 Okay, so welcome to the tutorial for capsule number 8. In this tutorial, we are going to look at some basic questions on aircraft stability and then we will move on to the quiz. So first we look at stability of flying wings. I thought this is a good occasion for me to talk something about flying wings because these are very unique kind of aircraft. So when you think of flying wings, you think of something like this. This is the Northrop N1M also called as Jeep which was designed in 1940s. One company Northrop has been a champion in the US of flying wings. So they have made many, many flying wings. I will show you four of them. So 1940 Northrop NM1, then in 45, five years later, they came up with this turboprop version of a four-ingent flying wing called as YB35. So this aircraft revolutionized the concept of flying wings because there were some flying wing designs also being taken up by the Germans and by people in England also. But this was the first one that actually took part in a lot of combat. In 1947, two years later, it was converted into a jet version. So those four turboprops were removed and jet engines were inserted. In place of the engines, we have these vertical tails and the intake of the engines is from the leading edge. So these are the buried engines. And then the most famous of them is the B2 Bomber which came in 1989. So this is the circle of just one company from 1940 to 45 to 47 to 89. But I want you to look at this old film that talks about flying wings. And it talks about one particular aircraft called as the HO2 flying wing. So this is called as a flying triangle. So you can see it is being pulled by a chaise aircraft. So a biplane is pulling a flying wing. A striking picture of the machine in flight before a fortless landing on its single wheel completes a successful experiment. Single wheel landing. Just see how it turned during the approach at very high angle. So this is a historical film. And these are the two brothers, the famous Holtang brothers from Germany. They were a part of Hitler's secret army. And it is said that in the first world war they learned flying during war. So if they learned flying on aircraft which took part in the war, they survived. And then later on they went on to join the Hitler's Youth Brigade. And they ended up designing many Delta or flying wing aircraft. One of them I have already shown you. This is the most famous of their flying wings, Holtang 2229. And this aircraft was designed to meet the requirements of 1000 by 1000 by 1000. Which were proposed by the German Luftwaffe in almost at the end of second world war. So the requirement was to carry 1000 kilograms over 1000 kilometers at 1000 kilometers per hour. And this was designed in keeping in mind these particular requirements. Very famous aircraft. Unfortunately it came into play only at almost the end of the second world war. So the Germans could not really use this to much effect in combat. So it was a wonderful aircraft that came too late. And in India also we are looking at flying wing configuration. So information regarding DRDO's unmanned combat aerial vehicle or UCAV is not very easily available. It is not yet being publicized too much. So some sources. So these numbers may be little bit off by here and there. It is basically the most complex and daunting mission taken up by DRDO. As far as aerospace engineering is concerned. DRDO likes to call this aircraft as a self defending high speed reconnaissance UAV. Which also has a weapon firing capability. So that sentence defines aura. So we have done some basic calculations and design studies related to aircraft like aura. One of my MTech students has published a paper also on this. So those of you who are interested in design of aircraft. Especially in this kind of aircraft you are welcome to look at this particular paper. Okay so let us come to tutorial now. So this is the first question. There will be three questions before we proceed to the quiz. So here is a flying wing therefore there is no tail. It has a center of gravity. It has an aerodynamic center at which we can assume the lift force to be acting. And also some moment about the nose. Okay so the question is that during cruise if the lift coefficient is known to be 0.8. And if the LCG that is the distance between the aerodynamic center and the center of gravity is 0.5 times the chord. The chord is from the nose to the end. And the pitching moment coefficient Cm is 0.02. If this is known the question is determine the coefficient of moment around the aerodynamic center or we call it as Cmac. Okay so this is your first question. So very simple question just by taking the moments about center of gravity you can solve this question. So the net moment is 0.02, net moment coefficient Cm is 0.02, Cl is 0.8, LCG is 0.5c. Find the location of Cmac. Okay I will give you a few minutes to solve this question. So please solve, draw this figure. So Cm is the moment coefficient about the center of gravity which is the net moment coefficient of the aircraft. Since there is a positive value we must do something to trim it out. Normally in a flying we do by giving a reflex at the trailing edge because there is no tail available. So you give a reflex at the trailing edge to take care of the moment, the net moment. Okay so if you have already drawn this figure then I think you can now proceed with the calculation. So very simple one line solution taking moments about the center of gravity. Yeah, yes minus 0.38. Okay we will confirm if somebody else gets the same answer, minus 0.38. Now what is your answer if the value of Cm is minus 0.02 not plus 0.02. Can you just do that calculation also? Because the value of Cm is not 0.02 plus minus 0.02. So it is just an addition basically. Yes everybody has the answer? Same answer? Okay let us see how we can solve it. So basically we have to take the sum of the moments around center of gravity which will be mac plus l times lcg, right? So if you divide both sides by half rho v square s into c to make it non-dimensional. Then you get Cm is equal to Cmac plus l by half rho v square lcg by c. Or in coefficient form Cm will be Cmac plus Cl into lcg by c. And now you put all the values. So Cmac is known, Cl is known, lcg upon c is also known because lcg is given in terms of c. So 0.02 is Cm and that is equal to Cmac plus 0.8 times 0.05c. Therefore 0.02 is equal to Cmac plus 0.8 into 0.05. Therefore Cmac will be 0.02 minus 0.8 times 0.5 so it is minus 0.02. So if you have assumed the pitching moment to be positive by looking at the figure, you have to keep in mind that figures are drawn for a general case. Yes? Oh, okay, okay, okay, that is the reason you got the wrong answer. Sorry, right then I made a mistake there. Okay, so if it is 0.05c then you can put the corresponding values. Yeah, so then it will be 5 into 840 minus 0.2 then you stand corrected. So I am wrong, I gave the wrong number in the tutorial. Okay, but I think you get the idea, it is a very simple moment balance. Okay, let us go to the next one. The next question talks about, yes, so what is your question? Yeah? The CG is of aircraft. Here there is nothing like aircraft and wing, it is a flying wing. So CG of the aircraft is CG of the wing? Right, yes. I have one question. The length of CG location is moving 50% at the core. Yeah, so that is a mistake. It was mentioned as 0.5c, it is actually 0.05c, 5% of the core. Okay. My question is not that clear. Because lift is acting at the aerodynamic center. Yeah. That is why we call it subsonic, we consider that one fourth of the core. So the distance between the CG and the aerodynamic center where the lift is acting will be half of 0.05. Why will it be half? So you just take the values about center of gravity. You do not have to worry about from the nose. Correct. But in two distance, so distance between agodynamic center and CG. You have already been given that LCG, the distance between aerodynamic center and CG is 0.05. Okay, that is not from nose. No, you see the figure. It is given. I have not that it is between from nose to CG. It is given in the figure as a distance from the aerodynamic center to the center of gravity. So a lot of time we use wind tunnel testing. WTT is wind tunnel testing to determine all these parameters. So in wind tunnel testing basically we make a model of the aircraft. In this case it is a wing and we put it in the wind tunnel and there is something called as a wind tunnel balance. So you can see in this case we have mounted the model and at some particular point there is this contraption coming in. So this particular contraption is called as a wind tunnel balance. The wind tunnel balance essentially works on the principle that if there are some forces or moments acting on a model because of the wind if you externally apply those forces and moments and balance out the values then the forces are in equilibrium and hence whatever you have applied is equal to what is acting. That is why we call it as a balance. So you externally outside the wind tunnel you balance the forces and moments coming by adding weights at some distances and use it to calculate the forces and moments. So in this experiment there is a wing mounted on a wind tunnel with a wind tunnel balance. Now we assume that the conditions are ISA conditions in this example and we assume that note down these numbers because you will need them. The area of the wing, reference area of the wing and now in this case it is a single wing. So if it is a single wing then the area of the wing exactly is geometrical area in the top view. So SW will be 1.5 square meters and chord of the model is 0.45 meters and assume that in this wind tunnel now that is a very high speed but assume that you are able to generate 100 meters per second in this wind tunnel. Normally wind tunnels only produce around 30 to 40 maybe 50 meters per second but this is a numerical example. We assume that the wind is 100 meters per second. Now we mounted the aircraft at some angle and we kept on changing the angle till we reach an angle at which there is no net lift acting. Notice that that angle could be negative. It may probably be minus 2 or minus 3 degrees for such a wing. So is it possible that you have a wing and the lift is 0? It is possible. It depends on how you orient it. So there is some alpha 0 lift, alpha OL. The numerical value is not important. The point is that at that condition the aircraft was producing no net lift. So in the wind tunnel balance the place where you experience the lift force you have 0. But under that condition, under that condition obviously the moment may not be 0 because the moment is basically dependent upon the angle and the summation of all the forces acting ahead and behind the center of gravity. So the moment about CG was not 0. It was recorded as minus 12.4 Newton meter. So how is it recorded? You create a moment of minus 12.4 Nm to match it and you find that the total lift acting is 0. So now you move it to some other angle alpha 1. The numerical value is not important. Some angle alpha 1. At that angle now the moment changes direction and now you have 20.67 Newton meter positive moment and you also have lift 3, 5, 7, 5 Newton. So actually the numbers are actually applicable to an actual aircraft. We are just showing a simulated wind tunnel. So you have to calculate the two parameters. The moment coefficient about aerodynamic center and location of the aerodynamic center from the nose. So go ahead, copy down this information and start calculating. Get the value of cmac and after that get the value of xac. So to make it easy for you I am going to show you I am going to show you this figure here for some more time. So draw the same figure that you drew last time because once again this is the flying wing kind of a thing because there is nothing like a tail here. So draw the same figure. Draw one moment about aerodynamic center. Mark the location of CG. CG is at some point. So try to calculate and give me the answer. So the same equation I will show you once again. Same equation will be applicable here also. Moments about CG will be mac plus l into lcg. So let us solve this question as we go. I can give you a couple of minutes more. So the first thing you can do is to calculate dynamic pressure which is half into rho infinity v infinity square. Since the conditions are c level density is 1.2256 or 1.225 kg per meter cube and the velocity is given to 100. So this is the dynamic pressure. So do not look at me please. You have to do this calculation yourself. Now when you have 0 lift at that time cmcg will be equal to l which is 0 into xac plus mcg. So mcg upon q infinity sc. So q infinity is known, s is known, c is known. So you can get the moment coefficient about center of gravity at 0 lift which is minus 0.003. Interestingly because this is the condition for 0 lift this also is cmac. If you put it in the expression you will find out it is the same. So with this you can calculate cmac. The next is at any other condition we have the lift available. So we calculate cl at that condition. And then cmcg will be mcg upon q infinity s into c. The moment is given as 20.67. So with this expression you can get the moment at center of gravity under that condition when lift is 3675. After that it is very straight forward. cmcg is known, cmac is to be calculated, cl is known. So lcg will be nothing but the difference in the moments upon cl which will be 0.02. Yes, yes it is actually how to find the, so how do you find that? You cannot find it because it could be any number. It could be any number. So what you can locate is only lcg, the location from between the cg and the aerodynamic center. That is all you can calculate. I do not think you can calculate anything more than this. Right? So again I will have to amend that question. Yes, what is aerodynamic center? So do not we assume, so at the center of pressure there will be some lift and some moment. At aerodynamic center we have the same lift and the fixed moment. That is what we do. That is how we define aerodynamic center. So that we do not have to worry about the changing location of the center of pressure with angle of attack. So the definition of aerodynamic center is such that we do not have to worry about, so that is why it is a theoretical point which takes care of this problem. Yeah, yeah, so how much is the chord? So that should be there. You are right, again you are right. It should be 0.005 into c. But you are calculating moments. So the whole of it should be multiplied by c. So I mean I am giving this in terms of c. I am giving the whole thing in terms of c only. So I will just write c there. I will put c there and c here. Okay. Right, now this is a nasty picture. But this is what happens in general. Okay, but we will simplify this picture and bring it to level flight. I wanted to put this picture because I wanted you to understand what kind of complicated figures we have to deal with when we study flight mechanics. So this is a general picture. Now the aircraft in this case may not be in study horizontal flight. The aircraft in this picture will be in a general condition. So what is the condition? The condition is that there is something called as the angle of attack. There is a flight path angle. There is an incidence angle. The drag force DW is not acting along the aircraft. It is acting at some direction perpendicular to that is a lift force. Okay. And even the tail is also at some angle. Things become easy when we look at level flight. In level flight this lift vector L will not have any angle. And hence the drag also will act along the flight path. So the question is that the aircraft is in study level flight and it is in trimmed flight. So what is the meaning of trimmed flight? When do you say that the aircraft is in a trimmed condition? Somebody here? Yes, what do you mean by trimmed condition? Control stick. Okay, so from the pilot point of view, a trimmed flight is when the pilot feels no net force on the control stick. But when will it happen from the aircraft designer's point of view? When will that happen? Okay. Take a mic. When will that happen? When there is no net movement above the center of gravity. Okay. So can you still have moment about center aerodynamic center when the aircraft is trimmed? Yes. You can. So about center of gravity there is no net movement. So if there is an imbalance in the aircraft and the moment about center of gravity is not zero, what does the pilot do to make it zero? So one way is the pilot can deflect the control surfaces intentionally up or down, left or right to create a moment that cancels the imbalance. But then the pilot has to continue flying like that. So imagine a situation where the left wing or the port wing is heavier than the starboard wing by 1 kg. Can it happen? How can it happen? So there could be some issue because of which there is more fuel in the left wing than in the right wing. Even this 1 kg imbalance will create some kind of a moment. So the aircraft will tend to fly with left wing down. So what does the pilot do? The pilot moves the control stick to the right, create differential ailerons so that the moment by the ailerons is equal to the moment by the imbalance. And now for the remaining flight, the pilot has to fly like this. Uneasy, not acceptable. So what do they do? What can you do? One way would be that I create a bias in the control system so that so much deflection of aileron is always there. So I create a bias and then I lock it. And now throughout the flight, whenever I give a control stick, it will be that plus bias. That plus bias will take care of the imbalance. So what is the problem with this? Yeah. You are already eating away some consumed. That is very good point. You are already eating away some amount of controllability. Correct. That is true. That is one problem. One problem is that the aileron is now not useful for that many degrees because it is consumed in balancing. Any other problem? They will be more drag. Because now throughout the flight, in level flight, you are going to have aileron so many degrees up and down. So how do you cancel it? How do you take care? By waving a hand like this? You are saying, as a Karlo. So what is your solution apart from waving hands? So there is something called as a trim tab. So trim tab is a very small device or a very small aerodynamic surface which the pilot can deflect to create an counter moment to the imbalance and then lock at that position so you can fly hands free. That is why, as he said, pilot does not feel any net moment during the flight. So in this case, we are saying that the aircraft is in steady level trimmed flight. So steady means what? You are just murmuring. Constant velocity and altitude. No, steady is only constant velocity. Level will be constant altitude. Trimmed is no net moment on center of gravity. So this is the ideal condition for passengers like you and me where aircraft is in still level flight. So this particular aircraft under this condition. So remember, all these three words are going to be used in the calculation. So in this condition, the aircraft is having a lift of 40,000 newtons and the moment about aerodynamic center, I do not know how they measured it. It is minus 20,000 newton meter. The wing has a chord of 5 meters. XAC is quarter chord. XCG is 0.45C. These are some typical values. And LT is the distance between the aerodynamic center and the aerodynamic center between the CG of the aircraft and the aerodynamic center of the tail. That is LT. You can see it is shown here in this figure. This is LT. From the center of gravity to the quarter chord or aerodynamic center of the horizontal tail. That is 10 meters. So if this is the case, first of all you must calculate the lift at the tail and then calculate the weight of the aircraft or mass of the aircraft in newtons. So go ahead. So what you can do is you can draw this figure but not all those lines and the angles. You just draw lift vertical at AC. You draw the center of gravity. You draw a tail. Put the length as LT. Mark XAC, mark XCG and mark the chord SC. So simplify this figure. 2000 Newton and 42000 Newton. Correct. That is percent. But is this tail load acting up or down? If it is acting up then why is the weight 42000? It will be because lift is 40 and 2000 is the tail load. So it is up. Normally that is not the case. Normally we have a download on the horizontal tail. So some people have already solved it very nicely and very quickly. So take moments about CG. That will be equal to 0 because it is trimmed. So that will be moment about AC plus lift into XCG minus XAC minus LT into the tail force keeping in mind the direction. So this is minus 20,000 plus 40,000 into 0.45C minus 0.25C minus 10 into LT. And C is 5. So LT is 2000 and W is 42,000. Okay. So with this now we come to the end of the tutorial.