PROJECTION OF LINES.ppseg

8/13/2019 PROJECTION OF LINES.ppseg

1/44

O DRAW PROJECTIONS OF ANY OBJECT,

NE MUST HAVE FOLLOWING INFORMATION) OBJECT

{ WITH ITS DESCRIPTION, WELL DEFINED.}

) OBSERVER{ ALWAYS OBSERVING PERPENDICULAR TO RESP. REF.PLANE}.

) LOCATION OF OBJECT ,{ MEANS ITS POSITION WITH REFFERENCE TO H.P. & V.P.}

TERMS ABOVE & BELOW WITH RESPECTIVE TO H.P.AND TERMS INFRONT & BEHIND WITH RESPECTIVE TO V.P

FORM 4 UADRANTS.OBJECTS CAN BE PLACED IN ANY ONE OF THESE 4 UADRANTS.

IT IS INTERESTING TO LEARN THE EFFECT ON THE POSITIONS OF VIEWS ! FV, TV "OF THE OBJECT WITH RESP. TO #$Y LINE, WHEN PLACED IN DIFFERENT UADRANTS.

ORTHOGRAPHIC PROJECTIONSOF POINTS, LINES, PLANES, AND SOLIDS .

STUDY ILLUSTRATIONS GIVEN ON HEXT PAGES AND NOTE THE RESULTS.TO MAKE IT EASYHERE A POINT A IS TAKEN AS AN OBJECT. BECAUSE ITS ALL VIEWS ARE JUST POINTS.


2/44

NOTATIONS

FOLLOWING NOTATIONS SHOULD BE FOLLOWED WHILE NAMEINGDIFFERENT VIEWS IN ORTHOGRAPHIC PROJECTIONS.

ITS FRONT VIEW % %

SAME SYSTEM OF NOTATIONS SHOULD BE FOLLOWEDINCASE NUMBERS, LIKE 1, 2, 3 ARE USED.

OBJECT POINT A LINE AB

ITS TOP VIEW % %

ITS SIDE VIEW %' %' '


3/44

X

Y

( ST )%*.+ * )%*.

- * )%*. 4 /0 )%*.

X Y

VP

HP

O 12 32

THIS UADRANT PATTERN,IF OBSERVED ALONG #$Y LINE ! IN RED ARROW DIRECTION"WILL E#ACTLY APPEAR AS SHOWN ON RIGHT SIDE AND HENCE,IT IS FURTHER USED TO UNDERSTAND ILLUSTRATION PROPERLLY.


4/44

HP

VPa

a

A

POINT A IN( ST UADRANT

OBSERVER

VP

HP

POINT A IN+ND UADRANT

OBSERVER

a

a

A

OBSERVER

a

a

POINT A IN- RD UADRANT

HP

VP

A

OBSERVER

a

aPOINT A IN4TH UADRANT

HP

VP

A

P 5 / A 51P6%72* I*5882 2 /

9)%* % /1% * 5/1 F3 & T3% 2 ):0/ 51%;2

VP. B)/ %1 T3 51 51 % 352> H >% * ?@ @,

I 76 7 >512*5 27/5 .T02I 8 / = 65 2 % * /02


5/44

FV & TV of a point always lie in the same ve ti!al line

FV of a point "P is ep esente# $y p. %t shows position of the pointwith espe!t to HP.

%f the point lies a$ove HP, p lies a$ove the XY line.

%f the point lies in the HP, p lies on the XY line.

%f the point lies $elow the HP, p lies $elow the XY line.

TV of a point "P is ep esente# $y p. %t shows position of the point withespe!t to VP.

%f the point lies in f ont of VP, p lies $elow the XY line.

%f the point lies in the VP, p lies on the XY line.

%f the point lies $ehin# the VP, p lies $elow the XY line.

B!"#$ $%&$'()" *%+ +!-#& (+%/'$)#%& %* (%#&)


6/44

A

a

a A

a

a

Aa

a

X

Y

X

Y

X

YF o r F v

For Tv

F o r F v

For Tv

For Tv

F o r F v

POINT A ABOVE HP& INFRONT OF VP

POINT A IN HP& INFRONT OF VP

POINT A ABOVE HP& IN VP

PROJECTIONS OF A POINT IN FIRST QUADRANT.

PICTORIALPRESENTATION

PICTORIALPRESENTATION

ORTHOGRAPHIC PRESENTATIONSOF ALL ABOVE CASES.

X Y

a

a

VP

HP

X Y

a

VP

HP

a X Y

a

VP

HP

a

F0 ! %0' ,T0 '4%- .

F0 ! %0' ,T0 %& .

F0 %& ,T0 '4%- .


7/44

SIMPLE CASES OF THE LINE

(. A VERTICAL LINE ! LINE PERPENDICULAR TO HP & TO VP"

+. LINE PARALLEL TO BOTH HP & VP.

-. LINE INCLINED TO HP & PARALLEL TO VP.

4. LINE INCLINED TO VP & PARALLEL TO HP.

. LINE INCLINED TO BOTH HP & VP.

STUDY ILLUSTRATIONS GIVEN ON NEXT PAGESHOWING CLEARLY THE NATURE OF FV & TVOF LINES LISTED ABOVE AND NOTE RESULTS .

PROJECTIONS OF STRAIGHT LINES.

INFORMATION REGARDING A LINE 5'!&" ITS LENGTH,

POSITION OF ITS ENDS WITH HP & VPITS INCLINATIONS WITH HP & VP WILL BE GIVEN.

AIM $ TO DRAW ITS PROJECTIONS $ MEANS FV & TV.


8/44

X

Y

V. P.

X

Y

V. P.

b

a

b

a

F . V .

T . V .

a b

a

b

B

A

TV

FV

A

B

X Y

H.P.

V.P.a

b

a b

Fv

Tv

X Y

H.P.

V.P.

a b

a bFv

Tv

F o F v

Fo Tv

Fo Tv

F o F v

N /2Fv is a ve ti!al linehowin' T (e en'th

&Tv is a point.

N /2Fv & Tv $oth a e

** to +y&

$oth show T. .

1.

2.

A inepe pen#i!(lato Hp

&** to Vp

A ine

** to Hp&** to Vp

O+)6% +!(6#$ P!))'+&

O+)6% +!(6#$ P!))'+&

!P57/ 5%6 P 212 /%/5 "

!P57/ 5%6 P 212 /%/5 "


9/44

A Line inclined to Hp

andparallel to Vp

(Pictorial presentation) X

Y

V. P.

A

B

b

a

b

a

F . V .

T . V .

A Line inclined to Vp

andparallel to Hp(Pictorial presentation)

V. P.

a b

a

b

BA

F . V .

T.V .

X Y

H.P.

V.P.

F . V .

T.V.a b

a

b

X Y

H.P.

V.P.

a

b

a b

Tv

Fv

Tv inclined to xyFv parallel to xy.

3.

4.

Fv inclined to xyTv parallel to xy.

O /0 : %


10/44

X

Y

V. P.

F o r F v a

b

a b

B

A

For Tv

F . V .

T.V.

X

Y

V. P.

a

b

a b

F . V .

T.V.

F o r F v

For Tv

B

A

X Y

H.P.

V.P.

a

b

FV

TV

a

b

A Line inclined to both Hp and Vp

(Pictorial presentation)

5.

N /2 T0212 F%7/1 $B /0 F3 & T3 % 2 5 765 2* / =.

-o view is pa allel to +y)B /0 F3 & T3 % 2 2*)72*

62 :/01 .-o view shows T (e en'th)

O /0 : %


11/44

X Y

H.P.

V.P.

X Y

H.P.

V.P.

a

b

TV

a

b

FV

TV

b 1

b1

TL

X Y

H.P.

V.P.

a

b

FV

TV

a

b

H'+' TV 9! : #" &%) ;; )% XY 4#&'H'&$' #)" $%++'"(%& #& FV

! #" &%) "6%-#&T+ ,

H > / 85 * T )2 L2 :/0.!V52>1 % 2 /%/2* / *2/2 ;5 2T )2 L2 :/0 & 5/1 5 765 %/5 1

>5/0 H< & V,>5/0 /025 %


12/44

The most impo tant #ia' am showin' ' aphi!al elationsamon' all impo tant pa amete s of this topi!.

t(#y an# memo i e it as a CIRCUIT DIAGRAM An# (se in solvin' va io(s p o$lems.

T )2 L2 :/0 51 232 /%/2*. I/1 0 5 /%6 7 ;< 2 /51 * %> & 5/ 51 8) /02 /%/2* / 6 7%/2 352>.

V52>1 % 2 %6>%=1 /%/2*, ;%*2 0 5 /%6 & 8) /02

2 /2 *2* / 6 7%/2 TL, &

A4"% R'5'5 '+

I;< /% / TEN 5/0 N /%/5 1

)12* 02 2 >% *

/) T (e en'th T ) 0 a $ / & a $

1) An'le of T with Hp 2

3) An'le of T with Vp 0

4) An'le of FV with +y 0

5) An'le of TV with +y 0

6) TV len'th of FV) 0 Component !%$("7) FV len'th of TV) 0 Component !%$("

8) Position of A2 D51/% 721 8 % & % 8 ; =

9) Position of B2 D51/% 721 8 & 8 ; =

/:) ;istan!e $etween


13/44

a

b

a

b

X Y

b 1

b1

GROUP (A)GENERAL CASES OF THE LINE INCLINED TO BOTH HP !P

( "ase# on $% para&eters)'PROBLEM ("ine AB is 75 mm lon' an# it is 3: : &

4: : %n!line# to Hp & Vp espe!tively.(a# ant.

SOLUTION STEPS?(" D %> = 65 2 % * 2 < 27/ .+" L 7%/2 % (+;; % 32 = 65 2

& % (@;; 26 > = 65 2.-" T% 2 -@ @ % :62 8 ; % & 4@ @ 8 ; % % * ;% TL I.2. ;; /0 65 21. N%;2 /0 12 < 5 /1 ( % * ( 21 0 5 /%6 65 21 !L 7)1" 8 ;

/0 < 5 /1." D %> 0 5 /%6 7 ;< 2 / 8 TL

% ( 8 ; < 5 / ( % * %;2 5/ (. ! /02 62 :/0 %$( :5321 62 :/0 8 F3

%1 >2 0%32 122 %6 2%*=."" E /2 * 5/ )< / 6 7)1 8 % % *

/%/5 : % %1 72 /2 6 7%/2

%1 10 > . J 5 % %1 F3." F ; * < % < 27/ * >

1FV

T

T

FV

TV


14/44

X y

a

a

b1

4 5

!

T

1

b 1 b

FV

F V T

55 !

b

T V

P?@B < 1ine AB 75mm lon' ma es 45 : in!lination with Vp while its Fv ma es 55 : .

(a# ant# aw its p o=e!tions an# fin# its in!lination with Hp.

"#$% "F b

"#$% "F

S 6)/5 S/2 $= 65 2.+.D %> 2 < 27/ 8 % & %-.L 7%/2 ! (@;; % 32 $= &

T3 a ( ;; 26 > =.4.D %> % 65 2 4 @ 5 765 2* / =

8 ; < 5 / ! % * 7)/ TL ;; 5/ % * %;2 /0%/ < 5 / 1

D %> 6 7)1 8 ; < 5 / 1.T% 2 @ % :62 8 ; ! 8 F3% 32 = 65 2.

.D %> % 32 /57%6 65 2 8 ; 1 )< / 6 7)1 8 % % * %;2 5/ 1.

I/ 51 0 5 /%6 7 ;< 2 / 8TL & 51 LFV.

.C /5 )2 5/ / 6 7)1 8 ! % */%/2 )% * )< / /02 65 2

8 F3 % * %;2 5/ .T051 ! 65 2 51 F3.. D < % < 27/ 8 ;

6 7)1 8 ; < 5 / 1 % *%;2 5 /2 127/5 : < 5 / .

L5 2 ! 51 T3 8 65 2 AB.?.D %> 6 7)1 8 ; % * 8 ; ! >5/0 TL *51/% 72 7)/ < 5 / 1@ (@.J 5 ! 1 %1 TL % * ;2%1) 2 5/1 % :62 %/ !. I/ >566 2 / )2 % :62 8 65 2 >5/0 HP.


15/44

X a y

a

b

F V

5! !

b

&!!

b1

T

b 1

T

P?@B < 3 Fvof line AB is 5: : in!line# to +y an# meas( es 55mm lon' while its Tv is 6: : in!line# to +y line. %fen# A is /: mm a$ove Hp an# /5 mm in f ont ofVp, # aw its p o=e!tions,fin# T , in!linations of linewith Hp & Vp.

SOLUTION STEPS/.; aw +y line an# one p o=e!to .1. o!ate a /: mm a$ove +y an#

a /5 mm $elow +y line.3.; aw lo!(s f om these points.4.; aw Fv 5: : to +y f om a an#

ma $ C(ttin' 55mm on it.5. imila ly # aw Tv 6: : to +yf om a & # awin' p o=e!to f om $

o!ate point $ an# =oin a $.6.Then otatin' views as shown, lo!ate T (e en'ths a$ / & a$ /

an# thei an'les with Hp an# Vp.


16/44

X Ya

1

a

b 1

TV

T

b1

1

b

b

FV

T V

F V

T

PROBLEM 4 $L5 2 AB 51 ;; 6 : .I/1 F3 % * T3 ;2%1) 2 @ ;; & @ ;; 6 : 215/0 H< % * V


17/44

TRACES OF THE LINE?

THESE ARE THE POINTS OF INTERSECTIONS OF A LINE ! OR ITS E#TENSION "WITH RESPECTIVE REFFERENCE PLANES .

A LINE ITSELF OR ITS EXTENSION, WHERE EVER TOUCHES H.P.,

THAT POINT IS CALLED TRACE OF THE LINE ON H.P.9 IT IS CALLED H.T.:

SIMILARLY, A LINE ITSELF OR ITS E#TENSION, WHERE EVER TOUCHES V.P.,THAT POINT IS CALLED TRACE OF THE LINE ON V.P.! IT IS CALLED V.T."

V.T . 2 %t is a point on V


18/44

. Begin with FV. Extend FV up to XY line.

. N !e thi" point h# " it i" F$ o% point in Hp

. () w one p)o*e+to) %)o! h.

. Now extend T$ to !eet thi" p)o*e+to) .Thi" point i" HT

STEPS TO LOCATE HT.#-HEN P /0E T /N3 4 E 5 VEN.

1. Begin with TV. Extend TV up to XY line.

2. N !e thi" point $# " it i" T$ o% point in Vp

'. () w one p)o*e+to) %)o! $.,. Now extend F$ to !eet thi" p)o*e+to) .

Thi" point i" VT

STEPS TO LOCATE VT.#-HEN P /0E T /N3 4 E 5 VEN.

'

(TVT

v

a

x y

a

F V

b

T V

Observe & note )*1. +oint, ' - v al ay, on x*y line.

2. VT - v al ay, on one pro/ector.

' . (T - ' al ay, on one pro/ector.

, . FV * '* VT al ay, co*linear.

6. TV * v * (T al ay, co*linear.

These points are used to

solve next three problems .


19/44

x y

b b 1

a

v

VT

a

( T

b

'

b1

3! !

45!

PROBLEM $ Fv o0 line AB a e, 45 ! an le it' XY line and ea, re, &! .ine, Tv a e, 3! ! it' XY line. nd A i, 15 above (p and it, VT i, 1!

belo (p. 6ra pro/ection, o0 line AB7deter ine inclination, it' (p - Vp and locate (T7 VT.

15

1!

SOLUTION STEPS? ; aw +y line, one p o=e!to an#lo!ate fv a /5 mm a$ove +y.Ta e 45 : an'le f om a an#ma in' 6: mm on it lo!ate point $.; aw lo!(s of VT, /: mm $elow +y& e+ten#in' Fv to this lo!(s lo!ate VT.

as fv2h2vt lie on one st.line.; aw p o=e!to f om vt, lo!ate v on +y.F om v ta e 3: : an'le #ownwa # asTv an# its in!lination !an $e'in with v.; aw p o=e!to f om $ an# lo!ate $ %.e.Tv point.-ow otatin' views as (s(al T an#its in!linations !an $e fo(n#.-ame e+tension of Fv, to(!hin' +y as han# $elow it, on e+tension of Tv, lo!ate HT.


20/44

a

b

F V

3!45

1!

"#$% "F b - b 1

X Y

45 !

VT

v

(T

'

"#$% "F b - b 1

1!!

a

b

T V

b 1

T

T

b1

PROBLEM O 2 2 * 8 65 2 AB 51 (@;; % 32 H< % * /02 2 * 51 (@@ ;; 5 $8 / 8 V0562 5/1 HT & VT % 2 4 ;; % * -@ ;; 26 > = 21 < 27/5 1 % * 85 * TL >5/0 5/1 5 765 %/5 1 >5/0 H< & VP.

SOLUTION STEPS $; aw +y line, one p o=e!to an#lo!ate a /: mm a$ove +y.; aw lo!(s /:: mm $elow +y fo points $ & $ /; aw lo!i fo VT an# HT, 3: mm & 45 mm$elow +y espe!tively.Ta e 45 : an'le f om a an# e+ten# that line $a! wa #to lo!ate h an# VT, & o!ate v on +y a$ove VT.

o!ate HT $elow h as shown.Then =oin v 0 HT 0 an# e+ten# to 'et top view en# $.; aw p o=e!to (pwa # an# lo!ate $ a e a $ & a$ #a .-ow as (s(al otatin' views fin# T an# its in!linations.


21/44

X y

(T

VT

'

a

v

b

a

b

8!

5!

b 1

T

T

F V

T V

b 1

1!

35

55

oc , o0 a

PROBLEM $ P o=e!to s # awn f om HT an# VT of a line ABa e 8: mm apa t an# those # awn f om its en#s a e 5: mm apa t.


22/44

b1

a

F V

VT

v

T V

X Y

b

a

b

b1

T 7

T 7

T02 8 ; < 5 / 3 & HT% :621 7% 2 * %> .

&F ; < 5 / VT & 0

% :621 7% 2 * %> .-

&

Instead of considering a & a as projections of first point,if v & VT are considered as first point , then true inclinations of line

withHp & Vp i.e. angles - can be constructed with points VT & V

respectively.

THIS CONCEPT IS USED TO SOL! NE T THR PROBLE S'


23/44

PROBLEM ? $ ine AB 1!! lon i, 3! ! and 45 ! inclined to (p - Vp re,pectively.nd A i, 1! above (p and it, VT i, 2! belo (p

.6ra pro/ection, o0 t'e line and it, (T.

X Y

VT

$1!

2!

oc , o0 a - a 1

#'8 8

#,6 8

a1

1 ! !

b1

b1

a1

1 ! !

b

a

b

a

FV

TV

HT

hSOLUTION STEPS $; aw +y, one p o=e!toan# lo!ate on it VT an# V.; aw lo!(s of a /: mm a$ove +y.Ta e 3: : f om VT an# # aw a line.Ehe e it inte se!ts with lo!(s of aname it a / as it is T of that pa t.F om a / !(t /:: mm T ) on it an# lo!ate point $ / -ow f om v ta e 45 : an# # aw a line #ownwa #s& a on it #istan!e VT2a/ %.e.T of e+tension & name it a /


24/44

PROBLEM (@ $ A line AB i, 95 lon . :t, Fv - Tv a e 45 ! and &! ! inclination, it' X*Y line re,p

nd A i, 15 above (p and VT i, 2! belo Xy line. ine i, in 0ir,t ; adrant.6ra pro/ection,7 0ind inclination, it' (p - Vp. Al,o locate (T.

X Y

VT

$15

2!

oc , o0 a - a 1 a1

9 5

b1

b1

a1

9 5

b

a

b

a

FV

TV

HT

h

45 !

&!!

SOLUTION STEPS $imila to the p evio(s only !han'e

is instea# of lines in!linations,views in!linations a e 'iven.

o fi st ta e those an'les f om VT & vP ope ly, !onst (!t Fv & Tv of e+tension,then #ete mine its T V2a / )an# on its e+tension ma T of linean# p o!ee# an# !omplete it.


25/44

PROBLEM (( $ T'e pro/ector, dra n 0ro VT - end A o0 line AB are 4! apart.nd A i, 15 above (p and 25 in 0ront o0 Vp. VT o0 line i, 2! belo (p.

:0 line i, 95 lon 7 dra it, pro/ection,7 0ind inclination, it' (+ - Vp

X Y

4!

15

2!25

v

VT

a

a

a1

b1 b

b

T V

F V

9 5 1 1

b1

; aw two p o=e!to s fo VT & en# Ao!ate these points an# then

YES YOU CAN COMPLETE IT.


26/44

X

A ' I ' P '

GROUP (C)CASES OF THE LINES IN A'!'P'* A'I'P' PROFILE PLANE'

a

bine AB i, in A:+ a, ,'o n in above 0i re no 1.

:t, FV


27/44

++V+

(+

a

b

a

b

a>

b>

X Y

FV

TV

73V

A

B

a

b

a

b

F o r F ' ! '

For T'!'

LINE IN A PROFILE PLANE ! MEANS IN A PLANE PERPENDICULAR TO BOTH HP & VP "

R'"


28/44

PROBLEM (+ $ ine AB 8! lon 7 a e, 3! ! an le it' (pand lie, in an A x.Vertical +lane 45 ! inclined to Vp.

nd A i, 15 above (p and VT i, 1! belo X*y line.6ra pro/ection,7 0ine an le it' Vp and (t.

VT

vX Y

a

b

a

b

a1

b1

oc , o0 b

oc , o0 b

1!

15

(T

'

b1

AV+ 45 ! to V+

45 !

L 7)1 8 % & % (

imply !onsi#e in!lination of AVPas in!lination of TV of o( line,well then

You sure can co !"e#e $#as !re%$ous !ro&"e s'

Go a(ead''


29/44

PROBLEM (- $ A line AB7 95 lon 7 'a, one end A in Vp. "t'er end B i, 15 above (pand 5! in 0ront o0 Vp.6ra t'e pro/ection, o0 t'e line 'en , o0 it,

:nclination, it' (+ - Vp i, ?! ! 7 ean, it i, lyin in a pro0ile plane.Find tr e an le, it' re0.plane, and it, trace,.

a

b

(T

VT

X Y

a

b

%ide Vie < Tr e en t' =

a>

b>


30/44

APPLICATIONS OF PRINCIPLES OF PROJECTIONS OF LINES IN SOLVING CASES OF DIFFERENT PRACTICAL SITUATIONS.

I /0212 /=566 2 *217 5 2* .I/1 26%/5 >5/0 G ) * ! HP "

A * % W%66 1 ;2 32 /57%6 27/ ! VP " >566 2 :532 .

I *5 27/6= 5 8 ;%/5 2:% *5 : F3 & T3 8 1 ;2 65 2 65 21,5 765 2* / /0 282 2 72 P6% 21 >566 2 :532% *

= ) % 2 1) 5/1 < 27/5 1% *

8) /02 / *2/2 ;5 2 5/1 / )2 L2 :/0 % * 5/1 5 765 %/5 1 >5/0 : ) *.

H2 2 3% 5 )1 < 62;1 %6 : >5/0%7/)%6 6 5 : 8 352>1 5 :532 ARROW *5 27/5 1,

YOU % 2 1) < 27/5 1 & 85 * % 1>2 1,O88 7 ) 12 = ) ;)1/ 351)%65 2 /02 15/)%/5 <


31/44

W ! 4 4 P

W!44

A

B

PROBLEM (4 $ Two o$=e!ts, a flowe A) an# an o an'e B) a e within a e!tan'(la !ompo(n# wall, whose P & I a e walls meetin' at 9: : . Flowe A is / & 5.5 f om walls P & I espe!tively.@ an'e B is 4 & /.5 f om walls P & I espe!tively. ; awin' p o=e!tion, fin# #istan!e $etween them%f flowe is /.5 an# o an'e is 3.5 a$ove the ' o(n#. Consi#e s(ita$le s!ale..

TV

FV

PROBLEM ( $ Two man'os on a t ee A & B a e / 5 m an# 3 :: m a$ove ' o(n#


32/44

PROBLEM ( $ Two man os on a t ee A & B a e /.5 m an# 3.:: m a$ove o(n#an# those a e /.1 m & /.5 m f om a :.3 m thi! wall $(t on opposite si#es of it.%f the #istan!e meas( e# $etween them alon' the ' o(n# an# pa allel to wall is 1.6 m,Then fin# eal #istan!e $etween them $y # awin' thei p o=e!tions.

F V

TV

A

B

:.3 TH%CJ


33/44

PROBLEM ( $ oa7 ob - oc are t'ree line,7 25 7 45 and &5lon re,pectively.All e; ally inclined and t'e ,'orte,t i, vertical.T'i, 0i . i, TV o0 t'ree rod, "A7 "B and "#

'o,e end, A7B - # are on ro nd and end " i, 1!!above ro nd. 6ra t'eir pro/ection, and 0ind len t' o0

eac' alon it' t'eir an le, it' ro nd.

1 5 m m

45 mm

6 5 m m

A

B

C

@

FV

TV


34/44

PROBLEM ( 2 A pipe line f om point A has a #ownwa # ' a#ient / 5 an# it (ns #(e


35/44


36/44

4. M

7.5

-@@

4 @

( @ M

( M

F V

TV

A

B

C

PROBLEM (? $ K(y opes of two poles fi+e# at 4.5m an# 7.5 m a$ove ' o(n#,a e atta!he# to a !o ne of a $(il#in' /5 hi'h, ma e 3:: an# 45: in!linationswith ' o(n# espe!tively.The poles a e /: apa t. ;ete mine $y # awin' theip o=e!tions, en'th of ea!h ope an# #istan!e of poles f om $(il#in'.


37/44


38/44

F V

2 M 1 . D M

D M

A

B

C

;

HooTV

PROBLEM +( $ A ho i ontal woo#en platfo m 1 lon' an# /.5 wi#e is s(ppo te# $y fo( !hainsf om its !o ne s an# !hains a e atta!he# to a hoo 5 a$ove the !ente of the platfo m.; aw p o=e!tions of the o$=e!ts an# #ete mine len'th of ea!h !hain alon' with its in!lination with ' o(n#.

H


39/44

PROBLEM ++. A oom is of si e 6.5m ,5m ;,3.5m hi'h. An ele!t i! $(l$ han's /m $elow the !ente of !eilin'. A swit!h is pla!e# in one of the !o ne s of the oom, /.5m a$ove the floo in'.; aw the p o=e!tions an #ete mine eal #istan!e $etween the $(l$ an# swit!h.

3wit+h

Bulb

F r o n t . a l l

#eilin

% i d e . a l l

O b s e r v e r

TV

LD

H


40/44

PROBLEM +- $ A PICTURE FRAME + M WIDE AND ( M TALL IS RESTING ON HORI ONTAL WALL RAILINGMAKES - @ INCLINATION WITH WALL. IT IS ATTAACHED TO A HOOK IN THE WALL BY TWO STRINGS.THE HOOK IS (. M ABOVE WALL RAILING. DETERMINE LENGTH OF EACH CHAIN AND TRUE ANGLE BETWEEN THEM

- @

1 .D M

( M

2 M

W%66 %565 :

F V

T V

P O!LE" #O.$%


41/44

X Y

c

c

"#$% "F d - d 1d d1

d d 1

T V

F V

T

T

"#$% "F d - d 1

T.V. o0 a 95 lon ine #67 ea, re, 5! .nd # i, 15 belo (p and 5! in 0ront o0 Vp.nd 6 i, 15 in 0ront o0 Vp and it i, above (p.

6ra pro/ection, o0 #6 and 0ind an le, it' (p and Vp.

SOME CASES OF THE LINEIN DIFFERENT UADRANTS.

REMEMBERBELOW HP$ M2% 1$ F3 26 > =

BEHIND V


42/44

X Y

a

a b

b

T V

F V

"#$% "F b - b 1

"#$% "F b - b 1

b 1

T

b1

T

9!

P O!LE" #O.$nd A o0 line AB i, in (p and 25 be'ind Vp.nd B in Vp.and 5! above (p.

6i,tance bet een pro/ector, i, 9! .6ra pro/ection, and 0ind it, inclination, it' (t7 Vt.

P O!LE" #O.$'


43/44

X y

a b 1

3! !

p 1

a

p

b

b b1

"#$% "F b - b 1

"#$% "F b - b 1

p

35

25

T

T

F V

T V

nd A o0 a line AB i, 25 belo (p and 35 be'ind Vp.ine i, 3!! inclined to (p.

T'ere i, a point + on AB contained by bot' (+ - V+.6ra pro/ection,7 0ind inclination it' Vp and trace,.

P O!LE" #O $(


44/44

a

b

a

b

b 1

T

T F V

T V

b1

95

35

Ht VtX Y

25

55

P O!LE #O.$( nd A o0 a line AB i, 25 above (p and end B i, 55 be'ind Vp.

T'e di,tance bet een end pro/ector, i, 95 .:0 bot' it, (T - VT coincide on xy in a point735 0ro pro/ector o0 A and it'in t o pro/ector,76ra pro/ection,7 0ind T and an le, and (T7 VT.

Date post:	04-Jun-2018
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