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That is a banked aircraft doing a turn. I doubt any pilot having done his/her BAK will have any problems with it. Resolution of forces using scaled vectors is fine  You can't freeze it in time and just  not consider one of those forces. ONE of them causes the aircraft to turn.  Nev

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I base my flying on Newtons 3rd law "for every action there is an equal and opposite reaction" When I push the throttle forward, my bank account diminishes, but my life satisfaction increase

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I have never been confused when banking an aircraft at 60°

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10 hours ago, old man emu said:

This is one of those diagrams that I say is causing confusion:

That diagram is a turn not a roll. The force vectors are totally different.

 

Nor is is a diagram of the result where the AOA remains unchanged.

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IF you don't increase the speed in a turn the AoA must increase to provide the extra lift(and your stall margin will consequently reduce).  That's where a lot of our problems come from. Nev

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16 minutes ago, facthunter said:

You can't freeze it in time and just  not consider one of those forces

Actually I don't give a tinker's cuss whether it's a banquet, a roll or a bloody pizza. I'm not interested at all in what happens when and aerofoil is attached to box. I am going right back to the basic generation of a Lift force due to the movement of an airmass across a modified plate. If you want to drift the topic into what happens when a machine applying the principle of aerodynamic Lift operates in practice, please start your own thread. I will follow it with great interest.

 

WHAT I WANTED TO SAY WAS THAT THE COMMON DIAGRAMS AND PICTURES WE USE TO UNDERSTAND LIFT FORCE GENERATION ARE ALWAYS SIDE ELEVATIONS OF THE AEROFOIL. THIS IMAGE IS IMPLANTED IN OUR MIND'S KNOWLEDGE CACHE, SO THAT WHEN WE TRY TO UNDERSTAND THE CONCEPT OF RELATIVE AIRFLOW IN RELATION TO THE WING IN PLAN VIEW, THE IMAGE IN OUR MENTAL CACHE COMES OUT AND CONFUSES US.

 

22 minutes ago, aro said:

That diagram is a turn not a roll

AGREED, but it was the quickest way I could find to illustrate the directions of the force vectors.

 

As for an aerofoil not producing any Lift force when at 90 degrees to the horizontal - look at the tail fin of an aircraft. Air moves across each side, which have the same shape on both sides. Therefore the aerodynamic Lift force must be equal on both sides, each side cancelling the other. It is only when the camber is changed by the movement of the rudder that these forces go out of balance and a desired result is achieved. 

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I think I said the wings won't produce any force against gravity when at 90 degrees to the horizontal.. Lift of an aerofoil always acts at right angles to the surfaces  because it's a result of pressure. (force per unit area) .  Gravity is the main law we all must obey . The reason we need wings.  (Well we could use jets or rockets but lets not go there ).

      Is lift the TERM we would use if it's downwards?   It would not be standard understanding to  do that , but it CAN be suggested as long as it's clearly understood and agreed at the time. Nev

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3 hours ago, Bruce Tuncks said:

60 degrees is some bank angle for sure. Once I watched 2 awestruck at 2 big open-class gliders climbing in a narrow thermal core at about 60 degrees of bank. The wings were noticeably curved from the g loading. They were climbing well.

45 degrees seems very steep when you actually do it too. The diagonal instrument screws are parallel with the horizon. The g forces are noticeable too. I reckon few power pilots have ever done much of this.  In a lightly-loaded 15m  glider you need to do 45 degrees at 45 knots accurately and this is an impossible goal, but like a perfect golf shot, you need to just keep trying.  

 

Agree Bruce, I used to say when you thought you were at 45* you were most likely about 30 to 35* in fact.  Yep you need to push them over for 45*.  Cheers

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2 hours ago, facthunter said:

Is lift the TERM we would use if it's downwards?

Yes we could. Since Lift is a vector, it has direction. The name we have given to the force is "Lift", so that is the term we should use. It gains its "direction" from the angle between the direction it acts and the vertical extension of the direction of action of the weight force of the aircraft. We say that the value of the Lift force vector upwards is positive. 

 

3 hours ago, facthunter said:

think I said the wings won't produce any force against gravity when at 90 degrees to the horizontal..

Yep. That's because, relative to the horizontal axis, the Lift force would be acting at either zero or 180 degrees relative to the horizontal  axis and the magnitude of the Lift force in the horizontal axis is Lift x sin 0, or  of Lift x Sin 180, which are both zero. 

 

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A roll is not a turn and to maintain the direction required the turn has to be stopped and to prevent altitude loss rudder has to be applied. This means that the airflow over the wing is not the same as it was when flying level or in balance. The airflow is diagonally across the wing, therefore the chord is increased and the depth ratio is reduced, so there will be less wing lift perpendicular to the wings and there will be some vertical lift from the fuselage and tailplane.

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IF there's any lift happening  (that force opposing gravity) other than that from the wings there is LESS required from the wings IF the plane stays at the level it is at. If the plane is falling from it's trajectory then the forces in total are not equal to the gravitational pull so again lift is LESS than wings level. These conditions must be understood  and allowed/ compensated for if we are to  analyse what's going on. To perform a turn without losing height OR Airspeed (or both) extra lift must be created by a speed increase or or an AoA increase. EITHER require added energy. Power or height loss if no extra power available. So much lift to hold the plane in equilibrium PLUS that to accelerate it in a new direction  during the period we are TURNING. But ROLLS are Not turns. (Unless you are climbing or descending near vertically).  This last point can be kept born in mind for some other situations'' Assumptions made can lead to erroneous conclusions .  It's a necessary complication to get what's going on right  or just fly the thing by Rote . "(Somebody said to do it this way).. and that will do me."  There are people who do just that but you won't do well when something unusual pops up. Nev

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19 hours ago, old man emu said:

This is one of those diagrams that I say is causing confusion:

Figure 4-14. Forces in a turn.

While this is a totally correct way to explain why the pilot has to take action to increase the resultant lift in order to complete the desired manoeuvre, it does not illustrate how the relative airflow moves around the wing.

 

Here is a scrappy diagram of what I am getting at.

image.png.38f2b6649bddc1b2704780c3839c0c58.png

The rectangles represent the leading edge of a flat plate. The red dots represent the airflow meeting the leading edge, and you can assume that the air diverts over and under the wing as we know it does.  In order to get a bit more Lift force we'll arbitrarily assign an AoA of 4 degrees. I don't have the CAD software to draw that in a diagram. Imagine that the plate is connected to some device that measures the magnitude of the Lift force, and that device allows us to tilt the plate around the halfway point of the span. I have drawn the lower rectangle with a 20 degree tilt down with respect to the top one - simply to fit the diagram on a page. 

 

Now if we pass an airstream (relative airflow), having a known velocity and density of the air, at the top rectangular plate, a Lift force will be generated, which we can depict as an arrow pointing upwards. Our measuring device can tell us the magnitude of that force. Next, without changing AoA , airflow velocity or air density, we tilt the plate away from the horizontal and again obtain a value for the magnitude of the Lift force. The two magnitudes will be the same. 

 

In other words, the angle of the plate relative to the horizontal datum will not affect the magnitude of Lift force generated by the plate. This is because no matter how the plate is orientated in relation to the horizontal datum the density and velocity of the relative airflow are the same. The plate is immersed in the fluid the same way that a fish is immersed in the sea.

 

It is incredibly hard to produce a two-dimensional diagram of this, that's why I said to use a three-dimensional object to help envisage what I am saying.

 

If this plate was set up at 90 degrees to the horizontal datum, would the Lift force be the same as in the other two cases? Yes, but when one wants to use that Lift force to balance the weight force of an aircraft, there is no vector component that is equal and opposite to the weight force. aro mentioned at one point that the horizontal vector of the Total Lift force is used to turn the aircraft. When the wing is at 90 degrees vertical, one would guess that it's all turn and no holding altitude. 

 

Phew!

 

 

So OME, you've got some great info about what happens to the lift of the wing when someone is doing a hesitation roll. Would you care to summarise it for us? 

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3 hours ago, APenNameAndThatA said:

So OME, you've got some great info about what happens to the lift of the wing when someone is doing a hesitation roll. Would you care to summarise it for us? 

Using the hesitation roll was a bad choice of illustration. What I really wanted to show was a wing generating lift at an angle the average pilot rarely goes to. I admit it screwed up what I was intending to discuss.

 

3 hours ago, APenNameAndThatA said:

One more question: what is "kilogram-weight"?

This is where the fight begins. Sorry, but this is going to involve some definitions.

 

I think we agree that the magnitude of a force is the result of the acceleration of a mass, which we can calculate using the equation F = m.a  

 

The biggest problem in understanding this is getting our heads around what exactly is a kilogram. The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.626 070 15 × 10^-34 when expressed in the unit Joules times s seconds, which is equal to kg m2 s -1 , where the metre and the second are defined in terms of c, speed of light and ∆νCs,( radiation produced by the transition between the two hyperfine ground states of caesium (in the absence of external influences such as the Earth's magnetic field) has a frequency, ΔνCs, of exactly 9192631770 Hz.)

 

I don't know about you, but I can't visualise that as a ball of matter. 

 

Let's look at something tangible. Suppose we had a box of a dozen golf balls. They are manufactured to a standard, so we can assume that each one has the same amount of matter in it as the others. That represents the "m" in the equation. If I take a spring balance and cover the inscribed scale with masking tape, I have a device that that obeys Hooke's Law - the force (F) needed to extend or compress a spring by some distance (x) scales linearly with respect to that distance. If I connect the box of golf balls to the spring balance, the spring will stretch and I can mark the position of the pointer that indicates the length the spring has stretched on the masking tape.

 

Let's say that we apply Hooke's law by taking the length of stretch of the spring and working out the force required to stretch the spring that far and plug that value into our original Force equation. For the sake of the discussion, let's say that the value of the force was 14 Newtons, and we know that the acceleration due to gravity where we did the weighing was 9.81 m/s/s. The units of mass  are kilograms because we are working in Newtons. So we have

F = m.a

14 = m. 9.81

Dividing both sides by 9.91

14/9.81 = m

1.427 kg = m

So the force registered by a scale is not really a measure of the amount of matter being weighed. The weight of an average adult exerts a force of about 608 N.

608 N = 62 kg × 9.80665 m/s2 (where 62 kg is the world average adult mass)

 

 

So, to finally answer your question, a one kilogram-weight is the force resulting from the from the acceleration of (1 divided by 9.81) kilograms of matter due to the Earths gravitational attraction acting on the mass. 1 divided by 9.81 = 0.102 kg of matter.

 

If you go to the greengrocer to buy a 1 kg mass of potatoes, his scale will be marked off with symbols that say kilograms, but you know that in reality, when he puts your potatoes into the weighing pan, the spring in the scale will stretch is response to the application of 9.81 Newtons of force.

 

Scales show Kilograms because that is what people understand best, but it is really just an estimate of the mass above them. Scales should really show Newtons, but that might confuse people!

 

Thanks for coming. I'm here 'til Thursday. Try the veal.

 

 

 

 

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9 hours ago, old man emu said:

I think we agree that the magnitude of a force is the result of the acceleration of a mass, which we can calculate using the equation F = m.a

 

We do not agree. That is an equation about acceleration. You can have a force where acceleration is zero.

 

If you push a box of bricks across the floor at a constant speed, you are exerting a force to overcome friction. You can measure the work done as force x the distance you moved the box. Acceleration (once the box is moving at a constant speed) is zero, but you have a force that must be used if you want to calculate e.g. work and power.

 

Likewise, when an aircraft is straight and level at a constant speed (i.e. no acceleration) we do not say that there is no force acting on the aircraft. If we add all the forces we get zero net force, true, but that does not help us define force.

9 hours ago, old man emu said:

The biggest problem in understanding this is getting our heads around what exactly is a kilogram. The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant....

You do like complex explanations.

 

That doesn't really answer "what is a kilogram?", it answers how much is a kilogram. Previously there was a lump of stuff with a mass of exactly 1 kg for comparison, but the problem is any physical object can gain or lose mass to the environment. So that is the answer they came up with for how do you define a kilogram without using something that can physically vary.

 

For our purposes, the mass of 1 litre of water, or the mass of a reference object of 1 kg is almost certainly sufficient.

 

A kilogram is the measurement of mass: how much "matter" is in an object. Effectively a measure of how many protons, neutrons etc. it contains in it's atoms. (Can we ignore relativity please!)

 

Kilogram-weight is a measure of force - the amount of force exerted by gravity on a mass of 1kg on earth. We want to measure kilograms (mass), but scales measure force. Calibrating them in kilogram-weight and calling it kilograms is convenient. We could measure newtons and divide by 9.8, that would be less convenient.

 

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Back in the olden days when I used to teach this stuff for a living, the kg was a block of platinum held in a vault in Paris. The unit of force was the newton, which would accelerate a kg mass by 1 metre per second per second.

The unit " kilogram weight" was frowned upon because it involved knowing just what the gravitational acceleration is. If you take this as 9.81 m/sec2. then AT THAT PLACE a kg-weight is 9.81 Newtons.

There is an instrument called a gravimeter which measures this 9.81 variation very accurately and for example will help you find buried minerals. A geophysics mate of mine mapped out some ancient buried rivers in the outback using a gravimeter. ( deviations from the 9.81 are called " anomalies".

Well its better than the old english system, where the unit of mass was the slug.  One pound-force would accelerate a slug at one foot per sec2. All our engineering loads were in kips, that is kilo-pounds. What a mixture of metric and olde world the kip was huh.

OME, you are sure right about the definition of the kg mass being incomprehensible to simple folk these days.

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Mass is how much STUFF is in it. (protons neutron=Klingons electrons add ons and so ons

. Weight is how hard the planet you are on's GRAVITATIONAL attraction is pulling it down,up sideways towards it..  Mg  A Balance works anywhere. just because it's all  affected equally . Nev

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53 minutes ago, aro said:

We do not agree. That is an equation about acceleration. You can have a force where acceleration is zero.

 

If you push a box of bricks across the floor at a constant speed, you are exerting a force to overcome friction. You can measure the work done as force x the distance you moved the box. Acceleration (once the box is moving at a constant speed) is zero, but you have a force that must be used if you want to calculate e.g. work and power.

aro, I am crying tears of blood. You keep saying that I know f-all about physics. I cry because you know f-nothing. 

 

In physics, a force is any interaction that, when unopposed, will change the motion of an object. A force can cause an object with mass to change its velocity (which includes to begin moving from a state of rest), i.e., to accelerate. Force can also be described intuitively as a push or a pull. A force has both magnitude and direction, making it a vector quantity. It is measured in the SI unit of newtons and represented by the symbol F.

 

Characteristics of Force:

Forces are due to an interaction of at least two objects.

It may change the state of motion of an object.

It may change the shape of an object.

 

The very definition of Force involves a change in velocity over time (acceleration)  F = m.a  Force is proportional to acceleration, which is defined as the rate of change of velocity. It is also proportional to mass for a particular acceleration, but we don't usually talk about varying the mass, except if we are talking about stall speeds at different amount of load a plane is carrying, but let's not go there right now.

 

Work is the energy transferred to or from an object via the application of force along a displacement. Work is a scalar quantity, so it has only magnitude and no direction. Work transfers energy from one place to another, or one form to another. The SI unit of work is the joule (J), the same unit as for energy.

 

To calculate the amount of work done we use the equation W = Force . distance (W = F.d = m.a.d) The units of Work are Joules which have the dimensions

image.png.39c979246cb1cde8522c179be72ed1b2.png

which is the result of multiplying mass by acceleration by distance

image.png.d1e57bab28ff108294ffee6b9c547679.png

 

In your example of a box being pushed across the floor, consider Newton's Third Law - equal and opposite forces.

Friction Lesson for Kids: Definition & Examples - Video & Lesson Transcript  | Study.com

If you want to move the box in a certain direction, you apply a force to it. Because of the surface of the box and the floor are not perfectly smooth (an impossible situation, ever an molecular level) they lock together.

image.jpeg.08bf235b05bf34e865f594fbea615bb4.jpeg

To get the box moving you have to overcome that mechanical interlocking. That resistance to movement due to mechanical interference is called Friction, which is a Force acting in the opposite direction to the desired direction of movement. The level of friction that different materials exhibit is measured by the coefficient of friction. The formula is µ = f / N, where µ is the coefficient of friction, f is the amount of force that resists motion, and N is the normal force, or force acting at right angles to the desired direction of motion. In this case, it is the force of gravity.

 

Moving the box involves exerting a Force that is equal to Force to overcome the Friction force. Once that force has been overcome, the box will move and Work will be done. To keep the box moving you have to apply a force that is greater that Friction force. Since the force due to gravity never changes while you are applying the force to overcome Friction, you can assign it a value of 1 in calculations and add the real value in at the end of the calculations. Perhaps this diagram will help.

Force 

 

 

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13 minutes ago, facthunter said:

Mass is how much STUFF is in it.

Too true. The big problem is that we really cannot work out the mass of an atom so that we can multiply that number by The Avogadro constant which is the proportionality factor that relates the number of constituent particles (usually molecules, atoms or ions) in a sample with the amount of substance in that sample. The numeric value of the Avogadro constant expressed in reciprocal mole, a dimensionless number, is called the Avogadro number, which is exactly 6.02214076×10^23 is thus the number of particles that are contained in one mole (not one of these). 

image.jpeg.5414dc56cc08a0260f9408c619c32a37.jpeg

The mole (symbol: mol) is the unit of measurement for amount of substance in the International System of Units (SI). It is defined as exactly 6.02214076×1023 particles. The trouble is that, for example, one mole of water contains 6.02214076×1023 molecules, whose total mass is about 18.015 grams. Notice the circular argument? The mass of 1 mole of a substance is equal to its relative atomic or molecular mass in grams. The relative atomic number is related to Carbon, whose atomic number is 6 and atomic weight is 12. You could work out the mass of an atom of carbon by the use of Einstein's equation, {\displaystyle E=m\,c^{2}} but that a real lot of energy since C^2 is a massive number.

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Surely an unopposed force is no force at all? IF your person had roller skates on it could not exert a force on the box. IF the box had the magic (in Physics) frictionless surface what force could be exerted on it?  ONLY  the force that accelerates it.

    THEN your F=ma applies. 

     You cover work there where something happens as a result of an applied force.. A force does some work in moving or changing something. . If it's frictionless almost NO work is needed to move a mass on a horizontal surface.

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It could be that nobody knows anything and are just getting stuff from Google. I see these same sorts of posts in a few other (non-aviation) forums. I'm one of those folks who know a lot about nothing and a little of everything. Just clever enough to get by and not lose an eye in the process.

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2 minutes ago, WayneL said:

I just fly the aeroplane........it works for me!!!!😁

I base my flying on Newtons 3rd law

"for every action there is an equal and opposite reaction"

When I push the throttle forward, my bank account diminishes, but my life satisfaction increases.

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13 hours ago, old man emu said:

Using the hesitation roll was a bad choice of illustration. What I really wanted to show was a wing generating lift at an angle the average pilot rarely goes to. I admit it screwed up what I was intending to discuss.

 

This is where the fight begins. Sorry, but this is going to involve some definitions.

 

I think we agree that the magnitude of a force is the result of the acceleration of a mass, which we can calculate using the equation F = m.a  

 

The biggest problem in understanding this is getting our heads around what exactly is a kilogram. The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.626 070 15 × 10^-34 when expressed in the unit Joules times s seconds, which is equal to kg m2 s -1 , where the metre and the second are defined in terms of c, speed of light and ∆νCs,( radiation produced by the transition between the two hyperfine ground states of caesium (in the absence of external influences such as the Earth's magnetic field) has a frequency, ΔνCs, of exactly 9192631770 Hz.)

 

I don't know about you, but I can't visualise that as a ball of matter. 

 

Let's look at something tangible. Suppose we had a box of a dozen golf balls. They are manufactured to a standard, so we can assume that each one has the same amount of matter in it as the others. That represents the "m" in the equation. If I take a spring balance and cover the inscribed scale with masking tape, I have a device that that obeys Hooke's Law - the force (F) needed to extend or compress a spring by some distance (x) scales linearly with respect to that distance. If I connect the box of golf balls to the spring balance, the spring will stretch and I can mark the position of the pointer that indicates the length the spring has stretched on the masking tape.

 

Let's say that we apply Hooke's law by taking the length of stretch of the spring and working out the force required to stretch the spring that far and plug that value into our original Force equation. For the sake of the discussion, let's say that the value of the force was 14 Newtons, and we know that the acceleration due to gravity where we did the weighing was 9.81 m/s/s. The units of mass  are kilograms because we are working in Newtons. So we have

F = m.a

14 = m. 9.81

Dividing both sides by 9.91

14/9.81 = m

1.427 kg = m

So the force registered by a scale is not really a measure of the amount of matter being weighed. The weight of an average adult exerts a force of about 608 N.

608 N = 62 kg × 9.80665 m/s2 (where 62 kg is the world average adult mass)

 

 

So, to finally answer your question, a one kilogram-weight is the force resulting from the from the acceleration of (1 divided by 9.81) kilograms of matter due to the Earths gravitational attraction acting on the mass. 1 divided by 9.81 = 0.102 kg of matter.

 

If you go to the greengrocer to buy a 1 kg mass of potatoes, his scale will be marked off with symbols that say kilograms, but you know that in reality, when he puts your potatoes into the weighing pan, the spring in the scale will stretch is response to the application of 9.81 Newtons of force.

 

Scales show Kilograms because that is what people understand best, but it is really just an estimate of the mass above them. Scales should really show Newtons, but that might confuse people!

 

Thanks for coming. I'm here 'til Thursday. Try the veal.

 

 

 

 

I have to say that I think that it is poor form to have everyone come up with interesting answers to your question and then decline to think about them because it turns out that the only thing that you were interested in was that sine 45 = 0.7. 

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3 hours ago, aro said:

 

 

We do not agree. That is an equation about acceleration. You can have a force where acceleration is zero.

 

If you push a box of bricks across the floor at a constant speed, you are exerting a force to overcome friction. You can measure the work done as force x the distance you moved the box. Acceleration (once the box is moving at a constant speed) is zero, but you have a force that must be used if you want to calculate e.g. work and power.

 

Likewise, when an aircraft is straight and level at a constant speed (i.e. no acceleration) we do not say that there is no force acting on the aircraft. If we add all the forces we get zero net force, true, but that does not help us define force.

You do like complex explanations.

 

That doesn't really answer "what is a kilogram?", it answers how much is a kilogram. Previously there was a lump of stuff with a mass of exactly 1 kg for comparison, but the problem is any physical object can gain or lose mass to the environment. So that is the answer they came up with for how do you define a kilogram without using something that can physically vary.

 

For our purposes, the mass of 1 litre of water, or the mass of a reference object of 1 kg is almost certainly sufficient.

 

A kilogram is the measurement of mass: how much "matter" is in an object. Effectively a measure of how many protons, neutrons etc. it contains in it's atoms. (Can we ignore relativity please!)

 

Kilogram-weight is a measure of force - the amount of force exerted by gravity on a mass of 1kg on earth. We want to measure kilograms (mass), but scales measure force. Calibrating them in kilogram-weight and calling it kilograms is convenient. We could measure newtons and divide by 9.8, that would be less convenient.

 

So, we can agree, finally, that something with a mass of 0.3 kg does NOT have a weight of 3 kg? 

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