 Hello, I am Mr. D. J. Doshi, working as assistant professor, department of mechanical engineering waltz in the field of technology, SOLAP. At the end of this session, students will be able to discuss the different positions of axis of solids. Now, if you consider this axis of square prism parallel to VP and perpendicular to HP. If you observe 3D sketch drawn towards left hand side of the screen, it is a square prism means base and top base will be a square whereas, its side faces will be rectangular. So, side faces are the faces, this face this total will be called as a face. So, these faces are rectangular whereas, the top surface and the bottom one is a square. This is the axis of the prism, now if you observe this is called HP, this is called VP, now this axis is parallel to VP and perpendicular to HP. So, the square prism placed on HP, when you observe is for a front view that will be a rectangle with a one more line at the centre vertical line. So, this is 1 dash, 1 dash, 1 dash, 2 dash, 2 dash, 4 dash, 4 dash overlapping each other and 3 dash, 3 dash and when you observe it from top, it will be a square of the given side that is 40 mm side whereas, axis is 60 mm. Now, the axis line and these two corner lines are overlapping each other. So, it will be drawn as a dark line. So, it is a line 2 dash, 2 dash and 4 dash, 4 dash whereas, while drawing the square at the top view, you have to take care that the corners will be towards your right and left side and the line drawn through 1, 3 must be parallel to X, Y and or parallel to the another plane. So, now what is given this axis of the square prism is parallel to VP and perpendicular to HP that we have drawn and the name it accordingly, now when you will draw a square here 1, 2, 3, 4, project 1 vertically upwards from 1 dash to 1 dash, this is the height of the prism or axis length. So, this will be 60 mm project 3, so it will be 3 dash, 3 dash project 2 and 4 which will be on the same line, the line drawn vertically upwards will be passing through 2 and 4. So, 2, 4 will be a vertical line again at the height of 60 mm. So, this is the completion of first stage projection when the square prism axis is perpendicular to HP and parallel to VP. Now, this one is axis of square pyramid parallel to VP and perpendicular to HP. In this case, if you observe the position of the pyramid is such that its axis which is present here is perpendicular to HP and it is parallel to VP. Now, in first case that is in square prism we have seen that it was kept in such a way that the corners are towards right hand side and towards left hand side, but here we will observe that base edge are parallel to VP whereas in that case those are angular to VP. So, these are the base edges that is 1, 2, 2, 2, 3, 3, 2, 4, these are 4 base edges which are out of that 2 are parallel to VP or parallel to XY line and 2 are perpendicular to VP that is perpendicular to XY line. So, while drawing you have to draw a square of 1, 2, 3, 4 of which 2 lines or 2 base edges will be parallel to XY line or parallel to VP and 2 base edges will be perpendicular to VP. Now, in case of prism we never joined 1, 3 and 2, 4 or 1, 2, 0, 2, 0, 3, 0 and 4, 0, but as this is a pyramid when you observe from the top that is from apex O, you will be observing these generators as O 1, O 2, O 3 and O 4 which are the diagonals of the square whereas in case of prism the generators are vertical one which are not slant one. So, here we will be while drawing pyramid we will take care that the apex will be joined to all the base corners and the axis will be O O dash. So, after drawing 1, 2, 3, 4 a square or a top view of a square pyramid and O projected upwards now 2 and 1 will be on the same line. So, you will get a 0.1 dash 2 dash on XY line then 3 and 4 are on the same projector. So, you will get 3 dash 4 dash as another point then join O dash 3 dash 4 dash, O dash 1 dash 2 dash and this will be the axis of the pyramid which is of 60 mm. Now, these O 2, 3, O these are called as faces of the square pyramid or faces of the pyramid which are triangular one in case of prism if you observe these are rectangular one because their top face is also a square in case of square prism in case of pyramid there will be apex. So, here you have to write down O dash O dash and join O dash 1 dash O dash 2 dash of the same line and so on. So, this is a axis projection drawn for axis of a square pyramid parallel to VP and perpendicular to HP. Now, axis of a cylinder parallel to VP and perpendicular to HP. So, if you observe it is a square sorry it is a cylinder which is placed on HP in such a way that the axis is perpendicular to HP at the same time it is parallel to VP. Now, if you observe here there are no slant lines only base is a circle top is also a circle and there are no actual generators observed like a prism or a pyramid. So, here while drawing what will be its top view and we will be observing from here it is a circle. So, we will draw a circle of 40 mm diameter and now we need some generators which will be projected upwards to get the front view of the cylinder. So, we will divide it the circle into 8 equal parts or 12 equal parts name accordingly 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 or 1, 2, 8. Now, project each point now 1 is a extreme right hand side point. So, projected upwards you will get a line 1 dash, 1 dash that will be equal to length of the cylinder axis of the cylinder. So, this is 60 mm similarly 2 dash, 2 dash projected upwards you will get 2 dash, 8 dash and accordingly this is a center line or it is again overlapping to 3 and 7. So, it is the generator you have drawn this one also same thing and also on. So, we will get 2, 3 and 2 5 lines of length 60 mm join those points at the bottom as well as a drop you will get a rectangular surface which is the front view of the cylinder when it is resting on HP. So, in this case in case of cylinder as well as in case of cone you need to divide the circle which may be either front view or which may be either top view into equal parts to get the correct sketch or correct image of the concern, concern solid. Now, this is a cone again same thing axis is perpendicular to HP and parallel to will be in this case you have to specifically show the apex as over here again the top view will be a circle. So, we will draw a circle here of the 40 mm diameter divide the circle into 8 or 12 parts projected upwards you will get 0.1 dash, 2 dash, 3 dash, 4 dash, 5 dash, 6 dash, 7 dash and 8 dash and as in case of cylinder we have drawn the vertical lines as a generator here as this is a cone like pyramid that will have a apex. So, join the apex with each point that is O 1 O dash 1 dash, O dash 2 dash and O dash A dash same line O 3 O 7 O 4 O 6 and O 5. So, this will be the front view with axis length given 50 mm or 60 mm whatever is and this is a top view as a base circle. So, here the procedure is similar to cylinder except here there will be apex whereas in case of cylinder there will be 2 parallel lines to each other at the length equal to axis length. Now, up till now we have studied when the solid is in HP. Now, here the solid is in VP because the axis is perpendicular to VP and parallel to HP. Now, if you observe this square prism this is the axis here which is parallel to HP and it is perpendicular to VP. So, its front view will be a square which is equal to the base or top surface and the top view will be the rectangular rectangle of base length and axis length. So, while drawing first you will draw in VP that is in a front view as 1 2 3 4 as a square as the sides 2 sides are parallel to HP and 2 sides of the base are perpendicular to HP. So, accordingly we will draw a square as shown in this case as a front view project it downwards you will get 1 dash 2 dash 3 dash 4 dash and so on. So, draw it axis length is 60 mm join it you will get the top view as a rectangle of 60 by 40 whereas front view will be a square 40 mm by 40 mm. Now, this is a square pyramid again axis parallel to HP and perpendicular to VP. So, here now if you observe this is inclined the base edges are inclined at 45 degree to XY line or 2 HP. So, here we will draw a square as a front view in such a way that the base edges will be equally inclined to HP or equally inclined to XY line. So, name accordingly 1 2 3 4 now in case of cylinder sorry in case of square prism we need not to draw apex, but in case of pyramid you have to draw apex. So, there will be O at the center which is the intersection point of the ribonus. So, O 1 O 1 dash O 2 dash O 3 dash and O 4 dash will be the generators of the square pyramid drawn in the front. Now, project 1 downwards again 2 O and 4 will be on the same line. So, draw 2 4 O and 3 join O 1 O 2 4 and O 3 where the apex will be nearer to the cone parallel to HP and perpendicular to VP. In this case this is a cone which is resting on VP. So, you have drawn a top view as a circle and divide it into 8 parts or 12 parts project it downwards same procedure, but here the top view will be a circle and sorry front view will be a circle sorry front view will be a circle and the top view will be a triangle where apex is going towards the observer or nearer to the observer procedure is same. Now, you can imagine or you will imagine for the object which is position as a cylinder in VP or axis parallel to HP and perpendicular to VP. So, if you will think it it is similar to the cone initially we have to study it and except the apex point there will be 2 surfaces. So, this is the actual drawing you have to draw that is a circle as a front view and the top view will be a rectangle of length equal to axis length and this will be equal to diameter of the cylinder. So, this is a cylinder resting on VP or in VP and axis is perpendicular to VP and parallel to HP. So, these are the reference books we have used. Thank you.