1. ENGINEERING CURVES
Part-II
(Point undergoing two types of displacements)
INVOLUTE CYCLOID SPIRAL HELIX
1. Involute of a circle 1. General Cycloid 1. Spiral of 1. On Cylinder
a)String Length = πD One Convolution.
2. Trochoid 2. On a Cone
b)String Length > πD ( superior) 2. Spiral of
3. Trochoid Two Convolutions.
c)String Length < πD ( Inferior)
4. Epi-Cycloid
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 SUPERIORTROCHOID:
ERIPHERY OF A CIRCLE WHICH IF THE POINT IN THE DEFINATION
OLLS ON A STRAIGHT LINE PATH. OF CYCLOID IS OUTSIDE THE
CIRCLE
NVOLUTE: INFERIOR TROCHOID.:
IF IT IS INSIDE THE CIRCLE
IS A LOCUS OF A FREE END OF A STRING
HEN IT IS WOUND ROUND A CIRCULAR POLE
EPI-CYCLOID
IF THE CIRCLE IS ROLLING ON
SPIRAL: ANOTHER CIRCLE FROM OUTSIDE
IS A CURVE GENERATED BY A POINT HYPO-CYCLOID.
HICH REVOLVES AROUND A FIXED POINT IF THE CIRCLE IS ROLLING FROM
ND AT THE SAME MOVES TOWARDS IT. INSIDE THE OTHER CIRCLE,
HELIX:
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)

3. Problem: Draw involute of an equilateral triangle of 35 mm sides.
35
3X
5
2X3
35 3X35
35

4. Problem: Draw involute of a square of 25 mm sides
75
0
10
50
25
25 100

5. Problem no 17: Draw Involute of a circle. INVOLUTE OF A CIRCLE
String length is equal to the circumference of circle.
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. P
3 2 to p
2) Divide πD (AP) distance into 8
P2
1
P
number of equal parts.
3) Divide circle also into 8 number
of equal parts.
4) Name after A, 1, 2, 3, 4, etc. up
3 to p
to 8 on πD line AP as well as on
circle (in anticlockwise direction).
1 to p
5) To radius C-1, C-2, C-3 up to C-8
draw tangents (from 1,2,3,4,etc to 4 to p
circle). P4
4
6) Take distance 1 to P in compass 3
and mark it on tangent from point 1 5
on circle (means one division less 2
5 to p
than distance AP). 6
7) Name this point P1 1
8) Take 2-P distance in compass 7 A 8
7 to p
and mark it on the tangent from P
point 2. Name it point P2. P8 1 2 3 4 5 6 7 8
5
P
P6
6 to p
9) Similarly take 3 to P, 4 to P, 5 to P
7
P up to 7 to P distance in compass πD
and mark on respective tangents
and locate P3, P4, P5 up to P8 (i.e.
A) points and join them in smooth
curve it is an INVOLUTE of a given
circle.

6. INVOLUTE OF A CIRCLE
Problem 18: Draw Involute of a circle.
String length MORE than πD
String length is MORE than the circumference of circle.
Solution Steps:
In this case string length is more
than Π D.
P
But remember!
2 to p
P2
3
Whatever may be the length of
1
P
string, mark Π D distance
horizontal i.e.along the string
and divide it in 8 number of
equal parts, and not any other
3 to p
distance. Rest all steps are same
1 to p
as previous INVOLUTE. Draw
the curve completely.
4 to p
P4 4
3
5
5 to p
2
6
1
7
8
P
5
P
p8 1
7 to p
2 3 4 5 6 7 8
πD
P6
P
6 to p
7 165 mm
(more than πD)

7. Problem 19: Draw Involute of a circle. INVOLUTE OF A CIRCLE
String length is LESS than the circumference of circle. String length LESS than πD
Solution Steps:
In this case string length is Less
than Π D. P
3
But remember! 2 to p
Whatever may be the length of
P2
1
P
string, mark Π D distance
horizontal i.e.along the string
and divide it in 8 number of
equal parts, and not any other
3 to p
distance. Rest all steps are same
as previous INVOLUTE. Draw
1 to p
the curve completely.
4 to p
P4 4
3
5
5 to p
2
6
1
7
7 to p
8 P
P6
5
P
6 to p
P 1 2 3 4 5 6 7 8
7
150 mm
(Less than πD)
πD

8. PROBLEM 21 : Rod AB 85 mm long rolls
over a semicircular pole without slipping
from it’s initially vertical position till it
becomes up-side-down vertical. B
Draw locus of both ends A & B.
A4
Solution Steps? 4
If you have studied previous problems B1
properly, you can surely solve this also.
Simply remember that this being a rod, A3
it will roll over the surface of pole. 3
Means when one end is approaching,
other end will move away from poll.
OBSERVE ILLUSTRATION CAREFULLY!
πD 2
A2
B2
2
1
3
1
A1 4
A
B3
B4

9. Problem 22: Draw locus of a point on the periphery of a circle which rolls on straight line path . CYCLOID
Take circle diameter as 50 mm. Draw normal and tangent on the curve at a point 40 mm
above the directing line.
6 p5 p6
7 5 p7
4 p4 p8
8
C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 p9 C12
9 p3 3
p2 p10
10 p1 2
p11
11 1 p12
12 P
πD
Solution Steps:
1) From center C draw a horizontal line equal to πD distance.
2) Divide πD distance into 12 number of equal parts and name them C1, C2, C3__ etc.
3) Divide the circle also into 12 number of equal parts and in anticlockwise direction, after P name 1, 2, 3 up to 12.
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 P.
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.

10. 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.
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 3600.
3) Construct angle θ with radius
OC and draw an arc by taking O as c8 c9 c10
center OC as radius and form c7 c11
sector of angle θ. c12
c6
4) Divide this sector into 12 number
of equal angular parts. And from C
c5
onward name them C1, C2, C3 up to 8 9 10
11
C12. 7
6 12
5) Divide smaller circle (Generating c4
circle) also in 12 number of equal 5
parts. And next to P in anticlockw- c3
ise direction name those 1, 2, 3, up 4
to 12.
6) With O as center, O-1 as radius c2
3
draw an arc in the sector. Take O-2,
O-3, O-4, O-5 up to O-12 distances 2
3’
with center O, draw all concentric 4’ 2’
arcs in sector. Take fixed distance c1
1
C-P in compass, C1 center, cut arc 5’ 1’
of 1 at P1.
θ
Repeat procedure and locate P 2, P3, 6’
C 12’
P4, P5 unto P12 (as in cycloid) and P
join them by smooth curve. This is O
7’ 11’ OP=Radius of directing circle=75mm
EPI – CYCLOID.
PC=Radius of generating circle=25mm
8’ 10’ θ=r/R X360º= 25/75 X360º=120º
9’

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.
Solution Steps:
1) Smaller circle is rolling
here, inside the larger
circle. It has to rotate
9
anticlockwise to move 7
8 10
11
ahead. 6 12
2) Same steps should be
taken as in case of EPI – 5
CYCLOID. Only change is 4
in numbering direction of c7
c8 c9 c10
c11
c12
12 number of equal parts 3
c6
on the smaller circle. c5
3) From next to P in c4
clockwise direction, name 2
2’
3’ c3
4’
1,2,3,4,5,6,7,8,9,10,11,12 c2
4) Further all steps are 1 1’ 5’
that of epi – cycloid. This c1
is called θ
HYPO – CYCLOID. 12’
C
6’
P
11’ 7’
O
10’ 8’
9’
OP=Radius of directing circle=75mm
PC=Radius of generating circle=25mm
θ=r/R X360º= 25/75 X360º=120º

12. Problem 27: Draw a spiral of one convolution. Take distance PO 40 mm.
SPIRAL
IMPORTANT APPROACH FOR CONSTRUCTION!
FIND TOTAL ANGULAR AND TOTAL LINEAR DISPLACEMENT
AND DIVIDE BOTH IN TO SAME NUMBER OF EQUAL PARTS.
2
P2
Solution Steps 3 1
P1
1. With PO radius draw a circle
and divide it in EIGHT parts. P3
Name those 1,2,3,4, etc. up to 8
2 .Similarly divided line PO also in
EIGHT parts and name those
4 P4 O P
1,2,3,-- as shown. 7 6 5 4 3 2 1
3. Take o-1 distance from op line P7
and draw an arc up to O1 radius
P5 P6
vector. Name the point P1
4. Similarly mark points P2, P3, P4
up to P8
5 7
And join those in a smooth curve.
It is a SPIRAL of one convolution.
6

13. Problem 28 SPIRAL
Point P is 80 mm from point O. It starts moving towards O and reaches it in two of
revolutions around.it Draw locus of point P (To draw a Spiral of TWO convolutions).
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
3,11 P1 1,9
SOLUTION STEPS: P3
Total angular displacement here
P10
is two revolutions And P9
Total Linear displacement here P11
is distance PO. 16 13 10 8 7 6 5 4 3 2 1 P
Just divide both in same parts i.e. 4,12
P4 P8 8,16
P12
Circle in EIGHT parts. P15
( means total angular displacement P13 P14
in SIXTEEN parts)
Divide PO also in SIXTEEN parts. P7
Rest steps are similar to the previous P5
problem.
P6
5,13 7,15
6,14

14. STEPS: Involute
DRAW INVOLUTE AS USUAL.
Method of Drawing
MARK POINT Q ON IT AS DIRECTED. Tangent & Normal
JOIN Q TO THE CENTER OF CIRCLE C.
CONSIDERING CQ DIAMETER, DRAW
A SEMICIRCLE AS SHOWN.
INVOLUTE OF A CIRCLE
l
ma
MARK POINT OF INTERSECTION OF
r
No
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 en
THIS LINE FROM Q.
t
IT WILL BE TANGENT TO INVOLUTE.
4
3
5
C 2
6
1
7
8
P
P8 1 2 3 4 5 6 7 8
π
D

15. STEPS:
DRAW CYCLOID AS USUAL. CYCLOID
MARK POINT Q ON IT AS DIRECTED.
Method of Drawing
WITH CP DISTANCE, FROM Q. CUT THE Tangent & Normal
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.
DRAW A LINE AT RIGHT ANGLE TO
THIS LINE FROM Q.
al
No r m
IT WILL BE TANGENT TO CYCLOID.
CYCLOID
Q
Tang
e nt
CP
C C1 C2 C3 C4 C5 C6 C7 C8
P N
πD

16. Spiral.
Method of Drawing
Tangent & Normal
SPIRAL (ONE CONVOLUSION.)
2
nt
ge
No
n
Ta
rm
P2
al
3 1 Difference in length of any radius vectors
Q P1 Constant of the Curve =
Angle between the corresponding
radius vector in radian.
P3
OP – OP2 OP – OP2
= =
π/2 1.57
4 P4 O P = 3.185 m.m.
7 6 5 4 3 2 1
P7 STEPS:
*DRAW SPIRAL AS USUAL.
P5 P6 DRAW A SMALL CIRCLE OF RADIUS EQUAL TO THE
CONSTANT OF CURVE CALCULATED ABOVE.
* LOCATE POINT Q AS DISCRIBED IN PROBLEM AND
5 7 THROUGH IT DRAW A TANGENTTO THIS SMALLER
CIRCLE.THIS IS A NORMAL TO THE SPIRAL.
*DRAW A LINE AT RIGHT ANGLE
6
*TO THIS LINE FROM Q.
IT WILL BE TANGENT TO CYCLOID.

17. Basic Locus Cases:
PROBLEM 1.: Point F is 50 mm from a vertical straight line AB.
Draw locus of point P, moving in a plane such that
it always remains equidistant from point F and line AB.
P7
A P5
SOLUTION STEPS:
1.Locate center of line, perpendicular to P3
AB from point F. This will be initial
point P.
P1
2.Mark 5 mm distance to its right side,
name those points 1,2,3,4 and from those
draw lines parallel to AB.
3.Mark 5 mm distance to its left of P and p
name it 1. 1 2 3 4
F
4 3 2 1
4.Take F-1 distance as radius and F as
center draw an arc
cutting first parallel line to AB. Name
upper point P1 and lower point P2. P2
5.Similarly repeat this process by taking
again 5mm to right and left and locate P4
P3 P4 .
6.Join all these points in smooth curve. P6
B P8
It will be the locus of P equidistance
from line AB and fixed point F.

18. Problem 5:-Two points A and B are 100 mm apart. Basic Locus Cases:
There is a point P, moving in a plane such that the
difference of it’s distances from A and B always
remains constant and equals to 40 mm.
Draw locus of point P.
p7
p5
p3
p1
Solution Steps:
1.Locate A & B points 100 mm apart.
2.Locate point P on AB line, P
A B
70 mm from A and 30 mm from B 4 3 2 1 1 2 3 4
As PA-PB=40 ( AB = 100 mm )
3.On both sides of P mark points 5
mm apart. Name those 1,2,3,4 as usual. p2
4.Now similar to steps of Problem 2,
p4
Draw different arcs taking A & B centers
and A-1, B-1, A-2, B-2 etc as radius. p6
5. Mark various positions of p i.e. and join p8
them in smooth possible curve.
It will be locus of P
70 mm 30 mm

19. Problem No.7: OSCILLATING LINK
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 p
O
p1
Solution Steps: 1 p2 p4
Point P- Reaches End A (Downwards) p3
1) Divide OA in EIGHT equal parts and from O to A after O 2
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.
3
p5 A4
(Initial A point).
3) Take center O, distance in compass O-1 draw an arc upto 4
OA1. Name this point as P1.
1) Similarly O center O-2 distance mark P2 on line O-A2. 5 p6
2) This way locate P3, P4, P5, P6, P7 and P8 and join them. A3
6 A5
( It will be thw desired locus of P )
7 p7 A2
A6
A8 A1
p8
A7
A8

20. 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
p1 p4
1 p2
Solution Steps: 14 p3
( P reaches A i.e. moving downwards. 2
& returns to O again i.e.moves upwards ) 13
1.Here distance traveled by point P is PA.plus A 3 p5
AP.Hence divide it into eight equal parts.( so
12
12 A4
total linear displacement gets divided in 16 4
parts) Name those as shown. 11
2.Link OA goes 600 to right, comes back to A 5 p6
A13 11 A3
original (Vertical) position, goes 600 to left A5
10
and returns to original vertical position. Hence 6
total angular displacement is 2400. A10 p7 A2
Divide this also in 16 parts. (150 each.) 9 7
A14 A6
Name as per previous problem.(A, A1 A2 etc)
A9 8 A1
3.Mark different positions of P as per the A15 A p8
procedure adopted in previous case. A7
A8
and complete the problem.
A16

21. 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. A2
1) AB Rod revolves around
center O for one revolution and
point P slides along AB rod and A1
reaches end B in one A3
revolution. p1
2) Divide circle in 8 number of p2 p6
p7
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 p5
number of equal parts. p3
p8
4) Initially P is on end A. When
A moves to A1, point P goes A B A4
P 1 4 5 6 7
one linear division (part) away 2 3 p4
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, A7
A5
P7 and P8 which will be eight
parts away from A8. [Means P
has reached B].
8) Join all P points by smooth A6
curve. It will be locus of P

22. Problem 10 : ROTATING LINK
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. A2
Solution Steps
1) AB Rod revolves around center O
A1
A3
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 p5
parts and name in arrow direction
p1
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. p4
4) Initially P is on end A. When A p2 A4
A
moves to A1, point P goes one P 1+7 2+6 p + 5
3 4 +B
linear division (part) away from A1. p8 6
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. p7 p3
6) From A3 mark P3 three parts
away from P3.
7) Similarly locate P4, P5, P6, P7
A7
and P8 which will be eight parts away A5
from A8. [Means P has reached B].
8) Join all P points by smooth curve.
It will be locus of P
The Locus will A6
follow the loop path two times in
one revolution.