 So, now let us apply all that we have learned today to look at a real life operating problem. There is an airship which has got an envelope volume of 5000 meter cube. At the ground level it carries 8500 Newton's of helium, its pressure height is 1500 meters but it exceeds it by 200 meters and goes to 1700 meters and it remains there for some time. This means it is not, it is a sustained case. It is not a transient case, it is a sustained case. It remains there for some time. So now the question is what would be the weight of the lifting gas remaining? First of all, how much lifting gas will be thrown out and therefore what will be the weight of the lifting gas remaining? And then what is the inflation fraction earlier when it goes to pressure altitude and what is the inflation fraction when it goes to higher than pressure altitude. So, to make life easy for you, I have given you the numerical values of the sigma at 1500 meters and sigma at 1700 meters. Notice that sigma is 0.8637 and 0.846 respectively. And for helium the value of K is 0.03416. So, let us see if you can solve this question. If you want me to show some formulae, I can scroll back and show you the formulae. You can note down this formulae if you wish. The last line in this slide is the one that is of relevance to you because this is applicable for a sustained, sustained exceedance. In fact, this is applicable for both the cases. However, you need to know the formula to be applied for IMA. So, I will go one step back. This is what you need. Delta I L is equal to sigma SPH upon sigma as MA minus 1 for a sustained exceedance. So, IMA will be equal to 1 plus this particular ratio or it will be simply I will be equal to the ratio of the 2 sigmas. So, let us see. Now, IMA is going to be the ratio of the 2 sigmas. So, it is going to be sigma SPH by sigma as MA 1.020, 1.0202. That is the value. So, those of you who do not have the calculator, you can follow the calculations with me. Others should do it so that you learn. Otherwise, you will stare at the screen and say yes, but you will not understand. Now, the rate of the lifting gas at the maximum altitude is going to be the rate of the lifting gas under the ground which is 8500 Newton divided by IMA which is 1.0202. So, the rate of the lifting gas is going to be 8322 Newton. Is there any mistakes in these calculations? You can correct me. So, this is the first thing that we got. Now, the airship has lost 2% of its original lifting gas. That is why it is 1.0202. Nearly 2.02% is gone. So, therefore, delta N will be the difference in the density ratios into P0 by T0 into K into volume minus WLG upon into 1 upon IMA minus 1. So, sigma as MA is 0.8466, sigma as PH is 0.8637, P0 101325, T0 288.16, K is 0.03416, V is 5000 meter cube already given in the question. Minus lifting gas weight at ground was 8500 Newton, 1 upon IMA, IMA is 1.0202 minus 1. So, calculate this number and tell me the value of delta L, delta LN, 168.3. Let us wait for somebody else to get the answer. Yes, how much? 858.49. So, 2 people have got 858.49. I got it at 895 and it is negative. So, the net static lift has been reduced because 8500 minus 8322 Newton's of gas has been thrown away. That much of gas has been thrown away. Now, the question number B was what are the previous and subsequent C level standardized inflation fractions? This is very straightforward. This is equal to sigma. So, the inflation fraction is decreased by 1.7%. We know that the inflation fraction is simply equal to the sigma. So, it was supposed to have 86.4% gas and remaining air. But because pressure height has gone to 1700 or operating altitude is going to 1700, the inflation fraction is decreased by 1.7%. It should actually carry more air to go there.