The document describes various engineering curves including involutes, cycloids, spirals, and helices. It provides definitions and examples of how to construct these curves. Specifically, it explains how to draw: - The involute of a circle by dividing the string length and circle into parts and connecting the points to form the curve. - A cycloid by having a smaller circle roll along a straight path, marking points on the smaller circle's circumference to connect into a looping curve. - A spiral by having a point revolve around a fixed point while also moving toward it, dividing the angular and linear displacements to mark points forming the spiral curve. - Tangents and normals to these curves using

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:
IT IS A LOCUS OF A POINT ON THE SUPERIORTROCHOID:
PERIPHERY OF A CIRCLE WHICH IF THE POINT IN THE
ROLLS ON A STRAIGHT LINE PATH. DEFINATION
OF CYCLOID IS OUTSIDE THE
CIRCLE
INVOLUTE: INFERIOR TROCHOID.:
IT IS A LOCUS OF A FREE END OF A STRING
IF IT IS INSIDE THE CIRCLE
WHEN IT IS WOUND ROUND A CIRCLE OR POLYGON
EPI-CYCLOID
SPIRAL: IF THE CIRCLE IS ROLLING ON
ANOTHER CIRCLE FROM
IT IS A CURVE GENERATED BY A POINT OUTSIDE
WHICH REVOLVES AROUND A FIXED POINT
AND AT THE SAME MOVES TOWARDS IT. HYPO-CYCLOID.
IF THE CIRCLE IS ROLLING
FROM INSIDE THE OTHER
HELIX: CIRCLE,
IT IS A CURVE GENERATED BY A POINT WHICH
MOVES AROUND THE SURFACE OF A RIGHT CIRCULAR
CYLINDER / CONE AND AT THE SAME TIME ADVANCES IN AXIAL DIRECTION
AT A SPEED BEARING A CONSTANT RATIO TO THE SPPED OF ROTATION.
( for problems refer topic Development of surfaces)

3. Problem: Draw involute of a square of 25 mm sides
75
C 0
B 10
50
D 25 A
25 100

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

5. Problem no 23: Draw Involute of a circle of 40 mm diameter. INVOLUTE OF A CIRCLE
Also draw normal and tangent to it at a point 100 mm from the
centre of the circle.
Solution Steps:
1) Point or end P of string AP is
P3
exactly πD distance away from A. Tange
Means if this string is wound round
P4
nt
the circle, it will completely cover
given circle. B will meet A after P2
winding.
2) Divide πD (AP) distance into 12
number of equal parts.
3) Divide circle also into 12 number
of equal parts. P5
al
4) Name after A, 1, 2, 3, 4, etc. up
Norm
to 12 on πD line AP as well as on
circle (in anticlockwise direction). P1
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’ 6’
on circle (means one division less 7’ 5’
than distance AP). 8’
4’
7) Name this point P1
8) Take 2-P distance in compass 9’ c 3’
and mark it on the tangent from
10’ 2’
P
point 2’. Name it point P2.7
9) Similarly take 3 to P, 4 to P, 5 to 11’ 1’
12’ P
P up to 11 to P distance in compass A 1 2 3 4 5 6 7 8 9 10 11 12
and mark on respective tangents P11
P
and locate P3, P4, P5 up to P12 (i.e. A) 8 πD
P10
points and join them in smooth P9
curve it is an INVOLUTE of a given
circle.

6. 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
al
MARK POINT OF INTERSECTION OF
rm
No
THIS SEMICIRCLE AND POLE CIRCLE
AND JOIN IT TO Q. Q
THIS WILL BE NORMAL TO INVOLUTE.
Ta
n ge
DRAW A LINE AT RIGHT ANGLE TO nt
THIS LINE FROM Q.
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

7. 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 t
p7
n
ge
an p8
8 4 p4 T
p9
C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12
9 C p3 3
No
40mm
p2 p10
rm
CP
al
10 2
p1 p11
11 1 p12
12 P 1’ 2’ 3’ 4’ 5’ 6’ 7’ 8’ 9’ 10’ 11’ 12’ Q
πD
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. 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
N o rm
IT WILL BE TANGENT TO CYCLOID.
CYCLOID
Q
Tang
ent
CP
C C1 C2 C3 C4 C5 C6 C7 C8
P N
πD

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.
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. c9 c10
c8
3) Construct angle θ with radius OC c7 c11
and draw an arc by taking O as c12
center OC as radius and form sector c6
of angle θ.
4) Divide this sector into 12 c5 8 9 10
number of equal angular parts. And Rolling circle or 7 11
from C onward name them C1, C2, generating c4 6 12
C3 up to C12. circle 5
5) Divide smaller circle (Generating
c3
circle) also in 12 number of equal 4
Tangent
Normal
parts. And next to P in anticlockw- Directing circle
ise direction name those 1, 2, 3, up 3
to 12. c2
6) With O as center, O-1 as radius 2
3’
draw an arc in the sector. Take O-2, 4’ 2’
O-3, O-4, O-5 up to O-12 distances c1
1
with center O, draw all concentric 5’ 1’
arcs in sector. Take fixed distance
C-P in compass, C1 center, cut arc θ
6’
of 1 at P1. C P
12’
Repeat procedure and locate P2, P3, O
P4, P5 unto P12 (as in cycloid) and 7’ 11’ OP=Radius of directing circle=75mm
join them by smooth curve. This is PC=Radius of generating circle=25mm
EPI – CYCLOID.
8’ 10’ θ=r/R X360º= 25/75 X360º=120º
9’

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
C O
Rolling circle or P θ=103º
generating
circle

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
9 10
anticlockwise to move 7
8
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
c5
on the smaller circle.
3) From next to P in c4
2
clockwise direction, name 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
Rolling circle or 10’
9’
8’
generating circle OP=Radius of directing circle=75mm
PC=Radius of generating circle=25mm
θ=r/R X360º= 25/75 X360º=120º

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 SPIRAL
by point P.
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
2’
With OP radius & O as center draw a
circle and divide it in EIGHT P2
parts. Name those 1’,2’,3’,4’, etc. 3’ 1’
P1
Similarly 8’
up to divided line PO also in
EIGHT parts and name those P3
1,2,3, starting from P.
Take O-1 distance from OP line and
draw an arc up to O1’ radius 4’ P4 O P
7 6 5 4 3 2 1
vector. Name the point P1 P7
Similarly mark points P2, P3, P4 up to P5 P6
P8
And join those in a smooth curve. It is
a SPIRAL of one convolution. 5’ 7’
6’

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
1. With PO & QO radii draw two
4’ 2’
P3
circles and divide them in P2
twelve equal parts. Name those P4
1’,2’,3’,4’, etc. up to 12’
2 .Similarly divided line PQ also in 5’ P1 1’
twelve parts and name those
P5
1,2,3,-- as shown.
Tan
3. Take O-1 distance from OP line
N
g
and draw an arc up to O1’ radius
en t
al
P6 Norm c Q P
vector. Name the point P1 6’
4. Similarly mark points P2, P3, P4 O 12 11 10 9 8 7 6 5 4 3 2 1
P11
12’
up to P12 P7 P10
And join those in a smooth curve. P8 P9
It is a SPIRAL of one convolution.
7’ 11’
C=(Rmax-Rmin)/No. of
convolutions in radians
= (115-15)/3.14 X 2 =15.92 8’ 10’
9’

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’ 2’14’
P3
2 Mark Q at 15 mm from O P2
P4
3 with O as centre draw two circles with OP
and OQ radius
4 Divide the circle in 12 equal divisions and
17’ 5’ P1 1’13’
mark the divisions as 1’,2’ and so on up to 18’ P
5
5 Divide the line PQ in 18 equal divis- P15 P14
ions as 1,2,3 and so on upto 18 P16 P13
6.Take O-1 distance from OP line and P17
draw an arc up to O1’ radius vector. P6 P12
Name the point P1 18’ 6’ Q P
7.Similarly mark points P2, P3, P4 up
P18
O
18 16 14 12 10 8 6 4 2 12’
to P18. P11
8. And join those in a smooth curve. P7
It is a SPIRAL of one and half P10
convolution. P8
P9
7’ 11’
C=(Rmax-Rmin)/No. of
convolutions in radians
= (115-15)/3.14 X3 =10.61
8’ 10’
9’

15. Spiral.
Method of Drawing
Tangent & Normal
SPIRAL (ONE CONVOLUSION.)
2
ent
No
ng
Ta
rm
P2
a
Difference in length of any radius vectors
l
3 1
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.

16. 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

17. 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
1 p2 p4
p3
Solution Steps: 2
Point P- Reaches End A (Downwards)
) Divide OA in EIGHT equal parts and from O to A after O
3
p5 A4
name 1, 2, 3, 4 up to 8. (i.e. up to point A).
4
) Divide 60 angle into four parts (15 each) and mark each
0 0
point by A1, A2, A3, A4 and for return A5, A6, A7 andA8. 5 p6
(Initial A point). A3
6 A5
) Take center O, distance in compass O-1 draw an arc upto
OA1. Name this point as P1.
7 p7 A2
) Similarly O center O-2 distance mark P2 on line O-A2. A6
A8 A1
) This way locate P3, P4, P5, P6, P7 and P8 and join them. p8
A7
A8
( It will be thw desired locus of P )

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
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

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. 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 revolution. A3
2) Divide circle in 8 number of p1
equal parts and name in arrow p2 p6
p7
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. p5
4) Initially P is on end A. When p3
p8
A moves to A1, point P goes
one linear division (part) away A B A4
P 1 4 5 6 7
from A1. Mark it from A1 and 2 3 p4
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 A7
A5
parts away from A8. [Means P
has reached B].
8) Join all P points by smooth
curve. It will be locus of P A6

20. 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 linear P 1+7 2+6 p + 5
3 4 +B
division (part) away from A1. Mark it p8 6
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 p7 p3
away from P3.
7) Similarly locate P4, P5, P6, P7
and P8 which will be eight parts away
A7
from A8. [Means P has reached B]. A5
8) Join all P points by smooth curve.
It will be locus of P
The Locus will
follow the loop path two times in A6
one revolution.

21. Problem 28: A link OA, 100 mm long rotates about O
in anti-clockwise direction. A point P on the link, 15 θ= 2/5 X 360º = 144º
mm away from O, moves and reaches the end A, Total angular movement = 144º
while the link has rotated through 2/5 of a revolution.
Total linear movement = 85 mm
Assuming that the movements of the link to be
uniform trace the path of point P. To divide both of them in equal
no. of parts ( say 8)
5’
6’ 4’
7’ 3’
P6
P7 P5
8’ 2’
P8 P4
P3
1’
P2
144º
P1
O P 1 2 3 4 5 6 7 8A
15
100

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 P A
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 OA1 ,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. B
P12
Draw two straight lines OA & OB at angle of 30º. P11
P10
Mark a point P on OA at 40 mm from O. P8
P9
P7
Calculate OP1 such that OP1/OP = 9/8. => OP1 = 45 mm P
P5 6
P4
Mark a OP1 on OB at 45 mm from O. P2P3
P1
Join P with P1. 45
P 4 P 3
Draw an arc of radius OP1 from OB to OA. P2 O 30º A
P
Draw a line parallel to PP1 from P1 on OA to intersect OB at P2.
5 P P1P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 P12
P
40
Repeat the steps to get the points P3,P4 and so on up to P12.
1
P6
P P12
P7
P11
P8
P9 P10