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PROJECTION OF SOLIDS
alphinms@ssn.edu.in
SOLIDS
To understand and remember various solids in this subject properly,
those are classified & arranged in to two major groups.
Group A
Solids having top and base of same shape
Cylinder
Prisms
Triangular Square Pentagonal Hexagonal
Cube
Triangular Square Pentagonal Hexagonal
Cone
Tetrahedron
Pyramids
( A solid having
six square faces)
( A solid having
Four triangular faces)
Group B
Solids having base of some shape
and just a point as a top, called
apex.
SOLIDS
Dimensional parameters of different solids.
Top
Rectangular
Face
Longer
Edge
Base
Edge
of
Base
Corner of
base
Corner of
base
Triangular
Face
Slant
Edge
Base
Apex
Square Prism Square Pyramid Cylinder Cone
Edge
of
Base
Base
Apex
Base
Generators
Imaginary lines
generating curved surface
of cylinder & cone.
Sections of solids( top & base not parallel) Frustum of cone & pyramids.
( top & base parallel to each other)
X Y
STANDING ON H.P
On it’s base.
RESTING ON H.P
On one point of base circle.
LYING ON H.P
On one generator.
(Axis perpendicular to Hp
And // to Vp.)
(Axis inclined to Hp
And // to Vp)
(Axis inclined to Hp
And // to Vp)
While observing Fv, x-y line represents Horizontal Plane. (Hp)
Axis perpendicular to Vp
And // to Hp
Axis inclined to Vp
And // to Hp
Axis inclined to Vp
And // to Hp
X Y
F.V. F.V. F.V.
T.V. T.V. T.V.
While observing Tv, x-y line represents Vertical Plane. (Vp)
STANDING ON V.P
On it’s base.
RESTING ON V.P
On one point of base circle.
LYING ON V.P
On one generator.
STEPS TO SOLVE PROBLEMS IN SOLIDS
Problem is solved in three steps:
STEP 1: ASSUME SOLID STANDING ON THE PLANE WITH WHICH IT IS MAKING INCLINATION.
( IF IT IS INCLINED TO HP, ASSUME IT STANDING ON HP)
( IF IT IS INCLINED TO VP, ASSUME IT STANDING ON VP)
IF STANDING ON HP - IT’S TV WILL BE TRUE SHAPE OF IT’S BASE OR TOP:
IF STANDING ON VP - IT’S FV WILL BE TRUE SHAPE OF IT’S BASE OR TOP.
BEGIN WITH THIS VIEW:
IT’S OTHER VIEW WILL BE A RECTANGLE ( IF SOLID IS CYLINDER OR ONE OF THE PRISMS):
IT’S OTHER VIEW WILL BE A TRIANGLE ( IF SOLID IS CONE OR ONE OF THE PYRAMIDS):
DRAW FV & TV OF THAT SOLID IN STANDING POSITION:
STEP 2: CONSIDERING SOLID’S INCLINATION ( AXIS POSITION ) DRAW IT’S FV & TV.
STEP 3: IN LAST STEP, CONSIDERING REMAINING INCLINATION, DRAW IT’S FINAL FV & TV.
AXIS
VERTICAL
AXIS
INCLINED HP
AXIS
VERTICAL
AXIS
INCLINED HP
AXIS TO VP
er
AXIS
INCLINED
VP
AXIS TO VP
er AXIS
INCLINED
VP
GENERAL PATTERN ( THREE STEPS ) OF SOLUTION:
GROUP B SOLID.
CONE
GROUPA SOLID.
CYLINDER
GROUP B SOLID.
CONE
GROUPA SOLID.
CYLINDER
X
a
b c
d
o
o’
d’
c’
b’
a’
o
’
d
’
c
’
b
’
a
’
o1
d1
b1
c1
a1
d’1
Problem 1. A square pyramid, 40
mm base sides and axis 60 mm long,
has a triangular face on the ground
and the vertical plane is parallel to
axis. Draw its projections. Take apex
nearer to VP
Solution Steps :
Triangular face on Hp , means it is lying on Hp:
1.Assume it standing on Hp.
2.It’s Tv will show True Shape of base( square)
3.Draw square of 40mm sides with one side vertical Tv &
taking 50 mm axis project Fv. ( a triangle)
4.Name all points as shown in illustration.
6.Make visible lines dark and hidden dotted, as per the procedure.
.
For dark and dotted lines
1.Draw proper outline of new view DARK. 2. Decide direction of an observer.
3. Select nearest point to observer and draw all lines starting from it-dark.
4. Select farthest point to observer and draw all lines (remaining)from it- dotted.
Problem 2:
A cone 40 mm diameter and 50 mm axis
is resting on one generator on Hp
Parallel with Vp
Draw it’s projections.
h
a
b
c
d
e
g
f
X a’ b’ d’ e’
c’ g
’
f’
h’
o’
a
’
h
’
b
’
e
’
c
’
g
’
d
’
f
’
o’
a1
h1
g1
f1
e1
d1
c1
b1
o1
Solution Steps:
Resting on Hp on one generator, means lying on Hp:
1.Assume it standing on Hp.
2.It’s Tv will show True Shape of base( circle )
3.Draw 40mm dia. Circle as Tv &
taking 50 mm axis project Fv. ( a triangle)
4.Name all points as shown in illustration.
5.Make visible lines dark and hidden dotted,
as per the procedure.
For dark and dotted lines
1.Draw proper outline of new vie
DARK.
2. Decide direction of an observer.
3. Select nearest point to observer
and draw all lines starting from
it-dark.
4. Select farthest point to observer
and draw all lines (remaining)
from it- dotted.
a
b
d
c
1
2
4
3
X
a b d c
1 2 4 3
a’
b’
c’
d’
1’
2’
3’
4’
450
4’
3’
2’
1’
d’
c’
b’
a’
350
Problem 3:
A cylinder 40 mm diameter and 50 mm
axis is resting on one point of a base
circle on Vp while it’s axis makes 450
with Vp and parallel to Hp. Draw
projections..
Solution Steps:
Resting on Vp on one point of base, means inclined to Vp:
1.Assume it standing on Vp
2.It’s Fv will show True Shape of base & top( circle )
3.Draw 40mm dia. Circle as Fv & taking 50 mm axis project Tv.
( a Rectangle)
4.Name all points as shown in illustration.
Problem 5: A cube of 50 mm long
edges is so placed on Hp on one
corner that a body diagonal is
parallel to Hp and parallel to
Vp Draw it’s projections.
X
b
c
d
a
a’ d’ c’
b’
a’
d’
c’
b’
a1
b1
d1
c1
p’
p’
Solution Steps:
1.Assuming standing on Hp, begin with Tv,a square with all sides
equally inclined to xy.Project Fv and name all points of FV & TV.
2.Draw a body-diagonal joining c’ with 3’( This can become // to xy)
3.From 1’ drop a perpendicular on this and name it p’
1’
3’ 1’
3’
Y
Problem 6:A tetrahedron of 50 mm
long edges is resting on one edge on
Hp while one triangular face containing
this edge is vertical and parallel to Vp.
Draw projections.
X
T L
a o
b
c
b’
a’ c’
o’
a’
a1
c1
o1
b1
900
b’ c’
o’
a’1
b’1
IMPORTANT:
Tetrahedron is a
special type
of triangular
pyramid in which
base sides &
slant edges are
equal in length.
Solid of four faces.
Like cube it is also
described by One
dimension only..
Axis length
generally not given.
Solution Steps
As it is resting assume it standing on Hp.
Begin with Tv , an equilateral triangle as side case as shown:
First project base points of Fv on xy, name those & axis line.
From a’ with TL of edge, 50 mm, cut on axis line & mark o’
(as axis is not known, o’ is finalized by slant edge length)
Then complete Fv.
.
FREELY SUSPENDED SOLIDS:
Positions of CG, on axis, from base, for different solids are shown below.
H
H/2
H/4
GROUPA SOLIDS
( Cylinder & Prisms)
GROUP B SOLIDS
( Cone & Pyramids)
CG
CG
alphinms@ssn.edu.in
X a’ d’
e’
c’
b’
o’
a
b
c
d
e
o
g’
H/4
H
LINE d’g’ VERTICAL
a’b’
c’
d’
e’
g’
a1
b1
o1
e1
d1
c1
e”
FOR SIDE VIEW
Problem 7: A pentagonal pyramid
30 mm base sides & 60 mm long axis,
is freely suspended from one corner of
base so that a plane containing it’s axis
remains parallel to Vp.
Draw it’s e views.
IMPORTANT:
When a solid is freely
suspended from a
corner, then line
joining point of
contact & C.G.
remains vertical.
( Here axis shows
inclination with Hp.)
So in all such cases,
assume solid standing
on Hp initially.)
Solution Steps:
In all suspended cases axis shows inclination with Hp.
1.Hence assuming it standing on Hp, drew Tv - a regular pentagon,corner case.
2.Project Fv & locate CG position on axis – ( ¼ H from base.) and name g’ and
Join it with corner d’
3.As 2nd
Fv, redraw first keeping line g’d’ vertical.
4.As usual project corresponding Tv and then Side View looking from.
alphinms@ssn.edu.in
a’ d’ c’
b’
b
c
d
a
a
’
d
’
c
’
b
’
a1
b
d1
c1
X Y
1’
1’
Problem 8:
A cube of 50 mm long edges is so placed
on Hp on one corner that a body diagonal
through this corner is perpendicular to Hp
and parallel to Vp Draw it’s three views.
Solution Steps:
1.Assuming it standing on Hp begin with Tv, a square of corner case.
2.Project corresponding Fv.& name all points as usual in both views.
3.Join a’1’ as body diagonal and draw 2nd
Fv making it vertical (I’ on xy)
4.Project it’s Tv drawing dark and dotted lines as per the procedure.
5.With standard method construct Left-hand side view.
( Draw a 450
inclined Line in Tv region ( below xy).
Project horizontally all points of Tv on this line and
reflect vertically upward, above xy.After this, draw
horizontal lines, from all points of Fv, to meet these
lines. Name points of intersections and join properly.
For dark & dotted lines
locate observer on left side of Fv as shown.)
alphinms@ssn.edu.in
Axis Tv Length
h
a
b
c
d
e
g
f
y
X a’ b’ d’ e’
c’ g’ f’
h’
o’
a
’
h
’
b
’
e
’
c
’
g
’
d
’
f
’
o
’
450
a1
h1 f1
e1
d1
c1
b1
g1
o1
1
Problem 9: A right circular cone,
40 mm base diameter and 60 mm
long axis is resting on Hp on one
point of base circle such that it’s
axis makes 450
inclination with
Hp and 400
inclination with Vp.
Draw it’s projections.
This case resembles to problem no.7 & 9 from projections of planes topic.
In previous all cases 2nd
inclination was done by a parameter not showing TL.Like
Tv of axis is inclined to Vp etc. But here it is clearly said that the axis is 400
inclined
to Vp. Means here TL inclination is expected.
So assuming it standing on HP begin as usual.
alphinms@ssn.edu.in

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projection of solids - Engineering Drawing - engineering graphics

  • 2. SOLIDS To understand and remember various solids in this subject properly, those are classified & arranged in to two major groups. Group A Solids having top and base of same shape Cylinder Prisms Triangular Square Pentagonal Hexagonal Cube Triangular Square Pentagonal Hexagonal Cone Tetrahedron Pyramids ( A solid having six square faces) ( A solid having Four triangular faces) Group B Solids having base of some shape and just a point as a top, called apex.
  • 3. SOLIDS Dimensional parameters of different solids. Top Rectangular Face Longer Edge Base Edge of Base Corner of base Corner of base Triangular Face Slant Edge Base Apex Square Prism Square Pyramid Cylinder Cone Edge of Base Base Apex Base Generators Imaginary lines generating curved surface of cylinder & cone. Sections of solids( top & base not parallel) Frustum of cone & pyramids. ( top & base parallel to each other)
  • 4. X Y STANDING ON H.P On it’s base. RESTING ON H.P On one point of base circle. LYING ON H.P On one generator. (Axis perpendicular to Hp And // to Vp.) (Axis inclined to Hp And // to Vp) (Axis inclined to Hp And // to Vp) While observing Fv, x-y line represents Horizontal Plane. (Hp) Axis perpendicular to Vp And // to Hp Axis inclined to Vp And // to Hp Axis inclined to Vp And // to Hp X Y F.V. F.V. F.V. T.V. T.V. T.V. While observing Tv, x-y line represents Vertical Plane. (Vp) STANDING ON V.P On it’s base. RESTING ON V.P On one point of base circle. LYING ON V.P On one generator.
  • 5. STEPS TO SOLVE PROBLEMS IN SOLIDS Problem is solved in three steps: STEP 1: ASSUME SOLID STANDING ON THE PLANE WITH WHICH IT IS MAKING INCLINATION. ( IF IT IS INCLINED TO HP, ASSUME IT STANDING ON HP) ( IF IT IS INCLINED TO VP, ASSUME IT STANDING ON VP) IF STANDING ON HP - IT’S TV WILL BE TRUE SHAPE OF IT’S BASE OR TOP: IF STANDING ON VP - IT’S FV WILL BE TRUE SHAPE OF IT’S BASE OR TOP. BEGIN WITH THIS VIEW: IT’S OTHER VIEW WILL BE A RECTANGLE ( IF SOLID IS CYLINDER OR ONE OF THE PRISMS): IT’S OTHER VIEW WILL BE A TRIANGLE ( IF SOLID IS CONE OR ONE OF THE PYRAMIDS): DRAW FV & TV OF THAT SOLID IN STANDING POSITION: STEP 2: CONSIDERING SOLID’S INCLINATION ( AXIS POSITION ) DRAW IT’S FV & TV. STEP 3: IN LAST STEP, CONSIDERING REMAINING INCLINATION, DRAW IT’S FINAL FV & TV. AXIS VERTICAL AXIS INCLINED HP AXIS VERTICAL AXIS INCLINED HP AXIS TO VP er AXIS INCLINED VP AXIS TO VP er AXIS INCLINED VP GENERAL PATTERN ( THREE STEPS ) OF SOLUTION: GROUP B SOLID. CONE GROUPA SOLID. CYLINDER GROUP B SOLID. CONE GROUPA SOLID. CYLINDER
  • 6. X a b c d o o’ d’ c’ b’ a’ o ’ d ’ c ’ b ’ a ’ o1 d1 b1 c1 a1 d’1 Problem 1. A square pyramid, 40 mm base sides and axis 60 mm long, has a triangular face on the ground and the vertical plane is parallel to axis. Draw its projections. Take apex nearer to VP Solution Steps : Triangular face on Hp , means it is lying on Hp: 1.Assume it standing on Hp. 2.It’s Tv will show True Shape of base( square) 3.Draw square of 40mm sides with one side vertical Tv & taking 50 mm axis project Fv. ( a triangle) 4.Name all points as shown in illustration. 6.Make visible lines dark and hidden dotted, as per the procedure. . For dark and dotted lines 1.Draw proper outline of new view DARK. 2. Decide direction of an observer. 3. Select nearest point to observer and draw all lines starting from it-dark. 4. Select farthest point to observer and draw all lines (remaining)from it- dotted.
  • 7. Problem 2: A cone 40 mm diameter and 50 mm axis is resting on one generator on Hp Parallel with Vp Draw it’s projections. h a b c d e g f X a’ b’ d’ e’ c’ g ’ f’ h’ o’ a ’ h ’ b ’ e ’ c ’ g ’ d ’ f ’ o’ a1 h1 g1 f1 e1 d1 c1 b1 o1 Solution Steps: Resting on Hp on one generator, means lying on Hp: 1.Assume it standing on Hp. 2.It’s Tv will show True Shape of base( circle ) 3.Draw 40mm dia. Circle as Tv & taking 50 mm axis project Fv. ( a triangle) 4.Name all points as shown in illustration. 5.Make visible lines dark and hidden dotted, as per the procedure. For dark and dotted lines 1.Draw proper outline of new vie DARK. 2. Decide direction of an observer. 3. Select nearest point to observer and draw all lines starting from it-dark. 4. Select farthest point to observer and draw all lines (remaining) from it- dotted.
  • 8. a b d c 1 2 4 3 X a b d c 1 2 4 3 a’ b’ c’ d’ 1’ 2’ 3’ 4’ 450 4’ 3’ 2’ 1’ d’ c’ b’ a’ 350 Problem 3: A cylinder 40 mm diameter and 50 mm axis is resting on one point of a base circle on Vp while it’s axis makes 450 with Vp and parallel to Hp. Draw projections.. Solution Steps: Resting on Vp on one point of base, means inclined to Vp: 1.Assume it standing on Vp 2.It’s Fv will show True Shape of base & top( circle ) 3.Draw 40mm dia. Circle as Fv & taking 50 mm axis project Tv. ( a Rectangle) 4.Name all points as shown in illustration.
  • 9. Problem 5: A cube of 50 mm long edges is so placed on Hp on one corner that a body diagonal is parallel to Hp and parallel to Vp Draw it’s projections. X b c d a a’ d’ c’ b’ a’ d’ c’ b’ a1 b1 d1 c1 p’ p’ Solution Steps: 1.Assuming standing on Hp, begin with Tv,a square with all sides equally inclined to xy.Project Fv and name all points of FV & TV. 2.Draw a body-diagonal joining c’ with 3’( This can become // to xy) 3.From 1’ drop a perpendicular on this and name it p’ 1’ 3’ 1’ 3’
  • 10. Y Problem 6:A tetrahedron of 50 mm long edges is resting on one edge on Hp while one triangular face containing this edge is vertical and parallel to Vp. Draw projections. X T L a o b c b’ a’ c’ o’ a’ a1 c1 o1 b1 900 b’ c’ o’ a’1 b’1 IMPORTANT: Tetrahedron is a special type of triangular pyramid in which base sides & slant edges are equal in length. Solid of four faces. Like cube it is also described by One dimension only.. Axis length generally not given. Solution Steps As it is resting assume it standing on Hp. Begin with Tv , an equilateral triangle as side case as shown: First project base points of Fv on xy, name those & axis line. From a’ with TL of edge, 50 mm, cut on axis line & mark o’ (as axis is not known, o’ is finalized by slant edge length) Then complete Fv. .
  • 11. FREELY SUSPENDED SOLIDS: Positions of CG, on axis, from base, for different solids are shown below. H H/2 H/4 GROUPA SOLIDS ( Cylinder & Prisms) GROUP B SOLIDS ( Cone & Pyramids) CG CG alphinms@ssn.edu.in
  • 12. X a’ d’ e’ c’ b’ o’ a b c d e o g’ H/4 H LINE d’g’ VERTICAL a’b’ c’ d’ e’ g’ a1 b1 o1 e1 d1 c1 e” FOR SIDE VIEW Problem 7: A pentagonal pyramid 30 mm base sides & 60 mm long axis, is freely suspended from one corner of base so that a plane containing it’s axis remains parallel to Vp. Draw it’s e views. IMPORTANT: When a solid is freely suspended from a corner, then line joining point of contact & C.G. remains vertical. ( Here axis shows inclination with Hp.) So in all such cases, assume solid standing on Hp initially.) Solution Steps: In all suspended cases axis shows inclination with Hp. 1.Hence assuming it standing on Hp, drew Tv - a regular pentagon,corner case. 2.Project Fv & locate CG position on axis – ( ¼ H from base.) and name g’ and Join it with corner d’ 3.As 2nd Fv, redraw first keeping line g’d’ vertical. 4.As usual project corresponding Tv and then Side View looking from. alphinms@ssn.edu.in
  • 13. a’ d’ c’ b’ b c d a a ’ d ’ c ’ b ’ a1 b d1 c1 X Y 1’ 1’ Problem 8: A cube of 50 mm long edges is so placed on Hp on one corner that a body diagonal through this corner is perpendicular to Hp and parallel to Vp Draw it’s three views. Solution Steps: 1.Assuming it standing on Hp begin with Tv, a square of corner case. 2.Project corresponding Fv.& name all points as usual in both views. 3.Join a’1’ as body diagonal and draw 2nd Fv making it vertical (I’ on xy) 4.Project it’s Tv drawing dark and dotted lines as per the procedure. 5.With standard method construct Left-hand side view. ( Draw a 450 inclined Line in Tv region ( below xy). Project horizontally all points of Tv on this line and reflect vertically upward, above xy.After this, draw horizontal lines, from all points of Fv, to meet these lines. Name points of intersections and join properly. For dark & dotted lines locate observer on left side of Fv as shown.) alphinms@ssn.edu.in
  • 14. Axis Tv Length h a b c d e g f y X a’ b’ d’ e’ c’ g’ f’ h’ o’ a ’ h ’ b ’ e ’ c ’ g ’ d ’ f ’ o ’ 450 a1 h1 f1 e1 d1 c1 b1 g1 o1 1 Problem 9: A right circular cone, 40 mm base diameter and 60 mm long axis is resting on Hp on one point of base circle such that it’s axis makes 450 inclination with Hp and 400 inclination with Vp. Draw it’s projections. This case resembles to problem no.7 & 9 from projections of planes topic. In previous all cases 2nd inclination was done by a parameter not showing TL.Like Tv of axis is inclined to Vp etc. But here it is clearly said that the axis is 400 inclined to Vp. Means here TL inclination is expected. So assuming it standing on HP begin as usual. alphinms@ssn.edu.in