3. Definitions
Isometric Projection :
ISO (in Greek) means equal
METRIC means measure.
Hence Isometric Projection means a system of
projection of equality of measure.
NOTE :
To distinguish Isometric Projection from
Orthographic Projection, use CAPITAL
LETTERS for the corners in the Isometric
projection. Ravi Sivaprakash AP/MECH AAMEC
Kovilvenni
6. Terminology
Isometric Axes : The three lines RQ, RS and RV
making 120° with each other are called the
isometric axes.
Isometric Lines : A line parallel to any of these
isometric axes is known as an isometric line.
Isometric Planes : The planes containing the
faces of the cube and all the planes parallel
to these planes are termed isometric planes.
Ravi Sivaprakash AP/MECH AAMEC
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8. Isometric Scale
15°
P 0 1 2 3 4 R
Q
0’
1’
2’
3’
4’
45°
ISOMETRIC LENGTHS
TRUE LENGTHS
Isometric Length = 0.82 X True length
Example :
True length = 10 mm
Isometric length = 0.82 X 10 = 8.2 mm
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9. Problem 1
Draw the isometric projection of a square
prism of base side 35 mm and height 65 mm.
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10. Problem 1
65
35
a (a1) b (b1)
c (c1)d (d1)
a’1 (d’1) b’1 (c’1)
b’(c’)a’(d’)
Draw the top and front views of the prism using actual
scale.
Draw the isometric scale.
15°
A 0 1 2 3 4 C
D
0’ 1’ 2’
3’
4’
45°
ISOMETRIC LENGTHS
TRUE LENGTHS
Isometric length = 0.82 X true length
5 6
5’
6’
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11. Isometric Length = 0.82 X True length
Example :
Isometric length = 0.82 X 35 = 28.7 mm
= 0.82 X 65 = 53.3 mm
65
35
a (a1) b (b1)
c (c1)d (d1)
a’1 (d’1) b’1 (c’1)
b’(c’)a’(d’)
Problem 1
A1
30°
30°
A
B
B1
D1
D
C
C1
No need to draw hidden lines in Isometric
drawings. Here it is only for understanding
purpose only.
Front View
Ravi Sivaprakash AP/MECH AAMEC Kovilvenni
12. Problem 2
A hexagonal prism of base side 20 mm and
height 45 mm has a square hole of side 16
mm at the centre. The axes of the square and
hexagon coincide. One of the faces of the
square hole is parallel to a face of the
hexagon. Draw the isometric projection of
the prism with hole to full scale.
Ravi Sivaprakash AP/MECH AAMEC
Kovilvenni
13. p
q r
s
tu
20
45
16
1 4
32
Isometric Length = 0.82 X True length
Example :
Isometric length = 0.82 X 20 = 16.4 mm
= 0.82 X 45 = 36.9 mm
1
30° 30°
U
T
4
2
Q
P
3
P1
Q1
U1
T1
R
S
d
b
c
a
A
B
C
D
+
Problem 2 Mark U1 , measure the distance “1u”
in top view . Convert it into
isometric length
(p1)
(q1) (r1)
(s1)
(t1)(u1)
p’ u’(q’) t’(r’) s’
(p1) u1(q1) t1(r1) s1
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14. Problem 3
A square prism of base 20 mm side and
height 40 mm rests on one of its ends on the
HP. All the base sides of the prism are equally
inclined to the VP. It is cut by a plane
perpendicular to the VP and inclined at 45°
to the HP that passes through a point on the
axis 7 mm from the top. Draw the isometric
projection of the prism. Page no 393
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15. V.P
.
X Y
H.P
.
Problem 3
p (t)
2
s(w)
q (u)
p’
t’
s’(q’)
w’(u’)
r’
403’
v’
1
1’
2’(4’)
r (v)
745°
3
4
m
n
5
a b
cd
B
30°
30°
C
A
D
Front View
P
T
W V
U
S R
Q
1
2
3
4
5
m1 = M1 ; n1 = N1
M
N
W2 = w’2’ ; V3 = v’3’ ; U4 = u’4’
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16. Problem 4
Draw the isometric projections of a cylinder
of base diameter 50 mm and axis 80 mm long
when it rests with its base on HP and on VP.
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17. p q
rs
a
b
c
d
φ50
Problem 4
(a1)
(b1)
(c1)
(d1)
d’1 b’1a’1(c’1)
a’(c’) b’d’
P
30° 30°
A
B
QS
R
D1
C1 B1
A1
D
C
O1 O2
Mark the mid-points of PQ, QR, RS
and SP as A1, B1, C1, D1
P as center and PC as radius draw
an arc C1B1
80
Ravi Sivaprakash AP/MECH AAMEC
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19. Problem 5
A pentagonal pyramid, 30 mm edge of base
and 65 mm height stands on HP such that an
edge of the base is parallel to VP and nearer
to it. A section plane perpendicular to VP and
inclined at 30° to HP cuts the pyramid
passing through a point on the axis at a
height of 35 mm from the base. Draw the
isometric projection of the truncated
pyramid, showing the cut surface.
Ravi Sivaprakash AP/MECH AAMEC
Kovilvenni
20. Problem 5
a
30
b
c
de
a’ b’ c’(d’)(e’)
o
65
o’
35
1’
2’
3’(4’)
(5’)
30°
1 3
45
2
p q
rs
Q
30° 30°
2
RP
S
B C
D
A
1
O1
E
O
O1O = isometric length of the axis
3
4
5
Front View
11
21
31
41
51
At 11 erect vertical to meet the
slant edge OA at 1.
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21. Problem 6
A cylinder 50 mm diameter and 60 mm
height stands on HP. A section plane
perpendicular to VP inclined at 55° to HP cuts
the cylinder and passes through a point on
the axis at a height of 45 mm above the base.
Draw the isometric projection of the
truncated portion of the cylinder, when the
cut surface is clearly visible to the observer.
Page no 19.26
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22. Problem 6
V.P
.
X Y
H.P
.
60
a b’ c’ d’ e’
a1’
e1’
55°
a(a1)
b(b1)
c(c1)
d(d1)
e(e1)
(f’)(g’)(h’)
d1’(f1’)
c1’(g1’)
b1’(h1’)
f(f1)h(h1)
g(g1)
φ50
45
1’
2’(7’)
3’(6’)
4’(5’)
1
2
3
4
56
7
p(p1) q(q1)
r(r1)s(s1)
P1
30° 30°
P
R
Q1
S1
R1
A1
G1
E1
C1
1
S
O1 O2
Q
2
3
4
5
6
7
Mark the section point 1 on the generator A1A in the
isometric projection such that A11 = isometric length of a1’1’
Ravi Sivaprakash AP/MECH AAMEC
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23. Problem 7
A cone of base diameter 50 mm and height
55 mm is resting on its base on the HP. It is
cut by a plane perpendicular to the VP and
inclined at 30° to the HP. The plane meets
the axis at a distance of 25 mm from the
apex. Draw the isometric view of the
truncated cone. Page no 401
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Kovilvenni
25. Problem 8
Draw the isometric view of a frustum of a
hexagonal pyramid when it is resting on its
base on the HP with two sides of the base
parallel to the VP. The side of base is 20 mm
and top 8 mm. The height of the frustum is
55 mm. page no 391
Ravi Sivaprakash AP/MECH AAMEC
Kovilvenni
28. Problem 9
A hexagonal shaped solid of base edge 20 mm
and height 40 mm lies centrally on a cylinder
of 60 mm diameter and 20 mm thick. Draw
the isometric projection of the solids if their
axes lie on the same line.
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Kovilvenni