 Hello and welcome. In the last lecture, if you recall, we had looked at the implication of the gravity along with the propellant burn rate on the overall terminal performance and we had also looked at the loss of energy due to gravity and its connection with the burn rate. Now, we are ready to bring in the third force that we have considered in our earlier discussion that is the aerodynamic drag and consider its implications. So, let us begin. So, let us begin our discussion on the impact of drag. If we look at the drag as an effect which is of a reasonably smaller amount, then we find that for most missions it will essentially be an order of magnitude lower than gravity and it is treated as a tertiary effect. While in the case of gravity, we made a simplified assumption about the value of the gravity because the drag is an even smaller order effect, we can use an even simpler model to capture the effect of drag which is primarily based on an energy concept which we will look at next. So, the simplified drag model essentially is aimed at capturing the overall loss of energy that will occur during the complete trajectory through the definition of the concept of a constant average deceleration that we can introduce in the equations of motion so that they remain linear and we will try to match the overall energy that gets lost because of such a constant average deceleration against the actual energy loss which is likely to happen and it is found that such an approach gives a reasonable performance estimate. So, we again assume that our trajectory is along a straight line that is a radial line. Under that condition, we can now rewrite our equations of motion which we borrow from our earlier discussion on the equations of motion for gravity and then we just subtract one more term that is d by m on the right hand side where d is the drag and m is the mass. Now, this d by m is what I call an acceleration or with a negative sign we can even treat it as a deceleration to do drag that the vehicle will experience while moving through atmosphere and it is just a pure subtraction from the value of the thrust. Now, in this model the ad term that I have introduced is going to be treated as a constant drag acceleration term similar to a constant gravitational term that we have already introduced in the last lecture. Of course, we know that d by m is not a constant because the drag varies in a particular manner as a function of altitude and velocity while the mass varies as per the burn rate introduced. So, obviously d by m which is the actual acceleration at each time instant will vary along the trajectory. So, how does it vary and here we have something to help us which we have already seen earlier. So, let us consider the generic drag versus altitude plot which we have seen earlier as dynamic pressure versus altitude plot but now only it is scaled version. Here you will notice that the x axis which earlier was only the dynamic pressure has now been multiplied with the reference area and the drag coefficient so that this represents the drag variation as well. You will realize that the variation will remain the same except that it will get scaled to the force units rather than the pressure units because the surface area at least through the atmospheric phase of the flight will remain constant and the drag coefficient as we have discussed earlier is 1 for a bluff body assumption that we have already made. So, you will realize that the same curve can adequately represent the variation of drag with respect to altitude. Of course, in the equation that we have written down in the previous slide the drag is a function of time but you would realize is that as altitude itself is a function of time drag versus altitude is an adequate representation in the present case. Now, let us bring in the idea of the energy loss. So, as drag is a force and altitude is a distance from our basic understanding of mechanics we realize that if I calculate the area under the curve that will represent the total energy contained in the drag term because the area under the curve is nothing but the integral of capital D that is drag into dh taken over the applicable altitude. In this case as we have noted earlier also beyond 40 kilometer altitude the actual drag is negligible so that the area under the curve can adequately represent the total energy that would be lost because of drag. Now, we introduce an approximation through the red dotted triangle that is shown in the picture. Now, the reason for introducing this is not very far to see because the shape of the curve is very nearly a triangle except that it is a curved triangle. By introducing the straight lines I introduce an approximation but I have another point. If you see the triangle closely you will find that the overestimate of energy in the upper part by the triangle is to some extent compensated by the underestimate of the energy or the area by the lower part of the triangle which obviously means that in some manner the triangle can reasonably capture the overall energy that might be lost. The reason for generating this triangle is that once I create this triangle I can match this triangle with the green rectangle as the next level of energy matching. So, I can say that the area under the triangle which is same as the area under the actual curve is also the area under the rectangle and once I assume that the area under the rectangle is same as the area under the curve then the width of the rectangle is nothing but my average drag acceleration that I have used in the expression. I hope this translation is clear to you. So, once we do this then it is extremely simple to go from the actual curve to the value of average drag just by looking at certain features of the curve. So, in this case because we are drawing the triangle at the tip of the actual curve which is the maximum value of the drag by dividing this value with the instantaneous mass at that altitude it will become an instantaneous drag acceleration value which is the peak value that the system will experience. So, once I introduce this idea of the peak value of the acceleration which is nothing but the point at which the triangle has the apex then we know that the area under the triangle and the area under the rectangle will be exactly matched if the width of the rectangle is half of the peak value. This is directly from our triangle relations of area calculation which is nothing but half base into height. So, half the height is nothing but the total area as long as I match the height average value of acceleration is just the half of the peak value by matching the area and you will realize that that can represent a reasonably good approximation to a smaller order effect of drag. So, that in the initial estimates we can always get the impact of the drag for a given trajectory without too much of computational effort which would otherwise be required for solving the complete nonlinear differential equation. With this let us rework the previous example that we have seen in our gravity loss case to see what is now the order of magnitude or the impact of the drag on the terminal performance using the simplified expression that we have generated. So, let us take the same problem the burn rate of 600 pages per second and now introduce the three aerodynamic parameter necessary that is the CD naught 1 the surface area assumed to be pi meter square and one last parameter which is introduced just to kind of use the computational effort at this point is to make a stipulation that the drag will be maximum at roughly around 50 seconds which is the half of the total burn time of 100 seconds in this case. Let us assume that it would be somewhere around this the reason why this number is chosen will become clearer as we look at the solution but I just wanted to keep in mind the fact that the peak of the drag is roughly around the 10 kilometer to 12 kilometer altitude range which means that if we peg the time at around a value which results in the altitude value of around 10 to 12 kilometers then the drag calculated at that altitude will essentially represent the maximum value of drag for that particular trajectory it is an approximation but it helps us to simplify our analysis. So, let us try and determine the impact of drag. So, let us first recall the non drag values that we have already seen earlier for this case that is the final burnout velocity is 2.3 kilometers per second and the altitude is 78 kilometers when there is no drag and we would like to compare these values with the value of the same parameters when we introduce the drag term. So, let us now look at the analysis step by step. So, let us first calculate the velocity that this particular vehicle will reach in 50 seconds. So, it is the application of the same expression that is the expression with gravity but without drag and we find that this results in the velocity of 616 meters per second the corresponding altitude that is reached is around 13 kilometers which is not bad. You will find that by and large the drag will peak around this altitude for most of the trajectories and then we go to the atmospheric tables that are readily available and from that atmospheric table we read out the value of atmospheric density at this altitude. So, which is 0.267 kg per meter cube. Now, the next step is to calculate the drag which is nothing but half rho v square at 13 kilometer altitude into cd into the surface area. So, if you do this calculation you will find that the drag at this altitude is 159 kilo Newton. Further because we have proceeded up to 50 percent of the total burn time we have consumed half the propellant which means about 30 tons of propellant has been consumed. So, the effective mass at this point is only 50 tons. Based on these two quantities we now simply calculate the peak value of the drag curve as 3.18 meters per second square and compare this number with the gravitational acceleration which is about 9.81 it is one third of that but more importantly because we are going to use an average value it is not a constant it is only the peak value the average value just half of this this comes out to be 1.59 meters per second square. Now, you can see that this number is practically one sixth almost an order of magnitude lower than the value of the gravitational acceleration. So, our original hypothesis that the drag is a tertiary effect is also justified and with that small term we can also say that our approximation that we have introduced using the energy balance methodology is also justifiable. And with this drag I suggest that you do this exercise yourself as per the equation that we have seen earlier you will find that the velocity which was 2.30 without the drag becomes 2.14. So, this is the amount of loss about 160 meters per second is the loss in velocity due to drag. And similarly the altitude which was 78 kilometers becomes 70 kilometers about 8 kilometer loss in the altitude together you can add the potential and the kinetic energy loss together and that will effectively tell you how much of energy loss has happened over and above the gravity loss. You will find that this energy loss is significantly lower than the energy loss due to gravity for the same case because the ideal velocity is 3.264 kilometers per second that represents a particular energy compared to that the gravity case gives an energy lower by about 36 percent. You will find that this will add another 8 to 10 percent in the loss so that the total loss is likely to be around 45 percent of the complete energy. Let me now introduce an idea that we have not seen earlier. If you recall we had given an expression for gravity energy loss as a function of beta but we had not actually characterized the impact of beta directly but we had made a mention that chances are that the drag loss is likely to be higher if you burn faster. Let us try and examine that idea now to the same example and this time we take a significantly higher burn rate of 3000 kgs per second which is 5 times more than what was there in the previous case. You will immediately realize that if you are going to burn at 3000 kgs per second all the propellant will get burnt in 20 seconds itself and that the peak would be lower than what happens at 20 seconds. So, in this case we make a stipulation that the drag peak will occur somewhere around 15 seconds which is in the lower atmosphere and we are going to assume that this should again be roughly around 10 to 12 kilometers altitude but now because we have changed the burn rate our non-drag values change. We have already realized that the implication of a higher burn rate is to reduce the gravity loss. So, we get a higher burnout velocity at 3.07 kilometers per second and a significantly lower altitude at 23.4 kilometers but the total energy is significantly higher because the loss is much smaller. Now assuming that T equal to 15 seconds we calculate the altitude which is achieved at this time and that altitude turns out to be 11.5 kilometers. So, our assumption is justifiable that we are still in the same ballpark of 10 to 12 kilometers. The velocity now at this altitude please note instead of 616 meters per second that is what you saw in the previous case is now 3 times it is almost 1800 meters per second very high velocity at nearly the same altitude. So, obviously you see that your drag term which was 159 kilo Newton is practically 10 times at 1713 kilo Newton because the drag is significantly higher and more importantly because you have burned faster you have consumed lot more propellant in 15 seconds in 15 seconds you have actually consumed 45 times because you have consumed 45 tons of propellant the residual mass is only 35 tons so mass is also lower. So, your peak of acceleration is significantly higher which was earlier 3.48 is now on already 49 meters per second square that is practically 5 times your gravity which means your drag loss if you do this is going to be 5 times the loss that you are going to get because of gravity of course we can also calculate the average drag acceleration that is half of 49 and based on that we calculate the velocity and altitude and now you see that the impact of this drag is significantly higher the 3.07 kilometer per second has become 2.58 which is almost 0.5 or 500 meters per second lower and effectively you can say about 20 percent loss 15 to 20 percent loss in velocity. Similarly, you have another 5 kilometer loss in altitude which is also close to about you can say 20 percent together you will find that this is going to represent a very large energy loss that brings us to the same point that we had seen earlier but had not this particular result in front of us to make any comments. So, now let us look at it that if we burn faster we reduce the gravity loss but then we increase the drag loss similarly if we burn slower we increase the gravity loss but then we reduce the drag loss. So, is there going to be a golden mean of burn rate where we might be able to keep both the losses to their minimum value and thus make the mission an optimal one. So, this is the nice hypothesis that we need to examine and we need to do this by modeling both the losses together so that we now get what is called a combined loss and see if we can get a minimum for the combined loss and map it to a burn rate which is going to be applicable for that trajectory to say that if you burn the propellant at this rate then this is the most efficient mission with the given propellant and the lift of mass and the propellant that you can carry out. Considering the loss due to gravity and the loss due to aerodynamic drag. So, let us take the same example but this time we do not implement any burn rate. We are now going to look at the combined loss. So, let us see if we can get combined energy loss as a function of burn rate and locate the minima and determine the corresponding optimal burn rate. Just to simplify our analysis and the various algebraic steps, let us again assume that the drag profile peak is going to be around 12 kilometer altitude so that we can fix the density value and let us also use the beta in the range of 400 to 1200. Let us see if that gives us something. Now, I want to show you something on the right hand side which is the approximate solution and then I will tell you how the solution has been obtained. So, this solution is the solution for the loss of combined energy as a function of burn rate starting from 400 kg per second to about 1200 kgs per second. And you see that this is a classic concave curve inverted parabola with its minima lying very close to the 600 kg per second that we have used in our examples earlier. Which means the number that we had used was very close to the actual minima that you are going to get. The above curve has been obtained through MATLAB by writing a small code. I suggest that you can also try that exercise. I will tell you the steps involved in the code that is to be developed that is you take the solution because of gravity. Based on that solution generate the velocity at 12 kilometer altitude and with that velocity and the density at 12 kilometer altitude get the value of drag. And once you get the value of drag you can find out the time taken to reach that altitude multiply that time taken with beta to find out what is the residual mass as that altitude as a function of beta. It is not going to be a constant it will now be a function of beta. Based on that obtain the drag peak which is going to be a function of beta and then go back to the expression and recalculate the velocity and the altitude under the action of this drag acceleration term as a function of beta. Add this loss to the loss because of gravity and that becomes the combined loss. This is how that solution has been obtained. Of course you need to realize that this is a highly simplified analysis so obviously it is an approximate presentation and you will probably need to do a more rigorous analysis to arrive at the actual optimal rate. But the idea that I am trying to propose is that by making reasonable assumptions we can simplify the analytical procedure and still get a reasonable understanding of the physics involved in the process. So you realize that there is a possibility of arriving at an optimal solution for the burn rate by trying to adjust such that the combined energy loss is a minimum. Of course such an analysis would always be carried out even for the simplified case of a constant burn rate in a more rigorous manner towards the end of the design through complete nonlinear simulation of equations and numerical solutions. We will find that at the initial stages of design when you are trying to size the rocket and you want to understand how much loss would be there because of gravity and drag put together so that you can appropriately put the propellant and the ISP correspondingly as part of your design solution, you need a kind of a gross estimate of the burn rate. A very, very crude thumb rule which can be commonly used in such situations is that an optimal beta would generally lie close to a trajectory for which the gravity and the drag loss are nearly equal in their magnitude. This is based on the fact that variation of these two effects are broadly similar in nature, completely inverse of each other and also that when gravity is a secondary effect the drag is tertiary and when the drag is secondary the gravity is tertiary so that they kind of complement each other and that when there are nearly of the same order of magnitude the chances are that that would be the trajectory where you would have the minimum loss. Just to understand this let us go back to the previous picture. If you look at this picture you will find that the effective combined loss is of the order of around 47 percent which is slightly more for more value of beta beyond 600 kg per second. You already know that for this burn rate the loss due to gravity is about 36 percent so which means there is another 10 percent which has got added because of the drag and that when you start increasing the burn rate further why the energy loss due to gravity will reduce it will not reduce at the same rate at which the loss due to drag will increase because of the square of velocity. So you will realize that the loss because of drag will quickly climb up so that within the same ballpark it would also reach a reasonably high value and that it will overtake the loss because of gravity. So you can clearly see the increasing trend as you note the curve on the right hand side. To summarize drag is a smaller order effect which can be captured from energy consideration. Further an optimal burn rate exists that results in the most efficient mission for a given vehicle from the point of view of minimizing the combined loss. Hi so with this we have established a reasonable analytical basis for understanding the implication of various forces which are present in the trajectory solution and also a broad idea of how to estimate such a loss and find its impact on the terminal parameters. With this we conclude our discussion on the motion along a straight line and we will move over to the more realistic case of motion along a curvilinear path that we will look at in the next lecture. So bye see you in the next lecture and thank you.