 Now let us look at this particular limitation of airships. We are operating from an origin location O which is at a height HO to another destination which is D at a height of HD. Obviously either the destination would be at the same altitude or at a lower altitude or at a high altitude. Now if we fly to a elevation, so let us say you take off from a place like Pune. At what elevation is Pune from the sea level? It is not equal to sea level. It is much higher. So or just to make it more dramatic let us look at Lonavala. We know it is at some height from sea level. So you design the airship to fly from Lonavala. You have taken care of the balloone volume that when it goes from Lonavala to some maximum pressure altitude the balloone will become flush. So now you so what is the amount of air in the balloone at ground level? Is maximum or minimum? They are full. At ground level the balloone is completely full as the airship goes to higher altitude you take out the air. What will happen if you operate from an altitude, go to higher altitude and then you come to a much lower altitude? Do you see any problem? Why will you distort? No, on the other hand, on the other hand the balloone will become full but you still have to go down. There will be no volume available in the balloone to take in the gas needed to go down. So you will have a problem. You will not be able to go down. So if an airship is designed to operate from Pune and to go only to some height, you bring it to Mumbai it cannot land because when it wants to come below, so what will you do? You will lose gas. One way is okay expel some gas. So it is important. If a airship flies from a location O that is origin to a location D that is destination such that the height of the origin is more than the height of the destination then the inflation fraction at destination that is id it is going to be lower than the fraction at origin. The value of id can be obtained by simply looking at the ratio of pressures and temperatures plus the super heat. This is a familiar formula to you. Only thing is I have added S at the rate O for ambient pressure at origin etc. etc. Now please note these are the barometric pressure heights at the origin and destination. T and A, T are the ambient temperatures and the super heat. Now you have a minimum inflation fraction already designed in the airship to take care of delta H or the altitude change. So for example, if I design the airship to take off from Pune to a height of 2 kilometers from Pune I will work out how much is the Boulogne volume needed as a percentage of the total volume 1 minus I. So it will be 25%. So if 1 fourth of the volume occupied is by Boulogne 3 fourth by the LTA gas at the ground level I will design the Boulogne for that size. So it will have a given inflation fraction. But the inflation fraction needed at destination is more. So therefore what will you do if you want to operate with a much higher delta H? Delta H being the height change which means if I want to go from the same thing from Pune to a higher altitude and come down to Mumbai, delta H required is from number to that height. What do I do? I have already given the answer to you in my initial discussion. The only option you have is to change the lifting gas. You have to add lifting gas before the takeoff to take care of the height difference that you have to. So let us look at this formula and go for some simplification. What simplification? Probably superheat will be ignored. Then under consumer ISE conditions. So under ISE conditions you have a fixed pressure at sea level case at least. So you can say that I D by I O that is the inflation fraction at the destination upon the inflation fraction at the beginning is equal to P S at O because that additional term delta P SH I have removed. So P S and T S. Now interestingly we have just now done this in the beginning of today's class. P S O by P S D is basically delta O by delta D by delta is the pressure ratio P by P naught and T by T naught is theta that is the temperature variation. And we also know that P by P naught is equal to T by T naught to power 5.253. So this will become 4.253. So therefore the inflation fraction I D by I O is simply the ratio of the sigma which is the density ratio. So therefore you look up the atmospheric chart, find out the density at any altitude, find density at the altitude from which you are operating, calculate the sigma values and the ratio of that gives you the required inflation fraction. So the change in the ballonet volume will be easily obtainable as the from the inflation fraction directly.