Question on forces acting on an aircraft in climb
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In a banked turn,weight is a constant,
Even this bit is arguably wrong. Weight is a force. F=M*A. In a turn the aircraft is subject to the sum of acceleration due to gravity AND the acceleration of the turn, so it's effective weight is increased. The wing needs to produce additional lift to support this, which is why it requires either a higher AofA or extra speed to maintain altitude.
In a banked turn,weight is a constant,
Even this bit is arguably wrong. Weight is a force. F=M*A. In a turn the aircraft is subject to the sum of acceleration due to gravity AND the acceleration of the turn, so it's effective weight is increased. The wing needs to produce additional lift to support this, which is why it requires either a higher AofA or extra speed to maintain altitude.
Weight is not equal to the F in F = MA. (unless the object is in frictionless free fall of course).
In straight and level flight the aircraft is NOT subject to an acceleration due to gravity. It IS subject to a force due to gravity, but since equal and opposite lift is being applied the resultant force is zero and the aircraft is not accelerating.
One common problem with F=MA is that people get the idea it implies a cause; i.e. that a force exists BECAUSE a mass is accelerated. Then they try and apply that concept to weight, mass, and acceleration due to gravity and get themselves in a muddle. Objects have weight whether they are being supported or not and whether they are being accelerated or not.
Its often handy to re-express F=MA as A=F/M which gives us a better pointer as to what's going on; accelerations are caused by unbalanced forces, not vice versa.
[ Additionally, regarding the OP, a purist might argue any of the following:
- Fuel is being burnt, therefore mass is decreasing so weight is decreasing. Techically correct, but perhaps a bit pedantic and missing the point of the question.
- Actually, even in straight and level flight, lift is fractionally less than weight, (due to Earth Curvature, the aircraft must be following a curved flight path and is not therefore in a 'uniform state of motion' as Newton would have said). This effect is pretty negligable for most aircraft, although it becomes detectable for fast moving things (e.g. SR71) or spacecraft. However this is a depth of analysis beyond most interviewers so I wouldn't go there, nevertheless, strictly speaking there is NO steady flight phase where lift>weight. ]
pb
CPB,
Yes, my explanation may have been a bit "Clunky". My point was an aircraft in a turn IS being accelerated, and this produces a force which is indistinguishable from weight. The wings must therefore produce additional lift to balance that force, and can only do so by increasing either speed or AofA.
It was mearly intended to show how barking the concept of "Reduced apparent wingspan" as laid out by ATPTQ is.
Yes, my explanation may have been a bit "Clunky". My point was an aircraft in a turn IS being accelerated, and this produces a force which is indistinguishable from weight. The wings must therefore produce additional lift to balance that force, and can only do so by increasing either speed or AofA.
It was mearly intended to show how barking the concept of "Reduced apparent wingspan" as laid out by ATPTQ is.
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Capt Pitt Bull,
Im still not very clear about that "weight remains constant during a banked turn" statement.
During a steep turn, when an aircraft pulls say 1.5 Gs, the pilot would definitely feel "heavier" than he usually does.
Is it not therefore correct to say that during a banked turn, the "apparent" weight of the aircraft Increases?? (By virtue of the centrifugal force that it counteracts ?? )
Further, is it correct to say ...
1. Weight of an aircraft remains constant during a banked turn, WITH RESPECT TO the Earth ? (for an outside observer)
2. Weight of an aircraft increases during a banked turn, with respect to the ROTATING FRAME OF REFERENCE ? (ie. if the observer is inside the turn)
Im still not very clear about that "weight remains constant during a banked turn" statement.
During a steep turn, when an aircraft pulls say 1.5 Gs, the pilot would definitely feel "heavier" than he usually does.
Is it not therefore correct to say that during a banked turn, the "apparent" weight of the aircraft Increases?? (By virtue of the centrifugal force that it counteracts ?? )
Further, is it correct to say ...
1. Weight of an aircraft remains constant during a banked turn, WITH RESPECT TO the Earth ? (for an outside observer)
2. Weight of an aircraft increases during a banked turn, with respect to the ROTATING FRAME OF REFERENCE ? (ie. if the observer is inside the turn)
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In order for a plane to climb lift must be increased. One does this by increasing thrust which increases the drag due to a higher angle of attack required to maintain the speed, which in turn increases the lift. Then as if by magic the airplane climbs...very simple.
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airbus,
simple-yes
correct-no!
In a steady state climb lift is actually reduced compared to S&L at the same speed. The vertical component of thrust puts the forces in equilibrium.
simple-yes
correct-no!
In a steady state climb lift is actually reduced compared to S&L at the same speed. The vertical component of thrust puts the forces in equilibrium.
The mass remains constant (disregarding fuel use, bits falling off, bug and bird strikes etc 
) The weight is the force exerted by a mass as a result of gravity - and changes with altitude and geographic location.
Most people have problems in defining their frame of reference when looking at forces, especially in cases of acceleration. There are two techniques - using an accelerated frame of reference (and thus producing a 'balanced' equation by including the "imaginary" centrifugal force) or a static frame - in which the forces are out of balance, and the resultant force is providing the observed acceleration. In either case the weight is unaffected.
) The weight is the force exerted by a mass as a result of gravity - and changes with altitude and geographic location.Most people have problems in defining their frame of reference when looking at forces, especially in cases of acceleration. There are two techniques - using an accelerated frame of reference (and thus producing a 'balanced' equation by including the "imaginary" centrifugal force) or a static frame - in which the forces are out of balance, and the resultant force is providing the observed acceleration. In either case the weight is unaffected.
Check,
We might simply be getting into semantics here, but an observer on board a turning aircraft feels an increase in force on the normal axis which is exactley the same as an increase in weight. From his frame of reference, his and the aircrafts weight has increased.
Astronoughts in space are still subject to gravity, but as their orbital acceleration exactley cancels it out are said to be weightless. I would think it therefore not incorrect to say someone experiencing an increased force due to acceleration is more "weightful" i.e has experienced an increase in weight.
We might simply be getting into semantics here, but an observer on board a turning aircraft feels an increase in force on the normal axis which is exactley the same as an increase in weight. From his frame of reference, his and the aircrafts weight has increased.
Astronoughts in space are still subject to gravity, but as their orbital acceleration exactley cancels it out are said to be weightless. I would think it therefore not incorrect to say someone experiencing an increased force due to acceleration is more "weightful" i.e has experienced an increase in weight.
We might simply be getting into semantics here,
but an observer on board a turning aircraft feels an increase in force on the normal axis which is exactley the same as an increase in weight. From his frame of reference, his and the aircrafts weight has increased.
I agree with the reason for the force, I disagree that this cannot be characterised as weight.
As I said, the observer in the aircraft would have no way of distinguishing which part of the forces he was feeling were due to gravity, and which to acceleration, just as an astronaught in free-fall has no way of sensing whether he is in orbit or in deep space.
Do you agree that an Astronaght in orbit is "weightless" even though he is still subjected to a force very close to 9.8N due to gravity, but neutralised by his orbital acceleration?
As I said, the observer in the aircraft would have no way of distinguishing which part of the forces he was feeling were due to gravity, and which to acceleration, just as an astronaught in free-fall has no way of sensing whether he is in orbit or in deep space.
Do you agree that an Astronaght in orbit is "weightless" even though he is still subjected to a force very close to 9.8N due to gravity, but neutralised by his orbital acceleration?
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Do you agree that an Astronaght in orbit is "weightless" even though he is still subjected to a force very close to 9.8N due to gravity, but neutralised by his orbital acceleration?
"weightless" astronauts in orbit are no such thing. They are in orbit because of their weight, not inspite of it. Rather, this situation should be referred to as 'free fall'.
What you are talking about is a conditon of 'apparent weight'.
We sense our weight by virtue of being held up against it
Gravity, being a force caused by a field, is applied universally across all the mass in our body. If you're floating around inside a space shuttle - in free fall - you have no way of feeling if you are subject to gravity. You could be falling under 1 G somewhere close to the Earth, or 2 ish G close to Jupiter or 1,000 G about to dissapear down a black hole (*note 1).
Every part of your body is subject to the same acceleration, so you feel nothing.
Whereas when you are being held up against a gravitational force (e.g. me sitting in my chair) this a contact force. i.e. its applied at distinct places on my body... right now, my butt cheeks and left foot (I'm sitting cross legged). This force is distributed to the other parts of me by squashing parts of me (e.g. cartlidge in my back and my blubbery... I mean perfectly toned gluteus maximus!).
Because parts of me are being squashed, I can perceive the force. If I'm accelerated, e.g. upwards, then the extra (unbalanced) force that accelerates me has to be transmitted from the points of contact to the rest of me. I feel that extra force as my connective tissues get squashed. So I may feel like my weight has increased, but it hasn't.
So we have 2 concepts, 'Weight' and 'Apparent Weight'. And although they sound similar they really aren't, and using them interchangably is a barrier to understanding.
Hope thats of interest / use.
pb
note 1 - discounting gravity gradient when R is small
Again agree with the mechanics of what you are saying, but not entirely with the conclution.
In order that an aircraft maintain altitude it must produce lift equal to its' apparent weight, not just the force due to gravity.
In terms of predicting the result of any physical interaction a bodies apparent weight is actually more important than it's actual weight. Objects in free-fall act exactley as if they were not in a gravity field. Objects subject to both gravity and a normal acceleration act exactley as if they weighed more. Weight is measured with a scale, and in both cases, that measurment would be different to 1g, and there would be no way for an observer to tell the difference between force due to gravity and force (or lack of it) due to acceleration.
In order that an aircraft maintain altitude it must produce lift equal to its' apparent weight, not just the force due to gravity.
In terms of predicting the result of any physical interaction a bodies apparent weight is actually more important than it's actual weight. Objects in free-fall act exactley as if they were not in a gravity field. Objects subject to both gravity and a normal acceleration act exactley as if they weighed more. Weight is measured with a scale, and in both cases, that measurment would be different to 1g, and there would be no way for an observer to tell the difference between force due to gravity and force (or lack of it) due to acceleration.
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oxford POF
I think u all guys need to look at oxford ATPL books which are availaible in every subject..........& take a look at oxford ATPL principles of flight.....& technical....then you may not need any other book for references..........tc
Yes, bodies react according to the resultant force which acts upon them.
Weight, however, is an individual vector with a technical definition.
If you use "common use" terms you end up with an inexact definition of exactly what you are attempting to describe, as the definition of "common use" terms is affected by the vagaries of a living language.
Weight, however, is an individual vector with a technical definition.
If you use "common use" terms you end up with an inexact definition of exactly what you are attempting to describe, as the definition of "common use" terms is affected by the vagaries of a living language.
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Wizo
I think you are in real danger of tying yourself in knots over this.
In order to maintain its existing vertical velocity, the vertical components of all forces acting on the aircraft must (by vector addition) sum zero.
Taking a few assumptions: slowish aircraft (i.e. not a spacecraft), lowish angle of attack (i.e. vertical thrust component negligable), in level flight to start off with.
In this situation we can say that the vertical component of lift must be equal and opposite to the weight.
Well, for calculating interactions between objects in the system, contact forces are obviously important. But for external 'wheres-the-damn-thing-going' kind of calculations there is a lot of virtue in recognising that the aircraft weight is constant (in the short term) and has a fixed orientation.
Agreed, its just newton 1,2 and 3
Their apparent weight has increased, their actual weight is unchanges (assuming no change in gravitational field strength.)
Apparent weight is measured with a scale
It is true (and crucially important) that an observer inside can not tell the difference. However, an external observer may well wish to quantify between them (weight and an acceleration caused by an unbalanced contact force) in order to relate the situation to his/her own frame of reference.
The danger in this statement is that people can get in a muddle about Newton 2. Forces cause accelerations, not vice versa. I get very concerned when someone starts talking about a force due to an acceleration.
F=MA tells us what force must have been present in order to cause the mass M to undergo acceleration A.
It does not imply that force F exists because object M accelerated.
In summary, I'm with Checkboard. Weight is a defined term. Its constant (in this context). Apparent weight is also a defined term, its variable (in this context). Use them interchangably at your peril.
pb
I think you are in real danger of tying yourself in knots over this.
In order that an aircraft maintain altitude it must produce lift equal to its' apparent weight, not just the force due to gravity.
Taking a few assumptions: slowish aircraft (i.e. not a spacecraft), lowish angle of attack (i.e. vertical thrust component negligable), in level flight to start off with.
In this situation we can say that the vertical component of lift must be equal and opposite to the weight.
In terms of predicting the result of any physical interaction a bodies apparent weight is actually more important than it's actual weight.
Objects in free-fall act exactley as if they were not in a gravity field.
Objects subject to both gravity and a normal acceleration act exactley as if they weighed more.
Weight is measured with a scale,
and in both cases, that measurment would be different to 1g, and there would be no way for an observer to tell the difference
between force due to gravity and force (or lack of it) due to acceleration.
F=MA tells us what force must have been present in order to cause the mass M to undergo acceleration A.
It does not imply that force F exists because object M accelerated.
In summary, I'm with Checkboard. Weight is a defined term. Its constant (in this context). Apparent weight is also a defined term, its variable (in this context). Use them interchangably at your peril.
pb
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airbus,
simple-yes
correct-no!
In a steady state climb lift is actually reduced compared to S&L at the same speed. The vertical component of thrust puts the forces in equilibrium.
simple-yes
correct-no!
In a steady state climb lift is actually reduced compared to S&L at the same speed. The vertical component of thrust puts the forces in equilibrium.
There are airplanes out there that can actually have a nose down pitch in level flight and for that matter can climb with a nose down pitch. How do you explain that.
Also if you are cruising a jetliner the attitude is somewhere around 2.5 degrees up. When you add additional thrust the attitude increases in order to prevent the airspeed from increasing which increases lift and the altitude increases.
As for the vertical component of thrust, it doesn`t matter where the lift comes from it is still lift. By your argument one could say that a hovering helicopter is not producing lift.
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Vibes, if you really want to understand aerodynamics I would suggest reading Kermode's Mechanics of Flight. If you want it more simple try Kermode's Flight Without Formula. They have both been around for more years than I care to remember but are still valid.
Happy reading!
Happy reading!
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Hmmm let's take another tack on this
Ok dokey to put this in simple straight forward language let's ask what determines an aircraft's ability to climb (for a sustained time period not for a few seconds by just pulling up the nose/dumping flaps)? In a nutshell climb is dependent upon having more power available than is required to maintain level flight in a given configuration. This is why multi-engine aircraft can loose most or even all of their ability to climb after an engine failure.
Also, let's not forget the difference between weight and mass. Mass is an object's resistance to acceleration. Weight is a the product of mass and the acceleration of gravity. The pound is a unit of force that equals a mass of one slug acted on by the local constant of gravity expressed in feet/second/second...32ft/sec^2 is a common value.
Hope this helps!
Jon
Also, let's not forget the difference between weight and mass. Mass is an object's resistance to acceleration. Weight is a the product of mass and the acceleration of gravity. The pound is a unit of force that equals a mass of one slug acted on by the local constant of gravity expressed in feet/second/second...32ft/sec^2 is a common value.
Hope this helps!
Jon
Airbus,
Good explanation here-
Pilot's Web The Aviators' Journal - Forces Acting on an Airplane
But first, you must get some basic concepts right.
Attitude is not directley related to angle of attack, and thus not directly related to lift, as attitude is angle reference the horizon whilst AofA is reference the relative airflow.
Vertical component of thrust is NOT lift. Lift is force perpendicular to airflow. Thrust is (more or less, depending on the thrust line of the engine) parralel to airflow.Weight always acts directly donwards, so a compponent of lift plus thrust must equal weight for equilibrium. All of this is true of a hovering helicopter.
If lift (or indeed, the sum of all forces acting vertically upwards) was greater than weight, the aircraft wouldn't just climb, it would continuously Accelerate vertically upwards, that is it would be under a constant g force of greater than one, with a constantly increasing climb rate. When you climb your 'bus, are you constantly pushed down in your seat by a force greater than your normal weight? Does it maintain a constant rate of climb?If not, your aircraft is in equilibrium, which couldn't happen if the vertical component of lift was greater than weight.
I think (and it's not uncommon) that you are confusing Lift with Co-efficient of lift, which is only one part of the lift equation.
Good explanation here-
Pilot's Web The Aviators' Journal - Forces Acting on an Airplane
But first, you must get some basic concepts right.
Attitude is not directley related to angle of attack, and thus not directly related to lift, as attitude is angle reference the horizon whilst AofA is reference the relative airflow.
Vertical component of thrust is NOT lift. Lift is force perpendicular to airflow. Thrust is (more or less, depending on the thrust line of the engine) parralel to airflow.Weight always acts directly donwards, so a compponent of lift plus thrust must equal weight for equilibrium. All of this is true of a hovering helicopter.
If lift (or indeed, the sum of all forces acting vertically upwards) was greater than weight, the aircraft wouldn't just climb, it would continuously Accelerate vertically upwards, that is it would be under a constant g force of greater than one, with a constantly increasing climb rate. When you climb your 'bus, are you constantly pushed down in your seat by a force greater than your normal weight? Does it maintain a constant rate of climb?If not, your aircraft is in equilibrium, which couldn't happen if the vertical component of lift was greater than weight.
I think (and it's not uncommon) that you are confusing Lift with Co-efficient of lift, which is only one part of the lift equation.