 So, welcome back to this lecture on rocket and spacecraft propulsion. For the last couple of lectures, we have been discussing the optimization of multistage rockets. First we talked about the performance of multistage rocket, then we have been discussing the optimization of those rockets. And we have stated that optimization essentially means that either if the delta v that is the velocity increment is given and the final payload mass is given, how do we redistribute the masses among the rockets, so that the initial mass is minimum or we can put it differently that for the given overall payload ratio that is the final payload mass and the initial mass, how do we get the maximum velocity increment. Then we had talked about three different cases. In the first case what we had said was that the ISP was constant for all the stages and the structural coefficient was also constant and for that we derived an expression for the velocity increment which will give us the optimum value. And what we have seen is that in that case all the stages must have same payload mass fraction in order to give the optimum performance. The second case which we looked at was when we still maintained same ISP for all the stages, but we allowed for variation in structural coefficient and for that case we optimized and got expression for the velocity increment. The third case we talked about was the most general one where we said that ISP is different for all the stages and structural coefficient is different from all the stages. And what we did was we looked at the second approach for optimization in this where we actually minimize the payload mass fraction or maximize the payload mass fraction, we minimize the inverse of payload mass fraction. So, for that we got an expression that L which is the payload ratio or sorry the overall payload ratio or overall payload mass fraction is equal to this. We derived this expression in the last lecture. Let me call this equation A because this is a very important equation. So, in this equation epsilon i represents the structural coefficient of various stages and ISP i represents the specific impulse for different stages. So, what is given to this are the given values to us and then L is the overall payload mass ratio mass fraction which is also given. Now this parameter beta which is defined as minus alpha where alpha is our Lagrange's multiplier. This is what we used to get the optimization. Then now let us look at this equation and see how do we actually solve problems. As I said this is the most generic case. Any other case can be derived out of this. Let us look at this expression. So, essentially the problem statement is we have given some parameters, some are known. What are known is the number of stages n is known to us. The overall mass fraction L is known to us. The structural coefficient for every stage epsilon i is known to us and the specific impulse for every stage ISP i is known. So, these are all known parameters. With this known parameters what we have to do is we have to find out the best possible combination of essentially the payload mass fraction. So, that we get that total overall payload is the distribution is the best. So, that we get the highest velocity essentially velocity increment. So, what we do is the procedure is if I look at now this equation everything else in this expression except beta is known. Only beta is the unknown and as I have said towards the end of the last lecture when we expand this depending on the number of stages we get polynomials in beta. If it is two stage we get second order polynomial. If it is three stage we get third order polynomial etcetera. So, we get a polynomial in beta. So, therefore this equation can be solved this equation A can be solved for beta. So, the procedure is first solve A for beta. Once we have solved A and got this value of beta which is the inverse other negative of the Lagrange multiplier. Now, what we do is we take this value of beta and put into the expression for L i. Now, what is L i? L i is the payload mass fraction for individual stages which is equal to this is the expression for L i. So, essentially it comes from the same expression here. This is how we get L i. Once we get L i then we go back to the expression for R i. R i we have proved earlier is equal to 1 upon L i times 1 minus epsilon i plus epsilon i. So, the values of epsilon i's are given. L i we have calculated from here after we have calculated beta by solving A. Now, we have all the values of R i. So, the next step then is to get the final velocity increment which is equal to sigma i equal to 1 to n i s p i g e L n R i. So, this is the expression we use to get the velocity increment, where i s p is the specific impulse for different stages R i we are estimating from here. Now, we can put all of them together to get the final velocity increment. So, this is the solution procedure. As I have said that this is the most general case that we have talked about. Now, let us see that we have started with this. We got this expression. This is the most general case. Can we get back the other two cases we had discussed that is case 1 and case 2 starting from here. So, then if we just do this everything else should fall into place. So, let us now look at the other cases going back to case 2. What was our case 2? Case 2 said that equivalent velocity is constant for all the stages which implies i s p i equal to i s p which is constant. But, epsilon i is given it is varying. This is what case 2 was. Then for this case we get L is equal to coming back to this equation, because this equation is also valid for this case also. So, L is equal to pi i equal to 1 to n epsilon is a variable for stage to stage. So, we will write it like this. Only thing now here is that this i s p is constant. So, once again let us see if the value of L is given in this case i s p is given different values of epsilon i are given. Again we got a polynomial in beta which we can solve and get the value of beta. Once we get the value of beta then what happens is that since here beta is a constant i s p is also a constant, what we can do is we can take it out of the product sign, because we know that beta is multiplied as a constant and i s p is constant. So, we can take it out of this product sign. If we do that then this will be equal to 1 upon beta i s p g e minus 1 times n, because this was repeating n times. So, we will have this constant times basically to the rest to the power n and then just this term, just this term. Now, with this now let us simplify it little more. What we can do now is if the value of L is given and the number of stages are given, we can actually get a closed form expression for beta by solving this. Let us look at this expression little more clearly, carefully. So, what we have proved there is that L is equal to 1 upon beta i s p g e minus 1 to the power n pi i equal to 1 to n epsilon i 1 minus epsilon i. Let us now take this to the left hand side and bring L to the right hand side. While doing that what I will get is equal to L to the power 1 upon n, because this was rest to the power n. So, we take n th root of this then this becomes L to the power 1 by n divided by once again n th root of this term. So, this is not n th root that is 1 by n. Now, let us look at this expression L is given to us epsilon i is given. Therefore, right hand side is a known quantity and now we no longer have to solve the polynomial, because this is say some constant a. Therefore, what it becomes is that we 1 is equal to beta i s p g e times a minus a. I can take a to this side. So, this is 1 plus a upon a i s p g e is equal to beta. So, I can directly solve for beta by using this expression. I do not need to solve for a polynomial now and once I have solved the value for beta. I can now come back here and get L i, because L i is equal to nothing but we take this sign of this portion is L i. So, once we have got this then we can get L i is equal to epsilon upon 1 upon epsilon 1 i L to the power 1 by n and then power 1 by i equal to 1 to n epsilon i 1 minus epsilon i to the power 1 upon n. So, we can directly get all those values. So, this is the advantage that we started off with a very complex thing. Then now we are simplifying. So, of course, we are getting simpler expressions. So, we get now the expression for the mass fraction for every stage, very pillowed mass fraction for every stage directly by just putting these values, because L and n are known. Now, this is case 2. So, what we have done is we have got case 2 back here. Next we go to the even simpler case which was case 1. So, now when we go to case 1 then we get what was our case 1? In case 1 what we considered is that ISP for all the stages were same and epsilon was also same. So, once again what were given to us was N, L, ISP and epsilon. These are the things that are given to us. What we have shown for this case? So, this is that L i is equal to L to the power 1 upon n. We have shown that before. So, now for this case we can directly get an expression for delta u by ISPGE is equal to N L n 1 upon 1 minus epsilon L to the power 1 by n plus epsilon. What we have done actually is we have taken that expression and then put epsilon equal to constant. So, we get this from there. So, this what we see is that this expression is an interesting and important expression. Let us take a little closer look in this expression. What we are seeing here is the delta u is expressed as a function of the structural coefficient ISPGE plus the number of stage. Now, that brings us to a question. So, far we have been saying that in order to improve the performance what we do is we go to more number of stages. But then the question arises that is it a monotonic improvement that we keep on increasing the number of stages and the performance keeps on improving or there is a limit beyond which we probably would not get much of improvement. Then as the because as we are increasing the number of stages the complexity is increasing because every time when you have to discard something that separation process itself is a very critical thing. Just let me take a little detour and talk a little bit about the separation process. When the separation happens what is happening? This is one rocket on top of this another thing is sitting around. These things gets burned and then it has to separate. Now, when this is burning all the major controls are with this stage because this is the primary driver now. This essentially sits there as a dump payload does not participate in anything. So, all the controls are here. Now, at the point of separation you have to be very careful few things happen. First of all when they are separated they are still moving with the same speed. So, the entire thing is moving with the same speed and this guy was carrying this. Suddenly this is separated immediately the load acting on this has gone down and some force was acting on it. So, suddenly the load goes down this tendency to accelerate. So, this vehicle will try to accelerate in this direction whereas for this it has separated but the engine is still not taken over. So, now what happens something was pushing it suddenly the pushing force is gone. So, now it will have a tendency to slow down. So, this will try to move in this direction and this will try to move in this direction and the result is if you do not do anything they are going to collide and then break apart. So, therefore during separation it becomes very important that you have to prevent this from going this side and prevent this from coming down this side. So, what you have to do is you have small rockets here. These rockets are fired in the opposing direction. So, this rocket when it fires it will push it downward. So, you create a opposing force. So, you are retrograding the rocket or the stage. So, these are called retro rockets. So, they slow down they slow down this portion does not allow it to go up. At the same time these guys this thing has to be pushed out but still your main thruster has not started yet. You have smaller thrusters which essentially provide little force to compensate for sudden loss of the force that was occurring. So, you push with this small rockets again attached to this in this direction. So, this too has to fire simultaneously at the time of separation so that they actually truly separate. Till they are truly separated and far off only then they can safely set the separation is complete. This is one problem. Another problem is so far your control was here. Now, during the separation process you have retro rocket firing or all rocket firing but your stage has not yet burn started to burn. Once this starts to burn by that time you should have the control transfer from here to here. There is a gap during the separation where the control remains here and then it has to go there. So, during this time this guy is in free fall it does not have any control or anything. Now, if your control transfer is not proper then it may go out of the control loop at completely out of the control completely. If it goes beyond a certain limit the control later cannot bring it back to the path. So, therefore this control transfer is also very critical because of this reasons even though we see that multistaging is good very good but we cannot depend blindly on multistaging because some these are very dangerous because is a very tricky manoeuvre. Let me put it this way very tricky manoeuvre to separate the stages and transfer the control and then this also has to relight. It has to relight after that and then move forward. So, because of that we would not like to have more number of stages than that is required. We would like to get an optimum number of stages which gives us the required benefits. So, for that let us look at this expression that is why now this expression becomes important because this gives an idea of this n the number of stages. So, if I look back at this what I will do is I will plot the payload fraction that we can achieve for with the velocity increment for different number of stages. Let us consider that we take a case where the structural coefficient is same for all the stages 0.01. Let us also consider that the I S P is constant. So, our governing equation is this which is dictating the performance. Now for this case let me plot the variation of delta u by I S P g e which is the left hand side of this expression plus as L. L is the overall payload mass fraction. So, for a given value of this we get a certain we can move certain mass fraction in the optimal manner. Now let us do that for different values of n. If I do that for different values of n this is what I am going to get. This is n equal to 1, n equal to 2, n equal to 3, n equal to 4, n is equal to infinity. Let us see what we have here that first of all we know that as we increase delta u L is going to increase first point. We had also in discussed that for a single stage rocket there is a limit to which you can increase it beyond that the acceleration becomes too high etcetera. So, you do not want to operate. So, n equal to 1 we can get let us say a particular payload fraction moved optimal way. So, in order to move go to higher more higher velocity for the same payload fraction then we go to multi staging. So, let us say we go to a two stage. So, let us say this is the payload that we are trying to move. So, from one stage we go to two stage immediately we can see that the same payload we can now move at a higher much higher velocity because of the fact that we are discarding some of the dead weight. So, we are able to move to much higher velocity. If we increase the number of stages we are still improving we are still going to higher velocity because we are discarding this weight also. But the increase that we achieve between two and one and between three and two the increase the incremental increase in between three and two is less than that between two and one. We are improving it, but now the incremental gain is less. We go to four stages again we are getting some improvement, but once again the incremental improvement is less. As we keep on increasing the number of stages we keep on getting some increment or other, but this advantage is reducing is like law of diminishing return. As we keep on increasing the number of stages we are getting advantage, but the quantum of advantage is reducing. So, beyond a particular number of stages we see that actually there is a negligible change in velocity when we keep on increasing the number of stages. So, at infinity there is a maximum velocity we can get. We cannot get more than that there is a maximum this is optimum value we cannot get more than that. Then the point is that we have to make a call going from four to infinity we are actually not gaining much, but we are increasing the complexity a lot. So, then is it worth going for this complex system instead we can just stop at four. So, now that question is how do we decide what is our optimum number? So, once again to summarize this when we have higher number of stages we get greater delta u, but the pay of diminishes with n. So, let us consider a parameter called terminal velocity. So, terminal velocity is the maximum velocity that we can get with a number of stages. Let us say that so this is that delta u n is the terminal velocity when everything is born there is a terminal velocity that we get. Let us look at the extreme case when n equal to infinity. So, let us look at u n or delta u n when n tends to infinity and all the stages are same. So, we are looking at case one case one all the stages were same structural coefficient was same ISP was same the penod mass fashion was same. So, same thing we are looking at same epsilon same ISP same l for all the stages only variation is the number of stages. So, we are considering a rocket where all the stages are same, but we have infinite number of stages then the velocity increment will be given as limit u n equal to u equivalent n l n equal to n 1 upon I will just write some more thing once let me write the expression. This is the general expression for the velocity increment with n number of stages. Now, as n goes to infinity then what happens we get limit n tending to infinity u n equal to u equivalent limit n tending to infinity this sign was not there delta u was only this the summation sign was not there limit n tending to infinity n l n 1 upon 1 minus epsilon l to the power 1 by n plus epsilon this is the one. Now, if I expand this and take the limit and once again I am not going into the details of the mathematics I will give you the final number final expression. The final expression comes out to be like this. So, now notice one thing that the maximum velocity increment that we get is actually independent of number of stages it does not have n anywhere. So, when it make n tending to infinity it becomes independent of number of stages. Now, here l is a constant given value epsilon is a given value u equivalent is a given value. So, therefore, this is the maximum velocity increment we can get no matter how many stages we put and then once we decide on this then the choice comes in that for a practical application where do we pitch ourselves how many number of stages we should keep that can be essentially decided on by looking at how much increment we are getting by increasing the number of stages. So, this gives us then the optimum number of stages as well. So, with this we come to the end of our discussion on multi stage optimization. Next what we will do is before we go to the space dynamics let me solve couple of problems. So, let me solve one or two problems on this optimization. So, we have already solved a problem on case one. Next let us solve a problem on case two what was case two? IHP was varying now IHP was constant epsilon was varying. So, let me solve a question on case two. Again we consider a three stage rocket this are the given conditions epsilon one is equal to 0.15, epsilon two is equal to 0.1, epsilon three is equal to 0.05 and IHP is same equal to 270 seconds and the overall mass fraction is equal to 0.5 fraction is also given 0.01. For that we have to get the maximum delta v free space delta v free space means we do not have gravity we do not have drag. So, this is the problem that we want to solve. So, here this is a quite simple a problem first of all what we do is we had already formulated this. So, now I will just go and I use the equations. We had just few minutes back derived an expression for this case for variation in l 1 l i. l i according to our estimations were epsilon i 1 minus epsilon i l to the power 1 by n by pi i equal to 1 to n epsilon i 1 minus epsilon i to the power 1 by n. This was our expression for l i for this case. Now, we have l 1 l 2 l 3 n is given to be 3 l is given to be 0.01 and first we put epsilon is equal to epsilon 1 we get the value of l 1 this is equal to once you solve it 0.3762. Next we put epsilon is equal to epsilon 2 taking epsilon 2 from there again we solve this equation we get 0.2369 the third again we put epsilon is equal to epsilon 3 and solve this this is equal to 0.1122. Now, what we do is we estimate delta v which is equal to u equivalent another is p g e then sigma i equal to 1 to n l n 1 upon l i 1 minus epsilon i plus epsilon i. We have proved this now what we do is we have got the value of l 1 l 2 l 3 i s p is given epsilon 1 epsilon 2 epsilon 3 are given we just put them one at a time and add them. We get the final value this is 32780 meter per second. So, you can see that it is not very difficult to get it. Let us now look at another case which is a combination of let us say case 1 and 2. What I will do now is that I will consider the same constant value of s or epsilon and I will allow i s p to vary and do the same thing because if you recall the case 3 was most generic. So, then now we can use the case 3 equations to take any type of variation. So, next what I will do is similar problem, but with little difference. So, once again we are talking about estimating the maximum free stage velocity for a free stage rocket these are the condition epsilon 1 is equal to epsilon 2 is equal to epsilon 3 is equal to 0.1 this is what I am going to consider, but now what I am saying is that my i s p values are different. So, what I have here is that the value of a structural coefficient for all the stages are same and they are equal to 0.1. Now, i s p for the first stage is 250 second for the second stage is 300 second for the third stage is 350 second and I maintain the same overall mass fraction l which I have been doing for the other problems 0.01. Now, for this I have to get the delta v. So, see we had not discussed this case what we had discussed was either everything was constant or this was constant this was varying or everything was varying. So, since this was not discussed let us consider the everything varying case and there we will put this. So, that is going to work for this case also. So, since we are talking about that case where everything is varying. So, now we are to deal with the optimization what we have is l is equal to pi i equal to 1 2 3 l i and l is equal to 0.0. 1 and l i is epsilon i 1 minus epsilon i to the power n because this is constant epsilon i is constant. So, this earlier it was product. Now, since it is constant it will be just be rest to the power n and then 1 upon i s p g e upon alpha minus 1. 1 upon i s p 1 i s p 2 g e upon alpha minus 1 and 1 upon i s p 3 upon g e upon alpha minus 1. We had done it in terms of beta right, but again that was a constant. Here we are writing beta as 1 upon alpha. So, that is also a constant. So, I can use this definition. Now, once I expand it I get a polynomial in alpha. After solving this polynomial we will get the value of alpha. Actually what I will do is I will get an expression for alpha by g e because g is also constant. So, I can get an expression for alpha by g e. So, alpha by g e after solving this comes out to be equal to 192.396. Once we have that now look at this term. This is if I put is equal to i this is essentially l i. So, now I can get an expression for l i. So, l i is say l 1 is only this portion without this l. l 2 is this times this, l 3 is this times this. So, I can now get the value of l 1 which is equal to 0.3711, l 2 0.1987 and l 3 equal to 0.1356. Once I have this now I can get the expression for delta v delta v equal to sigma i equal to 1 to n i s p g e i s p i g e l n 1 upon l i 1 minus epsilon i plus epsilon i. Now, everything here is known I just put them and get the value. So, 36051 meter per second. So, now I can get the expression for this. So, this gives me the velocity increment for this specific case. Now I would like to point out something here. We solved three problems. For all of them the mass ratio was same. The average i s p was also same. What we had was different cases for different structural mass distribution and payload mass distribution. And as you can see for all these cases we got different velocity increment. So, this shows that how we distribute the masses dictated what velocity increment we get. That is why the optimization we were talking about is very very critical. So, with this let us now complete our discussion on the multi stage rockets. So, essentially the flight dynamics part is more or less done now. We know now what should be the mass distribution for a given i s p and structural coefficient etcetera to get a certain velocity. So, from this discussion what is emerged is that the velocity is an important parameter. We have to attain a certain velocity. Next thing what we do is we will focus on that that why do we need certain velocity that depends on the space dynamics or the vehicle dynamics. So, the next topic that we are going to start is space dynamics where we will see that in order to complete a given mission which is let us say putting a satellite to an orbit a given orbit to do certain things. And not only putting it an orbit we have to also have to specify how much is going to be the period of that. So, in order to achieve that how much velocity is required that we will try to estimate from using laws of gravitation. So, that is what space dynamics is. So, the next topic that we are going to start is essentially space dynamics the rocket vehicle dynamics we have covered in great detail. So, that is why I will stop here now. So, next lecture we will start space dynamics.