 Hello friends. So, I am Mr. D. J. Doshi, Assistant Professor, Department of Mechanical Engineering, Walsh and Ipsu Technology, Sola. Up till now we have studied so many videos of projection of solids. This is again new type of projection of solids, where new type of solid will be considered. At the end of this session, students will be able to draw the projections of solids as per given conditions for the given type of solid. Now, see this is related to frustum. If you read the problem carefully, frustum of a cone, what is frustum? A frustum is the sketch of which a section parallel to the basis is usually taken and top portion will be removed. That is suppose it is a cone, cone has a apex. So, apex will be removed at the particular given height. Suppose the apex height is 50 mm and the frustum is of 32 mm, then 18 mm part of the cone will be removed by a horizontal parallel line to the base and top portion will be removed and the remaining part is called as frustum of a cone. Frustum of a pyramid, so the pentagonal pyramid, hexagonal pyramid, in that chaos also we will be drawing a horizontal sectional line or sectioning will be drawn horizontally parallel to the base and top portion will be removed depending upon the given height of the particular frustum or a solid. So now, we are drawing projections of a solid which is a frustum of a cone. So, frustum of a cone carries base diameter 50 mm and upper phase diameter 30 mm and height of the axis is 34 mm. Its axis is at 30 degree to HRP and parallel to PRP. Draws projection in the base is hidden, if the base is hidden. Now, if you observe frustum as I told you, what is frustum? It is a part of a solid, particularly pyramid or cone of which top portion is removed by drawing a section line or section line or parallel to the base. Now, here we have to draw a frustum of which base diameter is 50 mm and upper phase diameter is 30 mm. So, now, as it is resting on HP, of course, top view will be two concentric circles. See in case of cone, we were drawing only the one circle and a point which is a apex, which is projection of apex, but here as we have a section or we have taken out the part of apex, here we will get in two concentric circles of the base diameter. One is base diameter circle, another will upper phase diameter circle. So, draw two concentric circles. First one is 50 mm diameter, which is a base diameter and another will be 30 mm diameter, which is a upper phase diameter. So, two concentric circles are drawn. Now, project those upwards. So, project these two upwards. So, you will get a base, which is resting on HP is of length 50 mm and as 30 mm, 34 mm is the height of the frustum. At 34 mm, we will draw a parallel line to HP or parallel line to XY and project the smaller diameter circle, that is 30 mm diameter circle, which will intersect the horizontal line. Those will be the projections of the smaller circle or upper phase. So, upper phase will be of 30 mm length and lower phase or base diameter will be or base will be 50 mm and join those things. Now, when we are going for second stage, we need to divide the frustum with the help of generators into some parts. Now, here it is divided into 8 parts, but if the diameters are bit more, you can divide that circle or those circles into 12 parts. So, now as here the diameters are smaller, we have divided the circle into 8 parts. So, if you observe here, this is O, which is imaginary apex. From here, I will divide it into 8 parts, that is at 45 degrees all the lines. Project those for both the, which will be dividing or intersecting both the circle periphery. So, here you are getting 1, 2, 3, 4, 5, 6, 7, 8 and then it is started as A, B, C, D, E, F, G, H. So, why we have taken here A, B, C, D, E, F, G, H, because see if we are dividing it into 8 parts, smaller diameter, it is numbered as 1 to 8. So, there should not be repetition of the same numbers. So, we have taken A, B, C, D, E, F, G, H as the division of the smaller, larger diameter circle and 1 to 8 will be division of smaller diameter circle. Now, project 1 to 8 which will be upwards, which will be dividing this line into parts. So, those will be named as 1 dash, 2 dash and so on and project this A to H to intersect the base into equal parts. So, here it is A dash, B dash, C dash, D dash, E dash, F dash, G dash and H dash. Now, join accordingly. Now, see here 1 is joined with the A. So, 1 will be joining with the A, 2 will be joining with the B. So, 2 will be joining the B, C will be joining with 3, D will be joining with 4 and so on. So, these are all thin lines, these are all generators drawn for our convenience for drawing purpose. So, now this is a front view of the prism of diameter upper diameter 30 and base diameter 50 which is divided into of course, 8 parts. Now, what is given? Its axis is at 30 degree to H B. So, now, we have to incline this first term front view such that axis is making 30 degree, which is a bit difficult task. Now, see this is actually the axis. So, drawing the axis at 30 degree here and the C point C must be resting on H B is a difficult task. So, what we will do as this is a right angle triangle. If you observe C, O dash and some imaginary parts here imaginary point, this is a right angle triangle where this is at 90 degree, this is at 30 degree. So, this will be at 60 degree. So, angle C will be 60 degree. So, better for practical purpose for drawing will make the base at 60 degree, but while showing the diagram you have to show the axis as 30 degree. So, draw a line of 50 mm length as we know this is a 50 mm at 60 degree, divide according to these distances here. So, this is divided into 8 parts that is 8 to H. Now, then draw then bisect this and draw the axis line which will be making 30 degree. Then draw a parallel line to this parallel line to base front view at a height of what is the height given? 34 mm height I will draw a parallel line to the front view of the base and again from here I will start marking the distances because this is the axis line. So, on axis line I will mark the distances accordingly 1 to 8 will be the divided parts of the line. So, this distance is axis 34 mm, this is at 60 degree, this axis will be automatically at 30 degree and at a distance of 30 mm I have drawn a parallel line. Again from here I will mark the distances as per in the first stage front view and join those draw the generators. So, here you will get again the generators A 1, B 2, C 3 and so on. So, this is the frustrum front view drawn or redrawn such a way that the axis is making 30 degree with H V. Now, project one by one now let us start with the base front view to draw a base top view project A to H 1 by 1 that is now here A dash is projected downwards. Here we have point A we will get the point A here in the second stage top view then B projected up downwards project B horizontally you will get the point B here. So, C D E F and so on. So, now if you observe it is resting on point C and if the observer is observing from this side I want the arrow drawing the point away from the observer which is not visible is point C. So, the line passing through C curve line passing through C will be a dotted line. So, what we will do firstly we will plot all the eight points that A to H then now it is confirmed that this half will be visible that is this half or this half will be visible. So, I will make it dark. So, draw A to E dark line and similarly project 1 to 8 downwards project 1 to 8 downwards. Now, this is totally visible. So, this will be connected with the dark line draw a point join A 1 and join E 5 C 5 and E 5. So, now after joining this the remaining part of the ellipse will be drawn as a dotted line. So, C here C is away from the observer. So, the line passing through C curve line passing through C will be a dotted line. So, other line will be the dark other part will be the dark line. So, I have connected these lines as a dark line then connected A 1 and here these line this will be dark. So, other part will be drawn as dotted line connect the generators join the generators. Now, major part is draw its projection if the base is hidden base is hidden means base is away from the observer. So, you have to rotate it in a clockwise direction such that the base is away from the observer means base is hidden. So, base will be away from the observer. So, redraw it or reconstruct it similarly here where axis will be perpendicular to the VP. So, this is the axis perpendicular to VP and I have constructed all these things by drawing the different parallel lines horizontal and vertical. So, we will get the sketch or the top view of third stage such that the base is away from the observer base is hidden. Now, project it upwards project it horizontally you will get the view here now in this case the point away from the observer is G. So, the line related to G will be a dotted line here. So, this is important of drawing the dotted line. So, this part will be drawn as dotted whichever is not visible other things are joined with the dark line. So, similarly this one is copied here such that axis is perpendicular and base is away this is projected upwards this is projected horizontal you will get the final view of the sketch where the first term is axis is making 30 degree with HP and base is away from the observer or base is hidden. So, for this use the book Engineering Graphics by Mahalakshmi publication house. Thank you.