 So, welcome to the 12th lecture of this course which is the second of capsule number 6. Today we are going to move on to basic exposure to flight performance. And the most basic activity in flight performance is indeed the study level flight. So, our task today is to get familiar with the issues regarding study level flight both from the point of view as a performance engineer in the beginning, so we will derive some expressions. And then we will wear the cap of a pilot and try to understand how pilots are trained to maintain study and level flight. This particular presentation is based on the content generated by an intern called Shubham Panda. First, let us get an idea about what exactly is study and level flight. There are two words here, study and level. So, let me first show you a small simulation of a study level flight on a simulator. So, what do you see when you see this simulation? What do you observe in this particular simulation? Or in other words, how can you say that this flight is study and level? Yeah, anybody? Well, you have in front of you a simulation, what do you observe? Yes. So, the rate of flight is not climbing and it is maintained at the same level, so and so. So, you said something about rate of climb? Rate of climb. Being 0 and what else? And I think it is moving at constant velocity. Constant speed, right. So, what about the altitude? constant. Remains constant. So, study level flight is when you are operating at a constant speed at a fixed altitude. But whatever be the situation, whenever the aircraft is flying there are a few forces acting on it. So, today we will start by looking at the components of the forces acting on the aircraft in a study level flight. So, as you can see, let us assume there is an aircraft now in this animation or in this figure I have shown purposely the aircraft at a very high angle. Normally, aircraft do not fly at such high angle in study level flight, but this is just for the purpose of illustration. So, you have an aircraft which is flying and we assume that the flight is study and level. So there is a center of gravity of the aircraft which we assume is at the point shown in the figure. So, that is the point at which the net weight of the aircraft will be acting vertically downwards. The aircraft does not go on a horizontal flight path. It goes along a particular flight path which is at an angle theta with the horizontal. So, that is the flight path angle. In other words, the aircraft is going to move horizontally but along that particular line. Now, in this case it is not necessarily in study level flight. This is in general. So, in general the aircraft is going along a path where the angle is theta. So, there will be a lift force acting which will be perpendicular to the flight path. But there is an angle called as the angle of attack which is the angle between the flight path and the chord line of the aircraft or the main reference line of the aircraft. That angle is alpha. There is also drag acting on the aircraft which is along the flight path and there is also the thrust force acting on the aircraft which in general is at some angle alpha t compared to the flight path. Is there any other force acting on this aircraft? That is all. The four forces which act on it. Let us look at now the equilibrium of these forces in study flight. So, the same picture I am reproducing here. So, if you resolve the forces along the flight path, then you can get the thrust into cos times the angle of mounting of the engine minus weight times sin theta. These are going to be equal or minus the drag force will be equal to 0. And similarly normal to the flight path you will have a balance of the lift force, a component of the thrust force which will be acting upwards in this case and it will be opposed by W cos theta. In other words, now if you look at study level flight, let us neglect two things. Let us neglect the angle of mounting of the engine just to make the equations a bit simpler for you. So, alpha t is equal to 0. Similarly, let cos theta be a very small angle or let us say alpha t is not 0, alpha t is very small. Let us say a couple of degrees and let us also assume that theta also is very small 3, 4, 5 degrees. So, therefore the signs of these angles are going to be equal to 0 and the cosines of these angles are going to be, okay, so cos theta will be 1 actually not 0, okay. So these equations will become very simple, lift will be equal to weight and thrust will be equal to drag. If that is the case, then in study level flight it is a very simple expression that lift will be equal to weight and thrust will be equal to drag. But is this really true? So if it is not true, then what is the real story? Do you agree with this? Because many of you were nodding your heads when I was deriving this expression which means you agree with this explanation. So now I say that something is wrong, so let us see what is wrong, let us see how the forces are balanced, okay. So once again now, now because the aircraft is in study level flight I am not showing any angle, so the first force I will talk about will be the drag force which will oppose the flight. To cancel that we need to have the thrust force which will be equal to drag because if thrust is more than drag you cannot get steady flight, you will get level accelerated flight. Weight is the force acting vertically downwards but the lift force does not necessarily act at center of gravity, so the lift force typically will act at the quarter chord of the aircraft or at the aerodynamic center, therefore these forces are going to create a couple, okay. That couple has to be balanced by a force acting on the tail, in this case the orientation of lift and weight is such that there is a moment in the nose down direction, so there has to be a download on the tail to give a moment, so this download on the tail means the lift has to be more than weight, otherwise the aircraft will start sinking, in other words the lift force generated by the aircraft has to be equal to the weight of the aircraft plus whatever is the tail load to balance the aircraft in level flight. Now in case we have a situation where the center of gravity is behind the place where the lift acts, then the direction of the lift force or the tail load might reverse, in which case that is going to help, in which case lift will be less, similarly if we use the canard which is the tail mounted in the front, then the nose up or the nose down pitching moment can be cancelled only by a vertical load on the canard, so that much lift can be subtracted from the lift of the wing and that is one reason why we use the canard because it allows you to manage with lesser lift requirement because the tail load, in this case the canard load is helping you in carrying the aircraft up, so therefore in reality lift is not equal to weight, similarly in reality the thrust and the drag forces also they do not act along the same line, so they also create a moment, that moment can cancel or oppose or support the other moment, so it is a very simplistic assumption that lift is equal to weight thrust equal to drag and we will live with the assumption because if we do not do that then the equations becomes little bit more complicated, so from now on we take this statement with the pinch of salt, let us see how much thrust is required if you want to fly level, the answer is simple it is equal to the drag that is created, so the drag that is created is equal to the dynamic pressure half rho v square or q times s, wing reference area times the drag coefficient cd and the lift equal to weight ignoring the tail load is equal to q scl, q is again half rho v square dynamic pressure, so with this you can easily get an expression that the thrust required will be nothing but the l by d of the aircraft during the level flight divide by the weight of the aircraft, so you get a very simple logical understanding if you want to fly at a particular speed at a particular altitude which means dynamic pressure is fixed, speed and altitude are fixed therefore q is fixed and if you want to reduce the thrust required you have one of the two options, one option is make the aircraft lighter or reduce the weight, what should you do if you want less thrust required, but can you change the aircraft weight that is something which is already fixed and aircraft with a particular amount of payload, particular amount of fuel is going to have a particular weight, so you really cannot arbitrarily change w unless you add fuel or dump fuel etc or being very cruel dump people and add people, so let us not touch w therefore you have only l by d available to you, but can you change l by d of the aircraft, lift over drag ratio of the aircraft is it possible what do you think, what does it depend on for a given aircraft the angle of attack at which the aircraft is flying, at different angle of attacks the aircraft the different l by d, so if you want to fly at a condition where the thrust required is minimum why would you do that because you would like to consume less fuel, the pilot will then choose an angle at which the l by d is l by d max and for a given aircraft weight that would be the condition or the operating situation at which the thrust required will be minimum, yes take a mic shouldn't it be l into d, shouldn't it be l into d, why should it be l into d, d into w is equal to d into l, so you see it yourself see if t equal to d and l and w are same then I can multiply, so do you think that l by d should be more or less, sir in the formula it should be l into d upon w, you are right there is a mistake in the formula correct I didn't realize there is a mistake in the formula, so this formula we will correct okay, so basically it will be the other way around 1 by l by d, so when l by d is more t required is less okay, yeah so whether you say you depends on where you take it, so I want l to bring in l by d because l by d is more important, so as the l by d increases the thrust required is going to reduce okay right, so in an attempt to reduce the thrust required in flight one way is to reduce the drag and aspect ratio is one parameter of the aircraft which allows us to reduce the induced drag, so if you increase the aspect ratio such as in this particular aircraft it has a very high aspect ratio of 12.8, the l by d increases and hence the thrust required reduces okay, let us compare the level flight of a jet engine aircraft such as the one shown versus that of a piston engine aircraft, so first let us have a look at the two, is there any difference visible in the way they fly let us see, so here is a jet engine aircraft in level flight and here is a piston engine aircraft, so is this aircraft actually level flight, we cannot say because we assume so much that we are not able to see the relative location between the horizon and the aircraft okay, so we see that if you draw the operating graph or the operating envelope of these two aircraft, we notice that the range over which speed can be maintained or the range over which the level flight can be maintained is much, much larger for one aircraft type and smaller for the other aircraft type and also the effect of altitude and Mach number is also a bit different, so let us have a look at why there is a difference in the operating envelope okay, so I have borrowed some material now from the standard textbook that we use for this course the book by Anderson okay, in Anderson's textbook he has taken two examples, one of them is called as the CP-1 which is a propeller driven aircraft and then there is another one which is a jet engine aircraft and he has derived the entire performance formulae for these two aircraft, so you are also expected to go through that as self-study, this is important because you have to go beyond the classroom, the classroom is not the complete story the classroom is only a curtain raiser, so I am going to assume that you are going to read the performance portion from the book by Anderson for this and the next two capsules because in the next three capsules we are going to cover aircraft performance, so here is the data for an actual piston engine aircraft, we have one horizontal line which is approximation for the power available of this aircraft as the velocity changes, so here the assumption is that for a piston engine aircraft at a given altitude the horsepower available remains constant as velocity changes, it does not remain constant it changes slightly but assuming it as a straight line is easier for classroom discussions and the required horsepower to fly at a particular speed it is increasing in the way that is shown, so the intersection of these two lines will give you the speed at which you can fly maximum because beyond that speed the power available will be less than power required, so the intersection on the right hand side, now there is also typically an intersection on the left hand side but it is not shown here, so you can see that there would be some value of minimum speed which is the stalling speed actually and there is some value of maximum speed which is the speed at which power available equal to power required, so that determines the operating range or the possible operating range of the aircraft. There is a similar graph available for a jet engine aircraft, the difference is that in the case of jet engine aircraft is the thrust that remains almost constant with the speed at a given altitude not power, so if the thrust remains constant power is thrust into velocity, so if t remains constant and v is increasing then the line that is plotted between power and velocity is going to be basically something like a constant into velocity, so it will be a straight line, so this is the line for and the inclination of this line is a function of the power plant characteristics, so you will have a line starting from origin linearly increasing as the power required with velocity and you have a similar curve for sorry power available and this is the power required curve, here they have shown two intersections, the one on the left is the intersection that shows that there is a low speed also at which power available equal to power required, now if this speed is lower than the stalling speed it is meaningless because you cannot fly below stalling speed, in fact from safety point of view you actually cannot fly at maybe 1.1 times below the stalling speed, so but however if the stalling speed is much lower than this speed then this intersection has any meaning, so again we see that there is a range, now the range of operation here appears to be larger for the jet engine aircraft.