INDIAN INSTITUTE OF TECHNOLOGY ROORKEE

Engineering Drawing: Projections

S. P. HARSHA, PhD
PROFESSOR
MIED, IIT ROORKEE
Projection of Planes
What is a plane surface ?
• A plane surface or figure has only two dimension
• Length
• Breadth
• No thickness
Representation of plane
• Three non-collinear points
• A line and a point not in on the line
• Two intersecting lines
• Two parallel lines
Classification of planes
• Principal planes
• Horizontal plane
• Vertical plane
• Secondary planes
• Perpendicular planes
• Oblique planes
Perpendicular planes
• These planes are perpendicular to one or both the principal planes.
They may be Horizontal, vertical or Inclined
• Perpendicular to both the principal planes
• Perpendicular to one of the principal planes and parallel to the other
• Perpendicular to one of the principal planes and inclined to the others
Oblique planes
• Oblique planes is inclined to both the principal planes
Perpendicular planes
Perpendicular to both the principal planes
• Its horizontal trace(H.T) and vertical trace (V.T) are in a straight line
perpendicular to xy

Perpendicular to one of the principal planes and parallel to the other

• Plane Perpendicular to H.P and parallel to the V.P.
Perpendicular planes
Perpendicular to one of the principal planes and parallel to the other
• Plane Perpendicular to V.P and parallel to the H.P.
Perpendicular to one of the principal planes and inclined to the
other
• Plane Perpendicular to H.P and inclined to the V.P.
Perpendicular to one of the principal planes and inclined to the
other
• Plane Perpendicular to V.P and inclined to the H.P.
 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
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

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’ a 1’ 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 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 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 rasp plane & project it’s other view.
(Ref. 2nd pair B on previous page illustration )

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

APPLY SAME STEPS TO SOLVE NEXT PROBLEMS

SOLVED EXAMPLES
Ex 1: A regular pentagon of 25mm side has one side on the ground. Its plane is inclined at
45º to the HP and perpendicular to the VP. Draw its projections and show its traces

Hint: As the plane is inclined to HP, it should be kept

parallel to HP with one edge perpendicular to VP

a’ b’
e’ d’ c’
45º
X Y

b b1
a a1
25

c c1

e e1
d d1
Ex 2: Draw the projections of a regular hexagon of 25mm sides, having one of its
side in the H.P. and inclined at 60º to the V.P. and its surface making an angle of 45º
with the H.P.

Side on the H.P. making 60°

with the VP.
Plane inclined to HP
at 45°and ┴ to VP
Plane parallel to HP
e1’ d1’
f1’
c1’
a’ b’ c’ f’ 45º
d’e’ a1’
Y
X f1 60º b1’
f
a e a1 e1

b d b1 d1

c c1
Ex 3: 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.

Keep AC parallel to the H.P. Incline AC at 30º to the H.P.

& BD perpendicular to V.P. i.e. incline the edge view Incline BD at 45º to the V.P.
(considering inclination of (FV) at 30º to the HP
AC as inclination of the
plane)
c1’

b1’ d1’
b’
a’ d’ c’ 30º
X Y
45º 45º a1’
b b1

a c a1 c1

d d1
Ex 4: 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.

Keep AC parallel to the H.P. & Incline AC at 30º to the H.P. Make BD parallel to XY
BD perpendicular to V.P.
(considering inclination of AC
as inclination of the plane and
inclination of BD as inclination
of edge) c1’

d1’ b1’
b’
d’ c’
a’ 30º
X Y
a1’
125

b b1

b1
c1
a1
50

a c

c1
a1
d
d1

d1
Ex 5: A regular hexagon of 40mm side has a corner in the HP. Its surface inclined at 45° 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.

Top view of the diagonal

Plane inclined to HP making 60° with the VP.
Plane parallel to HP at 45°and ┴ to VP

d 1’

c1 ’
e 1’

b 1’
f 1’
b’ c’
a’ f’ e’ d’ 45° a 1’ Y
X 60°
f1
f1 e1 a1
f e

e1
b1

a d d1
a1

d1
c
1

b c
b1 c1
Ex 6: 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

A rectangle can be seen as a F.V. (square) is drawn first Incline a1’b1’ at 30º to the
square in the F.V. only when its H.P.
surface is inclined to VP. So for
the first view keep the plane //
to VP & shorter edge ┴ to HP
60

b’ c’ b1’ c1’
30

a’ d’ a1’ d1’ b1 a1 30º

X Y
a c a
b d b

c c1 d1
d
Ex 7: 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 and one side inclined at 45º to xy. Draw its top view

c1’
a’ a1’
a1’
c’ c1’
50

b’ b1’
b1’ 45º

a.b 70 c a.b a1 b1

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

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
Problem 9: A circle of 50 mm diameter is final TV was completed.Study illustration carefully.
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
Ex 8: 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.

c’ c1

b’d’ d1 b1

a’ b’d’ c’ a’
X Y
60 a1
b b1
a1

60
d1 b1
a c a1 c1
100

c1

d d1 100
100
 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 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 9: 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 x 1y 1 perpendicular to it.
3.Project view on x1y1.
a) First draw projectors from a’b’ & c’ on x 1y 1.
b) from xy take distances of a,b & c( Tv) mark on these projectors from x 1y1. Name points a1b1 & c1.
c) This line view is an Aux.Tv. Draw x 2y2 // to this line view and project Aux. Fv on it.
for that from x 1y1 take distances of a’b’ & c’ and mark from x 2y= 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 Y X2 a’1
c c’1

ALWAYS FOR NEW FV TAKE

DISTANCES OF PREVIOUS FV
300 650 AND FOR NEW TV, DISTANCES
a b OF PREVIOUS TV
50 mm REMEMBER!!
 IMPORTANT POINTS
Problem 10 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 11: 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
DETERMINE TRUE SHAPE. c1
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 12: 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 13 : 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
Problem 14 : draw a regular hexagon of 40 mm sides, with its two sides vertical. Draw a circle
of 40 mm diameter in its centre. The figure represents a hexagonal plate with a hole in it and
having its surfacre parallel to the VP. Draw its projections when the surface is vertical and
inclined at 30º to the VP.
aa’ a1’

1 1’
1’ f1’ 12 1’ 2 1’ b 1’
f’ 12’ 2’ b’

11’ 3’ 11 1’ 3 1’

10’ 4’ 10 1’ 4 1’

5’ 9 1’ 5 1’
9’
e 1’ c1’
e’ 8’ 6’ c’ 8 1’ 6 1’
7’ 7 1’

d’ 30º d 1’
X Y
e f

a d b c 10
e f 9 a d
11 8
10 9 8 1 2 3 4 12 1
11 12 7 6 5 7 2
6 3 b c
5 4
Problem 15: Draw the projections of a circle of 5 cm diameter having its plane vertical and
inclined at 30º to the V.P. Its centre is 3cm above the H.P. and 2cm in front of the V.P. Show
also its traces

50 Ø
4’ 41’
3’ 5’ 31’ 51’

2’ 6’ 61 ’
21’

1’ 7’ 11’ 71’

12’ 8’ 121’ 81’

30

91’
11’ 9’ 111’
X 10’ Y
101 ’
20

30º

1 2, 3, 4, 5, 6, 7
12 11 10 9 8
Problem 16 : Draw an equilateral triangle of 75 mm sides and inscribe a circle in it. Draw the
projections of the figure, when its plane is vertical and inclined at 30º to the VP and one of the
sides of the triangle is inclined at 45º to the HP.

a1’
a’
1 1’
1’ 12 1’ 2 1’
12’ 2’
c 1’
11’ 3’ c’ 11 1’
3 1’

10’ 4’ 10 1’ 4 1’

9’ 5’ 9 1’ 5 1’
8’ 6’ 8 1’ 6 1’
7’ 7 1’

b’

30º b1’
45º
X Y

a b c
10 9 8 1 2 3 4
11 12 7 6 5
P 17: A semicircular plate of 80mm diameter has its straight edge in the VP and inclined at
45 to HP.The surface of the plate makes an angle of 30 with the VP. Draw its projections.
Plane inclined at 30º to the V.P.
Plane in the V.P. with and straight edge in the H.P. St.edge in V.P. and
straight edge ┴ to H.P inclined at 45º to the H.P.

1’ 11’
2’ 21’

3’ 31’

4’
Ø 80
41’

5’ 51’

6’ 71’ 61’
7’ 45º 71 11
X 30º Y
1 2 3 4
21
7 6 5 61
31
51 41
Problem 18: 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 Incline the T.V. till the Incline the F.V. till the
surface is inclined to VP. So for distance between the end major axis becomes
the first view keep the plane // projectors is 30 mm horizontal
to VP.

50 Ø
4’ 41’

41 ’
3’ 5’ 31’ 51’

51 ’
31 ’
2’ 6’ 21’ 61’

21 ’

61 ’
11’ 71’
1’ 7’

71 ’
11’
12’ 8’ 121’ 81’

121’

81’
111’ 91’
11’ 9’ 101’

111’

91’
X 10’ Y
11’

101’
121’ 21’
30
31 ’
111’
1 2, 3, 4, 5, 6, 7
101’ 41 ’
12 11 10 9 8

91’ 51’
81’ 61’
Problem 19: The top view of a plate, the surface of which is inclined at 60º to the HP is a circle of
60 mm diameter. Draw its three views.
X1

7’ 7”
6’
8’ 6” 8”
5’
9’ 5” 9”
4’
10’
4” 10”
3’
11’ 3” 11”
2’
12’ 2” 12”
600
X 1’ Y
1”
4
3 5
2 6

1 7
12 8
11 9
10
60 Y1