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
USES OF ENGINEERING CURVES
1. CONICS
2. CYCLOIDAL CURVES
3. INVOLUTE
4. SPIRAL
5. HELIX
6. SINE & COSINE
CLASSIFICATION OF ENGG. CURVES
What is Cone ?
Vertex/Apex
90º
Base
Generator
Cone Axis
90º
Base
Vertex/Apex
CONICS
B - CIRCLE
A - TRIANGLE
CONICS
C - ELLIPSE
D – PARABOLA
E - HYPERBOLA
TRIANGLE
CIRCLE
Sec Plane
Circle
Definition :-
ELLIPSE
α
θ
α > θ
PARABOLA
θ
α
α = θ
HYPERBOLA
Definition :-
α < θ
α = 0
θ
θ
CONICS
M
C
F
V
P
Focus
Conic Curve
Directrix
Axis
M
C
F
V
P
Focus
Conic Curve
Directrix
Vertex
N
Q
Ratio =
Distance of a point from focus
Distance of a point from directrix
= Eccentricity
= PF/PM = QF/QN = VF/VC = e
M
P
F
Axis
C
V
Focus
Conic Curve
Directrix
Vertex
Vertex
ELLIPSE
M
N
Q
P
C
F
V
Axis
Focus
Ellipse
Directrix
Eccentricity=PF/PM = QF/QN
< 1.
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
O
A
B
C
D
Major Axis = 100 mm
Minor Axis = 60 mm
CF1 = ½ AB = AO
F1
F2
O
A
B
C
D
Major Axis = 100 mm
F1F2 = 60 mm
CF1 = ½ AB = AO
Uses :-
PARABOLA
Definition :-
Directrix
Axis
Vertex
M
C
N
Q
F
V
P
Focus
Parabola
Eccentricity = PF/PM = QF/QN
= 1.
Uses :-
Home
Eccentricity = PF/PM
Axis
Directrix
Hyperbola
M
C
N
Q
F
V
P
Focus
Vertex
HYPERBOLA
= QF/QN
> 1.
Uses :-
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
Normal
P2’
R =A1
Tangent
1
2
3
4
A
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’S
METHOD
Axis
Minor
A
B
Major Axis
7
8
9
10
11
9
8
7
6
5
4
3
2
1
12
11
P6
P5
P4
P3
P2`
P1
P12
P11
P10
P9
P8
P7
6
5
4
3
2
1
12
C
10
O
CONCENTRIC CIRCLE METHOD
F2
F1
D
CF1=CF2=1/2 AB
T
N
Q
e = AF1/AQ
Normal
0
1
2
3
4
1
2
3
4
1’
0’
2’
3’
4’
1’
2’
3’
4’
A
B
C
D
Major Axis
Minor Axis
F1
F2
Directrix
E
F
S
P
P1
P2
P3
P4
Tangent
P1’
P2’
P3’
P4’
Ø
Ø
R=AB/2
P0
P1’’
P2’’
P3’’
P4’’
P4
P3
P2
P1
OBLONG METHOD
B
A
P4
P0
D
C
60°
6
5
4
3
2
1
0
5
4
3
2
1
0
1
2
3
4
5
6
5
3
2
1
0
P1
P2
P3
Q1
Q2
Q3
Q4
Q5
P6
Q6
O
4
ELLIPSE IN PARALLELOGRAM
R4
R3
R2
R1
S1
S2
S3
S4
P5
G
H
I
K
J
Minor Axis
Major Axis
P6
Normal
P5’
P7’
P6’
P1
Tangent
P1’
N
N
T
T
V1
P5
P4’
P4
P3’
P2’
F1
D1
D1
R1
b
a
c
d
e
f
g
Q
P7
P3
P2
Directrix
R=6f`
90°
1
2
3
4
5
6
7
Eccentricity = 2/3
3
R1V1
QV1
=
R1V1
V1F1
=
2
Ellipse
ELLIPSE – DIRECTRIX FOCUS METHOD
R=1a
Dist. Between directrix & focus = 50 mm
1 part = 50/(2+3)=10 mm
V1F1 = 2 part = 20 mm
V1R1 = 3 part = 30 mm
θ < 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
METHOD
Normal
G2’
R =A1
Tangent
1
2
3
4
A
B
G
G’
G1
G3
G2
G4
G4
G3
G2
G1
G1’
G3’
G4’
G4’
G3’
G2’
G1’
F2
F1
R=B1
R=B2
`R=A2
O
°
°
90°
90°
directrix
100
140
GF1 + GF2 = MAJOR AXIS = 140
E
AF1
AE
e =
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.
O
A
B
C
75
45
1
D
100
1
2
2
3
3
4
4
5
5
6
6
7
7
P1
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
C
D
F1
F2
P
Q
R
S
50
I1
I4
A
B
I2
O
1
1
2
2
4
4
3
3
100
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
60
80
Q
1
2
1
2
P1
P2
Q2
Q1
R1
R2
S2
S1
PROBLEM :-2
Draw 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.
P
Q
R
80mm
40mm
89mm
A
C
B
D
MAJOR AXIS = PQ+QR = 129mm
θ
θ
R=AB/2
1
2
3
1’
2’
3’
1’’
2’’
3’’
O3
O3’
O1
O2
O2’
O1’
TANGENT
NORMAL
60º
30º
2
3
1
directrix
F1
F2
MAJOR AXIS
MINOR AXIS
S
ECCENTRICITY = 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
6
6
5
4
3
2
1
0
5
4
3
2
1
0
1
2
3
4
5
6
6
5
3
2
1
0
P2
P3
P4
P5
Q1
Q2
Q3
Q4
Q5
B
A
O
4
ELLIPSE
100
45
75
P0
P1
P6
Q6
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
2
3
4
5
0
1
2
3
4
5
6
1
1
5
4
3
2
6
0
1
2
3
4
5
0
V
D
C
A
B
P4
P4
P5
P5
P3
P3
P2
P2
P6
P6
P1
P1
PARABOLA –RECTANGLE METHOD
PARABOLA
B
0
2’
0
2
6
C
6’
V
5
P’
5
30°
A
X
D
1’
2’
4’
5’
3’
1
3
4
5’
4’
3’
1’
0
5
4
3
2
1
P1
P2
P3
P4
P5
P’
4
P’
3
P’
2
P’
1
P’
6
PARABOLA – IN PARALLELOGRAM
P
6
B
A
O
V
1
8
3
4
5
2
6
7
9
10
0
1
2
3
4
5
6
7
8
9
10
0
θ
θ
F
PARABOLA
TANGENT METHOD
D
D
DIRECTRIX
90°
2
3
4
T
T
N
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
ROOT OF TREE
BUILDING
REQD.DISTANCE
TOP OF TREE
3m
6m
F
A
STONE FALLS HERE
3m
6m
ROOT OF TREE
BUILDING
REQD.DISTANCE
GROUND
TOP OF TREE
3m
6m
1
2
3
1
2
3
3
2
1
4
5
6
5
6
E
F
A
B
C
D
P3
P4
P2
P1
P
P1
P2
P3
P4
P5
P6
3
2
1
0
STONE 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 mm
A
B
C
D
1
2
3
4
5
1
2
3
4
5
O
P1
P2
P3
P4
P5
M
1’
2’
3’
4’
5’
1’
2’
3’
4’
5’
P1’
P2’
P3’
P4’
P5’
90 mm
EXAMPLE
A shot is discharge from the ground level at an angle 60 to the horizontal at a point 80m away from the point of discharge. Draw the path trace by the shot. Use a scale 1:100
ground level
B
A
60º
gun
shot
80 M
parabola
ground level
B
A
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
60º
gun
shot
D
D
VF
VE
=
e = 1
E
Connect two given points A and B by a Parabolic curve, when:-
1.OA=OB=60mm and angle AOB=90°
2.OA=60mm,OB=80mm and angle AOB=110°
3.OA=OB=60mm and angle AOB=60°
60
60
1
2
3
4
5
Parabola
5
4
3
2
1
A
B
90 °
O
1.OA=OB=60mm and angle AOB=90°
A
B
80
60
5
4
3
2
1
1
2
3
4
5
110 °
Parabola
O
2.OA=60mm,OB=80mm and angle AOB=110°
5
4
3
2
1
5
4
3
2
1
A
B
O
Parabola
60
60
60 °
3.OA=OB=60mm and angle AOB=60°
example
Draw a parabola passing through three different points A, B and C such that AB = 100mm, BC=50mm and CA=80mm respectively.
B
A
C
100
50
80
1’
0
0
2’
0
2
6
6’
5
P’
5
1’
2’
4’
5’
3’
1
3
4
5’
4’
3’
5
4
3
2
1
P1
P2
P3
P4
P5
P’
4
P’
3
P’
2
P’
1
P’
6
P
6
A
B
C
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
P3
P4
P5
0
P6
P0
A
O
E
X
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
P3
P4
P5
0
P6
P0
A
O
E
X=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”.
750
P4
E
6’
2’
1’
P1
1
2
3
4
5
6
D
P6
P5
P3
P2
P0
7’
P7
7
C
B
F
O
Y = 30
X = 20
Given Point P0
A
AXIS
NORMAL
C
V
F1
DIRECTRIX
D
D
1
2
3
4
4’
3’
2’
1’
P1
P2
P3
P4
P1’
P2’
P3’
P4’
T1
T2
N2
N1
TANGENT
s
Directrix and focus method
CYCLOIDAL GROUP OF CURVES
When 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.
Superior
Hypotrochoid
Cycloidal Curves
Cycloid
Epy Cycloid
Hypo Cycloid
Superior
Trochoid
Inferior
Trochoid
Superior
Epytrochoid
Inferior
Epytrochoid
Inferior
Hypotrochoid
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 = 360º x r/Rd
Circumference of Generating Circle
Rd
Rolling Circle
r
O
Ø/2
Ø/2
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.`
Directing
Circle(R)
P
Ø /2
Ø /2
Ø =
360 x r
R
R
T
Rolling Circle
Radius (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
11
0
12
0
3
4
5
6
7
8
9
10
11
12
P1
P2
P3
P4
P5
P7
P8
P9
P11
P12
C0
C1
C2
C3
C4
C5
C6
C7
C8
C9
C10
C11
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
C0
P0
7
8
P6
4
P1
1
2
3
C2
C3
P2
C4
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.
5
6
C1
P3
P4
P5
P7
P8
7
0
C5
C6
C7
C8
1
2
3
4
5
6
D/2
πD/2
πD/2
D/2
Floor
Wall
CYCLOID
5
6
7
8
Take diameter of circle = 40mm
Initially 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.
P6
P4
r
P2
C1
C0
C2
C3
C4
C5
C6
C7
C8
1
0
2
3
4
5
6
7
O
Rd
Ø/2
Ø/2
P1
P0
P3
P5
P7
P8
r
r
Rolling Circle
r
Rd X Ø = 2πr
Ø = 360º x r/Rd
Arc P0P8 = Circumference of Generating Circle
EPICYCLOID
GIVEN:
Rad. Of Gen. Circle (r)
& Rad. Of dir. Circle (Rd)
S
º
U
N
Ø = 360º x 25/75
= 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
C4
C5
C6
C7
C8
C9
C10
C11
C12
P10
P8
0
1
2
3
4
5
6
7
8
9
10
11
12
P2
P3
P4
P5
P6
P9
P7
P12
/2
/2
=
360 x 4
12
=
360 x r
R
=
120°
R
T
T
N
S
N
Normal
r
r
Rolling Circle
Radias (r)
Directing
Circle
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
C4
C5
C6
C7
C9
C8
C10
C11
C12
P8
O
10
5
7
8
9
11
12
1
2
3
4
6
P1
P2
P3
P4
P5
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 un wounding the string’s one turn. String is kept tight during unwound. Draw tangent & normal to the curve at any point.
P12
P2
0
02
12
6
P1
1
2
0
9
10
3
4
6
8
11
5
7
12
πD
P3
P4
P5
P6
P7
P8
P9
P10
P11
1
2
3
4
5
7
8
9
10
11
03
04
05
06
07
08
09
010`
011
Tangent
N
N
Normal
T
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.
P
R=7toP
R=6toP
R21
0
0
1
2
3
4
5
6
7
8
P
11
0
1
2
3
4
5
6
7
8
9
10
P1
P2
P3
P4
P5
P6
P7
P8
L= 100 mm
R=1toP
R=2toP
R=3toP
R=4toP
R=5toP
INVOLUTE
9
ø
11 mm = 30°
Then 5 mm = ζ
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.
P13
P11
3
13
14
15
P0
P12
O
7
10
1
2
3
4
5
6
8
9
11
12
1
2
P1
P2
P3
P4
P5
P6
P7
P8
P9
P10
P14
P
L=160 mm
R=21mm
6
4
5
7
8
9
10
11
12
13
14
15
ø
PROBLEM:-
Draw an involute of a pantagon having side as 20 mm.
P5
R=01
R=2*01
P0
P1
P2
P3
P4
R=3*01
R=4*01
R=5*01
2
3
4
5
1
T
T
N
N
S
INVOLUTE
OF A POLYGON
Given :
Side of a polygon
0
PROBLEM:-
Draw an involute of a square having side as 20 mm.
P2
1
2
3
0
4
P0
P1
P3
P4
N
N
S
R=3*01
R=4*01
R=2*01
R=01
INVOLUTE OF A SQUARE
PROBLEM:-
Draw an involute of a string unwound from the given figure from point C in anticlockwise direction.
60°
A
B
C
R21
30°
R21
60°
A
B
C
30°
X
X+A1
X
X+A2
X+A3
X+A5
X+A4
X+AB
R =X+AB
X+66+BC
1
2
3
4
5
C0
C1
C2
C3
C4
C5
C6
C7
C8
A stick of length equal to the circumference of a semicircle, is initially tangent to the semicircle on the right of it. This stick now rolls over the circumference of a semicircle without sliding till it becomes tangent on the left side of the semicircle. Draw the loci of two end point of this stick. Name the curve. Take R= 42mm.
PROBLEM:-
A6
B6
5
A
B
C
B1
A1
B2
A2
B3
A3
B4
A4
B5
A5
1
2
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:
P10
1
2
3
4
5
6
7
8
9
10
11
12
0
8
7
0
1
2
3
4
5
6
9
11
12
P1
P2
P3
P4
P5
P6
P7
P8
P9
P11
P12
o
To construct an Archimedean Spiral of one convolutions, given the greatest & shortest(least) radii.
Say Greatest radius = 100 mm &
Shortest radius = 60 mm
To construct an Archimedean Spiral of one convolutions, given the largest radius vector & smallest radius vector.
OR
3
1
2
6
5
8
4
7
9
10
11
2
1
3
4
5
6
7
8
9
10
11
12
P1
P2
P3
P4
P5
P6
P7
P8
P9
P10
P11
P12
O
N
N
T
T
S
R min
R max
Diff. in length of any two radius vectors
Angle between them in radians
Constant 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.
O
A
B
40
25
B
A
O
1
2
3
4
5
6
7
8
9
10
11
P1
P2
P3
P4
P5
P6
P7
P8
P9
P10
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.
A
Initial Position of point P
PO
P1
P2
P3
P4
P5
P6
P7
P8
2
1
3
4
5
6
7
O
1
2
3
4
5
6
7
8
2/3 X 360°
= 240°
120º
A0
Linear Travel of point P on AB
= 96 =16x (6 div.)
EXAMPLE: A link AB, 96mm long initially is vertically upward w.r.t. its pinned end B, swings in clockwise direction for 180° and returns back in anticlockwise direction for 90°, during which a point P, slides from pole B to end A. Draw the locus of point P and name it. Draw tangent and normal at any point on the path of P.
P1’
A
B
A1
A2
A3
A4
A5
A6
P0
P1
P2
P3
P4
P5
P6
P2’
P3’
P4’
P5’
P6’
96
Link AB = 96
C
Tangent
Angular Swing of link AB =
180° + 90°
= 270 °
=45 °X 6 div.
ARCHIMEDIAN SPIRAL
D
NORMAL
M
N
Arch.Spiral Curve Constant BC
= 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
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
23
13
22
24
1
2
3
4
5
6
7
8
9
10
11
12
14
15
16
18
19
20
21
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 circle
P1
P2
P3
P4
P0
P6
P7
P8
P5
P9
P10
1
2
3
4
5
6
7
8
9
10
1
2
3
4
5
6
7
8
9
10
0
C0
C1
C2
C3
C4
Directing line
Length of directing line = 3ΠD/2
540 ° = 360° + 180°
540 ° = ΠD + ΠD/2
Total length for 540 ° rotation = 3ΠD/2
C5
C6
C7
C8
C9
C10
S
normal