PROJECTIONS OF PLANES

In this topic various plane figures are the objects.

What is usually asked in the problem?

To draw their projections means F.V, T.V. & S.V.

What will be given in the problem?

1. Description of the plane figure.
2. It’s position with HP and VP.

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

1.Inclination of it’s SURFACE with one of the reference planes will be given.
2. Inclination of one of it’s EDGES with other reference plane will be given
(Hence this will be a case of an object inclined to both reference Planes.)
Study the illustration showing
R.OYYARAVELU
surface & side inclination given on next page.
 CASE OF A RECTANGLE – OBSERVE AND NOTE ALL STEPS.

SURFACE PARALLEL TO HP SURFACE INCLINED TO HP ONE SMALL SIDE INCLINED TO VP

PICTORIAL PRESENTATION PICTORIAL PRESENTATION PICTORIAL PRESENTATION

FV-1 FV-2
FV-3

-1
TV

3
-2

T V-
TV

ORTHOGRAPHIC ORTHOGRAPHIC ORTHOGRAPHIC

TV-True Shape FV- Inclined to XY FV- Apparent Shape
FV- Line // to xy TV- Reduced Shape TV-Previous Shape
VP
VP VP
d1’ c1’

a’ d’ a1’ b1’
b’ c’

a d a1 d1

b c b1 c1
HP A HP B HP
C
 PROCEDURE OF SOLVING THE PROBLEM:
IN THREE STEPS EACH PROBLEM CAN BE SOLVED:( As Shown In Previous Illustration )
STEP 1. Assume suitable conditions & draw Fv & Tv of initial position.
STEP 2. Now consider surface inclination & draw 2nd Fv & Tv.
STEP 3. After this,consider side/edge inclination and draw 3rd ( final) Fv & Tv.

ASSUMPTIONS FOR INITIAL POSITION:

(Initial Position means assuming surface // to HP or VP)
1.If in problem surface is inclined to HP – assume it // HP
Or If 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 – It’s FV will show True Shape.
3. Hence begin with drawing 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 on previous page illustration ).

Now Complete STEP 2. By making surface inclined to the resp plane & project it’s other view.
(Ref. 2nd pair B on previous page illustration )

Now Complete STEP 3. By making side inclined to the resp plane & project it’s other view.
(Ref. 3nd pair C on previous page illustration )

APPLY SAME STEPS TO SOLVE NEXT ELEVEN PROBLEMS

Problem 1a: Read problem and answer following questions
Rectangle 30mm and 50mm 1. Surface inclined to which plane? ------- HP
sides is resting on HP on one 2. Assumption for initial position? ------// to HP
small side which is 300 inclined 3. So which view will show True shape? --- TV
to VP,while the surface of the 4. Which side will be vertical? ---One small side.
plane makes 450 inclination with Hence begin with TV, draw rectangle below X-Y
HP. Draw it’s projections. drawing one small side vertical.

Surface // to Hp Surface inclined to Hp

d’c’ c’1 d’1

a’b’ c’d’
a’ b’ 450 b’1 a’1 Y
X 300
a d a1 d1 Side
Inclined
to Vp

b c b1 c1
 General Rule of Projections of Planes
• General Rules:
➢ When a plane surface is held perpendicular to a plane of projection, obviously it will be parallel to
projectors. The view of the plane projected on the plane of projection will be a line (Fig.1).
➢ If a plane surface is held parallel to a plane of projection, obviously it will be perpendicular to the
projectors, then its view on that plane will be in true shape associated with true or actual dimension
(Fig.2).
➢ If a plane surface is held inclined to a plane of projection and thus inclined to the projectors, then its
view on that plane will be in apparent shape associated with apparent or reduced dimension (Fig.3).

View (true shape)

View (Apparent
shape)

c
View (line) Project
c
or
b b
Project
 or b C 
c
B
 C
a a a
  90
B
 90  B
90
C
A
A
Project A
Observe or Observe
Observe Plane r Plane r
Fig. 1 Plane Fig. 2 Fig. 3
r
 Initial Conditions
Case 1: Lamina is resting with one of its edges or sides on HP and its surface is inclined to HP:
x
y

Equilateral Triangle Square Pentagon Hexagon Rectangle Right angle Triangle Isosceles Triangle Semi-
Case 2: Lamina
circularis resting with one of its corners on HP and its surface is inclined to HP:

x
y

Case 3: Lamina is resting withSquare

Equilateral Triangle
one of its edges or sides inHexagon
Pentagon
VP and its surface
Rectangle
is inclined to VP:
Isosceles Triangle Circular Semi-
circular

Equilateral Triangle Square Pentagon Hexagon Rectangle Right angle Triangle Isosceles Triangle Semi-
circular

Case 4: Lamina
x is resting with one of its corners in VP and its surface is inclined to VP:
y

Equilateral Triangle Square Pentagon Hexagon Rectangle Isosceles Triangle Circular Semi-
circular

x
y
 Example Problems on Projections of Planes
1. A triangular plane lamina of sides 40 mm is resting on HP with one of its side such that the surface of the
plane lamina makes an angle of 60° with HP. Side on which the plane lamina rests makes an angle of 30°
with VP, draw the top and front views in this position.

c c 1

(b ) (b )
VP a c a 60 a 1 b1
x y
HP
b b1 30
a1

b1
40

c1
c
c1
a1
a
2. A pentagonal plate with edge of 25 mm length is resting on HP on one of its edges. This edge is inclined
to VP at 45° and the plate surface makes 30° angle with HP. Draw its projections.

d d1
(c ) 
e e 1
(c ) (b ) c 1
(b )
VP a e d a 30 a 1 b1
x  y
HP 45
c b1
b1 c1
b c1
a1
d
25

d1

a a1 d1
e e1 e1
3. A hexagonal lamina of sides 25 mm rests on one of its sides on HP. The side opposite to the side on which it rests is
15 mm above of HP and the side on which it rests makes 450 to VP. Draw its projections.

(d )
(c ) e e 1 d1
(c ) f  f1
(b ) (d )

25
(b ) a 1
VP a f e a b c 1
x y
HP  1
c1 45
c
b1
b b1 c1
d d1
a1
25

d1
a e a1 e1
f1
f f1 e1
4. A triangular plane lamina of sides 40 mm is resting on HP with one of its corners touching it such that the surface of the
plane lamina makes an angle of 60° with HP. If the side opposite to the comer on which the plane lamina rests makes
an angle of 30° with VP, draw the top and front views in this position.

(b ) c b1 c 1

(b )
60 a 1
VP a c a y
x
HP
b b1 b1 30

a1
a c1

a1
c c1
5. A mirror 30 mm x 50 mm is inclined to the wall at such an angle that its front view is a square of 30 mm side. The longer
sides of the mirror appear perpendicular to both HP and VP. Find the inclination of the mirror with the wall.

50 30
b c b1 c 1 a 1 b1
30

d a d1 d1 c 1
VP a 1
x a1 y
HP (a) (d) (a) b1
b c b 53

(d) d1 c1
c
6. A square 40 mm side resting on one of its corners on the HP such that, its diagonal associated with corner
on the HP inclined at 450 to HP and inclined at 300 to VP. Draw its projections.

r 1
r

s
(q) s 1
q1
s p 45
VP p (q) r p1
x y
HP
q q1

30 q1
p1
 = 50
p r r1
p1
Locus of r
r1 r
s1
s s1
7. A square 40 mm side resting on one of its corners on the HP such that, its diagonal associated with corner
on the HP inclined at 450 to HP and it appears to be inclined at 300 to VP. Draw its projections.

r r 1

s
(q) s 1 q1

s p 45
VP p (q) r p1
x y
HP
q q1 q1
=
p130

p r r1
p1 r1

s1
s s1
8. A square PQRS of 40 mm side has its diagonal PR inclined at 450 to HP and the diagonal QS inclined at
300 to VP and parallel to HP. Draw its projections.

r 1
r

s
(q) q1
s 1

s p 45 p1
VP p (q) r
x y
HP
q q1

30 r1
q1
p r r1
p1

p1 s1

s1
s
9. A Circular lamina inclined to VP appears in the front view as an ellipse of major axis 60 mm and minor axis 30 mm. The
major axis is parallel to both HP and VP. One end of the minor axis is in both HP and VP. Draw the projections of the
lamina and determine the inclinations of the lamina with HP and VP.
30
d d1
c e c 1 e 1

b f b1 f1

a g a 1 g1
+ +

f1 g1
l 1 h1
h h1 e 1 i 1
l
60 i k 1 i 1 d1 j 1

+
k
j j 1 c 1
b1 a1 l 1 k 1
VP
x HP y
a b c d e f g a a1
(l)b b1 l1
(l) (k) (j) (i) (h) 60 c1 k1
(k)c
d1
j1

+
(j)d
e1
(i) e i1
(h) f f1 h1
g g1
 10. An elliptical lamina of size 50 mm minor axis and 80 mm major axis inclined to VP appears in the front view as a circle.
One end of the major axis is in both HP and VP such that the major axis appears to be perpendicular to both HP and
VP. Draw the projections of the lamina and determine the inclinations of the lamina with VP.

d1 50
d c 1
c e e 1
b f b1 f1
g1
g
50

+ + g1 f1 h1

a a 1
h1 e 1 i 1
l h l 1
i 1 d1 j 1
i k 1

+
k
j j 1
80 c 1 k 1
VP b1 a 1 l 1
x y
HP a b c d e f g a b1 a1 k1
51
(l) (k) (j) (i) (h) (l)b c1
l1
(k)c d1

+
d j1
(j)
e1
(i) e i1

(h) f f1
h1
g g1
 IMPORTANT POINTS
Problem 13 1.In this case the plane of the figure always remains perpendicular to Hp.
:A semicircle of 100 mm diameter 2.It may remain parallel or inclined to Vp.
is suspended from a point on its 3.Hence TV in this case will be always a LINE view.
straight edge 30 mm from the midpoint 4.Assuming surface // to Vp, draw true shape in suspended position as FV.
of that edge so that the surface makes (Here keep line joining point of contact & centroid of fig. vertical )
an angle of 450 with VP. 5.Always begin with FV as a True Shape but in a suspended position.
Draw its projections. AS shown in 1st FV.

A
a’
20 mm
p’
P

G b’
CG g’

c’

e’
d’
X Y

First draw a given semicircle

With given diameter,
Locate it’s centroid position b c a p,g d e
And
join it with point of suspension.
Problem 2: Read problem and answer following questions
A 300 – 600 set square of longest side 1 .Surface inclined to which plane? ------- VP
100 mm long, is in VP and 300 inclined 2. Assumption for initial position? ------// to VP
to HP while it’s surface is 450 inclined 3. So which view will show True shape? --- FV
to VP.Draw it’s projections 4. Which side will be vertical? ------longest side.

(Surface & Side inclinations directly given)

Hence begin with FV, draw triangle above X-Y
keeping longest side vertical.
a’ a’1

c’ c’1
side inclined to Hp
c’1

a’1

b’1
b’1
b’
300
X a
b 450 a1 b1 Y
a c
b c1
c
Surface // to Vp Surface inclined to Vp
Problem 3: Read problem and answer following questions
A 300 – 600 set square of longest side 1 .Surface inclined to which plane? ------- VP
100 mm long is in VP and it’s surface 2. Assumption for initial position? ------// to VP
450 inclined to VP. One end of longest 3. So which view will show True shape? --- FV
side is 10 mm and other end is 35 mm 4. Which side will be vertical? ------longest side.
above HP. Draw it’s projections
Hence begin with FV, draw triangle above X-Y
(Surface inclination directly given.
keeping longest side vertical.
Side inclination indirectly given)

First TWO steps are similar to previous problem.

Note the manner in which side inclination is given.
a’ a’1 End A 35 mm above Hp & End B is 10 mm above Hp.
So redraw 2nd Fv as final Fv placing these ends as said.
c’ c’1
c’1

a’1

35
b’1
b’1
b’
X 10 Y
a a1
b 450 b1
a c
b c1
c
Problem 4: Read problem and answer following questions
A regular pentagon of 30 mm sides is 1. Surface inclined to which plane? ------- HP
resting on HP on one of it’s sides with it’s 2. Assumption for initial position? ------ // to HP
surface 450 inclined to HP. 3. So which view will show True shape? --- TV
Draw it’s projections when the side in HP 4. Which side will be vertical? -------- any side.
makes 300 angle with VP Hence begin with TV,draw pentagon below
SURFACE AND SIDE INCLINATIONS X-Y line, taking one side vertical.
ARE DIRECTLY GIVEN.
d’ d’1
c’e’
e’1 c’1
b’ a’
X b’ a’ c’e’ d’ b’1 Y
450 a’1
a1 300
e e1
e1
a a1 b1

d1
d
d1 c1
b b1

c c1
 Problem 5: Read problem and answer following questions
A regular pentagon of 30 mm sides is resting 1. Surface inclined to which plane? ------- HP
on HP on one of it’s sides while it’s opposite 2. Assumption for initial position? ------ // to HP
vertex (corner) is 30 mm above HP. 3. So which view will show True shape? --- TV
Draw projections when side in HP is 300 4. Which side will be vertical? --------any side.
inclined to VP. Hence begin with TV,draw pentagon below
SURFACE INCLINATION INDIRECTLY GIVEN
X-Y line, taking one side vertical.
SIDE INCLINATION DIRECTLY GIVEN:

ONLY CHANGE is
the manner in which surface inclination is described:
One side on Hp & it’s opposite corner 30 mm above Hp. d’ d’1
Hence redraw 1st Fv as a 2nd Fv making above arrangement.
Keep a’b’ on xy & d’ 30 mm above xy. c’e’ c’1
30 e’1

b’ a’ c’e’ d’ a’
X b’ a’1 b’1 Y
300
e1 a1
e
e1
a a1 b1

d d1
d1 c1
b b1

c c1
 c’1
b’1
Problem 8: A circle of 50 mm diameter is a’ b’ d’ c’
resting on Hp on end A of it’s diameter AC 300 a’1 d’1 Y
X
which is 300 inclined to Hp while it’s Tv d1
450
d
is 450 inclined to Vp.Draw it’s projections.

a ca c1
1

Read problem and answer following questions

1. Surface inclined to which plane? ------- HP b b1
2. Assumption for initial position? ------ // to HP
3. So which view will show True shape? --- TV The difference in these two problems is in step 3 only.
4. Which diameter horizontal? ---------- AC In problem no.8 inclination of Tv of that AC is
Hence begin with TV,draw rhombus below given,It could be drawn directly as shown in 3rd step.
X-Y line, taking longer diagonal // to X-Y While in no.9 angle of AC itself i.e. it’s 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 illustration carefully.
Problem 9: A circle of 50 mm diameter is
resting on Hp on end A of it’s diameter AC
which is 300 inclined to Hp while it makes c’1
450 inclined to Vp. Draw it’s projections. b’1
a’ b’ d’ c’
a’1 d’1
d d1
300
Note the difference in
a ca c1
construction of 3rd step 1

in both solutions.
b b1
 Read problem and answer following questions
Problem 10: End A of diameter AB of a circle is in HP 1. Surface inclined to which plane? ------- HP
A nd end B is in VP.Diameter AB, 50 mm long is 2. Assumption for initial position? ------ // to HP
300 & 600 inclined to HP & VP respectively. 3. So which view will show True shape? --- TV
Draw projections of circle. 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 previous problem of circle – no.9.

But in the 3rd step there is one more change.
Like 9th problem True Length inclination of dia.AB is definitely expected
but if you carefully note - the the SUM of it’s inclinations with HP & VP is 900.
Means Line AB lies in a Profile Plane.
Hence it’s both Tv & Fv must arrive on one single projector.
So do the construction accordingly AND note the case carefully..

300
X Y
600
SOLVE SEPARATELY
ON DRAWING SHEET
GIVING NAMES TO VARIOUS
POINTS AS USUAL,
AS THE CASE IS IMPORTANT
 Problem 11: Read problem and answer following questions
A hexagonal lamina has its one side in HP and 1. Surface inclined to which plane? ------- HP
Its apposite parallel side is 25mm above Hp and 2. Assumption for initial position? ------ // to HP
In Vp. Draw it’s projections.
Take side of hexagon 30 mm long.
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 & it’s opposite side 25 mm above Hp.
Hence redraw 1st Fv as a 2nd Fv making above arrangement.
Keep a’b’ on xy & d’e’ 25 mm above xy.

e’1 d’1

25 f’1 c1’
X a’b’ c’ f’ d’e’ a’1 b’1 Y
f f1 e1 d1
f1 c1
a e a1 e1
a1 b1 As 3rd step
b d b1 d1 redraw 2nd Tv keeping
c1
side DE on xy line.
c
Because it is in VP
as said in problem.
 FREELY SUSPENDED CASES.
IMPORTANT POINTS
1.In this case the plane of the figure always remains perpendicular to Hp.
Problem 12: 2.It may remain parallel or inclined to Vp.
An isosceles triangle of 40 mm long 3.Hence TV in this case will be always a LINE view.
base side, 60 mm long altitude Is 4.Assuming surface // to Vp, draw true shape in suspended position as FV.
freely suspended from one corner of (Here keep line joining point of contact & centroid of fig. vertical )
Base side.It’s plane is 450 inclined to 5.Always begin with FV as a True Shape but in a suspended position.
Vp. Draw it’s projections. AS shown in 1st FV.

a’1
a’
C

b’1
b’ g’ g’1
H
G c’ c’1
H/3
X Y

A B

b a,g c 450
First draw a given triangle
With given dimensions,
Locate it’s centroid position
And Similarly solve next problem
join it with point of suspension. of Semi-circle
 To determine true shape of plane figure when it’s projections are given.
BY USING AUXILIARY PLANE METHOD
WHAT WILL BE THE PROBLEM?
Description of final Fv & Tv will be given.
You are supposed to determine true shape of that plane figure.

Follow the below given steps:

1. Draw the given Fv & Tv as per the given information in problem.
2. Then among all lines of Fv & Tv select a line showing True Length (T.L.)
(It’s other view must be // to xy)
3. Draw x1-y1 perpendicular to this line showing T.L.
4. Project view on x1-y1 ( it must be a line view)
5. Draw x2-y2 // to this line view & project new view on it.
It will be the required answer i.e. True Shape.

The facts you must know:-

If you carefully study and observe the solutions of all previous problems,
You will find
IF ONE VIEW IS A LINE VIEW & THAT TOO PARALLEL TO XY LINE,
THEN AND THEN IT’S OTHER VIEW WILL SHOW TRUE SHAPE:

NOW FINAL VIEWS ARE ALWAYS SOME SHAPE, NOT LINE VIEWS:
SO APPLYING ABOVE METHOD: Study Next
WE FIRST CONVERT ONE VIEW IN INCLINED LINE VIEW .(By using x1y1 aux.plane) Four Cases
THEN BY MAKING IT // TO X2-Y2 WE GET TRUE SHAPE.
 Problem 14 Tv is a triangle abc. Ab is 50 mm long, angle cab is 300 and angle cba is 650.
a’b’c’ is a Fv. a’ is 25 mm, b’ is 40 mm and c’ is 10 mm above Hp respectively. Draw projections
of that figure and find it’s true shape.
As per the procedure-
1.First draw Fv & Tv as per the data.
2.In Tv line ab is // to xy hence it’s other view a’b’ is TL. So draw x1y1 perpendicular to it.
3.Project view on x1y1.
a) First draw projectors from a’b’ & c’ on x1y1.
b) from xy take distances of a,b & c( Tv) mark on these projectors from x1y1. Name points a1b1 & c1.
c) This line view is an Aux.Tv. Draw x2y2 // to this line view and project Aux. Fv on it.
for that from x1y1 take distances of a’b’ & c’ and mark from x2y= on new projectors.
4.Name points a’1 b’1 & c’1 and join them. This will be the required true shape.
Y1
a1b1 Y
2

b’ b’1
15
a’

15
C1
10 C’ X1
X X2 a’1
Y
c c’1

ALWAYS FOR NEW FV TAKE

DISTANCES OF PREVIOUS FV
300 650 AND FOR NEW TV, DISTANCES
a OF PREVIOUS TV
b
50 mm REMEMBER!!
 Problem 15: Fv & Tv of a triangular plate are shown.
Determine it’s true shape.

USE SAME PROCEDURE STEPS 50

OF PREVIOUS PROBLEM: 25
BUT THERE IS ONE DIFFICULTY: c’
15
NO LINE IS // TO XY IN ANY VIEW. a’
1’
MEANS NO TL IS AVAILABLE.
20
IN SUCH CASES DRAW ONE LINE b’
// TO XY IN ANY VIEW & IT’S OTHER 10
X Y
VIEW CAN BE CONSIDERED AS TL 15
x1
FOR THE PURPOSE. a c

HERE a’ 1’ line in Fv is drawn // to xy. 40

HENCE it’s Tv a-1 becomes TL. 1 c’1
a’1 y2
b
THEN FOLLOW SAME STEPS AND
c1
DETERMINE TRUE SHAPE. b’1
(STUDY THE ILLUSTRATION) y1
x2

ALWAYS FOR NEW FV TAKE b1

DISTANCES OF PREVIOUS FV
AND FOR NEW TV, DISTANCES d1
OF PREVIOUS TV
REMEMBER!!
PROBLEM 16: Fv & Tv both are circles of 50 mm diameter. Determine true shape of an elliptical plate.

ADOPT SAME PROCEDURE.

a c is considered as line // to xy.
Then a’c’ becomes TL for the purpose.
Using steps properly true shape can be
50D y1
Easily determined. b’ b1 y2
Study the illustration.
TL ac1 1
a’ c’ b’1
c’1
d’ d
X1 1
X d Y
X2

ALWAYS, FOR NEW FV

a’1
TAKE DISTANCES OF d’1
PREVIOUS FV AND a c TRUE SHAPE
FOR NEW TV, DISTANCES
OF PREVIOUS TV
REMEMBER!! 50 D. b
 Problem 17 : Draw a regular pentagon of
30 mm sides with one side 300 inclined to xy.
This figure is Tv of some plane whose Fv is
A line 450 inclined to xy.
Determine it’s true shape. b1
a1
c1
IN THIS CASE ALSO TRUE LENGTH
IS NOT AVAILABLE IN ANY VIEW. X1

BUT ACTUALLY WE DONOT REQUIRE

TL TO FIND IT’S TRUE SHAPE, AS ONE a’ e1 d1
VIEW (FV) IS ALREADY A LINE VIEW.
SO JUST BY DRAWING X1Y1 // TO THIS b’
VIEW WE CAN PROJECT VIEW ON IT e’
AND GET TRUE SHAPE:
c’ Y1
d’
STUDY THE ILLUSTRATION.. 450
X 300 Y
e

d
ALWAYS FOR NEW FV a
TAKE DISTANCES OF
PREVIOUS FV AND FOR
NEW TV, DISTANCES OF
PREVIOUS TV c
REMEMBER!! b
 REFERENCES

• K.R. Gopalakrshina, Engineering Graphics, Subhas Publications, Bangalore, 1999.

• N.D. Bhatt, Engineering Drawing, Charotar Publishing House Pvt. Ltd., 1991.