That’s not a brilliant day at the office.

I wonder if the 18kt tailwind had anything to do with the crash?

[A0283]

Extra ordinary pics, F50 fuselage seems to be a very rugged construction being seemingly intact after landing on its back and having both wings torn off. I suppose it is similar if not the same as the F27 /FH227 forebears designed and built in the 1950s where ruggedness for this type of aircraft was essential given the kind of places they would routinely operate to. Mogadishu sounds an incredibly scary place to operate into.

The Fokker 50 is very rugged indeed. One day at Schiphol I saw one landing with only its NLG and left MLG leg. Solid piloting skill gave it a silky landing. Replacing a piece of skin and part of a frame and it was ready to fly again.

It was certainly not the same, Wikipedia will give you some basic upgrade information. Much improved engines and modern cockpit and modern interiors are examples.

One thing that did not change was the general layout with the high wing and long MLG legs required for the rough and unimproved runways. These required for the original F27 Friendship being the successful DC3 replacement.

[hans brinker]

1992 former Lufthansa Cityline F50, been around the houses a bit since then.

* Informational post only, no judgement implied. :O

Flew that plane for Denim Air.

[hans brinker]

Extra ordinary pics, F50 fuselage seems to be a very rugged construction being seemingly intact after landing on its back and having both wings torn off. I suppose it is similar if not the same as the F27 /FH227 forebears designed and built in the 1950s where ruggedness for this type of aircraft was essential given the kind of places they would routinely operate to. Mogadishu sounds an incredibly scary place to operate into.

Built like a brick, flew like one too.

[bluesideoops]

Jubba Airways Fokker 50 by the looks of things https://www.bbc.com/news/av/world-africa-62212987

[DaveReidUK]

Jubba Airways Fokker 50 by the looks of things https://www.bbc.com/news/av/world-africa-62212987

See photo in post #1.

[Herod]

Built like a brick, flew like one too.

Hans: agree with the first staement, not the second. I never flew the '50, but have some 6,000 hours on the '27. An aeroplane that would never let you down.

Amazing that there were no serious injuries. Says a lot for that fuselage.

[flyems]

The 2013 Ethiopian Airforce AN12 incident was thought to have been the result of a tailwind gust.

[hans brinker]

Hans: agree with the first staement, not the second. I never flew the '50, but have some 6,000 hours on the '27. An aeroplane that would never let you down.

Amazing that there were no serious injuries. Says a lot for that fuselage.

Well, the way I normally said was flew like a truck. Rotate definitely required both hands. Engine failure required so much rudder pressure I would normally use both legs, full aileron into dead engine, so uncross armes and hold the wheel upside down. Extend flaps and don't trim the wheel 3 times around and you need both hands to keep the nose down. Flew the Dash 8 at the same time, and was so much easier to control, but the cockpit was lousy.
F50 Cockpit design is still better than anything else I have seen.

[avionimc]

full aileron into dead engineYou probably meant aileron into the good engine (bank into the good engine), in order to raise the dead engine.
https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcRIwfNF345NVMpPvLRNoiOCke-1iWJMkHeQYAn9NKN3LXaCF70jurzmZ9xmSBhwd5ofQA&usqp=CAU

https://youtu.be/Wbu6X0hSnBY

[hans brinker]

You probably meant aileron into the good engine (bank into the good engine), in order to raise the dead engine.
https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcRIwfNF345NVMpPvLRNoiOCke-1iWJMkHeQYAn9NKN3LXaCF70jurzmZ9xmSBhwd5ofQA&usqp=CAU

https://youtu.be/Wbu6X0hSnBY

well, that’s embarrassing. Yes, full aileron to raise the dead engine.
after accelerating much less control travel required to fly, but the first few hundred feet climbing at V2 were challenging.

[meleagertoo]

Jeez! Is Mog really still so unsafe that people accept 18Kt tailwinds to avoid overflying the city? That makes the place damn hazardous for half the year! I really didn't like that time of year, it did expose you to small-arms fire from the city no matter what you did and individual approaches like mine irritated the **** out of the US 'controllers' who wanted to drag you over half the country on a 20 mile DME arc and then over the entire city at 1500ft on finals! You couldn't make it up!

This was my verion of dealing with the SW monsoon wind, a month or so before the Blackhawk incident. Can't see why a little Fokker shouldn't do much the same on such a long runway but clearly not an option for jets.

Not much progress there in thirty years then. Makes me wonder why did we bother?


https://cimg8.ibsrv.net/gimg/pprune.org-vbulletin/1608x1080/left_base_476d5d4f0e24de2dbba4e6ce7154b9f5ebdda245.jpg

[wideman]

Video:
https://twitter.com/SomaliGuardian/status/1552195910112780289?s=20&t=GYN-3AJOPCHHOkYyizeWmg

tried to post video, but it might not show up. Check twitter feed for
@SomaliGuardian (https://twitter.com/SomaliGuardian) for video

[ItsonlyMeagain]

Looking at the video, I’d guess it’s not so much the tailwind, but the fact it appears to land short of the runway and the then higher surface disassembles the aircraft in short measure. Stops quickly though.

Me

[FUMR]

Not only does it land short but quite heavily too. I'd say that a tailwind could well have contributed to that outcome.

[ItsonlyMeagain]

Only if you are looking at ground speed, not airspeed. A tailwind will make you land long.

Me

[fdr]

Only if you are looking at ground speed, not airspeed. A tailwind will make you land long.

Me

er, why?
¿por qué?
sei?
mengapa?
なぜ？
لماذا ا؟
waarom?
perché?
Varför?
ทำไม
tại sao?

How does a tailwind affect the aiming point which is determined as a geometric point on the runway, and to which in general a vertical path is expected to be flown to intercept the same point?
If the statement refers to the rollout due to the higher groundspeed, then that is different.

just curious

[ItsonlyMeagain]

You are correct that a 3 degree flight path with a fixed aiming point is fixed in space; I did not dispute that. However to maintain that with any wind, the variable is power. In a strong tailwind there will come a point where this is not possible as the power will be at idle. Of course, the way to maintain the flight path now is to point at the aiming point as the tailwind tries to put you above the profile. Should one arrive at the aiming point at the correct height, you will certainly be fast and….. you will land long. Professional pilots call this an unstable approach.

Me

[WideScreen]

er, why?
¿por qué?
sei?
mengapa?
なぜ？
لماذا ا؟
waarom?
perché?
Varför?
ทำไม
tại sao?

How does a tailwind affect the aiming point which is determined as a geometric point on the runway, and to which in general a vertical path is expected to be flown to intercept the same point?
If the statement refers to the rollout due to the higher groundspeed, then that is different.

just curious
A headwind, reducing close to the ground, will let you drop out of the air due to insufficient airspeed, whereas a tailwind, reducing close to the ground, will increase the airspeed and as such, keeps you flying/floating longer, IE your landing spot automatically moves further down the runway.

[fdr]

A headwind, reducing close to the ground, will let you drop out of the air due to insufficient airspeed, whereas a tailwind, reducing close to the ground, will increase the airspeed and as such, keeps you flying/floating longer, IE your landing spot automatically moves further down the runway.

The dynamics of a shear associated with the turbulent boundary layer is quite straightforward, but the aiming point is the aiming point, and the flare/ thrust reduction is a process that is trained and mastered by the pilot in qualification. A rollout will be extended by the higher ground speed, nut nothing in Part 25 Subpart B lets the pilot off the hook for not being able to fly a path to an aiming point, flare/reduce power at an appropriate time and land.

In simple terms an into wind landing has an undershoot shear where the wind component reduces due to the TBL approaching the ground. Response is a pitch adjustment and thrust change while transiting the shear to maintain parameters. The immediate change is a reduction in IAS from inertial factors, while groundspeed increases... speed stability results in the attitude naturally lowering which matches the higher sink rate rest for the higher ground speed. Once sterilised conditions occur, thrust and pitch change to maintain the higher sink condition for a geometric glide path.

For the tailwind case, indeed the handling is more fun; the reducing tailwind in the TBL gives an inertial effect of an increase in IAS, which gives arose up moment from speed (alpha) stability), the GS reduces so the required VS for the geometric path then requires IAS to be stabilised, thrust initially comes off but then has to be set at slightly higher levels than that required prior to the shear.... and the stable conditions attained. It is an overshoot shear, and is one of the reasons that for high inertia aircraft like the 747/777/380 etc reference ground speed or GSmini was a nice briefing item, it gets a target in mind as to what is going to be the processes for the maintenance of a stable path.

The requirement is still to maintain a path, and to arrive at the flare point with the correct energy state, so my query stands. Are we saying that pilot training is inadequate to fly an aircraft within normal criteria for a stable approach?

In respect to this bingle, the LH MLG has impacted soft ground prior to the lip of the runway hard surface, which is commonly called landing short. If the premise is that in order to not land long on a landing with a tailwind, instead we ens up with landing short, effectively having a ramp strike, then we seem to have other problems. As a simple horizontal shear to cause an outcome like that in a closed loop control system would be an instantaneous loss of the 0.3 amount of the 1.3Vs that Is being applied on the reference speed, or a sudden/instantaneous loss of 25kts of tailwind component. That being the reason to have some inkling of GSmini (lé Busse), RGS (Billie Boing) etc... For low inertia aircraft, the effects are less pronounced, but mechanical turbulence can be more annoying.

Still, Fokker made a solid tube, plane handled that bingle rather well.

[Teddy Robinson]

Finally some sense !!
sticking the MLG into the sand short of the concrete will in most cases be detrimental to it's health.

[fdr]

You are correct that a 3 degree flight path with a fixed aiming point is fixed in space; I did not dispute that. However to maintain that with any wind, the variable is power. In a strong tailwind there will come a point where this is not possible as the power will be at idle. Of course, the way to maintain the flight path now is to point at the aiming point as the tailwind tries to put you above the profile. Should one arrive at the aiming point at the correct height, you will certainly be fast and….. you will land long. Professional pilots call this an unstable approach.

Me

Me; the effects of inertia means that IAS is affected unless the "shear" (change/time) changes are very slow rate, and/or the aircraft has low inertia. Within an airmass, the aircraft is moving but takes time to accelerate the inertial mass, so the GS does not instantaneously increase by 50kts in a sudden 50kt reducing headwind, instead, the IAS (CAS) alters immediately, it is not subject to inertia, and the ground speed alters rapidly but not instantaneously until reaching equilibrium (turning a B747 to the NE over Kushimoto with a 250kt jet stream, where the component would go from a 100kt head to about 250 tail was always worth a bet for the first beer in Narita, a 350kt shear in 45 seconds was always worth watching). Slow rates, no problem light planes, no problem.

Re "power is the variable", that is not the issue with most reverse shear cases. The attitude has to change as the first effect of the drop off of a tailwind is an increase in IAS, that causes a pitch up if open loop control due to speed stability. That acts to reduce sink rate, while at the same time, the rapidly increasing ground speed now necessitates an increase in sink rate, so the immediate effect open loop is to go above glide slope. Closed loop, the pilot has to add a nose down input to avoid the balloon from the additional IAS increment, which also needs further adjustment due to the increased sink rate requirement over time as GS increases to equilibrium condition. The increased speed requires a power reduction, (first power change, more to come). As the conditions stabilise, the IAS starts to reduce other than the increment that occurs from the lower attitude required for the higher sink rate, which then requires an addition of thrust to stabilise back at the equilibrium condition, which hopefully is back at Vref + additive, and with the power slightly lower than prior to shear entry, and with a slightly lower attitude to maintain path. The problems arise when the crew carry excess speed into a reverse shear, and cannot get the aircraft to return in time to stable conditions in order to be stable into the flare. Stable here is still within the generic requirements for a stable approach that companies will specify in their OM's.... The routine problem from the investigators view is to determine whether the crews actions were within normal boundaries for the conditions. Carrying excess speed into a reverse shear is a headline that the crew are not cognisant of the dynamics that will occur (the opposite is true for a normal shear event). A ballooning out of the slot during transition of a reverse shear is indicative of the crew not adjusting the attitude for the speed change, excess IAS correlates to inadequate thrust reduction during the transition; low IAS/energy into the flare is indicative of failure to restore thrust before the flare... A factor that often shows up on hard landing investigation is that the crew take the power off due to the excess IAS, and then do not add it back to somewhere near a normal thrust level before entry into the flare...

All of this takes more time to say than do.... but the sequences of actions are constants, and show deviations from desired controlled state.

Tin hat time.... the B767 that was climbing at idle as it entered a spectacular shear from the jet wind... climbing at MMO with speed brake out and thrust at idle. Impressive.
Or the B737-800W that missed its level off by 4000' going up, with a 1.72g pull up, a 0.2g over the top, and the subsequent 1.7g pull to level off at the assigned altitude. All at idle and with the speed brake out, at MMO. Neat jet stream.

All of the above barely rises above the point of boredom, but if the driver starts briefing about adding speed for mum and the kids for an expected reverse shear, then it is apparent they don't comprehend what the aircraft does in shears. We get the unfortunate lot to be presented with interesting conditions that appreciate preparation, and occasionally a rational "er, no thanks, I'll take the other runway...". As a group, we routinely try harder than we should to accomodate lunacy arising from bureaucracy, and the outcome occasionally is a desire to have a repeat of the last 10 seconds again while sitting in the wreckage.

This may seem like teaching to suck eggs, but the number of pretty shambolic approaches that get analysed suggests that we can occasionally do with clearing our thoughts. For the non flying driver, it is even more important that he or she understands what is going to happen next. The side kick doesn't have the input of the control feel to include in the process so can only rely on the performance gauges and comprehension of "plain fiziks" to be inside of the loop in a meaningful manner. In the really ugly events that come across the desk, the data shows frequently around 4-5 seconds of prior deterioration of flight kinematics before the day gets mussed up. (slightly less I guess for the Mega Deaf II, if I recall correctly).

The wise words of Mme 'Bussé and Mr Boing for all of their shiny toys invariably indicate that thrust is one of the factors in determining whether stable or not, and almost every nasty landing that I have had to analyse had the thrust well away from a normal fist position crossing the threshold. Being at idle already or TOGA thrust at that point may result in expenses. (results may vary...:\)

[ItsonlyMeagain]

Well, apart from tin hat, boredom and suck eggs, I didn’t understand any of that. However, I now know how to ask why in an additional 10 languages.

I have always found teaching people in language and at a rate they could understand produces sound results.

I guess none of your explanations were going through the pilot’s head as he hit the undershoot.

Me

[Teddy Robinson]

I have flown into Mogadishu on many many occasions, and certainly it can be interesting.

Tailwind is either within limits, or out of limits in which case the flight diverts, or is cancelled.

The runway is very long, which rather begs the question what an F50 was doing with its MLG

in the undershoot of a very long runway on a pretty average day?

It makes no sense, these things were daily in and out of LCY, without issues.


TR

[WideScreen]

I have flown into Mogadishu on many many occasions, and certainly it can be interesting.

Tailwind is either within limits, or out of limits in which case the flight diverts, or is cancelled.

The runway is very long, which rather begs the question what an F50 was doing with its MLG

in the undershoot of a very long runway on a pretty average day?

It makes no sense, these things were daily in and out of LCY, without issues.


TR
Sorry, I oversaw the base leg approach with short final for MGQ.

I don't think, this was an "undershoot".

When there is a landing tailwind of (for example) 20kts, and one does approach on the base leg with 90kts (both airspeed and ground speed), as fdr writes, you'll need military jet performance to pick up the kinetic energy to stay on 90kts airspeed and reach 110kts ground speed. Don't have an acceleration and your airspeed will effectively become 70kts, and (you'll) subsequently drop out of the sky (or at least your margins get tight).

Not to say, because of the (tight) turn to final, your stall speed goes up significantly, the decreasing from 90kts airspeed is no longer enough to fly and your inside turn wing will start to drop. That'll start to happen somewhere half-way your turn to final. Depending on the speeds/situation, you end up nose-down, or just somewhere halfway flipping over and ending up trying to land on the wingtip. Depending on how far your flip did go (before or already through the vertical), you end up with just scraping the wingtip, an extreme hard landing on one MLG (breaking off the whole wing on that side) or just landing upside down. When your wing breaks off, you'll end up upside down, since the other wing still wants to fly (creating a fuselage rotating force, even when the wing is vertical).

The final location off the right side of the beginning of the runway corresponds with the flying direction when the in-turn wing starts to stall somewhere during the turn to final. For this situation, it looks like the wing broke off due to an extreme hard landing, with a subsequent flip-over.

So, no landing gear issues, or an undershoot or so, just a (marginal) in-turn stall. Which fortunately left the fuselage intact and all survived.

[Teddy Robinson]

There is a video taken from adjacent to the runway 05 threshold.
Enter one F50 right to left, left wing down, nose pitched slightly up.
Left main gear strikes start of prepared surface followed immediately by failure of the left wing structure.
I would call that fairly conclusive.

Missed approach for 05 is a gentle right turn to avoid overflying the port area to rejoin either the visual circuit, or RNP to visual depending on other traffic and ATC.

TR

[megan]

Why can't you be bothered to provide a link? The only one I could find.

https://www.youtube.com/watch?v=P6zT0SUauu4

https://twitter.com/i/status/1552300391597047808

[Teddy Robinson]

Sorry, I was rather busy today.

The second of those clips is taken from the area just north of the 05 threshold
This is the clip that I am referring to.

Note: the aircraft has touched down hard in the undershoot, by 3rd second of the clip the wing has failed and the left engine is pointing towards the tarmac .... and the rest as they say is history.

[hans brinker]

Sorry, I oversaw the base leg approach with short final for MGQ.

I don't think, this was an "undershoot".

When there is a landing tailwind of (for example) 20kts, and one does approach on the base leg with 90kts (both airspeed and ground speed), as fdr writes, you'll need military jet performance to pick up the kinetic energy to stay on 90kts airspeed and reach 110kts ground speed. Don't have an acceleration and your airspeed will effectively become 70kts, and (you'll) subsequently drop out of the sky (or at least your margins get tight).

Not to say, because of the (tight) turn to final, your stall speed goes up significantly, the decreasing from 90kts airspeed is no longer enough to fly and your inside turn wing will start to drop. That'll start to happen somewhere half-way your turn to final. Depending on the speeds/situation, you end up nose-down, or just somewhere halfway flipping over and ending up trying to land on the wingtip. Depending on how far your flip did go (before or already through the vertical), you end up with just scraping the wingtip, an extreme hard landing on one MLG (breaking off the whole wing on that side) or just landing upside down. When your wing breaks off, you'll end up upside down, since the other wing still wants to fly (creating a fuselage rotating force, even when the wing is vertical).

The final location off the right side of the beginning of the runway corresponds with the flying direction when the in-turn wing starts to stall somewhere during the turn to final. For this situation, it looks like the wing broke off due to an extreme hard landing, with a subsequent flip-over.

So, no landing gear issues, or an undershoot or so, just a (marginal) in-turn stall. Which fortunately left the fuselage intact and all survived.

I'm sorry, WHAT?? Airplanes do not fly relative to the ground, they fly relative to the air. If you fly an airplane at 60kts, against a 60kts headwind, you will be stationary over the ground, but you will not stall. If you make a 180 degree turn your groundspeed will now be 120 kts, but you will not exceed any aircraft speed limit. This applies in the patterns as well. If you are on the base leg and you have a headwind for landing, your turn to final will be a bit less than 90 degrees, due to the crab angle on base. Conversely, if you are landing with a tailwind your turn to final will be bigger that 90 degrees. If your approach speed is 90 kts, and you turn final in either cease, your indicated speed will (should) stay at 90 kts. Your ground speed will go up for a tailwind and down for a headwind, but your airplane wont realize that, because it's only reference is the air around it. I cannot believe there are pilots that think aerodynamics are affected by ground speed. Yes, landing distances are. Yes turn radiuses across the ground are. YOU WILL NOT LOSE AIRSPEED TURNING INTO A "TAILWIND".

[172_driver]

The downwind turn tale die slowly ;)

[WideScreen]

I'm sorry, WHAT?? Airplanes do not fly relative to the ground, they fly relative to the air. If you fly an airplane at 60kts, against a 60kts headwind, you will be stationary over the ground, but you will not stall. If you make a 180 degree turn your groundspeed will now be 120 kts, but you will not exceed any aircraft speed limit. This applies in the patterns as well. If you are on the base leg and you have a headwind for landing, your turn to final will be a bit less than 90 degrees, due to the crab angle on base. Conversely, if you are landing with a tailwind your turn to final will be bigger that 90 degrees. If your approach speed is 90 kts, and you turn final in either cease, your indicated speed will (should) stay at 90 kts. Your ground speed will go up for a tailwind and down for a headwind, but your airplane wont realize that, because it's only reference is the air around it. I cannot believe there are pilots that think aerodynamics are affected by ground speed. Yes, landing distances are. Yes turn radiuses across the ground are. YOU WILL NOT LOSE AIRSPEED TURNING INTO A "TAILWIND".
Maybe read again, what I wrote.....

You think in speed. That's the wrong approach. You need to think in Kinetic Energy, and that is ground reference based. When you turn final with Kinetic Energy Ek, it represents a ground speed corresponding to that Ek. And you have the same Ek on your base leg, with the same ground speed. Kinetic Energy doesn't know about airspeed. Only wings know about airspeed.

IF you need a higher ground speed to keep flying (because of a tailwind), you will need to add Kinetic Energy. If you don't do that, by pointing the nose down (trading in potential energy) and/or use engine power (trading in chemical energy), you are loosing airspeed. The extra Kinetic Energy does come from somewhere. Really.

[WideScreen]

Why can't you be bothered to provide a link? The only one I could find.

https://www.youtube.com/watch?v=P6zT0SUauu4

https://twitter.com/i/status/1552300391597047808
Looks to me, the video shows happening what I described. Though just at the end of the turn to final, where I expected this to happen a little earlier, more at 2/3 of the turn to final.

At the very first few seconds of this video, the F50 seems to be in a shallow bank to the left at the end of a turn to final. And then suddenly the F50 starts falling out of the sky, just mushing down with wobbly wings.

The turn to final with lots of tailwind on landing is really dangerous. Don't believe those fairy tales about "not loosing airspeed, when turning final". It needs positive action to avoid losing airspeed.

(And the same applies for a quick turn-out to cross-wind when taking off with a significant headwind, before you know, you are upside down.)

[fdr]

Megan's first video is looking in the opposite direction of the approach, and shows the aircraft was quite stable, wings level for some time before the flight path deteriorated. There is not enough resolution to have any idea of the speed, but she sure does plummet below slope there towards the end of the flight.

Hans Brinker: I cannot believe there are pilots that think aerodynamics are affected by ground speed. Yes, landing distances are. Yes turn radiuses across the ground are. YOU WILL NOT LOSE AIRSPEED TURNING INTO A "TAILWIND".

Hans... we get taught to fly light aircraft, and do stuff like ground reference manouevers in what are generally light aircraft, low inertia. Where inertia is considered to be zero, there would be no such thing as wind shear... as you say the plane only sees the air. The only difference that would arise would be a change in the flight path angle, FPA, as whatever the vertical speed, VS is relative to Ground Speed, GS, would change, FPA being VS/GS. An aircraft with inertia cannot suddenly accelerate the inertial mass instantaneously from one kinetic energy state to another, except on star trek maybe.

If you are turning into an increasing tailwind in a high inertia aircraft, you will see CAS sag and the ATR system compensate up to MCT to maintain the commanded speed. In a Boeing, that will be seen as the thrust levers moving rapidly forward, and then once the rate of change of wind (shear; dv/dt) has stabilized, and on target speed, the thrust will rapidly recover to the pre-turn levels. In an airbus, you get to see the command arcs move, the thrust values go up, then everything goes back to normal, and the crew pokemon game is undisturbed by the Boeing racket. On a B737, you also of course get the rattle of the trim system for the speed variations (there is also usually a slight trim change to maintain altitude even in calm conditions, so a clack or two is pretty normal in an extended turn on a B737).

There are two general types of wind shears that we mess about with, that related to microbursts which drivers tend to avoid wherever possible, and the common or garden horizonal shear. (There are also thermal shears, stock shares etc... ) Microbursts provide vertical and horizontal shear components that get pretty nasty quickly, and drivers tend to avoid them where possible. Horizontal shears happen every day at all altitudes, and arise from the component change to the direction of motion of the aircraft. That can be a velocity change or a vector change. If the component changes at the aircraft, there is a shear occurring.

Shear is time dependent. The driver can alter that in most cases by controlling the mode that the aircraft is controlling to. For vertical shears, (and with the ususal caveats that the driver is responsible at all times to remain within limitations and to follow company policy), traversing a vertical shear can be easily managed by using a VS mode... a nice simple 500FPM up or down is pretty darn difficult to bust a limit, or to stall the plane (but not impossible... ) a shear of 100kts in 4000' vertical is entertaining in a speed mode for climb or descent, such as FLCH, dropping out of the wind, with a reducing headwind component in FLCH will cause the plane to increase the sink rate with a pitch down to maintain speed, and the steady state sink rate will rapidly dissapear. Most memorable data I can recollect of that was a B772ER going through a 180 kt jet core, from above in a descent. Guys started the descent with the speed brake out, and the tailwind increased as they entered the core from a lower tailwind condition (taking advantage of a lower tailwind on the day.... ) so the tail wind rapidly increased, and that led to a rapid pitch down and higher sink rate... that then led to entering the airmass under the core which had a reducing tailwind component with a sink rate of greater than 12,000FPM, and a major pitch up and overspeed as the APFD wasn't nearly capable of keeping up with the rates that the driver had set up... apparently was spectacular from the inside. had the descent been done in VS at a modest rate, no ones beer would have spilled, even the pilots beer. In a modest VS, the trust would have changed to maintain the speed, and the VS would have controlled pitch. The VS compared to shear rate would have kept it boring, and boring is good. This is the reverse sort of event to the B767 commented on by me before in the climb...

For the ground reference cases, if you are zipping about the cans at Reno, even around 380CAS the rate of turn is relatively low and the winds are almost always light, so there is not much shear, and the planes are not very heavy... Doing that even in a B747 at VMO while fun would probably not give a particularly high shear rate, but it would get tea and bikkies with the boss for the overstress, or with RARA for disqualification for running off the course.

Biggest horizontal shear I ever came across was going into a VFR forecast airport, in a big twin, and noting at about 20 miles to run for a gentle turn to final that we had a northerly wind of 56kts, and the reported wind on the ground was 36kts from the south. Asked the FO who was PF to slow down early as we were going to get a pretty good shear at some point, so we configured to approach flaps, and at 1200'AGL, we still had 55kts tailwind, and 35kts in the opposite direction on the ground. Neither the ground or the upper wind were on the forecast. At 400ft the wind started to change, and the FO blew straight out of the slot with the flaps load relief getting a fair work out. As we had min fuel and no alternate, it became fun to do the final approach. It could be done, but it took controlling the rate of change to make it manageable and to land with reduced flaps. All other inbounds were diverted for the next 4 hours, and we would have too, had we had gas.

If inertia wasn't a factor, shear would not do anything other than change the VS to maintain a path, the wrecks on finals dotted around the world suggest that inertia and kinetic energy is indeed a factor, irrespective of what physics we train pilots in, akin to teaching Bernoulli's theory to explain why wings work. The oracle Google gives a number of popular science level descriptions that are completely wrong but follow the guidance that pilots get encumbered with at the start of their flying career. When DFDR data is looked at, the inertial effects can be seen in the data relatively clearly.

If this is surprising, then you should try screening out coriolis effects on jet aircraft performance data, or the variation in SAR that occurs when flying from one airmass to another. In the old days that was a pain, but with GPS geometric altitude that becomes easy to resolve, and actually ends up giving some interesting info on the actual weight/drag count, with W/delta still being a factor in the background.

If this sound pedantic or boorish, a disproportionate amount of time in DFDR QAR analysis gets used up educating ops managers that the crew were in conditions that caused changes to the flight path, and that the problem is not one of compliance it is one of awareness. That is about 37 years on that subject and it seems to have no end in sight.

Note, the reactive windshear alert algorithm is the F factor:

F = Wh G - Vd As

where:
Wh is the rate of airspeed loss in kts per seconds,
g is gravity,
Vd is the vertical down draft rate, and
As is the aircraft airspeed.

The Wh/g term represents the rate of change in the horizontal winds. The Vd/As represents a measure of down draft strength. A way to think of F-factor is the rate of removal of energy from the aircraft.

Back in the day, I suggested to the design team that the reactive alert include a +/- value rather than just a minus, as what goes up usually goes down... The predictive alert came in soon enough to be a better bet.

Stated another way...:

given the equations of motion of:
https://cimg8.ibsrv.net/gimg/pprune.org-vbulletin/1188x770/screen_shot_2022_08_02_at_12_37_34_pm_7c68de4a045b0012707288 9196773057e0836eaa.png

(refer the model below)





The specific energy Es becomes:

https://cimg1.ibsrv.net/gimg/pprune.org-vbulletin/426x146/screen_shot_2022_08_02_at_12_35_46_pm_556f4afbe8b06b60ed94fc 591aa932654150497b.png

and can be expanded to:
https://cimg3.ibsrv.net/gimg/pprune.org-vbulletin/1066x186/screen_shot_2022_08_02_at_12_36_11_pm_0db4d7dc374b75f20a331c 30e884d79317eec1f0.png

The first component being the airplanes specific excess power, so the rest is the wind effects which is a restated F factor:

https://cimg5.ibsrv.net/gimg/pprune.org-vbulletin/684x152/screen_shot_2022_08_02_at_12_36_19_pm_fd73759bd429e69c97df54 995ac7f1375527432a.png


https://cimg0.ibsrv.net/gimg/pprune.org-vbulletin/996x544/screen_shot_2022_08_02_at_12_39_14_pm_d85b4d7c8bab856738bf49 5887a54c2530f78738.png



So, in the end, there are triggers for, vertical rates that arise from the vertical shear, and dCAS that comes from the horizontal shear, both items affecting the specific energy, Es, of the plane.


PS: if you want to quantify the rate of shear that your plane can encounter and not have a bit of fun, you can sit in the sim and try from a steady state in whatever configuration that you have to go to full thrust, and record the acceleration rate that you achieve. The sim will normally be within 10% or so of the specific excess thrust available, which is the thrust avaliable minus the drag. The drag curves are able to be determined to be representative by looking at the acceleration rates and rate of climb for the same weights, vs the aircraft, and the same for the approach etc, but when without that, it will be ruffly +/-10% or better in most sims that have passed QTGs.

A B747-200B has about 3.6kts/sec in landing configuration on a 3 degree slope.... so for that case, a change of component in excess of 3.6kts will result in a change of CAS. Of course the overshoot shear gets to add CAS initially, and then the driver has to deal with the speed trim change, thrust trim change, balloon, pitch, and then get the toy back to stable. Which is why the DFDR and QAR data gets to be interesting to look at when the boys have had a wobbly day out. Transition of a shear can be drawn happily on a white board, and that is the only point of this long winded comment, if the drivers can't draw the pitch, speed, thrust, trim, dGS, attitude, AOA changes that occur then there is a gap in the education, and that is systemic, not an individual failing.

Suggested reading:

Townsend, J., Low-Altitude Wind Shear and Its Hazard to Aviation, National Academy, Washington, DC, 1983.
Hinton, D. A., "Flight Management Strategies for Escape from Microburst Encounters," NASA TM-4057, Aug. 1988.
Bowles, R. L., "Reducing Windshear Risk Through Airborne Systems Technology," presented at the 17th Congress of the Intl. Council of the Aeronautical Sciences, Stockholm, Sweden, Sept. 1990.
Kupcis, E. A., "Manually Flown Windshear Recovery Technique," Proceedings of the 29th Conference on Decision and Control, Honolulu, HI Dec. 1990, pp. 758, 759.
Miele, A., "Optimal Trajectories and Guidance Trajectories for Aircraft Flight Through Windshezis" Proceedings of the 29th Conference on Decision and Control, Honolulu, HI, Dec. 1990, pp. 737-746.
Psiaki, M. L., and Stengel, R. F., "Analysis of Aircraft Control Strategies for Microburst Encounter," Journal of Guidance, Control, and Dynamics Vol. 8, No. 5, 1985, pp. 553-559
Psiaki, M. L., "Control of Flight Through Microburst Wind Shear Using Deterministic Trajectory Optimization," Ph.D. Dissertation, Princeton Univ., Princeton, NJ, 1987 (Rept. No. 1787-T)
Zhao, Y, and Bryson, A. E., "Optimal Paths Through Downbursts," Journal of Guidance, Control, and Dynamics, Vol. 13, No. 5,1990, pp. 813- 818.
Miele, A., Wang, T., and Melvin, W, "Guidance Strategies for NearOptimum Takeoff Performance in Wind Shear," Journal of Optimization Theory and Applications, Vol. 50, No. 1, 1986, pp. 1-47.
Miele, A., Wang, T., and Melvin, W., "Optimization and Gamma/Theta Guidance of Plight Trajectories in a Windshear," presented at the 15th Congress of the Intl. Council of the Aeronautical Sciences, London Sept 1986.
Morton, B. G., Elgersma, M. R., Harvey, C, and Hines, G., "Nonlinear Plying Quality Parameters Based on Dynamic Inversion," Honeywell Systems and Research Center, AFWAL-TR-87-3079, Minneapolis MN Oct 1987.
Lane, S., and Stengel, R. R, "Flight Control Using Non-Linear Inverse Dynamics," Automatica, Vol. 24, No. 4, 1988, pp. 471-483.
Menon, P. K., Badgett, M., and Walker, R., "Nonlinear Flight Test Controllers for Aircraft," Journal of Guidance, Control, and Dynamics, Vol. 10, No. 1,1987.
Meyer, G., and Cicolani, L., "Application of Nonlinear System Inverses to Automatic Flight Control Designs: System Concepts and Flight Evaluations," Theory and Application of Optimal Control in Aerospace Systems, AGARD, AG251, pp. 10.1-10.29.
Elgersma, M., and Morton, B., "Partial Inversion of Noninvertible Nonlinear Aircraft Models," Honeywell Systems Research Center, Minneapolis, MN, Aug. 1989.
Frost, W, and Bowles, R., "Wind Shear Terms in the Equations of Aircraft Motion" Journal of Aircraft, Vol. 21, No. 11, 1984, pp. 866-872
Stengel, R. P., "Course Notes for MAE 566: Aircraft Dynamics," Princeton Univ., Princeton, NJ, Jan. 1990.
Singh, S. N, and Rugh, W. J., "Decoupling in a Class of Nonlinear Systems by State Feedback," ASME Journal of Dynamic Systems, Measurement, and Control, Series G, Vol. 94, Dec. 1972, pp. 323-329.
Etkin, B., Dynamics of Atmospheric Flight, Wiley, New York, 1972
Mulgund, S. S., and Stengel, R. F, "Optimal Nonlinear Estimation for Aircraft Flight Control in Wind Shear," Proceedings of the 1994 Congress of the International Council ofAstronautical Sciences, Anaheim, CA, 1994, pp. 1747-1755.
Greene, R.A., The Effects of Low-Level Wind Shear on the Approach and Go-Around Performance of a Landing Jet Aircraft, SAE Transactions
Vol. 88, Section 3: 790527–790858 (1979), pp. 2009-2015 (7 pages)
Mulgund S.S. & Stengel, R.F.; Aircraft Flight Control in Wind Shear Using Sequential Dynamic Inversion, J. Guidance, Control & Dynamics, Vol 18, No. 5, September-October 1995.

and...

Ostroff, A. J., et al., "Evaluation of a Total Energy-Rate Sensor on a Transport Airplane," NASA TP-2212,1983.

[hans brinker]

Maybe read again, what I wrote.....

You think in speed. That's the wrong approach. You need to think in Kinetic Energy, and that is ground reference based. When you turn final with Kinetic Energy Ek, it represents a ground speed corresponding to that Ek. And you have the same Ek on your base leg, with the same ground speed. Kinetic Energy doesn't know about airspeed. Only wings know about airspeed.

IF you need a higher ground speed to keep flying (because of a tailwind), you will need to add Kinetic Energy. If you don't do that, by pointing the nose down (trading in potential energy) and/or use engine power (trading in chemical energy), you are loosing airspeed. The extra Kinetic Energy does come from somewhere. Really.

I really hope you are not a pilot. You have absolutely no idea what you are talking about. Kinetic energy is not ground reference based.
When an airplane is flying around in a constant bank it will make a circle relative to the surrounding air. If that air moves across the surface of the earth the pattern across the ground will look like curly hair (sorry, don’t want to start looking for pictures). Whatever the wind is, in the airplane you will not feel any difference between turning into the wind or out of the wind. That is because your frame of reference is the air. It’s like a fly in a train. It’s not flying at a 100 mph if the train is going a 100 mph. I was going to say more, but I ran out of potential energy. (facepalm)

[hans brinker]

Hans... we get taught to fly light aircraft, and do stuff like ground reference manouevers in what are generally light aircraft, low inertia. Where inertia is considered to be zero, there would be no such thing as wind shear... as you say the plane only sees the air...

Ostroff, A. J., et al., "Evaluation of a Total Energy-Rate Sensor on a Transport Airplane," NASA TP-2212,1983.

I want to clarify my statement about ground speed vs airspeed and kinetic energy was in reference to a steady wind. I was trying to keep it simple enough for widescreen, but I think that might be an impossible goal. Obviously if you descend/climb into a different wind direction or speed it will affect the performance until a new equilibrium has been reached. Even in light planes inertia isn’t zero, but that has nothing to do with the assertion that turning to final if there’s a steady state tailwind would require a power change.

[hans brinker]

Looks to me, the video shows happening what I described. Though just at the end of the turn to final, where I expected this to happen a little earlier, more at 2/3 of the turn to final.

At the very first few seconds of this video, the F50 seems to be in a shallow bank to the left at the end of a turn to final. And then suddenly the F50 starts falling out of the sky, just mushing down with wobbly wings.

The turn to final with lots of tailwind on landing is really dangerous. Don't believe those fairy tales about "not loosing airspeed, when turning final". It needs positive action to avoid losing airspeed.

(And the same applies for a quick turn-out to cross-wind when taking off with a significant headwind, before you know, you are upside down.)

So you think that in cruise, with a steady wind from the left and no power changes, if you turn left your airspeed will increase, and if you turn right it will decrease?

[fdr]

I want to clarify my statement about ground speed vs airspeed and kinetic energy was in reference to a steady wind. I was trying to keep it simple enough for widescreen, but I think that might be an impossible goal. Obviously if you descend/climb into a different wind direction or speed it will affect the performance until a new equilibrium has been reached. Even in light planes inertia isn’t zero, but that has nothing to do with the assertion that turning to final if there’s a steady state tailwind would require a power change.

Hans, Widescreen happens to be correct.

A change of vector is a shear. Period. If you change your track relative to a steady wind, you are introducing a shear through the change in vector, your inertial mass has to accelerate or decelerate. That shows up as an airspeed change. There is a lag in the change of GS for an instantaneous change, and then it is simply sematics.

[fdr]

So you think that in cruise, with a steady wind from the left and no power changes, if you turn left your airspeed will increase, and if you turn right it will decrease?

If your aircraft has a finite inertial mass, and there is a rate of change, then, yes.

Now, you can go on line to a web page like this one: https://www.planeandpilotmag.com/article/downwind-turn/
and amongst the pretty pictures you will note a statement that is made that is the logical limit of the problem, and yet, the wrong conclusion is made from the limit case.
If indeed you turned 180 degrees from a 50 kt headwind to a 50 kt tailwind and you started at, say 50kts, you would indeed have a problem, you have just had a 100kt instantaneous shear occur, so how does the aircraft, which includes tin, plastic, protoplasm and gas, e.g., has mass, how does that go from the zero ground speed to 100kt ground speed in zero time? That would be an acceleration required of....:
(50m/sec)/0.5= 100m/sec,
which is a shade over 10g horizontal. In the same case, the C-172 or Piper Cub can accelerate at.... around 0.15g longitudinally. A lear gets about 0.2g with TFEs, a B773W gets about 0.18g... An F-18 gets just around 1.0g, an F-16 light gets about... 1.1g down in the weeds. roughly.

For most flight training, the ground reference case is not done pulling 2g in a steep turn, and with a 50kt wind, but if it did, you would still be looking at a modest shear rate... at 100kts CAS, you would take 9.8 seconds to do the 180, and meet a shear of 100 kts, so that is about 10kts/sec, which is going to need an inertial acceleration of 0.5g. That would actually exceed most aircraft out there. take the normal ground reference cases at a 25-degree bank, and that equates to rate 2, or just on 35 seconds for the same 180 turn, and that gives just on 3 kts shear per second, a relatively trivial amount, which you would see as a need for a bit of thrust change, as in... 0.15g longitudinal, well within a Lycoming or Conti's capability. And that is with a 50 kt wind... Most training is done in winds around 15-20kts for the same case, and the effect is perpetuating the position that planes. have no inertial effects. (15kts... 0.05g, etc... barely enough to set off an entry interface for the space shuttle).

Why this is a factor mainly at high altitude or with rapid wind shear is where there is a considerable rate of change, limited excess thrust, and high inertia. It is observable with 200kt jetstreams, or high vertical shear rates when climbing and descending at high rates of VS. It also is why losing a 30kt headwind on an approach will result in an instantaneous loss of airspeed, and a delay as the GS increases....

If you need more background, you can PM me, and I will forward some reading for you.

[DaveReidUK]

Newton has a lot to answer for ...

[megan]

You chaps are confusing this humble helo driver, it used to occur that it was found necessary to hold in 70 kt winds occasionally, max endurance was 74 kt, turning downwind was no different than had there been zero wind. Close the eyes and there was no difference, visual experience was dramatic though.

[fdr]

Newton has a lot to answer for ...

"Give me leave, Sir, to insinuate that I cannot think it effectual for determining truth to examine the several ways by which phaenomena may be explained, unless where there can be a perfect enumeration of all those ways. You know that the proper method for inquiring after the property of things is to deduce them from experiments. And I told you that the theory that I proposed wa[s] evinced to me, not by inferring it is thus because not otherwise, that is, not by deducing it only from confutation of contrary suppositions, but by deriving it from experiments concluding positively and directly. The way therefore to examine it is by considering, whether the experiments which I proposed do prove those parts of the theory to which they are applied; or by prosecuting other experiments that the theory may suggest for its examination. And this I would have done in a due method...."

in "A Series of Quaeries Proposed by Mr. Isaac Newton to be determined by Experiments, positively and directly concluding his new Theory of Light and Colours, imparted to the Editor in a Letter of the said Mr. NEWTON'S, of July 8, 1672. No. 85 p. 5004. In The Philosophical Transactions of the Royal Society, London, from their commencement in 1665 to the year 1800 (abridged) Vol 1 p. 734

English was a second language to the pommies in 1672 apparently...

[EXDAC]

A change of vector is a shear. Period. If you change your track relative to a steady wind, you are introducing a shear through the change in vector, your inertial mass has to accelerate or decelerate. That shows up as an airspeed change.

Isn't is true that "track" is ground referenced? How is a change of track (ground referenced) of any significance at all to the movement of the aircraft in a uniform airmass (airmass referenced).

[hans brinker]

Imagine swinging a weight attached to a rope around in around in a horizontal circle. It's swung at such a speed the angle of the rope is about 60 degrees from vertical. The ball experiences gravity and a put from the string. If you sat on the ball you would feel twice as heavy.
Let's assume we make the rope long enough that the ball moves at 60 miles per hour, and we swing the ball on a truck that moves at 60 miles per hour. If you stand on the truck the ball moves in a circle. To someone standing by the side of the road it .makes the same movement that would be made by a point on the wheel of the truck: ome to a stop, and leap forward, come to a stop and leap forward with dramatic changes in horizontal velocity and strong acceleration forces. The ball doesn't feel that, because it's frame of reference moves at that steady speed of 60mph. It still feels that same constant 2g pull from the rope.
Now change ball into airplane flying in circles, and truck into the air the plane is flying in.

[fdr]

Isn't is true that "track" is ground referenced? How is a change of track (ground referenced) of any significance at all to the movement of the aircraft in a uniform airmass (airmass referenced).

UNDERSTANDING WIND SHEAR

Definitions

Wind shear

Wind shear can be defined as a sudden change in wind velocity and/or direction over a short distance. It can occur in all directions, but for convenience, it is considered along vertical and horizontal axis, thus introducing the concepts of vertical and horizontal wind shear:

Vertical wind shear consists of wind variations along the vertical axis of typically 20 to 30 knots per 1000 ft. The change in velocity or direction can drastically alter the aircraft lift, indicated airspeed, and thrust requirements when climbing or descending through the wind shear layers.
Horizontal wind shear consists of variations in the wind component along the horizontal axis – e.g. decreasing headwind or increasing tailwind, or a shift from a headwind to a tailwind – of up to 100 knots per nautical mile. (fig.1) shows how a penetration would appear as an aircraft crosses a cold front.

This weather phenomenon can occur at many different levels of the atmosphere; however it is most dangerous at the lower levels, as a sudden loss of airspeed and altitude can occur.

It is usually associated with the following weather conditions: jet streams, mountain waves or temperature inversion layers, frontal surfaces, thunderstorms and convective clouds or microbursts, occurring close to the ground.


quote from Airbus.

the only matter of concern to the aircraft is the instantaneous rate of change of a component to the steady inertial state of the aircraft. If you suddenly change your direction is the same situation to the wind suddenly changing its direction.

Is it relevant for most training cases? not really, the rate of change is modest, so the inertial accelerations are not noticeable.

Flight in a Non-steady Atmosphere (https://www.sciencedirect.com/science/article/pii/B9780080982427000146)Michael V. Cook BSc, MSc, CEng, FRAeS, CMath, FIMA, in Flight Dynamics Principles (Third Edition) (https://www.sciencedirect.com/book/9780080982427/flight-dynamics-principles), 201314.3.2 Wind shearWind shear is defined as a time rate of change in wind speed and direction lasting for 10 seconds or more, the assumption being that shears of less than 10 seconds are unlikely to constitute a flying qualities hazard. See, for example Hoh et al. (1982), which includes a useful discussion of wind shear in the context of flying qualities requirements (https://www.sciencedirect.com/topics/engineering/flying-quality-requirement).

As for steady wind, wind shear presents a problem during take-off and landing and, when encountered, the aircraft will rise or sink according to the relative direction and magnitude of the wind velocity gradient (https://www.sciencedirect.com/topics/engineering/velocity-gradient). Since the effect of wind shear on the aircraft is to cause an upset in normal acceleration, the requirements quantify the wind shear limits in terms of acceleration response (https://www.sciencedirect.com/topics/engineering/acceleration-response) as given in Table 14.3, where subscript γmax is the maximum power climb angle and subscript γmin is the flight idle glide angle.


For this thread on the F-50 short landing, this discussion is irrelevant, however, for a flight around jet streams, it is pertinent.

By definition, shear is determined to be a change in a component wind, and for low-level flight that is normally due to wind gust, turbulent boundary layer effects, gust lines or convective phenomena. That is as taught. At altitude and with extreme jet streams, a rate of change of the heading is a change of wind component, the aircraft has inertia and that has to change and takes a finite time to do so. The speed will alter, and recorded flight data shows that occurring. It is at that point that the assumption that the aircraft's inertial frame of reference is only to the airmass has an issue. For the climb and descent through a jet stream, the same issue arises. If the aircraft was bound inertially only to the airmass, a change in the wind would not change the IAS, it would only change the ground speed. That is not observed in the flight data.

For basic teaching we can ignore the fact that Newton wrote stuff in 1665, and assume that the plane is tied only to the airmass, and ignore the fact that it's motion is referenced to "the aether", not the universe/planet. However, if you look at chapter 1, THE KINEMATICS AND DYNAMICS OF AIRCRAFT MOTION in Stevens et al, 2015 (https://catalogimages.wiley.com/images/db/pdf/0471371459.excerpt.pdf), you will note that the frame of reference isn't to cyclone Bob or Irma, it is to the earth, as is an INS, GPS (kind of). p. 3 of the section which is a riveting read, states:

Inertial Frame: a frame of reference in which Newton’s laws apply. Our best inertial approximation is probably a “helio-astronomic” frame in which the center of mass (cm) of the sun is a fixed point, and fixed directions are established by the normal to the plane of the ecliptic and the projection on that plane of certain stars that appear to be fixed in position.

INS, IRU's & ADIRU etc use accelerometer measurements added or subtracted to the platform alignment and original position georeferenced to the earth, and uses that to determine over time the velocity, and over time gives the distance shift which gives actual position. It doesn't need a wind input, it in fact gives that as a calculated output, from the component accelerations to the aircraft from wind components. The wings see IAS for stuff like V squares etc... the aircraft sees the sum of all fears accelerations.

As difficult as this may be to consider, the assumption that the aircraft miraculously is unbound to the earth's frame of reference would also suggest that as soon as the wheels came off the ground at the equator the aircraft would be doing 900 knots to the west, as apparently its inertial frame of reference is not to the earth. Most times that doesn't happen. For flying around the pattern you can disregard everything and just look out the window and enjoy the view, as the approximation that we are taught from day 1 suffices. That does not suffice at limit cases, where inertial changes are non-trivial.

The aircraft is immersed in an airmass that is tied to the planet and then to whatever reference you wish to take into consideration. The inertial moment of the aircraft is measurable to the earth as an inertial frame of reference. The airmass is tied to the external frame of reference itself.

Stevens B.L., Lewis, F.L., Johnson, E.N., (2015) Aircraft Control and Simulation: Dynamics, Controls Design, and Autonomous Systems (3rd edition), Wiley and Sons.

[EXDAC]

quote from Airbus.

Why do you introduce a discussion of wind shear when you previously were arguing the case of steady wind and I specifically stated "uniform airmass"?

How about keeping it simple to start with.

Case 1. Aircraft makes continuous circles at constant altitude with no change in thrust. The aircraft is in a completely uniform airmass that has no motion relative to the surface of the earth. What happens to indicated airspeed during the turn?

Case 2. Aircraft makes continuous circles at constant altitude with no change in thrust. The aircraft is in a completely uniform airmass that is moving with contant speed and direction relative to the surface of the earth. What happens to indicated airspeed during the turn?

[fdr]

Why do you introduce a discussion of wind shear when you previously were arguing the case of steady wind and I specifically stated "uniform airmass"?

How about keeping it simple to start with.

Case 1. Aircraft makes continuous circles at constant altitude with no change in thrust. The aircraft is in a completely uniform airmass that has no motion relative to the surface of the earth. What happens to indicated airspeed during the turn?

Case 2. Aircraft makes continuous circles at constant altitude with no change in thrust. The aircraft is in a completely uniform airmass that is moving with contant speed and direction relative to the surface of the earth. What happens to indicated airspeed during the turn?

Good point. Ordinarily, would assume that the cases are identical, our experience and teachings suggest the cases are identical, we would not see an observable difference, as kinetic energy is irrelevant. The kinetic energy changes considerably but the IAS would essentially remain constant. for a wind of 20kts for a C-172 the difference in kinetic energy states is equivalent to a B747 in a 200Kt jet stream going from a tailwind to a headwind. For the B747 case, the IAS change that occurs is around 6kts, which is equivalent to around 1 kt for the C-172. 6kts is ~ 0.01M and wakes up the ATR.

Is there a transfer function? I have no idea. Is the B747 case observable at all times, it is repeatable and it suggests that the assumption that we have on changes of heading in an airmass is a useful approximation rather than an absolute truth. It has no relevance at low speeds and low inertia, it is observable and recordable on aircraft in a turn in the steady state jetstream where the only change is the track of the aircraft. It is additionally reversibile, going from a tailwind to a headwind results in the opposite effect. Do we read an IAS accurately enough on a C-172? no. The B747, B777 B787 A330/A380 etc, yes.

Before the first time I saw the aircraft behave in a steady state 200Kt + jetstream, in level flight with >90 degree turn, I did not believe it would make any difference. The most spectacular case is coming into southern Japan, but it is not the only case. Discount cases where the track is traversing the core of the jet in such a manner that it results in a shear, the observable curiosity is for the case where the wind is the same entering and exiting the turn, so wind shear is removed as a factor for the change in speed and subsequent thrust required. This happens to be memorable as getting to MCT limit thrust at high altitude and with decaying airspeed is attention-getting.

Clear as mud.

[EXDAC]

Clear as mud.

Yes.

Case 1 - Does the airspeed change, Yes or No?
Case 2 - Does the airspeed change, Yes or No?

[hans brinker]

Yes.

Case 1 - Does the airspeed change, Yes or No?
Case 2 - Does the airspeed change, Yes or No?

hard to believe people don’t understand that.

[fdr]

Yes.

Case 1 - Does the airspeed change, Yes or No?
Case 2 - Does the airspeed change, Yes or No?
1: No
2: imperceptibly, but yes, it does change

[sablatnic]

1: no
2: no
In both instances you are circling in an air mass moving up to 900 Kts, a few Kts more or less won't matter.

[megan]

I would have expected a speed loss in a turn if power remains constant, due to the lift, the vector no longer being vertical, having to be increased by increase by an increase in AoA and resultant increase in drag. Never flown a big boy, just an expectation.

[EXDAC]

I would have expected a speed loss in a turn if power remains constant, due to the lift, the vector no longer being vertical, having to be increased by increase by an increase in AoA and resultant increase in drag. Never flown a big boy, just an expectation.

Both Case 1 and Case 2, as proposed by myself, specified continuous circles. The roll into the steady state turn is outside the condition being examined.

[EXDAC]

1: No
2: imperceptibly, but yes, it does change

I understand your argument that the earth referenced kinetic energy of the aircraft will change as it circles in an airmass that is moving in that same frame of reference. However, I don't yet accept that this change in kinetic energy influences the airspeed.

The aircraft is assumed to be circling in a completely uniform volume of air. Suppose we place a gas balloon in that same volume of air and it carries an observer. The aircraft circles the balloon. The motion of the aircraft and its knetic energy can be referenced to the observer. (The frame of reference for kinetic energy can be whatever we define it to be. It does not have to be earth referenced). The observer will see the aircraft circling with constant angular velocity and constant altitude. In the observer's frame of reference the aircraft energy does not change. There is no reason for the aircraft indicated airspeed to change.

Why would the presence of the balloon and its observer change the way indicated airspeed behaves?

[WideScreen]

Wow, what a confusion.

Actually, when you "turn" through a wind field/direction, you get "blown" away. This blowing away, will take care, the wind will increase your kinetic energy. Though, this is not instantly. Have a heavy airplane with a lot of wind, and it may take enough time to get a stall surprise.

On a base leg, you will need to turn a little into the wind, to avoid, your base leg (ground track) is not perpendicular to the runway. When you start turning, your airplane starts to get caught by the wind and accelerated in the direction of the wind. Be heavy and turn fast, the wind does not have sufficient time to accelerate the airplane and the airspeed will drop. Have a light airplane, and the wind will accelerate the airplane within a few seconds.

For the example to make 360 turns in a steady wind, it is not defined, whether the circle should be ground based or relative to the (moving) air. With a ground based circle, you will have a constant kinetic energy, though juggling with the power (and varying airspeed) to make it a circle on the ground.

With a moving air circle, your airspeed will remain the same, though because your ground speed will continuously vary, the kinetic energy will continuously vary. To accomplish this, you will need to juggle with the power setting. The moment you turn "into the wind", you will need to bleed off ground speed, to avoid your airspeed going up (IE you will need a little less engine power at that moment). And the other way around.

Oh, and I fly and a often windy airport, where multiple 360s due to other traffic are pretty common. And flying a nice 360 ground based requires a constant power setting / trim juggle. Often, the 360s tend to become ovals.

[WideScreen]

...... Kinetic energy is not ground reference based......
Ehhhhhhh, maybe back to school ?

[WideScreen]

You chaps are confusing this humble helo driver, it used to occur that it was found necessary to hold in 70 kt winds occasionally, max endurance was 74 kt, turning downwind was no different than had there been zero wind. Close the eyes and there was no difference, visual experience was dramatic though.
For a helo, it doesn't matter, you have 360 wings.

[WideScreen]

I would have expected a speed loss in a turn if power remains constant, due to the lift, the vector no longer being vertical, having to be increased by increase by an increase in AoA and resultant increase in drag. Never flown a big boy, just an expectation.
Yep, indeed, though that's a different chapter !

[DaveReidUK]

For the example to make 360 turns in a steady wind, it is not defined, whether the circle should be ground based or relative to the (moving) air.

Is it not assumed, when you perform a 360, that you will finish the manoeuvre at the same point in space as you started it at ?

[WideScreen]

Is it not assumed, when you perform a 360, that you will finish the manoeuvre at the same point in space as you started it at ?
For holding, etc, it's ground based (fortunately). But now we are discussing the wind effects on airspeed and kinetic energy, based on assumed manoeuvres.

[fdr]

Why would the presence of the balloon and its observer change the way indicated airspeed behaves?

Honestly, I have absolutely no idea other than a transfer function of kinetic energy, which goes against the teachings we get. I would not have believed that it does so, until seeing it first in airmasses over Japan in B747-200s, and even then, the observation is meaningless unless it works in reverse sense, and that also was observed. After that, I didn't need to pay for beer on Japanese trips very often.

The oddity was so curious I got the QAR data pulled to give some confidence that it wasn't observer bias. But if the temp and wind vector remains the same, yet your CAS sags out and the thrust comes on, then I can only assume that our teachings are an approximation of the real world.

If there was no inertia-related lag in the system, then there would be no such thing as a wind shear, wouldn't matter that the wind component dropped off 50 kts or not... or rose etc, as the aircraft would then be completely isolated from the outside world, and only referenced to the airmass, and it doesn't matter what the airmass does then. The reality is, it does indeed matter, and then I can only surmise that of the wind vector change causes a shear to be noted, then a vector change of the aircraft in the airmass would also cause a shear. There is only one case where the aircraft vector change really makes a difference and that is an extreme case, but still interesting, as if the plane is short of excess thrust, and enters a condition that is going to call for more than is available to maintain acceptable speeds. This is not an item that has only come up on beer bets, we had a number of aircraft that had speed excursions which resulted in an analysis as to why they ran out of puff.

[DaveReidUK]

For holding, etc, it's ground based (fortunately).

Then you appear to have answered your own question.

For the example to make 360 turns in a steady wind, it is not defined, whether the circle should be ground based or relative to the (moving) air.

[EXDAC]

Is it not assumed, when you perform a 360, that you will finish the manoeuvre at the same point in space as you started it at ?

If this is in reference to my Case 2, and if "space" refers to the airmass, then that it what was intended. The manouver is not ground referenced and the path over the ground will not be circular.

(Holding is ground referenced and has no relevance to Case 2.)

[hans brinker]

Is it not assumed, when you perform a 360, that you will finish the manoeuvre at the same point in space as you started it at ?

No. That's why when you perform "turns around a point" for your CFI your bank angle varies depending on wind speed, but a 360 is just a rate one turn for 2 minutes.

[hans brinker]

Honestly, I have absolutely no idea other than a transfer function of kinetic energy, which goes against the teachings we get. I would not have believed that it does so, until seeing it first in airmasses over Japan in B747-200s, and even then, the observation is meaningless unless it works in reverse sense, and that also was observed. After that, I didn't need to pay for beer on Japanese trips very often.

The oddity was so curious I got the QAR data pulled to give some confidence that it wasn't observer bias. But if the temp and wind vector remains the same, yet your CAS sags out and the thrust comes on, then I can only assume that our teachings are an approximation of the real world.

If there was no inertia-related lag in the system, then there would be no such thing as a wind shear, wouldn't matter that the wind component dropped off 50 kts or not... or rose etc, as the aircraft would then be completely isolated from the outside world, and only referenced to the airmass, and it doesn't matter what the airmass does then. The reality is, it does indeed matter, and then I can only surmise that of the wind vector change causes a shear to be noted, then a vector change of the aircraft in the airmass would also cause a shear. There is only one case where the aircraft vector change really makes a difference and that is an extreme case, but still interesting, as if the plane is short of excess thrust, and enters a condition that is going to call for more than is available to maintain acceptable speeds. This is not an item that has only come up on beer bets, we had a number of aircraft that had speed excursions which resulted in an analysis as to why they ran out of puff.

Everything you have said is based on changes in wind speed and that is not the question.

[hans brinker]

Ehhhhhhh, maybe back to school ?

A levels in highschool, studied physics. Show me your PhD

[hans brinker]

1: No
2: imperceptibly, but yes, it does change

No to both.

[hans brinker]

and I fly and a often windy airport, where multiple 360s due to other traffic are pretty common. And flying a nice 360 ground based requires a constant power setting / trim juggle. Often, the 360s tend to become ovals.

Ground referenced 360s require power adjustment because the airplane flies in reference to the air, thank you for proving my point.

For the love of flying, next time you are out flying do a few 360s on instruments only while there's a decent wind blowing. You will feel absolutely no difference during the turn regardless where the wind comes from. Your frame of reference will be the airmass you are flying in. It has a constant speed, no acceleration. You are in a constant turn so only feel the extra push into the seat. If you were right, it would be impossible to make a 180 with a TAS of 50 flying in a 100 kts wind without stalling/ripping the wings off.

[DaveReidUK]

Is it not assumed, when you perform a 360, that you will finish the manoeuvre at the same point in space as you started it at ?No. That's why when you perform "turns around a point" for your CFI your bank angle varies depending on wind speed, but a 360 is just a rate one turn for 2 minutes.

I think you've actually agreed with my point - though I could have phrased it better. i.e. "you should finish the manoeuvre at the same point in space as you started it at" (if performed correctly as you describe, that is).

[WideScreen]

A levels in highschool, studied physics. Show me your PhD
No paper show to you, though I do have a PhD in physics......

Maybe consider the following situation. We all know, that to physically "destroy an airplane" against a solid surface, we need to start with a sufficient amount of Potential + Kinetic energy (so, either height or speed, or a combination of that).

Assume, have an airplane (stall speed 35Kts) 1 mm above the runway, have a headwind of 40 Kts (Alaska, can happen) and an airspeed V of 40 Kts. Engine powered to keep that airspeed. Your ground speed will be zero. Your potential energy (relative to the runway) is very small and can be ignored.

Do the thought-experiment to snap turnoff the headwind as well as the engine. Of course, the airplane will immediately settle, with minimal mechanical movement, since it was not moving compared to the runway, even, the wheels will not start turning.

Will the airplane bleed off a Kinetic energy of 0 (IE effectively nothing happens) ?

Or will it bleed off a Kinetic energy of 0.5 * m * V(airspeed) ^2 and effectively destroy the airplane ? Given common sense tells us, the airplane will not be destroyed, the presumed amount of Kinetic energy has to go "somewhere". When not knowing where Energy goes, it is usually turned into heat. So, will the airplane (or the ground, or the -now no longer moving- air) heat up ?

The reverse of this experiment: Airplane on the runway. Have a headwind of 40 Kts, use the engine to keep the airplane steady on (!) the runway, wheels not moving. Your airspeed will show 40 Kts. To avoid, the airplane climbs above the runway, hold the elevator down, to stay on the ground. What is the Kinetic energy of the airplane ?

Do the same, by holding the brakes (instead of the engine). What airspeed will be shown ? What is the Kinetic energy of the airplane ?

Now, return to the first experiment. The airplane not 1 mm above the runway, though on the ground, with non-compressed struts. What is the Kinetic energy of the airplane ?

The same, though now with struts compressed. What is the Kinetic energy of the airplane ?

Next step. No headwind. No ground speed. With increasing headwind, the airspeed will go live, with the airplane stationary using brakes. What will be the Kinetic energy of the airplane ? What will happen when we snap-turn-off the headwind ? Assuming the airspeed will be the measure for the Kinetic energy, where will the Kinetic energy go, when the headwind turns off ?

Gives me the feeling to argue with "The Truth"........

[WideScreen]

Ground referenced 360s require power adjustment because the airplane flies in reference to the air, thank you for proving my point.

For the love of flying, next time you are out flying do a few 360s on instruments only while there's a decent wind blowing. You will feel absolutely no difference during the turn regardless where the wind comes from. Your frame of reference will be the airmass you are flying in. It has a constant speed, no acceleration. You are in a constant turn so only feel the extra push into the seat. If you were right, it would be impossible to make a 180 with a TAS of 50 flying in a 100 kts wind without stalling/ripping the wings off.
1 Nop, the moment you make your turns relative to the air, your ground speed (vector) will vary and as such, you will experience (minimal) accelerations.
2 The forces (except the inertial ones) on the airplane are relative to the air around the airplane. With 50 Kts airspeed in 100 Kts wind, this accelleration effect is no longer "marginal", though, because the 360s aren't snap turns, still relative mild forces.

You do forget, that when your airplane mass is not unlimited, the moving air will accelerate the airplane relative to the ground in the direction of the wind. With a light airplane and sufficiently large 360s, this is hardly noticeable, though it happens. And because that effect is happening, you will need to juggle the engine power. The effect is the most, when you turn the nose through the direction of the wind.

[RatherBeFlying]

Much of the preceeding commentary suffers from ignorance of the various aerodynamic frames of reference. Ground school textbooks are woefully deficient in explaining this.

For a modest introduction: Frames of Reference (https://jsbsim-team.github.io/jsbsim-reference-manual/mypages/user-manual-frames-of-reference/#:~:text=Stability%2C%20or%20%E2%80%9CAerodynamic%E2%80%9D%2 0Frame,-This%20frame%20is&text=The%20frame%2C%20named%20FA,completes%20the%20right%2Dh and%20system.)

[EXDAC]

No paper show to you, though I do have a PhD in physics......

What is the Kinetic energy of the airplane ?

I would submit that a question asking "What is the Kinetic energy of the airplane ?" would require a specification of the frame of reference being used.

An aircraft stationary on the ground and experiencing a headwind has no kinetic energy with respect to the ground. However, if referenced to the airmass, it has kinetic energy proportional to the square of the wind speed.

For those struggling with this concept - an internet search for "frame of reference kinetic energy" will find lots of discussions including university texts, video presentations, and discussion in physics speciality groups. I found nothing that defined "the earth" or "ground" as the only valid frame of reference for kinetic energy. I found many examples that say kinetic energy depends on the frame of reference.

[hans brinker]

No paper show to you, though I do have a PhD in physics......

Maybe consider the following situation. We all know, that to physically "destroy an airplane" against a solid surface, we need to start with a sufficient amount of Potential + Kinetic energy (so, either height or speed, or a combination of that).

Assume, have an airplane (stall speed 35Kts) 1 mm above the runway, have a headwind of 40 Kts (Alaska, can happen) and an airspeed V of 40 Kts. Engine powered to keep that airspeed. Your ground speed will be zero. Your potential energy (relative to the runway) is very small and can be ignored.

Do the thought-experiment to snap turnoff the headwind as well as the engine. Of course, the airplane will immediately settle, with minimal mechanical movement, since it was not moving compared to the runway, even, the wheels will not start turning.

Will the airplane bleed off a Kinetic energy of 0 (IE effectively nothing happens) ?

Very flawed experiment/example. To "snap turn it off", you would have to decelerate those air particles instantly, and that is impossible.

Or will it bleed off a Kinetic energy of 0.5 * m * V(airspeed) ^2 and effectively destroy the airplane ? Given common sense tells us, the airplane will not be destroyed, the presumed amount of Kinetic energy has to go "somewhere". When not knowing where Energy goes, it is usually turned into heat. So, will the airplane (or the ground, or the -now no longer moving- air) heat up ?

The reverse of this experiment: Airplane on the runway. Have a headwind of 40 Kts, use the engine to keep the airplane steady on (!) the runway, wheels not moving. Your airspeed will show 40 Kts. To avoid, the airplane climbs above the runway, hold the elevator down, to stay on the ground. What is the Kinetic energy of the airplane ?

That depends on your frame of reference

Do the same, by holding the brakes (instead of the engine). What airspeed will be shown ? What is the Kinetic energy of the airplane ?

Now, return to the first experiment. The airplane not 1 mm above the runway, though on the ground, with non-compressed struts. What is the Kinetic energy of the airplane ?

The same, though now with struts compressed. What is the Kinetic energy of the airplane ?

Next step. No headwind. No ground speed. With increasing headwind, the airspeed will go live, with the airplane stationary using brakes. What will be the Kinetic energy of the airplane ? What will happen when we snap-turn-off the headwind ? Assuming the airspeed will be the measure for the Kinetic energy, where will the Kinetic energy go, when the headwind turns off ?

Gives me the feeling to argue with "The Truth"........

So, I will try to not make this personal and have omitted references to your PhD.

First of all. Kinetic energy is not absolute, but depends on your frame of reference. As I sit here in a chair typing I am not moving, so my kinetic energy (1/2 M x V^2) is 0, using my house as a reference. But that "V" is not an absolute value. If I use the center of the earth as my reference I am spinning at 1250kmph (40 degree latitude). If you were standing on the sun, you would have to add 100000kmph to that. and the sun itself spins around in the Milky Way, and that flies away from the center of the universe where the big bang took place. It is all relative. Same but easier to understand is potential energy. The height in there is normally defined, because you can go below the surface of the earth, and most understand it is from the starting point to the stopping point that counts. The same goes for unaccelerated motion, you have to define the reference. And there is a slope here, because my motion around the earth & sun are circular, and thus not un-accelerated, but close enough for flying.

If you sit in a swing ride, you make a circular motion around its central axis, and your seat gets swung out. you will feel a vertical acceleration but no lateral or horizontal acceleration once the ride reaches it operating speed using your seat as your horizontal reference. Now imagine you put that ride on a truck and drive at the same speed as you were spinning. Once the truck and swing-ride reach that speed, you will feel the excact same thing as when the ride was stationary and spinning, because of the unaccelerated motion of the truck (aside from the curvature of the earth). For a person standing on the side of the road, it will appear you make a motion where you decelerate to come to a stop on his side of the road, accelerate to twice his speed on the other side, and come to a stop on his side again and over and over until you throw up. Your kinetic energy will be a constant to the truck driver, but go from 0 to 4 times the value the truck driver sees to the pedestrian every rotation.

To get back to flying:

(the following is based on steady wind, no shear, disregards the curvature of the earth, and the rotation of the earth itself, and using the air as your reference grid)
While taking off you move in reference to the earth, and thus are affected by the wind. That is why in GA airplanes you hold controls into the wind, and have to use rudder to stay on the centerline (and correct for prop spin). As you get rotate there is a period where your frame of reference becomes the air. Once that has happened, the ground doesn't matter. Your plane only senses the air particles around you. What the ground does underneath does not matter. If you make a coordinated turn at a constant airspeed and bank angle, your kinetic energy will stay constant. Your acceleration will be perpendicular to you, so if you look at your energy as a vector it will change direction. Depending on the bank angle your load factor will change, but the only acceleration is in the vertical plane, with reference to the aircraft vertical.

If there are wind changes, like when descending, or turning out of the jetstream, there will be a period to reach a new equilibrium but that is not what we are discussing here.

Finally, getting back to ground reference based maneuvers. That doesn't mean the plane flies ground based, but that the pilot uses the ground for reference to adjust speed and turn rate to get to a certain point on the ground.
If you turn to final with a tailwind (but why would you?), the turn will be more than 90 degrees, as opposed to the more expected less than 90 degrees. Similar for the tailwind on base, your turn to final will, if un-anticipated have to be made at a higher rate. If both happen at the same time, sometimes accidents happen, that is why it is considered dangerous. The pilot using the ground as a reference, but the plane flying in reference to the air is what is the problem, never the plane flying in reference to the ground.

[hans brinker]

I think you've actually agreed with my point - though I could have phrased it better. i.e. "you should finish the manoeuvre at the same point in space as you started it at" (if performed correctly as you describe, that is).

I guess I will too have to clarify. I think it depends on the situation.
If the tower asks you to do a 360 on downwind, although it is the normal way of saying it, it is technically incorrect, because they will expect you to rejoin the downwind approximately on the same spot. So, he is actually expecting you to do a "turn around a point". When ATC asked me to make a 360, while on a heading at around FL300 last month, there definitely was no such expectation on my side, and hopefully not on theirs either. Took us close to 4 minutes to complete at 300kts TAS, and 25deg AOB, so at a 60kts wind we would have moved us 4 miles from the starting point. I was not assigned a hold, DME arc or a turn around a point, so did not take the ground into consideration. I did finish the turn at the same point in space with the air as my reference though.

[hans brinker]

1 Nop, the moment you make your turns relative to the air, your ground speed (vector) will vary and as such, you will experience (minimal) accelerations.
2 The forces (except the inertial ones) on the airplane are relative to the air around the airplane. With 50 Kts airspeed in 100 Kts wind, this accelleration effect is no longer "marginal", though, because the 360s aren't snap turns, still relative mild forces.

You do forget, that when your airplane mass is not unlimited, the moving air will accelerate the airplane relative to the ground in the direction of the wind. With a light airplane and sufficiently large 360s, this is hardly noticeable, though it happens. And because that effect is happening, you will need to juggle the engine power. The effect is the most, when you turn the nose through the direction of the wind.

I hope I addressed most of this already in my previous post, but if not, please educate yourself on how airplanes fly. There is absolutely no need to change power while turning other than to correct for the required increase of lift to achieve the aircraft vertical acceleration, that turns the aircraft while in a bank. Nothing else. The airplane doesn't feel wind, the ground moves under the air opposite to the wind, but your airplane will be unaware.

[Saintly]

Built like a brick, flew like one too.

Cruises up to 270 knots TAS....thats not too bad.

[hans brinker]

Cruises up to 270 knots TAS....thats not too bad.

But definitely heavy on the controls... Loved the cockpit tho. Flew the DHC8 as well, and it is totally Airbus vs Boeing.