So, we stopped here last class where we derived the energy equation across the discontinuity representing the reaction zone and we wanted to see how this looks like in what is now beginning to shape up as goth's called as a Hugonio plot. So this is what you would actually call okay to some extent a simplified Rankine Hugonio equation. So how do you look at this curve in the Hugonio plot, the Hugonio plot is a domain of P infinity but 1 over rho infinity and then you begin to look at how this looks like you now have a coefficient ? over ? – 1 times P infinity divided by rho infinity.  So this is beginning to look like if you are now looking at like an XY plane you are now beginning to look at a curve that is kind of like some AXY plus BX plus CY plus D equal to 0 right. So if you now have like a leading order term like XY what would that actually signify if XY equal to 1 if that is what you had okay that is like A is equal to 1 and D is equal to – 1 then you would simply and B and C are 0 in whatever format that I just template that I just explain you would be looking at a rectangular hyperbola but since you also have a P infinity divided by rho not term and a P not divided by rho infinity term we also have some B's and C's that are non-zero therefore it is not exactly rectangular hyperbola but it is going to be modified so it is going to be like a hyperbola here. So hyperbola is going to now look like if you now had something like that. Now we do not know exactly if it is going to be a rectangular hyperbola in this P infinity 1 over rho infinity X domain it is very likely that you could have this cross over to lower value okay. So I am not saying that this exactly is the curve at the moment but what we are trying to do is to sort of construct this curve as we go along okay well why did we want this curve because we wanted to solve these conservation equations across this continuity to locate react product properties given the reactant properties including U0 which is something that we are not very sure that we know to begin with and U0 is actually embedded in the slope of the Rayleigh line but the Hugonio so this is what is now called the so this is rectangular hyperbola is what is called as a Hugonio curve and the Hugonio does not involve any m dot in other words it is like a purely thermodynamic curve okay. So the flow information is actually buried in the Rayleigh line which is a mixture which is a combination of continuity and momentum together so that carries the flow information whereas this the energy the thermal energy equation effectively is not covering any flow information there and then we are what we are looking for is a intersection of these two the these two curves that is a Rayleigh line and the Hugonio curve and sure enough we find that there are a couple of intersection points which will now tell us that if you now started out with the P0 and 1 over rho 0 as the reactant conditions and this particular point is typically referred to as what is called as the origin of the Hugonio plot and then what we want to see is where is actually the ma the conservation equations where are they satisfied again right so what you will find is that you now get like about two points where they could be satisfied so you have like at least two possible solutions this is a great progress that we have done starting from when we did not do when we did not have any equations to solve we had like an entire plane to search the solution and then when we now had the Rayleigh line we now reduce the possible set of points to a line rather than a plane and since you now also have the energy equation in the form of the Hugonio curve we look for point point of intersection of course you are now beginning to see like about two points of intersection instead of just one right so we now have this look at what are the different possibilities for these two points of intersection whether it be only two or whether it be whenever it be one and if it is two what are the possible two values and so on right obviously this is a this is a curve where which has the same same sign for the curvature that means it is always like a like concave facing upwards and side upwards to the right therefore since the curvature does not change sign you do not have any points of inflection in the Hugonio curve when a straight line intersects with this you can have up to two intersection points you are not going to expect more than that right but you could look for one intersection point or none when would you get one intersection point if the Ray if the Rayleigh line were tangent to the Hugonio right or the Rayleigh line never really intersects with the Hugonio then you do not have any solutions right so there is some hope we are now looking at zero solutions one solution or two solutions not more than that good so with this picture we now see how to go about constructing this plot little bit more carefully right the first thing that we notice is the first thing that we notice is if your Q is greater than 0 okay so for Q greater than 0 the Hugonio curve passes to the right and above the origin that is there okay for Q equal to 0 you could also check this if you want by plugging in values for P infinity and rho infinity for Q equal to 0 the Hugonio curve would pass through the origin you see right now what is Q Q is essentially negative of what is called as H infinity not minus H not not and H infinity not and H not not or in turn sigma i equals 1 to n delta H of i yi not delta H of i not and sigma i equals 1 to n yi infinity H a delta H of i not right so it is only because of the compositions at yi at the not condition and the infinity condition the delta H of i not is the same for both because it is a standard heat of formation right so we are now getting these different hi not and hi infinity and then the difference between the two is essentially the chemical heat release with a negative sign okay so that is a chemical heat release if you Q is equal to 0 that means you are not having any chemical heat release so the chemical heat release is not there essentially we are talking about a non-reacting mixture right if you have a non-reacting mixture with a Hugonio curve that is somewhat similar to what you might have gone through in a basic gas dynamics course with the where you are still looking at as a matter of fact if you now think about how we had this equation we had like a H not plus half u not square is equal to H infinity half u infinity square that is exact without the Q right and that is exactly the same as what you would call as an adiabatic energy equation that you would that you would write for a energy conservation across a shock okay so across a shock you would find that the stagnation temperature remains the same the stagnation enthalpy particularly remains the same right that that is true when you do not have any heat release okay or any chemical heat release. So a curve that actually passes through the origin with Q equal to 0 is what is called as the shock Hugonio right now that tells us that if you now have a Q not equal to 0 and we expect typically the Q to be greater than 0 rather than less than 0 we are not looking at cooling kind of thing right so what is it when you now have a wave that is reacting in addition to being a shock right so you are now beginning to think away and then of course a shock is actually either facing a supersonic flow or it is traveling at supersonic flows this is supersonic speeds relative to like still reactance what is it I mean this is not something that we thought about we started talking about something like low Mach number conditions and all those things but now we are getting into something but of course in what we are doing now we are not really making the low Mach number assumption right we should be now game for any kind of Mach numbers that are approaching your wave but the language that we are beginning to use is now beginning to now admit the possibility of supersonic flows right okay when would that happen so let us just do this a little bit more carefully let me let me declutter this picture of all the terminology that we have that we are now used to and we just want to now redraw this only with things that we want to now take take so let us say you now have the origin again I am not going to name it now and let me first draw the Hugonio line a little bit farther away and let us now say that you go here and here and let us see what happens here your Rayleigh line can have only a negative slope right so if you now if this is your origin of the Hugonio the Rayleigh line can fill only the third quadrant sorry the second quadrant or the fourth quadrant right it cannot go like this so you cannot get into the first quadrant you cannot get into the third quadrant so this part of the Hugonio is never going to be used right so we now have to erase this part away from consideration we just do not want to worry about that part of the Hugonio at all we will worry about a part of the Hugonio that starts from where whatever rho infinity is equal to 1 over rho 0 okay another part of the Hugonio where P infinity is equal to P 0 what does that mean if you were to now have a solution that is somewhere here that means your Rayleigh line is sort of passing like this it means two things one your final pressure is about the same as the initial pressure okay the pressure hardly changed when the flow went past this wave second this Rayleigh line has a almost like a zero slope that means the wave is not moving much so it is possible that if you are talking about nothing yes it is like you do not have a wave so obviously the pressure will increase yeah that cannot be true because the density increase the density decreased so there was a density decrease that was primarily because if the pressure remains the same and the density decreased that is because the temperature should increase so you had a heat addition all right that is there is a reason why the Hugonio is shifted away from the origin in the first place so you had a heat addition correspondingly you had a temperature rise but you do not really have a wave that was moving fast enough and you did not have a corresponding pressure rise this is now beginning to look like what we were talking about for what is called as low Mach number conditions right the pressure is approximately a constant right so low Mach number simply means that the the wave is not moving too fast or relative to the wave the reactants are not moving in too fast right therefore the Mach number is very low so you now have one branch of the Hugonio that is now beginning to correspond to waves that are moving kind of very slowly okay and the fastest wave could possibly have a slope that is tangent to this at this point that is the fastest wave that is possible that is the highest m dot that this branch of your Hugonio will intersect okay if you for any faster waves the Hugo the Rayleigh line is now going to actually go downward and fail to intersect the this branch of the Hugonio will that intersect the other branch not not not immediately you now see that you have to actually go to a still steeper Rayleigh line that will begin to intersect the other side or if you are thinking about an experiment where you want to conduct a stabilization of a wave by sending in reactants at faster and faster flow rates right between this particular flow rate and that particular flow rate you just cannot stabilize a flame at all and then you again begin to stabilize flames for all Rayleigh lines that are now having steeper slopes than this you can have points of intersection until you get to this point which has an infinite slope right it is kind of again just like how a 0 slope is almost impossible and we are thinking about like very low velocities right infinite slope again is not something that we that we that we want to think about but very high very essentially very high slopes that means very high velocities for which the pressure jumps but the volume did not change a lot so this is more like a isochoric process that is a what you started out with was the isobaric process right so in the one extreme you started out with an isobaric process corresponding to very slow waves right and then you could think about a this little bit faster waves but with a certain decrease in pressure not a whole lot of decrease okay from here to here you see that the pressure is decreasing only a little bit okay but the expansion that is the density decreases as you go this way 1 over density increases so density actually decreases there is a significant amount of expansion that is going on right from the beginning for because of the heat addition and then nothing until you get to very fast waves and you can now get these very fast waves to go all the way to the other limit where you are getting close to isochoric process that is like a constant volume situation right and there again because of the heat addition you hardly have any change in volume but essentially the pressure increased right so what you then have is this is what is called as the lower branch of the Higoneo and this is what is called as the upper branch and we want to call these points the detangents the tangent relays intersecting at these points as the LCJ point and this is the UCJ point Cj stands for Chapman Juge so the LCJ indicates the highest velocity of what is called as deflagration waves so this is the lower branch corresponds to what is now called deflagration right. So LCJ the fastest deflagration wave and the upper branch of the Higoneo corresponds to what is called as detonation and UCJ corresponds to the slowest detonation wave right and the m dot UCJ is still significantly greater than m dot LCJ that means the slowest detonation wave is still much faster than the fastest deflagration wave that simply means that you now have two classes of two classes of waves that are significantly apart there is really no overlap at all in their wave speeds and in fact what I would actually what I should actually say is this would be much greater than so in terms of wave speeds like instead of writing m dots if I were to write you not assuming like they start out with the same row not right what this now and see I told you that I do not even I do not know what the u0 is right I am still not going to know the u0 until the end of this exercise but I am beginning to get some ideas about what the bounds of my u0 should be for two different kinds of processes that we are now beginning to think about namely deflagration and detonation right so what we are talking about here is for detonation for deflagration waves your u0s of the order of a few tens of centimeters per second under laminar conditions but your detonation waves so to a few tens of centimeters per second is still less than a meter per second but your detonation waves are typically of the order of a few kilometers per second so now see that there is like about three orders my three to four orders magnitude change difference a factor of three three to four orders magnitude difference between the detonation velocities and deflagration velocities right what we should further notice then is that we want to think about the deflagration is going as subsonic waves where as the detonations are going at supersonic waves but before we do that we want to now quickly go through some mathematical properties because we started out with mathematical equations for these like this one and then we notice that that that goes like a hyperbola and then we of course had a simpler mathematical expression representing a straight line very very easy to see this for the Rayleigh line and we are now looking at these intersections and so on what I would like to first of all before we proceed further is to give you expressions in these are not very difficult for you to derive and these are all like typical exam exercises okay so you can figure out for example depending upon the Q what is this point okay the lowest value of 1 over rho infinity possible for the detonation branch and where is this point for example what are the coordinates of your LCJ point that will also depend on Q and what are the maximum so what is the maximum value of 1 over rho infinity that is possible right and what are the maximum values of P0 that is possible that is of course equal to P infinity and what is the minimum value of P infinity that is possible what you will find is you will you will notice if you now try to actually look at the mathematics of this this would ultimately go and intersect the x-axis for the abscissa right and where it intersects is where you would get the maximum value of 1 over rho infinity and the corresponding value of P infinity would be 0 okay as a matter of fact what you will find is this rectangular hyperbola asymptotes to a value of P infinity that is negative so what you can actually do is to go here and find out the horizontal asymptote for this rectangular hyperbola and you should find that that is negative and since you cannot admit negative pressures in reality you do not worry about where exactly it asymptotes and you have to stop thinking about it at where it hits the abscissa that is corresponding to your P infinity equal to 0 and then there is like a corresponding 1 over rho infinity that is a maximum value to which the proper the products can expand right. Similarly you can look at the upper branch and find out that the lowest value of P infinity it can take will depend on Q but the corresponding 1 over rho infinity will be equal to 1 over rho 0 right then it can also locate this point the ucj ucj and then where does it go so what you will have to find out is there is a vertical asymptote which is having a positive value of 1 over rho infinity that is the lowest 1 over rho infinity that it can take but it is now going to asymptote to that that means the maximum value of P infinity that can take is infinity when it hits a asymptote eventually at infinity right. So these now give you limits for these pressures and densities that it can take in these different branches of the Hugoniot curve so we can go through that but I think even before that we can point out that so in the in the deflagration branch in the deflagration branch P infinity is less than or less than or equal to P not right and 1 over rho infinity is greater than it is certainly greater it is not greater than equal to because it would be greater than or equal it would be equal to 1 over rho not only for q equal to 0 for positive values of q you can never get your 1 over rho infinity to be equal to 1 over rho not so it is always greater than 1 over rho not for q greater than 0 that is that is rho infinity is less than rho not the density decreases and the pressure decreases okay for q greater than 0 you always have T infinity greater than T not this is heat addition so the temperature increases right so with the temperature increasing density decreasing is not news but pressure decreasing these two together then given this means you now have the products go through an expansion right so it is when you have an expansion is when your pressure decreases and your density decreases even with the temperature increase right so a deflagration branch corresponds to an expansion wave right on the other hand if you now look at the detonation wave in the detonation branch your P infinity P infinity can never be equal to P not for q greater than 0 right only for only only for q equal to 0 will the he going to pass through the origin and you have only one tangent relay right but for q greater than 0 P infinity is always greater than P not for q greater than 0 which implies again T infinity is greater than P not you always have a heat addition that you are having in mind in both cases right T infinity is greater than P T not all right so P infinity is always greater than P not and 1 over rho infinity can be less than or equal to 1 over rho not right so the highest value of 1 over rho infinity that you can hope for is this right that is equal to 1 over rho not all of the 1 over rho infinities are going to be less than that and this implies that is rho infinity can be greater than or equal to rho not that means you now have a density increase corresponding to a pressure increase accompanying a temperature increase when the temperature increases you would like to think that the density should actually decrease but if the density decreased sorry if the density increased accompanying a temperature increase that simply means that the pressure increased a lot more right sure sure enough you look at the way the curve goes it is going crazily in the along the pressure axis right so the pressure is obviously increasing a lot so taken taken together all these things mean that we now have a compression wave right so a detonation basically corresponds to a compression wave so we talked about a shock you go Neo we are now beginning to look at one of the branches actually corresponding to compression so it is all kind of going together you can see that there are elements of what we have done before in gas dynamics beginning to look like a special case of what we are what we are talking about right it is a special case in two ways one first of all it is a non-reactive case okay what we went through in gas dynamics is basically a non-reactive case as opposed to a reactive case here and the moment you have reactions then it belongs to only one part which is like the supersonic propagation as we will see and show corresponding to detonation whereas there is also another part which is corresponding to deflagration that subsonic propagation and where expansion happens rather than compression all right let me also say one more thing so now that we have also acclimatized ourselves with some more here we could further actually divide this into five parts I should say four parts rather so you now have you can you can now divide your upper branch as something that is above the UCJ and something that is below the UCJ you can you can divide the lower branch as something that is to the left of LCJ and something to the right of LCJ so let us now call this call this region 1 this is region 2 this is region 3 and this is region 4 so we essentially are looking at the upper branches a detonation branch the lower branches the deflagration branch so what we would like to call this as region 1 at the moment they are just names we just do not still understand them more completely but let us just give the names first and then start understanding because that will kind of aid us aid us in understanding so region 1 is essentially what is called a strong strong detonation and then of course you now have CJ detonation and then we have region 2 what corresponds to your weak detonation and then we have region 3 this corresponds to a weak deflagration and then you have the CJ deflagration and then we have region 4 that should by now be easier for you to figure that should be called strong deflagration for you to remember this at this stage let us just say that when you are now looking at a region that is above the CJ point here the UCJ point here what that simply means is we are now looking at a solution so that means your your your Rayleigh line has to actually pass somewhere there and then you are looking for a intersection somewhere there right that means the Rayleigh line is obviously steeper than the tangent really there that means the wave is propagating faster there right and it is now leading to a very high pressure and a very very high density that is a low 1 over rho infinity and a high P infinity say a very low 1 over rho infinity means a very high rho infinity so this is actually corresponding to a lot of pressurization and lot of compression so that is why we are basically saying strong but it is going to be a little bit more than that we will be able to see what is the propagating Mach number for the wave and what is the downstream Mach number for the products it is based on that that actually we now give this normal nature that this is strong detonation but at the moment you can clearly see that you this is pretty strong in the sense it is having a lot of compression whereas a weak detonation is just as fast if you were to now look at a solution that is here it is Rayleigh line is pretty steep as well just to speak just to steep as well so it is pretty fast all right but it is not leading to a very high pressure and the departure from the origin is not not a lot in the 1 over rho infinity axis so your 1 over rho infinity is not a lot lower therefore the rho infinity is not a lot higher therefore you are having a smaller amount of compression when compared to the strong detonation so relatively speaking we would like to think of this is a weak detonation okay what we will actually find is when you now have a flow that is going through a detonation wave just like in a normal shock you should expect that if the flow is approaching the wave at supersonic speeds if the strong if the shock is very strong you should expect the flow to actually become subsonic all the way that is a mark of its extent of compression so if it has been compressed a lot it also gets diesel the flow also gets decelerated right so a lot of compression would mean a lot of deceleration all the way down to subsonic levels whereas in a weak detonation you expect that the downstream product flow is still remaining supersonic although not as supersonic as the upstream reactants there is the deceleration all right but not down to subsonic levels correspondingly the Cj detonation should mean that of course you are going through a supersonic reactant flow reaching up to the wave but the react the products correspond exactly to sonic conditions that is what is dictated that is what is denoting a Cj the detonation similarly if you now begin to look at this side what it means is in a deflagration we are looking for an expansion so when you now looking for an expansion that means the pressure should decrease and the density should also decrease if you now look at this region the pressure decreases are not a lot and the density decrease the density decrease or the or should say 1 over rho infinity increase is also modest therefore a rho infinity decrease is also modest right so you are now not having a fairly large decrease in P infinity and rho infinity and therefore you are expecting like a weak deflagration okay. So you are now looking at a Rayleigh line that is corresponding to a velocity that is still lower than the Cj deflagration right it is slower and so correspondingly you are not really expecting a large decrease in pressure and density but for this low deflagration you still can intersect over there so this wave is just a slow but it is actually causing a large decrease in pressure and a correspondingly large decrease in density right that is a lot of expansion that is going on and that is why we want to term it as a strong deflagration right. So when you now have a large expansion that is going on proportionately you are now expecting the flow to accelerate so when I react and flow is actually going through a deflagration wave it tends to accelerate and this is actually common practice so in all these things we always try to relate to what we already have experience with so what I would like to suggest is when you are dealing with detonation waves you now recall what happens in gas dynamics what you have learnt in your one-dimensional gas dynamics from your undergraduate let us say if you have gone through a gas dynamics course in your undergraduate level you can recollect what happens across a shock and try to modify your thoughts to accommodate reactions and so on when you are thinking about deflagration waves this is actually more commonplace in terms of daily experience like what happens in like in a Bunsen burner for example right so if you now have a Bunsen burner and then you have like a cold react and flow that is approaching the Bunsen burner and then it goes through the flame and the flame is like a conical flame and the flow goes through this it now expands and then kind of goes like that okay and then you can also see that occasionally and this is not very difficult for you to think about as a thought experiment or even go back and see if you can do this and see for yourself when you now have like a inclined flame like this in a Bunsen burner and if you have like a like a little particle that is kind of glowing okay so we now see how this glow happens as it as it goes along when it passes through the flame it goes much faster when compared to when it tried to approach so it is like a little spark of some particle that is just glowing you can clearly see that it just goes faster right so when you now have a deflagration wave the flow accelerates when it goes past it because it is expanding right now while you are here you are not expanding a whole lot and that is because it is a weak deflagration and therefore you started out with the flow approaching you at subsonic speeds and you are accelerating to further faster but still subsonic product speeds that is the mark of weak deflagration but when you now go to go past the LCJ over here to the strong deflagration what that means is you are now having a very large amount of expansion that is going on correspondingly there must be a very significant amount of acceleration that goes on to the extent that we end up finding that the products actually have supersonic speeds by the time they are getting past this wave where although they started out with subsonic conditions for the reactants correspondingly what just like how we found in the case of UCJ what this means for the LCJ is the approach velocities for the reactants is subsonic all right but the flow accelerates through the wave to exactly sonic conditions beyond for the for the LCJ now if you really think about it this is getting a little bit more fascinating or it is unbelievable fantastic okay how can I have a wave that started out to be subsonic go through a flame and accelerate so much as to become supersonic is it possible for me to do so the answer is we do not come across that at all right so in boons and burners and stuff you do not really get supersonic flows so what is going on the answer there is outside the picture of what we are talking about we started out with the mass conservation equation we started out and then we went through the momentum conservation and then we went through the energy conservation we also talked about the species conservation but the energy conservation keep in mind is essentially coming out of the first law of thermodynamics but we never really considered the second law okay and and we will never do that actually except to state that the this part of the wave that is the strong deflagration is not possible if you now start taking second law into account and as a reason why you will never really find a wave starting from subsonic flows going through a strong deflagration to accelerate all the way to supersonic speeds that typically does not happen because it violates second law right and I think I may be able to show that briefly later on but let me also point out that I will try to show first of all the detonation waves all detonation waves in or in fact I should say all detonation waves travel at supersonic speeds and I should also show that all deflagration waves travel at subsonic speeds the way I would like to show this is to consider the Cj detonation and the Cj deflagration I would like to show that the Cj deflagration is supersonic which means the slowest detonation is supersonic that means all other detonations are going to be supersonic and I am going to show you that the fastest deflagration the C the lower Cj deflagration is going to be subsonic therefore all other deflagrations are subsonic right and then we will also consider the downstream Mach numbers for the ucj and lcj and show that they are actually one thereby we can clearly see how these things are demarcated right but before we do that let us go back and just write out the properties of the Hugonio curve one where they intersect what the asymptotes are and all those things in the next class. Thank you.


Hey everyone, welcome back to Alchemy Stars with P5R in the horizon probably next week. I think they're dropping on July 14th. They are actually giving us a taste of what they're gonna give us once they drop. So first we'll be discussing the skill stacking and weak status because all of them will be having this effect. So exploiting weakness, stacking and weak status, all P5R Orion skills can add or change the stacking of this special effect on enemies when the number of stack reaches 100. The target enemy will automatically enter the weak status. Okay, likewise when an enemy enters the weak status, the number of stacks will be changed to 100. So it's going to be vice versa. Special skill info bring any P5R Orion to the current collaboration campaign into the stages can trigger the special skill. So a dedicated skill button, so a separate button will be added with a cooldown of its own is designated for players. The skill deals damage to all enemies. Take note of that. The exact damage depends on the number of stacks, sorry, the number of stacks of exploiting weakness on enemies, whether the enemy is in the weak status and the number whether the enemy is in the weak status or and the number of P5R Orion deployed. So take note, there are two kickers here, weak status and the number P5R deployed in your lineup. So both weakness stack and weak status of all enemies in the field will be effective continuously. So it takes it takes effect simultaneously when special skill ends, all enemies weak status will be cleared. Okay, performance of the special skill, performance of the special skill will produce the classic all out attack animation from the original of P5R, which is the main traction for players and inspiration for the design. So hopefully this mechanic is is not going to be a gimmick. That is actually my my apprehension for the collab and for new mechanics sometimes or most of the time it's usually a gimmick, but we're going to be discussing Mona here. So Mona is a sniper guys. Let's enlarge this a bit. Forest with subpar stats here. So let's go through Mona's skills. Active skill first here is show your might zero. Cool down is three preemptive strike. Mona enters the battle with its special skill. Celestia tile deals 240 damage to tiles along the way and an adjacent and adjacent tiles inflicts five stacks of exploit weakness to each enemy. And when it's going to be thunder it's going they're going to receive 10 stacks instead. So they have she actually has a what they call this a bit of a grudge with thunder and deals damage to all enemies according to the number of P5R collaboration characters deployed. So that is it deals more damage to enemies with weak status and dispels all weak statuses. When all enemies have weak status a special skill displayed is triggered. I'm not sure if this the special skill that the button is automatically triggered or you have to trigger it by pressing on it. So that will have to see exploit weakness enter weak status after gaining 10 stacks of exploit weakness sorry weak status chain combo he has four nine and 13. So 155, 160, 170 kind of an average in terms of damage deals to nearest enemies within surrounding clusters and inflicts one stack of weakness to each enemy hit thunder enemies receive three so it just increases as you do more chain damage and also the number of enemies increases as well and the number of stacks of exploit weakness also increases. So equipment when Mona inflicts exploit weakness that stacks on enemies there is a 30% chance of doubling the effect and only a 30% chance of that's good kind of but it should be more than 30 kind of around 35 probably or 40 and doubling the effect and a 1% chance of inflicting weak status okay each normal attack inflicts one stack of exploit weakness thunder remains enemies receive two stacks instead when Mona first changes from team member to captain during battle Mona gains 20% increased normal attack damage in this round from team member okay so captain swap increase tends to increase his normal attack damage in this round okay so what do we see here as something good for Mona probably just the new mechanic I'm actually you know I'm excited to find out what it is as per being a sniper um I think I think Mona is going to be just you know your average sniper I don't see any impact um that um Mona will give that much but more or less if you bunch them up if you have several you know several several aurorians with p5r on your team probably this could work because the stacks increases quickly if you have more of them on your team so that is I think what Mona brings at this point um even even the equipment skill is really really um geared towards increasing stacks so that's it so hopefully again the the synergy that I'm looking for with other other new aurorians with p5r is actually it's actually making me curious on what synergy will they have how you can really make them work is is this new mechanic really viable at this point so I'm actually waiting for the others to pop out in terms of their skills but at this point a five star I don't think Mona is going to make that difference probably another gimmick at this point or probably a support in terms of the new mechanic that they have but at this point that is actually my opinion of Mona so we'll just have to wait for the others to drop and let's see what their kit is but for Mona decent enough probably average enough but I don't think uh will be a good standalone probably good in a team of p5r in your team or in your you know in your team of five so hopefully we could get more from the others as they come okay guys so what do you think with Mona is Mona good is the kit good what are your thoughts please put them down in the comment section below and hopefully we could have a good collab and probably I'm looking for actually I'm looking forward to a better mechanism this time in terms of a new bunch of aurorians coming through okay guys so thank you very much for staying this far so again for comments please do put them in the comment section below thank you guys stay safe take care this is the warden and I'm out of here


Hi there, and welcome back to the BSG Primer series. This primer is on the key documents that we use in ITUT study groups. Inputs to the study group and outputs of the study group. Let's get started. Key inputs that are considered during ITUT study group meetings. The first and most important input are contributions from membership. Contributions are written documents that are submitted by member states, sector members, associates, or academia participants in advance of a meeting. They are intended to move the work forward. So it's very important for them to include a clear proposal at the end of the contribution. The contributions are numbered sequentially during the four-year study period within each study group. So you might have a contribution C44 or even C305. The next important input document is the TD. TDs are submitted by a meeting official. That is to say either a member of the study group management team, a rapporteur, an editor, or by the TSB secretariat. These can be posted before and during a meeting. That's very different from contributions. These have to be received at least 12 calendar days before the study group starts. Examples of TDs are agendas, group reports generated during the meeting, the latest draft text for recommendations, or liaison statements. What are liaison statements? Well, liaison statements simply put are communications sent from the study group or to the study group from other bodies. Could be another study group, could be TSAG, or even an external body like another standards development organization. These are called LS, and the LSs always indicate the source of the liaison statement, the question or the body to which it is directed, and the action desired. All of the incoming LS to a particular study group meeting are posted as TDs, as are outgoing liaison statements when they've been agreed by the group. So those are your key input documents. What are the key outputs of ITUT study group meetings? Well, the most important is the ITUT standard, or what we call the ITUT recommendation. Recommendations, ITUT RECs, are international standards that define how telecommunication networks operate and interwork. And usually work on a recommendation starts with the approval within the study group of what is known as a work item, and the development of what we call base texts. Until their final approval. So that's ITUT recommendations. We also have annexes that can be approved separately. These annexes are integral to the recommendation in question, and they contain material that's necessary for us to understand it. So in that respect, an annex to a recommendation is a normative text. It's part of the standard. We also approve at the ITUT, however, non-normative documents. Examples of those are ITUT supplements. A supplement is a document that contains material that is supplementary to a recommendation associated with it, but it's not essential to the completeness of our understanding or implementation of that standard. That's why an ITUT supplement is non-normative and agreement by the study group is sufficient. We don't need to go through the traditional or alternative approval processes. The same thing goes for an appendix. An appendix to a recommendation also contains supplementary material, and it's associated with the subject matter. An appendix always relates to a particular recommendation, whereas a supplement can relate to a series of recommendations. Finally, ITUT technical papers are also an important output of study groups, and these are non-normative, and they provide information on various topics addressed by the study groups. They're free of charge, they involve small editorial overhead, they can cover various areas of study, and some examples of that are accessibility, economic and policy issues as you see here on international mobile roaming, eHealth, mobility, outside plant installation, and so on. So that is a brief primer on the key inputs and outputs of our ITUT study group meetings. Thanks for joining us. This is ITUT, Bridging the Standardization Gap Program.


Hello everyone! A.J. here. Remember in my last video how I said that I would be recording in different locations? No? Well that's because you're not subscribed, so hit that subscribe button, you dummy! And hit that notification bell too while you're at it, please. In any case, last video I said that I would be changing locations across my apartment to see what would work best for me, but I don't want to do that because this is so convenient! This is very convenient for me. So yeah. By the way, who cares about location when we have quality content? So let's get right to the video on how a neural network learns. If you've gotten started with deep learning research in any way, you would have heard of a famous algorithm called backpropagation, a very important phase in the machine learning process. Two things I've noticed while learning this concept is one, lots of math equations, and two, every single person says the same thing but just in different ways. This can be pretty difficult while learning because you can't connect the dots right. And so I'm going to explain neural networks such that you don't have to memorize any book formula, but I'll also use standard notation to ensure that you can map my explanation to whatever is given in any kind of source that you may read for neural networks. To start, let's not just jump into forward and backpropagation. There are hundreds of books out there that can tell you how to do exactly that and I will get to that in a bit. But for now, I think it's more important to understand the concept of how neural networks actually learn and also how to define an objective. Just a disclaimer, this video has a lot of technical details, so I may be throwing some words around without explaining them. If you find a word or a few words that you just don't get, Google your friend. A typical feed for neural network consists of an input layer of size D, one or more hidden layers, the number of neurons per layer is a tunable hyperparameter, and an output layer, the size of which depends on the type of problem we are dealing with. Classification or regression. Classification and regression are the two broad categories of problems that encompass supervised learning. A statistical learning method where we are given labels Y with training data X. When we say a neural network should learn, we mean that it should determine the weights of edges between the layers and biases for every layer. Since neural networks are parametric models, the problem of modeling the given data is reduced to just determining the fixed set of network parameters, that is the network's weights and biases. So how are they determined? Such values are determined by an estimator, and we consider the maximum likelihood estimator, a core ML concept. I'm illustrating estimation with MLE because it is one of the most fundamental estimators that drives basic algorithms like logistic regression, naive vase classification, mixture models, and can even drive linear regression. Because of its importance in core machine learning algorithms, I'm thinking of making an entire video dedicated to estimators, or at the very least maximum likelihood estimation. So let me know your thoughts about that in the comments down below. However, in this video, I will derive MLE in the context of neural networks. We have N samples X with corresponding labels Y used to train the neural network. How does maximum likelihood estimation work? In the supervised learning context, MLE determines the parameters which maximizes the probability of seeing data. Here, theta represents the neural network parameters, weights and biases, and theta hat MLE is the maximum likelihood estimate of these network parameters. Assuming the samples are independently and identically distributed, that is IID, we can multiply the probabilities for the end samples. We're going to evaluate this function in two ways. The first is when the neural network is used for classification, and the second is when it is used in regression. In the classification setting, the labels can be modeled as Bernoulli variables, as they can only take a small set of specific values, like in the flipping of a coin, only two values can be taken heads or tails, or in the rolling of a die, that's only six values, one to six. Regardless of the value of n, the number of values each sample can take is fixed and is equal to the number of classes. Here, Tnk is either zero or one and indicates the target class, that is the actual value. We get the matrix T by one hot encoding the given labels. Maximizing the likelihood is the same as maximizing the log likelihood. We take logarithms to convert product of products to sum of sums, which is much easier to compute. The negative log likelihood is the error we wish to minimize. For a single sample xn, doesn't this error kind of look like cross entropy loss? You've probably seen this term used as an objective function in classification problems. Hope you understand now why we use this form. Similarly, let's get the maximum likelihood estimation for regression. We have our likelihood function, the range of yn for regression problems is any real number. So we cannot model it as a Bernoulli variable as we did in classification. However, here, let us assume that the conditional distribution of the labels y, given the samples x is Gaussian, with the mean being the estimated label y hat, and the variance being some sigma squared. We then substitute this in our likelihood equation. The product of exponential terms is annoying to evaluate. So let's consider the log likelihood. Removing the constant terms with respect to data, which is every term without the y hat, we then get an equation that resembles residual sum of squares. Once again, I'm sure you have seen the RSS being used as a measure for cost in regression problems. And now you know why I'm keeping the half because it cancels out when I take the derivative with respect to y hat, it will still converge at the same point as it has the same minima. The maximum likelihood estimator is an unbiased estimator. So the expected value of the parameters predicted using MLE is equal to the true value that is theta star. This means that given an infinite amount of training data, theta MLE will converge to theta star. In other words, the estimated values of your parameters will take their optimal values. This is why in machine learning we claim data is king. If you have more data, then the chances of fitting a good model are much higher. I hope you understand now why we use RSS to measure regression loss and cross entropy to measure classification loss. We now need to find the value of theta that minimizes this cost. And this is where neural network training takes place. We'll use stochastic gradient descent for this training. This involves minimizing the cost every time we look at a random sample by using gradient of the cost with respect to the weights and biases. Hence, we need to find these gradients. Determining these gradients and updating weights involves a two step process, forward propagation and back propagation. Let's first introduce some notation for our neural network. xn is a d dimensional input. w superscript l subscript ji is the weight on the edge connecting the ith neuron in layer l minus one and the jth neuron in layer l. blsabi is the weight connecting the bias neuron with the ith neuron in the layer l. And here we let n subscript l represent the number of neurons in layer l. We can consider you as the inner product between the output activation of the previous layer a and the weight matrix w. And we have an activation a as a function of you, which is typically nonlinear. Through this video, the math may look a tad messy with all the superscript and subscript notation. And this can be easily solved through vectorization that is by converting these scalars into a vector representation. But in order to do that, I would have to make sure that I explain Jacobian matrices and ensure that the vector multiplications are in the right order. And I don't want you to be bogged down by mathematical notation. It will take away from the intuition. So I left it out for this video. We have some special cases in our notation. First, a of zero is equal to x that is the output of the zeroth layer is the input to the network. Then for the last layer l, the activation for the classification is a softmax. But for regression, it's just a linear activation. And third, we have the activation output of the last layer equal to the prediction. As a consequence of two, in the regression case, since there is a linear activation, we can say that a is equal to you. Now that we have defined the notation for any l layer neural network, we can now begin the neural network training process. The first stage of this is forward propagation. We start with a random initialization for the weights and biases, and then compute you and a for every neuron using these two formula alternatively, hence the values propagate forward in the network. This is used to compute the prediction y hat. y hat along with the actual label t is used to compute the total cost for this sample and represented as e subscript n. We consider the total cost e as a sum of the costs incurred by each sample, E n. This is a reasonable assumption. Now that we have the cost, we need to compute the gradient of the cost with respect to the weights and biases. This gradient computation is done with back propagation. There is a lot of toxic air surrounding the back propagation algorithm, specifically it being too difficult to understand without being a math whiz. But I'll tell you now that you really don't need to know much to know back propagation. In fact, back prop is just chain rule. Just like how forward propagation gets its name from the fact that the values of u and a propagate in the forward direction in the network, the back propagation algorithm gets its name from the fact that the gradients with respect to you and a propagate in the reverse direction. And so we determine the cost gradients with respect to you and a throughout the network. And this is used to determine the cost gradients with respect to the weights and biases, which is what we are set out to do. Starting with the last layer, the cost gradient with respect to the activation can be simplified into the difference between the target and predicted value in regression, or the negative of the ratio between the target and predicted value in classification. This can be obtained by simple differentiation. Next, we compute the cost derivative with respect to you for all neurons in the last layer. A well known convention using textbooks and blog posts is to let this gradient be some sort of delta. Note that you don't really need to know this delta convention. However, it makes the entire notation much easier to read, and is also very useful to read in the recursive form that I will describe soon. To compute these delta values for all neurons in the last layer, just apply chain rule with the activation a on the same neuron, we have both terms. So the deltas for every neuron in the last layer can be computed. So far, we have computed the cost as well as the gradients with respect to a and you for the last layer neurons. Now let's compute the value of the gradient with respect to the activation a for the L minus one layer neurons that is a layer before the last one. And how do we do this using a good friend chain rule with you? So in this form, why am I summing over all K? To understand this, we need to know what the term on the left hand side represents. It is the change in the cost with respect to the change in the activation of the jth neuron in the layer L minus one change in this neurons activation affects all neurons in the next layer, all of them. So when the gradients propagate backwards, we can visually see neurons in layer L converge on this jth neuron in layer L minus one. The overall effect of the gradients is the sum of products of the deltas and the corresponding weights on the edge. This is the exact information conveyed while using chain rule. Generalizing this for any layer L, it is just the dot product of delta and weights for the next layer. Similarly, let us try to express the cost gradient with respect to you for the jth neuron using the gradient with respect to a that we just computed. First start with good old chain rule. And that's it. We know the value of the gradient with respect to AJ from the previous step. And the second term is just the derivative of the activation function with respect to you. Like this, we just compute the cost gradients with respect to you and a for every neuron in the network. And that's pretty simple, right? What you see in many textbooks and other references is the same equations but written in the delta notation. Substituting the first and second terms with their corresponding values, we get the following form. Bring the derivative of the activation F outside the summation as it is independent of K. Now the left hand side corresponds to delta j for the lth layer. And lookie here, we have a recursion in delta. In the first case, we express the gradient of ul in terms of al, which in turn is expressed in terms of ul plus one. So that's two separate equations. But in the second case, we express ul directly in terms of ul plus one. As far as complexity, they both do the same thing. It's just the first one is easier to read because it's broken down into two equations or two phases. While the second one has a nice little recursion. Once we compute all the cost gradients with respect to you and a for every single neuron in the network. The next step is to compute the cost gradients with respect to the weights and biases. The derivative of the cost en with respect to wji represents how much this edge affects the cost. We can apply good old chain rule through you j. And we simplify substitute the first term with delta and the second term can easily be computed with differentiation. And what about the bias? We do the same thing apply chain rule with you to get the delta form and substitute the values. Once you determine all the weights and biases in the network, perform the stochastic gradient update and get the new weights for the network. This is repeated for every sample in our training set. To summarize the process, the forward propagation algorithm is just determining you and a for every layer by repeatedly applying these two forms. The back propagation algorithm is just chain rule to determine the cost gradients of you and a. Clearly, you don't need to memorize the formulae to accomplish it. But there are five basic ones that we derived, or only four if we combine two of them together. In case you question the authenticity of this formulae, I'll have you know that these are the same formulae given by Michael Nelson in his neural networks and deep learning book, but in scalar notation. In case of a batch update, consider the simple mean of the gradient for all m samples in a batch and substitute this value in the weight update. There are other topics of interest of how we can optimize these weight and bias estimates much better through like momentum. But I think that's a topic that will be left for a separate video. I don't want to overdose it on this one. Here's the simple algorithm for neural net learning using stochastic gradient descent. Pick a random sample xn compute you and a in the forward direction, find the cost incurred by the sample that is en determine the gradients with respect to a and you from the backward direction, compute the cost gradients with respect to weights and biases, and update w and b using the stochastic gradient descent update rule. We repeat this for a number of epics until convergence. Like I mentioned in the beginning of this video, different authors say the same thing in different ways. If you're going to learn any concept, pick a verifiable source and stick to that. Don't get caught up in the notation. Furthermore, there's nothing you need to memorize here. Everything from maximum likelihood estimation to determine the right cost to minimize to the core concepts of back propagation can be derived entirely by hand as we just did. And that's all I have for you today. If you like the video, hit that like button. If you like videos like this, hit that subscribe button. If you want to be notified every time I upload, hit that bell icon links to blog posts and papers referenced in this video down in the description below. So check them out links to my equipment and other resources are also in the description below. So check them out interested in other hot topics in the field. Click or tap one of the videos right there and I will see you in the next one.


I go now to the second tutorial, which basically continues with this same data set because we are not there yet. There are some layers in the wrong projection. We haven't clipped the data to Malawi and we also want a study area. So we need to define a study area. So this is the tutorial. Just drag it here to process groundwater data. So we're gonna do that one and I'll show you that in QGIS. So let's go to the first step. Well, if you use this tutorial, your first step is to check a video on the theory of factor geoprocessing. That's quite useful because we use a lot of these tools in this second tutorial. But the first thing that we are going to do is take this geopackage with all these layers in it and then going to reproject the layers that are in the wrong projection. And that's the geology and the hydrogeology layers are still in WGS84. So if I hover my mouse on it, you will see that it's 4326. And that one is the same file, same layer, basically, but differently styled. So I can just export it, Save Features S. And I'm going to save it to our geopackage because we have a geopackage now. Browse to the geopackage. And I'm going to call this one geology UTM. And I change here the projection to the one of the project, which is the UTM one. And then I click OK. And now it's added the layer, Malawi GIS data. That's the name of the geopackage. And then geology UTM, that's the name of the layer. If I put my mouse on it, then you will see there the projection and that it links to the geopackage. So that's correct. Now I can copy the style. That's a nice trick. Click right on the geology layer, styles, copy style, all style categories. And I click right on Malawi GIS geology UTM. I go to styles, base styles, all categories. And it takes the same colors then as geology. And I can simply now remove the geology layer. And I can repeat this to have also the hydrogeology in a geopackage. So there are two ways. I can duplicate or I can export it, but let's in this case just export it so you can see it again. It's a bit of duplication of the data, but it doesn't matter. It's not that big. So go here to exports, say features as. Geopackage, choose the geopackage that we have and we're going to store here the layer hydrogeology UTM. Don't forget to change the projection to that of the project to make it UTM. Click okay. And now this one is also copied to a geopackage layer with the correct projection. Then I copy the styles here and I paste it here. There it is. And then I can remove this layer. Click right and remove. Removing layers is not deleting it from the disk or from your geopackage. Simply removing it from the layers list. There it is. We have re-projected these two layers. Let's see if there are more. No, that's okay for now. Then we are going to define a study area. Let me just double check. This one is correct. No one is correct. Aquifers are correct. They're all in the correct projection. So I can just save the project in the meantime to the geopackage. And then we go to chapter number four of the tutorial that is to define the study area. And we want the study area to be an aquifer in Malawi. And we want to buffer to take care of the boundary effects. And we only want the part that is in Malawi and not the part that's outside. So these are a few conditions that we need to process in this tutorial. So let's look at this nice transboundary aquifers data set that we got from Sadek GIP. And that now is in our database, our geopackage. And let's open the attribute table. And here we see all the names of the aquifers. Now I'm looking for a specific aquifer that is called the Shire Valley Alluvial Aquifer. And I can use this button here to select features using an expression. And then I can choose here fields and values. This gives me access to the different fields in the attribute table. And here's the aquifer name column, text files. So this gives that it's text for the FID is in integer values, et cetera. Now here I can click on all unique. That gives me all the unique entries in the aquifer name database field. And here I can look for the Shire Valley Alluvial Aquifer. And I'm going to build an expression here to select from the aquifer name. So I double click and it comes here with double quotes. And I want it to be equal to Shire Valley Aquifer. So that's a way you can compose your expressions to select. And then I choose here, select features. And then I do close. And I can also close the attribute table. Well, you can also show selected features in the attribute table. And then you see that that one is selected. Go back to all. And you see in QGIS, it will make the selection yellow. So this is the selected aquifer that is part of our study area, but we only want the part in Malawi and we also want to buffer around it. But still this selection is part of the bigger worldwide aquifer file. So what we are going to do is to export the selected feature. So I'm going to click right and use this export function again. And I'm going to use here now, save selected features as geo package. I want everything together in our nice geo package where all the data will end up. Click save. And this one, let me stick to the names of the tutorial. We are now in step 4.2. And let's call this Shire, activate this. Shire Valley Alluvial Aquifer. You notice maybe that I'm not using underscores here. It's wise to always use underscores, but this is not a file folder name. This is a name of a table in a database. So here you are allowed to use any character that you want to use. So just to demonstrate that it works, you can do that here. You see that this box is now automatically checked, save only the selected features. We see the projection is correct. So I can proceed with this. And now it has created this new layer, data, Shire Valley Alluvial Aquifer. I see this little question mark here. If that happens, that is a little bug in some versions. And when I made the tutorial and did the trial, it wasn't happening, but now it's happening. But if this happens to you, keep an eye on it, click on the question mark and assign the correct projection here and click okay. So that's how you solve that little issue. Then you're sure that you're using the correct projection. I can copy the style and I can remove the worldwide dataset because we don't need that anymore.  And it only has one feature, Shire Valley Aquifer. The next step is to create a buffer to deal with boundary effects. And you can find the buffer functions under Vector, Geoprocessing Tools, and then Buffer. Make sure that you choose the correct input layer. Here it's correct. And what we are going to use, let me use the same as in the tutorial, is we're going to use a 10 kilometers buffer. So therefore your projection need to be in UTM because that uses meters and that's more accurate than using it in latitude, longitude. And then your distance can be defined here, 10 kilometers. Just as an example, you can use anything that you're used to use in your studies. And then I'm going to save it, a browse to the Geo package, save the file. And the popup will ask for a layer name. I'm going to call this Shire Buffer, 10 kilometers. Click okay. And then I'm going to run it. And then I can close it. And now you see the buffer. So if I put this one on top, you see the original and now you see the 10 kilometer buffer around the vector file. That's great. But that's not really what we want. We also want just the part that's covering Malawi, which is a big country as you see. So the next step is to get the buffer that is only covering Malawi. We're going to create a country boundary. We can of course download it from all kinds of data sets. But here in this case, it is more accurate to use the country boundary from the geology and hydrogeology data set because then it has a better match with our data. Otherwise you might get slivers that you've learned about because the boundaries might be too rough or finer than the ones that we use for geology. So what we can do is use this geology layer and dissolve it that all the features are then into one feature which covers the boundary of Malawi. So that's a function here under vector geo processing tools. And it's called dissolve. And I use here as an input layer, the geology. You can also use the hydrogeology, of course, it's the same. And I don't change the dissolve fields. Normally you dissolve certain classes. The features have the same name of a certain class. Then you can dissolve it and treat it as one feature. And I'm going to create the output, of course, again in our geo package. So we have everything together. And I'm going to save this one as Malawi country boundary. Click okay, goes to the geo package and I run it. Then it's dissolved. And now you see that the whole country has one color. So that is the boundary of Malawi. But then on top of that, we can still see our buffer. So the next step is that we intersect our buffer with the country. We cannot simply clip it. We really need to use a function that is called intersect. Because if we clip it, we also get the parts that are outside of Malawi or outside of the buffered area. So let's do the next step. There's the intersect function here under vector geo processing tools intersection. And in that little video at the start of the tutorial, you can find the definitions in a visual way of all these different tools that you have here. Here we choose intersection. That is to maintain the data that is an intersection of two layers. Here you see the main, the same problem as we had before. So we're going to fix it first. So Malawi country boundary here has an unknown projection which is a little bug. Just point to the correct projection, do okay. And now it's solved. So I go back vector geo processing tools intersection. And there as an input layer, I'm going to choose the shire buffer that we just created, the 10 kilometer buffer. And as an overlay layer, I choose the country boundary. I'm not going to change anything here because that's not necessary here. And I'm going to save the intersection. And I'm going to save it again to the geo package. And that's, you know, tutorial one folder and there I'm going to give it the name. Let's see, that's already our final study area, shire. Let's see how I called it there, just shire study area. So there it is and I run it close. And now you see, here you see the original aquifer. Here you see Malawi, then we had our buffer and now we have the study area. I'm also going to solve this question mark here. It might that you don't have it. I don't know. It's a bit irregular when it shows it up but simply correct it by choosing the correct projection and click okay. So there it is in orange in this case is the boundary. Now we need to clip the other layers to this study area boundary. That's the next step. So we'll first do that again for the geology layer. And there we can use the clip function. So I use vector geo-processing tools clip and I'm going there to clip the geology, this one, with the overlay layer, the study area. Note that these projections in brackets need to be the same. If that's not the case, the best practice is to correct it. Some tools automatically reproject to something that will work but I would not rely on that. So make sure that you always verify here that with overlays, vector processing and use multiple layers that they are in the same projection and preferred, of course, the one that you want to use in the end. I go to clipped, save it to a geo package. I'm going to save this one as shire geology. Click okay. I run it. Click close. And I'm going to copy the styles. There it is. And you see that now we have the geology only in the study area. Now I'm going to repeat that for the other layers. So let's do the hydrogeology in the same way. Vector geo-processing tools clip. You see here that the geology has the wrong projection again. So I'm our unknown projection. So I'm going to set it first. Okay. Also save the project in the meantime. Good practice to do that often. I'm going to repeat this clip. And the input layer is the hydrogeology UTM layer because we want to clip it now to the study area and save it to our geo package. And there we call it shire hydrogeology. I chose here to have all the shire study area layer starting with shire as a name to easily find it. And we see that we have Malawi for all the Malawi layers. So that keeps them a bit separated. Later we'll save it in a dedicated project. So we don't get confused. I run this close, change the projection here and copy the style. There it is. And now we can remove the geology and the hydrogeology that we had before. And we see that we have the hydrogeology and the geology clip to the study area. But we still have our monitoring wells and we still have the Ramsar sites. So let's also clip those and store them into the geo package. Same way, vector geo processing tools clip. So you can see, you can use this for points, lines and polygons that doesn't matter. But you can't use this for rusters that will come in another tutorial. So let's use now the groundwater monitoring wells and we clip it to the study area and we save it to the geo package. And then we call it shire groundwater monitoring wells. Then we run it, verify the projection and we can move it to the top. We can copy the style and we can remove the other one if you hover your mouse over it, you can be sure that you remove the correct one. So this one is called the groundwater monitoring wells. That's the wrong one. So we remove it and we keep the shire one. And we're going to do the same for the Ramsar sites vector geo processing tools clip. Ramsar sites Malawi. And then we use the study area safe to geo package. And then we create the output shire Ramsar sites and then we run it, close it, change the projection and copy the style and we can remove the other one. And then we end up with one Ramsar site, three boreholes, the geology and the hydrogeology of our study area. And we're going to remove the layers that we don't need anymore. So we only have the layers that we use in our groundwater project here. So we don't need this one anymore. I also don't need the buffer and I don't need the country boundary. So now it's nicely cleaned up and I can save now this whole project into the same geo package. So I'm going to project safe to geo package, connect to our existing geo package but I'm going to save it as a new project. I'm going to save it as shire groundwater to distinguish from the other one which had the Malawi data. So I can still use both and simply go to this project in the database to get all our shire study area data in our QGIS project, click okay. And now it's saved. There's still one thing that we need to do as an extra for preparation here. It will be also repeated in the last, or I think one of the last tutorials. We're going to add the geology and hydrogeology attributes to our groundwater monitoring well attribute table. I'm going to show you how to do that for point layers. Therefore we need to install a plugin which is the point sampling tool. So plugins can be installed here from the plugins menu. You go to manage and install plugins. You need an internet connection and it will pop up with this window and here you can use the search box to use the plugins under all. And I want the point sampling tool, this one. Click on it. You can read the description who made it and what it does. And you will find many useful plugins there. And then I install the plugin, takes it from the internet, click close. And it will add this button here to your toolbar. So I'm going to use that to sample on these boreholes, the geology and hydrogeology. So I click the button. And in this window, you can see here, Shire rams our sites, that's not the correct one. We want the Shire groundwater monitoring wells. That's what we want to add it to. And here you need to select which fields you want to add to the new layer that will be created. And we also want all the existing fields. So I select with shift, all the Shire groundwater monitoring wells fields. So they remain there. And then I'm going to with control, select the hydrogeology. Well, you know that these layers are in fact the same. So you can select both fields, hydrogeology and geology from one of those, but here I'll just get them from the different ones. So from hydrogeology, I choose this one. And for geology, I choose this one. And I can give it an output name here, go to browse. You can save it as a geo package, but here it will not add anything to an existing one. That's a limitation of this tool. So better to save it as a shapefile. So we can later add it to the geo package because it will make a copy. So go to the tutorials one folder. And I just choose a new shapefile name there. And let me see what I choose. And we call this one, the wells hydrogeology. Wells hydrogeology shapefile. Now there's also this field step that you can look at before pushing the okay button. Here you can rename the fields of the output layers. On the left side, you see the name of the layer and the name of the field after the column, how it's called. And here you can change this, for example, into hydrogeology or geology. Here we are not gonna do it, we keep it as default. So I can click okay and it will perform it. Click close and there it is. It's a copy and I can open the attribute table and check it. And there I see that it has all the fields of the original groundwater monitoring well file. And it has the hydrogeology and the geology now added to it, sampled from the polygons because these are polygons and we add it by sampling with the point sampling tool to the points. Now there's only one problem here. That's yeah, I'm creating a lot of files and this one is already in our geo database in our geo package. And I want to infect this information with the geology and hydrogeology added to the existing one instead of creating yet another layer in our geo database. So what I'm going to do is, of course, you can in the browser panel, drag this to the geo package that we have and then it will be entered as an extra layer. But I'm going to show you another method where we are going to join the layers. So I want to join the new layers to the existing monitoring wells layer. So you do that by clicking right on the layer, going to properties and then choose joins. Then you click the plus button and then we choose the join layer. So we have the properties of the groundwater monitoring wells and we want to join the well hydrogeology shape file layer. And what they have in common, both are many fields because we kept those, but we are going to use the number field as the common field based on which we are going to join these tables. Now what I also want is to select the join fields because we don't want to copy again of all these fields. We only want to join the hydrogeology and the geology field. And I want to not to have this prefix, it's too long. So I'm just going to remove it. So we don't get any prefix. And then I click okay. And then here, if you expand this, you will see the summary. And when I click okay, the join is performed. I go back here to the attribute table of the groundwater monitoring wells. So I see all the input fields and I see here the hydrogeology field. And they are now in a join, which means unfortunately that if I remove this layer here, that the joints will be gone. So I will lose those fields. So I need to make a copy first and then I can remove the shape file safely. So remember with joints, you don't get the data copied into the layer but you get a virtual connection in memory with the layer that you join. And that's a problem because I want all the data in our geopackage. So what I'm going to do is to make a copy and you can do it by using the field calculator. First I'm going to switch to editing mode. I go to the field calculator and I'm going to create a new field that's by default checked. And I call this one geology. The field is a text field. Keep the other things as default. And I'm going to look up here in fields and values, the geology field. And that's this one, GLG. When I double click, it will put in double quotes here which simply means use the values for each feature from that field. So basically it will create a copy and I do okay. And what you see now is that we have the geology copied. I repeat this for the hydrogeology. Go to the field calculator, name it hydrogeology. It's a text, go to fields and values, double click on this one. So we create a copy, do okay. Now we have the hydrogeology there. To go off editing, click save. So look at this table. It has now the copies. I close this and I'm gonna remove the shapefile. I'm going to reopen monitoring wells. And now you see that the join from the shapefile is destroyed because it's removed but we still preserve the copy and it's in our database. So we can simply save our geo package project. And that's basically the whole thing that I wanted to demonstrate. So now you have a geo package of the whole project that you can use in this virtual project on groundwater mapping.


napa kwa wa sayi di kwa injizu. Wana pa. Ayu, imi dati, anwanya. Tensisha. Tensisha. Tensisha. Tensisha. Nana kwa kwa? Tensisha. Tensisha. Tensisha. Tensisha. Tensisha. Tensisha. Tensisha. Tensisha. Tensisha. We have done this first at the record street because it didn't matter. I do work here for the second part for the first four months. I shall be assisted by a sister. I shall be assisted by Karanja and Murei. Who is Karanja? They are somewhere they are unable to get space. Mr. Karanja can't so come forward and Mr. Murei. You are the Karanja. Where is Murei? I also have Puyo, Puyo. He assisted me and so Puyo Puyo. I am not saying Murei. I think he is unable to get here. Murei, the Agua? I think he is not here. He is not here. I am not in word that he is present. Yes we can have Puyo and Karanja. You are not in word that he is not in word that he is present. I continue to appear for the second part. You are not in word that he is present.


The federal government has ordered the release, the immediate closure of one of its colleges, the federal government college Kuali, located in Kuali area council of the FCT. The director of press and public relations at the federal ministry of education Ben Gong announced this in a statement. This isn't a bit to prevent any attack on the school as there has been the security breach on Shedda and Lombatta villages in the suburbs of Kuali area council.