 Hello and welcome. In this lecture, we will look at the gravity turned trajectory solution under the constraint of constant specific thrust or constant T by M. So, let us begin. So, constant specific thrust solution is an important trajectory design strategy that provides important practical benefits as we will see now. So, to understand that, let us first recall the constant burn rate design that we have seen in our rectilinear trajectory solutions that we have obtained earlier. We know that a constant burn rate design is the simplest to implement, but there is a drawback that we have not noted earlier. But now it is important to bring it to the front that a constant burn results in constant thrust. But as you know, the mass is continuously depleting. So, as the vehicle becomes lighter and lighter, the same thrust generates a larger and larger forward acceleration. Now because the thrust is constant, if we look at the force equilibrium that is acting on the rocket, the same force is acting all the time. Even though its mass is reducing continuously, it is becoming lighter, but the force which is acting on it remains the same. The impact of this is that we continue to get large compressive force on the rocket at all points on the trajectory. One way to avoid this problem is to reduce the thrust as mass reduces so that net forward acceleration remains within acceptable bounds. There is another aspect which is useful from a practical perspective that a large acceleration would also result in much larger velocity increases which you may want to limit. You may not want to achieve a very high velocity at a particular point on the trajectory. So, from that perspective as well, there is a need to reduce the thrust as the mass reduces and this is the basic philosophy of constant specific thrust based trajectory in which the specific thrust which is defined as t by m and is the amount of acceleration that propulsion generates is kept constant and thereby the two objectives of keeping the compressive loads for higher stages at a lower magnitude and keeping the velocity increments within bounds are broadly met. Let us now look at the applicable equations. So, now, when we take the v dot equation, we have the thrust term for m dot g naught isp by m but that is effectively t by m term. So, we replace that with t by m which is going to be a constant. So, that constant is represented as n0 into g theta where n0 is a real number that is indicative of the number of g's or the amount of g that the propulsion is generating. Of course, the theta dot equilibrium remains the same, but now we realize that my t by m equation gets split into two parts. The first part gets into the v dot equation as n0 g minus g theta cos theta and the second part of the same equation is n0 g equal to minus m dot g naught isp by m and that gives directly a differential equation for mass in terms of n0 and time. And now you note an interesting feature of this particular formulation that the mass is now an explicit function of time as against the earlier two solutions where mass was always an implicit function of theta which means now we have a direct control over the burn profile or the solution that we are going to get from this differential equation is going to be the variation of mass as a function of time or we can directly control the dm by dt or m dot. Let us now proceed with the solution of these four differential equations. So, the solution for velocity is obtained from our v dot equation, the v dot as n0 g minus g cos theta. We do a little bit of calculus based jugglery. So, we take the ratio of two differential equations which explicitly removes the time variable. So, and we get dv by dv on the left hand side and we then can perform the integrations which are not that straightforward, but again of the trindometric form which are commonly tabulated in any books on calculus. So, you can refer to those books from which you will be able to perform these integrals and we now get a solution for velocity in terms of tan and n0 which is your real number of g is that the proportion is generating. So, the velocity now is a function of the angle theta and n0. Of course, by substituting the initial condition we can evaluate the constant of integration k dash. Let us now move over to the time solution. The time solution is the integral of v d theta by g tilde sin theta. And now you realize that v itself is a slightly complex function of theta. So, obviously this integral becomes a little bit more complicated than the integral for the velocity. However, we can still take recourse to trigonometric substitutions which are commonly done in the process of performing trigonometric integrals and are part of standard textbooks on integral calculus. We will find that the above integral reduces to the form of the tan n0 minus 2 theta by 2 plus tan n0 theta by 2 into 6 square theta by 2 d theta. And this integral again can be performed through substitutions as per the standard integral table. And now we get a solution for time in terms of theta which is now a transcendental relation. The meaning of transcendental relation is that now it is in terms of trigonometric functions and it is an implicit relation. Of course, if theta b and theta not are specified along with n0, then this is just an evaluation for delta t which means that for specified n0, theta not and theta b, we can find out what is the time that is going to be taken to perform this maneuver. However, if we want to find out what should be theta b or theta not for a specified time during which the maneuver is to be performed, we will need to solve this transcendental equation in an iterative fashion. So, it is going to become numerically a lot more intensive. Let us now go to the Burn profile solution from the fourth equation that we generated and which is a direct result of the assumption of t by m equal to n0 g. So, this integral is not very difficult to perform and we get the mass fraction as an exponential function of n0 and time and we now see that we have another root to evaluate delta t. If we specify an m0 by m and an n0, then we can calculate delta t. Conversely, for a specified delta t and a mass fraction, we can calculate n0 which is going to satisfy this constraint relation which means we can use this as a design equation where we may design for a specific forward acceleration by specifying the time to be taken for the trajectory and the mass fraction available. We can now go over to the altitude and the x solutions using the same two equations that is dh by dt as v cos theta and dx by dt as v sin theta. We realize that the altitude integral now involves v square and similarly the horizontal distance integral also involves v square. Obviously, these are going to become a little bit more complex than the time integral itself. So, while I have not given the solution here, I will mention here that these integrals are possible to be performed using the same strategy that we have used earlier of trigonometric substitution of tan and sag functions. So, my suggestion to all of you is that please perform these integrals and obtain the expressions for h and x just to understand the solution effort involved in generating the solution for h and x in comparison to the two other solution techniques that is constant q0 and a constant v. So, that it will give you a fair idea of the amount of effort involved in generating the solution for the case of constant t by m. Of course, I have given you these solutions in the next slide, but my suggestion would be that please go through this exercise for your own satisfaction to understand the actual effort involved in arriving at these two expressions and that will also help you verify these two expressions. You can see that these expressions are as complicated as the expression for the time solution and the velocity. We now come to a stage where we realize that n0 is the primary is the primary driver for all the four relations that is the velocity, the time or theta, the h and the x expressions. Now, of course, at this point the equations have been obtained without any specific consideration to what n0 is going to be. We are just saying that it can be any positive real number. In fact, you will note that maybe a negative number also might be possible from a mathematical perspective that the solution obtained may also be acceptable for a negative value of n0. There are certain tactical constraints that we now put into the solution. The first thing we say is that we would like the velocity to increase continuously and not decrease. Of course, there could be very specific situations where you may want to use this to also decrease the velocity, but by and large you will not waste energy for decreasing the velocity. So, if we accept this primary idea that velocity is going to increase continuously, then the net forward acceleration must also be positive all the time. So, this is the fundamental requirement that n0 should be set such that the net forward acceleration which is difference of n0 and cos theta should be positive all the time. And that gives us this basic condition that in order for you to have a continuous increase in the velocity, n0 must be greater than cos theta all the time. But here you need to note an interesting feature that as theta continues to increase from 0 to 90 degree, cos theta is going to decrease continuously from 1 to 0 which obviously means that as you move along the trajectory and your theta keeps on increasing, the same condition will be satisfied with a lower value of n0. So, you can actually also talk about an n0 profile even though the present solution has strictly not admitted that possibility, we are assuming n0 to be a constant. But in a more generic sense, you would realize that it is possible for us to consider this fact that n0 could also be varying as a function of time or a function of theta. Of course, this point we have already mentioned that if we keep n0 greater than 1 at all times, we will always be ensuring the constraint that n0 is greater than cos theta including the starting point. And now we realize that n0 being greater than 1 means that thrust must always be greater than instantaneous weight. So, this is going to be the constraint that we will have to implement in order for us to have a positive forward velocity increment at all times. Now there is a degenerate case that we need to pay attention to. If you go back and look at the expressions for all the four quantities that is velocity, the delta t, h and x, n0 equal to 1 represents a very, very special case and it also leads to singularity in some of the solutions which means for n0 equal to 1, the denominator becomes 0 so that the solution becomes unbounded. Obviously, that is not acceptable from a physical perspective so we need to kind of resolve this singularity which is done in the following manner. So, in the case of velocity, when we implement n0 equal to 1, the first term becomes 1 for all values of theta and the whole expression reduces to 1 plus tan square theta by 2 which is 6 square theta by 2. So, your velocity expression degenerates to k dash into 6 square theta by 2. So, this is the expression that you will use for n0 equal to 1 for velocity. Now we can obtain the time solution using this velocity expression. So, instead of trying to use the original expression and taking its limit as n tending to 1, we now use this as the velocity expression and perform the integration afresh. Now, this is the integral that is 6 square theta by 2 by g tilde sin theta d theta is the integral that we need to perform for time and I will leave you to use all the trigonometry substitution possibilities and arrive at the solution as I have given here which is nothing but 2 ln tan theta by 2 plus 6 square theta by 2 is the delta t solution. The same thing can be done for h and x profiles that you can use this modified velocity expression now to integrate for altitude and horizontal distance. I will leave this exercise to you all to perform and arrive at those expressions. The point which now I would like to mention is that if you take the original expression with n naught and try to take the limit of n0 tending to 1, there is a methodology in calculus which you might be familiar with called the hospital rule where the singular functions, their limits can be obtained by applying a particular procedure. My suggestion would be that you can also take that root on those expressions, take the limit as n0 tends to 1 and see if the limit reduces to the expression that we have obtained for delta t h and x in the present case that would tell you the different ways in which we can arrive at the solution for the degenerate case of n0 equal to 1. It is only degenerate case from a mathematical perspective but from a physical perspective n0 should always be greater than 1 which will ensure that we have positive velocity increases. So, now let us look at some of the features of the solution which we have obtained. So, we find that a larger n0 will require more time and more propellant to achieve the same burnout inclination. So, which means if you are interested in generating larger velocity, you will use a larger n naught but then it will also require a larger propellant and will last longer. So, if your design requirements are larger terminal velocity, you will go for a higher n0. Typically, n0 is a design solution which is derived from specific terminal parameters and this is under the overall constraint of the vehicle structure. Typical values which are commonly employed in the context of constant specific thrust solution are between 1 and 1.6. Rarely, you will go beyond 1.6 to 1.7 when you want to perform this particular trajectory maneuver. More often than not, they will be closer to about 1.1, 1.2 and they provide good solutions for trajectory in the assigned commission. Let us now understand the features of these solutions that we have obtained through an example that we have been considering. So, let us consider the following parameters for a rocket having 74 tons as a starting mass, 54 tons of propellant and the initial velocity at 90 meters per second and initial inclination of 3 degree. And as we have no better information, let us put n0 as 2. We know that this is outside the bound of the range that we have given of 1 to 1.6. But let us try and experiment with this number just to find out what happens if we give such a value and see why the constraint on the n0 value is being implemented explicitly. Let us try and determine the velocity when the vehicle becomes parallel to local horizon. Let us try and find out the time taken along with a possible feasibility of performing such a mission because we are not sure that this mission is going to be feasible. It might be you do not know. So, the solution is as follows. So, we start with calculation of k dash which tells up to be 3434.6 based on that we calculate db. So, that turns out to be 6869 meters per second clearly high velocity starting from 90 meters. So, you are able to go from 90 meters per second to almost 7000 meters per second using this option. The time taken is 457 seconds because n0 is a large value. We have already said that it will take a lot of time. But when we come to the mass fraction, we hit a road block. The mass fraction solution in this particular case says that you must have had 72.4 tons of propellant to complete this mission. But you only have 54 tons which obviously means that this mission is not going to be feasible unless you now go back and modify the rocket and say that I am going to carry so much of propellant. Maybe it is going to be feasible. Let us now turn this problem around and see for what value of n0 the mission becomes feasible. The reason why the problem is posed in this manner is that n0 was an arbitrarily chosen figure and more importantly it was taken to be significantly higher than the range specified. So, obviously it means that this is where the problem is. The solution is not feasible not because of any other issue but because n0 is not a consistent parameter in the given mission. So, now let me restate the problem as follows. Let us try and determine n0, the velocity at this point and the total time if 54 tons of propellant is to be consumed. So, now I am saying that in place of n0, I am specifying the amount of propellant that I want to burn and let me see what value of n0 I can use which will make this happen. Now, I am not going to do this exercise. My suggestion is you now set up this problem based on what we have presented in the previous example and try and arrive at a value of n0 which is going to be consistent with the parameters of the problem. I will only mention two points. One, you are going to require to solve a set of non-linear algebraic equations and the second point you might have to resort to an iterative procedure to arrive at the correct solution for n0. I will give you only this much of hint. I suggest you do this. At some point, I will be uploading the solution for this so that you can verify whether you are thinking along the right lines. So, to summarize, the constant specific thrust case is complex from the point of view of both solution and implementation. It is not only not very easy to solve but also not going to be very easy to implement. Also, we note that it is non-intuitive from a design perspective. What it means is that it is not very straightforward to interpret what is likely to be the impact of change in n0 on the trajectory behavior because the mathematical relations involved are complex trigonometric functions and they cannot be directly interpreted. So, obviously you are going to require rigorous analysis. However, from a practical perspective, it is going to be an extremely useful trajectory design option because it is going to give you a handle on managing your structural mass which if you remember was one of the benefits that was mentioned when we were talking about the various options for gravity turn trajectory. So, we said that we are going to be able to manage the structural mass better when we control n0 because if the compressive force is maintained, you can maintain your structural mass which will support the compressive weight.