1
ArulmiguMeenakshi Amman College of Engineering
Vadamavandal-604410.
Impotant questions and answers
ENGINEERING GRAPHICS
Prepared by
V.Umapathy,
Prof. /Mech.
S.No Questions and answers
.
Unit-I
Fundamentals
Basic Concepts
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ENGINEERING DRAWING: QUICK REVISION NOTES
1. Conic Sections & Eccentricity (e)
Golden Rule: The nature of a conic section is determined by its eccentricity value.
🟢 e < 1 ---> Ellipse
o Examples: e = 3/4, e = 4/5, e = 5/6, etc.
🔵 e = 1 ---> Parabola
o Example: e = 1
🔴 e > 1 ---> Hyperbola
o Examples: e = 3/2, e = 4/3, e = 5/4, etc.
2. Vertex & Focus Ratios
Fundamental Formula: VF / AV = e
Division Breakdown:
o Numerator (VF): Indicates right-side division from the vertex.
o Denominator (AV): Indicates left-side division from the vertex.
3. Calculation Variations for 'e', 'AV', and 'VF'
Depending on what is provided in your question paper, use these variations:
V.
U
1. Finding 'e': When both AV and VF values are given:
m
ap
e = VF / AV
at
hy
2. Finding 'AV': When 'e' and VF values are given:
,P
ro
f./
EC
N
ot
es
ArulmiguMeenakshi Amman College of Engineering
Vadamavandal-604410.
Impotant questions and answers
ENGINEERING GRAPHICS
Prepared by
V.Umapathy,
Prof. /Mech.
S.No Questions and answers
.
Unit-I
Fundamentals
Basic Concepts
V.
U
m
ap
at
hy
,P
ro
f./
EC
N
ot
es
,2
es
ot
N
EC
f./
ro
,P
hy
at
ap
m
U
V.
,3
es
ot
N
EC
f./
ro
,P
hy
at
ap
m
U
V.
, 4
ENGINEERING DRAWING: QUICK REVISION NOTES
1. Conic Sections & Eccentricity (e)
Golden Rule: The nature of a conic section is determined by its eccentricity value.
🟢 e < 1 ---> Ellipse
o Examples: e = 3/4, e = 4/5, e = 5/6, etc.
🔵 e = 1 ---> Parabola
o Example: e = 1
🔴 e > 1 ---> Hyperbola
o Examples: e = 3/2, e = 4/3, e = 5/4, etc.
2. Vertex & Focus Ratios
Fundamental Formula: VF / AV = e
Division Breakdown:
o Numerator (VF): Indicates right-side division from the vertex.
o Denominator (AV): Indicates left-side division from the vertex.
3. Calculation Variations for 'e', 'AV', and 'VF'
Depending on what is provided in your question paper, use these variations:
V.
U
1. Finding 'e': When both AV and VF values are given:
m
ap
e = VF / AV
at
hy
2. Finding 'AV': When 'e' and VF values are given:
,P
ro
f./
EC
N
ot
es
