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Mel110 part3
1. ENGINEERING CURVES
P i t d i t t f di l tPoint undergoing two types of displacements
INVOLUTE: Locus of a free end of a string when it isg
wound round a (circular) pole
CYCLOID L f i t th i h fCYCLOID: Locus of a point on the periphery of a
circle which rolls on a straight line path.
SPIRAL: Locus of a point which revolves around a
fixed point and at the same time moves towards it.p
HELIX: Locus of a point which moves around the
f f i ht i l li d / d t thsurface of a right circular cylinder / cone and at the
same time advances in axial direction at a speed
bearing a constant ratio to the speed of rotationbearing a constant ratio to the speed of rotation.
2. INVOLUTE OF A CIRCLEINVOLUTE OF A CIRCLE
• Problem:
Draw Involute of
P3
P1
Draw Involute of
a circle. String
length is equal to
the
circumference of
i l
3
4
5
P4
4 to p
circle.
P
1
2
6
7
8
P
A
P
P8
P5
P7
P6 πD
1 2 3 4 5 6 7 8
3. INVOLUTE OF A CIRCLE
String length MORE than πD
Problem: Draw Involute of a circle. String length is MORE
than the circumference of circle.
P2
Solution Steps:
In this case string length is more
than Π D.
P3 P1
But remember!
Whatever may be the length of
string, mark Π D distance
horizontal i.e.along the string
and divide it in 8 number ofd d v de 8 u be o
equal parts, and not any other
distance. Rest all steps are same
as previous INVOLUTE. Draw
the curve completely.
2
3
4
5
P4
4 to p
1 2 3 4 5 6 7 8
P
1
6
7
8P5
p8
1 2 3 4 5 6 7 8
P7
P6
165 mm
(more than πD)
πD
4. INVOLUTE OF A CIRCLE
String length LESS than πD
Problem: Draw Involute of a circle.
String length is LESS than the circumference of circle.
P2
String length LESS than πD
Solution Steps:
In this case string length is Less
than Π D.
P3
P1
But remember!
Whatever may be the length of
string, mark Π D distance
horizontal i.e.along the string
and divide it in 8 number of
4 t
d d v de 8 u be o
equal parts, and not any other
distance. Rest all steps are same
as previous INVOLUTE. Draw
the curve completely.
2
3
4
5
6
P4
4 to p
1 2 3 4 5 6 7 8
P
1
6
7
8P5
P 1 2 3 4 5 6 7 8P7
P6
150 mm
(Less than πD)
πD
6. INVOLUTE OF
COMPOSIT SHAPED POLE
Problem : A pole is of a shape of
half hexagon (side 30 mm) and COMPOSIT SHAPED POLEg ( )
semicircle (diameter 60 mm). A
string is to be wound having
length equal to the pole perimeter Calculate perimeter length
P1
P2
P
g q p p
draw path of free end P of string
when wound completely.
1toP
P2
P
1
2
34
5
P3
3 to P
6
1 2 3 4 5 6
A
P
πD/2P πD/2P4
P5
P6
7. PROBLEM : Rod AB 85 mm long rolls over a semicircular pole without slipping from it’s
initially vertical position till it becomes up-side-down vertical. Draw locus of both ends A & B.
4
B
A4
3
4
B1
A3
OBSERVE ILLUSTRATION CAREFULLY!
when one end is approaching,
other end will move away from
πR 2
3y
poll.
2
πR 2
A2 B2
1
2
3
4
1
A 4A
A1
B3
BB4
8. CYCLOID:
DEFINITIONS SUPERIOR TROCHOID:
If the point in the definition of
cycloid is outside the circleCYCLOID:
LOCUS OF A POINT ON THE
PERIPHERY OF A CIRCLE WHICH
ROLLS ON A STRAIGHT LINE PATH.
cycloid is outside the circle
INFERIOR TROCHOID.:
If it is inside the circleROLLS ON A STRAIGHT LINE PATH. If it is inside the circle
EPI-CYCLOID
If the circle is rolling on another
circle from outside
HYPO-CYCLOID.
If the circle is rolling from inside theIf the circle is rolling from inside the
other circle,
9. CYCLOID
PROBLEM: Draw locus (one cycle) of a point (P) on the periphery of a
circle (diameter=50 mm) which rolls on straight line path.
p
p4
p
4
C1 C2 C3 C4 C5 C6 C7 C8
p2
p3 p5
p6
2
3 5
6
C
p1
p71 7
P p8
πD
Point C (zero radius) will not rotate and it will traverse on straight line.
10. SUPERIOR TROCHOID
PROBLEM: Draw locus of a point (P), 5 mm away from the
periphery of a Circle (diameter=50 mm) which rolls on straight
line path.
Using 2H
i
p44
Using H
p3 p53 5
C1 C2 C3 C4 C5 C6 C7 C8
p2 p6
2 6
C
p1
p7
p8
1 7
DP p8πDP
11. INFERIOR
PROBLEM: Draw locus of a point , 5 mm inside the periphery of a
Ci l hi h ll i h li h T k i l di t 50 TROCHOIDCircle which rolls on straight line path. Take circle diameter as 50
mm
C1 C2 C3 C4 C5 C6 C7 C8
p2
p
3
p4
p5
p6
2
3
4
5
6
C
P
p1 p7
p8
1
2 6
7
πD
13. EPI CYCLOID :
PROBLEM: Draw locus of a point on the periphery of a circle
(dia=50mm) which rolls on a curved path (radius 75 mm).
Solution Steps:
Distance by smaller circle = Distance on larger circle
Solution Steps:
1. When smaller circle rolls on larger circle for one revolution it
covers Π D distance on arc and it will be decided by included arc
angle θangle θ.
2. Calculate θ by formula θ = (r/R) x 360°.
3. Construct a sector with angle θ and radius R.
4 Divide this sector into 8 number of equal angular parts4. Divide this sector into 8 number of equal angular parts.
14. EPI CYCLOID
C
4 5
Generating/
Rolling Circle
C2
2
3 6
7EPI-CYCLOID
If the circle is rolling on
Pr = CP
1
2If the circle is rolling on
another circle from outside
r
R
3600
=
Directing Circle
O
R
15. PROBLEM : Draw locus of a point on the periphery of a circle which
rolls from the inside of a curved path Take diameter of Rolling circle
HYPO CYCLOID
rolls from the inside of a curved path. Take diameter of Rolling circle
50 mm and radius of directing circle (curved path) 75 mm.
P 7
P1
P21
6
P3
P
2
5
4 P4
P5
P6 P7
P8
3
4
O
OC R ( R di f Di ti Ci l )
r
R
3600
=
OC = R ( Radius of Directing Circle)
CP = r (Radius of Generating Circle)
17. SPIRALProblem: Draw a spiral of one convolution. Take distance PO 40 mm.
IMPORTANT APPROACH FOR CONSTRUCTION!
FIND TOTAL ANGULAR AND TOTAL LINEAR DISPLACEMENT
AND DIVIDE BOTH IN TO SAME NUMBER OF EQUAL PARTS.
2
P
13
P2
P1
P3
P4
3
P4 O
7 6 5 4 3 2 1
P4
P6P5
P7
O
5 7
6
18. SPIRAL of
two convolutions
Problem: Point P is 80 mm from point O. It starts moving
towards O and reaches it in two revolutions around. It Draw locus
of point P (To draw a Spiral of TWO convolutions).
2,10
P2
1,93,11 P1
P3
P
16 13 10 8 7 6 5 4 3 2 1 P
4 12
8,16P4 P8
P9
P10
P11
4,12
,
P12
P13 P14
P15
5,13 7,15
P5
P6
P7
6,14
19. Problem: A link 60 mm long, swings on a point Oob e : 60 o g, sw gs o a po O
from its vertical position of the rest to the left through
60° and returns to its initial position at uniformp
velocity. During that period a point P moves at
uniform speed along the center line of the link from Op g
at reaches the end of link. Draw the locus of P.
O, P
N
M
20. P8
HELIX
(UPON A CYLINDER)
PROBLEM: Draw a helix of one
convolution, upon a cylinder. Given 80 mm
6
7
8
P6
P7
8
pitch and 50 mm diameter of a cylinder.
4
5
6
P4
P5
6
Pitch: Axial advance during one
2
3
P2
P3
4
complete revolution .
1 P1
P
2
6
F
T
57
P
1 3
4
2
3
21. P8 HELIX
(UPON A CONE)
PROBLEM: Draw a helix of one convolution,
upon a cone, diameter of base 70 mm, axis
P5
P6
P7
( )p , ,
90 mm and 90 mm pitch.
P4
P2
P3
6
P
P1
X Y
57
P5P6
P
P 4
P1
P3
P4
P7
P8
1
2
3
P2
22. I i f S lid /Interpenetration of Solids /
Intersection of Surfaces /Intersection of Surfaces /
Lines & Curves of Intersection
More points common to both the solids
Basic required knowledge:
~ Projections of Solid
~ Section of Solid
22
~ Section of Solid
23. Problem: A cone 40 mm diameter and 50 mm axis is
.
Projections of Solid
Problem: A cone 40 mm diameter and 50 mm axis is
resting on one generator on Hp which makes 300
inclination with Vp. Draw it’s projections.
More number of generators Better
approximation.
o’
a’1
h’ b’1
g’1
h’1
F
g
X Ya’ b’ d’ e’c’ g
’
f’h’ o’
g1
g
c’1
d’1e’1
f’1 o1
o1
30
h
a e
f
a1
h1f1
e1
a1
be
f1
g1
h1
o1
o1
T
o
23b
c
d d1
c1
b1
c1
b1
d1
e1
1st. Angle
24. For TVFor TVSection of Solid
SECTIONSECTION
PLANEPLANE
I t t ti f S lidLi & Cxx yy
Apparent ShapeApparent Shape
Interpenetration of Solids
Intersection of Surfaces
Lines & Curves
of Intersection
SECTION LINESSECTION LINES
Apparent ShapeApparent Shape
of sectionof section
(45(4500 to XY)to XY)
SECTIONAL T.V.SECTIONAL T.V.
29. CURVES OF INTERSECTIONS are shown by WHITE ARROWS.
Machine component having Intersection of a Cylindrical
two intersecting cylindrical
surfaces with the axis at
acute angle to each other.
y
main and Branch Pipe.Industrial Dust collector.
Intersection of two cylinders.
WHEN TWO OBJECTS ARE TO BE
JOINED TOGATHER, MAXIMUM
Pump lid having shape of a
h l P i d
SURFACE CONTACT BETWEEN
BOTH BECOMES A BASIC
REQUIREMENT FOR
STRONGEST & LEAK PROOF hexagonal Prism and
Hemi-sphere intersecting
each other.
Feeding Hopper.
STRONGEST & LEAK-PROOF
JOINT.
30. Minimum Surface Contact.
( Point Contact) (Maximum Surface Contact)
Li f I i C f I iLines of Intersections. Curves of Intersections.
Square Pipes. Circular Pipes. Square Pipes. Circular Pipes.
MAXIMUM SURFACE CONTACT BETWEEN BOTH BECOMES A BASIC
REQUIREMENT FOR STRONGEST & LEAK-PROOF JOINT.
Two plane surfaces (e-g. faces of prisms and pyramids)
intersect in a straight line
Q
intersect in a straight line.
The line of intersection between two curved surfaces (e-g.
of cylinders and cones) or between a plane surface and aof cylinders and cones) or between a plane surface and a
curved surface is a curve.
30
More points common to both the solids
31. How to find Lines/Curves of Intersection
Generator line Method Cutting Plane Method
T
F
T
FF
75
31
75
32. Problem: Find intersection curve .
Draw convenient number of lines on the surface of one of the solids.
Transfer point of intersection to their corresponding positions in other
When one solid completely penetrates another, there
will be two curves of intersection. 32
p p g p
views.
33. Problem: A cylinder 50mm dia. & 70mm axis is completely penetrated by another
of 40 mm dia. & 70 mm axis horizontally. Both axes intersect & bisect each other.
Draw projections showing curves of intersections
4” 1”3” 2”1’ 2’4’ 3’
Draw projections showing curves of intersections.
a”
b”h”
a’
b ’h’
g” c”
f” d”
c’g’
d’f’
X Y
e”
f” dd f
a’
4
1 3
2
33
34. Problem: CYLINDER (50mm dia.and 70mm axis ) STANDING & SQ.PRISM (25 mm
sides and 70 mm axis) PENETRATING. Both axes Intersect & bisect each other. All faces of
i ll i li d t H D j ti h i f i t ti
4” 1”3” 2”1’ 2’4’ 3’
prism are equally inclined to Hp. Draw projections showing curves of intersections.
a”
d”d” b”
Projections ofProjections ofFind lines of
X Y
c”
j
Solid 1.
j
Solid2.
Find lines of
intersectionMissing critical
points at which
curve changes
direction
4
direction.
1 3
2
34
35. Problem. SQ.PRISM (30 mm base sides and 70mm axis ) STANDING & SQ.PRISM (25
mm sides and 70 mm axis) PENETRATING. Both axes intersects & bisect each other. All
faces of prisms are equally inclined to Vp Draw projections showing curves of
1’ 2’4’ 3’
faces of prisms are equally inclined to Vp. Draw projections showing curves of
intersections.
a”a’ a’
1’ 2’4’ 3’ 4” 1”3” 2”
d” b”
d’
b’
d’
b’
X Y
c”
c’ c’
4
Preference of
object line over
dash line
1 3
dash line…
2 35
36. Problem. SQ.PRISM (30 mm base sides and 70mm axis ) STANDING & SQ.PRISM (25
mm sides and 70 mm axis) PENETRATING. Both axes Intersect & bisect each other. Two
faces of penetrating prism are 300 inclined to Hp. Draw projections showing curves of
1’ 2’4’ 3’ 4” 1”3” 2”
p g p p p j g
intersections in I angle projection system.
f”
a’
f’
e’
c”
c’
d’
b’
X Y
4
300
d
1 3
2
od"
36
Other possible arrangements ???? b"
37. Problem: A vertical cylinder 50mm dia. and 70mm axis is completely penetrated by
a triangular prism of 45 mm sides and 70 mm axis, horizontally. One flat face of
prism is parallel to Vp and Contains axis of cylinder Draw projections showingprism is parallel to Vp and Contains axis of cylinder. Draw projections showing
curves of intersections in I angle projection system.
4” 1”3” 2”1’ 2’4’ 3’
aaa
b
c
d
bb
c
d
X Y
e
d
f
e e
f f
4
1 3
2
37
38. Problem: A vertical cone, base diameter 75 mm and axis 100 mm long, is completely
penetrated by a cylinder of 45 mm diameter and axis 100 mm long. The axis of the cylinder is
parallel to Hp and Vp and intersects axis of the cone at a point 28 mm above the base. Draw
o”o’
projections showing curves of intersection in FV & TV. in I angle projection system
11
1
2
3
46
7
8
33
7,
8,22
4
g
a’ b’h’ c’g’ d’f’ e’ g” f”h” a”e” b”d” c”
4
5
6
X Y
5 5
64 4
28
h
g
f
a e
b
c
d 38
39. Problem: Vertical cylinder (80 mm diameter & 100 mm height) is completely penetrated by a
horizontal cone (80 mm diameter and 120 mm height). Both axes intersect & bisect each other.
Draw projections showing curve of intersections in I angle projection systemDraw projections showing curve of intersections in I angle projection system.
7’7
6’ 8’
1’ 5’
2’ 4’
X Y
3’
1
2 8 Intersection of a
curve with
3 7
curve with
another !!!!
Generator lines..
4 6
5 39
40. Problem: CONE STANDING & SQ.PRISM PENETRATING
(BOTH AXES VERTICAL)
2’
1’
5’
3’
X Ya’ b’h’ c’g’ d’f’ e’
4’
6’
Y
h
g
f
8
a b h c g d f e
a e
f
1
6
10
9 7
Problem: A cone70 mm base diameter and 90 mm axis
is completely penetrated by a square prism from top
b d
2
3
4
5
with it’s axis // to cone’s axis and 5 mm away from it.
A vertical plane containing both axes is parallel to Vp.
Take all faces of sq.prism equally inclined to Vp.
Base Side of prism is 30 mm and axis is 100 mm long
c
5 mm OFF-SET
Base Side of prism is 30 mm and axis is 100 mm long.
Draw projections showing curves of intersections.
40
41. Intersection of two cylindersIntersection of two cylinders
oblique to each other –Use PAVoblique to each other Use PAV
III angle
projection
41
42. Intersection of two cylindersIntersection of two cylinders
oblique to each other –Use AVoblique to each other Use AV
42
43. Intersection of Cone and ObliqueIntersection of Cone and Oblique
cylinder using PAVcylinder using PAV
43
48. DEVELOPMENT OF SURFACES OF SOLIDSDEVELOPMENT OF SURFACES OF SOLIDSDEVELOPMENT OF SURFACES OF SOLIDSDEVELOPMENT OF SURFACES OF SOLIDS
Solids are bounded by geometric surfaces:Solids are bounded by geometric surfaces:
LATERLAL SURFACELATERLAL SURFACE IS SURFACE EXCLUDING SOLID’S TOP & BASE.IS SURFACE EXCLUDING SOLID’S TOP & BASE.
Development ~ obtaining the area of the surfaces of a solid.
Development of the solid when folded or rolled gives the solid
- Plane Prism, Pyramid
Single curved Cone Cylinder
Development of the solid, when folded or rolled, gives the solid.
- Single curved Cone, Cylinder
- Double curved sphere 48
49. Examples
Prism – Made up of same number of rectangles as sides of the base
One side: Height of the prismg p
Other side: Side of the base
Cylinder – Rectangle
One side: Height of the cylinder
Other side: Circumference of the baseOther side: Circumference of the base
Pyramid – Number of triangles in contact
h
The base may be included
if present
πd
T. L.
49
50. Methods to Develop Surfacesp
1. Parallel-line development: Used for prisms (full or truncated),
c linders (f ll or tr ncated) Parallel lines are dra n along thecylinders (full or truncated). Parallel lines are drawn along the
surface and transferred to the development
Cylinder: A Rectangle
H
Cylinder: A Rectangle
πD
D
H= Height D= base diameterg
Prisms: No.of Rectangles
H
50SS H= Height S = Edge of base
51. Ex:Ex:
DOTTED LINES are neverDOTTED LINES are never
shown on developmentshown on development
51
shown on developmentshown on development
52. 4, d
2, b
Complete de elopment of c be c t b c tting plane (inclined to HP at
52
Complete development of cube cut by cutting plane (inclined to HP at
30 degrees and perpendicular to VP)
54. Cylinder cut by
h
i
j
k
φ50 Cylinder cut by
three planes
a g
hl
three planes
G
T
F
b
c
d e
f
Cc’
d'
e' f'
g'
D
E
F
G
45o
H
I
J
K IA B C
a'
b'
c
15o
100
AI
3 4
5 6 7 8
9
1041 51 61 71
54
πx50
11
2
3 4 10
11 12
11
21
31 41
55. Problem: Development of a solid is a parabolaProblem: Development of a solid is a parabola
with a 180 mm base and a 90 mm height. Draw the
projections of solidprojections of solid.
180=π D
D=57.3 mm
55
56. Methods to Develop Surfacesp
1. Parallel-line development
2. Radial-line development: Used for pyramids, cones etc. in which the
true length of the slant edge or generator is used as radius
Cone: (Sector of circle) Pyramids: (No.of triangles)
θ
R=Base circle radius. L= Slant edge
θ = R
L
3600
L=Slant height.
L Slant edge.
S = Edge of base
56
57. Parallel vs Radial line method
Parallel line method
Radial line method 57
58. DEVELOPMENT OF DEVELOPMENT OF
FRUSTUMSFRUSTUMS
Base side
Top side
DEVELOPMENT OF
FRUSTUM OF CONE
DEVELOPMENT OF
FRUSTUM OF SQUARE PYRAMID
Top side
θ
θ = R
L
3600
R= Base circle radius of cone
L= Slant height of cone
L= Slant edge of pyramid
L1 = Slant edge of cut part.
L1 = Slant height of cut part.
1.1. DevelopmentDevelopment isis a shape showing AREA, means it’s a 2a shape showing AREA, means it’s a 2--D plain drawing.D plain drawing.
22 AllAll dimensions of it must be TRUE dimensionsdimensions of it must be TRUE dimensions
ImportantImportant pointspoints..
2.2. AllAll dimensions of it must be TRUE dimensions.dimensions of it must be TRUE dimensions.
3. As it is representing shape of an un3. As it is representing shape of an un--folded sheet, no edges can remain hiddenfolded sheet, no edges can remain hidden
and hence DOTTED LINES are never shown on development.and hence DOTTED LINES are never shown on development.
59. Development of Sphere using
Frustum of Cones: Zone MethodFrustum of Cones: Zone Method
Zone 1: Cone
Zone 2: Frustum of cone
Zone 3: Frustum of cone
Z 4 F t fZone 4: Frustum of cone
θ = R
L
3600
θ
59
60. Development by Radial Method
Pyramids (full or Truncated) &
Cones (full or Truncated)Cones (full or Truncated).
If the slant height of a cone is equal to its diameter of base then its development isIf the slant height of a cone is equal to its diameter of base then its development is
a semicircle of radius equal to the slant height.
60
62. Development of Oblique Objects
• Right regular objects – Axis of object
perpendicular to base.
• Axis of any regular object (prismAxis of any regular object (prism,
pyramid, cylinder, cone, etc.) inclined at
angle other than right angle Obliqueangle other than right angle – Oblique
OBJECT. Use ARC method.
62
63. Oblique prismOb que p s
c
de
fg j
a b
cf
h i
g
Parallel
j
a b
f
a' b’f’ c'
f
h’ i'g' j'
h ig
gg
63
64. Draw the development of an oblique circular cylinder with base diameter 30 mm and
axis inclined at 75o with the base. Height of the cylinder is 50 mm
• Divide the surface of the cylinder into equal parts as shown, with
the generator lines parallel to the end generators
• Draw projection lines from top edge of cylinder
φ30
AGg
a
T
• Draw projection lines from top edge of cylinder
such that they are perpendicular to end
generator
• Mark distances AB, BC etc. from one projector
A
E
B
CD
F
g
a
F A
line to the next to complete the profile
• Do the similar process for the bottom edgeA’G’
B
C
A
75o
G
50
A1
75
g’ a' a
A1
64
A1
g a
66. Intersection of
Plane & Pyramid. 14y
Development of
resulting lateral
0
C
D
Develop
1-D-A-2-1
2-A-B-3-21
B D
g
truncated Pyramid
23
A
B 3-B-C-4-3
1-D-C-4-12
1
B
YX
B’
0
A
B
D’
4
23
1’
o
3
2
66
67. Methods to Develop Surfacesp
1. Parallel-line development:
2. Radial-line development:
3. Triangulation development: Complex shapes are divided into a
number of triangles and transferred into the development
Tetrahedron: Four Equilateral Triangles
EXAMPLES:EXAMPLES:--
Boiler Shells & chimneys,Boiler Shells & chimneys,
Pressure Vessels Shovels TraysPressure Vessels Shovels Trays
All sides
equal in length
Pressure Vessels, Shovels, Trays,Pressure Vessels, Shovels, Trays,
Boxes & Cartons, FeedingBoxes & Cartons, Feeding
Hoppers, Large Pipe sections,Hoppers, Large Pipe sections,
B d & P t f t tiB d & P t f t ti
equal in length
Body & Parts of automotives,Body & Parts of automotives,
Ships, Aero planes.Ships, Aero planes.
67
68. Transition
Connect two hollow objects having different base.
Pieces
Three surfaces
Triangulation Method:
Dividing a surface into aDividing a surface into a
number of triangles and
transfer them to the
developmentdevelopment.
68
69. Ex: In air conditioning system, a square duct of 50mm by 50mm is connected to
another square duct of 25mm by 25 mm by using a connector (transition piece) of
height 25mm Draw development of lateral surface of the connector (Neglectheight 25mm. Draw development of lateral surface of the connector (Neglect
thickness of connector). Pyramids: (No.of triangles)
O’O
L= Slant edge.
S = Edge of base
a’ A
g
b’ B
a
b
O
69
c
70. D l t f T iti Pi fDevelopment of Transition Piece for
Difference Shapes and Sizesp
• Connect a Square pipe with circular pipe.
• Ex: Imagine a transition piece (height = 25)
to connect a chimney of square crossto connect a chimney of square cross
section 50mm * 50 mm to circular pipe of
30mm diameter Draw the projections and30mm diameter. Draw the projections and
develop lateral surface of the transition
piece.
70
71. 1’2’, 8’
1/8 of
i fc
a’, b’
circumferencebc
2
3
4 1
1
2
5
8
I angle p ojectionad
6
7
8
18
A B
71
I angle projectionad 18
73. Summarizing DEVELOPMENT OF SPHERE
• All three views -> CIRCLE.All three views CIRCLE.
• Approximate development by dividing
h i t i fsphere into a series of zones.
Cone FrustumCone, Frustum
of cone,
Symmetry
73
74. Draw the development of a hemispherical bowl of radius 3 cm by any method12 such developments.
42
6 8 10
4
8
10
6
o a cb d e
1 3 5
42
7 9
O
E
1
3
5
7
9
4
2
d
a
c
b
o
e
4
2
8
10
6
0
This “Lune
method” is also
T
1
3
5
7
9
d
a
c
b
o
e
1
3
5
7
9
4
2
8
10
6
4
8
10
6
method” is also
known as
“Polycylindric
method” or
2
OE
Rπ
=
o'F
9
d
a
c
b
O
E
1
3
5
7
9
4
2
6
d
a
c
b
o
e
4
2
8
10
6
method or
“Gore method”.
a'
b'
c'
1
3
5
7
9
d
a
c
b
o
e
1
3
5
7
9
4
2
8
10
6
4
8
10
6
For a reasonable
accuracy circle
c
d'
e'
7
9
d
a
c
b
o
e
1
3
5
7
9
4
2
6
d
a
c
b
o
e
4
2
8
10
6
needs to be
divided in 16 or
more parts.
o
1
3
5
7
9
d
a
c
b
o
e
1
3
5
7
4
2
8
10
6
8
10
6
74
75. Draw the development of a hemispherical bowl of radius 3 cm by any method
o4
o4'
o a b c d e
o
360
'
0 ×=
oa
oa
θ
T
o3
o
360
'1
1 ×=
oc
bo
ob
θ
o '
o'F
a'
b'
c'
o2
o
o
360
360
'2
2
×
×=
od
co
oc
θ
θ
o'
o1'
o2'
o3
o 'c
d'
e'
o
360
360
'
4
3
3
×=
×=
oe
do
θ
θ o
a'
b'
'
o'
o1'
'4
4
eo c'
d'
e'
76. PROBLEM: A room is of size L=6.5m, D=5m, H=3.5m. An electric
bulb hangs 1m below the center of ceiling. A switch is placed in oneg g p
of the corners of the room, 1.5m above the flooring. Draw the
projections and determine real distance between the bulb & switch.
Ceiling
TV
Bulb
TV
Switch
DD
76
77. PROBLEM:- A picture frame 2 m wide and 1 m tall is resting on horizontal
wall railing makes 350 inclination with wall. It is attached to a hook in the
wall by two strings The hook is 1 5 m above wall railing Determine length of eachwall by two strings. The hook is 1.5 m above wall railing. Determine length of each
chain and true angle between them
350
Wall railingWall railing
77
78. PROBLEM: Line AB is 75 mm long and it is 300 & 400 Inclined to HP & VP
respectively. End A is 12mm above Hp and 10 mm in front of VP. Draw
b’ b’1
projections. Line is in 1st quadrant.
TL
FV
a’
θ
a
X Y
Ø 1
LFV
TLTV
b b1
79. PROBLEM: Line AB 75mm long makes 450 inclination with VP while it’s FV
makes 550. End A is 10 mm above HP and 15 mm in front of VP. If line is in 1st
b’1b’
quadrant draw it’s projections and find it’s inclination with HP.
LOCUS OF b1’
0
Y
a’
550
X Y
a
1
LFV
b1
b
LOCUS OF b
80. PROBLEM: FV of line AB is 500 inclined to XY and measures 55 mm long while
it’s TV is 600 inclined to XY line. If end A is 10 mm above HP and 15 mm in
front of VP draw it’s projections find TL inclinations of line with HP & VP
b’ b’1
front of VP, draw it’s projections,find TL, inclinations of line with HP & VP.
X
a’
Y
500
θ
X Y
a
Φ
600
Φ
b
b1
81. PROBLEM :- Line AB is 75 mm long . It’s FV and TV measure 50 mm & 60 mm
long respectively. End A is 10 mm above Hp and 15 mm in front of Vp. Draw
b’1
b’
g p y p p
projections of line AB if end B is in first quadrant. Find angle with HP and VP.
X Y
a’
1’LTVθ
X Y
a 1
LFV
Φ
b1
b
82. PROBLEM.
Length (L) depth (D) andLength (L), depth (D), and
height (H) of a room are
6.5m, 5m and 3.5m
especti el An elect ic b lbrespectively. An electric bulb
hangs 1m below the center
of ceiling. A switch is placed
Ceiling
TV
in one of the corners of the
room, 1.5m above the
flooring.
Bulb
g
Draw the projections and
determine real distance
between the bulb and
Switch
Dbetween the bulb and
switch.
D
83. PROBLEM: L=6.5m, D=5m,
H=3 5m An electric bulbH=3.5m. An electric bulb
hangs 1m below the center
of ceiling. A switch isof ceiling. A switch is
placed in one of the corners
of the room, 1.5m above
6.5m
the flooring. Draw the
projections and determine
real distance between the
a’
b’ b’1
3.5m
1m
real distance between the
bulb and switch.
a
x y
1.5
b5m b5m
Answer: a’ b’1
84. PROBLEM: A
PICTURE FRAME 2 mPICTURE FRAME, 2 m
WIDE & 1 m TALL,
RESTING ON
HORIZONTAL WALLHORIZONTAL WALL
RAILING
MAKES 350
INCLINATION WITH
WALL. IT IS
ATTAACHED TO A
350
A
B
HOOK IN THE WALL
BY TWO STRINGS.
THE HOOK IS 1.5 m
D
THE HOOK IS 1.5 m
ABOVE WALL RAILING.
DETERMINE LENGTH OF
EACH CHAIN AND TRUE
Wall railingC
ANGLE BETWEEN THEM
85. PROBLEM- A picture frame 2 m
wide & 1 m tall is resting onwide & 1 m tall is resting on
horizontal wall railing makes 350
inclination with wall. It is attached
t h k i th ll b t t i
h’
to a hook in the wall by two strings.
The hook is 1.5 m above the wall
railing.a’b’
(chains)
DETERMINE LENGTH OF EACH
CHAIN AND TRUE ANGLE BETWEEN
a b 1.5 m
1m
THEM
c’d’ (wall railing)
X Y
ad
h
a1
(frame)
h
(chains)
Answers:
bcb1
Answers:
Length of each chain= hb1
True angle between chains =
86. PROBLEM:- Two mangos on a tree A & B are 1.5 m and 3.0 m above ground and
those are 1.2 m & 1.5 m from a 0.3 m thick wall but on opposite sides of it. If thethose are 1.2 m & 1.5 m from a 0.3 m thick wall but on opposite sides of it. If the
distance measured between them along the ground and parallel to wall is 2.6 m,
Then find real distance between them by drawing their projections.
TV
B
A
0.3M THICK
86
87. PROBLEM:- Two mangos on a
tree A & B are 1 5 m and 3 00
b’
tree A & B are 1.5 m and 3.00
m above ground and those
are 1.2 m & 1.5 m from a 0.3
m thick wall but on opposite
3.0
a’
pp
sides of it. If the distance
measured between them
along the ground and parallel
to wall is 2 6 m Then find
1.5
to wall is 2.6 m, Then find
real distance between them
by drawing their projections.
b
2.6
B
1.5
1 2
a
1.2
88. PROBLEM:-Flower A is 1.5 m & 1 m from walls Q (parallel to reference line) & P
(perpendicular to reference line) respectively. Flower is 1.5 m above the ground.
Orange B is 3.5m & 5.5m from walls Q & P respectively. Drawing projection, find
distance between them If orange is 3.5 m above ground.
b’ b’b 1
3.5 m
a’
x y
1.5 m
Ground
a
1.5 m
Wall Q
B
1 m
3.5 m
b
5.5 m
Wall P
88
89. PROBLEM :- A top view of object (three rods OA, OB and OC whose endsp j ( ,
A,B & C are on ground and end O is 100mm above ground) contains three
lines oa, ob & oc having length equal to 25mm, 45mm and 65mm
respectively. These three lines are equally inclined and the shortest line isrespectively. These three lines are equally inclined and the shortest line is
vertical. Draw their projections and find length of each rod.
Tv
O
A
C
Fv
B
90. PROBLEM :- A top view of object (three rods OA, OB and OC whose ends A,B & C are on
ground and end O is 100mm above ground) contains three lines oa, ob & oc having length
l t 25 45 d 65 ti l Th th li ll i li d d thequal to 25mm, 45mm and 65mm respectively. These three lines are equally inclined and the
shortest line is vertical. Draw their projections and find length of each rod.
o’
TL1
TL2
a’b’ c’ c1’b1’ a1’
x y
a
o Answers:
b
c
Answers:
TL1 TL2 & TL3
91. PROBLEM:- A pipeline from point A has a downward gradient 1:5 and it runs
due South - East. Another Point B is 12 m from A and due East of A and in same
l l f i li f 0 f h d i li flevel of A. Pipe line from B runs 150 Due East of South and meets pipeline from
A at point C. Draw projections and find length of pipe line from B and it’s
inclination with ground.
1
5
Bearing of a LINE:
A
1 Bearing of a LINE:
Horizontal angle
between line &
idi ( th thA
B12 M
E
meridian (north south
line)----Measured in
DEGREES (0 to 90◦).
C
Measured in Top View.
S 45° E.
S 15° E.
92. PROBLEM:- A pipe line from point A has a downward gradient 1:5 and it runs
due South - East. Another Point B is 12 m from A and due East of A and in same
level of A. Pipe line from B runs 150 Due East of South and meets pipe line from
A at point C. Draw projections and find length of pipe line from B and it’s
inclination with ground.
12m
1
5a’ b’
FV
x y
N
c’2c’ c’1
a b
y
450
N
EAST
W
150
TV
c
= Inclination of pipe line BC
SOUTH
= Inclination of pipe line BC
93. PROBLEM: A person observes two objects, A & B, on the ground, from a tower,
15 M high, at the angles of depression 300 & 450 respectively. Object A is due
North-West direction of observer and object B is due West direction. Draw
projections of situation and find distance of objects from observer and from
tower also.
OO
300
450
A
W
S
B
94. PROBLEM: A person observes
two objects, A & B, on thetwo objects, A & B, on the
ground, from a tower, 15 M high,
at the angles of depression 300 &
450 Object A is due North-West
o’
300 450
45 . Object A is due North-West
direction of observer and object
B is due West direction. Draw
projections of situation and find
a’1
’
30
15M
projections of situation and find
distance of objects from
observer and from tower also.Na
b’a’
W E
ob
Answers:
S
Answers:
Distances of objects
from observe
o’a’1 & o’b’
From tower SFrom tower
oa & ob
95. PROBLEM:- A tank of 4 m height is to be strengthened by four rods from each corner by
fixing their other ends to the flooring, at a point 1.2 m and 0.7 m from two adjacent walls
respectively, as shown. Determine graphically length and angle of each rod with flooring.
TV
A
4 M
B
96. PROBLEM:- A tank of 4 m height is to be strengthened by four rods from
h b fi i th i th d t th fl i t i t 1 2 deach corner by fixing their other ends to the flooring, at a point 1.2 m and
0.7 m from two adjacent walls respectively, as shown. Determine
graphically length and angle of each rod with flooring.
a’
True Length
FV
True Length
Answers:
Length of each rod
= a’b’1
Angle with Hp.
a
b’b’1
=
X Y
a
b
TVTV
97. SHORTEST DISTANCE BETWEEN POINT and LINE
Measure distance between Point P and Line AB for given TV &
FV arrangement. aa
p
b
y
T
F
x
p
p’
ba
p
A BD
a’
b’
ba A B
q
D
pD is ⊥ to
ab
97
bq ab
98. a2, b2
Point view of line
a2, b2
Shortest distance a2 p2
a1
b1
p2
Point
x1
p1
a
1
y1b
y
p
T
F
x
’
p’
a’
b’
99. Shortest Distance between 2 skew Lines (AB & CD)Shortest Distance between 2 skew Lines (AB & CD)
Skew (oblique) lines Lines that are not parallel & do not intersect
c
•Find T.L. of one of the lines
and project its point viewDistance measured along
Line ⊥ to both
b
c p j p
using auxiliary plane
method
Line ⊥ to both.
a
b
d •Project the other line also
in each view
T
F
a
b
•The perpendicular distance
between the point view of
c
p
one line and the other line
is the required shortest
di t b t th t
99
d distance between the two
lines
100. Primary auxiliary viewc
TL
d d, c
Secondary auxiliary view
c
Required distance
b
c a
a
b
d T
F
d
a
b
ab B AP
c
q
dP is ⊥ to
ab
d In mines, this method might be used to
locate a connecting tunnel.
101. True Angle between 2 Skew Lines (AB & CD)
Draw P.A. V. such that one line (AB)
shows its True Length
Measure angle in view that
Shows both lines in true length
Draw S.A.V. view with reference line
perpendicular to the True Length of the
line (AB) to get the point view of the line
a
b
d
T
F
Draw a tertiary auxiliary view with
reference line parallel to the other line in
order to get its True Length
x y
d’
c
d
Since the secondary auxiliary view had
the point view of the first line, the
tertiary auxiliary view will have the True
L th f th fi t li l
a’
d
Length of the first line also.
c’ b’
101
102. TRUE LENGTH OF
TERTIARY AUXILIARY VIEW
ANGLE BETWEEN TWO
LINES
a3
c3
TRUE LENGTH OF
BOTH LINES
True Angle
b h
Parallel
LINES
b3
2
d3
PRIMARY
AUXILIARY VIEW
between the two
lines
a2 ,b2
c2
Point view of
one line
SECONDARY AUXILIARY VIEWTrue length
b1
c1
d2
one line
Parallel
a1
b
d1
T
Parallel
a
c
d Angle between two
nonintersecting linesT
F
x y
a’
d’
nonintersecting lines
is measurable in a
view that shows both
lines in true shapea’
c’ b’
lines in true shape.
103. Angle between 2 planesAngle between 2 planes
Line of intersection of the 2 planes (here it is True Length)
e fe f
a
c
d
x T
F
y
a’ b’
a
b • Obtain an auxiliary view such that the
reference line is perpendicular to the True
Length of the line of intersection of the
l
a
planes
• In this case, the intersection line is parallel
to both principle planes and hence is in True
L th i b th f t d t i
e’
d’ c’
f’
Length in both front and top views
• Both planes will be seen as edge views in
the auxiliary view.
e f’
• The angle between the edge views is the
angle between the planes
104. Line of intersection of the 2 planes (here it is True Length)
x1
e f f1, e1
PRIMARY AUXILIARY VIEW
e f ,
a
c
d
b1, a1
c1, d1
x •Obtain an auxiliary view such that the
reference line is perpendicular to the True
Length of the line of intersection of the
l
T
F
y
y1
a’ b’
a
b
b1, a1
planes
•In this case, the intersection line is parallel
to both principle planes and hence is in True
L th i b th f t d t i
1
a
Length in both front and top views
•Both planes will be seen as edge views in
the auxiliary view.
e’
d’ c’
f’
•The angle between the edge views is the
angle between the planes
e f’
105. Piercing point of a line with a plane
Edge view of the plane
True length of
T A1
p1
g
principal line
In a view showing the
Line
p
T
g
plane as an edge, the
piercing point appears
where the line
Planep’
F
where the line
intersects the edge
view.
Principal line
Part of the line
hidden by the plane
h ld b h
Draw auxiliary view to
get EDGE VIEW.
should be shown
dotted
106. Find the shortest distance of point P from the body diagonal AB of the cube of side 50
1
mm as shown
a1
b1
a1‘, b1’
d’g’ b’ e’
p1
p ’
10
1 , 1
p’
d g
10
50
b’, e’p1’Required
distance
Draw an auxiliary view to get the 50Draw an auxiliary view to get the
true length of the line
Draw an auxiliary view to get the
point view of the diagonal c’, d’F, A
c,bf,d
p
point view of the diagonal
Project the point P in these views to
get the required distance
d,ea,g
