 So if you have understood this, let us see if you can solve this problem. So now we have an airship, same airship, same ISA conditions and we want to calculate the lifting gas weight and the balloon air weight and the volume of balloon air if there is a 1 degree change in the temperature and it is slow. So because it is a slow change in temperature, we have to refer to the previous formulae. So this is the formulae for slow change in the temperature. So using this particular formulae, you will have to calculate for an airship that is flying at sea level under ISA conditions, TA will be 288, PS will be 101325, K you already must have calculated 0.03416, V is 6000, always ensure the units are correct. So 1 by 288 minus 1 by 289 minus 1 by 288, how much is Delta Lg, 249 plus or minus, so reduction. Then what about Delta Ln, that will be 0. And what about the volume of the weight of the balloon air, correct it will be the same and you can calculate the volume of the balloon air by looking at the balloon air weight and the density of the balloon air at sea level condition. What happens if this 1 degree is not because of ambient air temperature change but because of super heat? Should it differ or should it remain the same, whether you add 1 degree because of super heat keeping T s same or whether you make T s 1 degree more and keep same super heat. Should there be any change, the only difference that will happen is in the numerical value. In one case there is an increase, in one case there is a decrease but the number remains the same. Okay, what happens if this is rapidly done, not random, not slowly, so for rapid same or different? It is the same, the formula is the same. So on that note I think we should stop for today because already we have reached 1230 but I just want to make some announcements. We will continue from here in the next class.