 Hello and welcome. In this tutorial, we will consider problems related to generating a curvilinear trajectory using the three basic analytical procedures that is constant pitch rate, constant velocity and constant specific thrust solutions. We will assume that all these trajectories are in vacuum and that constant sea level gravity is applicable to simplify our analysis procedure. So, let us begin. So, let us first consider the constant pitch rate problems. So, let us consider a rocket with the following configuration that is it has a lift of mass of 100 tons is carrying 90 tons propellant of 260 second ISP. Further, it moves vertically for 5 seconds under constant bond rate of 600 kg at which point a pitch kick is given and it acquires a pitch rate of 0.1 degree per second. Let us try and determine all the terminal parameters in case the rocket executes the constant pitch rate maneuver from this point onwards that is t equal to 5 seconds. For a further 145 seconds, assuming sea level gravity and vacuum conditions. So, let us look at the solution. So, we first calculate the velocity at the end of 5 seconds which is under the vertical motion assumption and constant gravity assumption. So, we use the standard formula of g0 ISP ln m0 by mb and in this case because we are burning propellant at 600 kg per second for about 5 seconds we burn 3 tons of propellant. So, what is left at the end of 5 seconds is 97 tons and this term corresponds to the effect of gravity. If we perform this simple calculation, the velocity at the end of 5 seconds is 28.6 meters per second. Now, the pitch rate at 5 seconds is given as 0.1 degree per second which is converted into 0.0011 radians per second and now we employ the basic equation that sin theta at t equal to 5 is V0 Q0 by g0. When we do this sin theta is 0.0051. So, we get theta at t equal to 5 second as 0.29 degree or 0.0051 radian. So, as you would remember we said that among the three quantities that is velocity pitch rate and the pitch angle only two can be specified and third will be solved using the constant relation which is what we have done in this case. So, once we know the initial angle with respect to vertical then it is just a question of applying the basic formula and we can get the pitch angle at the end of 150 second that is additional 145 seconds at the rate of 0.1 degree per second. So, we get the angle at the end of 150 second as 14.79 degrees. Once we get this angle we can then talk about the velocity using the same constant relation that is g0 sin theta that is sin of 14.79 degree divided by Q0 which is 0.0017 and we find that the velocity at the end of 150 seconds will be 1473.1 meters per second. Let us now turn our attention to the mass fraction. So, we use the mass expression of ln of m0 mass at t equal to 5 second that is the starting point for the gravity term divided by the mass at the 150 second instant and we use the two angles that is sin of 14.79 degree and the sin of 0.29 degree and with that we find that we are talking about a mass ratio or mass at the 150 second as 39.4 tons. Similarly, now we can use the terminal parameters of 14.79 degree as the angle to calculate the altitude solution and we find that the altitude is 110 kilometers and we can do the same thing for the horizontal distance that is 19 kilometers. So, this trajectory is such that it is almost like a straight line vertical trajectory and the horizontal distance travelled is only 19 kilometers. So, it is possible that the flat earth approximation is going to be valid in this case. Let us now proceed further and consider the next problem. In this, let us try and find out if we can burn all the remaining propellant and reach the angle of 45 degree. Such a thing is feasible. Let us try and find out the burnout parameters but if it is not possible let us try and find out if we burn the remaining propellant what are the possible burnout inclination other parameters. So, in this case the applicable solution starts like this. We start with the mass ratio. We assume that our angle is 45 degree. So, sign 45 degree is 0.707. So, we substitute this and from this we find that the residual mass expected at theta equal to 45 degree is 4.05 tons. Now, we realize that we are carrying only 90 tons of propellant which means there is a 10 tons of mass which is inert or cannot be burned. But this solution is expecting that the inert mass would only be 4.05 tons. So, obviously it is not a feasible solution which indirectly means that by burning all the propellants we cannot reach an angle of 45 degree because we do not have that much of propellant. So, let us now invert the problem to say what is the propellant that we have. So, we have only now 87 tons of propellant left once we have finished the 3 tons of propellant in the vertical motion. So, with this propellant let us try and find out what is the inclination that is possible. So, we solve the same equation but now we assume theta B to be an unknown and with this mass fraction which is known to us we find that we can reach only about 30.5 degree of inclination from vertical we cannot reach 45 degree. And to perform this mission itself it will take 302 seconds and at the end of this phase of burning all the propellants the velocity will be 2928 meters per second. Now let us look at the corresponding altitude solution and we find that now altitude is going to be 437 kilometers and the horizontal distance is going to be 152 kilometers. Now if you remember we had mentioned that every 110 kilometer of horizontal distance is equal to 1 degree change in the gravity direction. So, in the present case we find that there is going to be change of about 1.5 degree in the direction of gravity. And we will find that because of that the gravitational acceleration will change marginally and to that extent there will be a correction to the final trajectory parameters because there is only a small component of the gravity which is acting in horizontal direction. So, maybe you can think in terms of finding out what is the impact of this 1.5 degree change in the direction of the gravity. Let us now move forward to the problem number 3 where we are now putting one more constraint and the constraint is that I want to burn all the propellant and I want to reach 45 degrees and now can I design the pitch kick. So, this problem concerns designing the pitch kick and let us see if by burning the remaining 87 tons of propellant can be achieved 45 degree burning and in the process what should be the pitch kick parameters and then also we will compare the burn rates in these two cases. The last case where we have reached the 30.5 degree with burning all the propellant and the requirement that we should reach 45 degree by burning all the propellant. So, in this case what we do is we have the velocity which is a constraint of 28.6 and we use the constraint relation which is now a function of both theta naught which is an unknown and q naught which is also an unknown. But what is known is this mass ratio which is 2.272 and this should be equal to 2 by q naught into 0.707 minus sin theta naught. So, now we have two simultaneous equations in two unknowns theta naught and q naught. We solve this equation the first equation for sin theta naught in terms of q naught and substitute that into the second equation which now becomes an equation only in q naught and by solving this we get a solution for q naught as 0.0023 radians per second as compared to 0.0017 radian per second that is what we had assumed. So, the first change that has occurred is that instead of 0.1 degree per second pitch rate we would probably need a slightly higher pitch rate and when we substitute this value of q naught into the sin theta naught expression we find that theta naught instead of being 0.29 degrees is now it is 0.396 degrees and with that we find that this combination of theta naught and q naught and the pitch kick parameter will ensure that we will be able to burn all the remaining 87 tons of propellant and that we will reach the angle at the end of this phase as 45 degrees. Of course, we find that in order to do this the time taken would be 343 seconds and this 343 seconds gives us an average burn rate of 0.253 tons per second while in the previous case it is 0.288 tons per second. So, it is marginally lower but they are of the same ballpark. With this we have come to the end of the problems that we would like to consider for a constant pitch rate case. My suggestion is that you try these problems independently to verify some of the solutions and in case there are anomalies you may please report them to the TAS. Let us now look at the constant velocity problems and in that let us consider a rocket with the following specifications. So, we have a rocket with 80 tons of lift off mass, 70 tons of propellant with having an ISP of 300 seconds and the initial conditions are such that you already are having an inclination of 1 degree from vertical. The rocket is firing at a velocity of 800 meters per second and it is desired that at the end of a constant velocity maneuver the burnout final angle becomes 90 degree. So, our terminal constraint is that the angle becomes 90 degree from vertical or that the velocity vector is parallel to the local horizon. Let us try and now determine the burnout conditions. So, the burnout solution is as follows. First let us talk about the time taken to complete this maneuver which is given in terms of the velocity V0 the burnout angle theta B and the initial angle theta 0 and we find that the time taken to complete this maneuver is going to be 386.6 seconds and in this time the burnout mass can be obtained using this formula and when we apply this formula we get the burnout mass of 22 tons which means that from 80 tons on the 22 tons are left. So, which means about 58 tons of propellant have been consumed out of 70 tons of propellant. So, you may note that a 12 ton propellant is still there and of course then we can calculate the altitude. So, if you look at the altitude it is 309 kilometers and the horizontal distance travelled during this time is 101 kilometers. So, again there is a change of roughly about 1 degree in the direction of gravity you may assess its impact on the overall performance in terms of the change in the gravitational vector direction and its impact on the solution. Let us now consider the constant specific thrust problem. In this case, let us consider a rocket of mass 74 tons carrying a propellant of 64 tons of ISP 350 seconds. It has an initial velocity of 500 meters per second an initial inclination of 2 degree and the constant specific thrust factor is 1.2 which means that T by mg is 1.2 g. Let us try and determine the velocity and the residual mass when it reaches the angle 90 degree and also the time taken for this maneuver. So, let us first begin by calculating k dash as 500 that is velocity divided by tan 1 to the power 0.2 and tan 1 to the power 2.2. If you perform this calculation you will find that the k dash turns out to be 1123.2 meters per second units and your VB at the end of burnout which is defined as theta B equal to 90 degree will be k dash into tan 45 to the power 0.2 and tan 45 to the power 2.2. Now, you will realize that because tan 45 is 1 no matter what is the power raised it will remain 1. So, this is 1 plus 1 which is 2. So, the VB is just twice of k dash and that is 2246.4. Similarly, we can now calculate the time taken for this maneuver which is 101.7 seconds and then we can talk about the mass fraction which in present case leads to the burnout mass as 52.21 times. Hi. So, in this particular tutorial we have considered the three cases of gravity turn maneuver that is constant pitch rate, the constant velocity and constant specific thrust 30 by m k s and we have looked at the application of various formulae in different forms to arrive at different solutions and you will note that it is very easy to convert these relations into a trajectory design context where you can arrive at a design solution based on certain terminal requirement.