The downwind turn myth

IK04 said:
If two objects of equal mass (airplanes) were traveling in opposite directions at the same airspeed, one with zero groundspeed and the other with 40 knots groundspeed and they collided head on, would the resultant impact favor the object with the greater release of kinetic energy (mass * velocity squared)?
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No. The impact energy absorbed by each object would be determined by the change in velocity, which, other things being equal, would be the same for each object.
 
IK04 said:
If two objects of equal mass (airplanes) were traveling in opposite directions at the same airspeed, one with zero groundspeed and the other with 40 knots groundspeed and they collided head on, would the resultant impact favor the object with the greater release of kinetic energy (mass * velocity squared)?
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since the frame of reference is with respect to the indicated speed NOT ground speed, the only thing that matters in this question is IAS and weight of each plane.
 
One object has zero groundspeed (is stationary) and the other in moving at a speed greater than zero (40KIAS). There is going to be some acceleration imparted on impact.

See the picture now?
 
IK04 said:
One object has zero groundspeed (is stationary) and the other in moving at a speed greater than zero (40KIAS). There is going to be some acceleration imparted on impact.
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Ground speed is irrelevant. This example implies a 20 kts tail wind for one and same as a head wind for the other. Do the same thing in zero wind, they are both moving at 20 kts. (which they were in all examples relative to the air)
 
Hang 4 said:
Ground speed is irrelevant. This example implies a 20 kts tail wind for one and same as a head wind for the other. Do the same thing in zero wind, they are both moving at 20 kts. (which they were in all examples relative to the air)
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What example are you referencing? In my example there is one stationary object and one object moving 40 knots.

Airspeed and movement in the airmass is not a factor. It would be the exact same forces if the two objects were vehicles moving on the ground. An impact force of 40 knots is going to impart some acceleration.
 
IK04 said:
What example are you referencing? In my example there is one stationary object and one object moving 40 knots.
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This one;
IK04 said:
If two objects of equal mass (airplanes) were traveling in opposite directions at the same airspeed, one with zero groundspeed and the other with 40 knots groundspeed and they collided head on, would the resultant impact favor the object with the greater release of kinetic energy (mass * velocity squared)?
Click to expand...
Again, it just doesn't matter which one is "Moving" relative to the ground. They are impacting each other with equal force. The "stationary" one will move backward relative to the ground, but just because the relative movement, not because the downwind one has more "energy".
 
IK04 said:
If two objects of equal mass (airplanes) were traveling in opposite directions at the same airspeed, one with zero groundspeed and the other with 40 knots groundspeed and they collided head on, would the resultant impact favor the object with the greater release of kinetic energy (mass * velocity squared)?
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They would suffer equal damage.

The ground has nothing to do with their velocity. It has everything to do with space itself. They are travelling through space, like the earth and solar system, at about 870,000 MPH. Their difference in groundspeed is completely negligible.
 
IK04 said:
One object has zero groundspeed (is stationary) and the other in moving at a speed greater than zero (40KIAS). There is going to be some acceleration imparted on impact.

See the picture now?
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They both will incur a 20 knot speed change. I'm just not clear on what point you're trying to make.
 
Capt. Geoffrey Thorpe said:
Not full scale aircraft, but it happened to me. Flying an RC glider (Goldberg Gentle Lady) just above the treetop level making the turn from upwind (after launch) back down wind, from my perspective it appeared that the airplane was accelerating so I started back on the stick. And more back. And, whoops, WTF?
By the time "stall/spin" registered in my brain it was too late.

Turning from base to final with a strong crosswind from behind, I can see the potential to try to make the turn look "right" and do the tighten up...
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@Capt. Geoffrey Thorpe @nauga

Ok, I'll take the rc bait... My son and I were flying a small/light/powered rc plane in a stiff breeze over the summer. Instinctively, I know that, to have any prayer of wind penetration, I have to keep it pointed directly upwind, ease off the throttle, and hold some down elevator, otherwise there is the strong tendency for it to pitch up and turn itself around and run downwind.

Why don't we see this in full scale? Reynolds number? Time constants? Scale of atmospheric effects? Wing loading / power loading?
 
GeorgeC said:
Ok, I'll take the rc bait... My son and I were flying a small/light/powered rc plane in a stiff breeze over the summer. Instinctively, I know that, to have any prayer of wind penetration, I have to keep it pointed directly upwind, ease off the throttle, and hold some down elevator, otherwise there is the strong tendency for it to pitch up and turn itself around and run downwind.
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Was all this immediately after takeoff/launch? Once airborne none of that seems to make sense, "Reynolds number? Time constants? Scale of atmospheric effects? Wing loading / power loading?" notwithstanding.
 
FastEddieB said:
Was all this immediately after takeoff/launch?
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Nope, just trying to fly a level-ish rectangular pattern, three mistakes (maybe 75'?) high. Above the trees, not much terrain, so not really downwind of anything that would generate much mechanical turbulence.
 
GeorgeC said:
Nope, just trying to fly a level-ish rectangular pattern, three mistakes (maybe 75'?) high. Above the trees, not much terrain, so not really downwind of anything that would generate much mechanical turbulence.
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Not doubting your experience or your honest perception of it. But…

Once in flight, there is no “wind”. A plane simply moves through an airmass which is itself moving. It simply cannot know from which direction the wind is blowing*. Hence it should behave identically regardless of which direction the airmass is moving relative to the ground. As such, I suspect something else - wind shear, updrafts/downdrafts, something - must have been in play.


*Unless you count looking at the ground or using ground referenced navaids, of course.
 
FastEddieB said:
Not doubting your experience or your honest perception of it. But…

Once in flight, there is no “wind”. A plane simply moves through an airmass which is itself moving. It simply cannot know from which direction the wind is blowing*. Hence it should behave identically regardless of which direction the airmass is moving relative to the ground. As such, I suspect something else - wind shear, updrafts/downdrafts, something - must have been in play.


*Unless you count looking at the ground or using ground referenced navaids, of course.
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To "penetrate" the wind he would have to add power, which would cause the nose to pitch up and the airplane to want to turn left.
And if there's turbulence, the model airplane's airspeed will be fluctuating wildly, and pitch control becomes difficult.
 
Dan Thomas said:
To "penetrate" the wind he would have to add power, which would cause the nose to pitch up and the airplane to want to turn left.
And if there's turbulence, the model airplane's airspeed will be fluctuating wildly, and pitch control becomes difficult.
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What does “penetrate the wind” even mean? But of course turbulence and shear can wreak havoc and be mistaken for a broader “wind” effect.
 
Hang 4 said:
They are impacting each other with equal force. The "stationary" one will move backward relative to the ground, but just because the relative movement, not because the downwind one has more "energy".
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Is that not the definition of acceleration?
 
FastEddieB said:
Once in flight, there is no “wind”. A plane simply moves through an airmass which is itself moving. It simply cannot know from which direction the wind is blowing*. Hence it should behave identically regardless of which direction the airmass is moving relative to the ground.


.
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This is what so many people can't get their head around, pilots should understand this.

Many boaters have a huge problem with this concept when they are in a channel with a current. Lots of stupid boater tricks as a result. It really screws them up when they have to make movements relative to a fix object.
 
IK04 said:
Is that not the definition of acceleration?
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not in the context that you are trying to twist the physics to continue to prove your false theory
 
Initial Fix said:
not in the context that you are trying to twist the physics to continue to prove your false theory
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So you are saying an object in motion striking an identical stationary object will not result in an equal and opposite reaction?

Mr. Newton disagrees:

Newton's first law of motion predicts the behavior of objects for which all existing forces are balanced. The first law - sometimes referred to as the law of inertia - states that if the forces acting upon an object are balanced, then the acceleration of that object will be 0 m/s/s. Objects at equilibrium (the condition in which all forces balance) will not accelerate. According to Newton, an object will only accelerate if there is a net or unbalanced force acting upon it. The presence of an unbalanced force will accelerate an object - changing its speed, its direction, or both its speed and direction.

u2l3a1.gif


Newton's second law of motion pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.

u2l3a2.gif


Newton's Third Law

A force is a push or a pull that acts upon an object as a results of its interaction with another object. Forces result from interactions! Some forces result from contact interactions (normal, frictional, tensional, and applied forces are examples of contact forces) and other forces are the result of action-at-a-distance interactions (gravitational, electrical, and magnetic forces). According to Newton, whenever objects A and B interact with each other, they exert forces upon each other. When you sit in your chair, your body exerts a downward force on the chair and the chair exerts an upward force on your body. There are two forces resulting from this interaction - a force on the chair and a force on your body. These two forces are called action and reaction forces and are the subject of Newton's third law of motion. Formally stated, Newton's third law is:

For every action, there is an equal and opposite reaction.

The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object. The direction of the force on the first object is opposite to the direction of the force on the second object. Forces always come in pairs - equal and opposite action-reaction force pairs.

["Borrowed" from The Physics Classroom, https://www.physicsclassroom.com/]
 
GeorgeC said:
Ok, I'll take the rc bait... My son and I were flying a small/light/powered rc plane in a stiff breeze over the summer. Instinctively, I know that, to have any prayer of wind penetration, I have to keep it pointed directly upwind, ease off the throttle, and hold some down elevator, otherwise there is the strong tendency for it to pitch up and turn itself around and run downwind.

Why don't we see this in full scale? Reynolds number? Time constants? Scale of atmospheric effects? Wing loading / power loading?
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I've seen some odd behavior in model rockets coming down under a 'chute - they tend to circle around under the 'chute, but go straight for a moment when pointed into the wind. Somewhere I have that on VHS :)
All I can figure is that they are well into the boundary layer where the wind is decreasing with what remains of altitude. There also is a lot of turbulence in the boundary layer and model aircraft are a LOT lighter and smaller so they will respond even to small scale disturbances.
 
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Capt. Geoffrey Thorpe said:
There also is a lot of turbulence in the boundary layer and model aircraft are a LOT lighter and smaller so they will respond even to small scale disturbances.
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Yes, I suspect there are accelerations in the air that are on small enough time and length scales to affect a model, but too small to affect a real plane. Wing loading may be at play; I seem to remember that the rc sailplane folks add ballast on windy days.
 
GeorgeC said:
I seem to remember that the rc sailplane folks add ballast on windy days.
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Full scale sailplanes, too. Usually in the form of water ballast that can be dumped as needed.
 
IK04 said:
So you are saying an object in motion striking an identical stationary object will not result in an equal and opposite reaction?
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Stationary relative to what? In your original example, there is no difference between a head on between two 20KT planes in still air and your "stationary" plane. There is an equal and opposite reaction on both planes. Relative to the ground, the "stationary" one moves back. If you are in either plane, it's the same energy vectors regardless of which one is moving relative to the ground.
 
Palmpilot said:
They both will incur a 20 knot speed change. I'm just not clear on what point you're trying to make.
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I obviously need to make a graphical explanation.

Hang 4 said:
Stationary relative to what? In your original example, there is no difference between a head on between two 20KT planes in still air and your "stationary" plane. There is an equal and opposite reaction on both planes. Relative to the ground, the "stationary" one moves back. If you are in either plane, it's the same energy vectors regardless of which one is moving relative to the ground.
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I'm not sure where you are getting 20 knots from, so I'll need to make a drawing:

Downwind_Turn.png
 
IK04 said:
If two objects of equal mass (airplanes) were traveling in opposite directions at the same airspeed, one with zero groundspeed and the other with 40 knots groundspeed and they collided head on, would the resultant impact favor the object with the greater release of kinetic energy (mass * velocity squared)?
Click to expand...

IK04 said:
One object has zero groundspeed (is stationary) and the other in moving at a speed greater than zero (40KIAS). There is going to be some acceleration imparted on impact.

See the picture now?
Click to expand...

You changed the scenario for the 40 knot object from ground speed to air speed. I overlooked that fact, so I was still responding to the first scenario. My mistake.
 
Think of it this way...

If the plane on the left flew into a wall, barn, tall building or anything connected to the Earth, the energy expended would be theoretically zero, because it is not moving. Some small amount of speed would be a better example, but this is only a notional example. If the plane on the right flies into the wall, there will be a significant exchange of energy and there would an acceleration of whatever it hits.

The Kinetic energy of the two airplanes is very different, but the pilot only sees and feels the 40 KIAS. The increase in stored energy during a 180 degree turn to a downwind is tremendous, as Doug describes in the Video right here at 3:19:

I suppose you could describe the change in stored energy in Joules, but that is going way beyond the scope of this discussion. :cool:
 
Palmpilot said:
You changed the scenario for the 40 knot object from ground speed to air speed. I overlooked that fact, so I was still responding to the first scenario. My mistake.
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Yeah, without the diagram, it is confusing. In order for the two scenarios to match, the first one should have been 80 knots groundspeed. I was more concerned with the theory than the numbers.;)
 
Put an accelerometer in both planes. Have them impact head on with the wind in your diagram and with no wind. Both accelerometers will have exactly the same reading in both scenarios. 40kts of thrust X2 or the same as 80 kts into an object fixed to the earth. Change your stationary plane to a building fixed to the earth. The accelerometer in the plane will read exactly the same as impacting the other plane. 40kts of thrust plus 40 knots of tailwind into an object fixed to the earth The wind only matters relative to the ground, not to objects in the air. Both planes have the same energy coming from thrust. The wind is adding no relative energy to objects in the air, only to the earth itself or objects fixed to the earth.
 
IK04 said:
Think of it this way...

If the plane on the left flew into a wall, barn, tall building or anything connected to the Earth, the energy expended would be theoretically zero, because it is not moving.
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Exactly. But we are not talking about colliding with anything attached to the earth. WE are not attached to the earth except through gravity, which only works in the vertical. The airplane's movement is relative to the wind and its momentum is relative to space to space, not the earth.

If I walk up forward up the aisle of the airliner in cruise, I am doing 600 MPH+2MPH. The flight attendant is walking aft at 600 MPH-2MPH. Difference is 4 MPH, and that's all. I don't bump into the attendant harder than she bumps into me because of my greater forward speed. If the forward speed made a difference, the math would tell me that I hit her with the force of a 69 MPH collision.
 
IK04 said:
The Kinetic energy of the two airplanes is very different…
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I was about to type “…but only in relation to the earth.” But Dan above already made my point. I was going to use two colliding model trains on a moving train, but bumping into flight attendants is more fun!

If you go ahead and add the earth back into to the diagram, as others have said why not also include the rotational velocity of the earth - up to 1,000 mph or so at the equator - or the speed of the earth orbiting around the sun or the speed of the solar system through the galaxy and so on? Each frame of reference may have its own place in a discussion, but here - and in the downwind turn (remember that?) the only meaningful frame of reference is the air mass the planes are in.
 
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This brings up something I initially found counterintuitive…

Would those two planes hitting head on have the same force as one of them hitting a stationary object at 80 kts? Pretty sure the answer in no.
 
FastEddieB said:
This brings up something I initially found counterintuitive…

Would those two planes hitting head on have the same force as one of them hitting a stationary object at 80 kts? Pretty sure the answer in no.
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They would each encounter enough force to reduce their speed by 40 knots.
 
FastEddieB said:
This brings up something I initially found counterintuitive…

Would those two planes hitting head on have the same force as one of them hitting a stationary object at 80 kts? Pretty sure the answer in no.
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They would encounter the same amount of energy, but there are different kinds of collisions. Running into a solid object is probably the worst type you can have. I'm too far away from my physics classes to explain, but it has to do with elastic versus inelastic collisions IIRC. I'm sure someone will be along to explain.
 
FastEddieB said:
This brings up something I initially found counterintuitive…

Would those two planes hitting head on have the same force as one of them hitting a stationary object at 80 kts? Pretty sure the answer in no.
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The Mythbuster answer...
 
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