engg curves
DESCRIPTION
Ellipsce, parabola, hyperbola, cycloid, hypocycloid, epicycloid,rectangular method,spiral,helixTRANSCRIPT
Engineering drawing is a language of all persons involved in engineering activity. Engineering ideas are recorded by preparing drawings and execution of work is also carried out on the basis of drawings. Communication in engineering field is done by drawings. It is called as a “Language of Engineers”.
CHAPTER – 2
ENGINEERING
CURVES
Useful by their nature & characteristics.
Laws of nature represented on graph.
Useful in engineering in
understanding laws,
manufacturing of various items,
designing mechanisms analysis of
forces, construction of bridges,
dams, water tanks etc.
USES OF ENGINEERING CURVES
1. CONICS
2. CYCLOIDAL CURVES3. INVOLUTE
4. SPIRAL
5. HELIX
6. SINE & COSINE
CLASSIFICATION OF ENGG. CURVES
It is a surface generated by moving a Straight line keeping one of its end fixed & other end makes a closed curve.
What is Cone ?
If the base/closed curve is a polygon, we get a pyramid.
If the base/closed curve is a circle, we get a cone.
The closed curve is known as base.
The fixed point is known as vertex or apex. Vertex/Apex
90º
Base
If axis of cone is not perpendicular to base, it is called as oblique cone.
The line joins vertex/ apex to the circumference of a cone is known as generator.
If axes is perpendicular to base, it is called as right circular cone.
Generato
r
Cone Axis
The line joins apex to the center of base is called axis.
90º
Base
Vertex/Apex
Definition :- The section obtained by the intersection of a right circular cone by a cutting plane in different position relative to the axis of the cone are called CONICS.
CONICS
B - CIRCLE
A - TRIANGLE
CONICS
C - ELLIPSE
D – PARABOLA
E - HYPERBOLA
When the cutting plane contains the apex, we get a triangle as the section.
TRIANGLE
When the cutting plane is perpendicular to the axis or parallel to the base in a right cone we get circle the section.
CIRCLE
Sec Plane
Circle
Definition :-When the cutting plane is inclined to the axis but not parallel to generator or the inclination of the cutting plane(α) is greater than the semi cone angle(θ), we get an ellipse as the section.
ELLIPSE
α
θ
α > θ
When the cutting plane is inclined to the axis and parallel to one of the generators of the cone or the inclination of the plane(α) is equal to semi cone angle(θ), we get a parabola as the section.
PARABOLA
θ
α
α = θ
When the cutting plane is parallel to the axis or the inclination of the plane with cone axis(α) is less than semi cone angle(θ), we get a hyperbola as the section.
HYPERBOLADefinition :-
α < θα = 0
θθ
CONICSDefinition :- The locus of point moves in a plane such a way that the ratio of its distance from fixed point (focus) to a fixed Straight line (Directrix) is always constant.
Fixed point is called as focus.Fixed straight line is called as directrix.
M
C FV
P
Focu
s
Conic
CurveDirectrix
The line passing through focus & perpendicular to directrix is called as axis.The intersection of conic curve with
axis is called as vertex.
AxisM
C FV
P
Focus
Conic
CurveDirectrix
Vertex
N Q
Ratio =Distance of a point from
focusDistance of a point from
directrix= Eccentricity= PF/PM = QF/QN = VF/VC = e
M P
F
Axis
CV
Focus
Conic
CurveDirectrix
Vertex
Vertex
Ellipse is the locus of a point which moves in a plane so that the ratio of its distance from a fixed point (focus) and a fixed straight line (Directrix) is a constant and less than one.
ELLIPSE
M
N Q
P
CF
V
Axi
s
Focu
s
Ellips
eDirectri
x
Eccentricity=PF/PM = QF/QN
< 1.
Ellipse is the locus of a point, which moves in a plane so that the sum of its distance from two fixed points, called focal points or foci, is a constant. The sum of distances is equal to the major axis of the ellipse.
ELLIPSE
F1
A B
P
F2
O
Q
C
D
F1
A B
C
D
P
F2
O
PF1 + PF2 = QF1 + QF2 = CF1 +CF2 =
constant
= Major Axis
Q
= F1A + F1B = F2A + F2B
But F1A = F2B
F1A + F1B = F2B + F1B = AB
CF1 +CF2 = AB
but CF1 = CF2
hence,
CF1=1/2AB
F1 F2
OA B
C
D
Major Axis = 100 mm
Minor Axis = 60 mm
CF1 = ½ AB = AO
F1 F2
OA B
C
D
Major Axis = 100 mm
F1F2 = 60 mm
CF1 = ½ AB = AO
Uses :-
Shape of a man-hole.
Flanges of pipes, glands and stuffing
boxes.
Shape of tank in a
tanker.
Shape used in bridges and arches.
Monuments.
Path of earth around the sun.
Shape of trays etc.
Ratio (known as eccentricity) of its distances from focus to that of directrix is constant and equal to one (1).
PARABOLAThe parabola is the locus of a point,
which moves in a plane so that its distance from a fixed point (focus) and a fixed straight line (directrix) are always equal.
Definition :-
Directrix AxisVertex
M
C
N Q
FV
P
Focus
Parabol
a
Eccentricity = PF/PM = QF/QN = 1.
Motor car head lamp
reflector.Sound reflector and
detector.
Shape of cooling towers.
Path of particle thrown at any angle with earth, etc.
Uses :-
Bridges and arches construction
Home
It is the locus of a point which moves in a plane so that the ratio of its distances from a fixed point (focus) and a fixed straight line (directrix) is constant and grater than one.
Eccentricity = PF/PM
AxisDirectrix
Hyperbol
aM
C
N Q
FV
P
Focu
s
Verte
x
HYPERBOLA
= QF/QN
> 1.
Nature of graph of Boyle’s
law
Shape of overhead water
tanks
Uses :-
Shape of cooling towers
etc.
METHODS FOR DRAWING ELLIPSE
2. Concentric Circle Method
3. Loop Method
4. Oblong Method
5. Ellipse in Parallelogram
6. Trammel Method
7. Parallel Ellipse
8. Directrix Focus Method
1. Arc of Circle’s Method
Nor
mal
P2’
R =
A1
Tangent
1 2 3 4A B
C
D
P1
P3
P2
P4 P4 P3
P2
P1
P1’
F2
P3’ P4’ P4’P3’
P2’
P1’
90°
F1
Rad =B1
R=B2
`R=
A2
O
ARC OF CIRCLE’SARC OF CIRCLE’S METHODMETHOD
Axi
sM
inor
A BMajor Axis 7
8
910
11
9
8
7
6
543
2
1
12
11
P6
P5P4
P3
P2`
P1
P12
P11
P10P9
P8
P7
6
54
3
2
1
12 C10
O
CONCENTRIC CONCENTRIC CIRCLE CIRCLE
METHODMETHOD
F2F1
D
CF1=CF2=1/2 AB
T
N
Q
e = AF1/AQ
Normal
Normal
00
11
22
33
44
11 22 33 44 1’1’0’0’
2’2’3’3’4’4’
1’1’
2’2’
3’3’
4’4’
AA BB
CC
DD
Major AxisMajor Axis
Min
or A
xis
Min
or A
xis
FF11 FF22
Dir
ectr
ixD
irec
trix
EE
FF
SS
PP
PP11
PP22
PP33
PP44
Tan
gent
Tan
gent
PP11’’
PP22’’
PP33’’PP44’’
ØØ
R=AB/2
R=AB/2
PP00
P1’’
P2’’
P3’’P4’’P4
P3
P2
P1
OBLONG METHODOBLONG METHOD
BA
P4
P0
D
C
60°
6
5432
10
5 4 3 2 1 0 1 2 3 4 5 65
3210P1P2
P3
Q1
Q2Q3Q4
Q5
P6 Q6O
4
ELLIPSE IN PARALLELOGRAM
R4
R3 R2
R1S1
S2
S3
S4
P5
G
H
I
K
JM
inor Axis
Major Axis
P6
Nor
mal
P5’ P7’P6’
P1
TangentP1’
N
N
T
T
V1
P5
P4’
P4
P3’P2’
F1
D1
D1
R1
ba
cd
ef
g
Q
P7P3P2
Dir
ectr
ix
R=6f`
90°
1 2 3 4 5 6 7
Eccentricity = 2/3
3R1V1
QV1 = R1V1
V1F1 = 2
Ellipse
ELLIPSE – DIRECTRIX FOCUS METHOD
R=1a
R=1a
Dist. Between directrix & focus = 50 mm
1 part = 50/(2+3)=10 mm V1F1 = 2 part = 20 mmV1R1 = 3 part = 30 mm
< 45º
S
PROBLEM :-
The distance between two coplanar fixed points is 100 mm. Trace the complete path of a point G moving in the same plane in such a way that the sum of the distance from the fixed points is always 140 mm.
Name the curve & find its eccentricity.
ARC OF CIRCLE’S ARC OF CIRCLE’S METHODMETHOD
Nor
mal
G2’
R =
A1
Tangent
1 2 3 4A B
G
G’
G1
G3
G2
G4 G4 G3
G2
G1
G1’
G3’ G4’ G4’G3’
G2’
G1’
F2F1
R=B1
R=B2
`R=
A2
O
90°
90°
dir
ectr
ixd
irec
trix
100100
140140
GF1 + GF2 = MAJOR AXIS = 140GF1 + GF2 = MAJOR AXIS = 140
EE
ee AF1AF1
AEAEe e = =
R=70R=70 R=70
R=70
PROBLEM :-3
Two points A & B are 100 mm apart. A point C is 75 mm from A and 45 mm from B. Draw an ellipse passing through points A, B, and C so that AB is a major axis.
OA B
C
7545
1
D
100
1
2 2
3
3
4
4
5
5
66
7
7P1
P2
P3
P4
P5
P6
P7
P8
E
8
8
PROBLEM :-5
ABCD is a rectangle of 100mm x 60mm. Draw an ellipse passing through all the four corners A, B, C and D of the rectangle considering mid – points of the smaller sides as focal points.
Use “Concentric circles” method and find its eccentricity.
I3
CD
F1 F2P Q
R
S
5050I1 I4
A BI2
O
1
1
2
2
4
4
3
3
100100
PROBLEM :-1
Three points A, B & P while lying along a horizontal line in order have AB = 60 mm and AP = 80 mm, while A & B are fixed points and P starts moving such a way that AP + BP remains always constant and when they form isosceles triangle, AP = BP = 50 mm. Draw the path traced out by the point P from the commencement of its motion back to its initial position and name the path of P.
A B P
R =
50
M
N
O
1
2
1 2 6080
Q
1
2
12
P1
P2 Q2
Q1
R1
R2 S2
S1
PROBLEM :-2Draw an ellipse passing through 60º corner Q of a 30º - 60º set square having smallest side PQ vertical & 40 mm long while the foci of the ellipse coincide with corners P & R of the set square.
Use “OBLONG METHOD”. Find its eccentricity.
ELLIPSEELLIPSEPP
R80mm
40m
m
89mm
AA
CC
BB
DD
MAJOR AXIS = PQ+QR = 129mmMAJOR AXIS = PQ+QR = 129mm
R=A
B/2
11
22
33
1’1’ 2’2’ 3’3’ 1’’1’’2’’2’’3’’3’’
OO33 OO33’’
OO11
OO22 OO22’’
OO11’’
TANGENT
TANGENTNO
RM
AL
NO
RM
AL
6060ºº
30º30º
22
33
11
dir
ectr
ixd
irec
trix
FF11 FF22MAJOR AXISMAJOR AXIS
MIN
OR
AX
ISM
INO
R A
XIS
SS
ECCENTRICITY = AP / ASECCENTRICITY = AP / AS
?? ??
PROBLEM :-4
Two points A & B are 100 mm apart. A point C is 75 mm from A and 45 mm from B. Draw an ellipse passing through points A, B, and C so that AB is not a major axis.
D
C
665
43
21
0
5 4 3 2 1 0 1 2 3 4 5 66
5
32
10
P2
P3
P4
P5
Q1
Q2Q3
Q4Q5
BA O
4
ELLIPSE
100
4575
P0
P1
P6Q6
G
H
I
K
J
PROBLEM :-
Draw an ellipse passing through A & B of an equilateral triangle of ABC of 50 mm edges with side AB as vertical and the corner C coincides with the focus of an ellipse. Assume eccentricity of the curve as 2/3. Draw tangent & normal at point A.
PROBLEM :-
Draw an ellipse passing through all the four corners A, B, C & D of a rhombus having diagonals AC=110mm and BD=70mm.
Use “Arcs of circles” Method and find its eccentricity.
METHODS FOR DRAWING PARABOLA
1. Rectangle Method
2. Parabola in Parallelogram
3. Tangent Method
4. Directrix Focus Method
22334455
00
11
22
33
44
55
66 11 11 55443322 66
00
11
22
33
44
55
00
VVDD CC
AA BB
PP44 PP44
PP55 PP55
PP33 PP33
PP22 PP22
PP66 PP66
PP11 PP11
PARABOLA –RECTANGLE METHODPARABOLA –RECTANGLE METHOD
PARABOLA PARABOLA
BB
00
2’2’
00
22
66
CC
6’6’
VV
55
P’P’55
3030°°
AA XX
DD
1’1’ 2’2’4’4’
5’5’
3’3’
11
3344
5’5’
4’4’
3’3’
1’1’
00
55
44
33
22
11
PP11
PP22
PP33
PP44
PP55
P’P’44
P’P’33
P’P’22
P’P’11
P’P’66
PARABOLA – IN PARALLELOGRAMPARABOLA – IN PARALLELOGRAM
PP66
BA O
V
1
8
3
4
5
2
6
7
9
10
0
1
2
3
4
5
6
7
8
9
10
0
F
PARABOLA
TANGENT METHOD
D
D
DIR
EC
TR
IX
90° 2 3 4T
TN
N
S
V 1
P1
P2
PF
P3
P4
P1’
P2’P3’
P4’
PF’
AXIS
RF
R2
R1
R3
R4
90°
R F
PARABOLA
DIRECTRIX FOCUS METHOD
PROBLEM:-A stone is thrown from a building 6 m high. It just crosses the top of a palm tree 12 m high. Trace the path of the projectile if the horizontal distance between the building and the palm tree is 3 m. Also find the distance of the point from the building where the stone falls on the ground.
6m
6m
ROOT OF TREEROOT OF TREE
BUILDINGBUILDING
REQD.DISTANCEREQD.DISTANCE
TOP OF TREETOP OF TREE
3m3m
6m
6m
FF
AA
STONE FALLS HERESTONE FALLS HERE
3m3m
6m
6m
ROOT OF TREEROOT OF TREE
BUILDINGBUILDING
REQD.DISTANCEREQD.DISTANCE
GROUNDGROUND
TOP OF TREETOP OF TREE
3m3m
6m
6m
11
22
33
11
22
33
332211 44 55 66
55
66
EEFF
AA BB
CCDD
PP33
PP44
PP22
PP11
PP
PP11
PP22
PP33
PP44
PP55
PP66
33 22 1100
STONE FALLS HERESTONE FALLS HERE
PROBLEM:-
In a rectangle of sides 150 mm and 90 mm, inscribe two parabola such that their axis bisect each other. Find out their focus points & positions of directrix.
150 mmAA
BB CC
DD11 22 33 44 55
11
22
33
44
55
OO
PP11
PP22
PP33
PP44
PP55
MM1’1’ 2’2’ 3’3’ 4’4’ 5’5’
1’1’
2’2’
3’3’
4’4’
5’5’
PP11’’
PP22’’
PP33’’ PP
44’’
PP55’’
90 m
m
EXAMPLEEXAMPLE
A shot is discharge from the ground A shot is discharge from the ground level at an angle 60 to the horizontal level at an angle 60 to the horizontal at a point 80m away from the point of at a point 80m away from the point of discharge. Draw the path trace by the discharge. Draw the path trace by the shot. Use a scale 1:100shot. Use a scale 1:100
ground levelground level BA
60ºgun gun shotshot
80 M
parabolaparabola
ground levelground level BA O
V
1
8
3
4
5
2
6
7
9
10
0
1
2
3
4
5
6
7
8
9
10
0
F
60ºgun gun shotshot
DD DD
VFVF
VEVE== e = 1e = 1
EE
Connect two given points A and B by a Connect two given points A and B by a Parabolic curve, when:-Parabolic curve, when:-
1.OA=OB=60mm and angle AOB=90°1.OA=OB=60mm and angle AOB=90°
2.OA=60mm,OB=80mm and angle 2.OA=60mm,OB=80mm and angle AOB=110°AOB=110°
3.OA=OB=60mm and angle AOB=60°3.OA=OB=60mm and angle AOB=60°
6600
66 00
1 2 3 4 5
Parabola
5
4
3
2
1
AA
BB90 90 °°
OO
1.OA=OB=60mm and angle 1.OA=OB=60mm and angle AOB=90AOB=90°°
AA
BB
8800
6600
5
4
3
2
1
1 2 3 4 5
110 °°
Parabola
OO
2.OA=60mm,OB=80mm and angle 2.OA=60mm,OB=80mm and angle AOB=110°AOB=110°
54321
5
4
3
2
1
AA
BBOO
Parabola
6600
6060
60 60 °°
3.OA=OB=60mm 3.OA=OB=60mm and angle and angle AOB=60°AOB=60°
exampleexample
Draw a parabola passing through three Draw a parabola passing through three
different points A, B and C such that AB = different points A, B and C such that AB =
100mm, BC=50mm and CA=80mm 100mm, BC=50mm and CA=80mm
respectively. respectively.
BBAA
CC
100100
5050
8080
1’1’
00
00
2’2’
00
2266
6’6’55
P’P’55
1’1’ 2’2’ 4’4’ 5’5’3’3’113344
5’5’
4’4’
3’3’
5544
33
2211
PP11PP
22
PP33
PP44
PP55
P’P’44
P’P’33
P’P’22
P’P’11
P’P’66
PP66
AA BB
CC
METHODS FOR DRAWING HYPERBOLA
1. Rectangle Method
2. Oblique Method
3. Directrix Focus Method
D
F
1 2 3 4 5
5’4’3’
2’
P1
P2
P3P4
P5
0
P6
P0
AO EX
B
C
Y
Given Point P0
90°
6
6’
Hyperbola
RECTANGULAR HYPERBOLA
AXIS
AXIS
When the asymptotes are at right angles to each other, the hyperbola is called rectangular or equilateral hyperbola
ASYMPTOTES X and Y
Problem:-
Two fixed straight lines OA and OB are at right angle to each other. A point “P” is at a distance of 20 mm from OA and 50 mm from OB. Draw a rectangular hyperbola passing through point “P”.
D
F
1 2 3 4 5
5’4’3’
2’
P1
P2
P3P4
P5
0
P6
P0
AO EX=20
B
C
Y =
50
Given Point P0
90°
6
6’
Hyperbola
RECTANGULAR HYPERBOLA
PROBLEM:-
Two straight lines OA and OB are at 75° to each other. A point P is at a distance of 20 mm from OA and 30 mm from OB. Draw a hyperbola passing through the point “P”.
75 0
P4
E
6’
2’1’ P1
1 2 3 4 5 6 D
P6P5
P3P2
P0
7’P7
7C
B F
O
Y =
30
X = 20
Given Point P0
A
AXISAXIS
NORMAL
NORMAL
CC VV FF11
DIR
EC
TR
IXD
IRE
CT
RIX
DDDD
11 22 33 44
4’4’
3’3’
2’2’
1’1’PP11
PP22
PP33
PP44
PP11’’
PP22’’
PP33’’
PP44’’
TT11
TT22
NN22
NN11
TAN
GE
NT
TAN
GE
NT
ss
Directrix and focus methodDirectrix and focus method
CYCLOIDAL GROUP OF CURVESWhen one curve rolls over another curve without slipping or sliding, the path Of any point of the rolling curve is called as ROULETTE.When rolling curve is a circle and the curve on which it rolls is a straight line Or a circle, we get CYCLOIDAL GROUP OF CURVES.
SuperiorHypotrochoid
Cycloidal Curves
Cycloid Epy Cycloid Hypo Cycloid
SuperiorTrochoid
Inferior
Trochoid
SuperiorEpytrochoid
InferiorEpytrochoid
InferiorHypotrochoid
Rolling Circle or Generator
CYCLOID:-
Cycloid is a locus of a point on the circumference of a rolling circle(generator), which rolls without slipping or sliding along a fixed straight line or a directing line or a director.
C
P P
P
R
C
Directing Line or Director
EPICYCLOID:-Epicycloid is a locus of a point(P) on the circumference of a rolling circle(generator), which rolls without slipping or sliding OUTSIDE another circle called Directing Circle.
2πr
Ø = 360Ø = 360ºº x r/R x r/Rdd
Circumference of Generating Circle
RR dd
Rolling Rolling CircleCircle
r
O
Ø/Ø/22
Ø/Ø/22
P0 P0
Arc P0P0 =
Rd x Ø =
P0
HYPOCYCLOID:-Hypocycloid is a locus of a point(P) on the circumference of a rolling circle(generator), which rolls without slipping or sliding INSIDE another circle called Directing Circle.`
DirectingCircle(R)
P
ØØ /2 ØØ /2
ØØ =360 x r
RR
T
Rolling CircleRadius (r)
O
Vertical
Hypocycloid
P P
If the point is inside the circumference of the
circle, it is called inferior trochoid.
If the point is outside the circumference of the
circle, it is called superior trochoid.
What is TROCHOID ? DEFINITION :- It is a locus of a point inside/outside the circumference of a rolling circle, which rolls without slipping or sliding along a fixed straight line or a fixed circle.
P0
2R or D 5
T
T
1
2
1 2 3 4 6 7 8 9 10 110 120
3
4
567
8
9
1011 12
P1
P2
P3
P4
P5 P7
P8
P9
P11
P12
C0 C1 C2 C3 C4 C5 C6 C7 C8 C9 C10C11
Directing Line
C12
N
N
S
S1
R
P6
R
P10
R
: Given Data :
Draw cycloid for one revolution of a rolling circle having diameter as 60mm.
Rolling Circle
D
CC00
PP00
7788
PP
6644
PP11
11
22
33
CC22 CC33
PP22 CC44
Problem 1:A circle of diameter D rolls without
slip on a horizontal surface (floor) by Half revolution and then it rolls up a vertical surface (wall) by another half revolution. Initially the point P is at the Bottom of circle touching the floor. Draw the path of the point P.
55
66 CC11
PP33
PP44
PP
55
PP
77
PP
88
7700
CC55
CC66
CC77
CC88
11 22 33 44
5566
DD/2/2
ππDD
/2/2
ππDD/2/2 DD/2/2FloorFloor
Wal
lW
all
CYCLOIDCYCLOID
5566
77
88
Take diameter of circle = 40mmInitially distance of centre of
circle from the wall 83mm (Hale circumference + D/2)
Problem : 2
A circle of 25 mm radius rolls on the
circumference of another circle of 150
mm diameter and outside it. Draw the
locus of the point P on the
circumference of the rolling circle for
one complete revolution of it. Name
the curve & draw tangent and normal
to the curve at a point 115 mm from
the centre of the bigger circle.
First Step : Find out the included angle by using the equation
360º x r / R = 360 x 25/75 = 120º.
Second step: Draw a vertical line & draw two lines at 60º on either sides.
Third step : at a distance of 75 mm from O, draw a part of the circle taking radius = 75 mm.
Fourth step : From the circle, mark point C outside the circle at distance of 25 mm & draw a circle taking the centre as point C.
PP66
PP44
rPP22
CC11
CC00
CC22
CC33CC44 CC55
CC66
CC77
CC88
1
0
23
4
5
6 7
O
RR dd
Ø/2Ø/2 Ø/2Ø/2
PP11PP00
PP33 PP55
PP77PP88
r rRolling Rolling CircleCircle
r
Rd X Ø = 2πrØ = 360Ø = 360ºº x x r/Rr/Rdd
Arc P0P8 = Circumference of Generating Circle
EPICYCLOIDEPICYCLOIDGIVEN:Rad. Of Gen. Circle (r)& Rad. Of dir. Circle (Rd) S
ºU
N
Ø = 360Ø = 360ºº x 25/75 x 25/75
= 120°= 120°
Problem :3
A circle of 80 mm diameter rolls on
the circumference of another circle of
120 mm radius and inside it. Draw the
locus of the point P on the
circumference of the rolling circle for
one complete revolution of it. Name
the curve & draw tangent and normal
to the curve at a point 100 mm from
the centre of the bigger circle.
P0 P1
Tangent
P11
r
C0
C1
C2
C3
C4C5
C6 C7 C8 C9
C10
C11
C12
P10P8
0
1 2 3
4
5
67
89
10
11
12P2 P3
P4P5 P6
P9P7
P12
/2
/2
= 360 x 412
= 360 x rR
= 120°
R
T
T
N
S
N
Norm
alr
r
Rolling Circle
Radias (r)
DirectingCircle
O
Vertical
Hypocycloid
Problem :
Show by means of drawing that when the diameter of rolling circle is half the diameter of directing circle, the hypocycloid is a straight line
C
C1
C2
C3
C4C5
C6 C7
C9
C8
C10
C11
C12P8O
10
5
7
89
11
12
1
2 34
6P1
P2
P3 P4P5 P6 P7 P9
P10
P11
P12
Directing Circle
Rolling Circle
HYPOCYCLOID
INVOLUTE
DEFINITION :- If a straight line is rolled
round a circle or a polygon without
slipping or sliding, points on line will
trace out INVOLUTES.OR
Uses :- Gears profile
Involute of a circle is a curve traced out
by a point on a tights string unwound or
wound from or on the surface of the
circle.
PROB:
A string is unwound from a circle of 20 mm diameter. Draw the locus of string P for unwounding the string’s one turn. String is kept tight during unwound. Draw tangent & normal to the curve at any point.
P12
P2
002 12
6
P11 20 9 103 4 6 8 115 7 12
DP3
P4
P5
P6
P7
P8
P9
P10
P11
123
45 7
8910
11
03
04
05
06
07
08
09
010`
011
Tangent
N
N
Norm
alT
T.
PROBLEM:-
Trace the path of end point of a thread when it is wound round a circle, the length of which is less than the circumference of the circle.
Say Radius of a circle = 21 mm & Length of the thread = 100 mm
Circumference of the circle = 2 π r = 2 x π x 21 = 132 mm
So, the length of the string is less than circumference of the circle.
PP
R=7toPR=7toPR=6toPR=6toP
R21
R21
00
00 11 22 33 44 55 66 77 88 PP1111 00
11
22
33
445566
77
88
99
1010
PP11
PP22
PP33
PP44
PP55
PP66
PP77
PP88
L= 100 mmL= 100 mm
R=1toPR=1toP
R=2
toP
R=2
toPR
=3
toP
R=
3to
PR=4toP
R=4toPR=5toP
R=5toP
INVOLUTEINVOLUTE
99
ø
11 mm = 30°
Then 5 mm = Ø = 30° x 5 /11 = 13.64 °
S = 2 x π x r /12
PROBLEM:-
Trace the path of end point of a thread when it is wound round a circle, the length of which is more than the circumference of the circle.
Say Radius of a circle = 21 mm & Length of the thread = 160 mm
Circumference of the circle = 2 π r = 2 x π x 21 = 132 mm
So, the length of the string is more than circumference of the circle.
PP1313
PP1111
313
14
15
PP00
PP1212
O
7
10 123
45689
1112 1 2
PP11
PP22
PP33PP44
PP55
PP66
PP77
PP88 PP99 PP1010
PP1414PP
L=160 L=160
mmmm
R=21mm
64 5 7 8 9 101112131415
øø
PROBLEM:-
Draw an involute of a pantagon having side as 20 mm.
P5
R=01
R=201
P0
P1
P2
P3
P4
R=3
01
R=401
R=501
23
4 51
T
TN
NS
INVOLUTEOF A POLYGON
Given : Side of a polygon 0
PROBLEM:-
Draw an involute of a square
having side as 20 mm.
P2
1
2 3
04
P0
P1
P3
P4
N
NS
R=3
01
R=401
R=201
R=0
1
INVOLUTE OF A SQUARE
PROBLEM:-
Draw an involute of a string unwound from the given figure from point C in anticlockwise direction.
60°60°
AA
BB
CC
R21R2130°30°
R2R2
11
60°60°
AA
BB
CC
30°30°XX
X+AX+A
11
XXX+
A2
X+
A2
X+
AX
+A
33
X+A5X+A5
X+
A4
X+
A4
X+AX+ABB
R R
=X+AB=X+AB
X+
66+
BX
+66+
BCC
1122
33
4455
C0
C1
C2C3
C4
C5
C6
C7
C8
A stick of length equal to the circumference of a A stick of length equal to the circumference of a
semicircle, is initially tangent to the semicircle semicircle, is initially tangent to the semicircle
on the right of it. This stick now rolls over the on the right of it. This stick now rolls over the
circumference of a semicircle without sliding till circumference of a semicircle without sliding till
it becomes tangent on the left side of the it becomes tangent on the left side of the
semicircle. Draw the loci of two end point of this semicircle. Draw the loci of two end point of this
stick. Name the curve. Take R= 42mm. stick. Name the curve. Take R= 42mm.
PROBLEM:-
A6
B6
5
A
B
C
B1
A1
B2
A2
B3A3
B4
A4
B5
A5
12 3
4
5
O
1
2
3
4
6
INVOLUTE
SPIRALS
If a line rotates in a plane about one of its
ends and if at the same time, a point moves
along the line continuously in one
direction, the curves traced out by the
moving point is called a SPIRAL.
The point about which the line rotates is called a POLE.
The line joining any point on the curve with the pole is called the RADIUS VECTOR.
The angle between the radius vector and the line in its initial position is called the VECTORIAL ANGLE.
Each complete revolution of the curve is termed as CONVOLUTION.
Spiral
Arche Median Spiral for Clock
Semicircle Quarter Circle Logarithmic
ARCHEMEDIAN SPIRAL
It is a curve traced out by a point moving in such a way that its movement towards or away from the pole is uniform with the increase of vectorial angle from the starting line.
USES :-
Teeth profile of Helical gears.
Profiles of cams etc.
To construct an Archemedian Spiral of one convolutions, given the radial movement of the point P during one convolution as 60 mm and the initial position of P is the farthest point on the line or free end of the line.
Greatest radius = 60 mm &
Shortest radius = 00 mm ( at centre or at pole)
PROBLEM:
PP1010
1
23
4
5
6
7
89
10
11
120
8 7 012345691112
PP11
PP22PP33
PP44
PP55
PP66
PP77PP88
PP99
PP1111PP1212
o
To construct an Archemedian Spiral of one convolutions, given the greatest & shortest(least) radii.
Say Greatest radius = 100 mm &
Shortest radius = 60 mm
To construct an Archemedian Spiral of one convolutions, given the largest radius vector & smallest radius vector.
OR
3 1
26
5
8 4
79
1011
2
1
34
5
6
7
89
10
11
12
P1
P2P3
P4
P5
P6
P7
P8 P9
P10
P11
P12
O
N
N
T
T
S
R m
in
R max
Diff. in length of any two radius vectors
Angle between them in radiansConstant of the curve =
=OP – OP3
Π/2
100 – 90=
Π/2
= 6.37 mm
PROBLEM:-
A slotted link, shown in fig rotates in the horizontal plane about a fixed point O, while a block is free to slide in the slot. If the center point P, of the block moves from A to B during one revolution of the link, draw the locus of point P. OAB
40 25
BB AA OO1234567891011
P1
P2
P3 P4
P5
P6
P7
P8
P9P10
P11
P12
11
21
31
41
51
61
71
81
91
101
111
40 25
PROBLEM:-
A link OA, 100 mm long rotates about O in clockwise direction. A point P on the link, initially at A, moves and reaches the other end O, while the link has rotated thorough 2/3 rd of the revolution. Assuming the movement of the link and the point to be uniform, trace the path of the point P.
AAInitial Position of point PInitial Position of point PPPOO
PP11
PP22
PP33
PP44PP55
PP66
PP77
PP88
2211
3344556677OO
11
22
33
44
5566
77
88
2/3 X 360°
= 240°
120120ºº
AA00
Linear Travel of point PP on ABAB
= 96 =16x (6 div.)
EXAMPLE: A link ABAB, 96mm long initially is vertically upward w.r.t. its pinned end BB, swings in clockwise direction for 180° and returns back in anticlockwise direction for 90°, during which a point PP, slides from pole BB to end AA. Draw the locus of point PP and name it. Draw tangent and normal at any point on the path of PP.
PP11’’
AA
BB
AA11
AA22
AA33
AA44
AA55
AA66PP00
PP11
PP22
PP33
PP44
PP55
PP66
PP22’’
PP33’’
PP44’’PP55’’
PP66’’
99 66
Link Link ABAB = 96 = 96
CC
TangentTangent
AngularAngular SwingSwing ofof linklink AB = AB = 180° + 90°180° + 90°
= = 270270 ° ° =45 °X 6 div. =45 °X 6 div.
ARCHIMEDIAN ARCHIMEDIAN SPIRAL SPIRAL
DD
NO
RM
AL
NO
RM
AL
MM
NN
Arch.Spiral Curve Constant BCBC
= Linear Travel ÷Angular Swing in Radians
= 96 ÷ (270º×π /180º)
=20.363636 mm / radian
PROBLEM :
A monkey at 20 m slides down from a rope. It swings 30° either sides of rope initially at vertical position. The monkey initially at top reaches at bottom, when the rope swings about two complete oscillations. Draw the path of the monkey sliding down assuming motion of the monkey and the rope as uniform.
θ
o
012
3
4 5 67
8
9
1011
121314
15
16 17 18 1920
21
222324
23
13
22
24
12
34
56
78
910
1112
1415
16
1819
2021
17
P3
P9
P15
Problem : 2
Draw a cycloid for a rolling circle, 60
mm diameter rolling along a straight
line without slipping for 540°
revolution. Take initial position of the
tracing point at the highest point on
the rolling circle. Draw tangent &
normal to the curve at a point 35 mm
above the directing line.
First Step : Draw a circle having diameter of 60 mm.
Second step: Draw a straight line tangential to the circle from bottom horizontally equal to
(540 x ) x 60 mm= 282.6 mm i.e. 1.5 x x 60 mm 360
Third step : take the point P at the top of the circle.
Rolling circleRolling circle
PP11
PP22
PP33
PP44
PP00
PP66
PP77
PP88
PP55
PP99
PP1010
11
22
3344
55
66
77 8899
1010
11 22 33 44 55 66 77 88 99 101000
CC00 CC11 CC22CC33 CC44
Directing lineDirecting line
Length of directing line = 3Length of directing line = 3
540540 = 360 = 360 + 180 + 180
540540 = = D + D + D/2 D/2 Total length for 540Total length for 540 rotation = 3 rotation = 3D/2D/2
CC55 CC66 CC77CC88 CC99 CC1010
SS
norm
al