5. PROJECTION OF LINES.ppt

7/24/2019 5. PROJECTION OF LINES.ppt

1/44

TO DRAW PROJECTIONS OF ANY OBJECT,

ONE MUST HAVE FOLLOWING INFORMATIONA)OBJECT

{ WITH ITS DESCRIPTION, WELL DEFINED.}

B) OBSERVER

{ ALWAYS OBSERVING PERPENDICULAR TO RESP. REF.PLANE}.C) LOCATION OF OBJECT,

{ MEANS ITS POSITION WITH REFFERENCE TO H.P. & V.P.}

TERMS ABOVE & BELOWWITH 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 NAMEING

DIFFERENT VIEWS IN ORTHOGRAPHIC PROJECTIONS.

ITS FRONT VIEW % %

SAME SYSTEM OF NOTATIONS SHOULD BE FOLLOWED

INCASE 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

O1232

THIS UADRANT PATTERN,

IF OBSERVED ALONG #$Y LINE ! INREDARROW DIRECTION"

WILL E#ACTLY APPEAR AS SHOWN ON RIGHT SIDE AND HENCE,

IT IS FURTHER USED TO UNDERSTAND ILLUSTRATION PROPERLLY.


4/44

HP

VP

a

a

A

POINT A IN

(STUADRANT

OBSERVER

VP

HP

POINT A IN

+NDUADRANT

OBSERVER

a

a

A

OBSERVER

a

a

POINT A IN

-RD UADRANT

HP

VP

A

OBSERVER

a

aPOINT A IN

4TH UADRANT

HP

VP

A

P5/ A 51

P6%72* I

*58822/

9)%*%/1

%* 5/1 F3 & T3

%2 ):0/ 5

1%;2

VP. B)/ %1 T3 51

51 % 352> H>%* ?@@,

I 767>512*527/5.T02

I 8/

= 652 %* /02


5/44

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

FV of a point "P is epesente# $y p. %t shows position of the point

with 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 epesente# $y p. %t shows position of the point with

espe!t to VP.

%f the point lies in font 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 $ehin# the VP, p lies a$ove the XY line.

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


6/44

A

a

aA

a

a

Aa

a

X

Y

X

Y

X

YForFv

For Tv

ForFv

For Tv

For Tv

ForFv

POINT A ABOVE HP

& INFRONT OF VP

POINT AIN HP

& INFRONT OF VPPOINT AABOVE HP

& IN VP

PROJECTIONS OF A POINT IN FIRST QUADRANT.

PICTORIALPRESENTATION

PICTORIALPRESENTATION

ORTHOGRAPHIC PRESENTATIONS

OF 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 PAGE

SHOWING CLEARLY THE NATURE OF FV & TV

OF LINES LISTED ABOVE AND NOTE RESULTS.

PROJECTIONS OF STRAIGHT LINES.

INFORMATION REGARDING A LINE 5'!&"

ITS LENGTH,

POSITION OF ITS ENDS WITH HP & VP

ITS 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

FoFv

Fo Tv

Fo Tv

FoFv

N/2

Fv is a veti!al line

howin' T(e en'th

&

Tv is a point.

N/2

Fv & Tv $oth ae

** to +y

&

$oth show T. .

1.

2.

A inepepen#i!(la

to Hp

&

** to Vp

A ine

** to Hp&

** to Vp

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

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

!P57/5%6 P212/%/5"

!P57/5%6 P212/%/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 xy

Fv parallel to xy.

3.

4.

Fv inclined to xy

Tv parallel to xy.

O/0:%


10/44

X

Y

V.P.

ForFv

a

b

a b

B

A

For Tv

F

.V.

T.V.

X

Y

V.P.

a

b

a b

F.V.

T.V.

ForFv

For Tv

B

A

X Y

H.P.

V.P.

a

b

FV

TV

a

b

A Line inclined to bothHp and Vp

(Pictorial presentation)

5.

N/2 T0212 F%7/1$

B/0 F3 & T3 %2 57652* / =.

-o view is paallel 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

b1

b1

TL

X Y

H.P.

V.P.

a

b

FV

TV

a

b

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

H'&$' #)" $%++'"(%& FV

! #" &%)"6%-#&

T+ / 85* T)2 L2:/0.

!V52>1 %2 /%/2* / *2/2;52

T)2 L2:/0 & 5/1 5765%/51

>5/0 H< & V,

>5/0 /025 %


12/44

The most impotant #ia'am showin' 'aphi!al elations

amon' all impotant paametes of this topi!.

t(#y an# memoie it as a CIRCUIT DIAGRAM

An# (se in solvin' vaio(s po$lems.

T)2 L2:/0 51 232 /%/2*. I/1 05/%6 7; & 5/ 51 8)/02 /%/2* / 67%/2 352>.

V52>1 %2 %6>%=1 /%/2*, ;%*2 05/%6 & 8)/02

2/2*2* / 67%/2 TL, &

A4"% R'5'5'+

I;%*

/) T(e en'th T) 0 a $/& a $1

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

b1

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%> = 652 %* 2 05/%6 6521 !L7)1" 8;

/0 %* & :2/


14/44

X y

a

a

b1

45!

T

1

b1b

FV

FV

T

55!

b

TV

P?@B


15/44

X a y

a

b

FV

5!!

b

&!!

b1

T

b1

T

P?@B


16/44

X Ya

1

a

b1

TV

T

b1

1

b

b

FV

TV

FV

T

PROBLEM 4 $

L52 AB 51 ;; 6: .I/1 F3 %* T3 ;2%1)2 @ ;; & @ ;; 6: 21


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

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

2. 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/0ET/N3 4E 5VEN.

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/0ET/N3 4E 5VEN.

'

(TVT

v

a

x y

a

FV

b

TV

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

2. VT - v alay, on one pro/ector.

'. (T - ' alay, on one pro/ector.

,. FV * '* VT alay, co*linear.

6. TV * v * (T alay, co*linear.

These points are used to

solve next threeproblems.


19/44

x y

b b1

a

v

VT

a

(T

b

'

b1

3!!

45!

PROBLEM $Fv o0 line AB ae, 45!anle it' XY line and ea,re, &! .

ine, Tv ae, 3!!it' XY line. nd A i, 15 above (p and it, VT i, 1!

belo (p. 6ra pro/ection, o0 line AB7deterine inclination, it' (p - Vp and locate (T7 VT.

15

1!

SOLUTION STEPS?;aw +y line, one po=e!to an#

lo!ate fv a/5 mm a$ove +y.

Tae 45:an'le fom aan#

main' 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 fv2h2vtlie on one st.line.;aw po=e!to fom vt, lo!atevon +y.

Fom v tae 3::an'le #ownwa# as

Tv an# its in!lination !an $e'inwith v.

;aw po=e!to fom $ an# lo!ate $ %.e.Tv point.

-ow otatin' views as (s(al Tan#

its in!linations !an $e fo(n#.

-ame e+tension of Fv, to(!hin' +y as h

an# $elow it, on e+tension of Tv, lo!ate HT.


20/44

a

b

FV

3!

45

1!

"#$% "F b - b1

X Y

45!

VT

v

(T

'

"#$% "F b - b1

1!!

a

b

TV

b1

T

T

b1

PROBLEM

O2 2* 8 652 AB 51 (@;; %32 H< %* /02 2* 51 (@@ ;; 5$8/ 8 V0562 5/1 HT & VT %2 4;; %* -@ ;; 26> = 21 5/0 5/1 5765%/51 >5/0 H< & VP.

SOLUTION STEPS$

;aw +y line, one po=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.

Tae 45:an'le fom a an# e+ten# that line $a!wa#

to lo!ate han# VT, & o!ate von +y a$ove VT.

o!ate HT $elow has shown.

Then =oin v 0 HT 0an# e+ten# to 'et top view en# $.

;aw po=e!to (pwa# an# lo!ate $ae 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!

b1

T

T

FV

TV

b 1

1!

35

55

oc, o0 a

PROBLEM $Po=e!tos #awn fom HT an# VT of a line AB

ae 8: mm apat an# those #awn fom its en#s ae 5: mm apat.


22/44

b1

a

FV

VT

v

TV

X Y

b

a

b

b1

T7

T7

T02 8; .

&F; .

-

&

Instead of considering a & a as projections of rst point,if v & VT are considered as rst 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!NET TH!! PROBLES'


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 - a1

#'88

#,68

a1

1!!

b1

b1

a1

1!!

b

a

b

a

FV

TV

HT

hSOLUTION STEPS$

;aw +y, one po=e!to

an# lo!ate on it VT an# V.

;aw lo!(s of a/: mm a$ove +y.

Tae 3::fom VT an# #aw a line.Ehee it intese!ts with lo!(s of a

name it a/ as it is T of that pat.

Fom a/!(t /:: mm T) on it an# lo!ate point $/

-ow fom vtae 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 ae 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 - a1 a1

95

b1

b1

a1

95

b

a

b

a

FV

TV

HT

h

45!

&!!

SOLUTION STEPS$

imila to the pevio(s only !han'e

is instea# of lines in!linations,

views in!linations ae 'iven.

o fist tae those an'les fom VT & v

Popely, !onst(!t Fv & Tv of e+tension,

then #etemine its T V2a/)

an# on its e+tension ma T of line

an# po!ee# an# !omplete it.


25/44

PROBLEM (( $T'e pro/ector, dran 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 lon7 dra it, pro/ection,7 0ind inclination, it' (+ - Vp

X Y

4!

15

2!25

v

VT

a

a

a1

b1b

b

TV

FV

9511

b1

;aw two po=e!tos fo VT & en# A

o!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

b

ine AB i, in A:+ a, ,'on in above 0ire no 1.

:t, FV


27/44

++V+

(+

a

b

a

b

a>

b>

X Y

FV

TV

73V

A

B

a

b

a

b

ForF'!'

For T'!'

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

R'"


28/44

PROBLEM (+ $ine AB 8! lon7 ae, 3!!anle it' (p

and lie, in an Ax.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 anle 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!

L7)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&"es'

Go a(ead''


29/44

PROBLEM (- $A line AB7 95 lon7 'a, one end A in Vp. "t'er end B i, 15 above (p

and 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 tre anle, it' re0.plane, and it, trace,.

a

b

(T

VT

X Y

a

b

%ide Vie

< Tre ent' =

a>

b>


30/44

APPLICATIONS OF PRINCIPLES OF PROJECTIONS OF LINESIN SOLVING CASES OF DIFFERENT PRACTICAL SITUATIONS.

I /0212 /=5/0 G)* ! HP "

A*

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

I*527/6= 58;%/5 2:%*5: F3 & T3 8 1;2 652 6521,57652* / /0 282272 P6%21 >566 2 :532

%*

=) %2 1)1 5 :532 ARROW *527/51,

YOU%2 1)


31/44

W!44P

W!44

A

B

PROBLEM (4$Two o$=e!ts, a flowe A) an# an oan'e B) ae within a e!tan'(la !ompo(n# wall,

whose P & I ae walls meetin' at 9::. Flowe A is / & 5.5 fom walls P & I espe!tively.

@an'e B is 4 & /.5 fom walls P & I espe!tively. ;awin' po=e!tion, fin# #istan!e $etween them

%f flowe is /.5 an# oan'e is 3.5 a$ove the 'o(n#. Consi#e s(ita$le s!ale..TV

FV

PROBLEM ( $ Two man'os on a tee A & B ae / 5 m an# 3 :: m a$ove 'o(n#


32/44

PROBLEM ( $Two man'os on a tee A & B ae /.5 m an# 3.:: m a$ove 'o(n#

an# those ae /.1 m & /.5 m fom 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# paallel to wall is 1.6 m,

Then fin# eal #istan!e $etween them $y #awin' thei po=e!tions.

FV

TV

A

B

:.3 TH%CJ


33/44

PROBLEM ( $oa7 ob - oc are t'ree line,7 257 45 and &5

lon 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 rond and end " i, 1!!

above rond. 6ra t'eir pro/ection, and 0ind lent' o0

eac' alon it' t'eir anle, it' rond.

15mm

45 mm

65mm

A

B

C

@

FV

TV


34/44

PROBLEM (2 A pipe line fom point Ahas a #ownwa# 'a#ient /5 an# it (ns #(e


35/44

N

@

%

PROBLEM ( A peson o$seves two o$=e!ts, A & B, on the 'o(n#, fom a towe, /5 hi'h,

At the an'les of #epession 3::& 45:. @$=e!t A is is #(e -oth2Eest #ie!tion of o$seve an#

o$=e!t B is #(e Eest #ie!tion. ;aw po=e!tions of sit(ation an# fin# #istan!e of o$=e!ts fom

o$seve an# fom towe also.

A

B

O

-@@

4@


36/44

4. M

7.5

-@@

4@

(@M

( M

FV

TV

A

B

C

PROBLEM (?$K(y opes of two poles fi+e# at 4.5m an# 7.5 m a$ove 'o(n#,

ae atta!he# to a !one of a $(il#in' /5 hi'h, mae 3:: an# 45: in!linations

with 'o(n# espe!tively.The poles ae /: apat. ;etemine $y #awin' thei

po=e!tions,en'th of ea!h ope an# #istan!e of poles fom $(il#in'.


37/44

(.+M

@

.HM

4 M

FV

TV

P/B7E9 28:; 4 tn< o% , 9 height i" to be "t)engthened b= %ou) "t= )od" %)o! e+h +o)ne)

b= %ixing thei) othe) end" to the %loo)ing> t point 1.2 9 nd 8.? 9 %)o! two d*+ent wll" )e"pe+ti$el=>

" "hown. (ete)!ine g)phi+ll= length nd ngle o% e+h )od with %loo)ing.


38/44

FV

2M1.D

M

DM

A

B

C

;

Hoo

TV

PROBLEM +($A hoiontal woo#en platfom 1 lon' an# /.5 wi#e is s(ppote# $y fo( !hains

fom its !ones an# !hains ae atta!he# to a hoo 5 a$ove the !ente of the platfom.

;aw po=e!tions of the o$=e!ts an# #etemine len'th of ea!h !hain alon' with its in!lination with 'o(n#.

H


39/44

PROBLEM ++.

A oom is of sie 6.5m ,5m ;,3.5m hi'h.

An ele!ti! $(l$ han's /m $elow the !ente of !eilin'.

A swit!h is pla!e# in one of the !ones of the oom, /.5m a$ove the flooin'.

;aw the po=e!tions an #etemine eal #istan!e $etween the $(l$ an# swit!h.

3wit+h

Bulb

Front.all

#eilin

%ide.all

Observer

TV

LD

H


40/44

PROBLEM +-$A PICTURE FRAME + M WIDE AND ( M TALL IS RESTING ON HORIONTAL WALL RAILING

MAKES -@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.DM

(M

2M

W%66 %565:

FV

TV

PO!LE" #O.$%


41/44

X Y

c

c

"#$% "Fd - d1d d1

d d1

TV

FV

T

T

"#$% "Fd - d1

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 anle, it' (p and Vp.

SOME CASES OF THE LINE

IN DIFFERENT UADRANTS.

REMEMBER

BELOW HP$ M2%1$ F3 26> =

BEHIND V


42/44

X Y

a

a b

b

TV

FV

"#$% "F b - b1

"#$% "F b - b1

b1

T

b1

T

9!

PO!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 beteen pro/ector, i, 9!.

6ra pro/ection, and 0ind it, inclination, it' (t7 Vt.

PO!LE" #O.$'


43/44

X y

ab1

3!!

p1

a

p

b

b b1

"#$% "F b - b1

"#$% "F b - b1

p

35

25

T

T

FV

TV

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,.

PO!LE" #O $(


44/44

a

b

a

b

b1

T

T

FV

TV

b1

95

35

HtVtX Y

25

55

PO!LE" #O.$(

nd A o0 a line AB i, 25 above (p and end B i, 55 be'ind Vp.

T'e di,tance beteen end pro/ector, i, 95.

:0 bot' it, (T - VT coincide on xy in a point7

35 0ro pro/ector o0 A and it'in to pro/ector,7

6ra pro/ection,7 0ind T and anle, and (T7 VT.

5. PROJECTION OF LINES.ppt

Documents

Transcript of 5. PROJECTION OF LINES.ppt