projections of Solids Examples

projections of Solids Examples

Categories of illustrated problems

 

Problem no. 1,2,3,4            General cases of solid inclined to HP & VP

Problem no. 5 & 6               Cases of cube & tetrahedron

Problem no. 7                      Case of freely suspended solid with side view.

Problem no. 8                      Case of Cube (with side view)

Problem no. 9                      Case of true length inclination with HP & VP.

Problem no. 10 & 11          Cases of composite solids. (Auxiliary plane)

Problem no.  12                  Case of a frustum (auxiliary plane)

1 ] A square pyramid, 40 mm base sides, and axis 60 mm long, has a triangle face on the ground and the vertical plane containing the axis makes an angle of 45^º with the VP. Draw its projections. Take apex nearer to VP

Solution steps:

Triangular face on HP, means it is lying on HP:

1. Assume it standing on HP.


2. Its Tv will show the true shape of the base (Square)


3. Draw square of 40 mm sides with one side vertical TV & taking 50 mm axis project FV. (a triangle)


4. Name all points as shown in the illustration.)


5. Draw 2^nd Fv in lying position I.e o'c'd' face on XY. And project its TV.


6. Make visible lines dark and hidden dotted, as per the procedure.


7. Then construct remaining inclination with VP ( VP containing axis ic the centerline of 2^nd TV. Make it 45^º to XY as shown take apex near to XY, as it is nearer to VP ) & project final FV.

2 ] A cone 40 mm diameter and 50 mm axis is resting on one generator on HP which makes 30^º inclination with VP Draw its projections.

Solution steps:

Resting on HP on one generator, means lying on HP:

1.  Assume it standing on Hp.


2. Its TV will show True shape of base (Circle)


3. Draw 40 mm die. circle as TV & taking 50 mm axis project FV.(a triangle)


4. Name all points as shown in the illustration.


5. Draw 2^nd FV in lying position I.e. o'e' on XY. And project its TV below XY.


6. Make visible lines dark and hidden dotted, as per the procedure.


7. Then construct remaining inclination with VP (Generator o₁ e₁ 30^º to XY as shown) & project final FV.

For dark and dotted lines

1. Draw proper outline of new view dark.


2. Decide the direction of an observer.


3. Select nearest point to the observer and Draw all lines starting from it-dark.


4. Select the farthest point to an observer and draw all lines (remaining) from it-dotted.

3 ] A cylinder 40 mm diameter and 50 mm axis is resting on one point of a base circle on VP while its axis makes 45^º with VP and FV of the axis 35^º with HP. Draw projections.

Solution steps:

Resting on VP on one point of the base, means inclined to VP:

1. Assume it standing on VP.


2. Its FV will show True shape of base & top (circle)


3. Draw 40 mm die. Circle as FV & taking 50 mm axis project TV. (a rectangle )


4. Name all points as shown in the illustration.


5. Draw 2^nd TV making axis 45^º to XY and project its FV above XY.


6. Make visible lines dark and hidden dotted, as per the procedure.


7. Then construct remaining inclination with HP (FV of axis i.e. centerline of view to XY as shown) & project final TV.

4 ] A square pyramid 30 mm base side and 50 mm long axis is resting on its apex on HP, such that its one slant edge is vertical and a triangular face through it is perpendicular to Vp. Draw its projections.

Solution steps:

1. Assume it standing on HP but as said on apex. (inverted).


2. Its TV will show the true shape of the base (Square)


3. Draw a corner case square of 30 mm sides as TV (as shown ) Showing all slant edges dotted, as those will not be visible from the top.


4. Taking 50 mm axis project FV. (a triangle)


5. Name all points as shown in the illustration.


6. Draw 2^nd FV keeping o' a' slant edge vertical & project its TV


7. Make visible lines dark and hidden dotted, as per the procedure.


8. Then redraw 2^nd TV as final TV keeping a₁ o₁ d₁ triangular face perpendicular to VP i.e. XY. Then as usual project final FV.

5 ] A cube of 50 mm long edge is so placed on HP on one corner that a body diagonal is parallel to HP and perpendicular to VP Draw its projections.

Solution steps:

1. Assuming standing on Hp, begin with Tv, a square with all sides equally inclined to XY. project Fv and name all points of FV &TV.


2. Draw a body-diagonal joining c' with 3' ( This can become // to XY )


3. From 1' drop a perpendicular on this and name it p'


4. Draw 2^nd  Fv in which 1' -p' line is vertical means c'-3 diagonal must be horizontal. .now as usual project Tv..


5. In final Tv draw same diagonal is perpendicular to Vp as said in the problem. then as usual project final FV.

6 ] A tetrahedron of 50 mm long edges is resting on one edge on Hp while one triangular face containing this edge is vertical and 45 inclined to Vp Draw projections.

Solution steps:

As it is resting assume it standing on HP. Begin with TV, an equilateral triangle as side case as shown: First project base points of FV on XY, name those & axis lines. From a' with TL of edge, 50 mm, cut on-axis line & mark o' (as axis is not known, o' is finalized by slant edge length) Then complete FV. In 2^nd FV make face o' b' c' vertical as said in problem. And like all previous problems solve completely.



Important:

Tetrahedron is a special type of triangular pyramid in which base sides & slant edges are equal in length. Solid of four faces. Like cube it is also described by one dimension only. Axis length generally not given.

7 ] A pentagonal pyramid 30 mm base sides & 60 mm long axis, is freely suspended from one corner of the base so that a plane containing its axis remains parallel to VP. Drae its three views.

Solution steps:

In all suspended cases axis shows inclination with HP.'

1. Hence assuming it standing on HP, draw TV - a regular pentagon, corner case.


2. Project FV & locate CG position on-axis - (1/4 H from base.) and name g' and join it with corner d'


3. As 2^nd FV, redraw first keeping line g' d' vertical.


4. As usual project corresponding TV and then side view looking from.

Important:

When a solid from a corner, then line a joining point of contact & C.G. remains vertical. (Here axis shows inclination with HP.) So in all such cases, assume solid standing on HP Initially.

8 ] A cube of 50 mm long edges is so placed on HP on one corner that a body diagonal through this corner is perpendicular to HP and parallel to VP Draw its three views.

Solution steps:

1. Assuming it standing on HP begin with TV, a square of a corner case.


2. Project corresponding FV & name all point as usual in both views.


3. Join a' 1' as body diagonal and draw 2^nd FV making it vertical (I' on XY)


4. Project its TV drawing dark and dotted lines as per the procedure.


5. With standard method construct Left-Hand side view

(Draw a 45^º inclined line in TV region (below XY.) project horizontally all points of TV on this line and reflect vertically upward, above XY. After this, draw horizontal lines, from all points of FV, to meet these lines. Name points of intersections and join properly. For dark & dotted lines locate observer on the left side of FV as shown.)

9 ] A right circular cone, 40 mm base diameter and 60 mm long axis is resting on HP on one point of base circle such that its axis makes 45^º inclination with HP and 40^º inclination with VP. Draw its projections.

Solution steps:

This case resembles to Problem No. 7 & 9 from projections of planes topic. In previous all cases 2^nd inclination was done by a parameter not showing TL. Like TV of axis is inclined to VP etc. But here it is clearly said that the axis is 40^º inclined to VP. Means here TL inclination is expected. So the same construction done in those problems is done here also. See carefully the final TV and inclination taken there.

So assuming it standing on HP begin as usual.

10 ] A triangular prism, 40 mm base side 60 mm axis is lying on HP on one rectangular face with axis perpendicular to VP. One square pyramid is learning on its face centrally with axis || to VP. its base side is 30 mm & axis is 60 mm long resting on HP on one edge of the base. Draw  FV & TV of both solids. Project another FV on an AVP 45^º inclined to VP. 

Steps:

Draw FV of lying prism ( an equilateral Triangle) And FV of a learning pyramid. Project TV of both solids. Draw x₁ y₁ 45^º inclined to XY and project aux. FV on it. Mark the distance of the first FV from XY for the distance of aux. FV from x₁ y₁ line. Note the observer's directions shown by arrows and further steps carefully.

11 ] A hexagonal prism of base side 30 mm long and axis 40 mm long, is standing on HP on its base with one base edge || to VP. A tetrahedron is placed centrally on the top of it. The base of the tetrahedron is a triangle formed by joining alternate corners of the top of the prism. Draw projections of both solids. Project an auxiliary TV on AIP 45^º inclined to HP. 

Steps:

Draw a regular hexagon as TV of standing prism with one side || to XY and name the top points. Project its FV a  rectangle and name its top. Now join its alternate corners a-c-e and the triangle formed is base of a tetrahedron as said. Locate center of this triangle & locate apex o Extending its axis line upward mark apex o' By cutting TL of edge of tetrahedron equal to a-c. and complete the FV of the tetrahedron. Draw an AIP (x1y1 ) 45^º inclined to XY and project aux. TV on it by using similar steps like the previous problems.

12 ] A frustum of the regular hexagonal pyramid is standing on its larger base On HP with one base side perpendicular to VP. Draw its FV & TV. Project its aux. TV on an AIP parallel to one of the slant edges showing TL. The base side is 50 mm long, top side 30 mm long and 50 mm is the height of frustum.

Please write comments if you find anything incorrect and if you have any queries about this topic or you want to share more information about the topic discussed above.