 Welcome to today's lecture, this is on design analysis of gear pumps and this is part one. Now external tooth gear pumps with involuntary or similar teeth are widely used in fluid powered application. Gear motors are not very common, although there are some models of gear motors, but this is not very common with in value teeth. Gear motors with different teeth such as orbit motor is very popular, however the involuntary gear motor is not that popular because it is not much beneficial to use such units for motor. Now units with external external gears are called as external gear pumps and motors and units with internal internal gear gears are called as internal gear pumps or motors or the together either it might be pumps and motors because they basically features will be same, they will look alike only there is a minor change in valves, they are usually called hydrostatic units. So, you may call external gear hydrostatic units or internal gear hydrostatic units. Although gear type hydrostatic pumps enjoy a high level of reliability and offer a low purchase cost to the end customer, they are often with performance characteristics that tend to create higher noise levels than other types of positive displacement units. These noise levels are associated with the substantial flow ripples of the unit which induces a pressure ripple and oscillating force within the system. Since the flow ripple in case of pump is considered to be the first cause of these oscillation forces, it is assumed that a smoother flow delivery of the pump will also attenuate the noise that is generated. In this lecture, the working principle flow ripple and reliable characteristics of an involuted gear pump are discussed. Now, first of all what is the working principle of gear pump? While considering the operation of a external gear pump, it is common mistake to assume that the fluid flow occurs through the center of the pump where teeth of machine gears comes in contact. Normally, if a gear pump this plan view is seen, then many people may think that as the gears are rotating in this directions, so this must be flow in and this must be flow out, but actually it is not. It is otherwise, if you look into this picture that is in that case you can see that it is rotating in the clockwise directions. Let us assume the blue one is the driving one. Among these two, one of them is driving one. In this case, let us consider this blue one is the driving one. Then what we observe? The oil is coming and from this side and then it is being trapped and it is going out. And if you look into this view, this oil is coming in like this and oil is going out like this. So, this should be, this is the direction of the flow. So, in fact, again what we think that direction of flow we have corrected. Now, how it is the oil is being pumped out? And we should remember that input size pump is usually atmospheric pressure whether in the output side pressure is high. This is the working pressure and as earlier we have learned it might be in the range of 10, 15 mega Pascal in case of gear pump. Not may not be as high as in case of piston pumps that is cylindrical piston pumps where it can be attained up to 30, 40 mega Pascal, but here the 10 to 15 mega Pascal are very common. Now, then again how it is being pumped? Because we need a lot of pressure to generate. Now again to looking into this figure one may be confused with this wheel pump, bucket wheel pump. You might have seen in the paddy field usually you will find a wheel with bucket fitted on the ferry, ferry that is used to take fetch water from the pond or water source to the ground. Usually ground level is slightly higher than the water level and in that case what is done the it is as if the scooping the water and then it is put into the ground. Now, that is something like a bucket elevator there the water is taken into the cavity of the bucket captured into the bucket and that is taken out. If we think of the pressure pressure is almost same. Now, in this case it is not that it is not that the fluid is being taken by this pocket and it is going simply out. There is the suction and compression action by these gears. Now, to learn that we must find out that where is the suction and compression chamber in that. To produce a flow we should say with a gear pump fluid is carried along the periphery of each gear within pockets between two consecutive teeth enclosed by the casing wall from the intake side to the discharge side of the pump. Now, in that case if we asked what is the how many chambers are there. We can say that within one regulations there are how many pockets this is exactly the double of the number of gear teeth. So, one may consider that is the number of chambers in a gear pump. So, number of chambers are high and as we have learned earlier that with the increases number of chambers actually ripple reduces. Then the first statement I have told that there will be high ripple that is due to the fact that here this compression and geometry of the volume discharge is such that we have very high peak of the ripples. So, it is like that within one regulations if there are say 10 teeth. So, 20 chambers. So, there will be, but there will be actually 20 10 ripples because two pockets are giving the flow together and you will find that ripple is not same as in case of the piston cylinder. So, there is difference we will show that curve later of this lecture. Now, on the inlet side the gear teeth are coming out of the mesh and thereby creating and expanding space among the two contacts at casing and the driving and driven gears. Say here if we consider this contact point that is separating the high pressure zone to low pressure zone and from this point to this point we will find when the gears are moving this space is increasing. So, there is a suction whereas in the other side the space is that area is decreasing. That we will derive that how much area is decreasing or increasing that has to derive to find out the flow rate of such a gear pump. Now, it creates a suction zone the inlet side it creates a suction zone then in the an intake volume is then trapped into two pockets. Here is one pocket here is another pocket of two gears and carried to the delivery or outlet side and in the outlet side gear teeth are coming into contact and thereby creating a contracting space. There by fluid is squeezed out through the delivery or outlet port. So, this is the pumping action in case of the gear pump and in case of motor if we think of the motor then this this process is just reverse. What will be in that case the oil will be pumped in from this side in case of motor and volume will expand. So, this will be then high pressure side and this will rotate in the same manner whereas the low pressure oil will go out, but it will transmit that this pressure will be converted into the force and torque and that can be taken out either from both the gears or from a single gears. It can be shown that the volume displays among three contact this is the three contacts one contact is here another contact is here another contact is here. And then two contacts of two gears with casing and meshing teeth at centers at both inlet and outlet. So, one chamber means just consider the instantaneous the last teeth in contact with that this side on this side and the contact point here. So, this is the compression chamber and similarly this action chamber is with this contact point here and the first contact point coming to the wall and this side the first contact coming to the wall. Exactly equal to the summation of active amount of trapped volumes between two consecutive teeth of gear of each gears this means that what is the total expansion or the compression that volume this volume will be exactly equal to this two pocket volume, but here is one confusing statement is there. If we just measure this area one pocket and here one pocket this will be same and if we multiply them then that would give the total volume, but it is not that it will be these two volumes minus this entrapped volume which is actually inactive. If you just if you stop this gear you will find there is one small volume the same weld is coming back to the suction side that we have to subtract that we should remember. So, while we are calculating this we have to consider this part. However, if we calculate the total displacement of this area during the suction and compression then automatically that will be deducted from the calculated area we need not bother about that how much pocket that means if we give more undercut to this gears that will not matter that root fillet if we increase the root fillet that will not really matter. This is one thing second thing which I am going to show now that one important factor is that actually each and every when it is passing through there will be some wire leakage through the even through the contacts that we should take care of in that way efficiency of this gear pump is lower than what is available with the cylindrical pump and pistons. Another part which I would like to describe here which I would like to discuss that is if we consider in case of pump here the pressure is low may be the atmospheric pressure whereas here the pressure is high pressure. Now, if we consider the pressure curve here you can see this pressure is not increasing not increasing not increasing and then when it is being exposed suddenly the pressure is increasing if it is in case of ideal ideal case but what actually happening that there is a small amount of leakage. So, what we will find whatever pressure here may be some pressure is increasing and then suddenly it is being exposed to high pressure. Now, that causes some sort of the fluid borne noise to reduce that as well as to balance this I mean imbalance in the torque what is done that this somewhere a pocket is made here and that is connected at the midpoint and then again another pocket is made here and then it is connected here and then ultimately pressure distribution is come something like this. This is high pressure and gradually it is decreasing that sort of pressure distribution will be better for the performance and then that is done by such grooving. However, we shall not discuss anything about such grooving because for that we need some more complicated analysis but anyway I will show you the flow between these two teeth. Now, look at this figure as you can see you see just how the flow is taking place in that case. So, this is rotating in the clockwise and this is so this must be which side is the delivery side this side or this side. So, this side is the delivery side as we look into this delivery side not much this flow phenomena is being looked in but what we find as if there is a flow is going taking place in this side say look at this at that moment what you find that fluid is sudden fluid is going out that is a leakage that leakage is very important and in designing such gear pump controlling such leakage or correct estimating of such leakage such leakage for the analysis of gear pump is very important. Now, of course, within this scope of this lecture we cannot analyze such thing only what we will analyze between this contact say this is look at this contact this is the separating chambers. Now, this point is separating the chambers now this point is separating the chambers. So, something is happening there analysis of these will be very important. Now, we shall try to find out what is the volume displacement what we would do if you can calculate the variation of this area with shaft rotation or with respect to time then we can multiply the width of this gear in case of piston pump what we do we multiply with a constant area with the variable stroke length that length is varying but the area remain constant in case of this pump this gear pump what we find that area is varying and the width remain constant. So, if we cannot analyze the variation of area with the shaft rotations we can easily find out the volume displacement and then considering the time rate we can find out the flow rate theoretical flow rate. Now before going into the analysis we have non-dimensionalized the variables that is important that we should understand how we have non-dimensionalized. Now, what is the benefit of that to know that in order to interpret results obtained for any physical size of pump it is conducive to non-dimensionalize the various variables this means that if we know the shape geometric shape of the teeth number of teeth few parameters should be remain same remain constant then what we can do irrespective of the size we can go for analysis then whatever result is available in non-dimensional form that we can simply multiply with the real size some size with respect to we have non-dimensionalized then we can get the actual size of this pump. Say for example, we have taken a pump of the gear 10 teeth for the driving and 10 teeth for the driven then may be module is 5 say then whatever flow we are getting if we reduce this module to 1 definitely flow rate will be less it may be 5 times less or may be something else that can be found out from the non-dimensionalized parameters. But once we design this pump with a particular module particular teeth number we can find out what will be the real output input etcetera. Now, how it is non-dimensionalized that we shall examine now. Before that furthermore this will also simplify the development of equations by eliminating various scale factors within each equations. Hence, the final results may be simply scaled according to the rules that were used to non-dimensionalize the specific quantity of interest. The dimensionless variable used in the analysis are given as follows we will see this. Now, here for linear parameter we have considered the reference parameter as the base circle radius of driving gear. Perhaps you know that the once the gears say involute gears we say that this is the module this is the teeth number then and this is the pressure angle of course pressure angle should be there then one dimension of that gear will be fixed that is normally we make a mistake perhaps the piece circle is the fixed dimension. But it is not the fixed dimension will be the base circle that on which the involute gear is generated. Now, if you introduce the correction, correction means it might be the profile is shifted. In that case we will have may have different pitch circle radius even with uncorrected gears if they mesh with a enlarge expanded center distance we will have a new pitch circle radius, but the base circle radius will remain constant that is why this parameter is taken as the reference parameter for non-dimensionalizing. Now, another information is there that we have used a carrot to denote the dimensionless quantities. Say in that way now L it is actually instantaneous length of action of the gear in mesh I will explain this again with a figure of the gears and non-dimensional means with this carrot or the hat you can say. So, L divided by r b 1 is the non-dimensional parameter. Similarly r is an arbitrary radial dimensions this is also divided by the same base circles. In that way any in the Cartesian coordinate x y coordinates are also non-dimensionalized and the rho that a radial dimensions it is not arbitrary we can say this is from the contact point actually radius from the contact point. This is purely arbitrary and this or general radius and this is the specific radius with respect to the point of contact. Now, this is the linear dimension we have non-dimensional in this way, but there are some other dimensions also for example, the volume etcetera. In that case what we consider the width we consider the real width of that machines or the width in the parameter we consider divide the volume by width and then also speed. Now, this volume volume is well discharge volume this is divided by the width w that means what the non-dimensional volume that is per width unit width and then area is divided by simply by r b 1 square. So, this volume is non-dimensionalized. Now, we can non-dimensionalize this volume by this we dividing by a volume width into r b 1 square, but again flow rate we have to introduce what we where we have introduced that omega 1 what is the angular speed of this or in other words if we carry out analysis with non-dimensional parameters then if we find some volume non-dimensional volume simply we will multiply the actual real parameter of the gear pump which we have considered we shall consider that width we shall consider that r b 1 that base circle radius and in that in case of flow rate we will consider the actual speed, but here one thing I must mention that this gear two gears mostly you will find they are width same number of teeth that means if this side is 20 other side is also 20, but it is also possible that one is 20 other is 18 other may be 16 this is also possible. In that case if we non-dimensionalize all this diameter with one base circle in that case base circle will be different base circle then we have to take care while we are considering the some real parameters, but it is possible because knowing the number of teeth we know the base circle of other one will be the simply the ratio of this gear teeth so that is not will not be difficult this is in case of external tooth gears. In case of internal tooth gears as you know that numbers will be different because if the ring gear is of say 20 teeth then internal external tooth internal gears of that one has to be at least may be 4, 5, 6 less than that. So, in that case you can make the parameters non-dimensional considering any one base circle radius, but while you are converting the real data you have to consider the other base circle also with proper proportion that is not difficult, but one should take care of that. Now so we will now go for this analysis in that case as you see this gears are rotating like this and this is flow in and this is flow out. Now here what we have written this is say look at this dimensions rA1 this is the teeth circle radius which is also addendum circle radius so rA1 in that case we have consider this rA2 this is gear 1 and as well we have consider this is the driving gear and this is gear 2 this is the driven gear. We have shown the same number of teeth here so these dimensions will be same, but we will while we are analyzing this we will consider that these are different. And then the theta 1 is the angle of rotation of this one and omega 1 is the speed of this gears and these two volume I shall discuss during the analysis. However, this rho 1 is the distance between the contact point to the centre and you should remember this rho 1 and rho 2 is varying as the gear rotates because this contact point is along the line of action and it is therefore when it is contact point moving from one side to other side then definitely this rho 1 and rho 2 are varying. Now we have made some assumptions these are the fluid is incompressible because to consider the change in area we have to if we consider the compressible then we have to consider that not only due to this area change, but also due to the compression the volume may change I mean the total quantity of the fluid may change. So, if we consider incompressible then we need not bother about that. Fluid leakage are negligibly small and neglected also components of the pump are rigid and inflexible that we have to consider although the in some analysis this pipe is considered the flexible that will complicate this analysis. Now the important thing is that to analyze this as I told that these three points will constitute the chamber. Now when this point is moving in this directions due to this clockwise rotation of that then this point is moving, but what you find that after a small amount of rotation then this chamber will be connected because this path will be open then immediately at that instant we have to consider this contact point and this point. So, this we have to be careful or we should take care while we are considering such area that there should not be change in certain opening of this area with when this one pair in is contact maybe when the second pair is coming in contact then again the same phenomena is there. If within this contact zone if one teeth come I mean the open passage is open then analysis will be complicated that also we have considered that means one when one contact one pair is contact is there then same teeth is here only this point is moving on the wall. Now if we consider the hatched area so that area if we can calculate that area including gears that means the metal part and then fluid part. If we consider the variation of this area with respect to this contact point that definitely will give us the variation of area during the rotation and from there we can find out the expansion and compression. At a particular instant of time the boundaries of this chamber define the control volume we consider this as the control volume. Since the fluid is incompressible the total volume entering the discharge chamber must equal the total volume leaving the discharge chamber that constitute the control volume. Then we consider dVd is the flow out and we for the analysis we consider a very small amount of that is crossing the boundaries of the control volume at a particular instant of time. The material within these volumes is irrelevant since everything is considered to be incompressible. However the infinitely small volumes generally consist of both gear pump material like steel and fluid. In this case we have considered steel might be something else also so but the thing is that one is that material solid material and another is the fluid material. The import volume to the control volume from the driving gear that is gear 1 is denoted by dVi1. So when this is very small amount say d theta 1 rotating what we find the dVi1 is going inside this control volume that means within this boundary what we have considered. This boundary itself is varying but we have to consider in terms of flow in and flow out the volume in and volume out. A certain amount of gear volume is also leaving the control volume. This existing volume from gear 1 is denoted by dV01 whatever volume is going out. Similarly the entering and exiting volumes from gear 2 are represented by dVi2 and dVo2. Since the import volume is equal to the discharge volume the dimensionless governing equation for this problem will be so dVd where we have considered non-dimensional parameters. So input minus output and here also input minus output. So total this volume whatever this volume will be there that must be going out. If there is no difference there will be no volume output. In fact in case of if we consider simply the bucket type of pump bucket wheel type of pump then this will be 0 both will be 0. There is no such expansion and compression in the control volume but in this case it will be we will find out that. Where this input we can take if you look into this figure half rAi into rAi sin d theta1. So sin part of this one d theta1 is the volume into W that will be the volume displacement. If we consider this infinitely small amount of shaft rotation so sin theta part will be something like this here that part into this length which is rAi into half of that we have considered the area of triangle and then multiplied with the width of this gear W. This is simply volume area of isoscale triangle and width. Considering angular rotation to be very small now we can assume sin d theta1 is equal to d theta1 and in that case we can express the dVi1 is equal to this much. Now if we non-dimensionalize this volume we will get dVi1 is equal to half into rAi hat square into d theta1. Similarly we have also the relations found out in the same way. You can look into the equations these are same. In that case in when we are calculating the volume out we have considered this radius because this is the radius here. We have to consider a small amount of rotation and then this triangle formed by 2 arm is equal to rho1 in this case rho2. So, this will be the equations. Now we can rewrite this equations. Here subscript 1 and 2 denotes the gear 1 and 2 we need not mention already we have mentioned it and rA is equal to the addendum deep circle rho is the instantaneous contact radius theta is the angle of rotations that we have already mentioned. And then from the fundamental law of Gehrim now we have written this 4 equations we will before substituting this into the main equation. We shall now from the fundamental law Gehrs we know again this relation is that d theta2 is equal to rPi by rP2 d theta1 and where rP indicates the pitch radius. Substituting equation 3, 4, 5 into equation 1 we get this relations. So, remember the first equations and then all the equations we have derived so far we will arrived into these equations. Now in this derivation they have considered the pitch circle, but I prefer normally or anyone can prefer just directly the base circle. You can consider the base circle instead of the pitch circle, but considering the pitch circle is that when on this point will come on the axis on the line joining the two centers then this rho1, rho2 will be equal to the rPi and rP2 remember that fact. So, maybe this will be better to use at this moment, but anyway this will give the same relations. Now introducing theta1 is equal to Tw1 time into omega1 and that that at theta1 is equal to non-dimensional. So, we can consider that this is this itself is dimensionless time the angle itself is the dimensionless time. This is a tricky things, but one can consider in this way because d theta1 will give you dT1 rate of change of time if any there. So, this expression you have to visualize this. Now we have that mean non-dimensional time is equal to simply the angle of rotations. The final equations of the theoretical flow rate of the gear pump in dimensionless form is obtained as. So, now what to find out the flow rate what we do? We do the with differentiate this volume with respect to the time and then we will find this final relation will be in this form. Then now in this equation what we find at any instant say this is a fixed dimension fixed parameter this is also a fixed parameter. Once the gear we know the centre distance these two are also fixed parameter only with time this rho1 and rho2 are varying. So, we have to calculate now these two parameters once we can calculate we can easily calculate this one the volume displacement. To determine the instantaneous flow rate of pump the instantaneous length of action is to be found out from mesh geometry. That means if you will go back to this figure that we will find we have to find out the instantaneous length of contacts means from the pitch point to the actual point of contact that length we must know. Also we have to find out rho1 and rho2 and then it will become very easy to estimate this flow rate. Now we consider this geometry of this gears in this gear say this is the driving gear we have consider also the contacts points. Now what we find that first of all we must consider the coordinate system to analyze the gears in that case what we have considered that the midpoint of the teeth will be on this axis when theta is equal to 0. No not midpoint sorry I have made a mixed case this is the pitch point, but we have considered that when one teeth is coming in contact first contact that is that theta is equal to 0. Now we have shown that here the first point of contact and then we have also shown some intermediate point of contact. Now what else are shown here that from the midpoint of a teeth say this is the angle theta and beta1 is angle from that line to the point of contact instantaneous point of contact and then theta1 is the angle of rotations. These we have considered and we have considered two axis system which I am explaining in coordinate system and mesh geometry X and Y capital X and Y the fixed Cartesian coordinate system that is the fixed reference and this angle and it is oriented by this angle with the line joining the center of gears. That means what we have considered that when this first contact point is there we have taken the axis through the midpoint of this one which is oriented at this angle and this coordinate system capital X and capital Y is a fixed coordinate and then we have considered the another XY coordinates. However, this will be determined later we will show that how it can be determined and XY is in the rotating coordinate system which is fixed to the driving gears. This X and Y that coordinate is fixed to the driving gear now theta is equal to 0 theta1 is equal to 0 when these two coordinates are coinciding that we have to remember. It is shown that while the tooth mesh begins this X coordinate passes through the midline of the driving left gear tooth in contact. The rotation XY defined by theta1 theta1 is the angle of rotation of the driving gear. At theta1 equals to 0 X axis coincide with the capital X axis small x axis that way I have already explained. Now using the law of cosine and the geometry with reference to figure 4B which means that we have considered the intermediate point of contact some point of contact. It can be shown that the instantaneous radii of tooth contact between two meshing gear is given by this rho1 is equal to L square plus rp1 minus 2 rpi L sin alpha. Now this derivation is shown here you can see this you can write down this square must be equal to rpi minus L sin alpha square plus L cos alpha square. Now if we simplify this we will arrive into this form which we have non-dimensionalized and used in this equation. This is not difficult to find out. Similarly, rho2 square will also will be expressed as like this, but remember this sin. While alpha where alpha is the pressure angle and L is the instantaneous length of action from pitch point P to the current point of contact that means this is the pitch point and this current contact is defined by L. Now if we substitute 8 and 9 into equation 7 we will get this equation in this form. Here comes this L square. So the instantaneous point of tooth contact which always lies on the line of action can be defined in polar coordinates by radius rho1 and theta1 with respect to x and y. Now the equation of line of action with respect to x y coordinates system is expressed as y dash sin alpha minus theta1 the angle of rotation minus this. If you just look into this geometry and you will arrived into these equations is equal to 1 minus x dash cos alpha plus theta1 minus zeta. This is the equation of this line of actions. Then from the gear geometry the base circle radius Rb1 is equal to Rp1 into cos alpha that is known. Now keep in mind that alpha in this case what alpha we have considered that is the working pressure angle not the actual pressure angle or the standard pressure angle that you have to remember. This means that while you are going for this analysis you have to know that what centre distance they have made and the gear correction if they are produced and from there we have to find out the actual working pressure angle that we can put. But in normally in case of this gear pumps we do not change the centre distance. So even if the corrections are introduced there the alpha remain constant that is remain standard one. The instantaneous truth contact point x is given by this, y is given by this, this is simple. Then we substitute this in equation 11 and rearranging the contact point is defined by this equation. Now again from the geometry of involute truth profile beta1 is defined as beta1 is this angle. This can be defined as by this. How? We will see in the next slide but we should look into this psi1 angle is pi divided by twice z1. Pi divided by twice z1 means we have taken this angle not this angle this psi1 is somewhere it is not shown in this figure but this is pi by twice z1. Pi by z1 is equal to the one half the pitch and this is half of this truth angle. So we have to show this one this angle to be defined it is given here. Equation 12 and 13 the earlier equation must be solved numerically for a given theta1. These equations you can you can see that there is no solution direct solution we have to solve it numerically. However this is also analytical solution will be available we will come into that but let us see this how this expression how this is derived. Now referring to this figure you can see this now you can see this angle here this angle is given. So this angle is this much this is we have considered this angle that means this is the half of the teeth this angle is the half of the teeth angle. Now here these angles are defined and then we can write down this equation this O is equal to R B into 1 plus this angle this is the R B and this is the angle. So we can define like this. Now next beta is equal to from the geometry you can find out this and then transforming these into Cartesian coordinates we can write down these equations and also R B is equal to rho cos gamma and this is equal to this is from the property of involumetry we can find out this also and then after substituting all this equation together we will arrive to this equation which we have used as a equation 13. Now next referring to the starting tooth contact point to this figure we will now find out the starting tooth point contact point. Now this coordinate again is given by this one if you look into this geometry we can have this equation. Now here S indicates the starting point this rho 1 at starting that is denoted by rho 1 S as well L S is the length of instantaneous contact point at starting and the starting length L S is expressed in this form from the geometry and then from equation 15 and 13 angle zeta angle xi now can be derived with respect to the starting contact point parameters and this angle is equal to is expressed by these equations. So now we can derive what is this angle if we know this angle of contacts this we can calculate this angle also where this rho 1 S is equal to rho 1 which are already I have defined. Now instantaneous length of actions again look again to this figure what we find sign of this angle is equal to given by this is of course in the dimensionless form. So this we can find also this L length we can find out from these equations which we have re-arranged and once we can calculate this we can find out this also. But still remember these relations are non-linear hence numerical solution to these equations will yield the most accurate results for instantaneous length of action. However a closed form approximation to the solution is proposed by this one ring this you will find in the reference which will be discussed in the next lectures. Now I would like to say this all such analysis were not available till dates although the gear pump may be of being used from the 1950s or so. But this analysis only recently this analysis was done which really showing how this area can be calculated. Anyway we will discuss this in the next lecture this paper is important if you have time then you can go through this paper unfortunately this year is not given. But I think it is in the 2004 or may be 2005 only this was this research work was done. However we will discuss also there is another method to find out such volume in the next lecture. And what we find that you can also go through this papers if you have interested if you want to learn more about this. But if you read this and these two paper you will mostly understand what I am trying to tell you and how to find out this volume displacement. This will be continued in the next lecture and our and I thank you very much for listening.