Understanding Static Stability in Aircraft Dynamics

And we realized and we understood when you define static stability, we are very clear that it is to do with initial tendency of that system. If it has initial tendency to come back to the equilibrium, then the body or the system is in static stability ok. And we try to understand this through a diagram, the spring and this is the mass. So, this is the equilibrium and equilibrium means all the net forces and moments are 0 and once we want to understand static stability, what we say we disrupt this, we stretch this mass and release it and we know that the moment I stretch it, there is a force. kx proportional to the displacement it tries to take it back to the equilibrium, it will overshoot it may oscillate that different issue. For static stability those things are not important, for static stability is only important that it should have initial tendency to go towards the equilibrium that is static stability. Now for this mass spraying system. Who is providing this static stability characteristics? It is the through the spring. The moment I stretch the spring, there is an opposing force which tries to take it back to the equilibrium. Now, let us come back to the aircraft. If I say for aircraft to the ground, it is the same. For time being if we concentrate on the angular stability that is suppose the apple is going like this and if there is a disturbance which tries to take the attitude up we have to check it will be statically stable if it has initial tendency to come back to the same equilibrium or to the equilibrium that initial tendency. That means, for a mass spring system it has a spring which does all this work, but for an aircraft. what is that agency will ensure that the moment I am deviating from the equilibrium, there will be a restoring moment which will try to take back it to the equilibrium. That is, if it is going like this, if it is disturbed and its attitude has increased, then the aircraft should have a mechanism which should try to give a nose down moment to ensure that it has static stability. Is it clear? Now, the question is for a mass spring system, there are physical system like spring, mass, spring is giving force or forces, however, For an aircraft all these forces are being generated through interaction between body and the medium which is air in this case right. For example, you know aircraft has major component as wings, as horizontal tail, vertical tail. So, whatever disturbance it experiences if at all any restoring moment has to be generated it has to come from the interaction of the horizontal tail, wings, vertical tail. wing, fuselage from their interaction it should generate a nose down moment as long as it generate nose down moment for a positive disturbance, angular disturbance, angle of attack disturbance we will say it has static stability. Now, we will try to understand where from it will generate restoring moment almost analogous to Let us see, we all are aware of aerofoil. Let us for simplicity assume that this is a symmetric aerofoil and also we know there is a point called aerodynamic centre. Aerodynamic center and what is the definition of aerodynamic center? It is a fictitious point mostly for low speed and moderately thickness to chord ratio aerofoil, this point is at c by 4 that is This location is c by 4 and we call it quarter chord point where we are now very clear what is a chord. And what is the technical definition of this aerodynamic centre? It is that point in the terrafoil about which the pitching moment is independent of angle of attack. Is it clear? Even if I change angle of attack, the pitching moment about that point will be 0, right. So, it is very convenient to represent all the forces and moment acting at c by 4. So, you will find that for an aerofoil I will represent like this cl, cd and in general there will be a cm. Now, let us try to understand. how such configuration can generate a restoring moment to ensure it has static stability. That is the question we are addressing. It is important to understand that in free flight, if there is any disturbance on the body, it will try to rotate about an axis passing through center of gravity. So, we would like to see whatever moment is being generated, all the rotation about an axis passing through center of gravity. passing through center of gravity. So, what do we do? We say suppose this is the aerofoil and I can now imagine that this aerofoil is a cross sectional contour of an wing. We are very clear. We have already discussed the aerofoil is a two dimensional concept where the flow is supposed to follow the contour. There are no lateral flows, but for a wing there will be lateral flows. But the wing contour is decided by the aerofoil also you know lifting characteristic of the wing also depends upon what type of aerofoil you have chosen right. But what we are trying to focus is on stability mostly static stability at this point. So, this is aerodynamic centre. Let us say this wing I want to fly ok. It has enough area. Now, there are two positions are possible one is let us say center of gravity of the wing is ahead of aerodynamic center correct. If this is the situation let us see what happens. Let us say this gentleman is flying at alpha equal to alpha star and that is the equilibrium and because of some disturbance delta alpha. the wing is going to respond, we are looking for its initial tendency. If there is a delta alpha disturbance, then the delta C L I say lift coefficient or lift if I multiply it with dynamic pressure will be generated and that can be represented effectively at aerodynamic center, we have agreed on this point. then what this delta C L will do? This will generate a moment I say delta C m about C g which will be nose down. So, if. there is a positive disturbance of delta alpha, this wing has a tendency to put the wing down to make this to 0 or to come back to the alpha star. So, we say it has initial tendency and to come back to equilibrium. So, we said if aerodynamic center is behind center of gravity, then it possesses static stability, is it clear? To be more elaborate suppose. The aircraft is streamed at alpha 2 degree, I call it alpha star and there is a delta alpha let us say 0.5 degree. If this is statically stable, then what will happen? Because of delta alpha, now the angle of attack became 2 plus 0.5, 2.5 degree. But if it is statically stable, the airplane will have initial tendency or produce a nose down moment so that finally this gets cancelled or there is a tendency to cancel it right I will not use the word finally because finally once I say implicitly talking about time which static stability has nothing to do with time that is dynamic stability so correct statement will be the moment there is a change in angle of attack from 2 plus 0.5 that is 2.5 degree the airplane will generate a nose down moment to ensure that it has initial tendency to Come back to alpha star 2 degree and this is possible for a wing alone configuration only when aerodynamic center is behind center of gravity. So, you could see that the moment delta Cm which is proportional to the angle is equivalent to the spring force k x, where also the spring force was proportional to a displacement that is a linear displacement and it is angular displacement, here both are in opposing nature. So, that is how static stability is assured, you can immediately see that if. This is the aerodynamic centre and if CG is behind aerodynamic centre or I say aerodynamic centre is ahead of centre of gravity, then what happens? If there is a disturbance delta alpha, there will be a force here, let us say in non-dimensional form I call delta C L, this will give nose up moment right. So, this will actually take this way nose up. So, further this angle will increase. So, such configuration does not have any initial tendency to come back to the equilibrium. So, we said this is statically unstable. So, what is our learning? If you want to make a wing alone configuration statically stable, I must ensure that aerodynamic center of the wing should be behind center of gravity of the wing right. And we also realize that this production of moment which is proportional to the angular displacement can be thought to be analogous to the production of force because of linear displacement for a spring mass case. So, I can always say this separation between aerodynamic center and center of gravity will have some relationship or some similarity with the stiffness of the system. Like for mass spring system, the spring constant k has some influence. stiffness of the system, ok. So, this is this needs to be understood correctly before you again revisit my lecture on static stability, ok. Now, if you recall the aircraft has mainly This is the horizontal mainly those stabilizer component and lifting component. What is the lifting component? Primary lifting component is the wing that is wing has a primary role to generate lift and these are this part is the stabilizer. And its role is this is typically horizontal stabilizer, its role is to give stability the airplane right. So, now if I draw a configuration like this. And let us say CG of the airplane is here and aerodynamic center of the wing is here and aerodynamic center of the tail is here. So, by seeing this diagram can we simply tell or can we not? Answer a question by seeing this description and neglecting the Fuselage effect. So, let us say Fuselage effect we are neglecting. With this assumption that we are neglecting field large effect by seeing the location of aerodynamic centre vis-a-vis CG of the airplane, this is CG of the airplane. Please understand when I am drawing here, here this notation is this is CG of the airplane. Earlier since it was only wing to I was talking about CG of the wing, but this is whole airplane this is a CG of the airplane. Why CG of the airplane? Because airplane will rotate in free space through an about an axis which passes through CG of the aircraft right. Once I have represented the location of aerodynamic center of wing as well as tail. We serve with the CG of the airplane. I can easily comment that this configuration will necessarily be statically stable. I am assuming. There are no fuselage effect or neglecting fuselage effect. Why I know? Because aerodynamic center of the wing is behind CG of the airplane. So, this will be stabilizing. Anywhere aerodynamic center of the tail is behind CG of the airplane. So, this will be stabilizing. So, this aircraft has to be statically stable. There is no issue. Now, think of a different case. Think of a case where CG of the airplane is behind aerodynamic center of the wing. However, aerodynamic centre of the tail is behind CG of the airplane. Can I directly say this aircraft will naturally be statically stable? I am assuming that fuel gauge effect is neglected. Can I directly tell? Let us investigate. As far as wing is concerned, since aerodynamic centre of the wing is ahead of CG, it is giving destabilizing. Okay. So, this is not giving stabilizing contribution because I know for stabilizing contribution aerodynamic center has to be behind center of gravity, but for tail it is of course, behind so it is stabilizing. So, whether the aircraft will have adequate static stability or not or to be more straight, whether the aircraft will be statically stable or not will be decided by the actual contribution of wing. is destabilizing and actual contribution of tail which is stabilizing and they will try to nullify each other and whosoever wins that will decide is statically stable or not if the contribution of tail is more than the contribution of wing, then I will say statically stable. For example, the restoring moment which is primary reason for static stability in this case. So, the restoring moment generated by the horizontal tail, if it is more than the disturbing moment or destabilizing moment generated by the wing, then only this airplane will be statically stable. The restoring moment is decided by what? You could see not only the horizontal tail, tail whatever lift is coming or area, but also the distance between horizontal tail aerodynamic center and the CG because it is a moment ok. So, soon you will understand that if we when you talk about static stability, we talk about the area, we talk about the momentum and we define some term called tail volume ratio. So, this is just to warming you up. This concept should be clear to you. Also please understand suppose this is an airplane, this is CG of the aircraft and let us see AC. Of wing and of course, this is AC of tail ok. Let us say this is the configuration again I am assuming fuel gauge contribution is neglected. If I see this. I could easily check yes AC of the wing is behind CG of the aircraft. So, this will give stabilizing contribution anyway AC of the tail is behind the CG of the aircraft this also will give stabilizing contribution. So, it is having naturally having static stability, but now suppose I add some surface here. which are typically called canard, then what will happen? The static stability will increase or decrease? How do I see it as a designer? Okay, I know that the moment it is like this, the aerodynamic center of this canard is ahead of CG of the aircraft. So, what it will do? It will give again destabilizing. Here. We know that aerodynamic center is ahead of center of gravity of the aircraft, so it will destabilizing. So, whatever static stability it had without this canard, now the moment I put canard, its static stability will reduce and if it is not properly designed, it may become unstable also, right, statically unstable also. So, this is what the scenario. And what is the gross learning for designer? That any lifting surface, please note this important, any lifting surface which are located behind the center of gravity of the airplane will generate stabilizing effect. Any lifting surface which are located ahead of center of gravity will generate destabilizing contribution. If you understand this, then major part of static stability is understood. It does not require more than knowledge of class 11th or 12th. You have to know how to find out the moments. Is it clear? Okay. Let us take another example before you go through those lectures. Let us say very simple case. Let us say this is a hinge point. Let us say I have put a horizontal surface like this and I have kept this in a wind tunnel. You are all aware of this nomenclature, you know what is a wind tunnel. When I say hinge point means, so this is a horizontal tail. So, it is located like this and it can rotate about this point in a pitch plane. And we say remember we call it longitudinal motion right, but it cannot go forward or backward only one degree moves like this. So, now if this is the case can we talk about whether this is. statically stable or not. What is the way to think? You first identify what are the lifting surface, is that only one lifting surface that is this one and you know the aerodynamic center will be c by 4 of this. So, let us say this is the aerodynamic center. center and now what is the check? Since it is rotating about hinge point, so it is like the center of gravity for a free space, but in free space also it rotates about center of gravity axis passing through center of gravity right. So, this is the hinge point. So, now, this you could see aerodynamic center is behind the hinge point. So, this will give you. Necessarily static stability ok as per our understanding, but mechanics wise let us check how it happens ok. Suppose, these are system and suppose I have given let me draw it little once you understand this is the tail. So, I am drawing the cross section ok assume that tail is something like this ok. So, now I want to check whether it has static stability or not, though we understand we have become expert, we know the aerodynamic centre is behind the CG or the hinge point in this case. So, it will definitely contribute towards static stability right. Let us see by mechanics by what happens. Let us introduce some disturbance of 2 degree. this is the equilibrium ok. Equilibrium is at let us say alpha equal to 0 degree. Now, I introduce 2 degree disturbance. So, equilibrium is at alpha equal to 0, I introduce a disturbance of 2 degree. So, what this 2 degree will do? At c by 4 it will generate lift I am representing by c. C L this coefficient and we know very well if I multiply with the local dynamic pressure and area I will get the total lift. Here this one assumption generally or as per definition the lift should be perpendicular to the pressure. particular to the velocity vector, but since alpha is small 2 degree we are assuming it to be straight that is all right. What the C L will do? The C L will immediately generate a moment about hinge point, so which will also depend upon what is the length L t. So, the moment it generates a nosedown moment meaning thereby what? It will try to nullify these two degree. So, it has a initial tendency to make it or to ensure it comes back to alpha equal to 0 which is the equilibrium. So, you say yes indeed it has static stability, is it clear? But there is a catch please understand what happens. Suppose, I have fired a rocket. which has thrust T and which is statically stable. This is the CG and this is the aerodynamic center or center pressure for this. It is statically stable and it was having alpha equal to let us say 0. Now, because of disturbance of 2 degree, what will happen? Because it is statically stable, it will be stable. stable it has initial tendency to ensure that is 2 degree becomes 0. So, that or the equilibrium alpha equals 0 is achieved. So, it gives a nosedown moment. The moment it gives a nosedown moment the rocket. Initially rocket turns like this, but the thrust is on already right. So, it will have an serious effect on the range of the rocket, because which is supposed to go like this. is now turn like this and thrust is on. So, it will the elevation angle will change. So, the range of the rocket will change that is why it is important when you talk about stability, how much stability. So, as far as static stability is concerned it talks only about initial tendency but it has larger influence on its sensitivity or performance in dispersion we saw his wind So, it is very very important we should also clearly know how much static stability we want and our lecture will address this question. As I told you these are all warm up for this course. to orient you towards stability to understand and to have a love and hate relationship with static stability you love it don't love it too much don't hate it okay so how much I should love how much I should hate that will be decided by what do you require and how much you understand. Ok, clear. So, that is why I always say static stability and control has a very tacit relationship, one has to respect it. If you do not respect, the whole relationship can break and you may lead to a disaster. Ok. Thank you.

So, this is the equilibrium and equilibrium means all the net forces and moments are 0 and once we want to understand static stability, what we say we disrupt this, we stretch this mass and release it and we know that the moment I stretch it, there is a force. kx proportional to the displacement it tries to take it back to the equilibrium, it will overshoot it may oscillate that different issue. For static stability those things are not important, for static stability is only important that it should have initial tendency to go towards the equilibrium that is static stability. Now for this mass spraying system.

Who is providing this static stability characteristics? It is the through the spring. The moment I stretch the spring, there is an opposing force which tries to take it back to the equilibrium. Now, let us come back to the aircraft. If I say for aircraft to the ground, it is the same.

For time being if we concentrate on the angular stability that is suppose the apple is going like this and if there is a disturbance which tries to take the attitude up we have to check it will be statically stable if it has initial tendency to come back to the same equilibrium or to the equilibrium that initial tendency. That means, for a mass spring system it has a spring which does all this work, but for an aircraft. what is that agency will ensure that the moment I am deviating from the equilibrium, there will be a restoring moment which will try to take back it to the equilibrium. That is, if it is going like this, if it is disturbed and its attitude has increased, then the aircraft should have a mechanism which should try to give a nose down moment to ensure that it has static stability. Is it clear?

Now, the question is for a mass spring system, there are physical system like spring, mass, spring is giving force or forces, however, For an aircraft all these forces are being generated through interaction between body and the medium which is air in this case right. For example, you know aircraft has major component as wings, as horizontal tail, vertical tail. So, whatever disturbance it experiences if at all any restoring moment has to be generated it has to come from the interaction of the horizontal tail, wings, vertical tail.

wing, fuselage from their interaction it should generate a nose down moment as long as it generate nose down moment for a positive disturbance, angular disturbance, angle of attack disturbance we will say it has static stability. Now, we will try to understand where from it will generate restoring moment almost analogous to Let us see, we all are aware of aerofoil. Let us for simplicity assume that this is a symmetric aerofoil and also we know there is a point called aerodynamic centre. Aerodynamic center and what is the definition of aerodynamic center?

It is a fictitious point mostly for low speed and moderately thickness to chord ratio aerofoil, this point is at c by 4 that is This location is c by 4 and we call it quarter chord point where we are now very clear what is a chord. And what is the technical definition of this aerodynamic centre? It is that point in the terrafoil about which the pitching moment is independent of angle of attack. Is it clear? Even if I change angle of attack, the pitching moment about that point will be 0, right.

So, it is very convenient to represent all the forces and moment acting at c by 4. So, you will find that for an aerofoil I will represent like this cl, cd and in general there will be a cm. Now, let us try to understand. how such configuration can generate a restoring moment to ensure it has static stability.

That is the question we are addressing. It is important to understand that in free flight, if there is any disturbance on the body, it will try to rotate about an axis passing through center of gravity. So, we would like to see whatever moment is being generated, all the rotation about an axis passing through center of gravity.

passing through center of gravity. So, what do we do? We say suppose this is the aerofoil and I can now imagine that this aerofoil is a cross sectional contour of an wing. We are very clear.

We have already discussed the aerofoil is a two dimensional concept where the flow is supposed to follow the contour. There are no lateral flows, but for a wing there will be lateral flows. But the wing contour is decided by the aerofoil also you know lifting characteristic of the wing also depends upon what type of aerofoil you have chosen right. But what we are trying to focus is on stability mostly static stability at this point. So, this is aerodynamic centre.

Let us say this wing I want to fly ok. It has enough area. Now, there are two positions are possible one is let us say center of gravity of the wing is ahead of aerodynamic center correct.

If this is the situation let us see what happens. Let us say this gentleman is flying at alpha equal to alpha star and that is the equilibrium and because of some disturbance delta alpha. the wing is going to respond, we are looking for its initial tendency.

If there is a delta alpha disturbance, then the delta C L I say lift coefficient or lift if I multiply it with dynamic pressure will be generated and that can be represented effectively at aerodynamic center, we have agreed on this point. then what this delta C L will do? This will generate a moment I say delta C m about C g which will be nose down.

So, if. there is a positive disturbance of delta alpha, this wing has a tendency to put the wing down to make this to 0 or to come back to the alpha star. So, we say it has initial tendency and to come back to equilibrium.

So, we said if aerodynamic center is behind center of gravity, then it possesses static stability, is it clear? To be more elaborate suppose. The aircraft is streamed at alpha 2 degree, I call it alpha star and there is a delta alpha let us say 0.5 degree. If this is statically stable, then what will happen?

Because of delta alpha, now the angle of attack became 2 plus 0.5, 2.5 degree. But if it is statically stable, the airplane will have initial tendency or produce a nose down moment so that finally this gets cancelled or there is a tendency to cancel it right I will not use the word finally because finally once I say implicitly talking about time which static stability has nothing to do with time that is dynamic stability so correct statement will be the moment there is a change in angle of attack from 2 plus 0.5 that is 2.5 degree the airplane will generate a nose down moment to ensure that it has initial tendency to Come back to alpha star 2 degree and this is possible for a wing alone configuration only when aerodynamic center is behind center of gravity. So, you could see that the moment delta Cm which is proportional to the angle is equivalent to the spring force k x, where also the spring force was proportional to a displacement that is a linear displacement and it is angular displacement, here both are in opposing nature. So, that is how static stability is assured, you can immediately see that if.

This is the aerodynamic centre and if CG is behind aerodynamic centre or I say aerodynamic centre is ahead of centre of gravity, then what happens? If there is a disturbance delta alpha, there will be a force here, let us say in non-dimensional form I call delta C L, this will give nose up moment right. So, this will actually take this way nose up. So, further this angle will increase. So, such configuration does not have any initial tendency to come back to the equilibrium.

So, we said this is statically unstable. So, what is our learning? If you want to make a wing alone configuration statically stable, I must ensure that aerodynamic center of the wing should be behind center of gravity of the wing right. And we also realize that this production of moment which is proportional to the angular displacement can be thought to be analogous to the production of force because of linear displacement for a spring mass case.

So, I can always say this separation between aerodynamic center and center of gravity will have some relationship or some similarity with the stiffness of the system. Like for mass spring system, the spring constant k has some influence. stiffness of the system, ok. So, this is this needs to be understood correctly before you again revisit my lecture on static stability, ok. Now, if you recall the aircraft has mainly This is the horizontal mainly those stabilizer component and lifting component.

What is the lifting component? Primary lifting component is the wing that is wing has a primary role to generate lift and these are this part is the stabilizer. And its role is this is typically horizontal stabilizer, its role is to give stability the airplane right.

So, now if I draw a configuration like this. And let us say CG of the airplane is here and aerodynamic center of the wing is here and aerodynamic center of the tail is here. So, by seeing this diagram can we simply tell or can we not?

Answer a question by seeing this description and neglecting the Fuselage effect. So, let us say Fuselage effect we are neglecting. With this assumption that we are neglecting field large effect by seeing the location of aerodynamic centre vis-a-vis CG of the airplane, this is CG of the airplane.

Please understand when I am drawing here, here this notation is this is CG of the airplane. Earlier since it was only wing to I was talking about CG of the wing, but this is whole airplane this is a CG of the airplane. Why CG of the airplane?

Because airplane will rotate in free space through an about an axis which passes through CG of the aircraft right. Once I have represented the location of aerodynamic center of wing as well as tail. We serve with the CG of the airplane.

I can easily comment that this configuration will necessarily be statically stable. I am assuming. There are no fuselage effect or neglecting fuselage effect. Why I know?

Because aerodynamic center of the wing is behind CG of the airplane. So, this will be stabilizing. Anywhere aerodynamic center of the tail is behind CG of the airplane. So, this will be stabilizing. So, this aircraft has to be statically stable.

There is no issue. Now, think of a different case. Think of a case where CG of the airplane is behind aerodynamic center of the wing.

However, aerodynamic centre of the tail is behind CG of the airplane. Can I directly say this aircraft will naturally be statically stable? I am assuming that fuel gauge effect is neglected. Can I directly tell?

Let us investigate. As far as wing is concerned, since aerodynamic centre of the wing is ahead of CG, it is giving destabilizing. Okay.

So, this is not giving stabilizing contribution because I know for stabilizing contribution aerodynamic center has to be behind center of gravity, but for tail it is of course, behind so it is stabilizing. So, whether the aircraft will have adequate static stability or not or to be more straight, whether the aircraft will be statically stable or not will be decided by the actual contribution of wing. is destabilizing and actual contribution of tail which is stabilizing and they will try to nullify each other and whosoever wins that will decide is statically stable or not if the contribution of tail is more than the contribution of wing, then I will say statically stable. For example, the restoring moment which is primary reason for static stability in this case. So, the restoring moment generated by the horizontal tail, if it is more than the disturbing moment or destabilizing moment generated by the wing, then only this airplane will be statically stable.

The restoring moment is decided by what? You could see not only the horizontal tail, tail whatever lift is coming or area, but also the distance between horizontal tail aerodynamic center and the CG because it is a moment ok. So, soon you will understand that if we when you talk about static stability, we talk about the area, we talk about the momentum and we define some term called tail volume ratio. So, this is just to warming you up.

This concept should be clear to you. Also please understand suppose this is an airplane, this is CG of the aircraft and let us see AC. Of wing and of course, this is AC of tail ok.

Let us say this is the configuration again I am assuming fuel gauge contribution is neglected. If I see this. I could easily check yes AC of the wing is behind CG of the aircraft. So, this will give stabilizing contribution anyway AC of the tail is behind the CG of the aircraft this also will give stabilizing contribution. So, it is having naturally having static stability, but now suppose I add some surface here.

which are typically called canard, then what will happen? The static stability will increase or decrease? How do I see it as a designer?

Okay, I know that the moment it is like this, the aerodynamic center of this canard is ahead of CG of the aircraft. So, what it will do? It will give again destabilizing.

Here. We know that aerodynamic center is ahead of center of gravity of the aircraft, so it will destabilizing. So, whatever static stability it had without this canard, now the moment I put canard, its static stability will reduce and if it is not properly designed, it may become unstable also, right, statically unstable also.

So, this is what the scenario. And what is the gross learning for designer? That any lifting surface, please note this important, any lifting surface which are located behind the center of gravity of the airplane will generate stabilizing effect.

Any lifting surface which are located ahead of center of gravity will generate destabilizing contribution. If you understand this, then major part of static stability is understood. It does not require more than knowledge of class 11th or 12th. You have to know how to find out the moments. Is it clear?

Okay. Let us take another example before you go through those lectures. Let us say very simple case.

Let us say this is a hinge point. Let us say I have put a horizontal surface like this and I have kept this in a wind tunnel. You are all aware of this nomenclature, you know what is a wind tunnel. When I say hinge point means, so this is a horizontal tail.

So, it is located like this and it can rotate about this point in a pitch plane. And we say remember we call it longitudinal motion right, but it cannot go forward or backward only one degree moves like this. So, now if this is the case can we talk about whether this is. statically stable or not. What is the way to think?

You first identify what are the lifting surface, is that only one lifting surface that is this one and you know the aerodynamic center will be c by 4 of this. So, let us say this is the aerodynamic center. center and now what is the check?

Since it is rotating about hinge point, so it is like the center of gravity for a free space, but in free space also it rotates about center of gravity axis passing through center of gravity right. So, this is the hinge point. So, now, this you could see aerodynamic center is behind the hinge point.

So, this will give you. Necessarily static stability ok as per our understanding, but mechanics wise let us check how it happens ok. Suppose, these are system and suppose I have given let me draw it little once you understand this is the tail.

So, I am drawing the cross section ok assume that tail is something like this ok. So, now I want to check whether it has static stability or not, though we understand we have become expert, we know the aerodynamic centre is behind the CG or the hinge point in this case. So, it will definitely contribute towards static stability right.

Let us see by mechanics by what happens. Let us introduce some disturbance of 2 degree. this is the equilibrium ok. Equilibrium is at let us say alpha equal to 0 degree.

Now, I introduce 2 degree disturbance. So, equilibrium is at alpha equal to 0, I introduce a disturbance of 2 degree. So, what this 2 degree will do? At c by 4 it will generate lift I am representing by c. C L this coefficient and we know very well if I multiply with the local dynamic pressure and area I will get the total lift.

Here this one assumption generally or as per definition the lift should be perpendicular to the pressure. particular to the velocity vector, but since alpha is small 2 degree we are assuming it to be straight that is all right. What the C L will do? The C L will immediately generate a moment about hinge point, so which will also depend upon what is the length L t.

So, the moment it generates a nosedown moment meaning thereby what? It will try to nullify these two degree. So, it has a initial tendency to make it or to ensure it comes back to alpha equal to 0 which is the equilibrium.

So, you say yes indeed it has static stability, is it clear? But there is a catch please understand what happens. Suppose, I have fired a rocket.

which has thrust T and which is statically stable. This is the CG and this is the aerodynamic center or center pressure for this. It is statically stable and it was having alpha equal to let us say 0. Now, because of disturbance of 2 degree, what will happen?

Because it is statically stable, it will be stable. stable it has initial tendency to ensure that is 2 degree becomes 0. So, that or the equilibrium alpha equals 0 is achieved. So, it gives a nosedown moment. The moment it gives a nosedown moment the rocket. Initially rocket turns like this, but the thrust is on already right.

So, it will have an serious effect on the range of the rocket, because which is supposed to go like this. is now turn like this and thrust is on. So, it will the elevation angle will change.

So, the range of the rocket will change that is why it is important when you talk about stability, how much stability. So, as far as static stability is concerned it talks only about initial tendency but it has larger influence on its sensitivity or performance in dispersion we saw his wind So, it is very very important we should also clearly know how much static stability we want and our lecture will address this question. As I told you these are all warm up for this course. to orient you towards stability to understand and to have a love and hate relationship with static stability you love it don't love it too much don't hate it okay so how much I should love how much I should hate that will be decided by what do you require and how much you understand. Ok, clear.

So, that is why I always say static stability and control has a very tacit relationship, one has to respect it. If you do not respect, the whole relationship can break and you may lead to a disaster. Ok.

Thank you.

Transcript for:Understanding Static Stability in Aircraft Dynamics

Transcript for:
Understanding Static Stability in Aircraft Dynamics