Projections Of Planes

Projections Of Planes

in this topic various plane figures are the objects.

What is usually asked in the problem?

To draw the projection means F.V, T.V. & S.V.

What will be given in the problem?

1. Description of the plane figure.


2. It's a position with HP  and VP.

In which manner it's position with HP & VP will be described?

1. Inclination of its SURFACE with one of the reference plane will be given


2. inclination of one of its EDGES  with another reference  plane will be given

(Hence this will be the case of an object inclined to both reference planes.)

Study the illustration showing surface &  side inclination given next...

Case of rectangle ( projection Of rectangle )

1 ] Surface parallel to HP 



2 ] Surface Inclined to HP



3 ] One small inclined to VP

PROCEDURE OF SOLVING THE PROBLEM:

IN THREE STEPS EACH PROBLEM CAN BE SOLVED: (As Show in the previous Illustration )

step 1: Assume suitable conditions & draw Fv & Tv of initial position .

step 2: Now consider surface inclination & draw 2^nd   Fv & Tv.

step 3: After this, consider side/edge inclination and draw 3^rd (final) Fv &Tv.

ASSUMPTIONS FOR INITIAL POSITION :

(Initial position means assuming surface || to HP or VP )

1. If in position surface is inclined to HP - assume it || HP or if the surface is inclined to VP -assume it || to VP


2. Now if surface is assumed || to HP- it's TV Will show True shape. And if surface is assumed || to VP -its FV will show True shape.


3. Hence begin withdrawing TV or FV as true shape.


4. While drawing this True shape- keep one side/edge (which is making inclination) perpendicular to XY line (similar to pair no. A previous page illustration.)

Now complete Step 2. By making surface inclined to the reps plane & project its other view. (Ref. 2^nd pair B on previous page illustration )

Now complete Step 3. By making side inclined to the reps plane & project its other view. (Ref. 3^rd pair  C on previous page illustration)

APPLY SAME STEPS TO SOLVE NEXT PROBLEMS.

1 ] Rectangle 30 mm and 50 mm sides is resting on HP on one small side which is 30^º inclined to VP, while the surface of the plane marks 45^º inclination with HP. draw its projections.

Read the problem and answer the following question

1. Surface inclined to which plane?..........  HP


2. Assumption for initial position? ...........|| to HP


3. So which view will show True shape? ......... TV


4. which side will be vertical? .........One small side.

Hence begin with TV, draw rectangle below X-Y drawing one small side vertical.



 

2 ] A 30^º-60^º set square of longest side 100 mm long, is in VP and 30^º inclined to HP while its surface is 45^º inclined to VP. Draw its projections

Read the problem and answer the following question

1. Surface inclined to which plane?..........  VP


2. Assumption for initial position? ...........|| to VP


3. So which view will show True shape? ......... FV


4. which side will be vertical? ........ Longest side.

Hence begin with FV, draw triangle above  X-Y keeping longest side vertical.



 

3 ] A 30^º-60^º set square of longest side 100 mm long is in VP and it's surface 45^º inclined to VP. One end of the longest side is 10 mm and the other end is 35 mm above HP. Draw its projections (Surface inclination directly given. Side inclination indirectly given )

Read the problem and answer the following question

1. Surface inclined to which plane?..........  VP


2. Assumption for initial position? ...........|| to VP


3. So which view will show True shape? ......... FV


4. which side will be vertical? .........Longest side.

Hence begin with FV, draw triangle above X-Y keeping longest side vertical.



PROBLEM 4:

A regular pentagon of 30 mm sides is resting on HP on one of its sides with its surface 45^º inclined to HP. Draw its projections when the side in HP makes 30^º angle with VP (Surface and side inclinations are directly given)

Read the problem and answer the following question

1. Surface inclined to which plane?..........  HP


2. Assumption for initial position? ...........|| to HP


3. So which view will show True shape? ......... TV


4. which side will be vertical? .........Any side.

Hence begin with TV, draw pentagon below the X-Y line, taking one side vertical.



 

5 ] A regular pentagon of 30 mm sides is resting on HP on one of sides its opposite vertex (corner) is 30 mm above Hp. Draw projections when side in HP is 30^º inclined to VP. 

Read the problem and answer the following question

1. Surface inclined to which plane?..........  HP


2. Assumption for initial position? ...........|| to HP


3. So which view will show True shape? ......... TV


4. which side will be vertical? .........Any side.

Hence begin with TV, draw pentagon below the X-Y line, taking one side vertical.



Only change is the manner in which surface inclination is described:

one side on HP & its opposite corner 30mm above HP. Hence redraw 1st FV as a 2nd FVmaking above arrangement. Keep a' b' on XY & d' 30 mm above XY.

 

6 ] A circle of 50 mm diameter is resting on HP on end A of its diameter AC which is 30^º inclined to HP while its TV is 45^º inclined to VP. Draw its projections.

Read the problem and answer the following question

1. Surface inclined to which plane?..........  HP


2. Assumption for initial position? ...........|| to HP


3. So which will show True shape? ......... TV


4. which diameter horizontal? .........AC

Hence begin with TV, draw rhombus below X-Y line, taking longer diagonal || to X-Y.



 

7 ] A circle of 50 mm diameter is resting on HP on end A its diameter AC which is 30^º inclined to HP while it makes 45^º inclined to VP. Draw its projections.

The difference between these two problems is in step 3 only.

In problem no.6 inclination of TV of that AC is given, it could be drawn directly as shown in the 3^rd steps.

While in no. 7 angle of AC itself i.e. its TL, is given. Hence here angle of TL is taken, locus of c₁ is drawn and then LTV i.e. a₁ c₁ is marked and final TV was completed. Study illustrations carefully.



 

8 ] End A of diameter AB of a circle is in HP and End B in VP. Diameter AB, 50 mm long is 30^º & 60^º inclined to HP & VP respectively. Draw projection of circle.

Read the problem and answer the following question

1. Surface inclined to which plane?..........  HP


2. Assumption for initial position? ...........|| to HP


3. So which will show True shape? ......... TV


4. which diameter horizontal? .........AB

Hence begin with TV, draw CIRCLE below X-Y line, taking DIA. AB || to X-Y.

The problem is similar to the previous problem of the circle -no. 7 But in the 3^rd step, there is more change.

Like 7^th problem True length inclination of die. AB is expected but if you carefully note- The sum of its inclinations with HP & Vp is 90^º Means line AB lies in a profile plane. Hence its both TV & FV must arrive on to one single projector.

So do the construction accordingly AND note the case carefully.



Give a name to various points as usual, As the case is important.

 

9 ] A hexagonal lamina has its one side in HP and its opposite side is 25 mm above Hp and in VP. Draw its projections. Take side of hexagon 30 mm long.

Read the problem and answer the following question

1. Surface inclined to which plane?..........  HP


2. Assumption for initial position? ...........|| to HP


3. So which view will show True shape? ......... TV


4. which diameter horizontal? .........AC

Hence begin with TV, draw rhombus below X-Y line, taking longer diagonal || to X-Y.



 

ONLY change is the manner in which surface inclination is described:

One side on HP & its opposite side 25 mm above HP. Hence redraw 1^st FV as a 2^nd FV making above arrangement keep a' b' on XY & d' e' 25 mm above XY.

As 3rd step redraw 2nd TV keeping side DE on XY line because it is in VP as said in the problem.

Projection Of Plane-Freely Suspended Cases

1 ] An isosceles triangle of 40 mm long base side, 60 mm long altitude is freely suspended from one corner of the base side. its plane is 45^º inclined to VP. Draw its projections.

IMPORTANCE POINTS:

1. In this case the plane of the figure always remains perpendicular to HP.


2. It may remain parallel or inclined Vp.


3. Hence Tv in this case will be always a LINE view.


4. Assuming surface || to Vp, draw the true shape in suspended position as FV. (here keep line joining point of contact & centroid of fig. vertical)


5. Always begin with FV as a True shape but in a suspended position. As shown in 1^st FV.

First draw a given triangle with given dimensions, Locate it's centroid position and join it with a point of suspension.

2 ] A semicircle of 100 mm diameter is suspended from a point on its straight edge 30 mm from the midpoint of that edge so that the surface makes an angle of 45^º with VP. Draw its projections.

IMPORTANT POINTS :

1. In the case the plane of the figure always remains perpendicular to HP.


2. It may remain parallel or inclined to VP.


3. Hence Tv in these cases will be always a LIne View.


4. Assuming surface // to VP, draw the true shape in suspended position as FV. (Here keep line joining point of contact & centroid of fig. vertical )


5. Always begin with FV as a True shape but in a suspended position. As shown in 1^st FV.

First draw a given semicircle with given dimensions, Locate it's centroid position and join it with a point of suspension.

 

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