digital natives_1.pdf

7/24/2019 digital natives_1.pdf

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INVOLUTE CYCLOID SPIRAL HELIX

ENGINEERING CURVESPart-II

(Point undergoing to t!"e# o$ di#"%a&e'ent#

)* In+o%ute o$ a &ir&%e

aString Lengt, D

.String Lengt, / D

&String Lengt, 0 D

1* Po%e ,a+ing Co'"o#ite

#,a"e*

2* Rod Ro%%ing o+er

a Se'i&ir&u%ar Po%e*

)* Genera% C!&%oid

1* Tro&,oid

( #u"erior2* Tro&,oid

( In$erior

3* E"i-C!&%oid

4* H!"o-C!&%oid

)* S"ira% o$

One Con+o%ution*

1* S"ira% o$To Con+o%ution#*

)* On C!%inder

1* On a Cone

5et,od# o$ Draing

Tangent# 6 Nor'a%#

To T,e#e Cur+e#*

AND


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CYCLOID:IT IS A LOCUS OF A POINT ON THEPERIPHERY OF A CIRCLE WHICHROLLS ON A STRAIGHT LINE PATH.

INVOLUTE:IT IS A LOCUS OF A FREE END OF A STRINGWHEN IT IS WOUND ROUND A CIRCULAR POLE

SPIRAL:IT IS A CURVE GENERATED BY A POINTWHICH REVOLVES AROUND A FIXED POINTAND AT THE SAME MOVES TOWARDS IT.

HELIX:IT IS A CURVE GENERATED BY A POINT WHICHMOVES AROUND THE SURFACE OF A RIGHT CIRCULARCYLINDER / CONE AND AT THE SAME TIME ADVANCES IN AXIAL DIRECTION

AT A SPEED BEARING A CONSTANT RATIO TO THE SPPED OF ROTATION.( for prob!"# r!f!r $op%& D!'!op"!$ of #)rf*&!#+

DEFINITIONS

SUPERIORTROCHOID:IF THE POINT IN THE DEFINATIONOF CYCLOID IS OUTSIDE THECIRCLE

INFERIOR TROCHOID.:IF IT IS INSIDE THE CIRCLE

EPI-CYCLOIDIF THE CIRCLE IS ROLLING ONANOTHER CIRCLE FROM OUTSIDE

HYPO-CYCLOID.IF THE CIRCLE IS ROLLING FROMINSIDE THE OTHER CIRCLE,


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INVOLUTE O7 A CIRCLEProblem no 17:Draw Involute of a circle.

String length is equal to the circumference of circle.

1 2 3 4 5 6 7 8P

P8

1

2

34

5

6

78

P3

3top

P44 to p

P5

5to

p

P7

7top

P6

6

to

p

P2

2

to

p

P1

1top

D

A

Solution Steps:1) Point or end P of string AP isexactly D distance away fromA. Means if this string is wound

round the circle, it willcompletely cover given circle. Bwill meet A after winding.) Divide D !AP) distance into" num#er of e$ual parts.%) Divide circle also into "num#er of e$ual parts.') (ame after A, 1, , %, ', etc.up to " on D line AP as well ason circle !in anticlocwisedirection).*) +o radius -1, -, -% up to-" draw tangents !from1,,%,',etc to circle).) +ae distance 1 to P incompass and mar it on tangentfrom point 1 on circle !means

one division less than distanceAP)./) (ame this point P1") +ae -B distance incompass and mar it on thetangent from point . (ame itpoint P.0) imilarly tae % to P, ' to P,* to P up to / to P distance in

compass and mar onrespective tangents and locateP%, P', P* up to P" !i.e. A)


4/32

INVOLUTE O7 A CIRCLE

String %engt, 5ORE t,an D

1 2 3 4 5 6 7 8P

1

2

34

5

6

7 8

P3

3top

P44 to p

P5

5to

p

P7

7top

P6

6

to

p

P2

2

to

p

P1

1to

p

165 mm(more than D)

D

p8

Solution Steps:

In this case string length is more

than D. 8ut re'e'.er!Whateer ma "e the length o#

string$ mar% D &istance

hori'ontal i.e.along the string

an& &ii&e it in 8 nm"er o#

eal parts$ an& not an other

&istance. *est all steps are same

as preios I+,-/0. Dra

the cre completel.

Pro.%e' )9: Draw Involute of a circle.String length is MORE than the circumference of circle.


5/32

1 2 3 4 5 6 7 8

P

1

2

34

5

6

7 8

P3

3top

P44 to p

P5

5to

p

P7

7top

P6

6

to

p

P2

2

to

p

P1

1to

p

15 mm(ess than D)

D

INVOLUTE O7 A CIRCLE

String %engt, LESS t,an D

Pro.%e' );: Draw Involute of a circle.String length is LESS than the circumference of circle.

Solution Steps:

In this case string length is ess

than D. 8ut re'e'.er!Whateer ma "e the length o#

string$ mar% D &istance

hori'ontal i.e.along the string

an& &ii&e it in 8 nm"er o#

eal parts$ an& not an other

&istance. *est all steps are same

as preios I+,-/0. Dra

the cre completel.


6/32

12

34

5

6

1 2 3 4 5 6

P

D2

P1

1to

P

P2

2toP

P33 to P

P4

4to

P

P

4t

oP

P5

5

to

P

P6

6to

P

INVOLUTE

O7

CO5POSIT SHAPED POLE

PRO8LE5 1< : P- I - P - 9:-+ +D ;I


7/32

)

1

2

3

D

)

1

2

3

A

8

1

:1

2 :2

3

:3

4

:4

PRO8LE5 1) : Rod "0 1+ mm long rolls

over a semicircular pole without slipping

from its initiall2 vertical position till it

/ecomes up3side3down vertical.

Draw locus of /oth ends " 0.

So%ution Ste"#=

I# o hae st&ie& preios pro"lems

properl$ o can srel sole this also.

impl remem"erthat this "eing a ro&$

it ill roll oer the sr#ace o# pole.

;eans hen one en& is approaching$

other en& ill moe aa #rom poll.

OBSERVE ILLUSTRATION CAREFULLY!


8/32

P


9/32


10/32

P


11/32

C

C)C1 C2 C

3

C4

C9

C@

CA

EPI CYCLOID :

P

O

R

rB CP

Cr*

36

1

2

3

4 5

6

7

GeneratingB

Ro%%ing Cir&%e

Dire&ting Cir&%e

PRO8LE5 14:D*W -


12/32

OP

OP=Rad" o# dre$%n& $r$le=7'mm

C

PC=Rad" o# &enera%n& $r$le=('mm

7

)=r*R +,-./= ('*7' +,-./=1(./

'

)

%

4

+

,

5

1 6 '&''

')

'

)%4

+

,

5

1

6

'&

''

')

$1

$(

$,

$0

$'

$-

$7

$$2 $1.

$11

$1(


13/32

HYPO CYCLOID

C

P)

P1

P2

P3

P4 P@

P

P9

P

)

1

2

@

4

3

C) C1 C2

C3

C4

C@

CA

C9

O

OC R ( Radiu# o$ Dire&ting Cir&%e

CP r (Radiu# o$ Generating Cir&%e

Cr

*36

PRO8LE5 1@:D*W -


14/32

OP

OP=Rad" o# dre$%n& $r$le=7'mm

C

PC=Rad" o# &enera%n& $r$le=('mm

7

)=r*R +,-./= ('*7' +,-./=1(./

'

)

%

4

+

,

51 6 '&

''

')

$(

$1

$,

$0

$'

$-

$7$ $2 $1. $11 $1(

'

)%

4

+

,

5

1

6'&

''

')


15/32

7 6 5 4 3 2 1P

1

2

3

4

5

6

7

P2

P6

P1

P3

P5

P7

P4 -

SPIRALProblem (7: 3ra4 a "5ral o# one $on6ol%on Ta8e d"%an$e PO 0. mm

Sol%on S%e5"1. With P- ra&is &ra a circle

an& &ii&e it in I=0 parts. +ame those 1$2$3$4$ etc. p to 8

2 .imilarl &ii&e& line P- also in

I=0 parts an& name those

1$2$3$ as shon.

3. 0a%e o1 &istance #rom op line

an& &ra an arc p to -1 ra&is ector. +ame the point P14. imilarl mar% points P2$ P3$ P4

p to P8 n& Eoin those in a smooth cre.

It is a PI* o# one conoltion.

I9PORTANT APPROAC FOR CONSTRUCTION!

FIN3 TOTAL AN;ULAR AN3 TOTAL LINEAR 3ISPLACE9ENT

AN3 3IVI3E BOT IN TO SA9E NU9BER OF E


16/32

16 13 1 8 7 6 5 4 3 2 1 P

1$A

2$1

3$11

4$12

5$13

6$14

7$15

8$16

P1

P2

P3

P4

P5

P6

P7

P8

PAP1

P11

P12

P13 P14

P15

SPIRALof

$1o &o'o)$%o#

Pro.%e' 19Point P is 8 mm #rom point -. It starts moing toar&s - an& reaches it in to

reoltions aron&.it Dra locs o# point P (0o &ra a piral o# 0W- conoltions).

I9PORTANT APPROAC FOR CONSTRUCTION!

FIN3 TOTAL AN;ULAR AN3 TOTAL LINEAR 3ISPLACE9ENT

AN3 3IVI3E BOT IN TO SA9E NU9BER OF E


17/32

1

2

3

4

5

6

7

8

P

P1P

P2

P3

P4

P5

P6

P7

P8

1

2

3

4

5

6

7

HELIX

UPON A CYLIN3ER)

PROBLE98 Draw a heli of one convolution* upon a c2linder.

9iven 1& mm pitch and +& mm diameter of a c2linder.

:;he aial advance during one complete revolution is called

;hepitchof the heli %ra$ed o% b? a 5on% mo6n& n a 5lane@

n a 5ar%$lar manner@ #or one $?$le o# o5era%on

T>e $a"e" are $la""#ed n TREE $a%e&ore" #or ea"? nder"%andn&

A Ba"$ Lo$" Ca"e"

B O"$lla%n& Ln8

C Ro%a%n& Ln8

Ba"$ Lo$" Ca"e":ere "ome &eome%r$al obe$%" l8e 5on%@ lne@ $r$le 4ll be de"$rbed 4%> %>ere rela%6e

Po"%on" T>en one 5on% 4ll be allo4ed %o mo6e n a 5lane man%ann& "5e$#$ rela%on

4%> abo6e obe$%" And "%d?n& "%a%on $are#ll? ?o 4ll be a"8ed %o dra4 %D" lo$"O"$lla%n& Ro%a%n& Ln8:ere a ln8 o"$lla%n& #rom one end or ro%a%n& arond %D" $en%er 4ll be de"$rbed

T>en a 5on% 4ll be allo4ed %o "lde alon& %>e ln8 n "5e$#$ manner And no4 "%d?n&

%>e "%a%on $are#ll? ?o 4ll be a"8ed %o dra4 %D" lo$"

STUDY TEN C SES GIVEN ON NEXT P GES


23/32

:

"4 3 2 1

71 2 3 4

SOLUTION STEPS:

1.ocate center o# line$ perpen&iclar to

: #rom point . 0his ill "e initial

point P.

2.;ar% 5 mm &istance to its right si&e$

name those points 1$2$3$4 an& #rom those

&ra lines parallel to :.

3.;ar% 5 mm &istance to its le#t o# P an&name it 1.

4.0a%e 1 &istance as ra&is an& as

center &ra an arc

ctting #irst parallel line to :. +ame

pper point P1an& loer point P

2.

5.imilarl repeat this process " ta%ing

again 5mm to right an& le#t an& locateP3P

4.

6.Foin all these points in smooth cre.

It i%% .e t,e %o&u# o$ P euidi#tan&e

$ro' %ine A8 and $iFed "oint 7*

P1

P2

P3

P4

P5

P6

P7

P8

PRO8LE5 )*:Point is 5 mm #rom a ertical straight line :.

Dra locs o# point P$ moing in a plane sch that

it alas remains ei&istant #rom point an& line :.

Ba"$ Lo$" Ca"e":


24/32

:

"4 3 2 1 1 2 3 4

P1

P2

P3

P4

P5

P6

P7

P8

C

SOLUTION STEPS:

1.ocate center o# line$ perpen&iclar to

: #rom the peripher o# circle. 0his

ill "e initial point P.

2.;ar% 5 mm &istance to its right si&e$

name those points 1$2$3$4 an& #rom those

&ra lines parallel to :.

3.;ar% 5 mm &istance to its le#t o# P an&name it 1$2$3$4.

4.0a%e


25/32

A5 mm

3 D

6 D

"4 3 2 1 1 2 3 4

C1


26/32

1C


27/32

P :

4 3 2 1 1 2 3 4

7 mm 3 mm

p1

p2

p3

p4

p5

p6

p7

p8

Pro.%e' 4:-0o points an& : are 1 mm apart.

0here is a point P$ moing in a plane sch that the

&i##erence o# itHs &istances #rom an& : alas

remains constant an& eals to 4 mm.

Dra locs o# point P.

Ba"$ Lo$" Ca"e":

So%ution Ste"#1.ocate M : points 1 mm apart.

2.ocate point P on : line$

7 mm #rom an& 3 mm #rom :

s PP:B4 ( : B 1 mm )3.-n "oth si&es o# P mar% points 5

mm apart. +ame those 1$2$3$4 as sal.

4.+o similar to steps o# Pro"lem 2$

Dra &i##erent arcs ta%ing M : centers

an& 1$ :1$ 2$ :2 etc as ra&is.

5. ;ar% arios positions o# p i.e. an& Eoin

them in smooth possi"le cre.

It i%% .e %o&u# o$ P


28/32

1) ;ar% loer most

position o# ; on e@tension

o# : (&onar&) " ta%ing

&istance ;+ (4 mm) #rom

point : ("ecase + cannot go "eon& : ).

2) Dii&e line (; initial

an& ; loer most ) into

eight to ten parts an& mar%

them ;1$ ;

2$ ;

3p to the

last position o# ; .

3) +o ta%e ;+ (4 mm)

as #i@e& &istance in compass$

;1center ct line


29/32

1

2

3

4

5

6

7

8

p

p1

p2p3

p4

p5

p6

p7

p8

-

1

2

3

4

5

6

7

8

Pro.%e' No*:

in% OA$ 8 mm long oscillates aron& O$

6to right si&e an& retrns to itHs initial ertical

Position ith ni#orm elocit.;ean hile point

P initiall on Ostarts sli&ing &onar&s an&

reaches en& Aith ni#orm elocit.

Dra locs o# point P

Solution Steps:Point P- Reaches End A (Downwards)1) Dii&e - in I=0 eal parts an& #rom - to a#ter -

name 1$ 2$ 3$ 4 p to 8. (i.e. p to point ).

2) Dii&e 6angle into #or parts (15each) an& mar% each

point " 1$

2$

3$

4an& #or retrn

5$

6$

7an&

8.

(Initial point).3) 0a%e center -$ &istance in compass -1 &ra an arc pto

-1. +ame this point as P

1.

1) imilarl - center -2 &istance mar% P2on line -

2.

2) 0his a locate P3$ P

4$ P

5$ P

6$ P

7an& P

8an& Eoin them.

( It ill "e th &esire& locs o# P )

OSCILLATING LINK


30/32

p

1

2

3

4

5

6

7

8

A

1

11

12

13

14

15

16-

Pro.%e' No 9:

in% OA$ 8 mm long oscillates aron& O$

6

to right si&e$ 12

to le#t an& retrns to itHs initialertical Position ith ni#orm elocit.;ean hile point

P initiall on Ostarts sli&ing &onar&s$ reaches en& A

an& retrns to Oagain ith ni#orm elocit.

Dra locs o# point P

So%ution Ste"#( P reaches i.e. moing &onar&s.

M retrns to - again i.e.moes par&s )

).ere &istance traele& " point P is P.pls

P.ence &ii&e it into eight eal parts.( so

total linear &isplacement gets &ii&e& in 16

parts) +ame those as shon.

1.in% - goes 6to right$ comes "ac% to

original (,ertical) position$ goes 6to le#t an&

retrns to original ertical position. encetotal anglar &isplacement is 24.

Dii&e this also in 16 parts. (15each.)

+ame as per preios pro"lem.($ 12etc)

2.;ar% &i##erent positions o# P as per the

proce&re a&opte& in preios case.

an& complete the pro"lem.

2

1

3

4

5

6

78

A

1

11

12

13

14

15

16

p8

p5

p6

p7

p2p4

p1p3

OSCILLATING LINK

ROTATING LINK


31/32

A 8

1

2

4

5

3

6

7

P

p1 p2

p3

p4

p5

p6 p7

p8

1 2 34 5 6 7

Problem 9:

*o& :$ 1 mm long$ reoles in cloc%ise &irection #or one reoltion.

;eanhile point P$ initiall on starts moing toar&s : an& reaches :.


ROTATING LINK

1) AB >od revolves around

center 5 for one revolutionand point P slides along ABrod and reaches end B inone revolution.) Divide circle in " num#erof e$ual parts and name inarrow direction after A-A1,A, A%, up to A".%) Distance traveled #y

point P is AB mm. Dividethis also into " num#er ofe$ual parts.') 3nitially P is on end A.


32/32

A 8

1

2

4

5

3

6

7

P

p1

p2

p3

p4

p5

p6

p7

p81 2 3 4567

Problem 10 :

*o& :$ 1 mm long$ reoles in cloc%ise &irection #or one reoltion.

;eanhile point P$ initiall on starts moing toar&s :$ reaches :

n& retrns to in one reoltion o# ro&.


Solution Steps

ROTATINGLINK

1) AB >od revolves aroundcenter 5 for one revolution andpoint P slides along rod ABreaches end B and returns to A.) Divide circle in " num#er ofe$ual parts and name in arrowdirection after A-A1, A, A%, up toA".

%) Distance traveled #y point Pis AB plus AB mm. Divide AB in 'parts so those will #e " e$ualparts on return.') 3nitially P is on end A.

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