1.
ENGINEERING CURVES
Part-II
(Point undergoing two types of displacements)
INVOLUTE
1. Involute of a circle
a)String Length = πD
b)String Length > πD
c)String Length < πD
CYCLOID
1. General Cycloid
2. Trochoid
( superior)
3. Trochoid
( Inferior)
4. Epi-Cycloid
SPIRAL
HELIX
1. Spiral of
One Convolution.
2. Spiral of
Two Convolutions.
1. On Cylinder
2. On a Cone
2. Pole having Composite
shape.
5. Hypo-Cycloid
3. Rod Rolling over
a Semicircular Pole.
AND
Methods of Drawing
Tangents & Normals
To These Curves.
2.
DEFINITIONS
CYCLOID:
IS A LOCUS OF A POINT ON THE
ERIPHERY OF A CIRCLE WHICH
OLLS ON A STRAIGHT LINE PATH.
NVOLUTE:
SUPERIORTROCHOID:
IF THE POINT IN THE
DEFINATION
OF CYCLOID IS OUTSIDE THE
CIRCLE
INFERIOR TROCHOID.:
IS A LOCUS OF A FREE END OF A STRING
IF IT IS INSIDE THE CIRCLE
HEN IT IS WOUND ROUND A CIRCLE OR POLYGON
SPIRAL:
IS A CURVE GENERATED BY A POINT
HICH REVOLVES AROUND A FIXED POINT
ND AT THE SAME MOVES TOWARDS IT.
EPI-CYCLOID
IF THE CIRCLE IS ROLLING
ON
ANOTHER CIRCLE FROM
OUTSIDE
HYPO-CYCLOID.
IF THE CIRCLE IS ROLLING
FROM INSIDE THE OTHER
CIRCLE,
IS A CURVE GENERATED BY A POINT WHICH
OVES AROUND THE SURFACE OF A RIGHT CIRCULAR
YLINDER / CONE AND AT THE SAME TIME ADVANCES IN AXIAL DIRECTION
T A SPEED BEARING A CONSTANT RATIO TO THE SPPED OF ROTATION.
or problems refer topic Development of surfaces)
HELIX:
3.
Problem: Draw involute of a square of 25 mm sides
75
C
B
0
10
50
D
25
25
A
100
4.
Problem: Draw involute of an equilateral triangle of 35 mm sides.
35
3X
B
5
2X3
C
A
35
35
3X35
5.
Problem no 23: Draw Involute of a circle of 40 mm diameter.
Also draw normal and tangent to it at a point 100 mm from the
centre of the circle.
P3
Ta n g e
nt
P4
al
P2
Norm
Solution Steps:
1) Point or end P of string AP is
exactly πD distance away from A.
Means if this string is wound round
the circle, it will completely cover
given circle. B will meet A after
winding.
2) Divide πD (AP) distance into 12
number of equal parts.
3) Divide circle also into 12 number
of equal parts.
P5
4) Name after A, 1, 2, 3, 4, etc. up
to 12 on πD line AP as well as on
circle (in anticlockwise direction).
5) To radius C-1’, C-2’, C-3’ up to
C-12’ draw tangents (from 1’,2’,3’,4’,
etc to circle).
6) Take distance 1 to P in compass
P6
and mark it on tangent from point 1’
on circle (means one division less
than distance AP).
7) Name this point P1
8) Take 2-P distance in compass
and mark it on the tangent from
P
point 2’. Name it point P2. 7
9) Similarly take 3 to P, 4 to P, 5 to
P up to 11 to P distance in compass
and mark on respective tangents
and locate P3, P4, P5 up to P12 (i.e. P8
A) points and join them in smooth
curve it is an INVOLUTE of a given
circle.
INVOLUTE OF A CIRCLE
6’
7’
8’
5’
4’
9’
c
3’
10’
2’
11’
12’
A
P11
P9
P1
P10
1’
1
P
2
3
4
5
6
πD
7
8
9
10
11
12
6.
Involute
Method of Drawing
Tangent & Normal
STEPS:
DRAW INVOLUTE AS USUAL.
MARK POINT Q ON IT AS DIRECTED.
INVOLUTE OF A CIRCLE
ma
l
JOIN Q TO THE CENTER OF CIRCLE C.
CONSIDERING CQ DIAMETER, DRAW
A SEMICIRCLE AS SHOWN.
No
r
MARK POINT OF INTERSECTION OF
THIS SEMICIRCLE AND POLE CIRCLE
AND JOIN IT TO Q.
Q
THIS WILL BE NORMAL TO INVOLUTE.
Ta
ng
DRAW A LINE AT RIGHT ANGLE TO
THIS LINE FROM Q.
en
t
IT WILL BE TANGENT TO INVOLUTE.
4
3
5
2
C
6
7
1
8
P8
1
2
3
4
π
D
5
6
7
P
8
7.
CYCLOID
Problem 15: Draw locus of a point on the periphery of a circle which rolls on straight line path .
Take circle diameter as 50 mm. Draw normal and tangent on the curve at a point 40 mm
above the directing line.
7
5
a
4 p4 T
8
C1
p3
CP
p2
10
11
p1
12 P
3
C3
2
1
1’
2’
p7
p8
C4
C5
C6
al
rm
No
C
9
C2
t
en
ng
p6
p5
C7
C8
C9
C10
C11
p9
40mm
6
C12
p10
p11
p12
3’
4’
5’
6’
πD
7’
8’
9’
10’
11’
12’ Q
Solution Steps:
1) From center C draw a circle of 50mm dia. and from point P draw a horizontal line PQ equal to πD length.
2) Divide the circle in 12 equal parts and in anticlockwise direction, after P name 1, 2, 3 up to 12.
3) Also divide the straight line PQ into 12 number of equal parts and after P name them 1’,2’,3’__ etc.
4) From all these points on circle draw horizontal lines. (parallel to locus of C)
5) With a fixed distance C-P in compass, C1 as center, mark a point on horizontal line from 1. Name it P1.
6) Repeat this procedure from C2, C3, C4 up to C12 as centers. Mark points P2, P3, P4, P5 up to P12 on the
horizontal lines drawn from 1,2, 3, 4, 5, 6, 7 respectively.
7) Join all these points by curve. It is Cycloid.
8.
STEPS:
DRAW CYCLOID AS USUAL.
MARK POINT Q ON IT AS DIRECTED.
CYCLOID
Method of Drawing
Tangent & Normal
WITH CP DISTANCE, FROM Q. CUT THE
POINT ON LOCUS OF C AND JOIN IT TO Q.
FROM THIS POINT DROP A PERPENDICULAR
ON GROUND LINE AND NAME IT N
JOIN N WITH Q.THIS WILL BE NORMAL TO
CYCLOID.
No r m
al
DRAW A LINE AT RIGHT ANGLE TO
THIS LINE FROM Q.
IT WILL BE TANGENT TO CYCLOID.
CYCLOID
Tang
e
nt
CP
Q
C
C1
C2
C3
C4
P
C5
N
πD
C6
C7
C8
9.
PROBLEM 25: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLE WHICH ROLLS ON A CURVED PATH. Take
diameter of rolling Circle 50 mm And radius of directing circle i.e. curved path, 75 mm.Also draw normal and
tangent on the curve at 110mm from the centre of directing circle.
Tangent
Solution Steps:
1) When smaller circle will roll on
larger circle for one revolution it will
cover πD distance on arc and it will
be decided by included arc angle θ.
2) Calculate θ by formula θ = (r/R)
x 360º.
3) Construct angle θ with radius
OC and draw an arc by taking O as
center OC as radius and form
sector of angle θ.
4) Divide this sector into 12 number
of equal angular parts. And from C
Rolling circle or
onward name them C1, C2, C3 up to
generating
circle
C12.
5) Divide smaller circle (Generating
circle) also in 12 number of equal
parts. And next to P in anticlockwise direction name those 1, 2, 3, up
c2
to 12.
6) With O as center, O-1 as radius
3’
4’
draw an arc in the sector. Take O-2,
O-3, O-4, O-5 up to O-12 distances
c1
5’
with center O, draw all concentric
arcs in sector. Take fixed distance
C-P in compass, C1 center, cut arc
6’
C
of 1 at P1.
Repeat procedure and locate P 2, P3,
7’
P4, P5 unto P12 (as in cycloid) and
join them by smooth curve. This is
8’
EPI – CYCLOID.
9’
c9
c8
c7
c10
c11
c12
c6
c5
8
9
10
7
12
6
c4
5
c3
Normal
4
3
Directing circle
2
2’
1
1’
θ
12’
P
11’
10’
11
O
OP=Radius of directing circle=75mm
PC=Radius of generating circle=25mm
θ=r/R X360º= 25/75 X360º=120º
10.
Problem 17: A circle of 50 mm diameter rolls on another circle of 175 mm diameter and outside it.
Draw the curve traced by a point P on its circumference for one complete revolution.Also draw normal
and tangent on the curve at 125 mm from the centre of directing circle.
Draw a horizontal line OP of 87.5 mm and draw an arc
with O as centre and PO as radius
Draw a horizontal line CP of 25 mm and draw a circle
with C as centre and CP as radius.
θ=(OP/PC) X 360º = (25/87.5) X 360º = 102.8º ≈103º
Divide the rolling circle in 8 equal parts
Also divide the angle in 8 equal parts using angle bisectors
Directing
circle
Rolling circle or
generating
circle
C
P
O
θ=103º
11.
PROBLEM 26: 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 50 mm and radius of directing circle (curved path) 75 mm. Also
draw normal and tangent on the curve at a point 40mm from the centre of
directing circle
Directing circle
Solution Steps:
1) Smaller circle is rolling
here, inside the larger
circle. It has to rotate
anticlockwise to move
ahead.
2) Same steps should be
taken as in case of EPI –
CYCLOID. Only change is
in numbering direction of
12 number of equal parts
on the smaller circle.
3) From next to P in
clockwise direction, name
1,2,3,4,5,6,7,8,9,10,11,12
4) Further all steps are
that of epi – cycloid. This
is called
HYPO – CYCLOID.
10
11
12
6
5
4
3
c5
c6
c7
c8
c9
c10
c11
c4
2
2’
3’
c3
4’
c2
1
1’
5’
c1
12’
θ
6’
C
P
11’
Rolling circle or
generating circle
9
8
7
7’
10’
O
8’
9’
OP=Radius of directing circle=75mm
PC=Radius of generating circle=25mm
θ=r/R X360º= 25/75 X360º=120º
c12
12.
Problem 28: A point P moves towards another point O, 75 mm from it
and reaches it while moving around it once. Its movement towards O
being uniform with its movement around it. Draw the curve traced out
by point P.
SPIRAL
Important approach for construction
Find total angular and total linear displacement and divide both in to same number of equal
parts. Total linear movement 75 mm. Total angular movement 360º
With OP radius & O as center draw a
circle and divide it in EIGHT
parts. Name those 1’,2’,3’,4’, etc.
up to divided line PO also in
Similarly 8’
EIGHT parts and name those
1,2,3, starting from P.
Take O-1 distance from OP line and
draw an arc up to O1’ radius
vector. Name the point P1
Similarly mark points P2, P3, P4 up to
P8
And join those in a smooth curve. It is
a SPIRAL of one convolution.
2’
P2
3’
P1
1’
P3
P4
4’
O
P5
7 6
P7
5 4 3 2 1
P6
7’
5’
6’
P
13.
Draw an Archemedian spiral of one convolution, greatest and least radii being 115mm and 15
mm respectively. Draw a normal and tangent to the spiral at a point 65 mm from the pole.
Important approach for construction!
Find total angular and total linear displacement and divide both in to same number of equal
parts. Angular displacement =360º, Linear displacement = 100mm
3’
Solution Steps
2’
4’
P3
P2
P4
5’
P1
t
gen
6’
P6
N
al
Norm
c Q
O
12 11 10 9 8 7 6 5 4 3 2 1
P11
P10
P7
P8
P9
7’
C=(Rmax-Rmin)/No. of
convolutions in radians
= (115-15)/3.14 X 2 =15.92
1’
P5
T an
1. With PO & QO radii draw two
circles and divide them in
twelve equal parts. Name those
1’,2’,3’,4’, etc. up to 12’
2 .Similarly divided line PQ also in
twelve parts and name those
1,2,3,-- as shown.
3. Take O-1 distance from OP line
and draw an arc up to O1’ radius
vector. Name the point P1
4. Similarly mark points P2, P3, P4
up to P12
And join those in a smooth curve.
It is a SPIRAL of one convolution.
11’
8’
9’
10’
P
12’
14.
Draw an Archemedian spiral of one and half convolution, greatest and least radii being
115mm and 15 mm respectively. Draw a normal and tangent to the spiral at a point 70 mm
from the pole.
Important approach for construction
Find total angular and total linear displacement and divide both in to same number of equal
parts. Total Angular displacement 540º. Total Linear displacement 100 mm
3’15’
1 Draw a 115 mm long line OP.
16’4’
2 Mark Q at 15 mm from O
3 with O as centre draw two circles with OP
and OQ radius
4 Divide the circle in 12 equal divisions and
17’5’
mark the divisions as 1’,2’ and so on up to 18’ P
5
5 Divide the line PQ in 18 equal divisions as 1,2,3 and so on upto 18
6.Take O-1 distance from OP line and
draw an arc up to O1’ radius vector.
P6
Name the point P1
18’ 6’
7.Similarly mark points P2, P3, P4 up
to P18.
P7
8. And join those in a smooth curve.
It is a SPIRAL of one and half
convolution.
7’
C=(Rmax-Rmin)/No. of
convolutions in radians
= (115-15)/3.14 X3 =10.61
8’
2’14’
P3
P2
P4
P1
P16
P15
1’13’
P14
P13
P17
P18
P12
Q
18
O
16
14
12
10
8
6
4
2
P
12’
P11
P10
P8
P9
9’
11’
10’
15.
Spiral.
Method of Drawing
Tangent & Normal
SPIRAL (ONE CONVOLUSION.)
2
No
al
rm
3
P2
nt
ge
n
Ta
P1
Q
1
Constant of the Curve =
P3
=
P4
4
O
P5
7 6
P7
5 4 3 2 1
7
6
Angle between the corresponding
radius vector in radian.
OP – OP2
π/2
=
OP – OP2
1.57
= 3.185 m.m.
STEPS:
*DRAW SPIRAL AS USUAL.
DRAW A SMALL CIRCLE OF RADIUS EQUAL TO THE
CONSTANT OF CURVE CALCULATED ABOVE.
P6
5
P
Difference in length of any radius vectors
* LOCATE POINT Q AS DISCRIBED IN PROBLEM AND
THROUGH IT DRAW A TANGENTTO THIS SMALLER
CIRCLE.THIS IS A NORMAL TO THE SPIRAL.
*DRAW A LINE AT RIGHT ANGLE
*TO THIS LINE FROM Q.
IT WILL BE TANGENT TO CYCLOID.
16.
SPIRAL
of
two convolutions
Problem 28
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).
IMPORTANT APPROACH FOR CONSTRUCTION!
FIND TOTAL ANGULAR AND TOTAL LINEAR DISPLACEMENT
AND DIVIDE BOTH IN TO SAME NUMBER OF EQUAL PARTS.
2,10
P2
P1
3,11
SOLUTION STEPS:
Total angular displacement here
is two revolutions And
Total Linear displacement here
is distance PO.
Just divide both in same parts i.e.
Circle in EIGHT parts.
( means total angular displacement
in SIXTEEN parts)
Divide PO also in SIXTEEN parts.
Rest steps are similar to the previous
problem.
1,9
P3
P10
P9
P11
4,12
P4
16
10
8 7 6 5 4 3 2 1 P
P8
P12
8,16
P15
P13
P14
P7
P5
5,13
13
P6
6,14
7,15
17.
Problem No.7:
A Link OA, 80 mm long oscillates around O,
600 to right side and returns to it’s initial vertical
Position with uniform velocity.Mean while point
P initially on O starts sliding downwards and
reaches end A with uniform velocity.
Draw locus of point P
Solution Steps:
Point P- Reaches End A (Downwards)
1) Divide OA in EIGHT equal parts and from O to A after O
name 1, 2, 3, 4 up to 8. (i.e. up to point A).
2) Divide 600 angle into four parts (150 each) and mark each
point by A1, A2, A3, A4 and for return A5, A6, A7 andA8.
(Initial A point).
3) Take center O, distance in compass O-1 draw an arc upto
OA1. Name this point as P1.
1) Similarly O center O-2 distance mark P2 on line O-A2.
2) This way locate P3, P4, P5, P6, P7 and P8 and join them.
( It will be thw desired locus of P )
OSCILLATING LINK
O
p
1
2
p1
p2
p3
p4
3
p5
A4
4
5
p6
A3
6
7
8
A
p8
A8
p7
A1
A7
A2
A6
A5
18.
OSCILLATING LINK
Problem No 8:
A Link OA, 80 mm long oscillates around O,
600 to right side, 1200 to left and returns to it’s initial
vertical Position with uniform velocity.Mean while point
P initially on O starts sliding downwards, reaches end A
and returns to O again with uniform velocity.
Draw locus of point P
Op
16
15
14
Solution Steps:
( P reaches A i.e. moving downwards.
& returns to O again i.e.moves upwards )
1.Here distance traveled by point P is PA.plus A
12
AP.Hence divide it into eight equal parts.( so
total linear displacement gets divided in 16
parts) Name those as shown.
2.Link OA goes 600 to right, comes back to
A
A13 11
original (Vertical) position, goes 600 to left
and returns to original vertical position. Hence
total angular displacement is 2400.
Divide this also in 16 parts. (150 each.)
Name as per previous problem.(A, A1 A2 etc)
3.Mark different positions of P as per the
procedure adopted in previous case.
and complete the problem.
13
1
p1
p2
p3
p4
2
p5
3
12
A4
4
11
10
A10
A14
A3
6
9 7
A9
A15
p6
5
8
A p8
A8
A16
p7
A1
A7
A2
A6
A5
19.
ROTATING LINK
Problem 9:
Rod AB, 100 mm long, revolves in clockwise direction for one revolution.
Meanwhile point P, initially on A starts moving towards B and reaches B.
Draw locus of point P.
1) AB Rod revolves around
center O for one revolution and
point P slides along AB rod and
reaches end B in one
revolution.
2) Divide circle in 8 number of
equal parts and name in arrow
direction after A-A1, A2, A3, up
to A8.
3) Distance traveled by point P
is AB mm. Divide this also into 8
number of equal parts.
4) Initially P is on end A. When
A moves to A1, point P goes
one linear division (part) away
from A1. Mark it from A1 and
name the point P1.
5) When A moves to A2, P will
be two parts away from A2
(Name it P2 ). Mark it as above
from A2.
6) From A3 mark P3 three
parts away from P3.
7) Similarly locate P4, P5, P6,
P7 and P8 which will be eight
parts away from A8. [Means P
has reached B].
8) Join all P points by smooth
curve. It will be locus of P
A2
A1
A3
p1
p2
p6
p5
A
P
1
2
3
p7
p3
p4
A7
4
5
6
7
A5
A6
p8
B A4
20.
Problem 10 :
Rod AB, 100 mm long, revolves in clockwise direction for one revolution.
Meanwhile point P, initially on A starts moving towards B, reaches B
And returns to A in one revolution of rod.
Draw locus of point P.
ROTATING LINK
A2
Solution Steps
1) AB Rod revolves around center O
for one revolution and point P slides
along rod AB reaches end B and
returns to A.
2) Divide circle in 8 number of equal
parts and name in arrow direction
after A-A1, A2, A3, up to A8.
3) Distance traveled by point P is AB
plus AB mm. Divide AB in 4 parts so
those will be 8 equal parts on return.
4) Initially P is on end A. When A
moves to A1, point P goes one
linear division (part) away from A1.
Mark it from A1 and name the point
P1.
5) When A moves to A2, P will be
two parts away from A2 (Name it
P2 ). Mark it as above from A2.
6) From A3 mark P3 three parts
away from P3.
7) Similarly locate P4, P5, P6, P7
and P8 which will be eight parts away
from A8. [Means P has reached B].
8) Join all P points by smooth curve.
It will be locus of P
The Locus will
follow the loop path two times in
one revolution.
A1
A3
p5
p1
p4
A
P
p8
p2
1+7
2+6 p
6
+ 5
3
4
p7 p3
A7
A5
A6
+B
A4
21.
Problem 28: A link OA, 100 mm long rotates about O
in anti-clockwise direction. A point P on the link, 15
mm away from O, moves and reaches the end A,
while the link has rotated through 2/5 of a revolution.
Assuming that the movements of the link to be
uniform trace the path of point P.
θ= 2/5 X 360º = 144º
Total angular movement = 144º
Total linear movement = 85 mm
To divide both of them in equal
no. of parts ( say 8)
5’
6’
4’
7’
3’
P7
8’
P6
P5
2’
P4
P8
P3
1’
P2
144º
P1
O
15
P
1
2
3
100
4
5
6
7
8A
22.
Logarithmic Spiral:
If a point moves around a pole in such a way that
The value of vectorial angle are in arithmatic progression and
The corresponding values of radius vectors are in geometric progression, then the curve
traced by the point is known as logarithmic spiral.
A3
A2
P3
A1
P2
θ
θ
P1
θ
O
A
P
Let OA be a straight line and P be a point on it at radius vector OP from O.
Now let the line moves at uniform angular speed to a new position OA 1 ,at vectorial angle θ
from OA and the point moves to a new position P1 , at radius vector OP1 from O.
The line now gradually moves to the new position OA2, OA3 at vectorial angle θ and the
point to P2 and P3 , at radius vectors OP2 and OP3 respectively.
In Logarithmic spiral OP3/OP2 =OP2/OP1=OP1/OP
23.
Problem37: In a logarithmic spiral, the shortest radius is 40mm. The length of adjacent
radius vectors enclosing 30º are in the ratio of 9:8 Construct one revolution of the spiral.
Draw tangent to the spiral at a point 70 mm from it.
First step is to draw logarithmic scale.
Draw two straight lines OA & OB at angle of 30º.
B
Mark a point P on OA at 40 mm from O.
Calculate OP1 such that OP1/OP = 9/8. => OP1 = 45 mm
Mark a OP1 on OB at 45 mm from O.
Join P with P1.
P
P
Draw an arc of radius OP1 from OB to OA.
P
O
P
Draw a line parallel to PP1 from P1 on OA to intersect OB at P2.
P
Repeat the steps to get the points P3,P4 and so on up to P12.
4
45
P1
P
P2 3
P4
P
P5 6
P7
P8
P9
P10
P11
P12
3
2
30º
5
40
1
P6
P
P12
P7
P11
P8
P9
P10
P P1P2 P3 P4 P5 P6 P7
P8 P9 P10 P11 P12
A
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