Solutions to Exercise problems

Brief Overview on Projections of Planes: Solutions to Exercise problems By now, all of us must be aware that a plane is any D figure having an enclosed surface area. In our subject point of view, any closed polygon, circle, semicircle, plates with holes drilled centrally, etc are treated as simple planes with negligible thickness. The projections usually involve three stages in most of the problems. The answer to the problems lies in deciding the following questions: ) Where to draw the given shape? If the true shape is drawn correctly in the st stage, the rest of the stages are very simple and solution is easy to attain. So, we have to first decide what the true shape is and then where it is to be drawn. To do this, we usually learn that there can be three types of problems. (a) The surface angle and shape angle (side/diagonal/diameter/major axis) will be given in the data. If surface angle is with HP, shape is seen in Top View and if surface angle is with VP, shape is seen in FV. (b) The true shape and the reduced shape will be given and also where the reduced shape is seen will also be given. For e.g., a square is seen as a rhombus in the top view or a circle is seen as an ellipse in the front view or a rectangle becomes a square in the front view, etc will be given. So in this case, we can directly know that the shape is being seen in top view, front view, etc. (c) The final rd stage figure will be stated and then we have to arrive at the figure based on the fact that in the first stages, the height of the figure always remains the same and only the width reduces. So only in this case, a little bit of deep study of the question is to be done to carefully decide what the st stage shape of the plane is. For e.g., draw a rhombus of 0 mm and 60 mm diagonals with longer diagonal horizontal. The figure is the top view of a square with a corner on the ground and diagonals 0mm long. show its projections and find plane angle. Here we are asked to draw a rhombus with longer diagonal horizontal and also given that the rhombus is the top view of a square with a corner on HP. Since corner is mentioned, square is to be drawn at 0 to x-y in the top view. We also know that the diagonals of a square are equal and hence in the nd stage, the vertical diagonal height) remains the same and the horizontal diagonal reduces to 60. So, when the question is to draw the rhombus with longer diagonal horizontal, it implies that first

we have to draw the rhombus to 0 & 60 with shorter diagonal horizontal and then rotate the figure in the rd stage to get the longer diagonal horizontal. Same is the case with ellipse also. If a circle becomes an ellipse with its major axis horizontal, it means that first its width decreases and the minor axis will be horizontal. Then when we want the major axis horizontal, we need to rotate the figure by 90 0 in the rd stage to get the solution. So, just a little bit of understanding is essential to know that width always decreases and height remains the same in nd stage. Based on this also sometimes questions can be asked. The solutions to the exercise problems have been divided into cases. They are: ) Case : Given data is Surface angle (plane angle) and shape angle (Side /diagonal /diameter / major axis / etc) ) Case : True shape and reduced shapes are given and where they are seen is also given (in Top view, front view, etc) ) Case : rd stage answer will be stated and the first stages have to be drawn. ) Case : Side view problems (when sum of angles is 90 0 and one end of plane is in HP and other end is in VP). Only problem of circle with its diameters in HP & VP is to be solved. This is very important problem as it is drawn using side view concept of turning the top view by 0.

The following are the problems and the cases to which they belong to. Case : Solved examples are Problems.6,.7,.8,.9 from page 9-. Exercise problems are,,,,7, & in pages -. (Remember that the logic for all the problems is the same. First see where the plane angle or surface angle is & decide where to draw the shape. Then last will be side angle for the rd stage). Only variation is shapes are going to be different in each case). Rule: () Surface angle (Plane angle) is with HP start with the true shape in Top View. () Surface angle (Plane angle) is with VP start with the true shape in Front View. () Side or edge is resting on HP/VP take the starting side of polygon as vertical or perpendicular to x-y. () Corner is resting on HP/VP take the starting side of the polygon as horizontal or parallel to x-y. Case & Case : Solved examples are Problems.0 &. from page -. Exercise problems are, 8, 9& 0 from page. (Remember that the logic for all these problems is the same. First see where the true shape and the reduced shape are seen; either in top view or in front views. Then last will be side angle for the rd stage). Only variation is that the shapes are going to be different in each case. Case : Solved example is Problem. from page. Exercise problem is 6 from page. As the detailed notes on planes have been sent earlier, the steps of each problem are not being stated here. Instead, only the logic is discussed and the answer is given based on first drawing the true shape. The steps of drawing are shown as,,,, & 6 to mean that they are the sequence in which the figures are drawn. For some problems, lettering and dimensioning have not been shown completely with the assumption that you will be able to do it by yourself, observing the first few problems.

EXERCISE XII (Page ) Case type problems ) Draw an equilateral triangle of 7 mm side and inscribe a circle in it. Draw the projections of the figure when the plane is vertical and inclined at 0 0 to the VP and one of the sides is inclined at 0 to the HP. A) Given data: Shape Equilateral Triangle of 7 mm with in circle Plane angle 0 0 to VP (Shape is seen in Front View) Edge / Corner Edge (side) is given; starting side is vertical. Side angle 0 to HP; ( rd stage side rotation) a True Shape Reduced Shape Turn side by 0 to HP. a a 0 ` 7 b b c 7 7 7 c c b x VP HP (7) a(c), () (Top View is Line) b a(c) 0 0 (7) Plane angle is 0 0 b c a 6 y b Final Projections

. A regular hexagon of 0 mm side has a corner in the HP. Its surface is inclined at 0 to the HP and the top view of the diagonal through the corner which is in the HP makes an angle of 60 0 with the VP. Draw its projections. A. Given Data: Shape Hexagon of 0 mm sides. Surface or Plane angle 0 to HP (Shape is seen in Top View) Edge / Corner Corner is given; starting side is horizontal. Diagonal or Side angle diagonal, 60 0 to VP; ( rd stage diagonal rotation) d c 6 d e a f e d a 0 f e b a 60 0 f f a - e a d a - d b d b b c c c 0 True Shape Reduced Shape Turn diagonal a d by 60 0 to VP.

. Draw the projections of a regular pentagon of 0 mm side, having its surface inclined at 0 0 to the HP and a side parallel to the HP and inclined at an angle of 60 0 to the VP. A. Given Data: Shape Pentagon of 0 mm sides. Surface or Plane angle 0 0 to HP (Shape is seen in Top View) Edge / Corner Side parallel; starting side is horizontal. Diagonal or Side angle Side, 60 0 to VP; ( rd stage side rotation) Pentagon, ab=0; Surface angle to HP; side angle to VP. 6 a e d c a c 0 0 e d d 60 0 c c e b a b a c a c d b b e a True Shape Reduced Shape Turn side d e by 60 0 to VP 6

. Draw a regular hexagon of 0 mm side, with its two sides vertical. Draw a circle of 0 mm diameter in its centre. The figure represents a hexagonal plate with a hole in it and having its surface parallel to the VP. Draw the projections when the surface is vertical and inclined at 0 0 to the VP. Assume the thickness of the plate to be equal to that of a line. A. Given Data: Shape Hexagon, 0 mm with a central hole of Ф 0. Surface or Plane angle 0 0 to VP (Shape is seen in front View) Edge / Corner Sides vertical; starting side is vertical. Diagonal or Side angle No rd stage. In this particular problem, there is no rd stage and hence the answer is to draw the stages only. True Shape Reduced Shape (Final Projection) 0 0 0 Locate the centre of the hexagon by intersection of the diagonals and then draw a circle of diameter 0 mm (radius = 0 mm) at the centre. Then, follow the usual procedure to get the solution. Do the labeling as per the usual rules followed earlier. It is left as an exercise. 7

7. A semi circular plate of 80 mm diameter has its straight edge in the VP & inclined at 0 to the HP. The surface of the plate makes an angle of 0 0 with the VP. Draw its projections. A. Given Data: Shape Semicircle of 80 mm diameter. Surface or Plane angle 0 0 to VP (Shape is seen in Front View) Edge / Corner Edge in VP; starting side is vertical. Diagonal or Side angle Side, 0 to VP; ( rd stage side rotation) True Shape Reduced Shape Turn edge by 0 to HP 0 Ф 80 7 7 7 (7) (7) 0 0 7 6 8

E GRAPHICS: PROJECTION OF PLANES S.RAMANATHAN ASST PROF Final MVSREC Projections. A composite plate of negligible thickness is made up of a rectangle 60 mm X 0 mm and a semi circle on its longer side. Draw its projections when the longer side is parallel to the HP & inclined at 0 to the VP, the surface of the plate making an angle of 0 0 with the HP. A. Given Data: Shape Rectangle & Semicircle (60X 0; D= 60) Surface or Plane angle 0 0 to HP (Shape is seen in Top View) Edge / Corner Side parallel; starting side is horizontal Diagonal or Side angle Side, 0 to VP; ( rd stage side rotation) Labeling for the first stage has been shown. Please complete for the other by using same notations as discussed in earlier problems. 6 a (d ) b (c ) 0 0 0 d c 0 a b 60 True Shape Reduced Shape Turn side by 0 to VP 9

. A 60 0 set-square of mm longest side is so kept that the longest side is in HP, making an angle of 0 0 with the VP & the set-square itself inclined at 0 to the HP. Draw the projections of the set square. A. Given Data: Shape Set Square, mm (Right angle Triangle) Surface or Plane angle 0 to HP (Shape is seen in Top View) Edge / Corner Side (edge) in HP; starting side is vertical Diagonal or Side angle Side, 0 0 to VP; ( rd stage side rotation) 6 b a (c ) b 0 a c c c 0 0 c 60 0 b b b a 0 0 a a True Shape Reduced Shape Turn side a c by 0 0 to VP Draw the triangle abc by taking intersection of lines at 0 0 at a & 60 0 at c to get b. ac=. 0

EXERCISE XII (Page ) Case & Case type problems. Draw the projections of a rhombus having diagonals mm and 0 mm long, the smaller diagonal of which is parallel to both the principal planes while the other is inclined at 0 0 to the HP. A. Given Data: Shape Rhombus, mm & 0 mm diagonals. Surface or Plane angle 0 0 to HP (Shape is seen in Top View) Edge / Corner Longer diagonal is taken horizontal first. Diagonal or Side angle smaller diagonal parallel to both VP & HP. ( rd stage details-diagonal turned by 90 0 ) Start with the longer diagonal horizontal in the top view so that its plane can be rotated by 0 0 in the front view. Since plane angle is not given and details are given only about the diagonals, treat one of them as plane and the other as data for side rotation in the rd stage. 6 ( ) 0 0 0 In the rd stage, rotate the rhombus by 90 0 so that the smaller diagonal becomes horizontal. In the first stages, width changes and height remains the same.

8. The top view of a plate, the surface of which is perpendicular to the VP & inclined at 60 0 to the HP is a circle of 60 mm diameter. Draw its three views. Ans) Given data: Shape --not given. Surface or Plane angle 60 0 to HP (Shape is seen in Top View) Edge / Corner ------ Diagonal or Side angle ---------- Reduced shape circle of 60 mm diameter. ) Draw a circle in top view with diameter 60 mm and project it (). ) In front view, draw a line at 60 0 to cut the projector of circle & find the plane length. ) Using the plane length of (), draw it horizontally on x-y line and mark as many points as there are in (). ) Project lines from the plane line and match it with projector from circle to get the final shape of an ellipse. (7 ) 60 0 (7 ) 7 7 Ф 60

9. A plate having the shape of an isosceles triangle has base of 0 mm long and altitude 70 mm. It is so placed that in the front view, it is seen as an equilateral triangle of 0 mm sides and one side is inclined at 0 to x-y. Draw its projections. Ans) Given Data: True Shape Isosceles Triangle. Reduced shape Equilateral Triangle Where is it seen Front view Edge / Corner Side (Edge); starting side vertical. Side angle side, 0 to HP ( rd stage rotation) Since the side of triangle remains same as 0 mm, the starting side of the triangle is taken as vertical so that the width (altitude) reduces in the nd stage to give an equilateral triangle. 0 c 70 b c 60 0 0 b c 0 60 0 0 a a a b c(a) b c(a) a c 6 b b

0. Draw a rhombus of diagonals 00 mm and 60 mm, with the longer diagonal horizontal. The figure is the top of a square of 00 mm long diagonals, with a corner on the ground. Draw its front view and determine the angle made by the plane (surface) with the ground. Ans) Given Data: True Shape Square with a corner on HP. Reduced shape Rhombus of 0 X 60 Where is it seen Top view Edge / Corner corner; starting side horizontal. Side angle 90 0 ; diagonal being tilted, (i) (ii) (iii) Since the square is the true shape, draw it first in top view & draw its projectors. Convert the square into a rhombus such that the longer diagonal remains unchanged vertically and the width reduces to 60 mm in the nd stage. In the rd stage, tilt the rhombus such that 60 mm side is made horizontal; match the projections to get the final views. c c 6 a b (d ) c a b (d ) θ b a d d d a b d 60 a 00 c a c b b 00 60 c 00

Side View Problems EXERCISE XII (Page ) Case type problems 6. Draw the projections of a circle of diameter 7 mm, having the end A of diameter AB in the H.P., the end B in the V.P., and the surface inclined at 0 0 to the H.P and at 60 0 to the V.P. A. Given Data: Shape Circle, 7 mm diameter. Surface or Plane angle 0 0 to HP & 60 0 to V.P (Shape s surface angle is seen in Side View) In the front view and top view, we see a reduced circle (representing an ellipse). Best Example of this case is a ladder standing on a wall with one end on the wall and other end on the floor. The inclinations of the ladder surface can be seen in the side view, assuming the wall as VP & the floor as HP. Also in this special case of problems, the sum of angles made by the surfaces with HP & VP is always 90 0. So, to identify this case of problem in projection of planes, we have to check data: (i) One end on HP & other end on VP. (ii) Sum of angles made by plane (surface) will be equal to 90 0. (iii) TV PP VP HP FV SV

TV 60 0 PP VP 0 0 HP FV SV The final three views of the plane surface are shown below. VP PP FV 60 0 SV (80 mm) 0 0 TV HP 6

Side View Problems EXERCISE XII (Page ) Case type problems Procedure to solve this problem: ) Assume the circle to be resting completely on the VP or HP in the first stage and draw its projections. Usually, we assume it to be resting on the VP. Hence the Front View (FV) will be a circle of 80 mm diameter and the Top View (TV) & Side View (SV) will be straight lines of 80 mm length. To draw the Side View (Left Side View), draw a vertical line of 80 mm to the right of FV at some distance, on the same x-y line. Do the labeling using the usual rules. For the SV use,,, etc. Also note that the views have to lie on x-y since A & B are in HP & VP. Ф 80 TV b b ( ) SV ( ) 8 6 x VP a y HP (8) b(a) (6) a 8 (6 ) ) Since the inclination is seen in the side view, tilt the line b a in the side view by an angle 0 0 to the HP. The length will be the same as 80 mm. ) On b, draw a vertical line, x y which represents the profile plane. ) On b a, mark the same points ( ), ( ), 8 (6 ) at the same distances as on the original line by using arcs or scale. 7

SV TV Ф 80 b x VP a 0 0 y HP 8 (8) b(a) (6) 6 x Side View b ( ) b ( ) ( ) 60 0 ( ) a 8 (6 ) y ) Now, draw projectors from b,,,, etc of side view to the front view. 6) On these lines, draw projectors vertically down from,, b,, etc of the circle in the original FV to get points of the Final Front View, which will be an ellipse. TV b x SV b ( ) 8 6 ( ) 8 (6 ) x VP a y HP a Joining the above points will give ellipse in the FV. y The FV will be shown in the final figure along with the TV. The above figure is for understanding only to identify how the points of projections are to be marked. 8

To obtain the final top view from the side view: (Turn the SV by 0 to get TV) 7) First, project all the points on the SV line (inclined line) onto x-y line. From there draw lines at 0 to meet x - y line. Then project these points horizontally. b x ( ) TV ( ) 8 (6 ) x (8) b(a) (6) b projector a y - projector 0 - projector 8-6 projector a projector y 8) Now, after getting the projectors of side view, match projectors from the top view, (8), b(a), etc of the original line in step to get the final top view, which will also be an ellipse. Thus, obtain the front view by projecting from side view horizontally and obtain the top view by projecting vertically, turning by 0 and then projecting horizontally. In this figure, another exception is that both the final FV and final TV are shown on the same original FV and TV. A little bit of practice is essential to perfect this problem. But the concept involved is very simple, to use side view to get the final projections. 9

The combined final figure with the constructions is shown below. This is the final figure which we have to show and not the above individual figures, which have been shown only for understanding. Side View Ф 80 x b b ( ) Final Front View 8 b 6 x VP a 0 0 y HP (8) b(a) (6) b 60 0 ( ) ( ) a ( ) 8 (6 ) a Final Top View y As a practice, refer to the Problem No.. on Page which is of the same model as above. The only difference is that the surface angles are 60 0 to the HP and 0 0 to the VP. Solve it by using the same concepts mentioned above. 0