Plane flips over after crash-landing in Somalia
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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.
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Yep, indeed, though that's a different chapter !
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 ?
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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.
Then you appear to have answered your own question.
(Holding is ground referenced and has no relevance to Case 2.)
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.
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.
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.
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.
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.
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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"........
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"........
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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.
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.
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.
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
For a modest introduction: Frames of Reference
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.
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.
Last edited by EXDAC; 5th Aug 2022 at 16:37.
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"........
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"........
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.
Last edited by hans brinker; 6th Aug 2022 at 03:03. Reason: spelling
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.
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.
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.