 to all of you. I am Mr. Shashikant B. Gosavi, Assistant Professor, Department of Civil Engineering, Valkchand Institute of Technology, Sulapur. Today I am presenting the uses of control maps, another online educational resource from my course. The learning outcomes desired for this course are listed as the students will be able to use control maps to draw control profile for selected alignment, to set control gradient of root alignment and to calculate the capacity of reservoir. To begin with, we will see what is a control profile. A control profile is normally a profile of the land surface corresponding to a selected alignment which is taken up on a control map. Here for example, I have taken a control map in which you can see there is a mention of control interval of 5 meter. There is a mention of scale graphically as well as numerically and the map is showing the values of the contour rls at every 25 meter interval. Between every two successive contours, vertical distance is 5 meter. So, it begins from the bottom, slowly rises up up to that particular circle which is drawn over there and then in the inner portion once again it falls down reaches to the bottom. So, it is indirectly creating a crater over here or a reservoir over here at the center. To draw a control profile, an alignment has to be chosen. Here we have chosen alignment connecting A to B and across this particular alignment wherever there is intersection of the control lines, all those points are of utmost significance to us. See here, there is an intersection at this point for a control of 7725 rls. There is an intersection at this point corresponding to the highest point available along this alignment. Then there is an intersection at this dark control of 7725 once again. Then there is an intersection of 7700 and successively at every 25 meter interval whatever are the intersections those have been marked. At each of these points what we have done is we have drawn vertical ordinates passing through these particular points. All these in all 13 number of lines are there which are passing through the points of our interest and those have been vertically projected down. In the next control profile you can see that I have drawn a graph of vertical ordinate of respective rls. We have shown 7550, 7600, 7700, 7750 and 7800 contours. Sorry, reduce the levels of contours. And if vertically projected lines are taken up along this particular horizontal lines you can find that the intersection of these two lines is happening at these points. So this is corresponding to 7725. This is corresponding to 7744. This is 7725 again. This is corresponding to 7700. This is corresponding to 7675, 7650. 7628 is the lowest point and again on the right hand side there is a rise 7650, 7675, 7700, 7725, 7750 and so on and so forth. So this is a classical profile which we can draw from a given control map which helps us in identifying how the ground surface varies along a given alignment. Another important application of the control map is that of drawing a contour gradient. See this is very much important when we are passing a road through a hilly terrain or we are passing a canal of desired gradient or railway line of desired gradient across land surface with varying terrain. See this particular contour map which we have referred earlier is having a contour interval of 5 meter throughout and contours have been labelled at every 25 meter interval. Here the scale has been given as 1 as to 10,000 that means 1 centimeter is equal to 100 meter. Graphically also scale has been shown here. You know that the maximum permissible ruling gradient can be 1 as to 40 that means in every 25 meter maximum rise permitted is 5 meter and if a road is to be passed through this particular terrain we will have to be cautious about that. Let us say that we are starting from here. There is a contour line which has been drawn at this location having a RL mentioned over here as 7725. So from here to here this particular subsequent successive 2 contours are having a difference of 5 meter. For getting a difference of 5 meter one has to travel a distance of 200 centimeter such that the gradient will be 1 as to 40 and therefore what I have done is by taking this as the center I have drawn a circle with 2 centimeter as the radius. Therefore automatically this particular distance comes out to be 200 and with the same respective radius with the subsequent point I have drawn another circle. You can see that this particular circle is intersecting to the successive contour at this location as well as at this location. We can choose the one which is most suitable for us. The another next distance to be traveled will have to be 200 meter for every 5 meter and therefore I have taken this much radius and intersection point has been chosen over here. I was having option of going for 200 meter in this direction also but in that particular case probably I might have gone in a wrong direction than what is desired by me. I want to reach to this end as early as possible and therefore I have chosen this one. Subsequent rise is again taken up along this particular direction horizontal distance 200 meter traveled for a rise of 5 meter. This is continued again along this distance, along this distance, along this distance everywhere for the successive 2 contours having the rise of 5 meter we have to travel 200 meter distance and thus the contours are moving along these particular alignments. From here I was supposed to take a exit from here so I have ended up my respective gradient or alignment at this particular location. This is utmost essential for us to draw the contour gradient particularly in the hilly area when hilly roads are being planned. Another classical application of the contour map is determination of capacity of reservoirs. All of us are familiar that a contour map is a drawing indicating the nature of the ground. When number of loops are around each other as has been shown in this particular case the contours are forming loops around each other it may either represent a hill or it may represent a pond, pond like land profile. What does it mean is towards inner part the reduced level will be lowest and on the outer part reduced level will be higher. So here from 7678 up to this 7748 continuously the reduced level is rising and from here on again it is falling on this side as well as on this side. Same thing true on this side as well as on this side. What we can see clearly is up to this 7740 continuous rise is there so we are having so many contours drawn which will help us in finding the area between any 2 successive contour lines. So a0 is the innermost area then between 2 successive contours it will be a1, a2, a3 and so on and so forth we can keep on calculating the areas between every 2 successive contours. These areas can be put up into this particular formula. The vertical distance between every 2 successive contour can be put up as h and then by using this particular trapezoidal rule we can find out the value. There is prismidal rule also which can also help us in finding the area. One out of these 2 rules we can definitely adopt and we can find the capacity of reservoir. This is a review of what we have learned by now. Answer the following multiple choice questions. For this you need to pause your video and answer them then correct answers will be elaborated by them. The first question is contour profile is to be drawn for selected first option is contour, alignment, gradient and all of these. The second question is contour gradient is useful to first option is find alignment for desired gradient, find gradient of desired alignment both of the above and above. Dear learners the answers are very very clear for the first question the answer is contour profile is to be drawn for selected alignment and for the second question the answer is contour gradient is useful to find alignment for desired gradient. I have used these references thank you very much.