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You might recall that I've been prattling on about Lift generation and relative airflow, and it's all been theoretical with diagrams and trigonometry and some smoke and mirrors. Basically I was saying that the wing doesn't care what angle it is to the relative airflow, as long as the angle between it and the airflow is less than, say 15 to 16 degrees, and the speed of the airflow over the wing is sufficient, it will produce enough lift to do what is required at that time - climb, fly level or descend.

 

A bloke told me of an interesting fact about his plane when it is on finals. The plane is a little single-seater, and if memory serves me, the aerofoil is a pretty high lift type. Other than that, the plane is light weight and fairly basic. He told me that at cruise revs ( about 6000 rpm) the plane flies along nicely, going where he points it. However, as he is coming down finals, using about 4500 rpm, and pointing at a normal sort of descent angle, the wing is producing enough extra lift that the horizontal component is drawing the plane forward faster than it can descend. The result is that if he chooses the piano keys as an aiming point, to reach them he has to steepen the descent angle, otherwise he will land long. It's sort of like he has a built-in tailwind in that situation.

 

Are there any other RAAus type aircraft that exhibit this "built-in tailwind" on finals?

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 OME .  Sounds like BS to me.  There is so much wrong with that description it defies analysis. Why keep power on if you are overshooting?  Why is it IN Social Australia?  I guess it's not meant to be serious. Nev

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15 hours ago, facthunter said:

Why is it IN Social Australia

Because I wanted only sensible people to see it.

 

15 hours ago, facthunter said:

Sounds like BS to me.

I didn't elicit that comment, it just came up in conversation about flying his aircraft. We weren't having a BS session.

 

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15 hours ago, facthunter said:

Why keep power on if you are overshooting?

He didn't say that he was overshooting. He simply said that when operating at the engine rpm dictated by the POH, he noticed this effect. If a pilot is flying his own plane, then he gets to know its idiosyncrasies and adjusts for them. 

 

His comment has triggered another line of thought in relation to the Lift/Weight ratio and AoA/Relative airflow. I'm going to need some time to think about that line of thought, which will involve some vector diagrams and a bit of trigonometry. I've got a few domestic tasks today which will leave my mind free to ponder, so I'll expand on this tonight.

 

Before going too much further, answer me this. Let's assume for the sake of the discussion that you have the OK to make a straight-in approach, and you are flying straight and level at 500'  AGL flying level. What control inputs do you make so that when you reach the usual place you would join final, you begin a normal descent flight path to the keys?

 

Also, since I don't fly, give me an airspeed at beginning of finals and a stall speed, so I have realistic numbers to work with.

 

See you tonight!

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52 minutes ago, facthunter said:

This whole thing has to be a joke , right?

No, Nev. I'm just setting up the conditions for a hypothetical. I am being serious, and would appreciate your input. I just needed a height AGL as a starting point.

 

So far I've got 500" AGL and an airspeed of 6080 feet/minute (60 kts). Also a condition is that it is a Nil Wind situation.

 

Now your input:  What control inputs do you make so that when you reach the usual place you would join final, you begin a normal descent flight path to the keys?

 

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it's completely variable depending on power  , airspeed and flap etc( configuration ) changes and how the particular aircraft responds to them regarding pitch control loads. The approach  slope angle will require less power than the equilibrium S&L setting because you are using some of your potential energy  (height loss) in lieu of some of the engine's energy to oppose the drag.. Nev

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In practice you are correct and I agree wholeheartedly with you. But introducing all those variables to a HYPOTHEITCAL at the very beginning makes the whole thing too complicated to go on with. So let's start off with the simplest case - a wing moving through a stationary airmass starting a distance above  ground level and flying at a constant angle towards the ground. In other words, we are defining the three sides of a Right Triangle.

 

Now, can you tell me, how do you make an aeroplane fly at a constant angle towards the ground? What controls of the aeroplane do you manipulate to get that result?

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Is this somewhat of a correct description of how to make an aeroplane descend?

 

1. Reduce power to reduce Thrust because we are going to pinch some of the gravity force to make the aeroplane go down.

2. Make the aeroplane go below the horizontal by moving the stick forward and then centralise the stick and use the elevator trim to maintain the downwards angle.

 

After you critique the above, then I'll talk about AoA.

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Essentially correct.. I'd reword 2 to adjust pitch (with elevator)  to desired nose down ATTITUDE to maintain/attain desired  (descent) speed. Trim as required is always done, (except when the change is of short duration) but is only used to RELIEVE stick pressure. . Nev

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OK. We are on the same page.

So we set the ATTITUDE with the elevator. Once the attitude is established, can we centralise the elevator if the attitude remains constant? 

 

Do we close the throttle because we don't need as much Thrust from the engine because gravity is adding to the forward motion that Thrust imparts, and that gives us the motion through the air that relates to the desired airspeed?

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I don't apply "Centralise" to the elevators like I would to the ailerons as the forward 1/2 stick is rarely used by the average pilot except  when checking control effect and freedom. pre  flight.. Depending on what changes happen to C of G , speed and lift the control can go to something like where it was pre  lowering the nose.

 2 YES as you would in a car going downhill. The need for the motors thrust is reduced as the Potential energy is available from height loss. to replace it.. IF you have sufficient nose down pitch change you may not need any engine thrust at all to have a suitable speed .Nev

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OK. Now for some numbers. I picked the numbers for ease of arithmetic.

 

Let's have our plane messing about at 3000' on QNH - just to get the discussion away from the circuit. We'll also have NO WIND to further simplify things. Then we will set up an airspeed on a descending path of 60 kts. That's an airspeed of 1 nautical mile per minute, or 6080 ft per minute. Then let's establish a rate of descent of 500 ft per minute. 

 

The angle of the descending path relative to the horizontal can be found by a a bit of trigonometry (sorry, but it won't hurt  bit). 

sine (angle) = 500/6080

sine (angle) = 0.0822

The angle whose sine has the value of 0.0822 is 4.71 degrees

 

For that angle of descent we can call it 5 degrees for simplicity. Does that sort of angle appear OK with you?

 

Just to put things straight. Before the aircraft started its descent it was flying straight and level. The longitudinal axis of the aircraft was parallel to the horizontal. Once the descent was established with the airspeed and rate of descent, the longitudinal axis of the aircraft is 5 degrees below the horizontal.

 

(Got to go to work right now. Will be back after dinner if you care to reply to this post. - OME)

 

 

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OK. Now we come to the make or break bit.

 

1. Do you agree that if I take a wing and move it through an airmass, a nett force will be created? 

2. Do you agree that the nett force created by the wing through an airmass can, in the very simplest of terms be said to exert its effect at 90 degrees to the direction of a  straight line joining the leading and trailing edges of the wing? (NOTE: For the sake of simplicity, I am moving a wing, not a wing and attached airframe)

image.png.e0fcf71eb1bcd7eb0eeb9fc419da15b7.png

 

 

 

 

 

 

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No. 2. I know that you have in your mind all the other variables like Centre of Pressure and AoA, but at this stage of the discussion, ignore those variables and stick to the very simplest explanation of the direction that the nett force of lift acts.

 

Have a good day and see you tonight.

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OME I am jumping in here half way through your discussion, but I am wondering about your "longitudinal axis " of the aircraft. You have changed from cruise to  60kt descent speed, which I assume to be a slower speed than cruise. Would not the longitudinal axis have a higher angle of attack, rather than lesser.

I think what you are using as longitudinal axis is really the slope of motion through the air.

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Just

 

32 minutes ago, Yenn said:

but I am wondering about your "longitudinal axis " of the aircraft.

 

Forget about anything to do with the wing attached to an airframe. I'm simply talking about an aerofoil shape moving around the "Y" axis.. Look at this diagram of 3D coordinates:

THE RIGHT HAND RULE FOR ACAD COORDINATE SYSTEM - YouTube  image.thumb.png.3b1662ccb8732fb29c63efdf3914eab6.png

X direction is the line from front to rear = chord

direction is from left to right = wing span

Z direction is up and down with the "down" direction representing the force of Gravity.

Where the three axes meet is defined as point (0,0,0)

The longitudinal axis of the wing is represented by the X direction. 

 

Put the fingers of your right hand as they are shown in the diagram. Then grasp your middle finger lightly and rotate your hand around your middle finger. You can see that your index finger, which represents the X direction dips up and down.

 

1 hour ago, Yenn said:

You have changed from cruise to  60kt descent speed, which I assume to be a slower speed than cruise.

I said that the aircraft (let me correct that to "wing") was moving  at 60 kts airspeed through the airmass at a constant height above the ground and going in a constant direction across it (straight and level). Then I said that the wing was put on a constant direction path (straight) but that path was at an angle to the level path and that path lead to a position closer to the ground. I also said that we were messing about at 3000' AGL. Isn't it possible to move a wing in that direction through an airmass at 60 kts airspeed?

 

If I moved an aircraft on an angled path towards the ground at 40 kts, we'd call it a "descent". If I moved the aircraft on the same path at 100 kts, we might call it a "dive". Beginning at a height of 4,600 m (15,000 ft), the Stuka would roll 180°, automatically nosing into a dive. The aircraft would then dive at a 60-90° angle, holding a constant speed of 500 to 600 km/h (270 to 320 kts; 310 to 370 mph), until it had gone some 90% of the way to the ground.


 

 

image.png

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When a wing produces lift it also produces drag to do it. You can't have that lift for nothing The nett force  acting on the wing  as a vector will be somewhat aft of your vertical line at an  angle..  As Bill Whitney says "the never ending hill". Nev

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Will you stop going too far at this stage, please!. I said that I was trying to look at the absolutely simplest situation. I know that the Nett Force acting on the wing is the vector sum of a number of sources. You didn't mention skin friction or wing tip vortices. Please ignore them at this stage. If we want to determine the Nett Forces later, then we can. They are the "trimmings on the Rosary" http://boreelog.com.au/the-poems/the-trimmins-on-the-rosary

 

Let's keep it to the simple thing that Lift is a force (depicted in the Z direction on 3D coordinates) that acts at 90 degrees to the wingspan (depicted in the Y direction on 3D coordinates).

 

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