 So, we have looked at some methods of converting a unit hydrograph which is for a given duration let us say 4 hour duration to some other unit hydrograph which is a duration multiple of 4 hours. So, we have seen that we can use the principle of superposition and add let us say we want an 8 hour unit hydrograph, we can add to 4 hour range find out the total hydrograph which is because of these 2 range of 4 hours duration each and intensity of 1 by 4 centimeter per hour because that is the intensity of the 4 hour unit hydrograph. Now, for the total 8 hour we have 2 centimeters of range. So, after we add these 2 hydrographs up we have to divide the ordinates by 2 and then we will get the unit hydrograph for 8 hour duration. For durations which are not even multiples of 4 hours or the given hydrograph duration we have seen that an S curve can be developed and then any required duration hydrograph can be developed by shifting the S curve by that particular amount and then dividing the resulting hydrograph by appropriate amount. So, we will look at the example where we are given a 4 hour unit hydrograph and we want to obtain a 2 hour unit hydrograph. So, let us look at this table which shows the 4 hour unit hydrograph at different times. You can notice here that the intervals are not regular. So, initially we have data at 2 hour intervals and then we have data at 4 hour intervals. So, it makes the computations a little bit trickier. If we have equal space data then the calculations are little easier, but for this also we can use the method of S hydrograph. The S curve is nothing but addition of a large number of unit hydrographs. So, if we have this 4 hour unit hydrograph let us assume that this is due to the first rain of 4 hours then we shifted by 4 hours. So, all these ordinates get shifted by 4 hours. So, this 0 corresponds to 0 hour here, but 4 hour for the second curve similarly for 6 we will take the value which corresponds to 2. So, any value here for example, this 24 will correspond to 20, 20 will correspond to 16 and 16 will correspond to 12. So, all the ordinates here are shifted by an amount of 4 hours. That means if we have the first rain 4 hour intensity 1 by 4 centimeter per hour then the first column gives the ordinate because of that then we have another rain of 4 hour same intensity, but now starting at t equal to 4. So, we have shifted everything by 4 hours. So, the second column this column represents the hydrograph due to this rain and similarly we can have a third rain which represents this and so on. So, we add a lot of these hydrographs together. So, for the first one 0 and 2 and then 5. So, these 3 ordinates remain the same as the 4 hour U H because no other rain has contributed yet, but when you go to the 6 hours then we have an ordinate of 7.5 hours due to the first rain and 2 for the second rain. So, 7.5 plus 2 gives us 9.5 meter cube per second as the ordinate of the S curve. So, ultimately it will reach a constant value 22.2 the plot is shown here. So, this value and as we have discussed already that this corresponds to a rain of 1 by 4 centimeter per hour occurring continuously over the cachement that the area is a kilometer square. Then we have seen that we can relate this cube as 2.78 a where a is the area in kilometer square and this 2.78 comes because of using the units of kilometer square and hours. So, this will be in hours. So, this 2.78 comes because of conversion from kilometer square per hour or centimeter into kilometer square per hour to meter cube per second. So, once we get this S curve you can notice here that this is not a very smooth curve and what we would then do is do some kind of a smoothing to get a smooth S curve hydrograph and we will use this smooth curve to obtain the unit hydrograph for any other duration. So, for example, here we have listed the ordinates which are after smoothing the curve and then we lag it by the required amount in this case we want a 2 hour hydrograph. So, we lag is 2 hours. So, that means that once we get this smooth S curve from this figure we get the ordinates and then we shift everything by 2 hours. So, we can see here everything is shifted by 2 hours. So, initially we do not have any problem till 14 goes to 16. Once we go to 20 however if the data is unequally spaced we have a problem that we do not have any value corresponding to 18. So, in that case we can take the mean of these and put it here or when we are smoothing the curve we can take ordinates at every 2 hours or every 4 hours whatever we desire. In this case we could have taken ordinates at every 2 hours and written it in the table, but even if we do not have regular spacing of data we can do some kind of interpolation to obtain these ordinates. Once we lag it then we take the difference. Now the difference of these what it means is that if we lag the S curve by 2 hours the difference of these ordinates as we have seen the first S curve is because of a rain which is of infinite duration intensity 1 by 4 centimeter per hour. The second curve is due to a similar rain, but starting at 2 hours and intensity is same 1 by 4 centimeter per hour. So, this means that the difference of these 2 the ordinates shown here will be because of a rain which is 2 hour duration and intensity of 1 by 4 centimeter. Now since we want unit hydrograph for 2 hours our rain should correspond to 2 hour duration and 1 by 2 centimeter per hour intensity. Comparing these 2 we see that whatever ordinate we get here has to be multiplied by 2 because the ordinate corresponds to 1 by 4 and we want intensity of half centimeter per hour. So, that is why when we take the difference we multiply it by 2 and this way we can get the ordinate of the 2 hour unit hydrograph. So, in this case it is quite straight forward difference of these 2 into 2 will give us 8 difference of these 2 is 4 times 2 will give us 8. So, any point we can take difference of these 2 into 2 will give us 0.8. So, this 2 hour unit hydrograph which we can obtain from the S curve can be plotted and in this case it looks like this the peak occurs at about 8. In this case these 2 points are as can be seen from this table 8 and therefore, when we draw the hydrograph it goes little more than 8. We can again draw a smooth curve. So, some curve like this may be better suited for the hydrograph. So, any duration other than the multiple of the given hydrograph duration we can obtain the unit hydrograph using the S curve and this can be used to obtain even for higher durations. For example, if we want 6 hour duration we can still do the same thing, but suppose we want 6 hour duration we may again do it by first obtaining a 2 hour unit hydrograph and then using the superposition superposition 3 times to get 6 hour unit hydrograph. So, it depends on what we find more convenient, but S curve can be used to derive unit hydrograph for any duration which is not a multiple of the given duration. So, we have looked at if unit hydrograph for 1 duration is available how to get unit hydrograph for any other duration, but now the question is how to get unit hydrograph for a particular duration. So, we will look at some methods of obtaining let us say a 4 hour unit hydrograph. So, in this case suppose we have a catchment which is of area 60 kilometer square. So, let us see this is our catchment and let us say that at this point a we have a gaging station at which the stream flow record is available to us and let us also say that there is a 4 hour drain which occurs in the catchment of some intensity which may not be equal to 1 by 4 centimeter per hour. In this case the rainfall which occurs on the catchment has an intensity of 5 millimeters per hour. So, this implies that total effective rain is 4 into 5 millimeter, so this is 2 centimeter and because of this rain which again there are some assumptions which we have to make for example, we will assume that this 4 hour effective rain occurs uniformly over the entire catchment of 60 kilometer square area. So, if it occurs uniformly over the entire area and we measure the discharge at point a this is the table which gives the values of q at different times. So, at 0 hours we start now this is 0 this is the direct runoff hydrograph d r h. So, we have already separated the base flow that is why it is starting from 0 the actual stream flow at a at the beginning of the rainfall may be some other value which comes from the base flow. So, the d r h is start from 0 and these are the ordinates at different times. So, in this case we have taken data at equal spacing of 2 hours. So, every 2 hours we have measured the discharge at the point a and listed that in this table. A plot of this d r h is given here. So, we can see that start from 0 6 18 30. So, 30 is the peak discharge. So, we have 6 18 30 and therefore, this hydrograph this is d r h due to 2 centimeter rain. Assumptions of course, are that it is uniformly occurring over the entire area of 60 kilometer square. Once we have this d r h finding out the unit hydrograph is quite a straight forward because this is due to 2 centimeters of rain and we want unit hydrograph which corresponds to 1 centimeter of rain. So, u h corresponds to 1 centimeter of rain this is already available to us for 2 centimeter of rain. So, we divide each ordinate by 2 and we will get the u h ordinate. So, this is nothing but d r h divide by 2. So, the peak instead of 30 becomes 15. So, this is a very straight forward way of obtaining the unit hydrograph, but the condition is that rainfall of almost uniform intensity and for that particular duration should be available. So, in this case we say that 4 hour unit hydrograph can be obtained if a 4 hour rain is available and corresponding discharge is also available. So, the 4 hour unit hydrograph looks like this. This is similar to the d r h simply dividing by 2 all ordinates we get this unit hydrograph. Now, most of the times the rainfall constant intensity rainfall for 4 hours may not be available. For example, we may have a rainfall which may be a different duration it may be like 6 hours or 2 hours. In that case we can derive a unit hydrograph for 2 hours or 6 hours and then get 4 hour unit hydrograph from that hydrograph. Now, sometimes uniform intensity rainfall may not be available sometimes the intensity may be varying. For example, we may have an e r h effective rainfall hydrograph as let us say 2 different 4 hour rains of varying intensity. For example, we may have intensity like this then we may have intensity like this. So, in this case how to obtain the 4 hour u h. So, this is the case which we will discuss now suppose we are given that rainfall of 5 millimeter per hour occurs for 4 hour and then it is followed by another rainfall which is of uniform intensity, but a different intensity 10 millimeter per hour again for 4 hours. So, the problem is that if we know the direct runoff hydrograph d r h which corresponds to a combination of these 2. So, we have 5 millimeter per hour and then 10 millimeter per hour for 4 hours. This is the e r h. So, that means we have 2 centimeter rainfall here 4 into 5 millimeter and then we have another 4 centimeter of rainfall here because this is 1 centimeter per hour for 4 hours. So, in total we have a 6 centimeter of rainfall over the area which is 60 kilometer square and the d r h which we have measured at a is given by these values where the peak value as can be seen here occurs at 10 hours and is equal to about 77 meter cube per second. This d r h is plotted and can be seen in this figure. This if you see this curve up to 4 hours there will be no change compared to the 4 hour u h, but then since another rainfall occurs we will have this variation which means that we will have a sum of rainfall per hour ordinate or the d r h ordinate. Suppose we are looking at 6 hours this 42 corresponds to the 6 hour value for the first rain and the 2 hour value for the second rain because the second rain has started after 4 hours. So, that is why this curve ordinate up to 4 hours there is no problem it is only because of the single rain. So, we can use the 4 hour u h ordinate directly, but after that we will have a combination of the first 4 hour 5 millimeter per hour rain and then we have some ordinate which comes from the second 4 hour 1 centimeter per hour rain. Not the ordinate at the same time, but they are shifted. So, using this d r h if we want to evaluate the unit hydrograph we will have to do an analysis like this. Suppose, the ordinate of the 4 hour unit hydrographs are u 1, u 2, u 3. So, d 0 then 2, 4, 6, 8 like this and suppose the 4 hour u h ordinate are u 1, u 2, u 3 like this. So, the d r h ordinate since the first rainfall is of total 2 centimeter effective rain 2 u 1 should give us the direct runoff because of this rain. So, d 1 this represents the d r h ordinate d 1 d 2 d 3 and so on. So, what it tells is that if we have unit hydrograph for 4 hour with ordinate u 1 the direct runoff will be 2 times u 1 because the rainfall is 2 centimeter compared to 1 centimeter for the u h. Now, d 1 is given to us d 1 is 6 as can be seen from this table. So, d which is the direct runoff hydrograph ordinate at 2 hours is equal to 6 and therefore, it gives us a very straight forward way of computing u 1 and u 1 will come out to be 3 from this equation. Similarly, at 4 hours since the second rain has not it is started we have a similar equation which tells us that 2 u 2 will be equal to d 2 since d 2 is given as 18 u 2 will be obtained as 9. Now, when we go to the next time which is 6 hours then we have a combination of ordinate coming from both the first and the second rain. For the first rain the ordinate will be u 3 which will be at 6 hours and multiplied by 2 for the second rain the ordinate will be u 1 which corresponds to 2 hours because we are computing at 6 hours means the second rain has started 2 hours before that time. Therefore, we have to take u 1 and multiplied by 4 because the second rain total rainfall is 4 centimeter. So, 2 u 3 plus 4 u 1 this should give us the direct runoff ordinate at the third point which is 6 hours and this is known to us 42 from the table of d r h. So, this 42 will be equal to the sum of 2 u 3 and 4 u 1. Now, u 1 is already known to us because of the first equation u 1 is known to us as 3 and therefore, we can solve this equation and get this 15 because 4 u 1 will be 12. So, 42 minus 12 divide by 2 will give us u 3. So, we get an ordinate u 3 equal to 15 same way we can write 2 u 4 plus 4 u 2 equal to d 4 which is 58 giving us u 4 of 11. So, in this way from a complex storm the condition is that both the all the storms should be of the same duration 4 hour in this case. There can be multiple storms not only 2 there can be 3 4 any number of storms provided they are all of the same duration. We can use this the only thing is that after a few hours we will have sum of 3 different ordinates. So, here we had 2 ordinates because there were only 2 reigns of 4 hour duration if we have 3 then after 8 hours. So, 0 to 4 we will have a single ordinate 6 to 8 then we will have 2 and then after that we will have 3 ordinates sum of 3 ordinates. But again sequentially we can obtain all the values in this way the problem with this is that if we make some if there is some small error in the data it gets magnified because we are doing it sequentially. So, any small error in one of the ordinates will get magnified as we move away and sometimes it may happen that the values here get negative. So, or some oscillations may be there in this case since we have taken exact data this resulting unit hydrograph this resulting unit hydrograph which we have obtained by computing this u 1 u 2 u 3 and so on it turns out to be quite smooth. So, we say that this is a 4 hour u h which we have derived based on the direct runoff hydrograph corresponding to 2 different storms in this case of same duration 4 hours, but different intensities. So, obtaining the unit hydrograph for a gage casement is therefore, not very difficult if we have the required rainfall available and corresponding runoff also available. So, we can first separate the base flow obtain the d r h from the d r h we can obtain the u h depending on whether we have a single storm or whether we have complex storms and then using that 4 hour unit hydrograph or whatever duration we want using that duration then we can obtain unit hydrograph for any other duration by using a curve or superposition, but suppose we do not have a gaging station available that means we do not have a d r h available to us for any given e r h. In that case we go for synthetic unit hydrographs in which we use the geometric parameters of the casement for example, the length or length to centroid the slope and so on. And using those geometric parameters we can obtain a unit hydrograph which is not based on any observed runoff, but it is just a synthetic hydrograph. So, we look at the Snyder's method. So, synthetic u h means we do not have any data available for the casement Snyder's method is the one which is Snyder has proposed and there are some equations which are given by Snyder. For example, he said that if we have a casement the length of the casement is l along the stream centroid of the casement is somewhere here and the length to centroid is l c area is a in kilometer square lengths are typically although in the original Snyder method they were used in miles that we will be using kilometers. So, l and l c would be in kilometers a will be in kilometer square. So, what Snyder has in his synthetic hydrograph he has related the characteristics of the hydrographs suppose this is the duration of the rainfall t r effective duration from the center of this rain to the peak he called the lag as t p. So, Snyder related this t p with the casement characteristics and he says that this t p is standard value some constant into l l c. To the power 0.3 and for this is standard lag the duration t r is t p divided by 5.5. So, if we have the data available for let us say a gage casement. So, in this case we will take an example c t is a constant which depends on the basin properties in this case we will assume that there are two casements one is gaged. So, casement one is gaged and gaged casement means that we can obtain the unit hydrograph for this easily casement two is un gaged. So, using the data, but they are very similar the assumption has to be made that they are very similar that means whatever the value of c t is for this casement same value of c t can be used for casement two for the casement a the data is given as area 60 kilometer square length 15 and up to centroid 9. So, this l c is 9 l is 15 and the area is 60 kilometer square a 4 hour unit hydrograph for the first casement has been obtained because it is gaged. So, we have obtained the unit hydrograph the peak is given as 15 meter cube per second and time to peak is 4 hours from the centre of the rainfall. So, this t r is 4 hours. So, centre means that we have 2 hours and then further 4 hours from that. So, t p is 4 hours from here and 6 hours from the time 0 of the starting of the rainfall. So, t p in this case is 4 and the duration of the rain is 4 hour so we can see that t r and t p do not satisfy this relationship. So, this 4 hour is therefore not a standard duration as far as Snyder's definition is concerned. So, we would have to find out the coefficient c t and c p knowing that this t t p and t r are not a standard values and that is case Snyder has given a relationship. If t r is not a standard then for any other rain for duration of t capital R Snyder has given this relationship which can also be written as because this t r we have seen is t p divided by 5.5. So, let us use this relationship to obtain the value of c t for the catchment which is gauged and for which unit hydrograph is available to us. The other relationship which has been used to obtain the discharge the peak discharge is given here in which the peak discharge q p is given as c p into 2.78 a over. So, c p is also another constant which we can obtain for the catchment one. So, our aim would be to predict the value of c t and c p from this catchment and since they are similar we can use the same c t and c p for this catchment and predict the value of peak and time to peak for the second catchment. So, the computations are shown here for the gauged basin t p star is 4 hours and t r star is 4 hours star denotes the value for the gauged catchment. The standard lag as we have seen t p can be obtained using this equation as 22 by 21 t p star minus t r star by 4 and comes out to be 3.14 hours. So, this means that for the given basin the standard lag should be 3.14 hours and since we know the equation for standard lag we can obtain the value of c t as standard lag divide by l star l c star to the power 0.3 for the catchment one l is 15 l star l c is 9. So, we have 15 into 9 to the power 0.3 and t p is 3.14 hours. So, this is gives us c t of 0.72. Similarly, since we know the peak discharge for the catchment we also know the peak lag the basin lag we also know the area we can find out the c p value as 0.36. So, these two values are now known to us c t and c p and using these values we can derive a synthetic value of the unit hydrograph for the catchment two which is not gauged and for that let us assume that we want a 4 hour unit hydrograph. So, given a 4 hour unit hydrograph for catchment one we want a 4 hour u h for catchment two. This 4 hour again is not the standard duration as far as Snyder's definition is concerned. So, we can write this equation where c t into l l c 0.3 this is the standard lag 21 by 22 plus t r by 4 should be equal to 1 the basin lag for this corresponding rainfall of 4 hours. So, this gives us a lag of 4.56 hours we can see that the time to peak or the lag in this case was 4 hours the area was 60 kilometer square since this is a larger basin 100 kilometer square area l and l c are also larger we expect the lag will also be larger. So, in this case compared to 4 we get a lag of 4.56 hours the peak discharge is obtained c p we have already seen from the previous basin as 0.36 c t we have already seen as 0.72. So, using these values of c p and c t we get the peak discharge as 21.95 meter cube per second area of the catchment 100 and t p r of 4.56 per unit area. The discharge the peak discharge per unit area can be written as since the area is 100 21.95 divided by 100 meter cube per second per kilometer square. Then to help draw the hydrograph there are some equations which are proposed for example, the base time base of the hydrograph can be obtained by approximating it by a triangle and equating the area to be unit. So, t b can be given by 5.56 divided by q p r and it comes out to be 25 hours. In this case 25 hours seems to be a little small because we have a earlier seen that the unit hydrograph for the catchment 1 which is gauged and is a smaller has a time base of about 40 hours. So, this 25 hours seems to be a little small and we will increase it later when we plot the curve. There are two other significant values one is what is the width at 50 percent of the discharge peak discharge and what is the width at 75 percent of the discharge. The equations which are given are w 50 which is the width at 50 percent of the peak discharge is 2.14 divided by the peak discharge to the power 1.08 comes out to be 11 hours w 75 comes out to be 6.3 hours. It has also been proposed that one third of this should be before the peak and two thirds after the peak. So, out of this 11 hours one third should go before the peak one two thirds after the peak similarly for this. So, using this data time base w 50 w 75 peak discharge the lag 4.6 hours using all this data we can plot a few points and then join them by a smooth curve to obtain the synthetic unit graph. In this case 0 and 25 these two are based on the time base of 25 hours then we have the peak this should be 4.6 hours from the center of the rain. Center of the rain is at 2 hours because rain actually is occurring up to 4 hours and from the center this 4.56 hours based on the value computed here. So, this total should be 6.56 hours. Now, the peak is also known to us let us say that this is 22 roughly because the value which came out from the calculation is 21.95. So, we can say that this is peak is 22 at 11 which is 50 percent of peak and then 16.5 means which is 75 percent of the peak. We know that this total width is 11 hours and this is 6.3 hours half of this one third of 6.3 2.1 goes here and 4.2 goes here. So, knowing the peak we can obtain these two points similarly from the peak 11 by 3 here 22 by 3 here. So, we can obtain these. So, 1 2 3 4 5 6 7 points are known on the Snyder's curve the area under the curve will be 1 centimeter of rain over 100 kilometer square area, but if we look at this curve it is not smooth. So, what we will probably do in most cases is join it by a smooth curve and also we have seen that 25 seems to be little small. So, we would like to have it increase up to 40. So, we will try to make it such that the area remains similar. So, whatever area we have excluded here can be included here and so on. So, we can get a smooth curve based on neighboring catchment a similar catchment and we can obtain this synthetic unit hydrograph. So, the synthetic unit hydrograph can be derived for a catchment for which we have geometric or geomorphological data available and we can estimate the values of the coefficient C T and C P for a similar catchment which is gauged and therefore, we can obtain those values from a unit hydrograph derived for that other gauged catchment. Using those values of C T and C P we can estimate the synthetic unit graph for the basin which is not gauged, but this synthetic unit hydrograph or whatever hydrographs we have discussed till now they are of a particular duration. That means that we can use them to predict the DRH for a rainfall of the same duration and of course, this duration can be changed we can derive a rainfall runoff relationship for a different duration of rain by using a unit hydrograph for that particular duration and we have already seen that given any duration of unit hydrograph we can derive any other duration hydrograph easily using S curve or using superposition, but sometimes the rainfall which occurs may not have a uniform intensity. The intensity may be varying quite a bit for example, we have let us say a rainfall which has intensity like this. In this case making the assumption of a constant intensity over finite intervals may be valid, but it may not be very accurate. So, in this cases where the intensity varies continuously with time and uniform intensity assumption cannot be made, we should look for a unit graph which corresponds to instantaneous rainfall and this is known as the IUH or instantaneous unit hydrograph. We have seen some of the methods for example, we have seen the Nash model in which the instantaneous unit hydrograph which is defined as the instantaneous unit hydrograph T versus Q. The rain which causes so this is the IUH the effective rainfall for which causes this instantaneous unit hydrograph is a rainfall of very high intensity and so very small suppose this is delta T and this is intensity is let us call it 1 over delta T. So, the volume of rain which occurs is 1 or the depth of rain into the catchment area. So, depth of rain is 1 centimeter, but it occurs at almost 0 times as the limit delta T tends to 0 it becomes instantaneous rainfall of very large intensity very small duration in such a way that the depth is 1 centimeter. In other words the entire 1 centimeter of rainfall occurs instantaneously at the catchment and because of that it causes a rain a runoff like this. We have seen that Nash model is one of the models which is used for instantaneous unit graph calculation. The equation which is used for Nash model comes from linear combination in series of reservoirs which is so we have reservoir in which there is some rain here some outflow goes into second reservoir there is some outflow which goes into third reservoir and so on. So, n is the number of reservoirs and k is the time constant in this case we have taken to illustrate the Nash hydrograph n equal to 4 and k equal to 3 hours for a catchment of area 60 kilometer square. This value u value is an ordinate which is expressed in per hour and what it really means is centimeter per hour per centimeter rain. So, since we are saying that this is unit hydrograph it will have 1 centimeter of rain therefore, u can be written in terms of centimeter per hour and that is what we have done in this figure in this table which shows the value of u computed from this equation with k equal to 3 and n equal to 4. The ordinates are expressed centimeter per hour here and then based on the catchment area we can convert it into meter cube per second. These ordinates can be seen here with the peak occurring at about 10 hours of almost 12 meter cube per second and the plot of the IUH this indicates that over a catchment of 60 kilometer square if a 1 centimeter rain occurs at time t equal to 0 instantaneously. Then this would be the resulting hydrograph with the peak of almost 12 meter cube per second and as we have already seen that this IUH can be used to obtain the DRH because of rain fall of varying intensity by using the convolution theorem. So, here we would look at some uses of this IUH suppose we are given the IUH and we want a 2 hour unit hydrograph derived from the IUH now since the IUH represents instantaneous rain if we take any time interval delta t the average of these 2 ordinates will give us the ordinate of the unit hydrograph suppose this is 2 hours then if we want the 2 hour unit hydrograph the value 0.7 here is the average of these 2 mean of the 0 and 1.4 will give us 0.7. Similarly, the ordinate of 2 hour unit hydrograph at 4 hours would be the mean of 1.4 and 5.8 and so on. So, all the ordinates can be obtained in such a way. So, this 0.13 is the mean of 0.16 and 0.1. So, the 2 hour unit hydrograph can be derived this is the 2 hour because the difference of these times is 2 hours and it can be plotted like this with a peak of about 12 meter cube per second. Now, once 2 hour unit hydrograph is available to us we can use this to obtain S curve. So, suppose this 2 hour unit hydrograph is available to us this can be obtained from an I U H or it can be obtained directly from if we have a gaged basin suppose we have a 2 hour rain we can get the corresponding D R H and get the 2 hour U H from there. So, from I U H also we can obtain the unit hydrograph for a given duration, but what we would look at is for a given hydrograph how to obtain the I U H. So, suppose directly we know the 2 hour U H based on the 2 hour unit hydrograph we can obtain the 2 hour S curve. So, those values are shown here and as we have seen the relationship between the ordinate of the I U H and this is the S curve ordinate the S curve looks like this what this equation tells us that the slope of the S curve at any point d S by d t divide by the intensity of rain will give us the ordinate of the instantaneous unit hydrograph. Because S curve represents the rainfall due to infinite value if we shift it suppose we shift it by delta t and suppose this is delta S. So, when we shift it by an amount delta t delta S represents the q value due to a rain of intensity I for period. So, this is what delta S represents it represents since we have shifted by an amount delta t it represents the amount of runoff which occurs because of 2 rains 1 of intensity I continuously the other of same intensity, but shifted by delta t. So, any change here delta S represents the q due to intensity of I period of shifting of delta t and as delta t becomes 0 delta S by delta t will tend to will tend to d S by d t. So, this I rainfall for period delta t means the volume is I delta t since we want a volume of unity delta S over I delta t will give us the unit hydrograph instantaneous unit hydrograph ordinate u. So, u will be equal to delta S divide by the volume of rain which is I delta t and therefore, in the limit delta t tending to 0 delta S by delta t will tend to d S by d t and therefore, u will be 1 over I d S by d t which is what is written here. So, once we derive 2 hour S curve we can obtain the slope of the S curve at any point divided by the intensity in this case since it is a 2 hour S curve I will be 1 by 2 centimeter. So, the procedure is obtaining the S curve first S curve of course, is very easy to obtain because once you have the 2 hour u h we have already seen how to lag them by 2 hours add them up and then get the 2 hour S curve. So, in this case by doing that we have obtained the 2 hour S curve ordinates maximum value of 83.35 which corresponds to intensity half centimeter per hour over an area which in this case is given as 60 kilometer square. So, area is given as 60 kilometer square if the intensity is taken as half centimeter per hour it will result in an ordinate of 83.35 at the end or the asymptotic value. So, we can plot this S curve and as it is shown here this is the 2 hour S curve if you look at the slope of this S curve initially the slope is almost 0 at the end also it is almost 0 because it attains a constant value. So, the slope is 0 here increases then there is a point of inflection after that it will start decreasing and then ultimately the slope will become 0 and the instantaneous unit hydrograph can be obtained as 1 over i which in this case will come out to be 2 since i is half into d s by d t. So, slope of this we can obtain either numerically. So, we can use central difference or we can try to fit a curve and then take the slope linear interpolation or quadratic interpolation. Here what we have done is suppose we want the slope here we have used a central difference approximation. So, the slope will be obtained by joining the two surrounding points and divide by delta t delta t in this case since we have every 2 hour intervals delta t will be 4 hours and delta t delta q will be the difference in the next q and the previous q. So, using that we have here the value of 0 and 0.7 and 4.3 we want to find out the slope at this point at 2 hours. So, what we will do is take the difference of the two neighboring points the next point 4.3 previous point 0. So, 4.3 divide by delta t which is 4 gives us a slope of 1.08 intensity as we have seen. So, the u the i u h ordinate is 1 by i d s by d t i is 0.5. Centimeter per hour d s by d t we have obtained. So, 1.08 is 4.3 minus 0 divide by 4 similarly 2.88 is the difference of 12.2 2 and 0.7 divide by 4 and then i u h will be multiplied by 2. So, this is into 2 will give us 2.15 similarly all these values are multiplied by 2. In this way given 2 hour unit hydrograph we can derive an i u h for that particular basin and in this case the i u h turns out to be like this we can join them smooth with a smooth curve and this i u h comes out to be very similar to the i u h which we had used to derive the 2 hour u h. So, this was the i u h which we had used to derive the 2 hour u h then we use the 2 hour u h to derive the s curve for 2 hours. So, the 2 hour u h is given by this curve using this 2 hour u h we have derived the 2 hour s curve and then from this 2 hour s curve we can obtain the i u h which should be similar to what we had started with, but this just gives an idea of how to obtain an i u h for a given unit hydrograph. Typically duration of 2 hours we have taken, but 1 hour duration would be better more than 1 hour taking the slope by this finite difference approximation may not work very well, but for 1 hour it should be quite good less than 1 hour is, but 2 hours we have taken for example, it may not be very good from accuracy point of view. So, we have looked at some of the techniques in this case to obtain the unit hydrograph for a duration which is not a multiple of the given hydrograph duration. We have also looked at if a catchment is not gauged, but a similar gauged catchment is available to us nearby how to obtain the unit hydrograph synthetically using the Snyder's method. Then we have seen that if the rainfall is not of uniform intensity over the whole period if it is varying widely then we may not be able to use the unit hydrograph theory because that assumes rainfall of constant intensity for a given duration. In that case we can go for instantaneous unit graph which can be used to obtain DRH for any varying rainfall by using the convolution theorem. The i u h can be derived based on hash model which we have seen is based on a gamma function and it is a combination of linear reservoirs in series. There are other methods also of obtaining i u h. We have seen one method which is based on obtaining the s curve for a given unit graph and then taking the derivative of the s curve to obtain the unit hydrograph i u h ordinates. There are some other methods also for example there is a Clark's model in which we use what is known as time area diagram. So, we divide the catchment area into sub areas which all have almost same time of travel for reaching the outlet point and then these different areas can be plotted with time and we get a time area diagram and that time area diagram we can use to obtain the i u h because we assume that instantaneously the water is falling over the entire area. So, it will have some lag time for reaching the outlet and those time area diagrams can be prepared and they can be used for i u h, but we will not discuss them here. So, we have looked at the two methods which are Nash model and then using any other u h of preferably a short duration for example 1 hour duration or 2 hour duration we can obtain using the s curve from the slope of the s curve we can obtain the i u h.