Aerodynamics Discuss the concepts of aerodynamics here

Downwind Turns

Old 03-09-2009, 01:43 AM
  #151  
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Originally Posted by HX3D014 View Post
You do know that the car example is not helping you Downwind turn Upwind turn discussion. It contradicts it

the car example is just like a plane doing a 360deg circles in no wind. there is no KE change in that example (KE relative to the Ground) and the point is. your math to try and show the amount of change in KE while doing an upwind turn V a down wind turn is Flawed.

Re Read my edit to your math. Try to understand that the Direction is a part of the Change.


Re Read the Math. Forget the car example.

or if you want to do the car example. then do it with the same Velocities weights and Headwind downwind speed (treadmill speed).

Bryce.
Car example simply proves that change of direction by itself causes no change in KE. Change in KE from upwind state to downwind state is the difference between the KE. (Same change in KE when a 1 KG car changes speeds from 10 MPH to 30 MPH)
Upwind state:
Plane flying 20 MPH into 10 MPH headwind (10 MPH = 4.98 M/S)
KE = 1/2 m x V^2
KE = 1/2 x 1 x 4.98 ^2 = 12.4 Jules

Downwind state:
Plane flying 20 MPH with 10 MPH tailwind (30 MPH = 14.9 M/S)
KE = 1/2 m x V^2
KE = 1/2 x 1 x 14.9 ^2 = 112 Jules

Change in KE
112 - 12.4 = 99.4 Joules


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Old 03-09-2009, 01:56 AM
  #152  
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Originally Posted by cbatters View Post
Car example simply proves that change of direction by itself causes no change in KE. Change in KE from upwind state to downwind state is the difference between the KE. (Same change in KE when a 1 KG car changes speeds from 10 MPH to 30 MPH)
Upwind state:
Plane flying 20 MPH into 10 MPH headwind (10 MPH = 4.98 M/S)
KE = 1/2 m x V^2
KE = 1/2 x 1 x 4.98 ^2 = 12.4 Jules

Downwind state:
Plane flying 20 MPH with 10 MPH tailwind (30 MPH = 14.9 M/S)
KE = 1/2 m x V^2
KE = 1/2 x 1 x 14.9 ^2 = 112 Jules

Change in KE
112 - 12.4 = 99.4 Joules


I must say, That's a poor effort there, You loosing steam ?


Against the treadmill state:
Car Driving North 20 MPH on a treadmill Running south 10 MPH
The car is moving North 10 MPH (relative to the People not on the treadmill)
= 4.98 M/S North (relative to the People not on the treadmill)

KE = 1/2 x 1 x 4.98 ^2 = 12.4 Jules

With the treadmill state:
Car Driving South 20 MPH on a treadmill Running south 10 MPH
The car is moving South 30 MPH (relative to the People not on the treadmill)
= 4.98 M/S North (relative to the People not on the treadmill)

KE = 1/2 x 1 x 4.98 ^2 = 12.4 Jules



Change in KE
112 + 12.4 = 124.4 Joules North to south and South to north or to put that another way.
=124.4 Joules with the treadmill to against the treadmill

and

=124.4 Joules Against the treadmill to with the treadmill

I would like to add that the Acceleration forces felt within the care are the same through out the Circles. That is;
If you were in the car and had your eyes closed<obviously as a passenger >, You would not feel any difference. just one constant force through out the Circles .

Bryce.
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Old 03-09-2009, 02:32 AM
  #153  
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Originally Posted by Sparky Paul View Post
"When the plane changes from downwind to upwind, there is no change in kinetic energy"
Period!
Correct (Relative to the air <Which is all that matters>. Provided you do the Turn with Constant speed IAS.

In that case the plane is still doing 20mph IAS through the AIR in both up wind and downwind turns.

But if you do the turn with constant power. you will get a reduced airspeed IAS through the turn.

Because it takes a force to turn the plane around.

So, If you do the turn at a constant speed IAS, you will need more power.

Notice the similarity between the KE and Lift formula.

cbatters should look at the formula for Acceleration and look up the physics definition. (Or revise it again if it is not new to him)

Bryce.
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Old 03-09-2009, 02:46 AM
  #154  
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The wind pushing against the plane and with the plane banked this force(wind) will be pushing the plane up .
It will also be pushing against the fuselage like rubbing you hand down the side as it turn's. This energy is stored in the body until it can be released.
As the plane pass 90 degree's to the wind it releases and pushes the plane faster.
Depending on what kind of aeroplane you fly the out come will be different.
Something sleek will let this energy out fast and gain speed fast.
Something fat and slow flying may not gain as much speed and may stuggle to release the energy fast enough which will cause some kind of airframe failure.
A swept wing or low drag airfoil would be better in this situation than a straight wing or high lift airfoil.
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Old 03-09-2009, 04:54 AM
  #155  
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Originally Posted by cbatters View Post
Car example simply proves that change of direction by itself causes no change in KE. Change in KE from upwind state to downwind state is the difference between the KE. (Same change in KE when a 1 KG car changes speeds from 10 MPH to 30 MPH)
Upwind state:
Plane flying 20 MPH into 10 MPH headwind (10 MPH = 4.98 M/S)
KE = 1/2 m x V^2
KE = 1/2 x 1 x 4.98 ^2 = 12.4 Jules

Downwind state:
Plane flying 20 MPH with 10 MPH tailwind (30 MPH = 14.9 M/S)
KE = 1/2 m x V^2
KE = 1/2 x 1 x 14.9 ^2 = 112 Jules

Change in KE
112 - 12.4 = 99.4 Joules


You left out a term. The work done by the wind.
You've accounted for the work done by the engine (if it's an airplane), or the loss in altitude (if it's a glider).
But if we put that airplane on the ground, engine off, and assume the ever-popular-in-gedanken-experiments level frictionless surface, we find that it accelerates until it's going the same speed as the wind. That work isn't being done by the engine -- it's off. It isn't due to loss in altitude -- it's on the ground, it can't lose any more. It's being done by the wind, which has energy relative to the ground, and is using it to do work on the airplane -- relative to the ground.
There's nothing special about touching the ground that allows the wind to perform work on the airplane -- it's a frictionless surface, just as frictionless as it is when the airplane is not touching it.

You should recalculate, including the work being done on the airplane by the wind.

ETA: I phrased that badly. Actually, you've calculated the sum of the two energies, without explicitly acknowledging the work done by the wind. The energy you have assumed must be provided by a loss in altitude is actually provided by the wind.
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Old 03-09-2009, 05:14 AM
  #156  
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Yes I think this thread has gone terribly south, south as in back down on the ground. It needs to stay in the air, and thats just what our planes will do if we keep all this KE PE and all that out of our minds.
Not that any of it isnt true, for it is. But its engineer's jargon and equations that to the pilot is totally useless.

What is useful to a pilot is getting out of his head at the outset that a plane is some sort of air going car. And that anything he knows about driving a car on soild ground needs to be thrown out the window.

Ok, back to those three key ideas I mentioned. The second idea, is that motion is relative. And you have to limber up your ideas of what motion really is. Skipping all the fancy business, it would be useful to observe a man who is walking about inside a moving train. This is because he is just like an airplane flying in a wind, as will be shown. While the airplane ina wind is puzzling, you can easily try out just exactly what happens to a man who walks about in a moving train.
That man can walk about in the train just as if it were standing still. Except for some jolting and bouncing, it's headlong speed of say, 60mph does not affect him. He find it requires no more effort to walk to the front of the train, than to walk to the rear, in fact if he did not look at the moving scenery outside, he could not even tell which way the train is going. If he drew the blinds he could easily persuade himself that the train is moving in a reverse direction. It is important that you believe this; if you dont try it and see. If the track were smooth, he could not even tell the train is moving at all! Proof of this you have sometimes experienced while in a station, when the train on an adjacent track began to slide out, and you, in mommentary confusion, had to make a conscious effort to determine whether it was your train moving, or the other train, or perhaps both of them!
We disregard here the forces that pull on the passengers while a train starts up and accelerates, and again while it brakes; and also disregard the jolting due to roughness of of track. We are talking here about steady motion. As long as the train maintains a steady speed, the passenger can step fro the right hand to the left hand side of the coach(walking "cross-train" as it were) without making any allowance for the motion of the coach itself-entirely as if he and the coach were standing still.
When a pilot flies in a wind, the wind is his railroad coach.

In another respect, the train's motion does very much concern the passenger,- the little matter of being carried to Chicago at 60mph. Regardless of whether he walks, runs, sleeps, dances, or stands on his head within the train, the train carries him right along.
The results if you put them bluntly are quite paradoxical. For example he might walk all the way forward to the dinning car, have dinner and then walk all the way back to his seat, and then he might say,"Well here I am, back where I started from." In one sense he is right, and yet in another he would be wrong. The place he has returned to has itself in the meantime moved something like 30 miles. Whether or not he is truly back where he started from depends, in highbrow words, on his "Frame of reference", that is which side of the facts he chooses to regard, and which side he chooses to disregard, whether hjudges his position by reference to the train, or the outside stationary world.
You could keep on constructing complicated examples of this sort. Suppose the train were sliding slowly out of the station. The fellows best girl is is on the platform seeing him off; he waks toward the rear of the train and temporarily succeeds in standing still realative to his girl. There are many similar examples. Nor is this realtivity of motion peculiar to railroad trains, you find the same story again when a ferry crosses a river or when a man swims in a current, and perhaps you used to have fun(I did) when you walked down the up-escalator in your local department store, watching yourself walk down yet stay where you are.

The third idea : You're "in the air".
Here is the third key idea,- an airplane that flies in moving air is like a man who walks within a moving train. This, mark well, is not a simile, not a figure of speech, but a precise statement of fact. The two cases are not only pretty nearly alike, but they are just plain alike. the mathematics, physics, and logic of the two cases are identical.
Just as the passenger is "contained" by the railroad coach and cut off from the scenery outside, so is the airplane, once in flight, entirely "contained" by the surrounding air and cut off from all connection with the ground. It's propeller propels it, not by acting on the ground, but by acting on the surrounding air. It's wings hold it up, not by acting on the ground but by acting on the surrounding air. Thus the cubic mile or so in which the airplane flies, is to it exactly what the railroad coach is to the passenger The only difference is that the coach surrounds the passenger visibly, while the air surrounds the plane invisibly.
Hence, whatever is true of a passenger who walks within a moving train is equaly true of an airplane that flies within a mass of air which is itself in motion. The airplane too, has two motions, both at the same time. It has motion "through" the air, which corresponds to the walking of the train passenger through the train. And it has a motion "with" the air, called "Drift", which corresponds to the travel of the train. Just like the train passenger he can fix his attention on the one sort of motion or the other, as he chooses; and after he has become at home in the air, he can watch both motions at the same time and not get confused.

All that above is why that to us, on the ground look at it all from a fixed position can fool us into thinking we must do something differntly on the sticks to manage the plane in a turn in winds. We are watching two motions at once. But dont let your eyes decieve you, nothing is wrong up there, the plane is flying normally, so fly it normally.

Alright, geez that was long..........
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Old 03-09-2009, 05:47 AM
  #157  
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That train example really wrecked my brain.
Because RC planes are small I think the only way to get some closure would be to tow a streamer behind a couple of different planes like they do in "combat" and see how the streamer reacts behind!
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Old 03-09-2009, 06:45 AM
  #158  
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Steamers wouldnt tell anything, other than if the pilot is using his controls properly, for if he isnt they would certainly indicate it by streaming sideways if say, he were'nt applying enough, or too much rudder in a turn allowing the airplane to "slip", or "skid".
Though they will react to banking while the wings are being banked. But once banking stops, and are in a bank, the streamers would again flow straight back away.
But that too is two motions at once, the wingtip swinging downward is also moving forward. So the streamer's whould stream up & back and down & back for the other tip. (streamers attached to the wingtips)

Nothing complicated about the train example that should wreck one's brain though, its as simple an idea as can be I think. It's the best way I've seen it portrayed actually.
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Old 03-09-2009, 07:18 AM
  #159  
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OK, but what if you put the plane on a big flying conveyor belt within a even bigger train?
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Old 03-09-2009, 09:25 AM
  #160  
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If the man was on top of the train it would be different. if the train was doing 60mph ,he would be doing 60mph. With the forces of nature acting with him weight,gravity and traction he stays on the train.
If he jumps in the air he loose these forces he looses speed from drag so he land further down the train.
Still this is not a good example.
Automobile and train use traction and track's to aid their movement but they still use aerodynamic's where it is too stick to there medium or to make them pass through a less solid medium "air".
An airplane is different it doesn't stick to the medium it flow's through it.
Air is a solid but it is a viscous and changes density ,heat ,cool,altitude, etc
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Old 03-09-2009, 09:47 AM
  #161  
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If the man was on top of the train it would be different. if the train was doing 60mph ,he would be doing 60mph. With the forces of nature acting with him weight,gravity and traction he stays on the train.
If he jumps in the air he loose these forces he looses speed from drag so he land further down the train.
Not if his name is Bond .... James Bond.
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Old 03-09-2009, 09:55 AM
  #162  
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Here are three pages from my ultralight intsruction manual's.
1st is about load factor while banking
2nd is gross weight take distances, IAS speeds, ground speed etc.
3rd is about banking, side slip, airflow etc
I hope it help's
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Old 03-09-2009, 12:02 PM
  #163  
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Originally Posted by TjW View Post
You left out a term. The work done by the wind.
You've accounted for the work done by the engine (if it's an airplane), or the loss in altitude (if it's a glider).
But if we put that airplane on the ground, engine off, and assume the ever-popular-in-gedanken-experiments level frictionless surface, we find that it accelerates until it's going the same speed as the wind. That work isn't being done by the engine -- it's off. It isn't due to loss in altitude -- it's on the ground, it can't lose any more. It's being done by the wind, which has energy relative to the ground, and is using it to do work on the airplane -- relative to the ground.
There's nothing special about touching the ground that allows the wind to perform work on the airplane -- it's a frictionless surface, just as frictionless as it is when the airplane is not touching it.

You should recalculate, including the work being done on the airplane by the wind.

ETA: I phrased that badly. Actually, you've calculated the sum of the two energies, without explicitly acknowledging the work done by the wind. The energy you have assumed must be provided by a loss in altitude is actually provided by the wind.
I agree. We need to consider the work done by the wind to accelerate the plane up to the same speed as the wind. (and the force on the bottom of the plane when banking which may help keep it aloft.) However, additional energy is necessary to continue accelerating the plane to an airspeed of 20 MPH.


Clint
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Old 03-09-2009, 03:18 PM
  #164  
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Originally Posted by cbatters View Post
However, additional energy is necessary to continue accelerating the plane to an airspeed of 20 MPH.
The plane started with an airspeed of 20mph. If it needs to accelerate to get back to 20mph how did it lose the airspeed it started off with ? Perhaps you could show what it's minimum airspeed was and when that occurred ? Also where all the energy that the plane "lost" as it slowed down went to.

Steve
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Old 03-09-2009, 04:05 PM
  #165  
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Originally Posted by cbatters View Post
I agree. We need to consider the work done by the wind to accelerate the plane up to the same speed as the wind. (and the force on the bottom of the plane when banking which may help keep it aloft.) However, additional energy is necessary to continue accelerating the plane to an airspeed of 20 MPH.


Clint
.
There is no "force on the bottom of the plane" .
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Old 03-10-2009, 01:17 AM
  #166  
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There must be some thing wrong with the Math. I have concluded that it must be a + (but not sure about the math anyway now)

I had a hard time to walk to my Toilet last night.

Last night I did it (Just about midnight) and my KE relative to the sun while moving forward was 913 638 280perKG (Just using an average earth orbital speed of 29.780km/s) and I weigh 90kg so it was about 41 11 372 260 joules to get me moving that way.

Today at mid day I will only need 860 457 156 per KG so that's about 38 720 572 020 Joules
the difference is phenomenal.

I think the fault lay with the way you calculate the Two KE components.

If you want to see the change in KE state. you should look at the change in m/s one direction to the other first.
so in the Example where the plane flys 20 into a 10 head wind with a 10 ground speed
and then flys 20 with a 10 tail wind with a 30 ground speed.
then the difference is 40

so work the KE of 1kg at 40mph

if you do not believe me.

then do your same math "CBatters" and have a look at your answer for the KE change when you do a circuit in a 200mph wind flying 20mph IAS all the way (apparently that is possible if you fly from a gondola in a Jet stream (Not to far away though eh ).

I dare not tell you what happened to me when I urinated 500g @ about 1.5m/s
add that to my speed of 30226.45/ms around the sun at midnight and you get 30227.95
Now If do the math on that 500g
Before Urination its 1/2*0.5kg*30226.45^2 = 228 409 570
After Urination its 1/2*0.5kg*30227.95^2 = 228 432 240
the difference is painful (22 670 joules)
Consequently, I was blasted back out the bathroom and Off the balcony. Took me an hour to crawl back home. The Doctor says We will survive.

Last edited by HX3D014; 03-10-2009 at 01:35 AM.
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Old 03-10-2009, 06:28 AM
  #167  
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If you want to do calculations, you must consider every little detail, every little gain and loss of energy and how it act on the plane. A simple flawed calculation like this, considering only one fraction of the whole system can never simulate what is happening in flight. If you have ever the slightest knowledge of what is happening in the latest flight simulator programs, (which is far from perfect to simulate the real world) you would not even consider giving such a simple calculation on its own as a explanation of a turn in flight.

Look what happened because of calculations, we could have lost Bryce. Leave alone the calculations, get a plane and go fly and observe what is happening.
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Old 03-10-2009, 11:07 AM
  #168  
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Hi Bryce, and all. Just checking in.. Wow caused quite a stir havent you...

C'mon guys cars, trains haven't read all of this yet but has anyone mentioned a boat going up or down stream. I know a little about that as i was a barge operator for several years over here... Lol im not even going to bother, I will say one thing, certainly had to allow for the current... While you guys are all getting your degrees out im flyin....Cheers Darren
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Old 03-10-2009, 11:53 AM
  #169  
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Originally Posted by Sparky Paul View Post
.
There is no "force on the bottom of the plane" .
I belive we can calculate the pressure in PSI on bottom of the wing when the plane is banking based on the cross sectional area of the wing presented towards the wind.


Clint
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Old 03-10-2009, 03:38 PM
  #170  
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Originally Posted by cbatters View Post
I belive we can calculate the pressure in PSI on bottom of the wing when the plane is banking based on the cross sectional area of the wing presented towards the wind.


Clint
.You believe wrong.
An airplane flies IN the airmass. The wind does NOT push against the bottom of the wing.
Your continual refusal to dissociate the airplane's flight in the air from it's motion relative to the ground prevents you from comprehending how airplanes fly.
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Old 03-10-2009, 04:03 PM
  #171  
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Originally Posted by cbatters View Post
I belive we can calculate the pressure in PSI on bottom of the wing when the plane is banking based on the cross sectional area of the wing presented towards the wind.


Clint
I'm certain we can -- though not based on the cross sectional area of the wing. Why would we do that?
The wind does work on the airplane all the time.
That's why it flies more slowly relative to the ground going upwind than downwind.
In the upwind case, it opposes the work being done by the engine.
In the downwind case, it aids it. Crosswind, it's pushing it sideways.
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Old 03-10-2009, 05:01 PM
  #172  
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Originally Posted by TjW View Post
I'm certain we can -- though not based on the cross sectional area of the wing. Why would we do that?
The wind does work on the airplane all the time.
That's why it flies more slowly relative to the ground going upwind than downwind.
In the upwind case, it opposes the work being done by the engine.
In the downwind case, it aids it. Crosswind, it's pushing it sideways.
.
Only relative to the ground!
Only relative to the ground!
The airplane is IN/carried by/flies with the air mass, whether the air mass is moving or not.
This is extremely basic aerodynamics.
You ( I won't waste the time ) can "calculate" the plane's motion relative to anything you want.
The airplane (and I) won't care.
It flies IN the airmass.
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Old 03-10-2009, 05:45 PM
  #173  
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sparky is correct here.
and the fact that the planes behaviour is dependant on the aircrafts speed relative to the surrounding air mass, and NOT the ground, this imples that if you are flying upwind at near stall speed, and then make a turn without changing your absolute speed (speed relative to ground), then the surrounding airmasses of the aircraft has turned 180 degrees, and your delta windspeed (aircraft speed + wind speed) has suddenly changed to (aircraft speed - wind speed), and you can suddely find yourself in the situation that the speed that was just sufficent for flying upwind, is way below stallspeed in downwind. now, keep in mind that even though the aircrafts aerodynamic behaviours is purely related to aircraft speed relating to air masses, the aircrafts mechanical behaviour (the inertia caused by the planes mass) is still not affected by surrounding airspeed.
so its not as simple as saying the aircrafts begaviour is purely dependant on the speed of the surrounding air either.
as you turn from upwind to downwind, you can say that the speed of the wind will accelrate the aircraft to the absoute speed that correlates to the same speed relative to wind when flying upwind. for an infinite timespan, this is true. but for a short timespan (for example a sharp turn) the aircrafts inertia will resist this acceleration, and you will indeed get strange behaviours when turning sharply into downwind directon, in strong winds.
i guess most guys in here are flying aircrafts with low wing loadings (more typical for electric planes), and therefore dont feel this effect too much (low wing loading = low mass relative wing area, wich in practice is a face for aerodynamic forces to work on).
if you try flying fast jets of heavy wing loaded fuel guzzlers, you will feel this effect very much.
just yesterday, i was doing this very turn with a very heavy wing loaded cap 232, while doing aerial photography. i flew upwind, gave full throttle and headed downwind in a half-loop. and the aircraft is overpowered as heck, so it was not a "ordinary stall". but still, i lost total control of the aircraft and it went into a nose down behaviour, until it quickly gained speed again and i regained control. this is caught on onboard video, and you find link to it under my thread "my first aerial video". i dont remember wich one of them right now, but its one of the videos from the second day.
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Old 03-10-2009, 08:20 PM
  #174  
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Originally Posted by Moxus View Post
sparky is correct here.
and the fact that the planes behaviour is dependant on the aircrafts speed relative to the surrounding air mass, and NOT the ground, this imples that if you are flying upwind at near stall speed, and then make a turn without changing your absolute speed (speed relative to ground), then the surrounding airmasses of the aircraft has turned 180 degrees, and your delta windspeed (aircraft speed + wind speed) has suddenly changed to (aircraft speed - wind speed), and you can suddely find yourself in the situation that the speed that was just sufficent for flying upwind, is way below stallspeed in downwind. now, keep in mind that even though the aircrafts aerodynamic behaviours is purely related to aircraft speed relating to air masses, the aircrafts mechanical behaviour (the inertia caused by the planes mass) is still not affected by surrounding airspeed.
so its not as simple as saying the aircrafts begaviour is purely dependant on the speed of the surrounding air either.
as you turn from upwind to downwind, you can say that the speed of the wind will accelrate the aircraft to the absoute speed that correlates to the same speed relative to wind when flying upwind. for an infinite timespan, this is true. but for a short timespan (for example a sharp turn) the aircrafts inertia will resist this acceleration, and you will indeed get strange behaviours when turning sharply into downwind directon, in strong winds.
i guess most guys in here are flying aircrafts with low wing loadings (more typical for electric planes), and therefore dont feel this effect too much (low wing loading = low mass relative wing area, wich in practice is a face for aerodynamic forces to work on).
if you try flying fast jets of heavy wing loaded fuel guzzlers, you will feel this effect very much.
just yesterday, i was doing this very turn with a very heavy wing loaded cap 232, while doing aerial photography. i flew upwind, gave full throttle and headed downwind in a half-loop. and the aircraft is overpowered as heck, so it was not a "ordinary stall". but still, i lost total control of the aircraft and it went into a nose down behaviour, until it quickly gained speed again and i regained control. this is caught on onboard video, and you find link to it under my thread "my first aerial video". i dont remember wich one of them right now, but its one of the videos from the second day.
Expect some questions when I come back.

Off to work now.

Bryce.
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Old 03-10-2009, 10:59 PM
  #175  
Moxus
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Join Date: Jun 2008
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and in general, to all of you, i think there is a lot of would-be engineers here. being an engineering student myself, i know very good how tempting it is to overcomplicate something that should be done simple (for example steam powered computers. how gneious isnt that!)
but, lets face it: simple solutions usually works better.
a 100% mathematical correct answer is impossible to get here simple because gases is too hard to work with exept for with computer numerical methods. now, i COULD build up an aircraft in solidworks and then prove to you all is cosmosflo virtual wind tunnell that lift is severely reduced in a turn like that, but its a lot of work. and such a model would only predict the outcome, it wouldnt explain WHY. i guess what veryone wants to know here is why.
try to forget the conservation of energy angle of attack, thats pointless.
the wings sole purpose is to produce a force to oppose gravity.
as an aircraft flies at constant altitude NO ENERGY IS LOST OR GAINED, due to the simple fact that E = F*deltaP, and deltaP (altitude) is unchanged.
kinetic energy in the form of speed is totally irrelevant, as that would be a product of the aircrafts absolute speed (speed relative to ground). thats not interesting, its not the speed relative to ground that determines a stall. its speed relative to surrounding air masses. so forget the whole KE thing.
so, back to speeds, forces and stuff. we all agree that the winds speed relative to the surrounding airmasses is what determines lift. NOT speed relative to the ground? right?
good. when the aircraft turns, its inertia wants to keep the aircraft moving along its original path.
to change its path takes force (not neccessarily energy. if speed is kept constant, and only direction of travel is changed, no energy is used. only a force applied. keep that in the back of your head).
so, we assume that the aircraft takes a 180 degree turn, and no energy is spent or gained, because has been at constant absolute velocity.
now, when you fly upwind, the speed of the air flowing over your wing is:
(ground speed + wind speed)
when you fly downwind, the speed of the air flowing over your wind is:
(ground speed - wind speed)

that clearly shows that your wing has less velocity relative to the surrounding air when flying downwind that upwind, when keeping absolute speed constant.
we all agreed earlier that speed relative to the surrounding air masses determines lift, so this means you get less lift when you turn from upwind to downwind.

now, you can always point out that the wind will accelerate the aircraft so the aircraft speed wont be constant, and that is true. but because of the aircrafts inertia, this wont happend instantly. it has a lag, and for the period it takes the wind to accelerate the aircraft, thats a period where you got reduced lift.

complicated problem made simple. good engineering.
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