This document discusses various types of engineering curves defined by the motion of a point or object along a path. It provides definitions and examples of involutes, cycloids, trochoids, spirals and helices. Methods for drawing tangents and normals to these curves are also mentioned. Specific problems are given to illustrate how to draw different types of curves step-by-step, including involutes with various string lengths, composite pole shapes, loci of rod ends rolling on a semicircular pole, standard and superior/inferior cycloids and trochoids, and epicycloids and hypocycloids defined by a smaller circle rolling on the outside or inside of a larger curved path, respectively.

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 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 P2
given circle. B will meet A after
winding.
2) Divide πD (AP) distance into 8 P3
number of equal parts. P1
3) Divide circle also into 8 number
2 to p
of equal parts.
3
4) Name after A, 1, 2, 3, 4, etc. up to
p
to 8 on πD line AP as well as on
p
circle (in anticlockwise direction).
o
1t
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
than distance AP). 6
p
o
7) Name this point P1
5t
1
8) Take 2-B distance in compass 7 A 8
6 to p
and mark it on the tangent from 7
to P
point 2. Name it point P2.
P5 p P8 1 2 3 4 5 6 7 8
9) Similarly take 3 to P, 4 to P, 5 to P7
P up to 7 to P distance in compass P6 π
and mark on respective tangents
and locate P3, P4, P5 up to P8 (i.e. D
A) points and join them in smooth
curve it is an INVOLUTE of a given
circle.

4. 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: P2
In this case string length is more
than Π D.
But remember!
Whatever may be the length of P3 P1
string, mark Π D distance
2 to p
horizontal i.e.along the string
and divide it in 8 number of 3
to
equal parts, and not any other p
p
distance. Rest all steps are same
o
1t
as previous INVOLUTE. Draw
the curve completely.
4 to p
P4 4
3
5
2
p
o
5t
6
1
P5 7
8
7 p8 1 P
6 to p
to
p
2 3 4 5 6 7 8
P7
165 mm
P6 (more than πD)
πD

5. 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: P2
In this case string length is Less
than Π D.
But remember!
Whatever may be the length of P3
P1
string, mark Π D distance
horizontal i.e.along the string
and divide it in 8 number of
2 to p
3
to
equal parts, and not any other p
distance. Rest all steps are same
as previous INVOLUTE. Draw
op
1t
the curve completely.
4 to p
P4 4
3
5
2
p
o
6
5t
1
6 to p
P5
7
to 7 P
p 8
P7 1 2 3 4 5 6 7 8
P6
150 mm
(Less than πD)
πD

6. PROBLEM 20 : A POLE IS OF A SHAPE OF HALF HEXABON AND SEMICIRCLE.
ASTRING IS TO BE WOUND HAVING LENGTH EQUAL TO THE POLE PERIMETER
INVOLUTE
DRAW PATH OF FREE END P OF STRING WHEN WOUND COMPLETELY. OF
(Take hex 30 mm sides and semicircle of 60 mm diameter.) COMPOSIT SHAPED POLE
SOLUTION STEPS:
Draw pole shape as per
dimensions. P1
Divide semicircle in 4
parts and name those
P
along with corners of
P2
hexagon.
Calculate perimeter
length.
1 to P
Show it as string AP.
On this line mark 30mm
2
to
from A
P
oP
Mark and name it 1
At
Mark πD/2 distance on it
from 1
And dividing it in 4 parts P3
name 2,3,4,5. 3 to P 3
Mark point 6 on line 30 4
2
mm from 5
Now draw tangents from
5 1
all points of pole
oP
and proper lengths as
A
4t
done in all previous 6
5 to P
involute’s problems and 1 2 3 4 5 6 P
6t
oP
complete the curve. πD/2
P4
P6
P5

7. 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

8. CYCLOID
PROBLEM 22: DRAW LOCUS OF A POINT ON THE PERIPHERY OF A CIRCLE
WHICH ROLLS ON STRAIGHT LINE PATH. Take Circle diameter as 50 mm
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 clock wise 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 upto C12 as centers. Mark points P2, P3, P4, P5 up to P8 on the
horizontal lines drawn from 1,2, 3, 4, 5, 6, 7 respectively.
7) Join all these points by curve. It is Cycloid.

9. PROBLEM 23: DRAW LOCUS OF A POINT , 5 MM AWAY FROM THE PERIPHERY OF A SUPERIOR TROCHOID
CIRCLE WHICH ROLLS ON STRAIGHT LINE PATH. Take Circle diameter as 50 mm
4 p4
p3 p5
3 5
p2 C C1 C C3 C4 C5 C6 C7 C8 p 6
2 6 2
p7
1 p1 7
P πD p8
Solution Steps:
1) Draw circle of given diameter and draw a horizontal line from it’s center C of length Π D and divide it
in 8 number of equal parts and name them C1, C2, C3, up to C8.
2) Draw circle by CP radius, as in this case CP is larger than radius of circle.
3) Now repeat steps as per the previous problem of cycloid, by dividing this new circle into 8 number of
equal parts and drawing lines from all these points parallel to locus of C and taking CP radius wit
different positions of C as centers, cut these lines and get different positions of P and join
4) This curve is called Superior Trochoid.

10. PROBLEM 24: DRAW LOCUS OF A POINT , 5 MM INSIDE THE PERIPHERY OF A
INFERIOR TROCHOID
CIRCLE WHICH ROLLS ON STRAIGHT LINE PATH. Take Circle diameter as 50 mm
p4
4
p3 p5
3 5
p2
C C1 C2 C3 C4 C5 C6 C7 p6 C8
2 6
p1 p7
1 7
P p8
πD
Solution Steps:
1) Draw circle of given diameter and draw a horizontal line from it’s center C of length Π D and divide it
in 8 number of equal parts and name them C1, C2, C3, up to C8.
2) Draw circle by CP radius, as in this case CP is SHORTER than radius of circle.
3) Now repeat steps as per the previous problem of cycloid, by dividing this new circle into 8 number
of equal parts and drawing lines from all these points parallel to locus of C and taking CP radius
with different positions of C as centers, cut these lines and get different positions of P and join
those in curvature.
4) This curve is called Inferior Trochoid.

11. 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 EPI CYCLOID :
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 Π D distance on arc and it will be
decided by included arc angle θ.
2) Calculate θ by formula θ = (r/R) x
Generating/
3600. Rolling Circle
3) Construct angle θ with radius OC 4 5
and draw an arc by taking O as center C2 C3
C1 C4
OC as radius and form sector of angle 3 6
θ. C C
5
4) Divide this sector into 8 number of 7
equal angular parts. And from C 2
C6
onward name them C1, C2, C3 up to
C8.
1 P
r = CP
C7
5) Divide smaller circle (Generating
circle) also in 8 number of equal parts.
And next to P in clockwise direction Directing Circle
name those 1, 2, 3, up to 8. R C
8
6) With O as center, O-1 as radius
draw an arc in the sector. Take O-2, O- = r 3600
+
R
3, O-4, O-5 up to O-8 distances with
center O, draw all concentric arcs in O
sector. Take fixed distance C-P in
compass, C1 center, cut arc of 1 at P1.
Repeat procedure and locate P2, P3,
P4, P5 unto P8 (as in cycloid) and join
them by smooth curve. This is EPI –
CYCLOID.

12. c8 c9 c10
c7 c11
c12
c6
c5
8 9 10
7 11
c4 6 12
5
c3
4
3
c2
2
4’ 3’ 2’
c1
5’ 1
1’
θ
6’
C P
12’
O
7’ 11’
OP=Radius of directing circle=75mm
8’ 10’
9’ PC=Radius of generating circle=25mm
θ=r/R X360º= 25/75 X360º=120º

13. 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 HYPO CYCLOID
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
anticlockwise to move P 7
ahead.
2) Same steps should be P1
6
taken as in case of EPI –
CYCLOID. Only change is 1 P2 C2
C1 C3
in numbering direction of 8 C4
number of equal parts on C C
P3 5 5
the smaller circle.
2 C
3) From next to P in 6
anticlockwise direction,
4 P4 C
name 1,2,3,4,5,6,7,8. 3 7
4) Further all steps are P5
P8
that of epi – cycloid. This P6 P7
is called
C8
HYPO – CYCLOID.
r 3600
=
R
+
O
OC = R ( Radius of Directing Circle)
CP = r (Radius of Generating Circle)

14. 8 9 10
7 11
6 12
5
4
c8 c9 c10
c7 c11
c12
c6
3
c5
c4
2 3’ c3
2’ 4’
c2
1 1’
c1 5’
θ
12’ 6’
P C
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

15. 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

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. HELIX
(UPON A CYLINDER)
PROBLEM: Draw a helix of one convolution, upon a cylinder. P8
Given 80 mm pitch and 50 mm diameter of a cylinder. 8
(The axial advance during one complete revolution is called P7
The pitch of the helix) 7
P6
6
P5
SOLUTION: 5
Draw projections of a cylinder.
Divide circle and axis in to same no. of equal parts. ( 8 ) 4 P4
Name those as shown.
3
Mark initial position of point ‘P’ P3
Mark various positions of P as shown in animation. 2 P2
Join all points by smooth possible curve.
Make upper half dotted, as it is going behind the solid 1 P1
and hence will not be seen from front side.
P
6
7 5
P 4
1 3
2

18. HELIX
PROBLEM: Draw a helix of one convolution, upon a cone, 8 P8 (UPON A CONE)
diameter of base 70 mm, axis 90 mm and 90 mm pitch.
(The axial advance during one complete revolution is called 7 P7
The pitch of the helix)
6 P6
P5
SOLUTION: 5
Draw projections of a cone
Divide circle and axis in to same no. of equal parts. ( 8 ) 4 P4
Name those as shown.
Mark initial position of point ‘P’ 3
P3
Mark various positions of P as shown in animation.
Join all points by smooth possible curve. 2
P2
Make upper half dotted, as it is going behind the solid
1
and hence will not be seen from front side. P1
X P Y
6
7 5
P6 P5
P7 P4
P 4
P8
P1 P3
1 3
P2
2

19. 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

20. 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

21. 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.

22. LOCUS
It is a path traced out by a point moving in a plane,
in a particular manner, for one cycle of operation.
The cases are classified in THREE categories for easy understanding.
A} Basic Locus Cases.
B} Oscillating Link……
C} Rotating Link………
Basic Locus Cases:
Here some geometrical objects like point, line, circle will be described with there relative
Positions. Then one point will be allowed to move in a plane maintaining specific relation
with above objects. And studying situation carefully you will be asked to draw it’s locus.
Oscillating & Rotating Link:
Here a link oscillating from one end or rotating around it’s center will be described.
Then a point will be allowed to slide along the link in specific manner. And now studying
the situation carefully you will be asked to draw it’s locus.
STUDY TEN CASES GIVEN ON NEXT PAGES

23. 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.

24. Basic Locus Cases:
PROBLEM 2 :
A circle of 50 mm diameter has it’s center 75 mm from a vertical
line AB.. Draw locus of point P, moving in a plane such that
it always remains equidistant from given circle and line AB. P7
P5
A
SOLUTION STEPS: P3
1.Locate center of line, perpendicular to 50 D
AB from the periphery of circle. 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 1,2,3,4. C
4 3 2 1 1 2 3 4
4.Take C-1 distance as radius and C 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
P3 P4 . P4
6.Join all these points in smooth curve.
B P6
It will be the locus of P equidistance P8
from line AB and given circle.
75 mm

25. PROBLEM 3 : Basic Locus Cases:
Center of a circle of 30 mm diameter is 90 mm away from center of another circle of 60 mm diameter.
Draw locus of point P, moving in a plane such that it always remains equidistant from given two circles.
SOLUTION STEPS:
1.Locate center of line,joining two 60 D
centers but part in between periphery P7
of two circles.Name it P. This will be P5 30 D
initial point P.
P3
2.Mark 5 mm distance to its right
side, name those points 1,2,3,4 and P1
from those draw arcs from C1
As center.
3. Mark 5 mm distance to its right p
side, name those points 1,2,3,4 and C1 C2
4 3 2 1 1 2 3 4
from those draw arcs from C2 As
center. P2
4.Mark various positions of P as per
previous problems and name those P4
similarly. P6
5.Join all these points in smooth
curve. P8
It will be the locus of P
equidistance from given two
circles. 95 mm

26. Basic Locus Cases:
Problem 4:In the given situation there are two circles of
different diameters and one inclined line AB, as shown.
Draw one circle touching these three objects.
60 D
Solution Steps: 30 D
1) Here consider two pairs,
one is a case of two circles
with centres C1 and C2 and
draw locus of point P
equidistance from them. C1
C
(As per solution of case D 1 C2 350
above).
2) Consider second case
that of fixed circle (C1) and
fixed line AB and draw
locus of point P
equidistance from them.
(as per solution of case B
above).
3) Locate the point where
these two loci intersect
each other. Name it x. It
will be the point
equidistance from given
two circles and line AB.
4) Take x as centre and its
perpendicular distance on
AB as radius, draw a circle
which will touch given two
circles and line AB.

27. 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

28. Problem 6:-Two points A and B are 100 mm apart. FORK & SLIDER
There is a point P, moving in a plane such that the
A
difference of it’s distances from A and B always
remains constant and equals to 40 mm. M
Draw locus of point P.
p M1
p1 M2
Solution Steps: C p2
N3 N5 M3
1) Mark lower most N6 p3
position of M on extension N2
of AB (downward) by taking
N4 p4 M4
N1
distance MN (40 mm) from N7 N p5
N8 9
point B (because N can N10 90 0 M5
p 6
not go beyond B ). N N11 M6
2) Divide line (M initial p7 60 0
and M lower most ) into p8
N12
eight to ten parts and mark N13
B M7
them M1, M2, M3 up to the p9
last position of M . M8
3) Now take MN (40 mm) p10
as fixed distance in compass, M9
M1 center cut line CB in N1. p11
4) Mark point P1 on M1N1 M10
with same distance of MP p12
M11
from M1.
5) Similarly locate M2P2, p13
M12
M3P3, M4P4 and join all P
points.
M13
It will be
locus of P.
D

29. 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

30. 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

31. 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

32. 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.