engineering graphics · 2019. 1. 30. · 1. a 40 mm above hp and 55 mm in front of vp. 2. b 10 mm...
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Engineering Graphics
Universal language for engineers.
A drawing that contains all information
of an object
Drawing is important for all branches of engineering.
-
Roll of engineering graphics
Visualization
Ability to mentally picture things
that do not exist.
Communication
The design solution should be communicated without ambiguity.
Documentation
Permanent record of the solution.
-
3-D objects are represented on a 2-D media.
The act of obtaining the image of an object
is termed “projection”.
The image obtained by projection is known
as a “view”.
Projection theory
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Observer at finite distance from the object
A simple Projection system
Observer
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ORTHOGRAPHIC PROJECTION
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Quadrant pattern observed along XY
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Opening the quadrants
HP is rotated clockwise and brought in the plane of VP
• Q1 and Q3 open outward
• Q2 and Q4 open inward
After rotation
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FOR T.V.Different views
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X
Y
X Y
VP
HP
PP
FV LSV
TV
Z
Y
Z
Y
ZPosition of views
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Possible positions of an object
1. In 1st quadrant: above HP & in front of VP
2. In 2nd quadrant: above HP & behind VP
3. In 3rd quadrant: below HP & behind VP
4. In 4th quadrant: below HP & in front of VP
5. In plane: on HP & in front of VP
6. In plane: on HP & behind VP
7. In plane: on VP & above HP
8. In plane: on VP & below HP
9. In planes: on HP & on VP
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VP
XHP
Y
A in First quadrant
A
a’
a
For
TvLo
S
Pro
jecto
r
-
v
-
VP
HP
POINT A IN
2ND QUADRANT
OBSERVER
a’
a
A
XY
x Y
a’
a
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22
OBSERVER
a
a’
POINT A IN3RD QUADRANT
HP
VP
A
a
a’
Convention: Horizontal plane is always rotated clockwise
XY X y
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25
OBSERVER
a
a’ POINT A IN4TH QUADRANT
HP
VP A
Convention: Horizontal plane is always rotated clockwise
XY
X Y
a
a’
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POINT ON HP
Point on HP, its FV (elevation) on XY
1 - on HP & in front of VP
1’ – on XY
1 - below XY
2 – on HP & behind VP
2’ – on XY
2– above XY
1’
2
1
2’
X Y
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P - on VP & above HP
P – on XY
p’ - above XY
Q – on VP & below HP
q – on XY
q’ – below XY
POINT ON VP
Point on VP, its TV (plan) on XY
X YP
q
p’
q’
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POINT ON BOTH HP & VP
X Ye,e’
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A in First quadrantA above HP ………… a’ above XYA in front of VP …...a below XY
A in Second quadrantA above HP …………a’ above XYA behind VP ………..a above XY
A in Third quadrantA below HP …………a’ below XYA behind VP ………..a above XY
A in Fourth quadrantA below HP …………a’ below XYA in front of VP ……a below XY
PROJECTION OF POINTS
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A above HP a’ above XY
A on HP a’ on XY
A below HP a’ below XY
A in front of VP a below XY
A on VP a on XY
A behind VP a above XY
PROJECTIONS OF POINTS
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PROJECTIONS OF POINTS
Draw the projections of the following points.
1. A 40 mm above HP and 55 mm in front of VP.
2. B 10 mm above HP and 25 mm behind VP.
3. C 35 mm below HP and 20 mm behind VP.
4. D 10 mm below HP and 40 mm in front of VP.
5. E on HP and 50 mm in front of VP.
6. F on HP and 80 mm behind VP.
7. G on VP and 75 mm above HP.
8. H on VP and 30 mm below HP.
9. I on both HP and VP.
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Draw the projections of the following points.
1. A 40 mm above HP and 55 mm in front of VP.
X Y
a
a’4
05
5
-
XY
f’
f
c
c
a’
avv
c
b’
b
c
c’
c c
c
d’
d
vvvvvvvv
c
e’
e c
cv
g’
gc
c
h’
h
c
ci’
ic
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(1). Line AB has its end A 25 mm above HP and 30 mm in front of VP. End B is 50 mm above HP and 70 mm in front of VP. Distance between the end projectors is 80 mm. Draw the projections of the line.
YX
a’
b’
b
a
70
50
30
25
80
β
αPositions of A
Positions of B
Length of FV
Length of TV
Inclination of FV to XY …….…()
Inclination of TV to XY ……..…()
Distance b/w projectors… ( proj)
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X Y
a
b
FV
TV
a’
b’A Line inclined to both
HP and VP
X
Y
a’
b’
a b
T.V.
Positions of A
Positions of B
Length of FV
Length of TV
Inclination of FV to XY…….…()
Inclination of TV to XY …()Distance b/w projectors… ( proj)
B
A
V. T.
a’
b’
a bX
Y
-
B
A
V. T.
a’
b’
a b
X
Y
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Two trapeziums through the line AB
(1). ABba
(2). ABb’a’
Surface inclined to HP. Base on VP.Base a’b’ represents FV of AB
Surface perpendicular to VP.
Surface inclined to VP. Base on HP.Base ab represents TV of AB
Surface perpendicular to HP.
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𝜃𝜙
VT(on VP)
HT(on HP)
Line AB in space.TV (ab) on HP.
FV (a’b’) on VP.
𝜃 –inclination of AB to HP.--angle b/w True length & its TV.--angle b/w AB & ab.
𝜙 --inclination of AB to VP.--angle b/w True length & its FV.--angle b/w AB & a’b’.
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ab inclined to XY
Length of ab1 = length of ab
Height of b’1 = Height of b’
TV parallel to XY; FV represents the True length
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𝜃
𝜙
b
b’
YX
a
a’ α
β
LINE AB INCLINED TO HP & VP
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(1). Line AB has its end A 25 mm above HP and 30 mm in front of VP. End B is 50 mm above HP and 70 mm in front of VP. Distance between the end projectors is 80 mm. Draw the projections of the line.
YX
a’
b’
b
a
70
50
30
25
80
β
αPositions of A
Positions of B
Length of FV
Length of TV
Inclination of FV to XY …….…()
Inclination of TV to XY ……..…()
Distance b/w projectors…( proj)
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b
YX
a’
b’
a
ht’vt
ht (HT)
vt’ (VT)
TRACES OF A LINE
All Elevation points are collinear; ht’ on xy
All Plan points are collinear; ht’ on xy
-
b
YX
a’
b’
a
ht’ vt
ht (HT) vt’
(VT)
-
vt’
b
b’
a’
a b1
Y
b1’
ht’ vt
ht
X
1’θ
β
(a’- 1’) = Length of (a-b)
XY component of (a’-b1’) = length of (a-b)
𝜃 line (a’-b1’) from β line(a-b)
Three θ lines
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vt’
b
b’
a’
a
Yb3
b3’
ht’ vt
ht
X
2. 𝜃 line (vt’-b3’) from line(vt-b)
3’θ(vt’- 3’) = Length of (vt-b)
β
XY component of (vt’-b3’) = length of (vt-b)
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vt’
b
b’
a’
a
Y
b5
b5’
ht’ vt
ht
X
3. 𝜃 line (ht’-b5’) from line(ht-b)
5’θ (ht’- 5’) = Length of (ht-b)β
XY component of (ht’-b5’) = length of (ht-b)
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YX
b’
b
a’
a
2
b2
𝜙
b2’
vt’
vt ht’
ht
vv
vv b4’
b4
4
6
b6
b6’
Three 𝜙 lines from Three lines
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vt’
b
b’
a’
a b1
Y
b5
b3
b5’b1’ b3’
ht’ vt
ht
X
Three 𝜃 lines from Three lines
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YX
b’
b
a’
a
2
b1’
𝜃
b2
𝜙
b2’
αTrue length from line
Triangle formed
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X Y
a
b
b’
a’
80
20
40
70
50
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X Y
a
b
b’
a’ Locus of a’
Locus of b
Locus of a
Locus of b’
b1
b1’
1’
-
vt
b6
6
4
b4
hta
b2
2
Base length of each 𝜙 triangle equals Length of corresponding lines
Three 𝜙 Triangles
Length of α line (a’-b’)
Length of α line (ht’-b’)
Length of α line (vt’-b’)
b2, b4, b6, on the locus of b
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VT(on VP)
𝜃𝜙
HT(on HP) 𝜃 Triangle
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X Y
a
b
a’
b’
ht’
htvt’
vt
b1’
b1
2
b2
-
vt
b6
6
4
b4
hta
b2
2
Base length of each 𝜙 triangle equals Length of corresponding lines
Three 𝜙 TrianglesLength of α line (a’-b’)
Length of α line (ht’-b’)
Length of α line (vt’-b’)
b2, b4, b6, on the locus of b
-
b1’
a' 1’
ht'
b5’
5’
3’vt'
b3’
Base length of each 𝜃 triangle equals Length of corresponding β lines
Length of β line (a-b) Length of β line (vt-b)
Length of β line (ht-b)
Three 𝜃 Triangles
b1’, b3’ b5’, on the locus of b’
-
X Y
a
b
a’
b’
ht’
htvt’
vt
b1’
b1
2
b2
A TYPICAL SOLUTION
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VT(on VP)
𝜃𝜙
HT(on HP)
-
𝜃
𝜃
𝜃
-
𝜃
𝜃
b(b1)
b’(b1’)
YX
a
a’ θ
-
vt’
b
b’
a’
a b1
Y
b1’
ht’ vt
ht
X
𝜃 line (a’-b1’) from line(a-b)
1’θ
β
(a’- 1’) = Length of Top View((a-b)
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YX
b’
b
a’
a
b1
1’
b1’
𝜃
b2
𝜙
𝜃 Triangle formed
True length from β line
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BA
a’
b’
a bX
Y
φ
φ
-
φY
φ
X
a
b
b’a’
-
vt’
b
b’
a’
a
Y
b6b2
b2’
b4’
ht’ vt
ht
X
Three φ lines from Three lines
b4
b6’
2
4
6
Three φ Triangles
(a – b2 – 2)……… from (a’-b’)
( vt – b4 – 4)…… from (vt’-b’)
( ht – b6 – 6)…... from (ht’-b’)
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LINES.
Projection of lines.
Locate Traces of the line.
Find True length of the line.
True length from Plan.
True length from Elevation.
Obtain inclinations of the line.
Master solution.
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YX
b’
b
a’
a
b1
1’
b1’
𝜃
b2
𝜙
𝜃 Triangle formed
True length from β line
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Inclinations of AB to HP and VP
(1).To HP = angle between AB and ab(Trapezium on HP provides θ )
(2).To VP = angle between AB and a’b’(Trapezium on VP provides φ )
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VP
XHP
Y
A in First quadrant
A
a’
a
For
TvLo
S
Pro
jecto
r
-
3’
a'
b1’
vt'
b3’
ht'
b5’
5’
1’
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