 Hello and welcome. So, this particular session is tutorial number 1 in which we will go through some of the problems in the context of a rectilinear trajectory and we will also look at some of the important solution steps and aspects of the various formulations that we have done in this context with more details. So, let us begin. Let us first explore the idealized burnout solution that we had started off at the beginning of our discussion on launch vehicle trajectories. So, let us take the problem. So, a rocket has the following configuration that is its propellant loading is 0.9 that is mp by m0 as a specific impulse of 260 seconds. Let us try and determine the ideal burnout velocity. So, we recall the expression for the ideal burnout velocity as g0 into isp natural logarithm of m0 by mb where mb is your burnout mass. We know that this mb is nothing but m0 minus mp where mp is your propellant mass. So, what we can now do is we can bring in the symbology that we have defined earlier and we can show that the expression for vb in the context of ideal burnout is minus g0 isp into ln or natural logarithm of 1 minus lambda which is defined in the problem. So, now we have all the parameters available for this expression and which we now use to evaluate. So, it is a simple task to substitute these numbers. So, 9.81 into 260 that is our isp and the natural logarithm of 1 minus 0.9 which is of 0.1 and this results in a number 5873 meters per second. My suggestion is you can also independently carry out this task once you have gone through this particular session completely so that you also become familiar and comfortable with these expressions. Let us now move over to the problems of rectilinear motion under gravity which means let us now include the effect of gravity and let us look at the nature of problems which will get solved. So, let us take the same rocket that we defined earlier that is it has a propellant loading of 0.9 but isp is only 240 seconds and lift of thrust is 1.5 times g0 in the given in the form of thrust per unit mass. So, what we are saying is that at the lift of the thrust is 1.5 times g0 and that remains constant and the motion is along a local vertical which means it is a constant thrust case. Please note a constant thrust case is same as a constant boundary case because our thrust is m dot g0 isp. So, if thrust is constant we already know g0 and isp are constant for practical purposes. So, obviously your m dot has to be a constant. Let us try and determine the burnout velocity, the altitude. First, assuming the gravitational acceleration to be the sea level value and let us compare these with the ideal burnout solution. So, the velocity solution is as your burnout time is nothing but the ratio of mp by beta and please note we have not specified the beta what we have specified indirectly is the lift of thrust which remains constant. So, which means we have specified thrust and from thrust we must discover the burn rate m dot. So, we bring in that idea so beta is nothing but our m dot and that is nothing but thrust divided by g0 isp and thrust is nothing but 1.5 times m0 into g0 because 1.5 times g0 was the acceleration given per unit lift of mass. So, the total thrust would be 1.5 into m0 into g0. Of course, we can cancel g0 which I have done here. So, I am only writing 1.5 into m0 divide that by isp this is my beta. So, what is TV? It is mp by beta. Now, what is mp? mp is lambda into m0 this is our definition. So, I substitute that expression here lambda into m0 divide that by beta and now I get an expression for burnout time in terms of lambda isp and just the ratio 1.5 and it directly tells me that it will take 156 seconds to complete the burning. Now, with this TV I just go back to my VB expression under gravity which is nothing but V ideal minus g0 TV and if I do this I find that my velocity is going to be 4342.7 meters per second as against 5873 for the ideal burnout case. Let us now move over to the altitude solution. So, in this case we recall the altitude expression which is m0 g0 isp by beta 1 minus lambda into ln 1 minus lambda plus lambda minus half g0 TV square. Now, as beta is a function of isp and m0 I substitute that 1.5 m0 by isp m0 will cancel and what will be left will be g0 isp square by 1.5 lambda is already specified as 0.9. So, this will become 0.1 ln 0.1 plus 0.9 and time is 156 seconds. So, TV square I do this simple calculation and it shows that the altitude reached in this case will be 176.7 kilometers. So, we now have the velocity which is go back 4342.7 meters per second and we have altitude which is 176.7 kilometer in the present case. Now, let us extend this problem to problem number 3 where let us try and correct the g value applicable for that altitude. Please note the altitude is almost 180 kilometers. So, it is very large. So, obviously the C level gravity value is no longer applicable and we must now correct the gravitational value and recalculate the burnout parameters. And let us see what difference does it make if we correct the gravitational acceleration value because of the change in the altitude. So, we bring in our gravitational acceleration expression in terms of the altitude and radius of earth and we find that at the burnout altitude the gravitational acceleration will only be 9.29 meters per second square. And now we bring in the simplification and the approximation that we have seen in the lecture that because this variation is small over a large altitude an average value of gravitational acceleration can reasonably capture the effect of change in gravity. So, I define that g tilde as an average of the C level value and the value of gravitational acceleration at 180 kilometer altitude. So, that turns out to be 9.55 meters per second square. So, now I go back and use this value of gravity to recalculate the velocity and the altitude. So, I find that the velocity is 4383 meters per second which is roughly about 40 meters per second higher than what we have predicted from our conservative formulation based on C level gravity whereas the altitude is higher by roughly about 3 kilometers as compared to the 176 kilometer altitude under the constant C level gravitational assumption. So, you find that even with a drastic change in the altitude of about 180 kilometers the effect of this on the terminal performance is still only marginal. For example, if you look at the correction due to velocity it is typically of the order of 1%. If you look at the altitude correction it is even less than that. So, what it means is that for most cases of ascent mission which are going to end around 180 to 200 kilometers. Even if we do not correct for altitude the conservative estimate is not very much off and if we want a more realistic a simplified correction based on the average gravitational acceleration is more than adequate. Let us now look at the gravity loss part because of the presence of gravitational acceleration. So, let us take the problem number 2 that we have considered and estimate the loss of mechanical energy due to gravity that is as compared to ideal burnout. What is the loss of mechanical energy under the action of gravitational acceleration? So, we bring in the idea of energy per unit mass of burnout. We say that the velocity under ideal condition is 5873. So, energy of ideal burnout is 1.7246 10 to the power 7 in appropriate units. Now, we have the velocity under gravity as 4342.7 meters per second altitude as 176727 meters consistent units. So, we calculate the energy as half Vg square plus g into lg that comes out to be 1.1163 into 10 to the power 7 in the same units. So, as a percentage the loss turns out to be 35.2 percent quite large very significant loss that we have incurred. But then there is a problem that we have also seen earlier that for a given burn rate if it is low you are going to incur a large loss of energy due to gravity. And if you want to reduce this you must increase the burn rate. We will leave this problem at this point and then we will go over to the next solution of straight line trajectory and bring in the effect of aerodynamic drag and the loss that happens because of the aerodynamic drag. So, let us consider the altitude corrected solution given in problem number 3. So, in problem number 3, we have corrected the altitude because of the change in gravity and that is the new gravity value is what we have used for terminal velocity. So, we have a velocity of 4380 odd meters per second and the altitude of close to 180 kilometer that is 179 odd kilometers. And now we bring in the idea of a mass rocket and lift off of 500 kg. We say that it has a diameter of 2 meters and we are going to use the bluff body drag value of cd drag coefficient value of 1. Now, let us assume that the peak of the drag acceleration is going to occur around 50 seconds and based on that we are going to bring in our simplified drag modeling and then we will try to find the impact of this on the terminal performance and find out in what way is the approximation valid or applicable and what is the overall order of magnitude of the impact of the drag. So, let us first obtain the velocity and altitude solutions at t equal to 50 seconds. So, at t equal to 50 seconds, the amount of propellant that we are going to consume will be given by because it is a constant burning rate. I know that I will burn all the propellant in 156 seconds. So, in 56 seconds, I would be burning only a small amount of propellant. So, propellant burn ratio becomes only 0.2885. Now, with this I can go and calculate the velocity which is nothing but the ideal velocity at t equal to 50 seconds minus the gravity adjusted velocity. So, this turns out to be 868 meters per second. So, at t equal to 50 seconds, the rocket would have acquired a velocity of 868 meters per second. Let us do the same thing for the altitude. So, again substitute the same expression of lambda at t equal to 50 into this. The time is already specified as 50 seconds. And by doing that, we show that it will reach an altitude of roughly about 8.5 kilometers. So, these are the performance parameters under the corrected gravity value of 9.55 meters per second square at t equal to 50 seconds, where we have assumed that the peak of the dynamic pressure or the acceleration would occur. Once that happens, let us now use this information to find out the value of the drag, the drag acceleration and the average drag acceleration which is going to be used to calculate the modified terminal performance. So, first we get the density from our atmospheric tables. So, at this altitude of 8.5 kilometers, the density is 0.442 kg per meter cube. Using this density, the velocity and the surface area because of the diameter of 2 meters, that is radius of 1 meter and Cd equal to 1.0. The total drag in this case is 523.1 Newton. Now, the mass at this point is M0 into 1 minus the burn fraction at t equal to 50. So, that is 70.7115 times the M0 which is 500 kg. So, this turns out to be 355.7 kg. Now, the acceleration due to drag is nothing but the value of drag divided by this mass. So, turns out to be 1.47 meters per second square that is the peak value and from our rectangular approximation that we have been using, the average value of drag which is a constant is 0.735 meters per second square. Now, with this acceleration value, when we recalculate our velocity due to drag, it is velocity due to gravitational part minus 0.735 into the total trajectory duration which is 156 seconds and what we get is 4268.5. Now, I want to point out an interesting observation. If we had not corrected for the gravity and drag, suppose we had neglected both the effects which means we had not corrected the gravity for altitude which means we had used the C level value of gravity and we had not included the drag. This is the velocity that you would have predicted 4243. But if we correct the gravity value for altitude, it is a plus for us because performance improves and then if we include the drag, the deterioration of performance in terms of velocity is still not sufficient and that we get a velocity which is still higher than what you would get if you did not consider the correction because of altitude and drag. So, you will realize that even if we do not consider the impact of correction to gravity because of altitude and if we ignore the drag in the first initial sizing of the rocket, we would get a fairly good idea of the rocket terminal performance from a design perspective. The change in altitude is from 179 to 170 kilometers, the actual altitude that we have seen is of the order of about 176 kilometers. So, we find that if we were to use these values of 4268 and 170.9, this would be more or less the performance that would be close to the performance which you are going to get because of only C-level gravity modeling and nothing else. So, in this lecture or other tutorial, we have gone through a set of 5 cases starting from the ideal burnout, the impact of drag and we have noted how the terminal performance gets influenced as we introduce the various corrections such as gravitational correction, correction to gravitational acceleration due to altitude and the correction in terminal performance due to the presence of atmospheric drag. And we find that in the initial stages, if we are looking at a gross estimate of the terminal performance just by introducing the C-level gravitational model along a straight line motion, the performance that we generate is fairly representative of what we are going to get under various realistic scenarios. So, with that, we come to the end of this tutorial. Bye and see you in our next tutorial. Thank you.