 So, now we look at the first simple problem, this should be a very simple problem for all of you. So, we want to predict how much is the gross static lift of an airship, envelope volume is 7000 meter cube, operating condition is ISAC level and we assume dry atmosphere. There is a reason for this, the effect of atmosphere being wet, which means humidity. We will spend quite some time in the next few slides trying to model it. So, right now I am assuming air is completely dry, helium, the gas is perfectly pure and we may also assume that there is no ballonet, it cannot be simpler than this. So, now what do you need for this? What is the formula will you apply for this? Okay, rho A into V envelope into G. So, what is the value of rho A? No, from now onwards, please remember, the value of rho under ISAC conditions at C level or rho 0 will be always taken as 1.2256 kg per meter cube, write it down somewhere. This is the correct number, third decimal place etc., be careful because in these calculations sometimes decimal places will matter a lot. So 1.2256 kg per meter cube is rho A but only under ISAC level condition and V ENV 7000 meter cube, G is 9.807. Please tell me how much is the net, what is the gross static lift, 84.16 kilo Newtons. The number seems correct, 84, so note down, so the net lift is 84.16 kilo Newtons. Okay, everybody got it? Any problem anybody has? Okay, so if that is the number, now if the empty weight of this airship is 6575 kg, then please tell me what will be the net useful weight and this empty weight includes the payload, the fuel, the ballast, crew, all the items, oh I am sorry it does not include all the items sorry, the items will be coming in that useful weight, it does not include fuel and payload but it includes ballast and the structure and the operating requirements. So the weight of the pilots will be a part of the operating empty weight. Now I want the answer is kilograms, so you will have to convert your 84.16 kilo Newtons into kilograms, 2004.2 kg, everybody got that number? So approximately 2000 kilograms correct, so you divide, divide that by g, so you will get the kg force and then you subtract 6575, so you get around 2000, that is my field. Around 2000 kg will be net useful weight which will be a combination of the payload and the fuel, very any problem got it okay. So now this airship wants to carry passengers and cargo of 1500 kg, transfer will be in 10 kilometers, Pratik how far can it travel? You can do it orally, very simple, 2000 kg is available, 1500 is gone, remaining is 500 and each kg of fuel gives you 10 kilometers, so it is 5000 kilometers, simple. So we have started now looking at basic airship performance okay. Now we look at one of the most complicated things that you will encounter in static lift calculation, yeah I am surprised you asked this question because I have already explained in the calculations that do not do double accounting of the weight of the gas, when you calculate rho A minus rho G into V and V into G you have already taken care of that is what you have to understand that rho A into V and V into G is the vertical force, you subtract from that rho G, rho gas into V and V into G, you have taken care of the weight of the helium there, you do not have to subtract my friend that is what I am saying. The force you will get this is the concept between the static lift and the gross lift, gross lift will be rho A into V envelope into G, net lift will be rho A minus rho G into, so now that is interesting, so that is interesting, so you have only calculated net lift, sorry gross lift, so you forgot to calculate net lift, so density of helium you need, yes density of helium 0.1639 if I remember rightly, 0.1664 kg per meter cube okay, so now the calculations will change then accordingly.