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 planes
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 planes
Perpendicular to one of the principal planes and
inclined to the other
– Plane Perpendicular to H.P and inclined to the
V.P.
 Perpendicular planes
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’ 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 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

c c1
25

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’
X f1 Y
f 60º b1’
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’
b1
b

a c a1 c1

d d1
Ex 4: 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

d1’

c1’
e1’

b1’
f1’
b’ c’
a’ f’ e’ d’ 45° a1’ Y
X 60°
f1
f1 e1 a1
f e

e1
b1

a d d1
a1

d1
c
1

b c
b1 c1
 Problem 6: A rhombus of diagonals 40 mm c’ c’1
and 70 mm long respectively has one end
of it’s longer diagonal in HP while that d’ b’1
b’
diagonal is 450 inclined to HP. If the top- d’1
b’d’
view of the same diagonal makes 300 X a’ c’ a’ 450 a’1 Y
inclination with VP, draw it’s projections. 300 d1
d d1 a0
1
a a c1
Read problem and answer following questions c
b1
1 c1
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.
In problem no.6 inclination of Tv of that diagonal is
4. Which diagonal horizontal? ---------- Longer
given,It could be drawn directly as shown in 3rd step.
Hence begin with TV,draw rhombus below While in no.7 angle of diagonal itself I.e. it’s TL, is
X-Y line, taking longer diagonal // to X-Y 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 7: A rhombus of diagonals 40
mm and 70 mm long respectively having c’ c’1
one end of it’s longer diagonal in HP while
d’ b’1
that diagonal is 450 inclined to HP and b’
d’1
makes 300 inclination with VP. Draw it’s b’d’
X a’ c’ a’ 450 a’ Y
projections. 300
1
d d1
Note the difference in a c a c1
construction of 3rd step 1 c2
b b1
in both solutions.
 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’
300 a’1 d’1
d d1
450
Note the difference in
a ca c1
construction of 3rd step 1

in both solutions.
b b1
Thank You