 So, by putting the moment, so what we do is, you take the CG location and CP as variables. That means the central gravity is not fixed, by ballasting you can change it. So first you assume at some place, no ballast and assume the CP location to be equal to diameter for X and Z. Then what you do is, in this calculation you will see that you have these forces L and D which will be half rho V square. Similarly M will also have half rho V square L. So there is a V there which is the velocity. So what you can do is, vary the velocity from 0 to V max which is expected and calculate the alpha at which it trims. For every velocity, there will be a different alpha at which it trims. So I will show you in the next slide, it is possible for you to invert the relationship in such a way that you get, you can take a derivative of Cm by du, dou Cm by dou u equal to 0. That means whatever be the velocity, my alpha remains the same because my Cm remains the same, it is possible. So to answer this question in detail, there is a very interesting paper written by a B Tech student called Ashok Rajani. Interestingly last week he got married and I got a card from him. So Ashok Rajani did his B Tech here in 2008 and 2009 I think and the V process with Vakar as the co-author, we did some work on the dynamic stability analysis of a tethered aerostat system and it appeared in journal of aircraft. So it is a big achievement for a B Tech student to publish a paper in journal of aircraft. So I am very proud of this guy. So you can see what he has done. You will see many things which are already discussed with you. So he has done a very good work on the historical developments in modeling and simulation of aerostats. A very nice piece of work where all tethers were reviewed very nicely. Then he looked at the ADLD aerostat and he did the equilibrium analysis but this analysis was done based on the method given by Professor Krishnamurthy and Panda. Now Krishnamurthy and Panda, this paper has come from NAL Bangalore. Professor Krishnamurthy was a professor of aerospace engineering at IIT Kanpur and then towards the end of his career he shifted to Bangalore to NAL and there he supervised some scientist called Panda to work on the, to derive the expression which I just told you. So Krishnamurthy and Panda's report is a very famous report which gives basically the picture which I showed you. Then it gives equations for the forces, the moment equation and the force equation and finally if I just save some time in the end you will see there is a very interesting chart which he gives. So I will upload this paper and you can read it and derive the expressions and use them. So this is the picture from that particular paper which is very similar to what we saw but there are some minor changes. Now the locations B, G etc are all variable, do not worry about them. You, by looking at the value of X, J and XC you can always change the location. So the forces are weight buoyancy and aerodynamic forces. These forces can be converted into X force, Y force and a moment and the tension. So if you take the balance of the forces along the X direction you will see that Fax will be equal to this force La and this will be alpha. So the vertical force, horizontal force will be basically L, Cl sin alpha and Cd cos alpha and if you take the vertical force will be Cl cos alpha plus Cd sin alpha times half rho V square. A is the volume power 2 by 3 here. And moment as I have already showed you is half rho V square A into L into Cm naught and then it defines the differences as Cx and Cz which are the locations. So coming back to the same equation, let us do this derivation ourselves. I think it is important for us to get the feeling. So can you look at the figure and help me in force balance in the XZ plane. So can you get this expression please? Now do not copy it down blindly. Do the force balance. So on your notebooks I want you to carry out the force balance. There you go. Are you able to see? Okay. Now you can see. So take the balance of forces along the X axis or along the centre line of the envelope. So remember the forces that you will consider will be B, Bf or the buoyancy force. It will have 2 components Bf sin theta and Bf cos theta. Similarly you have W cos alpha W sin alpha. Then you have L again sin alpha and cos alpha and then you have D sin alpha and cos alpha and then you have this Tv and Th which are the vertical components of the tension and horizontal component. This is also at an angle theta. So along X for example you will have Th cos theta acting along the right side which is the positive side and you will have Tv sin theta acting in the opposite direction. Similarly you will have La sin theta acting on the left direction, Bf sin theta acting this way and W sin alpha but D cos theta will be along the direction. So please take these moments and get the expression. Similarly do it for the vertical components. In the vertical components remember the downward force is positive. Upward is negative, downward is positive because from centre line down is the positive axis. So you will have negative component of Bf, Bf cos theta, negative of La cos theta, you will have positive of Da sin theta, positive of W cos alpha, you will have negative of Th sin theta sorry positive of Th sin theta and negative of Tv cos theta. So with this please get me the expressions for the two forces. The moments are to be taken along the X direction and normal to it. So along the X direction you will have Fax and vertical will be Faz. Those of you who have finished you must start the moment balance now. So for the moment balance there are these distances Xa, Xg, Xb and for the for the tether we have Xc and Zc. The moments are to be taken about the confluence point for which you will need the distance between the confluence point and B, constant points and G, confluence point and A. So for that what they have done is they have given you a term called as for example you have a term Xg minus Xc which is Cx. Similarly for the gravity you have Zg minus Zc, yeah that is Xc and Zc. From the nose the location is X and Z for the confluence point. So it is Xc below and Zc ahead. So you do not see it to be ahead but the distance is so Zc will be draw a vertical line from here up to here and different between that and the nose will be Zc. Ready? You have done it. Let me show it to you. So these are the two force equations you will get. You can see them in the figure now. Fax is the force minus Bf sin alpha it will go on that side to become positive. So please notice that Fax is the net force which will be equal to 0. So you take all the terms other than Fx on that side you will get the correct signs. For example W cos theta is shown negative here because it is brought on this side. Fax and Fz are actually lift and ride. These are basically the aerodynamic forces acting which are being equated to other forces for equilibrium. So Fax is actually Fax is actually your D and Fz is basically L only yes Fax is an angle. So if you see what we are doing is now we are basically so basically you know along this direction you have Fax actually this direction is Fax this direction is Fax along the line and normal to the line. Now let us balance the moment about the center of conference point. So balance the moments about Cp okay I will just go ahead because you will be able to get these numbers only and only if you do the calculations okay. From this point we will do a small hand calculation. So you will have the notes with you so you can derive it yourself. Important point is I have copied the equation from the previous slide here to give you the moment balance and then I always bring in the fact that Fax is this, Fz is this, Ma is this, Cl is basically lift curve slope A v into sin alpha and Cd is not plus K Cl square but Cl and alpha are related by alpha Cl curve so it could be equal to alpha square. So this K is not 1 by pi Ae this K is a constant which gives you the link between Cd and alpha square understand that Cl versus alpha is a straight line so for every alpha there is a corresponding Cl. So when I have Cl square I can have K dash alpha square where K dash takes care of K and the relationship between Cl by D alpha. So if you put all the expressions for example if you replace F a z with this expression okay I have done this myself you get expression given in my paper. So right now at the moment those of you who are interested to verify go home comfortably paper is already going to be uploaded look at the equations see what happens now. Now let us see how we use this expression to get the value of the optimal location. Now if you look at this expression Cl is the net moment coefficient which should be 0 at equilibrium Cl not is the aerodynamic moment of the body which is a fixed number based on the geometry of the body it is available to you from aerodynamic data. A b is the lift curve slope which is fixed unavailable then you have Cd drag coefficient which is known from aerodynamics x a bar x g bar x c x bar are basically x a upon l x g upon l the momentum will have a l term when you divide it it becomes bar this is all geometry it is known to you. So the only variable in this are u that is the velocity and alpha. So what you can do is you can solve this expression Cn equal to 0 for various velocities you will get various alphas but if you take a differentiation of this with u those same by the u then you get an and if you say that is equal to 0 that will give you an expression which is invariant with velocity. So then at every velocity you will have the same value of alpha which is the equilibrium of trim alpha and that is a good thing for you because you do not care what velocity are you want to keep changing the alpha with time. So the same expression I have copied here and I want to show you the variation of the equilibrium angle of attack alpha with velocity for two conditions one is the constant Cd condition that means in this expression you assume Cd to be constant basically k is 0. So therefore there is no change with alpha so you get this to what we see is at 0 velocities or low velocities the arrow start will trim at a high angle and at a velocity increases this angle will keep reducing and a time will come when there will be hardly any change. So at around 10, 12 meter per second onwards the angle is not changing much. So the arrow start will trim at this angle at low speeds and then come and then it will become almost fixed. Now this particular information has come from this Krishnamurti and Pandas report. I will just show you I have scanned this report because it is an old report from NAL and in this they have done the analysis. This particular work was sponsored by ADRD Agra. So in the end they have actually done the analysis for one ADRD arrow start and they have given the pictures which I showed you. Those two figures are from the same report only.