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
In which manner it’s VP.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 the
Study a case of an showing
illustration object inclined to both reference Planes.)
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

For T.
For T.

V.
For
Tv
V.

For
For Fo
Fv rF
F
V. . V. .

ORTHOGRAPHI ORTHOGRAPHIC ORTHOGRAPHIC

C FV- Inclined to XY FV- Apparent
TV-True Shape TV- Reduced Shape
FV- Line // to xy Shape d V TV-Previous Shape
V V
c’ P d1 c1
P P
’ ’ ’
a d a1 b1
b’ c’’ a
b’ ’ ’
’ ’ d
a d a d 1

1 1
c
1
b c b c a
H A H
1 1
B H
1
C
b
P P P 1
 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. 2 nd pair B on previous page illustration )

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

APPLY SAME STEPS TO SOLVE NEXT ELEVEN PROBLEMS

Problem 1: Read problem and answer following
Rectangle 30mm and questions
50mm sides is resting on 1. Surface inclined to which plane? -------
HP on one small side HP
which is 300 inclined to VP, 2. Assumption for initial position? ------// to
while the surface of the HP
plane makes 450 3. So which view will show True shape? ---
inclination with HP. Draw TV
it’s projections.
Surface // to 4. inclined
Surface Whichtoside will be vertical? ---One
Hp Hp
d’c’small side. c’1 d’1
Hence begin with TV, draw rectangle
c’d’ below X-Y
ab drawing one small side vertical.
’ ’ b
a 450 b’ a’ 1 Y
X ’ 30 0
’ 1
a a1 d1

a1
d Side
Incline
d

b
to Vp
b c b c11

d1
1

c1
Problem 2: Read problem and answer following questions
A 30 0 – 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

Hence begin with FV, draw triangle above X-Y
given)
keeping longest side vertical.
a’ a’
1
c’ c’ 1
side inclined to
c’Hp
1

a’
1

b’
b’
b 1
’ 1 300
X a
b 450 a b1 Y
a c 1
b c
c 1

Surface // to Surface inclined to

Vp Vp
Problem 3: Read problem and answer following questions
A 30 0 – 60 0 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
45 0 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

(Surface inclination directly given. Hence begin with FV, draw triangle above X-Y
Side inclination indirectly given) keeping longest side vertical.

First TWO steps are similar to previous problem.

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

a’
1

35
b’
b’
b
X ’ 1
1 1 Y
a a
0
b 450 b1
a c 1
b c
c 1
Problem 4: Read problem and answer following
A regular pentagon of 30 mm sides questions
is resting on HP on one of it’s sides 1. Surface inclined to which plane? ------- HP
with it’s surface 450 inclined to HP. 2. Assumption for initial position? ------ // to
Draw it’s projections when the side HP
in HP makes 300 angle with VP 3. So which view will show True shape? ---
SURFACE AND SIDE INCLINATIONS TV
ARE DIRECTLY GIVEN. 4. Which side will be vertical? -------- any
side.d d’
’
c’e’Hence begin1 with TV,draw pentagon
belowe’ c’ 1

b a
X-Y1line, taking one side vertical.
X b a c’e’ d 45 b’ Y
’ ’ a’
’ ’ ’ 0
a 1
e e 1 30
0
1
1 e
a a 1 b
1 1
d
d
1
d c
b b 1 1
1
c c
1
Problem 5: Read problem and answer following
A regular pentagon of 30 mm sides is questions
resting 1. Surface inclined to which plane? -------
on HP on one of it’s sides while it’s HP
opposite 2. Assumption for initial position? ------ //
vertex (corner) is 30 mm above HP. to HP
Draw projections
SURFACE INCLINATIONwhen side in HPGIVEN
INDIRECTLY is 3. So which view will show True shape? -
300SIDE
inclined to VP. DIRECTLY GIVEN:
INCLINATION -- TV
4. Which side will be vertical? --------
ONLY CHANGE is any side.
the manner in which surface inclination is described: Hence begin with TV,draw pentagon
One side on Hp & it’s opposite corner 30 mm above Hp. d d’
Hence redraw 1 st Fv as a 2 nd Fv making above
below
’
c’e’ one side verticalc’.
1
arrangement. X-Y line, taking
3 e’ 1
Keep a’b’ on xy & d’ 30 mm above xy.
0 1

X b a c’e’ d a
a’ b’ Y
’ ’ ’ ’
b’ 30 1
e 1
a0
e
1 e 1
a a 1 b
1 1

d d
1
d c
b b 1 1

1
c c
1
 Problem 6: A rhombus of diagonals 40 c’ c’ 1
mm and 70 mm long respectively has one
end of it’s longer diagonal in HP while that d b’
b
diagonal is 35 0 inclined to HP. If the top- ’ d’
bd ’ 1
view of the same diagonal makes 40 0 X a c’ 0 a’ Y
’ ’ a 45 1

inclination with VP, draw it’s projections. ’ 30

d ’ d 1
a0 d
1
1 1
Read problem and answer following a c a c
b c
questions b
1
b
1
1 1
1. Surface inclined to which plane? ------- 1
HP The difference in these two problems is in step 3
only.
2. Assumption for initial position? ------ // In problem no.6 inclination of Tv of that diagonal is
to HP given,It could be drawn directly as shown in 3rd step.
3. So which view will show True shape? --- While in no.7 angle of diagonal itself I.e. it’s TL, is
TV given. Hence here angle of TL is taken,locus of c 1
4. Which diagonal horizontal? ---------- Is drawn and then LTV I.e. a1 c1 is marked and
Problem 7: A rhombus of diagonals 40 final TV was completed.Study illustration carefully.
Longer
mm and 70 mm long respectively having c’ c’ 1
Hence
one end begin withdiagonal
of it’s longer TV,draw rhombus
in HP while
below 0 d b’
that diagonal is 35 inclined to HP and b
’ d’
makes 40 0 inclination
X-Y line, taking longer
with VP. diagonal
Draw it’s // to a bd ’ 1

projections. X c’ a 450 a’ 1
Y
X-Y ’ ’ ’ 300
’ 1a1
d d d
Note the difference in a c a
1
c
1
T
construction of 3rd step 1
b L c
b
1
b c 2

in both solutions. 1
1
1
 c c’1
d ’ b’1
Problem 8: A circle of 50 mm diameter is a b d c b ’
resting on Hp on end A of it’s diameter ’ ’ ’ ’ a 300’ a’1 d’1 Y
X
AC ’ 45
d d1 0 a d
which is 300 inclined to Hp while it’s Tv 1 1

is 45 0 inclined to Vp.Draw it’s projections.

a ca c1
1

b c1
Read problem and answer following 1

questions b b1
1. Surface inclined to which plane? -------
HP The difference in these two problems is in step 3
2. Assumption for initial position? ------ // only.
to HP In problem no.8 inclination of Tv of that AC is
3. So which view will show True shape? --- given,It could be drawn directly as shown in 3rd step.
TV While in no.9 angle of AC itself i.e. it’s TL, is
given. Hence here angle of TL is taken,locus of c 1
4. Which diameter horizontal? ---------- Is drawn and then LTV I.e. a1 c1 is marked and
ACProblem 9: A circle of 50 mm diameter is final TV was completed.Study illustration carefully.
resting on
Hence Hp onwith
begin end ATV,draw
of it’s diameter
rhombus
AC c c’1
below
which is 300 inclined to Hp while it makes d ’ b’1
a b d c b ’
X-Y
45 0 line, taking
inclined longer
to Vp. Draw diagonal // to ’
it’s projections. ’ a ’ d’1
’ ’ a’1
X-Y d ’ d1 d
a 30
1
1
Note the difference in 0
T
a ca c1
construction of 3rd step 1 L
c1
b
in both solutions. 1
b b1
 Read problem and answer following
Problem 10: End A of diameter AB of a circle is in HP questions
A nd end B is in VP.Diameter AB, 50 mm long is 1. Surface inclined to which plane? -------
0 0
30 & 60 inclined to HP & VP respectively. HP
Draw projections of circle. 2. Assumption for initial position? ------ //
to HP
3. So which view will show True shape? ---
TV
The problem is similar to previous problem of circle – no.9.
4. Which diameter horizontal? ----------
But in the 3 rd step there is one more change.
Like 9th problem True Length inclination of dia.AB is AB definitely expected
but if you carefully note - the the SUM of it’s inclinationsHence
with HPbegin
& VP with
is TV,draw CIRCLE
0
90 . below
Means Line AB lies in a Profile Plane. X-Y line, taking DIA. AB // to X-Y
Hence it’s both Tv & Fv must arrive on one single projector.
So do the construction accordingly AND note the case carefully ..

30 0
X Y
60
0 SOLVE SEPARATELY
ON DRAWING SHEET
T GIVING NAMES TO VARIOUS
L

POINTS AS USUAL,
AS THE CASE IS IMPORTANT
 Problem 11: Read problem and answer following
A hexagonal lamina has its one side in HP and questions
Its apposite parallel side is 25mm above Hp
1. Surface inclined to which plane? -------
and
In Vp. Draw it’s projections. HP
Take side of hexagon 30 mm long. 2. Assumption for initial position? ------ //
to HP
3. So which view will show True shape? ---
ONLY CHANGE is the manner in which surface inclination TV
is described: 4. Which diameter horizontal? ----------
One side on Hp & it’s opposite side 25 mm above Hp. AC
Hence redraw 1 st Fv as a 2 nd Fv making above
arrangement.
Hence begin with TV,draw rhombus
Keep a’b’ on xy & d’e’ 25 mm above xy. below
X-Ye line,e’taking
1 d’ 1longer diagonal // to
d ’
X-Yf ’
2 c ’ f’ 1 c1’
’ Y
X a’b cf de
5
b a’ 1 b’ 1
’ a’ ’
f’ ’ ’ ’ f1 e d
f1 1 1 c1
a e a e
1 1 a b As 3 rd step
1
b d b d 1 redraw 2 nd Tv
1
c1
1 keeping
c
side DE on xy line.
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
Problem 12: perpendicular to Hp.
An isosceles triangle of 40 mm 2.It may remain parallel or inclined to Vp.
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
freely suspended from one corner position as FV.
of (Here keep line joining point of contact & centroid of fig. vertical )
Base side.It’s plane is 45 0 inclined
to 5.Always begin with FV as a True Shape
Vp. Draw it’s projections. a’1 but in a suspended
position. a
C AS shown ’ in 1st FV.

b’1
b g g’1
’ ’
H
G c c’1
X ’ Y
H/
3
A B b
c
a,
b a,g g 45
First draw a given triangle c
0

With given dimensions,

Locate it’s centroid position
And Similarly solve next
join it with point of problem
suspension. of Semi-circle
 IMPORTANT
Problem 13 POINTS
1.In this case the plane of the figure always remains
:A semicircle of 100 mm diameter perpendicular to Hp.
is suspended from a point on its 2.It may remain parallel or inclined to Vp.
straight edge 30 mm from the 3.Hence TV in this case will be always a LINE view.
midpoint 4.Assuming surface // to Vp, draw true shape in suspended
of that edge so that the surface makes position as FV.
an angle of 450 with VP. (Here keep line joining point of contact & centroid of fig. vertical )
Draw its projections.
5.Always begin with FV as a True Shape but in a suspended
position.
A AS shown in 1st FV.
a
20 ’
mm p’
P

G b
’
CG g’

c
’
e
d ’
X Y
’
0.414

b
e
c
R

First draw a given semicircle

a
With given diameter,
b c a p,g d

p,
Locate it’s centroid position

g
And e

d
join it with point of suspension.
 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 x 1-y1 perpendicular to this line showing T.L.
4. Project view on x1-y1 ( it must be a line view)
5. Draw x 2-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. Four
plane)
THEN BY MAKING IT // TO X2-Y2 WE GET TRUE SHAPE.
Cases
 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 x 1y1 perpendicular to it.
3.Project view on x1y1.
a) First draw projectors from a’b’ & c’ on x 1y1.
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 x 2y2 // 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
Y1 them. This will be the required true shape.
a1b1 Y2

90
b 0 b’1
15 TL ’
a
’
1
C1
5
10 C X1
X ’ X2 a’1
Y
c c’1 TRUE
SHAPE
ALWAYS FOR NEW FV TAKE
DISTANCES OF PREVIOUS FV
30 0 65 0 AND FOR NEW TV,
a DISTANCES OF PREVIOUS TV
b
50 REMEMBER!!
mm
 Problem 15: Fv & Tv of a triangular plate are shown.

Determine it’s true shape.

USE SAME PROCEDURE STEPS 5
OF PREVIOUS PROBLEM: 2 0
BUT THERE IS ONE DIFFICULTY: 5 c
1 ’
NO LINE IS // TO XY IN ANY VIEW. 5 a
MEANS NO TL IS AVAILABLE. ’ 1
2 ’
IN SUCH CASES DRAW ONE LINE 0 b
// TO XY IN ANY VIEW & IT’S OTHER 1
X ’ Y
VIEW CAN BE CONSIDERED AS TL 10
x1
FOR THE PURPOSE. 5 a c
TL
HERE a’ 1’ line in Fv is drawn // to xy. 4 90
HENCE it’s Tv a-1 becomes TL. 1 0 c’1
0
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, TRUE d1
DISTANCES OF PREVIOUS TV SHAP
REMEMBER!! E
 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
50 y1
be b
Easily determined. D b1 y2
’

Study the illustration.

T ac1 1
a c b’1
’ L ’ 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
FOR NEW TV, DISTANCES SHAPE
OF PREVIOUS TV
REMEMBER!! 50 b
D.
 Problem 17 : Draw a regular pentagon of
30 mm sides with one side 30 0 inclined to xy.
This figure is Tv of some plane whose Fv is
A line 450 inclined to xy. TR
SH UE
Determine it’s true shape. b1 A
PE
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 a e1 d1
ONE ’
VIEW (FV) IS ALREADY A LINE VIEW. b
SO JUST BY DRAWING X1Y1 // TO THIS ’
e
VIEW WE CAN PROJECT VIEW ON IT
’ Y1
AND GET TRUE SHAPE: c
’ d
STUDY THE ILLUSTRATION.. X ’ 45
0 Y
30
0
e

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