engineering-graphics.pdf

1.
1. SECTIONS OF SOLIDS.
SECTIONS OF SOLIDS.
2.
2. DEVELOPMENT.
DEVELOPMENT.
3.
3. INTERSECTIONS.
INTERSECTIONS.
ENGINEERING APPLICATIONS
ENGINEERING APPLICATIONS
OF
OF
THE PRINCIPLES
THE PRINCIPLES
OF
OF
PROJECTIONS OF SOLIDES.
PROJECTIONS OF SOLIDES.
STUDY CAREFULLY
STUDY CAREFULLY
THE ILLUSTRATIONS GIVEN ON
THE ILLUSTRATIONS GIVEN ON
NEXT
NEXT SIX
SIX PAGES !
PAGES !

SECTIONING A SOLID.
SECTIONING A SOLID.
An object ( here a solid ) is cut by
An object ( here a solid ) is cut by
some imaginary cutting plane
some imaginary cutting plane
to understand internal details of that
to understand internal details of that
object.
object.
The action of cutting is called
The action of cutting is called
SECTIONING
SECTIONING a solid
a solid
&
&
The plane of cutting is called
The plane of cutting is called
SECTION PLANE.
SECTION PLANE.
wo cutting actions means section planes are recommended
wo cutting actions means section planes are recommended.
.
Section Plane perpendicular to Vp and inclined to Hp.
Section Plane perpendicular to Vp and inclined to Hp.
( This is a definition of an Aux. Inclined Plane i.e. A.I.P.)
( This is a definition of an Aux. Inclined Plane i.e. A.I.P.)
NOTE:- This section plane appears
as a straight line in FV.
as a straight line in FV.
Section Plane perpendicular to Hp and inclined to Vp.
Section Plane perpendicular to Hp and inclined to Vp.
( This is a definition of an Aux. Vertical Plane i.e. A.V.P.)
( This is a definition of an Aux. Vertical Plane i.e. A.V.P.)
as a straight line in TV.
as a straight line in TV.
emember:-
emember:-
After launching a section plane
After launching a section plane
either in FV or TV, the part towards observer
either in FV or TV, the part towards observer
is assumed to be removed.
is assumed to be removed.
As far as possible the smaller part is
As far as possible the smaller part is
assumed to be removed.
assumed to be removed.
OBSERVER
OBSERVER
ASSUME
ASSUME
UPPER PART
UPPER PART
REMOVED
REMOVED SECTON
PLANE
SECTON
PLANE
IN
FV.
IN
FV.
OBSERVER
OBSERVER
ASSUME
ASSUME
LOWER PART
LOWER PART
REMOVED
REMOVED
SECTON PLANE
SECTON PLANE
IN TV.
IN TV.
(A)
(A)
(B)
(B)

ILLUSTRATION SHOWING
ILLUSTRATION SHOWING
IMPORTANT TERMS
IMPORTANT TERMS
IN SECTIONING.
IN SECTIONING.
x
x y
y
TRUE SHAPE
TRUE SHAPE
Of SECTION
Of SECTION
SECTION
SECTION
PLANE
PLANE
SECTION LINES
SECTION LINES
(45
(450
0
to XY)
to XY)
Apparent Shape
Apparent Shape
of section
of section
SECTIONAL T.V.
SECTIONAL T.V.
For TV
For TV
For True Shape
For True Shape

Section Plane
Section Plane
Through Apex
Through Apex
Section Plane
Section Plane
Through Generators
Through Generators
Section Plane Parallel
Section Plane Parallel
to end generator.
to end generator.
Section Plane
Section Plane
Parallel to Axis.
Parallel to Axis.
Triangle
Triangle Ellipse
Ellipse
P
a
r
a
b
o
l
a
P
a
r
a
b
o
l
a
Hyperbola
Hyperbola
Ellipse
Ellipse
Cylinder through
Cylinder through
generators.
generators.
Sq. Pyramid through
Sq. Pyramid through
all slant edges
all slant edges
Trapezium
Trapezium
Typical Section Planes
Typical Section Planes
&
&
Typical Shapes
Typical Shapes
Of
Of
Sections
Sections.
.

DEVELOPMENT OF SURFACES OF SOLIDS.
DEVELOPMENT OF SURFACES OF SOLIDS.
MEANING:-
MEANING:-
ASSUME OBJECT HOLLOW AND MADE-UP OF THIN SHEET. CUT OPEN IT FROM ONE SIDE AND
ASSUME OBJECT HOLLOW AND MADE-UP OF THIN SHEET. CUT OPEN IT FROM ONE SIDE AND
UNFOLD THE SHEET COMPLETELY. THEN THE
UNFOLD THE SHEET COMPLETELY. THEN THE SHAPE OF THAT UNFOLDED SHEET IS CALLED
SHAPE OF THAT UNFOLDED SHEET IS CALLED
DEVELOPMENT OF LATERLAL SUEFACES
DEVELOPMENT OF LATERLAL SUEFACES OF THAT OBJECT OR SOLID.
OF THAT OBJECT OR SOLID.
LATERLAL SURFACE
LATERLAL SURFACE IS THE SURFACE EXCLUDING SOLID’S TOP & BASE.
IS THE SURFACE EXCLUDING SOLID’S TOP & BASE.
ENGINEERING APLICATION
ENGINEERING APLICATION:
:
THERE ARE SO MANY PRODUCTS OR OBJECTS WHICH ARE DIFFICULT TO MANUFACTURE BY
THERE ARE SO MANY PRODUCTS OR OBJECTS WHICH ARE DIFFICULT TO MANUFACTURE BY
CONVENTIONAL MANUFACTURING PROCESSES, BECAUSE OF THEIR SHAPES AND SIZES.
CONVENTIONAL MANUFACTURING PROCESSES, BECAUSE OF THEIR SHAPES AND SIZES.
THOSE ARE FABRICATED IN SHEET METAL INDUSTRY BY USING
THOSE ARE FABRICATED IN SHEET METAL INDUSTRY BY USING
DEVELOPMENT TECHNIQUE. THERE IS A VAST RANGE OF SUCH OBJECTS.
DEVELOPMENT TECHNIQUE. THERE IS A VAST RANGE OF SUCH OBJECTS.
EXAMPLES:-
EXAMPLES:-
Boiler Shells & chimneys, Pressure Vessels, Shovels, Trays, Boxes & Cartons, Feeding Hoppers,
Boiler Shells & chimneys, Pressure Vessels, Shovels, Trays, Boxes & Cartons, Feeding Hoppers,
Large Pipe sections, Body & Parts of automotives, Ships, Aeroplanes and many more.
Large Pipe sections, Body & Parts of automotives, Ships, Aeroplanes and many more.
WHAT IS
WHAT IS
OUR OBJECTIVE
OUR OBJECTIVE
IN THIS TOPIC ?
IN THIS TOPIC ?
To learn methods of development of surfaces of
To learn methods of development of surfaces of
different solids, their sections and frustums
different solids, their sections and frustums.
.
1. Development is different drawing than PROJECTIONS.
1. Development is different drawing than PROJECTIONS.
2. It is a shape showing AREA, means it’s a 2-D plain drawing.
2. It is a shape showing AREA, means it’s a 2-D plain drawing.
3. Hence all dimensions of it must be TRUE dimensions.
3. Hence all dimensions of it must be TRUE dimensions.
4. As it is representing shape of an un-folded sheet, no edges can remain hidden
4. As it is representing shape of an un-folded sheet, no edges can remain hidden
And hence DOTTED LINES are never shown on development.
And hence DOTTED LINES are never shown on development.
But before going ahead,
But before going ahead,
note following
note following
Important
Important points
points.
.
Study illustrations given on next page carefully.
Study illustrations given on next page carefully.

Q 14.11: A square pyramid, base 40 mm side and axis 65 mm long, has its base on the HP and all
the edges of the base equally inclined to the VP. It is cut by a section plane, perpendicular to the
VP, inclined at 45º to the HP and bisecting the axis. Draw its sectional top view, sectional side
view and true shape of the section.
X Y
45º
a
b
c
d
o
a’
b’
c’
d’
o’
1
2
3
4
1’
2’
3’
4’
11
41
21 31
X1
Y1
d” a”c” b”
o”
3”
2”
4”
1”

Q 14.14: A pentagonal pyramid , base 30mm side and axis 60 mm long is lying on one of its triangular faces
on the HP with the axis parallel to the VP. A vertical section plane, whose HT bisects the top view of the axis
and makes an angle of 30º with the reference line, cuts the pyramid removing its top part. Draw the top view,
sectional front view and true shape of the section and development of the surface of the remaining portion of
the pyramid.
X Y
a
b
c
d
e
o
a’ b’e’
c’d’
o’
60
30
c’d’ o’
a’
b’e’
a1
b1
c1
d1
e1
o1
1’
2’
3’
4’
5’
6’
1
2
3
4
5
6 31’
41’
21’
11’
61’
51’

Q 14.6: A Hexagonal prism has a face on the H.P. and the axis parallel to the V.P. It is cut by a vertical
section plane the H.T. of which makes an angle of 45 with XY and which cuts the axis at a point 20 mm from
one of its ends. Draw its sectional front view and the true shape of the section. Side of base 25 mm long
height 65mm.
X Y
a
b
c
d
e
f
a’ b’ c’
d’
e’
f’
25
65
a’ b’ c’
d’
e’
f’
a’
b’
c’d’
e’
f’
a’
b’
c’d’
e’
f’
d1
a1
b1
c1
e1
f1
d1
a1
b1
c1
e1
f1
20
1’
2’
3’
4’
5’
6’ 7’
1 2
3
4
5
6
7
X1
Y1
31’
41’
21’
11’
71’
61’
51’

X Y
1
2
3
4
5
6
7
8
9
10
11
12
Q 14.24: A Cone base 75 mm diameter and axis 80 mm long is resting on its base on H.P. It is cut by a section
plane perpendicular to the V.P., inclined at 45º to the H.P. and cutting the axis at a point 35 mm from the
apex. Draw the front view, sectional top view, sectional side view and true shape of the section.
1
2
12
3
11
4
10
5
9
6
8 7
o
o’
35
a
b
k
c
d
l
e
f
g
h
i
j
a’
b’
k’
c’
d’
l’
e’
f’
g’
h’
i’
j’
a
1
b
1
c
1
d
1
e
1
f
1
g
1
h
1
i
1
j
1
k
1
l
1
X1
Y1
4” 5” 6” 7” 8” 9”10”
11”
12”
1”
2”
3”
o”
a”
b”
c”
d”
e”
f”
g”
h”
i”
j”
k”
l”

πD
H
D
S
S
H
L
θ
θ = R
L
+
3600
R=Base circle radius.
L=Slant height.
L= Slant edge.
S = Edge of base
L
S
S
H= Height S = Edge of base
H= Height D= base diameter
Development of lateral surfaces of different solids.
(Lateral surface is the surface excluding top & base)
Prisms: No.of Rectangles
Cylinder: A Rectangle
Cone: (Sector of circle) Pyramids: (No.of triangles)
Tetrahedron: Four Equilateral Triangles
All sides
equal in length
Cube: Six Squares.

L L
θ
θ = R
L
+
3600
R= Base circle radius of cone
L= Slant height of cone
L1 = Slant height of cut part.
Base side
Top side
L1 L1
L= Slant edge of pyramid
L1 = Slant edge of cut part.
DEVELOPMENT OF
FRUSTUM OF CONE
DEVELOPMENT OF
FRUSTUM OF SQUARE PYRAMID
STUDY NEXT
STUDY NEXT NINE
NINE PROBLEMS OF
PROBLEMS OF
SECTIONS & DEVELOPMENT
SECTIONS & DEVELOPMENT
FRUSTUMS
FRUSTUMS

X Y
X1
Y1
a’
b’ e’
c’ d’
A
B
C
E
D
a
e
d
b
c
TRUE
SHAPE
A B C D E A
DEVELOPMENT
a”
b”
c”
d”
e”
Problem 1:
Problem 1: A pentagonal prism , 30 mm base side & 50 mm axis
A pentagonal prism , 30 mm base side & 50 mm axis
is standing on Hp on it’s base whose one side is perpendicular to Vp.
is standing on Hp on it’s base whose one side is perpendicular to Vp.
It is cut by a section plane 45
It is cut by a section plane 450
0
inclined to Hp, through mid point of axis.
inclined to Hp, through mid point of axis.
Draw Fv, sec.Tv & sec. Side view. Also draw true shape of section and
Draw Fv, sec.Tv & sec. Side view. Also draw true shape of section and
Development of surface of remaining solid.
Development of surface of remaining solid.
Solution Steps:
Solution Steps:for sectional views:
for sectional views:
Draw three views of standing prism.
Draw three views of standing prism.
Locate sec.plane in Fv as described.
Project points where edges are getting
Project points where edges are getting
Cut on Tv & Sv as shown in illustration.
Cut on Tv & Sv as shown in illustration.
Join those points in sequence and show
Join those points in sequence and show
Section lines in it.
Section lines in it.
Make remaining part of solid dark.
For True Shape:
For True Shape:
Draw x
Draw x1
1y
y1
1 // to sec. plane
// to sec. plane
Draw projectors on it from
cut points.
cut points.
Mark distances of points
of Sectioned part from Tv,
on above projectors from
x
x1
1y
y1
1 and join in sequence.
and join in sequence.
Draw section lines in it.
It is required true shape.
For Development:
For Development:
Draw development of entire solid. Name from
Draw development of entire solid. Name from
cut-open edge I.e. A. in sequence as shown.
cut-open edge I.e. A. in sequence as shown.
Mark the cut points on respective edges.
Mark the cut points on respective edges.
Join them in sequence in st. lines.
Join them in sequence in st. lines.
Make existing parts dev.dark.
Make existing parts dev.dark.

Y
h
a
b
c
d
e
g
f
X a’ b’ d’ e’
c’ g’ f’
h’
o’
X1
Y1
g” h”f” a”e” b”d” c”
A
B
C
D
E
F
A
G
H
SECTIONAL T.V
SECTIONAL S.V
TRUE
SH
APE
OF
SECTIO
N
DEVELOPMENT
S
E
C
T
I
O
N
P
L
A
N
E
Problem 2:
Problem 2: A cone, 50 mm base diameter and 70 mm axis is
A cone, 50 mm base diameter and 70 mm axis is
standing on it’s base on Hp. It cut by a section plane 45
standing on it’s base on Hp. It cut by a section plane 450
0
inclined
inclined
to Hp through base end of end generator.Draw projections,
to Hp through base end of end generator.Draw projections,
sectional views, true shape of section and development of surfaces
sectional views, true shape of section and development of surfaces
of remaining solid.
of remaining solid.
Solution Steps:
Solution Steps:for sectional views:
for sectional views:
Draw three views of standing cone.
Draw three views of standing cone.
Project points where generators are
Project points where generators are
getting Cut on Tv & Sv as shown in
getting Cut on Tv & Sv as shown in
illustration.Join those points in
illustration.Join those points in
sequence and show Section lines in it.
sequence and show Section lines in it.
For True Shape:
For True Shape:
Draw x
Draw x1
1y
y1
1 // to sec. plane
// to sec. plane
cut points.
cut points.
x
x1
1y
y1
1 and join in sequence.
and join in sequence.
For Development:
For Development:
Draw development of entire solid.
Draw development of entire solid.
Name from cut-open edge i.e. A.
Name from cut-open edge i.e. A.
in sequence as shown.Mark the cut
in sequence as shown.Mark the cut
points on respective edges.
points on respective edges.
Join them in sequence in
Join them in sequence in
curvature. Make existing parts
curvature. Make existing parts
dev.dark.
dev.dark.

X Y
e’
a’ b’ d’
c’ g’ f’
h’
a
’
h
’
b
’
e
’
c
’
g
’
d
’
f
’
o’
o’
Problem 3: A cone 40mm diameter and 50 mm axis is resting on one generator on Hp( lying on Hp)
which is // to Vp.. Draw it’s projections.It is cut by a horizontal section plane through it’s base
center. Draw sectional TV, development of the surface of the remaining part of cone.
A
B
C
D
E
F
A
G
H
O
a1
h1
g1
f1
e1
d1
c1
b1
o1
SECTIONAL T.V
DEVELOPMENT
(SHOWING TRUE SHAPE OF SECTION)
HORIZONTAL
SECTION PLANE
h
a
b
c
d
e
g
f
O
Follow similar solution steps for Sec.views - True shape – Development as per previous problem!
Follow similar solution steps for Sec.views - True shape – Development as per previous problem!

A.V.P300
inclined to Vp
Through mid-point of axis.
X Y
1,2
3,8
4,7
5,6
1
2
3 4
5
6
7
8
2
1
8
7
6
5
4
3
b’ f’
a’ e’
c’ d’
a
b
c
d
e
f
b’
f’
a’
e’
c’
d’
a1
d1
b1
e1
c1
f1
X1
Y1
AS SECTION PLANE IS IN T.V.,
CUT OPEN FROM BOUNDRY EDGE C1 FOR DEVELOPMENT.
TRUE SHAPE OF SECTION
C D E F A B C
DEVELOPMENT
SECTIONAL F.V.
Problem 4:
Problem 4: A hexagonal prism. 30 mm base side &
A hexagonal prism. 30 mm base side &
55 mm axis is lying on Hp on it’s rect.face with axis
55 mm axis is lying on Hp on it’s rect.face with axis
// to Vp. It is cut by a section plane normal to Hp and
// to Vp. It is cut by a section plane normal to Hp and
30
300
0
inclined to Vp bisecting axis.
inclined to Vp bisecting axis.
Draw sec. Views, true shape & development.
Draw sec. Views, true shape & development.
Use similar steps for sec.views & true shape.
Use similar steps for sec.views & true shape.
NOTE:
NOTE: for development, always cut open object from
for development, always cut open object from
From an edge in the boundary of the view in which
From an edge in the boundary of the view in which
sec.plane appears as a line.
sec.plane appears as a line.
Here it is Tv and in boundary, there is c1 edge.Hence
Here it is Tv and in boundary, there is c1 edge.Hence
it is opened from c and named C,D,E,F,A,B,C.
it is opened from c and named C,D,E,F,A,B,C.
Note
Note the steps to locate
the steps to locate
Points 1, 2 , 5, 6 in sec.Fv:
Points 1, 2 , 5, 6 in sec.Fv:
Those are transferred to
Those are transferred to
1
1st
st
TV, then to 1
TV, then to 1st
st
Fv and
Fv and
Then on 2
Then on 2nd
nd
Fv.
Fv.

1’
2’
3’
4’
5’
6’
7’
7
1
5
4
3
2
6
7
1
6
5
4
3
2
a
b
c
d
e
f
g
4
4 5
3
6
2
7
1
A
B
C
D
E
A
F
G
O
O’
d’e’ c’f’ g’b’ a’
X Y
X1
Y1
TRUE SHAPE
F.V.
SECTIONAL
TOP VIEW.
D
E
V
E
L
O
P
M
E
N
T
Problem 5:A solid composed of a half-cone and half- hexagonal pyramid is
shown in figure.It is cut by a section plane 450
inclined to Hp, passing through
mid-point of axis.Draw F.v., sectional T.v.,true shape of section and
development of remaining part of the solid.
( take radius of cone and each side of hexagon 30mm long and axis 70mm.)
Note:
Note:
Fv & TV 8f two solids
Fv & TV 8f two solids
sandwiched
sandwiched
Section lines style in both:
Section lines style in both:
Development of
Development of
half cone & half pyramid:
half cone & half pyramid:

o’
h
a
b
c
d
g
f
o e
a’ b’ c’ g’ d’f’ e’
h’
X Y
θ = R
L
+
3600
L=Slant height.
θ
A
B
C
D
E
F
G
H
A
O
1
3
2
4
7
6
5
L
1
1
2
2
3
3
4
4
5
5
6
6
7
7
1’
1’
2’
2’
3’
3’ 4’
4’
5’
5’
6’
6’
7’
7’
Problem 6:
Problem 6: Draw a semicircle 0f 100 mm diameter and inscribe in it a largest
Draw a semicircle 0f 100 mm diameter and inscribe in it a largest
circle.If the semicircle is development of a cone and inscribed circle is some
circle.If the semicircle is development of a cone and inscribed circle is some
curve on it, then draw the projections of cone showing that curve.
curve on it, then draw the projections of cone showing that curve.
Solution Steps:
Solution Steps:
Draw semicircle of given diameter, divide it in 8 Parts and inscribe in it
Draw semicircle of given diameter, divide it in 8 Parts and inscribe in it
a largest circle as shown.Name intersecting points 1, 2, 3 etc.
a largest circle as shown.Name intersecting points 1, 2, 3 etc.
Semicircle being dev.of a cone it’s radius is slant height of cone.( L )
Semicircle being dev.of a cone it’s radius is slant height of cone.( L )
Then using above formula find R of base of cone. Using this data
Then using above formula find R of base of cone. Using this data
draw Fv & Tv of cone and form 8 generators and name.
draw Fv & Tv of cone and form 8 generators and name.
Take o -1 distance from dev.,mark on TL i.e.o’a’ on Fv & bring on o’b’
Take o -1 distance from dev.,mark on TL i.e.o’a’ on Fv & bring on o’b’
and name 1’ Similarly locate all points on Fv. Then project all on Tv
and name 1’ Similarly locate all points on Fv. Then project all on Tv
on respective generators and join by smooth curve.
on respective generators and join by smooth curve.
L
L
TO DRAW PRINCIPAL
TO DRAW PRINCIPAL
VIEWS FROM GIVEN
VIEWS FROM GIVEN
DEVELOPMENT.
DEVELOPMENT.

h
a
b
c
d
g
f
e
o’
a’ b’ d’
c’ g’ f’
h’ e’
X Y
A
B
C
D
E
F
G
H
A
O L
1
2
3
4
5
6
7
θ = R
L
+
3600
L=Slant height.
θ
1’
1’
2’
2’ 3’
3’
4’
4’
5’
5’
6’
6’
7’
7’
1
1
2
2
3
3
4
4
5
5
6
6
7
7
Problem 7:
Problem 7:Draw a semicircle 0f 100 mm diameter and inscribe in it a largest
Draw a semicircle 0f 100 mm diameter and inscribe in it a largest
rhombus
rhombus.If the semicircle is development of a cone and rhombus is some curve
.If the semicircle is development of a cone and rhombus is some curve
on it, then draw the projections of cone showing that curve.
on it, then draw the projections of cone showing that curve.
TO DRAW PRINCIPAL
TO DRAW PRINCIPAL
VIEWS FROM GIVEN
VIEWS FROM GIVEN
DEVELOPMENT.
DEVELOPMENT.
Solution Steps:
Solution Steps:
Similar to previous
Similar to previous
Problem:
Problem:

a’
a’ b’
b’ c’
c’ d’
d’
o’
o’
e’
e’
a
a
b
b
c
c
d
d
o
o e
e
X
X Y
Y
A
A
B
B
C
C
D
D
E
E
A
A
O
O
2
2
3
3
4
4
1
1
Problem 8:
Problem 8: A half cone of 50 mm base diameter, 70 mm axis, is standing on it’s half base on HP with it’s flat face
A half cone of 50 mm base diameter, 70 mm axis, is standing on it’s half base on HP with it’s flat face
parallel and nearer to VP.An inextensible string is wound round it’s surface from one point of base circle and
parallel and nearer to VP.An inextensible string is wound round it’s surface from one point of base circle and
brought back to the same point.If the string is of
brought back to the same point.If the string is of shortest length
shortest length, find it and show it on the projections of the cone.
, find it and show it on the projections of the cone.
1
1 2
2
3
3
4
4
1’
1’
2’
2’ 3’
3’ 4’
4’
TO DRAW A CURVE ON
TO DRAW A CURVE ON
PRINCIPAL VIEWS
PRINCIPAL VIEWS
FROM DEVELOPMENT.
FROM DEVELOPMENT. Concept:
Concept: A string wound
A string wound
from a point up to the same
from a point up to the same
Point, of shortest length
Point, of shortest length
Must appear st. line on it’s
Must appear st. line on it’s
Development.
Development.
Solution steps:
Solution steps:
Hence draw development,
Hence draw development,
Name it as usual and join
Name it as usual and join
A to A This is shortest
A to A This is shortest
Length of that string.
Length of that string.
Further steps are as usual.
Further steps are as usual.
On dev. Name the points of
On dev. Name the points of
Intersections of this line with
Intersections of this line with
Different generators.Bring
Different generators.Bring
Those on Fv & Tv and join
Those on Fv & Tv and join
by smooth curves.
by smooth curves.
Draw 4’ a’ part of string dotted
Draw 4’ a’ part of string dotted
As it is on back side of cone.
As it is on back side of cone.

X Y
e’
a’ b’ d’
c’ g’ f’
h’
o’
h
a
b
c
d
e
g
f
O
DEVELOPMENT
A
B
C
D
E
F
A
G
H
O
O
1
1
2
2
3
3
4
4
6
6 5
5
7
7
1’
1’
2’
2’
3’
3’
4’
4’
5’
5’
6’
6’
7’
7’
1
1
2
2
3
3
4
4
5
5
6
6
7
7
HELIX CURVE
HELIX CURVE
Problem 9:
Problem 9: A particle which is initially on base circle of a cone, standing
A particle which is initially on base circle of a cone, standing
on Hp, moves upwards and reaches apex in one complete turn around the cone.
on Hp, moves upwards and reaches apex in one complete turn around the cone.
Draw it’s path on projections of cone as well as on it’s development.
Draw it’s path on projections of cone as well as on it’s development.
Take base circle diameter 50 mm and axis 70 mm long.
Take base circle diameter 50 mm and axis 70 mm long.
It’s a construction of curve
It’s a construction of curve
Helix of one turn on cone
Helix of one turn on cone:
:
Draw Fv & Tv & dev.as usual
Draw Fv & Tv & dev.as usual
On all form generators & name.
On all form generators & name.
Construction of curve Helix::
Construction of curve Helix::
Show 8 generators on both views
Show 8 generators on both views
Divide axis also in same parts.
Divide axis also in same parts.
Draw horizontal lines from those
Draw horizontal lines from those
points on both end generators.
points on both end generators.
1’ is a point where first horizontal
1’ is a point where first horizontal
Line & gen. b’o’ intersect.
Line & gen. b’o’ intersect.
2’ is a point where second horiz.
2’ is a point where second horiz.
Line & gen. c’o’ intersect.
Line & gen. c’o’ intersect.
In this way locate all points on Fv.
In this way locate all points on Fv.
Project all on Tv.Join in curvature.
Project all on Tv.Join in curvature.
For Development:
For Development:
Then taking each points true
Then taking each points true
Distance From resp.generator
Distance From resp.generator
from apex, Mark on development
from apex, Mark on development
& join.
& join.

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