 Hello, I am Mr. D. J. Doshi, Assistant Professor, Department of Mechanical Engineering Networks, Walthian Institute of Technology, SolarPort. We will be today learning about projection of solids. At the end of this session, students will be able to draw the projections of solids as per given conditions. Up till now, we have studied projection of solids related to simple problems or simple sketches. But during this video, we will study about bit difficult projections of the solids. So, up till now, we have studied two stage problems as well as three stage problems in which we have already discussed that first stage drawing is important by locating whether it is resting on base or resting on corner. And then we will go on producing for projection related to HP and projections related to VP or inclination related to HP and inclination related to VP. Whether the object is resting on VP or HP, depending upon that we went on drawing the two stage problem as well as three stage problems. In this problem, we observe a hexagonal base prism with base side 18 mm and axis height 30 mm based on one of its bottom edges on HP such that its axis is tilted to HP by 30 degree and tilted to VP by 45 degree complete its projections. Now, if you read first line that is hexagonal base prism is resting on HP. So, it is clear from this that top view will be observed as a hexagon whereas, front view will be observed as a rectangle. Now, another thing it is clear here that the prism is resting on base edge. So, base edge will be towards your right side because we have practice of tilting the sorry towards right side or in the clockwise direction. So, when we are rotating or we are inclined in the front view or in the clockwise direction we must rest the prism in such a way that the base edge will be towards your right hand side. So, accordingly we will draw the top view. So, top view will be hexagon with the side base side parallel to HP and perpendicular to VP. So, if you observe here we have drawn a hexagon here in such a way that this base edge and this base edge is perpendicular to VP whereas, parallel to HP. So, it is named accordingly A B C D E F and according to 1 2 3 4 and so on. Now, project this base edge vertically upwards project this base edge C B vertically upwards the height is given as 32 mm and the base side is given 80 mm. So, each of this side will be 80 mm base side will be 18 mm and when we project it upwards for the front view height of this that is height of the axis will be 32 mm. So, draw a hexagon of 18 mm side such that the base edges will be perpendicular to VP and parallel to HP project it upwards to get the front view of height 32 mm. So, accordingly we will name it now this is C B projected here. So, it will be B B dash C dash this is D and A projected upwards. So, it will be A dash D dash and this is E F projected upwards. So, it will be E dash F dash. So, this is the base on which it is resting on HP. Now, another phase that is top phase will be 32. So, here it will be 2 dash 3 dash 1 4 this will be 1 4 and this is 5 6 and the axis of course, will be overlapping by overlapped by A 1 or D 4. Now, we have completed the first stage that is a rectangle as a front view and hexagon as a top. Now, as the rectangle is in the front view we will be inclining this in such a way as given in the problem that is the axis is tilted to HP by 30 degree. So, we will make it inclination as 30 degree with HP. So, axis is along the line A 1 or D 4. So, this axis will be making 30 degree with HP, but well while drawing this case we must ensure that E F should not leave the base or X Y. So, it should be resting E F should be resting on HP as it is given the problem that is one of its base H is resting on HP. So, E F should not leave the X Y line or do the plane HP. So, now as it is given that axis is making 30 degree if you consider this as a right angle triangle where this is the right angle here and this angle will be 30 degree as given in the problem. So, this will be 60 degree. So, better way practically we will draw a 60 degree angle base H or base phase by keeping E F on the X Y line or E F on the HP make a 60 degree angle and redraw this cage here. So, this total prism or front view of the prism with axis making 30 degree will of course will be make base will be making 60 degree we will redraw it. So, after redrawing it we have to project now we will project BC downwards like this. So, this is BC we have projected downwards this is point B projected here you will get a point B here then it is followed by C. So, we will project C here we will get BC here then it is AD point. So, A will be projected downwards A will be projected horizontally. So, you will get A point similarly you will get D point that is projecting D horizontally projecting D vertically then E F similarly we will plot the point E and F E is projected horizontally and vertically F is also projected horizontally and vertically. So, you will get A B C D E F it is the view of the base when the axis of the prism is making 30 degree with HP. Now, what top phase? So, when we project 1 2 3 4 5 6 like this. So, 2 and 3 is projected downwards horizontally 2 will be projected you will get point 2 then 3 point is projected horizontally 3 point will be projected vertically you will get point 3. So, it is a 2 3 line then 1 will be projected horizontally 1 will be projected vertically you will get point 1. Similarly, on the same line you will get point 4 by projecting 4 horizontally and 4 vertically and about 5 6. So, 5 dash and 6 dash will be projected downwards 5 will be projected horizontally here and you will get point 5 and 6. Now, the question arises that which of the lines will be dotted line and which of the lines will be dark line. Now, the observer is here now the observer is observing vertically downwards from here to here from the observer E F is away from it from the observer E F is away. So, the lines related to point E and F will be dotted line. So, see here we have point E. So, point E is a common point that is E D E F and E 5 these 3 lines must be dotted line whereas, E F is here. So, line F A line F 6 and line F E will be dotted lines. So, in all if you observe 3 lines of this and 3 lines related to F total 5 lines will be there because E F is common. So, these 5 lines must be dotted line. So, while drawing we will join A B B C C D as the dark lines A F as the dotted line F E as the dotted line and E D as the dotted line. Similarly, here as this phase is totally visible. So, all the lines related to this phase will be dark lines. So, we will connect join 1 2 3 3 4 4 5 and 4 6. Now, as the external area or external lines will be dark lines. So, we will connect 1 A 4 D and this will be the hexagon. Now, as A E E 6 and E F A F F 6 and F E will be dotted lines. We will join those with dotted line, but as a B 2 is a dark line. See if you observe this is a B 2 line. So, B 2 is a dark line. So, B 2 will be drawn as a dark line and then later on remaining part of F 6. Which is to be expected as a dotted line. So, F 2 will not be dotted, but B 2 when we join with dark line it will be overlapping the dotted line. So, remaining part of F 6 will be dotted. Similarly, here E 2 E 5 is expected to be a dark line dotted line sorry, but E 3 that is the part of C 3 will be a dark line. So, 3 5 will be a dotted line. So, if you observe this visibility of this, when you are observing from here point E F is away from you. So, the lines related to E and related to E F will be dotted line. So, accordingly we have drawn it here up till here it was similar to the video we studied priorly. Now, the next part is the axis is inclined to VP or tilted to VP by 45 degrees. So, now we have to rotate this or we have to rotate in such a way that the axis of this will be at 45 degree. But, if you observe here the axis is 32 mm, later on when the prism was inclined or the axis was inclined at 30 degree to HP. This is short term or this line will not be a two length line. So, 4 dash which is axis of the prism will be reduced by length. So, this is a apparent length, but when we are saying that axis is making 45 degree to VP means the true length of the axis or true length of the axis of the prism will be making 45 degree. So, here we will initially draw a 45 degree line which is expected to be a true length line mark the true length on this draw a locus of this. And on that locus we will mark the apparent length or top view length of axis here. So, it is O O dash which is a two length of the axis line that is 32 mm. O O dash is a locus or locus of O dash or O 1 dash on which we will mark this as a top view length. And then we will redraw these things here such that again A B C D will be towards VP and 1 2 3 4 5 6 will be away from the VP. So, redrawing this by drawing these parallel lines 1 2 3 4 parallel lines parallel to the apparent axis we have drawn. And according to the by marking with the compass the distances here will complete the view. Similar view we will be copying here. So, dotted line shown here will be dotted here also. Now, when we are completed this prism view we are observing from here. So, the line or the points away from we are A B. So, here when we project these things that is A B C D we will project upwards A B C D we will project horizontally. The points related to B and A which is away from the observer will be dotted. So, here again similar thing like here. So, lines related to A and B will be dotted lines other lines we will complete as a dark line. And finally, we will get the view of the prism when it is axis is making 30 degree with HP and 45 degree to VP. So, this part is new for you where we are drawing a two length line at particular angle and later on the apparent length of the axis marking it on locus of the O and completing the view here or rotating this total sketch at that particular angle and project it we will get the final view. So, for this I have referred engineering graphics by Mahalakshmi Publications House. Thank you.