 Let us look at how we can get the limit on the thrust to weight ratio from the constraint on the mist approached gradient. As we can see from the table, the specified value of mist approach gradient is 2.1 percent. This happens at the maximum landing weight of 165608 kilograms under sea level ISA conditions. So once again, this is the constraint, the formula for constraint on the mist approach gradient, only thing is I have inserted the term MA wherever there is mist approach and MAG stands for the mist approach gradient. Now here any stands for number of engines which is 2, L by D MAG is the L by D of the aircraft in the mist approach configuration. As you know during the mist approach, the flaps are in the approach configuration. So they are at a higher angle than in takeoff and the landing gear is down. So from the aircraft data related to this particular aircraft, we have got this number of 0.1135. So the delta CD because of the deflection of flaps and the landing gear is a very large number 0.1135 and the requirement on the mist approach gradient as specified is 2.1 percent. Let us now calculate L by D in the mist approach just like we did it for the previous case of second stage climb gradient. The procedure is identical, only the numbers will change slightly. So the steps are very similar. First we calculate the atmospheric parameters during MA. In this case, we do not have to do this step because the mist approach is happening under ISA conditions. So therefore, we already know the values of density and temperature and sonic speed under ISA conditions. Then we calculate the Mach number of the aircraft because once you know the speed and the sonic speed, you can get the Mach number. Once you get the Mach number, then you can get the Oswald efficiency. Once we have the Oswald efficiency, we can get the induced drag coefficient. Once we have the induced drag coefficient, we can get the total drag coefficient. We already know the lift coefficient. So therefore, we can get the L by D and let us do these steps for Boeing 787 now. It is specified that the mist approach occurs at sea level ISA conditions. So therefore, the parameters of the atmosphere are very well known. T is 288.16, rho is P by RT which is 101325 divided by 287 into 288.16. Although all of us know this value, but still it makes sense to calculate it. The number comes out to be 1.225 kg per meter square. The sonic speed will be obtained by a root of gamma RT for these standard values. Once again, this number is known to us, but still it is important for us to calculate it and assure ourselves that we are on the right track. Now the value of sigma will be equal to 1 because rho and rho naught are the same in this case. Now to calculate the Mach number, we first use the fact that the V stall landing as specified in the requirements is 102 knots EAS at ISA. Now under ISA conditions at sea level EAS is equal to TAS because sigma is equal to 1. Therefore A and also we have just calculated that A is equal to 340.27. So just convert the value of speed 102 knots into meters per second by as I said multiplying 1852 divided by 3600 or simply multiplying by 0.514444. Now during missed approach, we assume that the speed is not 1.2, but 1.3 times V stall. So that will come as 1.3 times 52.47. Pause the video and please calculate this value. The value is 68.2 meters per second. And hence since we know the value of VMA and we also know the value of sonic speed A, we can get the Mach number by taking a simple ratio. Please pause the video calculate the value. The value comes out to be 0.20. Once we know the Mach number, we can calculate the asphalt efficiency factor using the information that is available already using the same formula that we have used earlier for the value estimation of the value of E during the second stage climb. So calculate the value of E missed approach. Please pause the video here and calculate the value by this simple expression. The value comes to be 0.7276. So with this we can now get the induced drive coefficient in missed approach. So we know the value of CL max landing at 2.66 is specified in the aerodynamic data and we have just calculated the value of asphalt efficiency in the missed approach EMA as 0.7276. We know that in missed approach condition the stall velocity is 1.3 times V stall. Hence CLMA will be CL max landing by 1.69. Please calculate this value. It will be 2.66 by 1.69 that is 1.574. Once the value of CL is known, therefore you can get the induced drive coefficient because now we have all the information we have CL missed approach and we have the value of aspect ratio 10.58 and E missed approach 0.7276. So by including these values inside the formula, now you can pause the video and calculate the value of CDI MA. The value comes out to be 0.0997. Moving ahead, now you can calculate the lift to drag ratio in the missed approach. For this we know from the previous calculations the value of CDI SSC and CL missed approach. And now we also know that we have to add additional term because of the extra drag of the flaps and the landing gear during this condition. So CDO plus CDI thus Lyd will be CL missed approach by CDI MA and that will be 6.965. Let us look at the T by W missed approach. For that we insert the corresponding values of T, W and L by D and the gamma in the formula. Now we have already been given by the requirements that the missed approach gradient is 2.1 percent and we know that this aircraft Boeing 787-8 has 2 engines. The L by D missed approach has been calculated previously as 6.965. So therefore, once you insert these numerical values in the expression, it becomes very easy for us to calculate. So pause at this stage and calculate the values. We see that the value of T MA by W MA has to be more than 0.3291 during the missed approach condition. But remember that the missed approach condition occurs during landing whereas we are interested in the value of T by W under the C level static conditions and for maximum takeoff weight. So therefore, we have to convert this value and to convert this value we will have to incorporate some factors. So to get the value of missed approach T by W limitations at C level static conditions and max takeoff weight, we just reproduce from the previous slide the values that we know. We also know that the thrust during missed approach is equal to thrust at the C level condition because the missed approach constraint is supposed to be met at C level conditions and also we can assume that the weight of the aircraft during the missed approach is almost equal to the weight at landing. So keeping these 2 in mind, we can first calculate the value of beta which is the ratio of weight of the aircraft during the missed approach condition and the max takeoff weight. These 2 numbers are known to us and hence it can be easily estimated. Please pause the video and calculate this ratio. The ratio turns out to be 0.7668. So now we know that TSL by WTO is equal to TMA by WMA into beta. So you just multiply this number by beta and you will get the value TSL by WTO missed approach will be more than or equal to 0.2524. This is one mistake which many students make they forget to convert the values obtained for a constraint to the standard conditions of C level static thrust and weight of the aircraft at takeoff. So be very careful. Let us see how this maps into the constraint diagram as a line because the constraint was directly on T by W there is no role of W by S here. So therefore, this will appear as a horizontal line in the graph as shown. So we also arrive at an interesting conclusion that the thrust to weight ratio of the aircraft cannot be less than 0.2524 from the constraint on the missed approach gradient. Let us look at the constraints that depend on wing loading alone. In our table of constraints there are three such constraints the one on takeoff stalling speed, landing stalling speed and the landing ground roll. So let us see how these three appear in the constraint diagram. So since they depend on W by S alone it is clear that they will appear as vertical lines. First let us look at the limit on wing loading from the constraint on takeoff stalling speed. As per the specified requirements the value has to be 71 meters per second equivalent air speed or less at the max takeoff weight under C level conditions but under ISA plus 15 conditions. This is the requirement which you have to meet. The formula that we will use is very straightforward. We know that lift is equal to weight during a 1g stall. So from there you can easily get an expression between the wing loading required versus the density of the air at takeoff condition, the maximum value of CL at takeoff condition and the specified value of stalling speed square. So let us see how this constraint can be incorporated. First of all there is a specified requirement that the V stall should be less than 138 EES which I have already converted 2 meter per second. Now previously we have already calculated that the density of air at ISA plus 15 at C level is 1.1646 kg per meter cube and we also know from the data that CL max of the aircraft at takeoff conditions is 1.91. So therefore you can get the value of the stall speed at ISA plus 15 conditions. And once we obtain the value we just have to insert the numbers in this formula which means we have to replace rho takeoff by 1.1646, CL max takeoff by 1.91 and V stall takeoff by 72.82. Pause at this stage and calculate the value of wing loading required. We see that the wing loading has to be less than or equal to 5897.7 Newton per meter square. This is what you get by directly inserting the numbers in this formula. But remember that the answer you get would be in Newton per meter square which has to be converted into kg per meter square so that we can by dividing with G so that we can have consistent units in our constraint diagram. So with this let us see how the constraint line appears on the diagram because of takeoff stalling speed. As I said it will be a vertical line which is stationed at 601.37 kg per meter square and the area on the left of this line is going to be feasible because the wing loading cannot be more than this number it has to be less than or equal to this number. So the area on the right of this line is infeasible on the left is feasible. Moving on let us look at the limit imposed on wing loading from the constraint on landing stalling speed. Here the requirement is to calculate at the landing weight at sea level ISE conditions there will be no need to convert the densities we can use the standard sea level values and the units are also the value is specified as 52.46 it was 102 knots EAS which can be converted to meter per second using a simple formula. Once again we will use the standard formula just like we used for the takeoff condition the only difference is that we will put subscript land for wing loading density CL max and V stall. So let us see how this number pans out this is the requirement which I am repeating here for ease in memory CL max at landing is given as 2.66 from the aerodynamic data for this aircraft. So you have a simple expression and now if you substitute in it the values you get 1.225 in place of row land CL max land is replaced by 2.66 and V stall land by 52.46. So pause at this stage and please do the number. We find that the value is 4483.8 Newton per meter square or 457.2 kg per meter square. So this is a much stricter limit but remember again this is at landing so it has to be converted to the values corresponding to W takeoff. So W takeoff by S is what is our on the constraint line. So during landing condition it is supposed to be less than 457.2 and we know that the takeoff weight is 215971 and W land is 165608 specified. So we can calculate the value of beta which we got in the last expression also as 0.7668. So with this beta you can easily estimate what will be the value of WTO by S at landing. It will be W land by beta land by S if you want to convert W land to WTO and since both are in the denominator you can get it as W by S landing by beta landing. So we know these values insert the values pause the video and try to get the constraint limit as shown. So get this number this number is going to be less than 596.24 kg per meter square. So it turns out that the landing stalling speed constraint is slightly more harsher than the constraint due to the takeoff stalling speed but that is only for this particular example it is not a general statement. So once again we see that the line will map on the constraint diagram as a vertical line with the area on the left of the line as feasible and the area on the right as infeasible. Moving ahead let us look at the third constraint which depends only on W by S that is the landing ground roll. The requirements specify that the ground roll should be less than 621 meters at maximum landing weight of 165608 kg under sea level ISA conditions. So for that we will use this particular formula. If you are not familiar with this formula I would request you to go back and look at the video of clips of the lecture on constraint analysis for transport aircraft where all these formulae have been explained. So this formula will be used by us. So the specifications are that the atmosphere is ISA the altitude is sea level and the required distance is less than 621 meters. The aerodynamic data is that the CL max is 2.66. We have to assume a few values in this case we do not assume any afterburner therefore alpha is equal to 1. We assume that the operations are from some standard airports therefore the mu roll is going to be 0.4. Since the thrust force you know at landing is such and also just after landing during the ground roll we put the spoilers therefore we can assume that the lift at landing is completely killed and hence we also neglect the drag at landing. So and the sea level condition density is 1.225. So with this what we need to now do is just calculate beta. Beta is already known actually as 0.7668 we have done it just a few minutes ago. So we can just remember that value. So this is the formula that I mentioned. Now if you look at neglecting the values of D land and L land then the formula simplifies and with a simple rearrangement of W by S on the left hand side you can see that it is just a formula in terms of landing distance S land density CL land and mu roll and beta. So inserting these values we can calculate the W by S constraint. So please pause and insert the values we all know these values from the last slide. So the number turns out to be less than 624.6 kg per meter square. So it is interesting to note that the landing ground roll constraint is not really that harsh in this case. And also please remember that we have not worried about converting it to take off conditions because already in the formula we have included those parameters in the form of beta. So therefore the constraint because of the landing ground roll turns out to be 624.6 kg per meter square and appears as a vertical line with the area on the left being feasible and area on the right being infeasible. So this is how we looked at the constraints that depended only on W by S. Let us look at constraints that depend on both W by S and T by W. There are two such constraints in our table. The first one is the constraint due to the residual climb rate at cruise conditions and the second is the balance field length. So let us see the first one the one which is a constraint due to the climb rate at cruise. This particular requirement is 2.2 meters per second or 429 feet per minute which is to be calculated when the aircraft is just at the beginning of the cruise and the altitude is the cruising altitude of 37000 feet or 11278 meters under ISA conditions. For this we will use this master equation and if you have if you do not understand this equation or if it seems very unfamiliar to you I would request you to go back and have a look at the video clips of the lecture on constraint analysis for military aircraft because we have used master equation mostly in that particular application. But we will use it also for the current requirement. So in this in this large expression which relates the TSL by WTO and WTO by S for several operating conditions, we notice that the climb rate is actually DH by DT. So we are given the value of DH by DT and we have to calculate the link between or the relationship between relationship between TSL and WTO by S for we have to calculate the relationship between TSL by WTO and WTO by S for a given value of DH by DT and the other parameters will be obtained from either the known value or from the specified constraints. Let us see how it is done. So the data for this requirement is that the ROC is 2.2 meters per second under ISA conditions since it is a level flight n is equal to 1 and also it is a study flight. So therefore, there is no acceleration. So the aircraft is in a study level flight. The cruise Mach number is 0.85 and the cruise altitude is 37,000 feet or 11 to 78 meters. And we have obtained some information about the fuel consumed in the warm-up, take out, take off, climb, etc. So through that we come to know that the aircraft weighs approximately 203457 kg. When it comes in at the beginning of the cruise, the condition at which this climb rate is to be estimated. There are some values which we will assume. First of all, we will assume that the takeoff weight of the aircraft is 215971. This is given in the aircraft data. We also know from the aerodynamic information that the CDO of this aircraft is 0.01277. The wing aspect ratio is 10.87. And from the engine data, we come to know that the value of alpha at this cruise Mach number is 0.1789. And we have also calculated earlier that the value of the Oswald efficiency at cruise Mach number is 0.6961. So with this information, we now have to calculate few parameters which are required in the master equation. These are the induced drag coefficient K1, the value of aircraft weight ratio beta, the cruise speed and the dynamic pressure during cruise. These can be straightaway calculated using standard formula. Let us see how it is done. First of all, for the induced drag factor, it is actually 1 upon pi AR into E. So A wing is known, ECR is known. So pause the video and calculate the value. This value turns out to be 0.04207. The next information we need is beta, which is the ratio of the weight of the aircraft at the cruise condition or beginning of the cruise divide by the weight at the maximum takeoff condition or the maximum takeoff weight. This also can be easily calculated. The number is 0.9421. Then the cruise velocity at the cruising altitude can be obtained either by a numerical calculation or you can also pick it up from the ISA table because it is an ISA condition and that number is 0.3494 kg per meter cube. We can also calculate the temperature at the cruising altitude and the value of sonic speed. The temperature we all know, temperature at cruising altitude is already known because it is a standard value, it is 216.66 at an altitude between 1125 kilometers under ISA conditions. And hence the value of sonic speed can be easily calculated as root of gamma RT, where gamma is 1.4, R is 287 joules per kg degree Kelvin and T is 216.66 degree Kelvin. So with that the value comes out to be straight away as 295.05. Now since we know the value of cruise Mach number and now we know the cruise, the sonic velocity at cruise conditions, you can calculate the value of V and once you know V, you know rho and you know V. So half rho V square is Q, please pause the video and calculate these two numbers. So VCR is 0.85 into 295.08 or 250.8 and QCR is half into rho into V square, which if I would request you to pause and calculate comes out to be 1120.6 kg per meter square. So we now have all the numbers that we need k1, beta, V, CR, Q and HCR, which we need in the master equation. So we can go ahead. The master equation is written there on the top of the screen. So we know that dV by dt is 0 therefore the last term will vanish. We know that dH by dt is 2.2 and V is 250.8. So with that you can actually substitute the values and you can get a long expression and here what we have done is we have inserted the values of beta, alpha, Q, CD0, k1, N, V and dH by dt in the expression. So this can be easily solved. In fact, if this turns out to be an expression like TSL by WTO is equal to this whole constant, this whole number divided by WTO by S and then this whole number divided by into W by S plus a constant C which will be the terms outside the bracket, these terms and these terms. So I think we should pause the video at this stage and calculate the values of these three constants A, B and C. The values come out to be as shown on the screen. Now with these values you can replace now in this equation. So you will get relationship between TSL by WTO because you can put A here, you can put B here, you can put C here and for various numerical values of WTO by S ranging from 300 kg plus meter square to 650 kg by meter square, you can see how it pans out into the diagram. So this constraint on climb rate at cruise depends both on WTO by S and TSL by WTO, therefore it will come up as a line in the constraint diagram. So if you draw this line in the constraint landscape, we see that it appears as a curved line with the area below the line as being infeasible. Moving ahead to the last constraint which is on balance field length, the requirement given is that the balance field length should be 2 weight, 1, 2 meters at maximum takeoff weight and under sea level and IC plus 15 conditions. For this we are going to use the takeoff parameter as explained in the textbook by Daniel Rehmer. So the takeoff parameter is a parameter that is obtained from a graph and from there you can get a direct link between T by W and W by S. Let us look at the constraint on takeoff balance field length. This is an image taken from the famous textbook by Dan Rehmer which shows the correlation between various types of takeoff distances and the takeoff parameter. So we know from our specifications that the BFL has to be 9 to 25 feet at IC plus 15 and the data is that there are only 2 engines in this aircraft. So with that if you proceed from the Y axis with 9 to 25 feet and hit the line for BFL or balance field length for twin engine jet aircraft, you can go down and obtain the value of takeoff parameter as 233 pounds per square feet. This has to be converted into SI units and after that the value of sigma at IC plus 15 has already been calculated it is 0.9502. So let us look at how this constraint can be calculated. So the takeoff parameter is parameter defined as W by S divided by sigma CL max takeoff by T by W. So by rearrangement you can get a link between T by W and W by S. Sigma CL max takeoff and TOP are known. CL max is specified by the aerodynamic data. Sigma and TOP have already been calculated. So T by W can be obtained simply as W by S divided by 3 constant terms or the expression is T by W is equal to A times W by S. So I think you can pause the video and calculate the value of A. A turns out to be 4.8 410 power minus 4 and with this you can get various combinations between W by S and T by W as shown. So this pans up into the constraint diagram as a straight line and this line the area below this line is infeasible and the above the line is feasible. So now let us look at how do we plot the constraint diagram by suppurposing all the constraint lines one by one. First let us look at the constraint that depend on T by W alone which are the second stage climb gradient and the mist approach gradient. These are appearing as horizontal lines. The first one is SSCG and the second one is MAG. So for this problem we see that the mist approach gradient is more severe than the second stage climb gradient of constraint. Then let us look at the constraint that depend on W by S alone. There were 3 constraints in our diagram takeoff stalling speed, landing stalling speed and landing ground roll. These will appear as vertical lines. So the first one is the one on stall takeoff. Then the next one is on stall at landing and the third one is because of the landing ground roll. Area on the left of these lines is feasible. So we notice here that the constraints because of the stalling speed at takeoff and at landing ground roll are weaker than the constraint on the stalling speed at landing. Finally let us look at the 2 constraints that depend on both W by S and T by W which are the climb rate at cruise and balance field length. These will appear as curves or lines in the constraint diagram. The first one that we bring here is the line due to the climb rate which is a curved line and the next one is the constraint due to the landing due to the balance field length. So the intersection of all these lines this is the final constraint diagram. So this area above this purple line and on the left of this red line this is going to be the feasible area because the area below this line is infeasible the area on the right of this line is infeasible. So therefore this area is the one which is the feasible area. So now moving on let us see how do you find the design point. This is obtained by taking the intersection of the most critical constraints which is as shown in the screen 596.6 kg per meter square and T by W of 0.291. The design point will be taken slightly above and slightly on the left of this constraint this intersection point because of the requirement to give some margin for growth of the aircraft and for changes in the design as the analysis procedure progresses. So it is always good to give some margin. So what I have done is I have given little bit of margin on the left hand side and little bit margin on the vertical side. So the design point. Now let us see how far we are from reality. Let us just check. So this is our constraint diagram and the values that are given in piano for Boeing 787 are 600 and 0.28. So what we see is that we are very much near the values which are available in piano. So it shows that our constraint analysis calculations is very much near reality. Hope your calculations were matching with these. Thanks for your attention.