 So, in this lecture, we will look at some of the details of the trade-off ratio concept including its formulation and the solution in order to understand the implication of trade-off ratio as an important design tool. So, let us begin. So, let us proceed with our discussion on trade-off ratio for variant design. As I had mentioned in the previous lecture, trade-off ratios are nothing but the partial derivatives of the rocket performance equations with respect to two configuration parameters that is the structural mass or the propellant mass which we know directly influence the rocket performance in terms of either the burnout velocity or in terms of the mission payload mass fraction. Now, while there can be requirements on performing a different mission with the same m star, in most cases the variants are required to perform the same mission for a different mass as we have seen in the examples in the previous lecture. So, at least to understand the idea, we use the basic constraint that v star is an invariant which represents the spacecraft mission which is to be performed while we want to look at the implication of changes in the stage to the change in the mission payload mass or payload mass fraction by star. So, in the present study, we will talk about a philosophy for trade-off ratios which will assume that we do not want the v star to change and we would like to find out how a small change in the stage mass will influence the change in the payload mass. To do this, we make use of the v star expression to examine the applicable sensitivities. So, let v star be the g0 i equal to 1 to n sum ispi ln of m0 i by mfi is treated as a variant. You note that m0 i and mfi have already been defined while discussing the multi-stage rocket configurations. So, we just recall those definitions for mfi and m0 i for ith stage as m0 i equal to m star plus mpi plus msi that is the configuration of the stage plus j equal to i plus 1 which means the stages which are higher than the ith stage right up to the nth stage and their mass configuration in terms of msj plus mpj. We know that mfi is going to be m0 i minus mpi so that it is m star plus msi plus the same sum. Now, we can write the ideal burnout velocity in terms of the above expressions as follows. So, v star can be expressed as g0 i equal to 1 to n ispi ln and now I open up the expression for m0 i and mfi it is the ratio of m star, mpi and msi what you realize is that now this v star we would like to keep invariant while making changes to msi, mpi and m star and this is the primary objective of this whole exercise. But this is not a straightforward exercise because of the fact that v star is a discrete sum of individual stage contributions. This is the first thing that we need to note that it is a discrete sum. So, what it means is that this is essentially a piecewise continuous function it is not a continuous function which means as you go from stages i equal to 1 to n the numbers change and because of which there are certain jumps in the masses at the interface points and we need to generate the partial derivatives to understand the sensitivities of a piecewise continuous function. Of course, from the calculus you would have probably dealt with these kind of functions for arriving at their derivatives. So, I presume you know how this can be done and in case there are certain gaps I would suggest that you go through and refresh this material again just to understand the implication of taking a derivative of a piecewise continuous function. There is another calculus concept that we are going to use. As v star is a function of many variables, instead of talking about a derivative of v star, we talk about variation of v star. So, we introduce the concept of a variation which is kind of a total derivative. Total derivative in the calculus is defined as a linear combination of the individual partial derivatives with respect to all the variables of the function. In the present case, there are three variables that influence the v star that is m star, MPI and MSI and that is why our v star will be defined in the context of variation as partial derivatives corresponding to these three quantities. So, let us bring in this idea through this simple formulation. So, we define dv star as a small variation in the v star, the total variation as a linear combination of the partial derivative of v star with respect to MSI into a small change in MSI plus a partial derivative of v star with respect to m star for a small change in the m star that is delta m star. Now, even though there are three variables, what we will do is that we will take two at a time, which means that we will take m star and MSI as one group and m star and MPI as another group. And because we are using partial derivatives, it is a linear formulation so that in case we need to change both MSI and MPI, it would just be a linear combination of these two separately. Once we make use of this philosophy, we just say that if v star is a constant, then the small variation in v star must be 0, which means that the dv star is equal to 0. And that gives us a relation between a small change in m star with respect to a small change in MSI as the ratio of two partial derivatives dv star by d MSI and dv star by d m star with a minus sign. So, we immediately see that the sensitivity of the payload mass for a small change or a unit change in the structural mass of ith stage is nothing but negative of the ratio of the two partial derivatives, one with respect to MSI and other with respect to m star. We now repeat this exercise for the MPI and we similarly find that the sensitivity of m star with respect to MPI is again negative of the ratio of the two partial derivatives of v star on the numerator with respect to MPI and in the denominator with respect to m star. The above expression for sensitivity establishes the possible changes in m star due to the change in MSI and MPI one at a time for a constant v star. Here, it is also worth noting that I could easily have done this exercise for keeping m star equal to constant and then taken the m star derivative with respect to MSI and MPI and then I could have defined the partial derivatives for the sensitivities which would give me a different v star for keeping the m star constant. So, the same formulation strategy is entirely applicable for the other case as well. Now, we need to understand how we can evaluate these partial derivatives. Of course, we realize that these partial derivatives two for each stage please remember that for each stage there are two partial derivatives one with respect to the structural mass, one with respect to the propellant mass. And if there are n stages there are going to be two n such parameters which will establish the sensitivity of the velocity to both m star and MSI and MPI. And again, let me reemphasize that the evaluation of these partial derivatives has to be carried out in the context of velocity being a discrete function or a piecewise continuous function of MSI, MPI and m star. So, let us try and demonstrate this for a simple case of a two stage rocket to understand the procedure involved and how these piecewise continuous functions can be differentiated to give the sensitivities. So, let me just for the sake of demonstration open up the sum and write v star as the sum of velocities coming from the two stages that is the first stage and the second stage. Similarly, let me write m naught 1 which is the starting mass of the first stage in the longhand fashion of m star plus MS1 plus MP1 plus MS2 plus MP2. The reason why I am writing like this is that my process of differentiating a piecewise continuous function will become simpler when I express it in this form and will be clearly visible as to what the steps are involved. Similarly, I write mf1 by subtracting MP1 from m naught 1 so that I get m star plus MS1 plus MS2 plus MP2. And similarly, the starting mass for second stage that is m naught 2 as m star plus MS2 plus MP2 and mf2 as m star plus MS2. As there are only two stages, the final mass for the second stage would be just the m star and the structural mass of the second stage at which the final velocity we would have achieved. Let us now take the partial derivatives of the velocity equation the way it is written and immediately we realize that as both m naught 1 and mf1 and m naught 2 and mf2 contain the m star and it is a logarithmic function of m star. We know that when we differentiate a logarithmic function we get 1 by that function. So, I can easily write ln of m naught 1 by mf1 as ln of m naught 1 minus ln of mf1 and then when I differentiate these two ln of m naught 1 when I differentiate I get 1 by m naught 1 and m star differentiated with respect to m star gives me unity, so no problem. And similarly, I get 1 by mf1 when I differentiate ln of mf1. Similarly, when I differentiate m naught 2 and mf2 I get 1 by m naught 2 and 1 by mf2 and this becomes my dv star by dm star expression. When I look at v star expression I immediately realize that when I use ms1 the variation is only with respect to the structural mass of first stage. So, the structural mass of second stage is constant. So, the partial derivative terms will contain only the term corresponding to ms1 which appears in m naught 1 and mf1 because in m naught 2 and mf2 ms1 does not exist because we have removed that mass and this is the implication of a piecewise continuous function being differentiated. I hope you have understood the philosophy but maybe you can work with this a little more just to understand the step. Once we understand this we do the same thing for ms2 and now you realize that ms2 is part of both m naught 1 and m naught 2 as well as part of mf1 and mf2. So, you get the same expression as what you would get for differentiating with respect to m star. So, you can see that dv star by ms2 is same as dv star by dm star while dv star by ms1 is different. With this let us now go back to our sensitivity definition that is ratio of delta m star to delta ms1 and we find that the ratio is as ratio of two partial derivatives and when I take the ratio of delta m star by delta ms2 I suddenly find that both numerator and denominator are the same so I get minus 1. We will talk about this idea a little more but at this point it is sufficient to mention that a small change in ms2 will generate an equal and opposite change in m star which means that if I reduce the structural mass by 1 kg it immediately allows me to add 1 kg in the payload mass and that tells you that this is the best possible efficiency that one can get that you are trading of 1 kg of structural mass with 1 kg of payload mass. Conversely if you want to increase 1 kg payload mass you must reduce 1 kg of structure from the second stage of course this number is going to be less than 1 in case of ms1. So obviously the efficiency is lower so for the same change desired in m star you may need to make larger changes in this structure of the first stage as compared to the second stage and this establishes the fact that from a structural perspective the second stage is 100% efficient or the final stage. Now the same logic could be extended to nth stage so that the last stage is the most efficient from structural point of view. Let us now do this exercise for the propulsion mass. So I am not going to go through the details we can again generate the derivative dv star by d mp1 and dv star by dmp2 and we already have dv star by m star and so we can take the ratios directly as we have seen earlier. So now we are going to get the two efficiencies that is delta m star by delta mp1 and delta m star by delta mp2 in terms of the m01, mf1, m02, mf2 and an interesting feature is that isp also directly appears in these two expressions. Now of course I could have independently generated the same sensitivities with respect to isp by keeping both v star and m star in a particular manner but even without doing that I realize that if m01, m02 these quantities do not change which means if my structural mass does not change then if there is a change in isp I can use this partial derivative expression directly to show how a small change in isp will affect for a constant propulsion mass the change in m star. It is possible for me to reinterpret these algebraically. I suggest that you try this on your own just to understand the implication of the discussion. Once we have that we can now directly talk about the sensitivities and in a generic sense we now see that dm star by the msn stages minus 1 is always less than 0 and dm star by dpi is this expression it is always a positive quantity and this is another point that you need to understand that the trade-off ratio for structural mass is inversely related that in order to increase the payload mass I must reduce the structural mass that is the relation whereas in order to increase the payload mass I must increase the propellant mass. So, if I increase the propellant mass or if I increase the isp I will get a higher m star whereas if I decrease the structural mass I will get a higher m star or if I increase the structural mass I get a lower payload mass. Let us just understand these relations through a simple example for a rocket which has the following mass configuration. So, it is two-stage rocket with having first stage propellant of 21000 kgs and a structure of close to 1300 kgs with an isp of 261 seconds the second stage has propellant of 3850 kgs the structure of 360 kg and isp of 324 seconds and the payload mass of 668 kg. Let us try and obtain this trade-off ratio now we just do the substitution in the expression I suggest that you do that exercise yourself just to confirm I am just giving you the m01, mf1, m02 and mf2 which can be obtained from the data that is given in the previous sheet and with those I can evaluate the partial derivatives and then take the ratio of the partial derivatives and I find that delta m star by delta ms1 is minus 0.116. What it means is that if I reduce the structural mass of the first stage by 1 kg I will be able to add only 116 grams of payload but if I make the same change in the second stage structural mass I can add full 1 kg or 1000 grams so you can see that second stage is 100 percent efficient. With regard to propulsion find that the first change is highly inefficient that for 1 kg propellant mass increase I am only able to add on 34 grams of payload of course the second stage is definitely better than the first stage so that the same 1 kg propellant gives me 119 grams of payload this is because of two reasons one the second stage is more efficient and more importantly it also has a higher ISP compared to first stage and now you will realize that most launch vehicles try to use a higher ISP fuel in the higher stage and now you will understand why why you do not use solid propellant in the higher stages because it is a lower ISP fuel it will give you a lower efficiency of the stage so that in case you want to do a tradeoff it is not going to be a very efficient design whereas if you use the cryogenic fuel in the higher stages then even a small saving of structural mass or a small increase in the propulsion mass is going to significantly add to the payload mass and becomes an extremely useful way of creating a variant. So, to summarize we see that tradeoff ratios are an elegant mechanism to understand the launch vehicle state sensitivities we also note that a small change in configuration based on the sensitivity in the vicinity of the parent configuration has the potential to preserve the optimality of the solution. So, in this lecture we have seen a simple mechanism through which we can set up the solution for the sensitivities of the stage under the constraint that the V star is a constant and we have obtained the solutions and understood that within the limitations of the assumptions that we have made it is still an extremely useful idea for designing a modified mission with minimal computational effort just by looking at the stages which are more efficient and the amount of saving in the mass can lead to improvement in the payload mass capability. With this we close our discussion on the serial, the series or what is also called the restricted staging as a concept for multi-stage rocket design. We will now conclude this idea in the next lecture by looking at the concepts of parallel staging that is if you add a booster stage then in what way the configurations change what are the issues involved with the booster stage and what kind of benefits we can derive by making use of parallel staging as compared to a serial staging or a restricted staging. So, bye see you in the next lecture and thank you.