Observation

Candidates were asked to draw the true shape of the given views of a triangular laminae.  Candidates were expected to find the true lengths of the lines.  Many of the candidates who attempted this question did not get it right.

Candidates were expected to use either of the two methods provided below:                                 
A.         CONSTRUCTION OF HYPERBOLA (METHOD 1)
(i)         draw the directrix  d-d and axis A-B at 900 to each other;
(ii)         locate the focus f 30 from A;
(iii)        locate the vertex V using ratio 4:5;
(iv)        draw perpendicular lines E,F,G,H, etc at equal distances (in multiple of 8) on axis ab;
(v)        use f as centre and radius 1.25 times the distances of each line from directrix dd and draw arcs to give 1,2,3,4, etc.;

1. join the points to form a smooth curve.

 

 

The solution is shown below:     

OR

 

 

 

 

 

 

CONSTRUCTION OF HYPERBOLA (METHOD 2)

(i)         draw the directrix and axis at 900 to each other;
(ii)         locate the focus 30 from the directrix;
(iii)        divide DF in a ratio 4:5 to locate vertex V;
(iv)        draw vertical lines across the axis at equal distances and parallel to the directrix;
(v)        draw a right angled triangle ABC based on ratio horizontal : vertical = 4:5;
(vi)        use F as centre and radius on ratio 5:4 on each corresponding vertical and horizontal lines to give points 1, 2, 3, 4 etc;
(vii)       join the points to give the curve – hyperbola.

The solution is shown below:

(B)        CONSTRUCTION OF LAMINA (METHOD 1)

            (i)         copy the given views;
(ii)         use B’C’ as radius and describe an arc to locate point D on A’B’;
(iii)        draw horizontal line EF at C;

1. project vertical line from D to locate C2 on EF;
2. join C2 B as true length of C’B’;
3. use A’C’ as radius and describe an arc to locate point G on A’B’;
4. project vertical line from G to locate C3 on EF;
5. join AC3 as true length of A’C’;
6. draw true shape of triangular lamina using lengths AB, AC3 and BC3.

The solution is shown below:

OR

 

CONSTRUCTION OF LAMINA (METHOD 2)

1. copy the given views;
2. extend BA to any convenient length at E;
3. draw a perpendicular at E to M;
4. draw a line from C parallel to EB to locate F;
5. draw a perpendicular on A’B’ from C’;
6. transfer the vertical height of triangle A’B’C’ = DC’ to draw FG;
7. use GE as radius and draw an arc to touch EM at H;
8. draw a perpendicular from AB through C to J;
9. draw line HP parallel to EB to intercept NJ at C2;
10. join AC2, C2B and AB to give the true shape of triangular lamina.

 

The solution is shown below: