Wind affecting handling ?

[Myron Beaumont]

Tony

Dead right ! That's what I've been trying to explain . Yes ,I do understand kinetic & other forms of energy But in my book (& yours ) it has nothing to do with it

[Biggles' Elder Brother - Moderator]

I give up! You can lead a horse to water etc.

 

BEB

[Myron Beaumont]

BEB

'er indoors's horses /sheep/chickens/sheep dogs  know exactly where water is .What do you mean ?

[Martin Harris - Moderator]

Personally, I believe BEB is proposing a situation which cannot physically occur in the real world when using aerodynamic effects to initiate the turn. He may very well be correct hypothetically but I freely admit that it's beyond my understanding to apply his reasoning to the real world.  All I know is what I've experienced and my basic and empirical understanding of the dynamics of flight.

 

Not my words but I think this sums up the situation as far as my limited maths skills can tell:

 

Please feel free to discuss!

 

The problem of the downwind turn is entirely associated with the effects of wind speed. Here are four simple calculated examples based on 4 different wind speeds: zero wind, wind at 1/4 airspeed, wind at 1/2 airspeed and 'wind' experienced inside a large aircraft cruising at 570 mph.

Before we go any further it will be necessary to set some parameters. All values will be in metric units, i.e. metres, and seconds; there is no need to consider a numerical value for weight because at all times in a level turn it equals lift, which simplifies it a bit. It will be referred to as 'wt'. Also, we need to lay down a few definitions:

Kinetic energy. This is a red herring and does not concern us. The common mistake is to assume that it affects turns as if there were a law of conservation of kinetic energy. There isn't, so we can forget about it. For the record kinetic energy is 1/2 mass * velocity squared.

Momentum. This is the one that matters. There IS a law of conservation of momentum, both of linear momentum and angular momentum (rotation). The snag is that this too is not relevant to effects produced by the aerodynamic forces that cause turns. The point of the examples will be to demonstrate that the change in momentum during a turn is the result of the reversal of airspeed and that this is independent of any effects of windspeed. Momentum is defined as mass * velocity.

Because velocity is a directional quantity we have to be very careful with the number values. When the direction reverses, velocity changes from a positive value to a negative one. It is probably neglect of the sign (+ or -), more than any other factor, that leads to confusion when the physics of the turn are considered.

In all the examples the initial condition is flight into wind. The airspeed in all cases will be 20 m/sec, so the groundspeed in each case will be airspeed - wind speed. The term 'speed' is not as precise as 'velocity' but it will be used for convenience because of its familiarity.

Example 1 Windspeed 0 m/sec, model airspeed 20 m/sec.

Initial state: Groundspeed(1) = airspeed(1) - windspeed = 20 - 0

End state: Groundspeed(2) = airspeed(2) - windspeed = -20 - 0 Note, the reversal of the direction of flight has made airspeed(2) a negative value. The change in momentum is groundspeed(1)m - groundspeed(2) and again the sign of each value is vital.

Change in momentum = wt * groundspeed(1) - wt * groundspeed(2) which is wt * 20 - wt * (-20) or

[Martin Harris - Moderator]

(continued)

wt * 40

Example 2 Windspeed 5 m/sec, model airspeed 20 m/sec.

Initial state: Groundspeed(1) = airspeed(1) - windspeed = 20 - 5 or 15 m/sec

End state: Groundspeed(2) = airspeed(2) - windspeed = -20 - 5 or -25 m/sec

Change in momentum = wt * groundspeed(1) - wt * groundspeed(2) which is

wt * 15 - (wt * -25) which becomes wt * 40 same as before.

Example 3 Windspeed 10 m/sec, model airspeed 20 m/sec.

Initial state: Groundspeed(1) = airspeed(1) - windspeed = 20 - 10 or 10 m/sec

End state: Groundspeed(2) = airspeed(2) - windspeed = -20 - 10 or -30 m/sec

Change in momentum = wt * groundspeed(1) - wt * groundspeed(2) which is

wt * 10 - (wt * -30) which again becomes wt * 40

Example 4 Windspeed' 570 mph, or 250 m/sec (well, almost).

Inside the transport aircraft within which the model is to fly, it has to be launched towards

the tail, so that the mass of moving air will be coming in the opposite direction to

the model's flight and thus constituting a headwind. The speed of the big aircraft therefore

has to be entered as -250 m/sec.

Initial state: Groundspeed(1) = airspeed(1) - windspeed = 20 - 250 or -230 m/sec

End state: Groundspeed(2) = airspeed(2) - windspeed = -20 - 250 or -270 m/sec

Change in momentum = wt * groundspeed(1) - wt * groundspeed(2) which is

wt * -230 - (wt * -270) which once more becomes wt * 40

Conclusion No matter how you cut it the calculation, done with proper respect for vector quantities, brings the same result.

The values of windspeed, however great they may be, cancel out and only the aerodynamic flight effects contribute to the change in momentum.

Edited By Martin Harris on 04/10/2010 00:14:06

[Dan Hedges]

Well, while that seems to make sense, and I do get that we are travelling in a medium you have to take into account the effectrs of drag etc on the model... ummm trying to think of a way to put it... Ah I know... If I hand launch my model downwind in a reasonable breeze I am going to find that it is not so responsive to the controls and if the breeze is strong enough I may even stall and crash.

 

Now, I think the reason for the above is that although my model is in a moving body of air and does not care what it's groundspeed is the air will need a certain amount of time acting on my model to bring it up to speed.

 

If I fly into wind and make a TIGHT turn downwind I will have the same problem, for a certain amount of time the air will be blowing accross my plane from behind until it has pushed my aircraft up to speed to match the wind +  the thrust from my prop. However while my aircraft is accellerating to match the wind I will have less airflow over the wings and could concievably stall. 

 

Hmmmm... At least that is how it logically works in my head (which is not always a great place to be ) but.... on giving it a lot of further thought (hey, there is nothing else to do when in a meeting...) I am actually not so sure. As the plane is moving in relation to me I might be trying to slow it down without thinking about it... Also, if you are an aerobatic pilot you are trying to make your manouevers look consistent.. to the judges on the ground.

 

I will conceed that actually, it might be me causing the issue... I will try a few experiments next time I am flying in a breeze, I might even learn something

 

 

Edited By Dan Hedges on 04/10/2010 15:18:17 - poor spelling!

Edited By Dan Hedges on 04/10/2010 15:19:22

[Martin Harris - Moderator]

Dan,

 

I'm glad you're approaching this with an open mind and I'm sure you'll see where we're coming from because of it. You seem to have pretty much got there in your later paragraphs anyway!

 

With your downwind hand launch it takes time to gain airspeed because you're starting with minus airspeed because you are attached to the ground so you need to make this up - as opposed to into wind where before you do a thing, you've already got some airspeed.  Try the same from each side of a free floating hot air balloon and there will be no difference between those launches.

 

With your "crosswind" turn, it's only crosswind from your static viewpoint.  The model continues to drift in it's "block of air" and doesn't feel the effect of the wind that you do.

Edited By Martin Harris on 04/10/2010 15:30:44

[Dan Hedges]

Thanks Martin,

 

I am always prepared to learn something and this has been quite an enlightening discussion. I have always assumed that the wind works on the model as stated in my opening paragraphs above, but now I think I may have been wrong about that.

 

I also emailed a freind who is a full size aerobatic pilot and he agrees with everyone on here  -(it is no different turining upwind than downwind) with one exception... when he is flying in a contest he will throttle the aircraft up turning into wind and down turning away from it as he tells me his aim is to keep groundpeed consistent so that manouevers look correct to the judges on the ground, so in this case he really does have a different airspeed depending on his direction relative to the wind, but that is on purpose.

 

Oh, I also round this link which was helpfull in getting my head around this.

 

Dan.

 

[DH 82A]

 From Dan hodges' link

 

  " The wind direction can also affect a pilot's behavior--when a pilot is consciously or unconsciously modifying the aircraft's bank angle and airspeed to make the flight path follow some preconceived radius of curvature and forward velocity, IN RELATION TO THE GROUND, rather than judging his flight entirely by the airspeed indicator, the sound of the airflow, and the feel of the controls, then the wind direction certainly does matter, and the pilot risks stalling the aircraft during a downwind turn.  A conscious or unconscious desire to make the flight path follow a fixed radius of curvature and forward velocity, in relation to the ground, will tempt the pilot to slow down too much during a downwind turn, and this is most likely to be a problem when the pilot is maneuvering in relation to an interesting target (such as a desired landing spot) at low altitude."

 

   Which is exactly what we are doing!

      

[Biggles' Elder Brother - Moderator]

Posted by Dan Hedges on 04/10/2010 15:00:50:

 

 

If I fly into wind and make a TIGHT turn downwind I will have the same problem, for a certain amount of time the air will be blowing accross my plane from behind until it has pushed my aircraft up to speed to match the wind +  the thrust from my prop. However while my aircraft is accellerating to match the wind I will have less airflow over the wings and could concievably stall. 

 

 That is exactly my point. Thank you Dan you expressed concisely what I was trying to say.

 

Posted by Martin Harris on 04/10/2010 00:08:33:

Personally, I believe BEB is proposing a situation which cannot physically occur in the real world when using aerodynamic effects to initiate the turn.  

Agreed that the point is certainly hyothetical. But I don't agree the turn could not be initiated by aerodynamic means. It could certainly be initiated by such means - whether it was completed would depend on the relative size of the model's polar moment of inertia vs its translation or linear moment of inertia.

 

 

Posted by Martin Harris on 04/10/2010 00:08:33:

Kinetic energy. This is a red herring and does not concern us. The common mistake is to assume that it affects turns as if there were a law of conservation of kinetic energy. There isn't, so we can forget about it. For the record kinetic energy is 1/2 mass * velocity squared.

 

Sorry Martin - I can't let this go. The Conservation of Energy is one of the most fundemental laws of Physics. You can't escape from it. Pretty well every single mathematical proof in aerodynamics is based on three fundemental laws: the conservation of mass, the conservation of momentum and the conservation of energy.

 

What I genuinely don't understand is that you appear to agree with what Dan says - which I quote above - yet disagree with what I said. The thing is I belive they are the same thing! Now I'm quite prepared to accept that its my poor communication but I'm genuinely puzzled!

 

All I've every said is that at the instant of the turn - if the model can turn faster than it can gain bulk speed (because of inertia effects) there could be point when the model is either deficient or has a surplus of airspeed. Which is what I understand Dan to be saying as well. Confusing

 

BEB

 

Edited By Biggles' Elder Brother on 04/10/2010 21:06:16

Edited By Biggles' Elder Brother on 04/10/2010 21:07:37

[John Privett]

Posted by Biggles' Elder Brother on 04/10/2010 21:05:35:

Posted by Dan Hedges on 04/10/2010 15:00:50:

 

 

If I fly into wind and make a TIGHT turn downwind I will have the same problem, for a certain amount of time the air will be blowing accross my plane from behind until it has pushed my aircraft up to speed to match the wind +  the thrust from my prop. However while my aircraft is accellerating to match the wind I will have less airflow over the wings and could concievably stall. 

 

 That is exactly my point. Thank you Dan you expressed concisely what I was trying to say.

 

 

BEB - the difficulty is that you are (usually!) using aerodynamic forces to turn the plane, and the point in the air about which those forces act is moving with the block of air.  I guess you could put sideways-pointing rockets on the nose and tail (catherine-wheel style!) to cause the plane to very suddenly turn 180 degrees,  but I don't see that the behaviour would be any different going "upwind" or "downwind."  In both cases the plane would suddenly have a negative airspeed and drop out of the sky!

 

Sorry Martin - I can't let this go. The Conservation of Energy is one of the most fundemental laws of Physics. You can't escape from it. Pretty well every single mathematical proof in aerodynamics is based on three fundemental laws: the conservation of mass, the conservation of momentum and the conservation of energy.

 

 

Overall, energy is conserved,  but that's in the entire "system" not just the plane.  Does the kinetic energy of my car remain constant when I accellerate from standstill to 100mph? (Obviously not really 100 officer...)

 

Momentum too is only conserved within a closed system with no external forces acting on it.

 

[Biggles' Elder Brother - Moderator]

Now that's the sort of plane I'm talking about

 

Do you think Webbies have one in stock?

 

BEB

[Geoff N]

Go sloping flying. Flying a glider without an engine in a 40 mph wind soon sorts out your ideas about updwind and down wind turns, and air speed and ground speed. Engines just complicates matters.

[Martin Harris - Moderator]

Perhaps you should build a bit lighter?

Edited By Martin Harris on 04/10/2010 22:58:33

[Biggles' Elder Brother - Moderator]

Posted by Phil Wood on 04/10/2010 23:45:16:

If it picks up to 70mph we launch Timbo. ........He defies the laws of physics.

 

Pol.

 

LOL Why don't I find that difficult to believe

 

Can he turn really fast?

 

BEB

 

[Former Member]

[This posting has been removed]

[Dan Hedges]

Hmmmm... So, if I were to pull back REALLY hard on the stick while flying would my plane do a quantum loop? and do all control line flyers believe in string theory???

 

My head hurts!

[Tim Mackey]

It wont be just YOUR head that hurts if pollycheek keeps this libel up