 Hello everybody, let us look at how we estimate the drag polar of a military aircraft. And to illustrate this procedure, we have chosen F16C fighting Falcon as the base aircraft on which we will do these calculations. So the color scheme in this presentation is that the general instructions like these will be given in brown color. If there are any values which are specified in any reference source, they will be shown in black color. Any values that we assume will be in brown blue color. The places where you should do calculations will be highlighted in red color with question marks and there will be this symbol, the pause button. So wherever you see this button, you have to remember that that is the place where you should stop the video and do some calculations and then match with the values obtained by us. The calculated values will be shown in this dark blue color and wherever we compare our numbers with existing aircraft, we are going to generally use the green color. Let us look at the source of the data and comparison for this particular tutorial. As I said we are going to use F16C and the textbook by Brandt, Stiles, Burton and Wittford contains a detailed description of the procedure for estimation of the drag polar and that procedure has been borrowed by us and used in this tutorial. Many aircraft have a nonparabolic drag polar. Here is an example. So the CD for such aircraft can be expressed in terms of CD0 plus K1Cl square plus K2Cl. There may also be actually some higher order terms but those are normally neglected. So this becomes a nonparabolic drag polar because there is a linear term K2 into Cl there, the first order term K2 into Cl. So we know that here CD stands for the drag coefficient, CD0 stands for the parasite drag coefficient also called CDO and Cl is the lift coefficient, K1 is the coefficient which is obtained in terms of the aspect ratio AR and the Oswald efficiency factor E0 which can be determined using this formula in terms of the wing aspect ratio and the leading at sweep. Let us look at the second parameter K2. Now K2 is used to model the effect of camber on drag polar among other things. There are many aircraft which may generate a minimum drag not at an angle at which Cl is equal to 0 but at a slightly different angle. There are many reasons for this because an aircraft is actually not just a wing, aircraft is basically wing plus fuselage plus tail. So there are certain situations in which the angle at which the aircraft flies and has a minimum drag does not correspond to the condition where the Cl is 0. So here is an example of an airfoil which has a nonparabolic drag polar. So that is the top blue line that is a nonparabolic drag polar which has got two components that is a parabolic component which is modeled by the coefficient K1 and then there is a nonparabolic component that is modeled by this equation. So in such a craft the profile drag is minimum at some small positive value of Cl generally. So this additional K2 Cl term in this equation models this particular effect. Otherwise if this term was absent then we would actually get a parabolic drag polar. So we will see today how we can get the equivalent parabolic drag polar also for an aircraft. Now if you look at this expression CD in terms of K1 Cl square and K2 Cl and let us say if you want to find the value at which the drag is minimum. So if you differentiate this expression with respect to Cl take dou CD by dou Cl put it equal to 0 then take the second derivative and confirm that that number is negative. Then you can easily derive the condition that K2 will be equal to minus 2 times K1 into Cl and that Cl will be called as the Cl minimum drag. So that is how the coefficient K2 can be calculated. But to do that we need to get the value of Cl minimum drag that means we need to know K1 we already know from the previous slide but to know K2 we need to know the Cl at which the drag is minimum. So for that we have to make some assumptions. So we assume that this minimum drag occurs the aerofoil has a minimum drag at alpha equal to 0 and we look at the equivalent skin friction coefficient that will be CFE is equal to CD0 into S by SWET and the parasite drag coefficient will be then obtained as CDO equal to CFE into SWET by S just by reverting the formula where S is the wing reference area and SWET is the aircraft wetted area. So now let us see how we get the equivalent parabolic drag polar. So this is what we would like to have we would like to have CDO equal to CD min plus K1 Cl min drag square this is the equivalent drag polar. So how do we obtain the value of Cl minimum drag that is the question. So what we do is since the aerofoil is assumed to generate minimum drag at alpha equal to 0 therefore what we assume is that the induced drag actually does nothing but moves this Cl which gives minimum drag to a mean value and this value is assumed to be halfway between 0 and the value of Cl when alpha equal to 0. So Cl alpha equal to 0 and 0 half of it if you take that would be the assumption of moving of Cl minimum drag. In other words we know that the Cl at alpha equal to 0 will be absolute angle of attack into the lift curve slope and this absolute angle of attack is actually going to be minus of the lift equal to 0. So that means the angle at which lift is equal to 0. So therefore Cl minimum drag will be this value plus 0 upon 2 or it will be Cl alpha into minus alpha L0 by 2. So knowing the aerofoil you can get this value and you know also the value of Cl alpha in fact we have a separate tutorial on calculating Cl alpha of an aircraft. So therefore now we can get CD min which is the requirement here as CFE into S fed by S. Let us look at the typical values of the equivalent skin friction coefficient that is CFE for a jet bomber and civil transport it is 0.0030 for military jet transport it is 0.0035 for air force jet fighter 0.0035 again for a carrier based navy jet fighter it is slightly higher for a supersonic cruise aircraft it is lower and for a single seat propeller aircraft it is very much higher for a light twin propeller aircraft it is a little bit lower and for propeller C plane it is the highest value. So in our case and of course there is also a jet C plane. So we do have in our case the problem that we are looking at is the air force jet fighter. So the typical value of skin friction coefficient equivalent skin friction coefficient would be 0.0035. So let us start the calculation for an aircraft in level flight where all the four forces are in balance but while operating at a cruise Mach number. So we are going to go to calculations at cruise Mach number under the ISA sea level conditions. Here is the aircraft geometry and this geometry is then we just note on the value of various parameters. So for this particular aircraft we know that the wingspan is 30 feet as specified and the conversion is 9.144 meters. The wing reference area is specified as 300 square feet which converts to 27.87 square meters. Similarly the tail span is specified as 18 feet which converts to 5.49 meters. The tail reference area 108 square feet which converts to 10.033 square meters. Streg surface area 20 square foot streaks 1.85 square meters. Root chord at the wing at the center of the fuselage 16.5 feet which is 5.03 meters. Tip chord is 3.5 feet or 1.07 meters. So what we do is we look at some more parameters. For example the sweep of the hinge line of the flaps is 10 degrees. The leading at sweep is 40 degrees. The sweep of the quarter chord line is 30 degrees and the sweep angle of the maximum thickness line is 24 degrees. So the aerofoil maximum thickness line is a 24 degrees sweep. We also need some data from the side view. For example we need the distance from the quarter chord of the wing to the quarter out of the tail the so called tail arm and we also need the value of the lateral or vertical displacement of the horizontal tail from the plane of the wing which is 1 foot or 0.3048 meters in this case. So with armed with this information we can now do the estimation of the wetted area of this aircraft. You can estimate this by either making a CAD model and then the CAD software gives you the wetted area and that is what can be done by using a software such as OpenVSP. We have already covered a detailed description of the software OpenVSP and we hope that by now you have already tried our hands on OpenVSP. The other option that we have is that you convert the whole geometry into some standard and simple shapes like cylinders, cones, rectangles, half cylinders etc. And then calculate the wetted area of each component. So we have done the same thing. So first we looked at CAD model by using OpenVSP and we took a model available from the VSP hanger. This is that model but if you notice this model has these two bombs already loaded and also these two additional missiles loaded. So we do not need this. So this is how you render the shape of the model and then you can see there are so many components which are there including the missile, missile stabilizer and pods etc. But in our case we have to modify the model by removing the pods or the external tanks and the missiles. So we get a simple clean model and for this simple clean model we just have these various components which are going to be considered. So using this particular CAD model it is possible to actually get the wetted area and also the parasite drag coefficients directly in the software. So this is the wetted area of each component as calculated by OpenVSP and the total wetted area if you add these numbers comes out to be 180.2 square meters. Now let us look at the second method where we look at the aircraft geometry and convert that into standard simple shapes. So here is a 3D diagram of the aircraft and what we do is we look at the geometrical data as specified in the book by Brent, Stiles, Burl, Tin and Whitford and then what we do is we approximate the geometry in simple shapes. So for instance you have these views of the aircraft so you can see that in the top view you just create some simple surfaces and also in the side view you also try to create some simple surfaces try your best possible to match the geometry with that given. And then you can do rendering of those surfaces to remove the internal details and hence you can get some idea about the various components. Now what we do is we look at this geometry and calculate the value of a sweat for each component using some equations. Now these equations and procedures are already explained in the textbook by Brent et al. The total area comes out to be 1418 square feet but we are going to now explain to you one by one how this is done. So what we did is that we first tried to compare the values between the numbers that we got using open VSP with the standard model available and the values which are quoted in square feet and then in converted in square meters. So we noticed that there is a large amount of error okay. Around 25 percent 26 percent for wings and 19.25 percent for vertical tail around the same value almost for that but huge variation in wing streaks horizontal tail and canopy. As a result there is a 37 percent increase in the value of a sweat as against quoted values. So now the problem is that if you continue with this a sweat in your calculations then you are going to introduce this much error automatically in all the calculations. So what we did is we calculated this wetted area using the geometry. So for any surface like wing or any stabilizing surface okay. So the formula is given here for S exposed and then using that to get the value of S wet. So for the wing which is surface 1 and surface 2 we know the wingspan from the geometry we know the root chord we know the tip chord and we know the t by c max. So we can calculate the value of S wet ref for each of the trapezium and then multiply it by 2. Similarly for the horizontal tail we have two small trapezia which are identical in geometry because the same aspect ratio same sweeps and then we look at the streaks. So these are also simple triangular devices. So maybe you can do half of base into height once the geometry is known to you. Vertical tail is also a simple trapezium because the root chord is known to you tip chord is known to you and the span is known to you. The dorsal fin happens to be the fin which is mounted on the below the vertical tail that is also a standard geometrical construct and then there are two hidden surfaces 9 and 10 which also contributes slightly to the geometry. So similarly if we now look at the fuselage and the canopy or the mid part we can replace it with these we can calculate the wetted area using these standard formulae depending on whether the cross section is elliptical or rectangular. So for a fuselage for example it is a rectangular cross section so you can get the cylinder height, cylinder width. So it is a mix between it is actually more towards elliptical and then we have fuselage sides these are two semicircular cylinders. Then we have a mid part of the canopy which is a pure cylinder and then we have a fuselage bottom. So all these are the formulae which can be used by you to calculate the actual wetted area of the aircraft. So these numbers have been given here only for comparison purposes for you actually you should be doing these calculations yourself. Moving ahead if you look at the nose the canopy front and the rear part because the mid part is a cylinder and the nozzle we get cones or frustums of the cones and for that we can use standard formulae available. So nose can be considered to be conical and with a nose cone of length, height and width. The and no end the rear portion can be a frustum of a cone so there are two cones here with l1, h1 and w1 and l2, h2, h2, w2 then for the canopy front it is half cone or a small cone and for the rear canopy also it is considered to be conical. So using these numbers you can calculate the areas of various components and after that here is the full table and as per this full table if you add these numbers the number comes out to be 139.31 square meters. The value of S-Wet given is 1418 square feet and S-Wet calculated is 131.73 by converting this number. So what has happened is throughout in these calculations we have done rounding off because each quantity given in feet has been converted into meters rounded off and used in the calculation that is why there was an error of approximately you know 8 and a half in 130. So we will go ahead assuming this to be the value because we do not want to introduce errors unnecessarily in our calculation. So coming back to the nonparabolic drag polar estimation we know now the wetted area of the aircraft is 131.73 and the wing reference area is 27.87 square meter. The CFE for this aircraft is 0.0035 and the hinge line sweep is 40 degrees wing span is 9.144 meter. With this information you can now calculate the value of CD min which is CFE by S-Wet and there is a formula for CFE but since we know the value of S-Wet and SRF and also the value of CL alpha we can always get the value of CD min which will be 0.0035 that is CD alpha CFE into sorry CFE into 131.73 which is the aircraft wetted area and 27.87 which is the wing wetted area. So in terms of the wing you can get the value of CL minimum for the profile drag coefficient of this aircraft to be 0.1654. Similarly, the aspect ratio is known as B square by S so it is 3 and E0 is the efficiency factor E which already was described and explained earlier. So you put the values in this put AR S3 and lambda LE S40 degrees please calculate these values do not just look at the screen do these calculations yourself and only then you will be able to learn and experience design. So the value comes out to be 0.9086. Let us continue with the nonparabolic drag power estimation k1 is 1 upon pi E0 AR now we know the value of E0 and also the AR so we know the value of k1 please calculate this value. The value is 0.1167 that is the value of k1 and average chord will be spanned by aspect ratio or 9.144 by 3 that value comes out to be 3.048 meters. So this is for a standard now let us look at standard sea level conditions for m equal to 0.2 for these conditions the Reynolds number is equal to rho into V into C bar upon mu. So rho is the density V is taken as 0.2 times this particular number because 0.2 is the mark number this is the speed of sound and C bar is the chord upon this is the mu value this number comes out to be 14203467 or 14.2 million. So the Reynolds number at which the aircraft is operating at mark number 0.2 at sea level is coming to be 14.2 million. So if you look at the wing alone drag coefficient variation of airfoil NACA 64A204 this is the table which shows up and this is the graph that shows up it shows very clearly that the minima is not at 0 the minima of the drag is at some point this is the equivalent parabolic polar and this is the k2 terms the k2 cl term. So by this by inspection we find that the for the airfoil the minimum drag occurs at cl I mean minimum drag equal to 0.04 approximately. Here is a closer look at the same graph which I just showed you. So here we see that the aircraft has a nonparabolic drag polar which is not centered at 00 but actually it is centered at some point here because it consists of two summations it consists of k1 cl which is this particular term sorry which is this particular term and k2 cl square which is this particular term. So CD naught plus k1 cl plus k2 cl square CD naught being a constant quantity you get this particular line and from there you can get the minimum as 0.04. Moving ahead let us look at some data. So k1 is now known as 0.1167 cl for minimum drag is known as 0.04 CD for minimum drag is 0.01654. So CD naught is equal to CD min plus k1 cl min square. So it is just a simple substitution of the equation there and CDO will be called will be known as 0.0167 if CDO is equal to a function of cl. So k2 will become minus 2 k1 times cl minimum drag. At this point we have calculated cl minimum drag and we already know the value of k2 we should keep in mind that there is a minus sign here by definition of k2. So k2 comes minus 0.00, 0.009 or nearly 0.01 nearly 0.01 and this is the formula for CD as a function of CD naught and k1 and k2. So if you put the numbers you can just replace CDO by 0.01667 you can replace k1 here by 0.1167 into 0.04 that is 0.1167 and you can put the constant term here. So you can notice that the value here is coming more accurate than the value written here that is because we are stopping here at the fourth decimal place. So when you plot this polar this is how you get. So not very apparent but it is clear that this polar is not having the minimum at cl equal to 0. There is a shift in the cl and this is the formula. Now let us move on to the last aspect which is the supersonic drag or the wave drag. So for this the maximum T by C is 0.04, hinge sweep line is 40 degrees, the quarter quad sweep is at 30 degrees. Now we can estimate the critical mark number for insert wings using this standard formula and that number comes out to be. So can you please confirm the value what number do you get by taking the claim? It is 1 minus 0.6, 0.65 into 100 into 0.004, power 0.6 it is 0.85 and the critical mark number is 0.85. So maximum mark number for cruise is occur at some point but that point may not really be of much use to you for your day-to-day calculation. So the formula is M critical is equal to the same number and just insert the values of the parameters except that here we have got the 1 minus M critical which is 0.85. So therefore MCDO max is 1 upon cost to the power 0.2 leading it. So it will become 1 upon cost to the power of 0.240. So if you calculate this value you should get the value of approximately 1.05. So the maximum mark number will occur, the maximum drag will occur at mark 1.05 then it will fall down again. So for wave drag again we look at these standard numbers which are available. Here we have a one more new parameter called as a wave drag efficiency parameter and now we already know the maximum mark number at max CD naught. So in this expression you have to calculate M minus MCD maximum taken then subtract it with the cruise mark number and multiply and then do this long expression you will get the value of EWD. So for our case mark number 1.05 we get the value of CD wave as 0.0261 and then there are some numbers at which you have calculated. Now let us look at the supersonic drag due to lift. Our assumption there will be that K2 equal to 0 and K1 is given by AR into M square minus 1 upon 4 AR root of M square minus 1 minus 2 into cost CD. So with that calculation we are getting the values as follows and we need to now know how does our value compare with the actual values. So let us check. So first thing is that here is a table that plots the estimated drag polar for various mark numbers right from 0.3 to 2 with the knowledge that it is maximum at 1.05 and then it comes down and compare this with the data given for the actual aircraft. You can see that the numbers are broadly agreeing at least for the values of the initial CDO at those conditions. So here is a graph taken from a source which talks about the wind tunnel testing and flight test data of YF16 which was basically the predecessor of F16. And on this particular graph if we superimpose our values we see that the graphs are for mark number 0.9, 1.2 and 1.6 and when we calculate our values for mark number 0.85 we get a curve which is fairly parallel to this aircraft okay. So this is how we can show that the numbers we have got are also quite reasonable and near to reality. You can notice that the divergence happens more at higher values of CD or even here we can say also higher values of CD. So here is the quoted value of the variation of CD0 with mark number. These are the points at which we have calculated the values. The graph could actually not be linear like this straight lines but it could be curved. But I have just plotted the points which I have calculated. So this is the quoted value and this is the estimated CD0 variation with mark number. And if you superimpose both of them you see that the values calculated shown in the dark blue line are quite matching with the values which have been available in the open literature. Similarly if you look at the quoted values of K1 and K2 their variation of mark number we notice that K1 remains constant up to some mark number I think at the critical mark number 0.85 and then it starts rising whereas the change in the slope of K2 is far minimum as compared to K1. So what we did is we also estimated the values of K1 and K2 using our methods and plotted them against the same graph and then we superimpose the values. So we realize that we are under predicting, we are under predicting obviously our values have to be properly determined. So before I close I would like to acknowledge a few people Brandt, Stiles, Bertin and Wittford for their lovely textbook on Introduction to Aeronautics Design Perspective from which I have used lot of formulae and lot of data for the existing F-16 aircraft. Daniel Ramer needs to be thanked for his seminal textbook on aircraft design which we use a lot and last but not the least I would like to sincerely thank my teaching assistant Namanuddin for putting in sleepless nights and creating this tutorial. Thank you.