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11 minutes ago, RFguy said:

My point is that ignoring something intentionally (but being aware of its presence)  is done to simplify an analysis that could otherwise be too complex to take into account every possibility.

A turn is a change of momentum. The change of momentum is what is being analyzed. You can ignore some things e.g. kinetic energy and even gravity and get a useful answer. But not momentum.

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Wind-shear aside,  if we take it that we're flying within a moving parcel of air, then, is 'turning into a tailwind' really a thing, up-there, as it is, on the ground, say, while taxiing?  Of cou

Skidding from base to final. The low speed is not much relevant as is angle of attack. This video may help in explaining differences between skidding and slipping. https://www.youtube.com/wa

I think a basic understanding of lift and how a plane gets airborne is essential, but the deep physics isn't needed and doesn't make you a better pilot. 

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I'm not analysing that.  anyway, if the momentum , in POLAR vector (lets say 2D space)  notation is  1 angle 90 kgms-1 and changes to 1 angle 270 kgms-1, then the modulus is not changed, it is still 1 .   but you probably want 3D space...  

 

anyway, enough.  I agree with your strict interpretation of the matter.  that's fine     :-) 

 

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Yenn, what  for you would define the limit of uphill slope you might land on ? 

 

I have done very few slope landings. really really minor stuff most probably wouldnt consider as a slope

 

Crookwell was probably the most slope. land downhill, takeoff downhill. (high grass was very slow to accelerate) 

 

Harden landed uphill, took off uphill.  (clay/dirt)

 

 

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11 minutes ago, Yenn said:

I don't know what you mean by "use your shoulder" but I always adjust the start of my turn from downwind according to the wind direction. A strong cross wind will either extend my downwind if it is going to be a tailwind on base, or reduce my downwind if it will be a headwind on base. If I am going like a rocket on downwind I will start the turn early, so that I don't undershoot.

 

Yenn on downwind to get myself positioned to the strip for the run downwind I use my shoulder as a guide..it will be different for everyone I suppose but right from my gliding days until now thats the way I do it. once i am abeam the end of the runway I totally pull the throttle and then hit the trim to slow down then flaps at the right speed then a nice curved turn at about 20 to 25deg of bank in a decending turn to final. Regulating my speed. I cant explain it any further...my shoulder is a reference point I use...the strip is not hidden or online with my shoulder its so long as I can see the strip above my shoulder and I remain in the approved circuit area...

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Velocity - This is a value that is described in terms of magnitude and direction. It is a VECTOR quantity

Mass - This is a value that is described in terms of magnitude only. It is a SCALAR quantity.

Acceleration - This is a value described in terms of CHANGE in the magnitude of Velocity after an elapsed period of time. Because velocity is a Vector, acceleration is a vector.

Time - Since we usually consider time to go from the past to the future, we consider it linear and having magnitude only, therefore it is a scalar.

Distance - Has magnitude only. It is a scalar.

Momentum - is the product of a scalar and a vector, mass and velocity, therefore, momentum is a vector.

Force - is the product of a scalar and a vector, mass and acceleration, so it is a vector.

 

Scalar quantities of the same item can be treated by simple arithmetic. Vector quantities, having both direction and magnitude are treated by the method of addition of vectors (Pythagoras). 

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14 minutes ago, Garfly said:

But I'm here to learn

Trouble is that if people start using terms in the wrong way, we'll end up only learning how to confuse.

 

It's hard to carry on a conversation about a topic that is strongly practical in application, but has an equally strong theory background. Half the people get bored by the theory and the other half by the practical. You've just got to decide how deeply you will dive in.

 

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Lets go back to the original statement and look at it in more detail:

1 hour ago, RFguy said:

1) The aircraft is travelling at 100 K over the ground headed south 

2) and there is a 50 K headwind from the south   

3) therefore IAS is 150 K. yes, agree ?

OK

 

1 hour ago, aro said:

now, the aircraft 'instantly' turns left 90 degrees (and we assume the turn does not cost any energy)

The laws of physics don't allow it. It's like dividing by zero in the middle of a sequence of calculations. Everything that follows is invalid.

 

1 hour ago, RFguy said:

4) the ground speed is still 100 K (momentum was conserved )

After a turn, if kinetic energy is unchanged, airspeed must still be 150kt. If momentum is unchanged you are still travelling at 150kt in the original direction while pointing 90 degrees to the original path. (In a helicopter maybe?)

 

1 hour ago, RFguy said:

6) Now ground speed = airspeed = 100K. We lost 50 K of airspeed by turning.

It is important that people understand this is wrong. Because airspeed, velocity, momentum, lift are all relative to the air, not the ground.

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"It is important that people understand this is wrong. Because airspeed, velocity, momentum, lift are all relative to the air, not the ground."

YES I AGREE, to be sure no one goes down this same track ,   my example was flawed and with respect to THE GROUND which was NOT the subject of this discussion and I confused myself and everyone else....

 

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It just amazes me how I can have a brain explosion and completely forget the parcel of air.... 

 

I am clearly human. Ahh it is good to be brought back to earth occasionally with a thud.

 

 

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OME - you wanted some circuit figures (I wont mention the aircraft make or model, someone might accuse me of making a sales pitch): This diagram forms part of the POH for this aircraft. Not on the schematic, is the stall speed of 27 knots, so to minimise ground role, expect to touch down at sub 30 knots to minimise float/ground effect. Max fuel load for this aircraft is 100L. Typical empty weight sub 300 kg. Max TO weight  (in Au) 600 kg

 

Okruh%20AJ-1.jpeg

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6 hours ago, RFguy said:

yes we are flying within a parcel of air sure, but my point is (without a shear etc) the airplane does not go instantaneously from a headwind to a tailwind . turning the aircraft takes time, and as long as you know your airspeed often enough (by looking at the ASI or feeling it through your hair) , you wont be caught out.  if you are  S&L 10 kts about the stall with a 30 kt headwind, and turn 180 "instantly" and are flying S&L   then now you are 20kts below the stall speed. But it wont happen that way on a 30 deg AOB turn at base/final speeds of rec aircraft.

 

and yes, I agree  the wing is relatively unloaded in a moderate speed steep descending < 30 deg turn. Not much required of the wing...  very different to the high AoA / flow separation stall condition. wing will be at low AoA for high flying descent rates.

  

No. If you are straight and level into a 30 kt headwind and are 10 kt above stall, and you turn instantly 180 degrees, you are 10 kt above stall

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5 hours ago, RFguy said:

(my reference is the ground)
"making a 180 degree turn within our parcel (all else being equal) has no effect on the airspeed - and thus, the approach of the stall speed - whatsoever."
 

I disagree. Let me provide an example : 

 

1) The aircraft is travelling at 100 K over the ground headed south 

2) and there is a 50 K headwind from the south   

3) therefore IAS is 150 K. yes, agree ?

now, the aircraft 'instantly' turns left 90 degrees (and we assume the turn does not cost any energy) 

4) the ground speed is still 100 K (momentum was conserved ) 

5) the headwind is now ZERO because the wind is now off the wing .

6) Now ground speed = airspeed = 100K. We lost 50 K of airspeed by turning.

 

 


 

4) is wrong. Ground speed increases. 

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4 hours ago, RFguy said:

No, the quantity of momentum does not change. 

the forces acting on the plane are unchanged because gravity is directly below us. 

so for these purposes, momentum can be a scalar.

 

yes- in real life  there will be force required to make that change  but I specifically said that was not  going to be taken into account. 

 

Momentum is a vector. Otherwise, two blocks running into each other on a frictionless plane would have the same momentum after the crash regardless of the directions they were going before the crash. 

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43 minutes ago, Kyle Communications said:

Stall of 27kts at 450kg  not 600kg. The savannah can stall at 27 kts too...at 450kg but 30 kts at 600kg with full flap

 

Seems the same as a Foxbat, at idle. At about 575 kg, at full throttle, at 30 kt indicated, the Foxbat climbs at 400 fpm.

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Your pitot is telling you lies. If you are at full throttle on climb doing 30kts indicated ...the AOA would be very steep and you would be like a helicopter on its way down not going up

 

 

 

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There is a general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the total momentum of a system remains constant.

 

Basically this says that if two objects, J & K will collide in the future, the sum of the individual momentum of J + the individual momentum of K has a value. After the collision the individual momenta can change, but the sum of the changed momenta will be the same as the total pre-collision momentum of the system. In analysing the collision, one accepted assumption is that all forces act through the centre of mass, as if all the mass of the object was condensed into a point.

 

If the two blocks of equal mass m, were both travelling at equal velocity v, then each would have the same momentum, mv. However if we assign positive velocity to velocity in the left to right direction, then if the other block is travelling right to left, its velocity is -v.

 

If the blocks collide the final momentum will be mv - mv = 0. The blocks will stop at the point of impact. Note also that the pre-impact total momentum was also mv - mv = 0. This is as unique case in the real world. Collisions usually occur with both objects approaching the point if impact from an angle. This means that to calculate the post-impact momentum of each object, you have to do a vector analysis in two dimensions (plot the movements on an X-Y coordinate system.

 

I'm not going to write out the equation for this vector analysis as it involves using the sine and cosine of approach and departure angles, and a lot of readers here wouldn't be able to see the forest for the trees without experience working with X-Y coordinate systems.

 

Now here's a confuser. Momentum involves a vector - velocity - which has both size and direction. So if an aircraft of mass m, is flying along, straight and level, at velocity, v, then it has momentum of mv. If anything happens to change v, then the momentum of the aircraft changes. Making a turn is, in physics, a change in the vector value of velocity, so we say that the aircraft's momentum has changed. It's a bit hard to spot the source of the Force that makes the change, but it could be change in engine power (chemical energy) or use of controls (mechanical energy)

 

Phew! I need a Bex and good lie down.

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1 hour ago, Kyle Communications said:

Your pitot is telling you lies. If you are at full throttle on climb doing 30kts indicated ...the AOA would be very steep and you would be like a helicopter on its way down not going up

 

Stall of 27kts at 450kg  not 600kg. The savannah can stall at 27 kts too...at 450kg but 30 kts at 600kg with full flap

 

 

 

2 hours ago, APenNameAndThatA said:

Seems the same as a Foxbat, at idle. At about 575 kg, at full throttle, at 30 kt indicated, the Foxbat climbs at 400 fpm.

Now now Kyle, its not like you to be making such dogmatic comments.

 

I flew a Foxbat, for the first time, last weekend.  Being between aircraft & needing some practice/currency, I felt that the Foxbat has a circuit performance, very similar to the aircraft I am most familiar with and not to dissimilar to the one that is in the above schematic - I was correct. The training was most helpful and I will repeat it periodically until such time as I have found a new mount.

 

These figures may not sit well with you but they have been demonstrated in the real world, not just factory "gloss" and anyhow this conversation is about Turns at Low Air Speed not a contest between aircraft  & their supporters.

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3 hours ago, Kyle Communications said:

Stall of 27kts at 450kg  not 600kg. The savannah can stall at 27 kts too...at 450kg but 30 kts at 600kg with full flap

 

 

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Here is a point of confusion which I have noticed in this discussion... the velocity reference appears by some to be assumed to be the ground. In fact it is the air.

So a 50 knot asi helicopter in a 50 knot wind on the nose has zero groundspeed...  so what? If it turns to fly downwind still at 50 knots it will have 100 knots groundspeed but unless it interacts with the ground ( like trying to land) , this is irrelevant. the momentum of the helicopter has changed dramatically of course, and this is the result of the force times time applied to make it turn. The momentum has reversed.

We should however always state the frame of reference we are using.

Absolute velocity? There is no such thing. The planet Earth has about 100,000 knots speed with respect to the galactic center. Just as well we don't notice that in our Jabirus huh.

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11 hours ago, Kyle Communications said:

Your pitot is telling you lies. If you are at full throttle on climb doing 30kts indicated ...the AOA would be very steep and you would be like a helicopter on its way down not going up

 

 

 

I’m not saying its not saying lies. It’s still weird. The plane shudders a bit, as if, maybe, part of the wing is stalled.

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12 hours ago, old man emu said:

Making a turn is, in physics, a change in the vector value of velocity, so we say that the aircraft's momentum has changed. It's a bit hard to spot the source of the Force that makes the change

The source of the force that makes the change is the wing.

 

We bank the aircraft, which means that there is a sideways component to the lift force. This sideways force turns the aircraft.

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