 Hello, I am Professor D.J. Doshi working as Assistant Professor Department of Mechanical Engineering, Valchin Institute of Technology, Solapur. At the end of this session, students will be able to draw the projection solid as per given conditions. So, during last session or last video, we have studied the solid resting on HP. Now, we will study the solids which are resting on VP or sometimes it is called as nailed on VP or fixed on VP. So, this problem is related to cone, a cone of base 60 mm diameter and axis 66 mm long, relying on one of its generator on VP with its front view of an axis making an angle of 50 degree with HP, draw its projection considering the apex nearer to the observer. So, now first thing as the axis is making or resting on HP or VP. So, initially what we will do, we will draw the projection of solids that is the top view and front view of the solid considering it is resting on VP. Now, as it is resting on VP, its front view will be the shape of the base only. So, what will be the base of the cone, it is a circle. So, when it is resting on VP, the front view will be visible as a circle, whereas the top view of the cone will be visible as a triangle, but as it is given that the apex is nearer to the observer, it should be away from XY line because observer will be observing from the front that is from the apex. So, apex will be nearer to the observer and away from the XY line. So, while drawing first stage, you have to take care, draw a circular cone, circular circle as a base as a front view and consider the, this is the circle you have drawn as a front view and as a top view, we will draw a triangle with apex away from XY line or this is towards the observer because observer is observing from this side. So, it will be the observer from here, so it is nearer to the observer. So, apex will be towards the observer, observer is here and it is resting on VP. So, triangle will be, triangle will be drawn in such a way that apex is towards the observer. Now, for the sake of redrawing, what we have to do, divide the circle into 12 equal parts, project those things downwards, name it accordingly. So, it is divided into 12 parts, 1 to 12, naming is given and we are projected. So, this is 1, this is 2 and 12, this is 11 and 3, 4 and 10, 5 and 9, 6 and 8 and 7. Now, what next part is given? One of its generator on VP. So, now, which is the nearest one? 7 O or O7 is the generator on which it will be resting on VP. So, tilt the cone or tilt the triangle in such a way that O7 will be lying on XY line or lying on the VP. So, we have drawn O7. Now, complete or copy the same view that is this triangle, copy is the same thing here around this O7. So, O7 is lying on VP and we have completed the triangle which is a top view of the cone in the first stage. Now, this is second stage top view. Now, project the points upwards. Now, O is projected from here, from the center O is projected you will get O dash. Then, this 1 to 12 parts will be projecting also 1 is projected upwards accordingly 1 and 7 are the points here which are on the same line projected here you will get 2 points 1 and 7. Similarly, 12 and 8 you will get here 12 and 8, then 11 and 9 and 10. Similarly, 4, 3 and 5, 2 and 6 and these points we will join with a smooth curve which will be an ellipse. Draw the tangential lines or join O dash 10 and O dash 4 dash. So, we will get a second stage front view of the cone of 60 mm diameter and 66 mm axis. But here the length of the axis will go on changing as it is a virtual view of the axis. So, now here you will get O dash, here you will get O dash. This is the second stage problem, second stage of the problem. Now, next step is given as axis is making 50 degree with HP. Axis is making an angle at 50 degree to HP. So, now here we will draw the line at 50 degree to HP. Redraw this sketch against around it. So, while taking redraw, redrawing the sketch you have to take care you will get easily the point O dash because you will be imagining the point as here, then you will be drawing axis. But for that you need all these 12 points. So, this length is known that is axis length is known you will get the point here O dash again. From there draw a perpendicular line which will give you point 10 dash and 4 dash at that particular distance which we will measure from here. Then draw the parallel lines to it. Draw the perpendicular lines at the major distance you will get all the 1 to 12 points. Join points 1 to 12 you will get the ellipse again join O dash 10 dash and join O dash 4 dash. So, here you are drawing the third stage front view of the cone which is making axis of which is making 50 degree to the HP. So, you have drawn a 50 degree line axis and you have reconstructed or redrawn the cone here. Now, again project vertically downwards the points 1 to 12 of the drawn ellipse. Draw 1 to 12 projectors from 1 to 12 numbers horizontally you will get the intersection points 12 here you will get intersection points accordingly you have to draw. Now, here you will feel difficulty in the imagining which line or which part of the ellipse will be a dotted line. So, if you observe from here the away point is this one. The away from the observer this is the point. So, the point here will be not visible or will not be visible. So, that point that is from here to point 4 you will be drawing a dotted line. So, those points you will be joining with dotted line whereas, another point other points will be you will be joining with the dark line. Again O7 as it is stated in the second stage that O7 is on the VP and now HP observer here the apex is away from towards the observer. So, it is towards the observer because observer is observing from here to get the top view and for drawing the front view the observer is observing from here. So, it is away from the observer it is towards the observer accordingly we have constructed this. So, this is first stage, second stage, third stage here you have to take care while drawing that the apex should be nearer to the observer and the dotted lines if you take care properly you will be drawing the correct third stage problem for solving this one. Now, next problem is the pentagonal pyramid of 30 mm H of base 60 mm axis height is lying on one of the triangular surfaces in the VP so that the axis is inclined at an angle of 45 degree to HP draw its front view. Now, again this is resting in VP so you will be drawing a pentagon as a front view and the triangle as a top view you are aware of the drawing the generators also. So, initially you will draw a pentagon now it is resting when it is tilted it will be tilted it is resting on the triangular face. So, the base side will be towards your right hand side because you will be tilting it towards your right hand side or clockwise side. So, we have taken the base towards right hand side so 3 4 is towards right hand side so that why after tilting it will be resting on 3 4 or O 3 4 as a triangular face these are the triangular faces. So, while drawing pentagon take care that it will be the base edge will be towards your right side and complete the pentagon. Now, in case of pyramid you have to join all the base points to the apex O which will be visible. So, these are all generators O 4 O 3 O 5 O 2 all these are generators you will be joining this one whereas, in case of prism you are not joining all these things here again accordingly visibility you will complete it. Then our next part is generator one of the generator is on VP so we will tilt it or rotate it in such a way that O 3 4 which is a triangular face which is resting on VP. So, O 3 4 will be on the x y line redraw this to complete the top view of the first stage projected upwards again horizontal projections and vertical projections you will give the 1 2 3 4 points join those things because everything is visible from here to the observer and as the point 3 4 is away from the observer O 3 and O 4 will not be visible which will be lying on the VP. So, O 3 and O 4 will be dotted lines and 1 2 3 4 5 which is a pentagonal shape will be dark line and other generators will be dark line. So, only O 3 and O 4 will be dotted line. Next part is axis is making 45 degree with HP so draw a 45 degree line draw a 45 degree line and now here as this is the imaginary line or this is the apparent length. So, we will mark the apparent length here on 45 degree line then it is called as locus of O O and with true length now as axis is making 45 degree with HP you have to take care that true length of the axis or true length of the pyramid is making 45 degree. So, draw a 45 degree line draw a locus of point O horizontal line parallel to x y take the true length or original length of the axis mark it here and reconstruct the pyramid around it that is redraw these things here projected downwards. Now, when you are observing from here you will observe that point O 2 is away from the observer. So, O 2 and O 1 will be the dotted lines other things other lines will be dark line because point 2 is away from the observer and 1 is also not visible. So, O 1 and O 2 will be dotted lines and accordingly complete the projections thank you.