 So, now this I have done last time but I want to revisit once again because these are very important expressions and very soon you are going to do an assignment in which you will need these equations. So, there is something called as a force weight form which focuses only on the forces acting and the weights of the various system components. So, as for this formulation the net lift is equal to the gross lift which is equal to weight of the displaced air minus weight of the lifting gas minus weight of the air in the ballonet. This is very straight forward right nothing to discuss here. Similarly, the lift the gross lift we have seen the expression last in the last page is going to be obtainable in terms of the pressure P s minus the humidity effect upon T a into K into V e and V. So, therefore now if you look at the weight of the lifting gas because we are looking at the force weight form we will now bring in the infraction fraction y or sorry we will bring in the purity fraction y in the previous calculations we have actually not looked at the purity issues. So, if you want to bring in the purity then the weight of the lifting gas will be equal to the density of the lifting gas into the volume occupied by lifting gas. The volume occupied by lifting gas will be equal to the amount that is there in the envelope minus the ballonet. So, that is why we have to use the infraction fraction I. So, weight of the air in the ballonet will be using the expression derived last time for the same expression as just copied here. Now we can go for some simplification. So, what do you do you say let us neglect the value of y we say y is equal to 100 percent or 1 again get rid of P sp delta P sp get rid of delta T sh also get rid of e. So, with that you get a much simpler expression which says that the net shattered lift is equal to 1 minus i d P g times P s divided by T a i into k into V e and V. So, V e and V is the envelope volume which is the total volume inside it there are 2 volumes the volume of the air in the ballonet and volume of the lifting gas k is the constant which we have already seen P s is the pressure acting outside 1 minus r d into P g takes care of the fact that not the entire air inside is equal to the lifting gas there is some sorry ambient air there is some lifting gas and some ambient air. So, r d P g is the relative density of the pure gas and T a is the ambient temperature. Now the same expressions we can also look at from the density point of view keeping in mind that the lift available is equal to the difference of density times the volume. So, what will happen in that case is that the net lift will be density of the ambient air rho a minus density of the air of the lifting gas inside the envelope which will be density of lifting gas times i where i is the infraction fraction minus rho b a density of the ballonet air bracket 1 minus i because only that much is occupied by the ballonet. So, this is basically the difference of density the classical rho a minus rho g that multiplied by the envelope volume into g this is another way of expressing the net static lift. In this we are now going to insert those equations those expressions for the various terms. So, for density of air keeping in mind the effect of humidity for density of lifting gas keeping in mind the effect of super pressure and super heat and the gas purity fraction and for the density of the ballonet air keeping in mind super pressure super heat and the presence of humidity. So, all three areas the ambient air the lifting gas inside and the ballonet the air inside the ballonet all three of them have their own densities and those are obtained by this expression. Now it is your task because you will keep staring at the board and say yes yes yes your task is now to insert these expressions and get me the expression for net lift. You will be surprised it will be a very small expression because many terms will cancel out. So, please do it and when you finish please raise your hand. What I need is I need an expression for L n net static lift by replacing the three terms yes. P s is essentially the ambient pressure under standard conditions in the atmosphere at that altitude. P a is the actual pressure of the ambient air because you may operate in the atmosphere which is not I s a it may be I s a plus 10 look I s a is a standard atmosphere where the sea level temperature is 15 degree centigrade. But if you go to sea level in Mumbai the temperature is not 15 degrees it may be 30 degrees which means we are operating in I s a plus 15 correct. What it means is at every altitude above Mumbai the pressure temperature density etc or pressure and I should say temperature the temperature of the ambient air will be equal to temperature at the I s a plus 15 degrees. So, when you operate under non I s a conditions the pressure and density of the air is not standard it is P a P a therefore the density is rho a. So, please understand P s is the standard ambient air pressure at that altitude P a is the actual ambient air pressure at that altitude and these two will be same if you are operating under standard I s a conditions these two will be different in case you operate under non standard. Now, the presence of water vapor can be there at under any atmosphere that depends on the value of T or it shows itself in terms of E okay right. Does somebody have the expression? So, all you need to do is let us try to let us try to achieve it here do it here if we can. So, essentially L n is equal to rho a minus rho l g i minus rho b a 1 minus i into v e n v g. So, this is equal to rho a can be replaced by P s minus 1 minus T naught by this minus rho l g. Now, rho l g is 1 minus 1 minus r d w v times e not e it will be y in this case where y is the all of it into rho l g. P s plus delta P s p over T a plus delta T s h again you have T naught rho naught by P naught then again you have minus rho b a. So, it is P s plus delta P s p minus 1 minus r d w v times e times same T naught rho naught by P naught by T a plus T delta s h. Now, let us assume that there is no super heat which means this will go this will go to a relative density of water vapor or relative density of pure gas. No r d is the relative density of the pure gas with respect to the gas air outside there is no water vapor term there. See the water vapor term is taken care in terms in terms of e that is the humidity effect. In the case of rho b a in the case of rho l g. So, in the case of rho l g what is y y is the purity of the lift of the of the lifting gas. Now, r d P g means what the relative density of the lifting gas with respect to air correct. So, this is the rho of the lifting gas upon rho ambient air. So, you are saying what should be here in which expression in this expression rho l g here sorry I am looking here and I am sorry you are you are you are right. Now, we also remove we also remove the super pressure that means we need to knock out this term we knock out this term then suppose we have been pure lifting gas that means this i y is equal to 1 and let us also ignore humidity. So, that means this term goes this term goes and also this term goes. So, what do you get here you get P s by T a p naught rho naught by P naught minus this will be this is again P s no this is 1 minus 1 so, this will be minus 1 minus 0 that is 1 again it will be P s by T a T naught rho naught by P naught. So, these 2 terms cancel out minus P s by T a T naught rho naught by P naught. So, I knocked off 1 term by mistake 1 minus r d P g. This term I knocked off by mistake. So, that term will remain here and basically if you look at this expression this expression is basically equal to 1 by r and P by r T is equal to rho density. So, that is why it will become rho a which is the ambient air density and sigma is basically rho into rho by rho 0. So, if I take sigma rho 0 that is equal to rho. So, that is how we are getting rho a sigma rho a sigma 0 here and this term of course will is going to remain. So, it simplifies a lot if you go for assumptions. Now, not every LTA vehicle is basically going to be an airship with balloon air. There are many other LTA vehicles. So, can you name some other LTA vehicle which will not have just an envelope and balloon air. So, for example, rigid airships. What do you have rigid airships? Did you have a balloon air? You do not have a balloon air rigid airships. What happens in the rigid airships is that the gas is stored in individual gas bags many of them and these gas bags are inside the structure. There is a there is a structure, there is a covering over the structure and inside the structure there are independent gas bags. So, therefore the volume occupied by each gas bag will be a partial volume of the total envelope volume and that too because of the presence of the structure. The total volume of the outside envelope is not available for the gas. Although the volume of the air displaced is equal to the outer volume of the body, but the volume occupied by the lifting gas if you add it up it is not going to be same because there are other things inside. So, in rigid airships we assume that there is some volume V and each of these gas bags are going to occupy some volume with its own inflation fraction because the radius of the gas bag will be less than the radius of the airship envelope at any point. So, that effect is taken care by individual IN and as you go from most to the tip because there is a curvature in the envelope shape even the infraction fraction will not be the same. You will have for instance less value of I at the front and the back and more value of I at the constant diameter of central portions. So, therefore the infraction fraction I can be replaced by summation of individual infraction fractions times individual volumes of the end gas bags upon the total envelope. Second, there are no balloonets in there are no balloonets in a rigid airship. You do not need balloonets in the rigid airship because you are not supposed to maintain an internal pressure. In a rigid airship there is nothing like a balloonet needed because you have an external envelope which does not deform under the loads coming on the envelope are taken care by the structure. So, it will not allow the balloon to compress or expand and inside a large volume of the envelope is filled with gas bags and they are individual because if one of them leaks it does not lead to catastrophic failure or loss of lift. So, we can always bring back the same equations as last time. These equations we have seen so many times now. So, I have just copied and pasted them. How will they be changed when you go for rigid airships? So, what will happen in rigid airships? Please tell me one by one. Which terms will drop off? Which terms will change if you have a rigid airship?