 Hello and welcome. In the last lecture, we had established the basis for generating a curvilinear trajectory through consideration of a two-dimensional motion model and we had also examined the set of equations for the possibility of extracting the closed form solutions and we had mentioned that there are three possible solutions that we can obtain which also have a practical utility. So, we are now going to look at these solutions one by one. The first one that we are going to consider in this lecture is a solution that we obtain through the eruption of pitch rate being constant. So, let us begin. So, let us see what are the features of such a solution and how we can obtain it. In this case, that is the case in which we assume the pitch rate to remain constant throughout the trajectory. The rocket is commanded to track a specified pitch rate d theta by dt so that velocity solution is obtained directly as shown below. We know that theta dot which is the pitch rate is a constant. So, we give a symbol q naught that directly gives us theta solution as q naught d plus theta naught where theta naught is the angle at the initial time. Then we go to the second equilibrium equation that is theta dot equal to g a tilde sin theta by v and we directly get the v solution as g sin theta by q naught. So, we see that v is a sinusoidal function of theta and is also inversely proportional to q naught indicating that we are going to get a higher velocity for a higher theta and for a lower q naught value. So, what are the features of this solution that we have obtained just now? So, we note that if q naught is a constant throughout the trajectory including the starting point at t equal to 0. It has to be valid at all the points. So, obviously the same relation is also valid at the initial point or the start of the gravity turn. So, we can write down the applicable relation for velocity at the start of the gravity turn as shown below and this relation is nothing but q naught equal to g sin theta naught by v naught where v naught and theta naught are the initial values of velocity and the pitch angle theta. Now, as q naught is a finite quantity it automatically means that there are going to be certain restrictions that theta naught and v naught will have to satisfy. For example, if you start from theta naught equal to 0 even though v naught is not 0 your q naught will become 0 and if q naught is 0 that is not an admissible solution. Similarly, if we start from a finite theta naught but if we make v naught equal to 0 it results in finite q naught which is also not an admissible solution. So, we come to the conclusion that both v naught and theta naught have to be non-zero values before we can start our gravity turn. Of course, the reason for this is not far to seek because our gravity turn depends on the normal component of gravity to the velocity vector and for theta naught equal to 0 the gravity vector does not have any component normal to the velocity vector for a vertical motion which obviously means that the gravity turn can also not be started when the vehicle is vertical and that it must have some finite inclination from vertical before the gravity turn can start. Moreover, we also note that not only gravity turn can be started only after it acquires a certain attitude with respect to vertical it also needs to acquire a finite velocity and this is where we need to take a view of how such a thing is going to be achieved particularly because our lift off is basically vertical. The velocity vector is pointing in the vertical direction even though once the rocket clears the launch tower it will have a finite velocity but if it does not have an inclination with respect to vertical you cannot execute a gravity turn maneuver as we have established and this requirement is usually met by giving a pitch kick or executing what is called a pitch down maneuver on the vehicle at an appropriate time. The implication of this pitch kick or a pitch down maneuver is to give a lateral impulse generates a pitching moment instantaneously and this generates a pitch disturbance that starts the pitching motion and because of the pitching motion it acquires a finite theta it is given in the form of an impulse actually so that we acquire a constant pitch at the end of the maneuver which becomes our initial pitch angle at that velocity v0 which is what we treat as our initial time t0. We also need to note from the relation that we have seen for q0, theta0 and v0 that as this is a constraint relation only two of these three can be actually specified and third one will be determined from the constraint for example if you specify q0 and v0 you can determine theta from that relation or if you specify v0 and theta0 you can determine q0 so this also has to be kept in mind while setting up the configuration for a gravity turn with the constant pitch rate maneuver as we have discussed just now. Now let us go to the next equation in the context of the gravity turn that is the velocity equation but you immediately realize that we already have the velocity solution so what is it that the velocity solution is going to give us the velocity solution now is going to give us the mass profile because from the second equation we already have the velocity for a given pitch rate and we already have theta solution as a function of time so the only unknown now left is the mass solution and the tangential equilibrium is going to be used to write a differential equation for mass as shown below. So we are going to convert this into a differential equation for mass through basic calculus based manipulations change of variables so by doing that we can show that the tangential equilibrium differential equation can be rewritten as dm by m equal to minus 2g tilde cos theta dt divided by g0 isp and what we do now is also the change of variable of dt which is converted to d theta through the basic expression of d theta by dt so using theta dot expression we change the variable from dt to d theta now this differential equation if we integrate we are going to get the solution of mass as a function of theta and not an explicit function of time but we also realize that as theta is already an explicit function of time by substituting that solution into the mass solution we can also generate an explicit solution for mass in as a function of time so we are going to have the velocity and mass and theta all as an explicit function of time so let us go through with the integration process and we find that this leads to the basic result that ln of m0 by m is 2g tilde by q0 g0 isp into sin theta minus sin theta naught so this is the solution for mass fraction that is executed under a constant pitch rate maneuver so we find that similar to velocity mass is also a sinusoidal function of angle theta and also inversely proportional to q0 so if q0 is small the mass fraction is going to be larger now what is the implication of m0 by m being larger the implication is that m by m0 will be smaller so that we will require a larger propellant to support a larger velocity which is going to automatically result for a smaller q0 that we have already seen we also note the converse of it that if you use a large q0 we are going to get a smaller terminal velocity and if you realize an important fact that by appropriately choosing a q0 value it is possible for us to design a trajectory that is going to have different terminal conditions and conversely it means that if you want to achieve different terminal conditions you just need to tweak your q0 value which is going to be not very difficult as this would be just the part of your control algorithm where you may want to define a different reference for a constant pitch rate solution you need to also realize that as the trajectory solution is extremely sensitive to this q0 value you need to be very careful with the selection of this number as we will see through an example that we will conduct next but let us now look at the solution for time and solution of time also as a function of mass so because now there is an interplay of variables we can express different variables in terms of different quantities so if we are given the starting and the terminal condition along with q0 we can directly get the solution for the flight time that is how long the mission will last on the other hand if we are given the mass and initial condition it is possible for us to determine the terminal inclination through the mass profile specification so if we know what is the burn out mass and what is the lift off mass there is a mass fraction if we know the initial angle and if we know the pitch rate we can directly calculate the terminal angle through this trigonometric expression the last two solutions correspond to the trajectory parameters that is the profile in terms of the altitude and the horizontal distance so let us first take the altitude equation which is nothing but our dh by dt equal to v cos theta that is if I take the vertical component of the velocity and integrate it I should get my altitude so again we do the same trick of changing variable from t to theta and then we get a differential equation for h which is a simple differential equation of g tilde sin 2 theta divided by 2 q0 square if we integrate this I leave this for you to verify the integrating complicated expression you get h as a function of theta and as you know theta is a function of time so we can also obtain h as a function of time now we find that h is a cosine function of 2 theta and we know an interesting feature that for theta equal to 90 degree the altitude will reach its maximum value which is a result that we already know from our projectile motion this is something similar to the projectile motion that we have seen in the basic rigid body mechanics let us now look at the horizontal distance solution as the solution of the differential equation dx by dt equal to v sin theta so again we do the same change of variable of dt into d theta and arrive at the integrand that is g tilde sin square theta by q0 square this is a simple trigonometric integration and now we get a solution for x as containing two terms theta minus theta naught and sin 2 theta minus sin 2 theta naught let us now see all these relations through an example that we have been following from our first idealized solution case but the fact remains now that we cannot really start our gravity term with velocity and theta equal to 0 so we assume that the vertical lift off has only occurred for some time and at the end of that vertical lift off a small pitch kick has been given which has resulted in an inclination from vertical off about 5 degree and by this time the vehicle has achieved a velocity of 85.4 meters per second of course in the process it has consumed 6 tons of propellant and it has also acquired an altitude of about 415 meters let us now assume that from this point forward we are going to conduct the gravity term maneuver and let us try and find out what would be the terminal parameters if we conduct a constant pitch rate maneuver for the next 90 seconds assuming the sea level gravity so in this case I have specified the time I have specified the initial angle and I have specified the velocity which are the only parameters that I can specify all other parameters now we can calculate using the expression so let us see how those parameters can be calculated and what their values are so q naught which is g tilde sin theta naught by v naught comes out to be 0.01 radians per second which can be converted to degrees per second as 0.57 v degrees per second the moment I know the q naught I only know theta naught so I know theta at 90 seconds that is theta naught plus q naught into the delta t which is 90 and if I do this I get an angle of 56.6 degrees so this particular trajectory when executed will result in the inclination from vertical as 56.6 degree at the end of 90 seconds we can also use the velocity expression that we have derived for obtaining the velocity at the end of 90 seconds which uses the theta at the 90 second of 56.6 degree and gives the value of 890 meters per second so that is the velocity at that point now with theta 90 known and q naught known we can solve for the mass fraction m naught by m 90 which comes out to be 0.62 and through that we find that mass at 90 seconds would be 39.7 tons so please note we have started from 74 tons and at the end of 90 seconds 40 tons of mass is left so which means that we have approximately consumed 34 tons of propellant during this process which effectively means that we still have about 20 tons of propellant left and during this time we have acquired an altitude of around 34 kilometers and have traveled about 26 kilometer over the surface of the earth which is not a very large distance so possibly our flat earth approximation is still valid now let us explore this problem a little further as I had mentioned there is still about 20 tons of propellant left but and we have not reached 90 degree which in many cases would be a requirement from terminal conditions so let us hypothetically say that we would like to reach 90 degree and we want to find out if by burning the remaining 20 ton propellant we can reach 90 degree obviously we know that the time will not be 90 seconds but it will take a little longer time but we are not sure we don't know whether it will happen or not so if it happens let us try and find out what are those burnout parameters which we are going to get but if it does not happen then let us find out the reason as well as to say that when it reaches 90 degree what would be the final burnout mass I will leave you to work with the formulae that we have derived now that we have seen in the previous example those formulae I suggest that you become a little bit more familiar by practicing them and you will find that it does not happen in fact you see a very very surprising result that from 56.6 degree that is for 90 seconds to 90 degree it just requires an additional 2 tons of propellant from 39.7 tons to 37.4 tons by burning additional little more than 2 tons of propellant it has covered an angle of practically 34 degrees so by the time it reaches 90 degree you still have about 17 tons of propellant left which obviously means that we will not be in a position to consume all the propellant by the time we reach 90 degree so this brings us to the conclusion that there is some disconnect between the rocket mass configuration the propellant and some of the trajectory parameters that we have assumed so please note an important parameter that we had assumed was theta naught because that is something which was to be given as a pitch kick and it is something which is under our control and it was somewhat given in an arbitrary fashion there was no reason that theta naught would be applicable in the present case so now let us invert the problem and try and find out that theta naught which will make this particular mission feasible from a design perspective that it will burn all the propellants it will reach an angle of 90 degree let us try and find out what is going to be the theta naught that will make it happen and that is going to be a design solution so that that is the theta naught that you are going to give at the pitch kick point and with that you should be able to execute the trajectory as desired so this is the exercise that we now conduct of course we know the final mass fraction because we want to consume all the propellant so we know that the mass fraction is going to be 4 that is m naught by m b is going to be 4 and final angle also we have specified as 90 degree so now we are required to determine the theta naught for which the solution is going to be possible and that can be obtained through simultaneous solution of two non-linear equations one corresponding to the mass fraction and other one corresponding to the final angle so if we do this in fact I suggest that you do this exercise yourself you will find that for theta naught equal to 3.01 degree such a mission would be feasible but it may take 252 seconds to complete and in this case your q naught is going to be 0.345 degree per second instead of 0.547 degree per second that is what we had taken earlier so which means we are reducing q naught and as we have seen a reduced q naught automatically results in a higher burnout velocity so now we have instead of 816 meters per second we have double the velocity that is 1.63 kilometers per second also the altitude which was around 35 kilometers has gone to 135 kilometers and horizontal distance which was around 25 kilometers has become 212 kilometers so you realize that just by a small tweak of the initial angle from 5 degree to 3 degree there is a great change in the trajectory with the same propellant mass and now you realize the value and the potential of the gravity term trajectory through a constant pitch rate maneuver realize that the possibilities of generating large number of trajectories that you may desire from the same rocket just by suitably tweaking your q naught and ensuring that q naught is maintained throughout the trajectory therefore to summarize it is possible to obtain closed form solutions for trajectory under the assumption of constant pitch rate we also note that the solutions so obtained fixed at the trajectory time once the starting and terminal angles are specified. Hi so I hope you have realized that gravity term maneuver is an extremely powerful tool for designing the ascent mission trajectories what we have also not explicitly stated but can be seen from the various relations that we have obtained that for given terminal conditions it is possible to also determine the rocket configuration that would achieve a particular trajectory configuration so you realize that a simple constant pitch rate assumption has provided us with an extremely powerful tool in the next lecture we will look at the other two tools that is the constant t by m and a constant velocity so why see you in the next lecture and thank you