Eg projection of plane

PROJECTIONS OF PLANES
In this topic various plane figures are the objects.
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
In which manner it’s position with HP & VP will be described?
1. Description of the plane figure.
2. It’s position with HP and VP.
To draw their projections means F.V, T.V. &
S.V.
What will be given in the problem?
What is usually asked in the problem?

TYPES OF PLANE FIGURES
SQUARE RECTANGLE TRIANGLE
CIRCLE TRAPEZOID PARALLELOGRAM
PENTAGON HEXAGON DIAMOND

TYPES OF PLANES
PERPENDICULAR PLANES
Planes which are perpendicular to one of the principal planes
of projection and inclined or parallel to the other
OBLIQUE PLANES
Planes inclined to both the reference planes
TRACE OF PLANE
HORIZONTAL TRACE
The intersections line of the plane surface with
H.P
VERTICAL TRACE
The intersections line of the plane surface with V.P

Depending on its orientation with respect to the reference
planes, a plane can be classified as
Perpendicular plane
Perpendicular to both the reference planes (profile
plane)
Perpendicular to one and parallel to the other
Perpendicular to H.P. and parallel to V.P.
Perpendicular to V.P. and parallel to H.P.
Perpendicular to one and inclined to the other
Perpendicular to H.P. and inclined to V.P.
Perpendicular to V.P. and inclined to H.P.
Oblique plane
Inclined to both the reference plane
TYPES OF PLANES

Summary: Projection of a plane inclined to one reference
plane and perpendicular to the other reference plane
1. Assume the plane parallel to that reference plane to which it is to
be made inclined
2. Tilt the plane to the required inclination
3. Rotate the appropriate view so that the edge makes the required
angle with the reference axis. Make sure that the angle of rotation is
drawn taking into account the true length of the edge. The size and
shape of this view does not change during the rotation. Project the
other view so that the distances of the corners from the xy axis are
preserved

1. If the true shape and size of a plane is seen in its front (top) view,
then it is parallel to the V.P. (H.P.).
2. If the front (top) view is not the true shape, then the plane is inclined
to the V.P. (H.P.).
3. If a line is in the V.P. (H.P.), its true length is seen in the front(top)
view
4. If a plane or line does not change its relation with the reference
plane the projection on that reference plane does not change in size
and shape.
5. The angle of inclinations of the plane can be obtained by drawing
edge view using the front view and the top view. The edge views are the
auxiliary front and auxiliary top views.
6. To obtain the true shape of a plane given its projections, one needs to
proceed by first drawing the edge view of the plane. Then the true
shape is obtained by drawing an auxiliary view using a reference line
drawn parallel to the edge view.
POINTS TO REMEMBER

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 rest plane & project it’s other
view.(Ref. 2nd
pair B on previous page illustration )
Now Complete STEP 3. By making side inclined to the rest plane & project it’s other view.
(Ref. 3nd
pair C on previous page illustration )

Problem: A thin rectangular circular plate of 50 mm diameter appears
as an ellipse with major
axis 50 mm and minor axis 30 mm in the top view. Draw the projections when
the major axis
appears horizontal

EXAMPLE:-
Draw the projections of a regular hexagon of 20 mm side, having one
of its sides in the H.P and inclined at 45° to the V.P; and its surface
making an angle of 30° with the H.P

X Y
a
b c
d
a’
b’
c’d’
a1
b1 c1
d1
a1
b1
c1
d1
a’b’
d’c’ c’1 d’1
b’1 a’1
450
300
Problem 1:
Rectangle 30mm and 50mm
sides is resting on HP on one
small side which is 300
inclined
to VP,while the surface of the
plane makes 450
inclination with
HP. Draw it’s projections.
Read problem and answer following questions
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.
Surface // to Hp Surface inclined to Hp
Side
Inclined
to Vp

Problem 2:
A 300
– 600
set square of longest
side
100 mm long, is in VP and 300
inclined
to HP while it’s surface is 450
inclined
to VP.Draw it’s projections
(Surface & Side inclinations directly given)
Read problem and answer following
questions
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.
c1
X
Y
300
450
a’1
b’1
c’1
a
c
a’
a
b1
b’
b
a1b
c
a’1
b’1
c’1
c’
Hence begin with FV, draw triangle above X-Y
keeping longest side vertical.
Surface // to Vp Surface inclined to Vp
side inclined to Hp

c
c1
X Y
450
a’1
b’1
c’1
a
c
a’
a
b1
b’
b
a1b
a’1
b’1
c’1
c’
35
10
Problem 3:
A 300
– 600
set square of longest side
100 mm long is in VP and it’s surface
450
inclined to VP. One end of longest
side is 10 mm and other end is 35 mm
above HP. Draw it’s projections
(Surface inclination directly given.
Side inclination indirectly given)
questions
1 .Surface inclined to which plane? -------
VP
2. Assumption for initial position? ------// to
VP
FV
4. Which side will be vertical? ------longest
side.
Hence begin with FV, draw triangle above XY
First TWO steps are similar to previous problem.
Note the manner in which side inclination is given.
End A 35 mm above Hp & End B is 10 mm above Hp
So redraw 2nd
Fv as final Fv placing these ends as sa

2. Assumption for initial position? ------ // to HP
4. Which side will be vertical? -------- any side.
Hence begin with TV, draw pentagon below
X-Y line, taking one side vertical.
Problem 4:
A regular pentagon of 30 mm sides is
resting on HP on one of it’s sides with
it’s surface 450
inclined to HP.
Draw it’s projections when the side in
HP makes 300
angle with VP
a’b’ d’
b1
d
c1
a
c’e’
b
c
d1
b’1
a1
e’1 c’1
d’1
a1
b1
c1d1
d’
a’b’
c’e’
e1
e1
a’1
X Y450
300
e
SURFACE AND SIDE INCLINATIONS
ARE DIRECTLY GIVEN.

Problem 5:
A regular pentagon of 30 mm sides is resting
on HP on one of it’s sides while it’s opposite
vertex (corner) is 30 mm above HP.
Draw projections when side in HP is 300
inclined to VP.
4. Which side will be vertical? --------any side.
b’
d’
a’
c’e’
a1
b1
c1d1
e1
b1
c1
d1
a1
e1
b’1
e’1 c’1
d’1
a’1
X Ya’b’ d’c’e’
30
a
b
c
d
e
300
SURFACE INCLINATION INDIRECTLY GIVEN
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.
Hence redraw 1st
Fv as a 2nd
Fv making above arrangement.
Keep a’b’ on xy & d’ 30 mm above xy.

4. Which diagonal horizontal? ---------- Longer
Problem 6: A rhombus of diagonals 40 mm and 70 mm long
respectively has one end of it’s longer diagonal in HP while that
diagonal is 350
inclined to HP. If the top-view of the same diagonal
makes 400
inclination with VP, draw it’s projections.
In problem no.6 inclination of Tv of that
diagonal is
given, It could be drawn directly as
shown in 3rd
step.
Hence begin with TV, draw rhombus
below
X-Y line, taking longer diagonal // to X-Y

Q12.7: Draw the projections of a regular hexagon of 25mm sides, having one of its
side in the H.P. and inclined at 60 to the V.P. and its surface making an angle of 45º
with the H.P.
X Y
a
b
c
d
e
f
f’ e’d’c’b’a’
f’
e’
d’
c’
b’
a’
a1
b1
c1
d1
e1
f1
f1
a1
c1
b1
d1
e1
45º
60º
a1’
b1’
c1’
d1’e1’
f1’
Plane parallel to HP
Plane inclined to HP
at 45°and ┴ to VP
Side on the H.P. making 60°
with the VP.

Q12.6: A square ABCD of 50 mm side has its corner A in the H.P., its diagonal AC inclined at
30º to the H.P. and the diagonal BD inclined at 45º to the V.P. and parallel to the H.P. Draw its
projections.
X Y
a
b
c
d
45º
50
a’
b’
d’ c’ a’
b’
d’
c’
30º
a1
c1
b1
d1
45º
a1
c1
b1
d1
b1’
c1’
d1’
Keep AC parallel to the H.P.
& BD perpendicular to V.P.
(considering inclination of
AC as inclination of the
plane)
Incline AC at 30º to the H.P.
i.e. incline the edge view
(FV) at 30º to the HP
Incline BD at 45º to the V.P.
a1’

Q: Draw a rhombus of 100 mm and 60 mm long diagonals with longer diagonal horizontal. The figure
is the top view of a square having 100 mm long diagonals. Draw its front view.
X Y
100
100
a
b
c
d
a’ b’ c’d’
60
a’
b’
c’
d’
a1
b1
c1
d1
a1
c1
d1 b1
100
60
a1
c1
d1 b1

Q4: Draw projections of a rhombus having diagonals 125 mm and 50 mm long, the smaller
diagonal of which is parallel to both the principal planes, while the other is inclined at 30º to
the H.P.
X Y
a
b
c
d
125
a’
b’
d’ c’
Keep AC parallel to the H.P. &
BD perpendicular to V.P.
(considering inclination of AC
as inclination of the plane and
inclination of BD as inclination
of edge)
Incline AC at 30º to the H.P. Make BD parallel to XY
a’
b’
d’
c’
30º
a1
b1
c1
d1
a1
b1
c1
d1
a1’
b1’
c1’
d1’
50

a
b c
d
ef
a’
b’
f’
c’
e’ d’
a’
b’
f’
c’
e’
d’
45°
a1
f1 e1
d1
b1 c1
60°
a1
b1
c1
d1
e1
f1
a1’
b1’
c1’
d1’
e1’
f1’
Q 2:A regular hexagon of 40mm side has a corner in the HP. Its surface inclined at45° to
the HP and the top view of the diagonal through the corner which is in the HP makes an
angle of 60° with the VP. Draw its projections.
Plane parallel to HP
Plane inclined to HP
at 45°and ┴ to VP
Top view of the diagonal
making 60° with the VP.
Y
X

Q12.10: A thin rectangular plate of sides 60 mm X 30 mm has its shorter side in the V.P. and
inclined at 30º to the H.P. Project its top view if its front view is a square of 30 mm long sides
X Y
a
b
30
a’
b’ c’
A rectangle can be seen as a
square in the F.V. only when its
surface is inclined to VP. So
for the first view keep the
plane // to VP & shorter edge ┴
to HP
F.V. (square) is drawn first Incline a1’b1’ at 30º to the
H.P.
d’
60
c
d
a
b
c
d
a1’
b1’ c1’
d1’ 30º
a1’
b1’
c1’
d1’
a1b1
c1 d1
60

Q12.11: A circular plate of negligible thickness and 50 mm diameter appears as an ellipse in
the front view, having its major axis 50 mm long and minor axis 30 mm long. Draw its top
view when the major axis of the ellipse is horizontal.
A circle can be seen as a
ellipse in the F.V. only when its
surface is inclined to VP. So
for the first view keep the
plane // to VP.
X Y
1’
2’
3’
4’
5’
6’
7’
8’
9’
10’
11’
12’
1 2,
12
3,
11
4,
10
5,
9
6,
8
7
1
2,
12
3,
11
4,
10
5,
9
6,
8
7
50 Ø
Incline the T.V. till the
distance between the end
projectors is 30 mm
30
11’
21’
31’
41’
51’
61’
71’
81’
91’
101’
111’
121’
11’21’
31’
41’
51’
61’
71’81’
91’
101’
111’
121’
Incline the F.V. till the
major axis becomes
horizontal
11’
21’
31’
41’
51’
61’81’
91’
101’
111’
121’

a’
b’
c’
a.b c
50
70 a.b
c
a1’
b1’
c1’
a1’
b1’
c1’
a1
b1
c1
45º
Problem 9 : A plate having shape of an isosceles triangle has base 50 mm long and
altitude 70 mm. It is so placed that in the front view it is seen as an equilateral triangle of
50 mm sides an done side inclined at 45º to xy. Draw its top view

Problem 12.9:
A 300
– 600
100 mm long, is in VP and 300
inclined
to HP while it’s surface is 450
inclined
to VP.Draw it’s projections
(Surface & Side inclinations directly
given)
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.
c1
X Y
300
450
a’1
b’1
c’1
a
c
a’
a
b1
b’
b
a1b
c
a’1
b’1
c’1
c’
Surface // to Vp Surface inclined to Vp
side inclined to Hp

c
c1
X Y
450
a’1
b’1
c’1
a
c
a’
a
b1
b’
b
a1b
a’1
b’1
c’1
c’
35
10
Problem 3:
A 300
– 600
100 mm long is in VP and it’s surface
450
inclined to VP. One end of longest
side is 10 mm and other end is 35 mm
above HP. Draw it’s projections
(Surface inclination directly given.
Side inclination indirectly given)
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.
First TWO steps are similar to previous problem.
Note the manner in which side inclination is given.
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.

4. Which side will be vertical? -------- any side.
Problem 4:
A regular pentagon of 30 mm sides is
resting on HP on one of it’s sides with it’s
surface 450
inclined to HP.
Draw it’s projections when the side in HP
makes 300
angle with VP
a’b’ d’
b1
d
c1
a
c’e’
b
c
d1
b’1
a1
e’1
c’1
d’1
a1
b1
c1d1
d’
a’b’
c’e’
e1
e1
a’1
X Y450
300
e
SURFACE AND SIDE INCLINATIONS
ARE DIRECTLY GIVEN.

Problem 5:
A regular pentagon of 30 mm sides is resting
on HP on one of it’s sides while it’s opposite
vertex (corner) is 30 mm above HP.
Draw projections when side in HP is 300
inclined to VP.
4. Which side will be vertical? --------any side.
b’
d’
a’
c’e’
a1
b1
c1d1
e1
b1
c1
d1
a1
e1
b’1
e’1
c’1
d’1
a’1
X Ya’b’ d’c’e’
30
a
b
c
d
e
300
SURFACE INCLINATION INDIRECTLY GIVEN
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.
Hence redraw 1st
Fv as a 2nd
Keep a’b’ on xy & d’ 30 mm above xy.

T L
a
d
c
b
a’ b’ d’ c’
X Ya’
b’ d’
c’
a1
b1
d1
c1
a
1
b
1
d
1
c
1
450
300 a’1
b’1
c’1
d’1
a1
b1
d1
c1
a
d
c
b
a’ b’ d’ c’
a’
b’ d’
c’
a
1
b
1
d
1
c
1
300
a’1
b’1
c’1
d’1
Problem 8: 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’s Tv
is 450
inclined to Vp. Draw it’s projections.
which is 300
inclined to Hp while it makes
450
4. Which diameter horizontal? ---------- AC
Hence begin with TV, draw rhombus below
The difference in these two problems is in step 3 only.
In problem no.8 inclination of Tv of that AC is
given, It could be drawn directly as shown in 3rd
step.
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.
Note the difference in
construction of 3rd
step
in both solutions.

Problem 10: End A of diameter AB of a circle is in HP
A nd end B is in VP.Diameter AB, 50 mm long is
300
& 600
inclined to HP & VP respectively.
Draw projections of circle.
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..
SOLVE SEPARATELY
ON DRAWING SHEET
GIVING NAMES TO VARIOUS
POINTS AS USUAL,
AS THE CASE IS IMPORTANT
TL
X Y
300
600
4. Which diameter horizontal? ---------- AB
Hence begin with TV, draw CIRCLE below
X-Y line, taking DIA. AB // to X-Y

As 3rd
step
redraw 2nd
Tv keeping
side DE on xy line.
Because it is in VP
as said in problem.
X Y
a
b
c
d
e
f
Problem 11:
A hexagonal lamina has its one side in HP and
Its apposite parallel side is 25mm above Hp and
In Vp. Draw it’s projections.
Take side of hexagon 30 mm long.
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
Keep a’b’ on xy & d’e’ 25 mm above xy.
25
f’
e’
d’
c’
b’
a’
a1
b1
c1
d1
e1
f1
c1
’
b’1a’1
f’1
d’1
e’1
f1
a1
c1
b1
d1e1
4. Which diameter horizontal? ---------- AC
Hence begin with TV, draw rhombus below

X Y
which is 300
inclined to Hp while it’s Tv
is 450
Read problem and answer
following questions
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

which is 300
inclined to Hp while it makes
450
Note the difference in
construction of 3rd
step
in both solutions.
The difference in these two problems is in step 3 only.
In problem no.8 inclination of Tv of that AC is
given, It could be drawn directly as shown in 3rd
step.
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 10: End A of diameter AB of a circle is in HP
A nd end B is in VP.Diameter AB, 50 mm long is
300
& 600
inclined to HP & VP respectively.
Draw projections of circle.
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..
SOLVE SEPARATELY
ON DRAWING SHEET
GIVING NAMES TO VARIOUS
POINTS AS USUAL,
AS THE CASE IS IMPORTANT
TL
X Y
300
600
questions
1. Surface inclined to which plane? -------
HP
2. Assumption for initial position? ------ // to
HP
TV
4. Which diameter horizontal? ----------
AB
Hence begin with TV, draw CIRCLE
below
X-Y line, taking DIA. AB // to X-Y

As 3rd
step
redraw 2nd
Tv keeping
side DE on xy line.
Because it is in VP
as said in problem.
X Y
a
b
c
d
e
f
Problem 11:
A hexagonal lamina has its one side in HP and
Its apposite parallel side is 25mm above Hp and
In Vp. Draw it’s projections.
Take side of hexagon 30 mm long.
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
Keep a’b’ on xy & d’e’ 25 mm above xy.
25
f’
e’
d’
c’
b’
a’
a1
b1
c1
d1
e1
f1
c1
’
b’1a’1
f’1
d’1
e’1
f1
a1
c1
b1
d1e1
questions
1. Surface inclined to which plane? -------
HP
2. Assumption for initial position? ------ // to
HP
TV
4. Which diameter horizontal? ----------
AC
Hence begin with TV, draw rhombus
below

A B
C
H
H/3
G
X Y
a’
b’
c’
g’
b a,g c
b
a,g
c
450
a’1
c’1
b’1
g’1
FREELY SUSPENDED CASES.
1.In this case the plane of the figure always remains perpendicular to Hp.
2.It may remain parallel or inclined to Vp.
3.Hence TV in this case will be always a LINE view.
4.Assuming surface // to Vp, draw 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 1st
FV.
IMPORTANT POINTS
Problem 12:
An isosceles triangle of 40 mm long
base side, 60 mm long altitude Is
freely suspended from one corner of
Base side. It's plane is 450
inclined to
Vp. Draw it’s projections.
Similarly solve next problem
of Semi-circle
First draw a given triangle
With given dimensions,
Locate it’s centroid position
And
join it with point of suspension.

Eg projection of plane

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Eg projection of plane