David Brooks can he get Bournemouth off to a positive start in this shootout? He will be left-witted against his Welsh international colleague Wayne Hennessy and Brooks finds the bottom left corner with a smart spot kick even though the keeper went the right way. No problems there, Captain Hardin low, Brooksy. Well, Luka Milovoyevich will be making his way forward next for Crystal Palace to try and level it up immediately. Remember last season, Jerry's saved all the penalties fired at them by Forrest Green. Can Asmyr Begovic possibly repeat the heroics of Mark Travers? As Milovoyevich, the Serb, up against the Bosnian Begovic. Here he comes now, right-footed and it's 1-1 as Milovoyevich confidently slots it into the bottom right. Yeah, that was a pretty positive penalty, wasn't it? Hardin low, almost in the side netting. Even if you go the right way, unlikely you're going to save that one. Sam Seridge will go next for the Cherries away to a right and a bright game this evening. Again, good willingness in industry and made some good runs as the out-and-out striker had the ball in the net earlier on, of course, ruled offside possibly wrongly. Can he put the ball in the net now when it counts legally for the Cherries to put them ahead in the shootout? He does. Good penalty to the left-hand side. This time Hennessy went the wrong way. Cherries 2-1 up after taking two. Yeah, nice pen by Sam. Firmly hit. Do you like it when the ball hits that side netting? Usually means the keeper's got pretty much no chance. Mishy Batshuay will stroll forward for Palace, way to our right. Back on loan from Chelsea, of course, when he's down the pecking order. He didn't have too many opportunities himself in the game to worry as Milovoyevich. So Palace, so far 100% both teams. This to level it up at 2-2 after two each. Batshuay against Begovic. And this time Begovic goes the right way. But again, it's a very good penalty. You've got to say four very good penalties all round. Well, again, that one's hit the side net and even Begovic picked the right corner to go for. Well, you just can't get near it. So Batshuay will be replaced by Billing, who level at 2-2. Cherries have got that advantage, first of all, of course. Don't forget, Palace were eliminated on penalties last season by lower league opposition Colchester after a 0-0 draw, also in the second round. Cherries came out the right side of a shootout after a 0-0 draw in the second round. Can Billing give them the advantage again here with his left foot and then an excellent penalty from Billing. Lashes it into the bottom right corner. Absolutely. These are high-class penalties. I don't know, here somebody asking, is it different taking penalties with no crowd there? Andros Townsend will come forward now for Crystal Palace, way to our right. So 3-2, the Cherries lead. No advantage either way at the moment. Nothing to choose between these two sides as Begovic gives the crossbar in the net. One last rattle. Townsend waits for it to settle down, and he finds the corner as well. Begovic wants more, going the right way, but unable to do anything about it. 3-3 after 3 each. Yeah, not quite as firm as some of the other penalties, but he still found the bottom corner. Townsend. Cherries captain Lloyd Kelly taking the opportunity and the responsibility. Wait to our right here. Manchester City, you remember. Wait, next weekend. Sorry, next Thursday, I should say. Here's Kelly then with the armband on, with the Cherries fourth. Everyone has scored so far. Kelly scores, keeps that line going. 4-3 Bournemouth, having taken one more. The best seven penalties I've seen for ages. Keepers haven't had a threat. They're getting near them. They're right in the corner, hitting the side net in. And they've got pace on them as well. Ebera Eze will come forward next for Crystal Palace. We're getting towards crunch time now, because outside the top five, there's one or two in this Cherries team who have taken a lot of penalties in a pressure situation like this before. But Ebera Eze up against Begovic to level it at 4-4 after 4 for Crystal Palace. Eze takes his time, takes a deep breath. A couple of steps to the side, a couple of little feints, and then rolls it and somehow finds the corner. How did that get in? Begovic in the end almost apologetically escorted it into the corner. I've got to say, Eze, that is the coolest one you like. It's one of those where you can easily look silly, but he found the net. 4-4. I still thought Begovic was going to save it. Well, I did too. It didn't look like it had that much pace on it now. The fifth penalty comes from Bournemouth-dom Solankey there. 4-4, four each taken. The final of the regular takers. Solankey up against Hennessy. Still not a blot on anyone's copybook here at the Vitality Stadium. Solankey, short steps, finds the bottom corner again, and Hennessy again goes the right way. And the keepers are wondering, what do we have to do here? We're going the right way. We still can't get near any of them. Well, they've all been excellent penalties. Apart from that little hop step and a jump running up to the ball. I'm not sure about that one. So 5-4 Bournemouth lead. The final palace penalty of this regular part of the shootout will come from Jordan Ayu. The Ghanaian leading scorer last season came on as a substitute this evening. Has to score. Otherwise Bournemouth are off to the air he had next Thursday. Begovic stands in his way. Jordan Ayu with a relatively short run-up. 3 or 4 steps. Here he comes and he finds the corner as well. And we have a full house willow, 10 out of 10, 5-5. And we move to sudden death. Yeah. This is where it gets interesting. People who didn't really want to take a penalty are being forced into it now. Jack Simpson. Next. So where were you on the penalty taker roster willow? I'd be about three more and you'd get to me. Desperation measures. Jack Simpson, number six on the roster. Everyone has scored so far. Somebody eventually will miss. Let's hope it's not Jack Simpson right now. We're into sudden death then. Here he comes, Jack Simpson. He scored as well. Very good. Had a good game, Jack, as well. Just captures his performance. One hard and low into the bottom left-hand corner. We might still be here. Max Meyer will come next. We are starting to get... If we get to the goalkeepers, I can't remember a shootout for a long time that's had the goalkeepers taking them as the 11th taker. You see some smart aleck goalkeepers who get themselves earlier in the order, don't you? Very confident. Anyway, here comes Max Meyer for Palace. Simple scoreline, has to score. Has to keep Palace in it by beating Begovic, which he does as well. To the left-hand side of the goal, Begovic goes the wrong way. Already into his walk forward is Dan Gosling. What do you think? Dan might have been in the original five. Sixth responsibility. Did a good lead out. We're up to number seven in the order of the 11 who finished this game. What's the record for the most amount of penalties taken? I haven't got that. I just have not got that stat. I haven't got it. No, he won't be able to find that either. Anyway, Dan Gosling. Both teams have had six. Cherries are taking first. Gosling to restore the advantage. Oh, the Palace players, as it goes into the back of the net, the Palace players were celebrating, jumping around. They thought Hennessy had saved it. Off the goalkeeper's fingertips onto the post and into the net. Oh, no, not on the post. The goalkeeper's flat hand. I thought he'd hit the post. Well, it went in. 48 penalties, isn't the record. The Namibian cut, that's got to be. 48 penalties. So Gosling scored. He was a bit lucky. Hennessy had saved it. Here is Mamadou Sacco who has to score. 7-6 Bournemouth lead. And he scores. It is left foot into the high side of the left edge of the net. 7-7. Yeah, that was getting into the realms of top corner there. Another good penalty. Jefferson Lerma scored on Saturday here. Opening his account for the season. Corner's a cucumber. Normally, Jefferson when he's getting hot under the collar, of course. Keith Stroud, the referee underarms the ball to Jefferson Lerma. I'm pleased to see no messing about from the goalkeeper's either. Hennessy stands big with his arms up. Jefferson Lerma, 7-7 in the shootout to put Bournemouth in front once again. And he scores as well into the right hand side of the goal. They must have been practicing penalties all in lockdown. Do you know what I have to say? Because there isn't a crowd behind the goal, I think that makes a difference. Obviously, the hard bit comes when you've got 30,000 by Anagol and you've got to take a penalty. Certainly does. So, here comes Martin Kelly with Pallas' 8th. We really are starting to get towards the guys who wouldn't really want to be taking them. And again, the equation as it has been for the last three penalties is that Pallas have to score. Martin Kelly against Begavitch with Pallas' 8th. Top corner from Martin Kelly. Top left corner. Unbelievable standard. 8-8. Well, the goalkeepers of both, I would say three-quarters picked the right way to go, but the ball has just been too firmly hit or too high in the net to get a hand on it. Well, there are three penalty takers left for the Cherries. The next is Nandi Offervor who will take one, still to take one in the team. Asmir Begavitch of course, the keeper and Jordan Zamora. But this one falls to Nandi Offervor. The ninth for the Cherries. 8-8 it stands. Here's Nandi Offervor. Scores as well. Police from as well because you don't want it really to be one of the youngsters who makes the faux pas if one is to come, but Offervor that was confident. Drove it in. Well, you have to remember Pallas missed the penalty once. I'm sure you missed one once as well. I'm sure it's easily done. So Pallas, their 10th penalty, sorry, 9th penalty and this is Big Ryan Innis, the big centre half with their 9th. He has to score up against Begavitch and he's put it in the top corner as well. The right top corner with his right foot, 9-9. Very, very confident. Well, here is the Zimbabwean Jordan Zamora full 90 minutes on his Bournemouth debut and he has the 10th penalty of this lead cup shootout with the score on the scoreboard away to our right saying 9-9. So Zamora scoring squad number 33 on his back with his orange boots away to our right in front of the south end with a little posse of cameras and ball boys just away to the left. He is up against Wayne Hennessy. What's the young man got? Here comes Jordan Zamora and he's somehow found the net with an unconvincing penalty but he's got it in and Bournemouth have 10. They went up again. They thought he'd got you. They picked the right way to go. There wasn't a lot of pace on the ball now there but it crept in. I'm not sure if he had a free shot of goal. He could strike of all less convincingly again than he did there. Scuffed it with his instep but it bobbled its way almost apologetically into the corner. So here we go with the final penalty of the outfield players and it's Jaroslav Jak who spots it down. The Polish player away to our right and Bournemouth are we heading for the keepers or is Begavits going to have a say? He does have a hand but unfortunately not to stop it going into the net and it is 10-10 and we're down to the goal keepers John Williams. Yeah, he did get a hand on it. Well, I've got to say what I don't think I've ever been at a shootout that's had the keepers taking them after a full house of 10-H. Nobody has missed with our 20 successful penalties. Bournemouth Palace we're just tuning in. In the second round of the Carabao Cup the first goalkeeper will be Asmir Begavits of Bournemouth against his opposite number. Right, but he goes for power! Asmir Begavits just went with the kitchen sick there rather than any kind of placement and Hennessy gets right and he saves it and Palace have the advantage and the Welshman Hennessy can win it. Well, that's kind of unfortunate because he's got a good strike but once it's off the ground most of the other penalties if they weren't in the bottom corner they were in the top corner and that little two-foot height always gives the keeper a chance. So Asmir Begavits and his first Bournemouth game for 18 months or so now has to make up for it by saving the one from his opposite number. Hennessy! A disastrous penalty from Hennessy. He missed by almost the whole postcode over the bar. Absolute disaster and it remains 10-10 and we're coming round again with David Brooks back for a second go! Did he go out the ground? Nearly! On the beach, look out! That was miles over the bar! Goodness me, I think he forgot he'd put concrete in his boots before he took that one. I've never been at a shootout when it's gone round for a second time. So David Brooks is back. There's people on the halfway line putting their boots back on. David Brooks then. He accepted the keeper's score and then the keeper's both missed. So here's David Brooks. 10-10 if you're still keeping count. Brooks left-footed scores again in the top corner. Another excellent penalty from David Brooks. 11-10 to Bournemouth. Absolutely fantastic penalty. See my Brooksy smashed it high into the top corner. No goalkeepers getting that one. Goodness me.  I've now got a second column going. Anything could happen here. 11-10 to Bournemouth. And of course, we remain in sudden death. So Milovoyevich comes forward for his second penalty. And he has to score to keep Bournemouth... I'm sorry, to keep Palace in it. Here we go. The 24th penalty for Milovoyevich! It's all over! Bigovic has saved it to his left-hand side. A good, strong hand from the big goalkeeper who thought he'd blown it a moment ago when he missed his own penalty. But he dives to his left. He saves Milovoyevich his penalty. Bournemouth have beaten Crystal Palace 11-10 on penalties. And they will go to Manchester City in the third round next Thursday night.


Hello friends, today we will look at the constraint analysis of Boeing 787-8 Dreamliner aircraft. This aircraft is a very path-breaking aircraft and the picture that you see is of the first 787 which was delivered to Air Nippon Airways, the launch customer. They placed an order for 50 aircraft in 2009 and that is how the journey of this aircraft started. I am going to follow a particular color scheme in this presentation which I would like you to make note of. The general instructions regarding calculations are going to be given in this brown color which is the basic background color. Here there are values specified as the parameters of the constraints. Those values will be shown in black color. Any values which are assumed based on past information or data about the aircraft will be shown in light blue color. At places where I would like you to do some calculations, I am going to use the red color and followed by two vertical lines either two black lines or a pause button. The purpose of this is to alert you while watching the video that at that stage you have to pause the video and you have to do some calculations and then resume. Finally, the calculated values that you obtain using the formulae will be shown in the dark blue color. You have to match the values obtained by you with these values and in the end I have some comparison between our estimated values, calculated values and the values which are prescribed for Boeing 787. Only at one place in the whole presentation towards the end we will show that data in green color. So I hope you appreciate this color scheme and it will help you in understanding the scheme of things. Let us look at the list of constraints for the Boeing 787 Dreamliner that we will look at today. You can notice there are 7 constraints in all as specified by the user and you can also observe that the values specified are in the British system of units. So this is the reality in aviation. Most of the aviation experts still work in the FPS system and hence I have decided to reproduce the values as it is. But as you know we are supposed to work in the SI system. So the values in this table have been converted into the SI system in this table. Now the numbers are not rounded but they are all in the SI units. So I think it is important for you to note down these values. In fact, we will use them, we will need them as we go ahead. So we will show them before we do the calculations. Let us have a look at some useful data about Boeing 787. It will be assumed that either this data is supplied to you as input or you have obtained these numbers prior to this tutorial through previous tutorials and calculations. Let us first look at the aircraft and performance data. This aircraft cruises at a Mach number of 0.85 and the maximum Mach number in cruise is 0.9. The cruising altitude is 37000 feet which becomes 11278 meters. The value of thrust-lapse ratio alpha at the cruising altitude is 0.1789. This information is not readily available. We have obtained this information by looking at some specific data regarding the Boeing 787. The maximum takeoff mass of the aircraft is 476000 pound which converts to 215971 kilograms. The mass of this aircraft when it ends its climb and begins the cruise is 448,049 pounds which converts to 203,457 kilograms. This information also is not easily available but it has been obtained by us by looking at some data. The maximum landing mass of the aircraft is 365000 pounds which converts to 165608 kg. The taper ratio of the wing is 0.1528. The aspect ratio of the wing is 10.58 and the thickness quad ratio of the wing is 9.4%. In reality the taper ratio, the thickness ratio of the wing varies from root to mid span to tip but we have taken some medium value. In fact, I have taken the value at the break. The quarter quad sweep of the aircraft is 32.2 degrees. And this aircraft has two engines. This information is related to the aircraft and its performance. Now let us look at some aerodynamic data. The zero lift dry coefficient of this aircraft when it is in cruising flight is 0.01277. We can assume it to be the same in all phases of flight as long as we do not go at a very high Mach number. The CL max at takeoff is 1.91 with flaps deflected and the CL max at landing is higher it is 2.66. The additional drag due to deflection of flaps at takeoff is 0.01 and the additional drag due to deflection of flaps during landing plus the landing gear that is 0.1135. We assume that during the initial climb stage the landing gear is retracted so there is no additional drag coefficient because of the landing gear. Now what is remaining is an important parameter which is the Oswald efficiency and that too we would like to calculate the value at various Mach numbers but to start with we would like to get the value at Mach number of 0.85 which is the cruise Mach number. So at this stage I would request you to please note down this data or you can take a picture of the screen and use it while you do the calculations because these blue numbers that you see are going to be required by you in the calculations that follow. So let us start our first calculation by trying to obtain the value of the Oswald efficiency of this aircraft. Now there are many methods available for estimating the value of Oswald efficiency as you know it is a very important parameter and there is a huge amount of literature available on which formula to use. The one that is very popular and most suitable for transport type aircraft is the one that you see on the screen given by Professor Dennis Howe of Cranfield University.  Let us look at it one by one. The first and the foremost parameter that you need is the Mach number MA because the value of E depends upon the sixth power of Mach number. The next parameter is the taper ratio and once you know the taper ratio you have to calculate a function of taper ratio called F lambda wing. Then we have to also use the wing aspect ratio and then the T by C wing. So here the T by C wing will be not expressed in percentages but if it is for example if it is 10 percent it will be 0.10. C as you know stands for number of engines and lambda 25 wing is the quarter chord wing sweep. So all these parameters are needed before we can estimate the value of the Oswald efficiency. As I mentioned we need to calculate a function related to lambda wing it is called as F lambda and the value of F lambda is as shown on the screen. Now using this formula once we obtain the value of E for any Mach number we can estimate the value at any other Mach number simply by dividing EM2 by EM1. Now you can notice that the quantities inside the square bracket are all constant with respect to Mach number. Therefore if you want to obtain the Oswald efficiency at some other Mach number M2 and you know it at M1 it is very easy EM2 will be EM1 times just the ratio of the first term. Let us proceed now this is the formula that we will use let us try to calculate the Oswald efficiency factor of Boeing 787 at the cruise Mach number using this formula. We will divide this formula into some 3 components and do it slowly. So first of all the data so MA is MCR which is 0.85, lambda wing is 0.1528, A wing is 10.58, T by C wing is 9.4% or 0.094, NE is equal to 2 and the quarter chord sweep is 32.2 degrees but remember to convert it into radiance because in the formula that you see on the screen where you calculate the cos in the denominator you have to use the value of quarter chord wing sweep in terms of radiance. So what we can do is we can divide this formula into 3 parts. So E will be equal to a term A into bracket 1 plus B plus C where A, B and C are highlighted as shown in the screen. So therefore first thing we need to do is we need to calculate the value of F wing which is given as per the formula shown in the bottom in the green background. Then we need to calculate the values of A, B, C and once you know A, B and C you will be able to calculate the value of EMCR. So at this point please pause the video and do these calculations and watch ahead only after you have done the calculations so that you can match. If you really want to learn aircraft design you need to solve the tutorial along with me and not just watch the video because then you will not learn. So please pause the video do these calculations once you are ready then you can play it again. Assuming that you have obtained these values by calculation let us compare. So the F lambda wing is 0.0065 the value of the first parameter A is 1.0453 the value of parameter B is 0.2924 and the value of parameter C is 0.0821. So using these values of A, B, C in the formula E is equal to A times bracket 1 plus B plus C bracket close and the whole of it 1 upon that or power minus 1 we can estimate the value of E or the Oswald efficiency to be 0.6961 remember this value is only for the cruise mark number and for any other mark number you need to use the formula as I described earlier to estimate the value. Moving ahead let us start our constraint analysis in which we determine the values of T by W and W by S that meet all the requirements. But to begin with we will have a first cut estimation of thrust to weight ratio and for this we will use the considerations for maximum mark number. The typical values of T by W and P by W for various types of aircraft we need to have an idea about it. So let us first look at jet engine aircraft or aircraft which have a turbofan or turbojet engines. In these cases we calculate the thrust to weight ratio T by W this ratio this dimension this is the dimensionless parameter in case of jet engines because both thrust and weight are in the same units. So for civil transport aircraft it is very easy to remember if it is a 2 engine aircraft it is around 0.2 if it is 3 engine it is around 0.3 if it is 4 engine it helps 0.4. But when you go to the military aircraft then depending on the type of the aircraft whether it is a strategic bomber or a tactical bomber or a trainer or a fighter or a interceptor or air superiority aircraft the value comes as shown in the screen. So for an aircraft like F-16 the value of T by W is approximately 1. If their craft is powered by a turboprop or a piston prop then we calculate power to weight ratio because there is nothing like thrust to weight ratio defined in that because turboprop and piston prop engines they develop power and that power is then extracted in the case through a propeller to generate the thrust force. So the engine has only P by W which has to be converted to T by W. The units in this case would be watts per gram this is the standard SI unit for P by W where P is the power in watts and W is the mass in grams. So for civil aircraft again depending on the type there are various values which are the ballpark recommended or first cut values and if you are looking at the military aircraft the values tend to be slightly on the higher side. This data shows that for our case which is a twin engine aircraft we can expect the value to be nearly 0.2 or between 0.2 and 0.3 and this data has actually come from the textbook by Dan Raymer. Moving ahead let us also estimate based on some suggestions by Dan Raymer in his book on how to estimate or get a better estimate for thrust to weight ratio as a function of the maximum Mach number because we are looking at a jet engine aircraft. So Dan Raymer has given a table there is one table for aircraft powered with jet engine aircraft and there is another one for prop engine aircraft we are looking only at the one with jet engines and in our table our aircraft is a jet transport aircraft. So therefore the T by W 0 that is the thrust produced at sea level static conditions divide by the design gross weight W 0 is expected to be A into M max power C where M max is the maximum Mach number and A and C are constants to be determined from this table. For example for jet transport aircraft Raymer recommends to use A as 0.267 and C as 0.363. So for Boeing 787-8 the maximum Mach number as specified is 0.9 and therefore if you can estimate the T by W 0 using this particular simple approximation. So let us do that calculation. So the calculation will be very simple it is just 0.267 into 0.9 to the power 0.363 the value comes out to be 0.257. So please remember as per the previous slide the approximate value was 0.2 between 0.2 and 0.3 because it is a twin engine aircraft and now because we have the maximum Mach number information available we get a slightly better estimate. So if you get your answer very far away from this number suppose you get the answer as 0.4 or 0.5 etc or 0.1 you should start distrusting your calculations. If you get a value near 0.257 then you can be quite confident that the answer is correct. Moving on now let us start the process of doing constraint analysis. So what we will do is first we will look at constraints that depend only on thrust to weight ratio. There are two such constraints as we know in the list. The first one is the second stage climb gradient and the second one is the missed approach gradient and we will start by looking at how the thrust to weight ratio is limited by constraints on these parameters. So the first one is the second stage climb gradient as specified in the constraint table the second stage climb gradient is expected to be 3.49 percent minimum when the aircraft is at the maximum takeoff weight of 215971 kilograms under ISA plus 15 conditions at sea level and to calculate this we will use this particular formula. If this formula seems unfamiliar to you I would request that you stop here go back and watch the lecture on constraint analysis for civil aircraft where we have explained how this formula is obtained and what it stands for. So proceed further only if you are familiar with the formulae and you have watched the lecture regarding constraint analysis procedure for civil transport aircraft otherwise you will not be able to appreciate the numbers. Moving on when we are looking at the second stage climb gradient then we are using the subscript SSC for second stage climb and SSCG stands for second stage climb gradient. So the formula which I just flashed I have just inserted the corresponding values for the thrust for the weight for the L by D and for the gamma in this particular formula. So here N E stands for number of engines which is specified or given as 2 the L by D in the second stage climb is the L by D to be calculated for the second stage climb configuration. Now recall from our previous lecture that during the second stage of the climb flaps are in the takeoff configuration and for that we have already been given that the additional drag because of the deflection of flaps in the takeoff configuration for this aircraft can be assumed to be 0.01 and also the landing gear is up or retracted and therefore there will be no additional drag because of landing gear deflection. So the only additional drag which will occur would be only because of the flaps. The required second stage climb gradient as specified in the constraint diagram constraint table was 3.49 percent. So with this information now we need to calculate the L by D in the second stage climb. So that is our next step and let us see how we can calculate it. So this calculation happens over I think 5 steps. First because the second stage climb is happening at ISA plus 15 degrees we need to do some atmospheric parameter calculation. Then we calculate the Mach number in the second stage climb which we call it as MSSC. Now recall we have seen a formula to get the value of Oswald efficiency as a function of Mach number. So the next step will be to calculate the Oswald efficiency in the second stage climb call it as ESSC. Once you know the value of ESSC then the induced drag coefficient which is 1 upon pi into AR into E can be estimated. So that is the next step. Now once you calculate the induced drag coefficient you already know the value of CDO and you can add the additional drag coefficient because of the flap deflection. So with that you can get the value of CD and once you get the value of CD you already know the value of CL therefore you can get the value of L by D. Let us do these calculations for Boeing 787-8. It is specified that the second stage climb occurs at ISA plus 15 degrees. So first thing to do is to calculate the value of the temperature at ISA plus 15 degrees centigrade that is 288.16 which is the ISA temperature at C level plus 15 degrees. And using this information we can calculate the value of density which will be P by RT, P remains the same 101325 at C level whatever be the ISA value 287 is the gas constant and the units are joules per kg degree Kelvin and 3 is not 3.16 is the ambient air temperature at ISA plus 15 conditions. So pause here and do the calculations for rho. The value comes out to be 1.1646 kg per meter cube. Next we need to calculate the sonic speed at this particular condition. The sonic speed is basically root of gamma RT where gamma is the ratio of the specific heats 1.4 for air 287 is again the universal gas constants and 303.16 you know is the ambient air temperature. So calculate this value by pausing the video. The value turns out to be 349 meters per second and the value of sigma which is the density ratio would be the density under these conditions divided by the standard C level density of 1.225. Pause the video do the calculation the value of sigma turns out to be 0.9502. The next step is to calculate the Mach number in the second stage climb. For this the V stall is already specified as 138 knots EAS under ISA conditions. So first what we do is we convert this into TAS. So EAS you know is equal to TAS into root of sigma therefore TAS will be EAS into divided by root sigma. So it will be 138 upon the value of sigma which has been obtained in the previous slide as 0.9502. Please get this value pause the video and let us continue. The value comes out to be 141.57 knots or 72.83 meters per second. To convert knots into meters per second you should multiply by 1852 and divide by 3600 or you can also multiply by 0.514444. Now we assume that the velocity of the aircraft in the second stage climb would be 1.2 times V stall to keep some safety. So that would be 1.2 times 72.83. Please calculate this value by pausing the video. The value turns out to be 87 per 4 meter per second and since we know the velocity now and we also know the sonic speed obtained in the previous slide we can calculate the value of the Mach number in cruise which is 87.4 divided by 349 or 0.2504. So the Mach number in the climb for this the Mach number in the second stage climb is 0.2504. The next step would be to calculate the value of the Oswald efficiency in the second stage climb. You know that the Mach number is 0.2504 during the second stage climb. We also know that the cruise Mach number is 0.85 and the value of E at cruise was derived earlier as 0.6961. We have a formula which can be used to calculate the value of E at any other condition. We need to calculate the E at the second stage climb. So we can use this formula. It is a simple ratio of two terms multiplied by the value of E at the cruise condition. Calculate the value. It turns out to be 0.7275. Now once you know the value of E you can go ahead and calculate the induced drag coefficient during the second stage climb because CL max takeoff is known as 1.91 specified. ESSC we have just now calculated as 0.7275. Now since V SSC is 1.2 ton V stall therefore CL in SSC will be CL max by 1.44 which will be 1.91 by 1.44 that would be 1.326. I hope you are pausing the video whenever you see this two vertical lines and doing the calculations yourself. Once again I would like to repeat that otherwise you will not be able to really get a hang of this procedure you will simply nod your head and go ahead and you will not be able to appreciate. So please spend some time wherever you see two vertical lines like this you should pause the video do the calculation that has been asked and then check your values in the next. So now once we know CL SSC we can get the value of CDI which will be CL square by pi AE now everything is known to us. So I put the values in the formula for you for your ease in calculation but please pause the video and calculate the value of CDI SSC. The value comes out to be 0.0708. Now finally once you know the lift to drag ratio once you know the values of these parameters and you know that the additional drag because of flap deflection is 0.01 and additional delta drag because of the landing gear is not to be considered.  Please calculate this number by putting in the corresponding values CDO is 0.0201277 CDI is all known. So with all these values you can get the value of CD at SSC as 0.0936. So L by D would be CL SSC by CD SSC both the values are known to you. So therefore please pause the video and calculate L by D SSC. The L by D SSC turns out to be 14.17. Compare this with the L by D of approximately 19 to 20 that you get in a cruise condition. So we see that during the second stage climb because the landing gear is up and only the flaps are deflected you lose little bit from 19 to 20 you just come to around 14. So there is definitely a reduction but it is not too much. Later on when I show you the calculations for the missed approach gradient you will notice that the L by D is going to fall to a much lower value. So with this we can calculate the thrust to weight ratio for second stage climb gradient again I am just flashing the old formula. But now we have all the values available SSCG is specified as 3.49%. We can assume that during the second stage the weight is equal to almost the takeoff weight. So therefore the weight parameter beta will be almost equal to 1 the number of engines is equal to 2 L by D SSC is just calculated. So therefore the value of T by W in SSC condition or T SSC by WTO can be easily calculated using this formula. This value pause the video and calculate please the value comes out to be 0.2109. However this is not the answer because this is T SSC by WTO what we want is T SL by WTO. So this is at an ISA plus 15 condition whereas we need to get the values at ISA T ISA upon WTO. So for this we need to use the value of alpha that is obtainable which I will show you shortly. So what you do is because we know the value of T SSC by WTO and we also know value of sigma we can always say so the T SSC is basically the thrust under ISA plus 15 condition and T SL is the thrust under ISA conditions. And we know from our basic engine model that for a subsonic turbofan engine the value of thrust lapse ratio alpha when there is no afterburner is simply the density ratio rho by rho SL or sigma. So therefore T SL will be equal to T SSC divided by sigma. So you can show that T SL by WTO which is what we want should be more than equal to T SSC by sigma divided by WTO. But since both are in denominator you can exchange them and you can get T SSC by WTO divided by sigma. Now T SSC by WTO is known as 0.2109 and sigma is known as 0.9505. Therefore please pause the video and calculate the value of T SL by WTO it will be 0.2109 divided by 0.9509 or 0.2219. Hence we come to the first conclusion that thrust to weight ratio of the aircraft should be more than 0.2219 from second stage climb gradient consideration. So now we are going to do similar calculations for other constraints. Let us look at how we can calculate the constraint line due to the second stage climb gradient. So this appears as a single horizontal line, single horizontal line because for there is no role of wing loading in this case we got a direct constraint on T by W. So the T by W value has to be always more than 0.2219. So therefore this is the horizontal line. So the area which is shaded below this line is infeasible and the area above this line is feasible. In the same way we can obtain the lines on the constraint diagram due to several other constraints. Thank you so much for your attention. We will move on to the next section now.