St line(new)

TO DRAW PROJECTIONS OF ANY OBJECT, ONE MUST HAVE FOLLOWING INFORMATIONA) OBJECT

{ WITH IT’S DESCRIPTION, WELL DEFINED.}

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

C) LOCATION OF OBJECT, { MEANS IT’S 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 QUADRANTS. OBJECTS CAN BE PLACED IN ANY ONE OF THESE 4 QUADRANTS.

IT IS INTERESTING TO LEARN THE EFFECT ON THE POSITIONS OF VIEWS ( FV, TV ) OF THE OBJECT WITH RESP. TO X-Y LINE, WHEN PLACED IN DIFFERENT QUADRANTS.

ORTHOGRAPHIC PROJECTIONSOF POINTS, LINES, PLANES, AND SOLIDS.

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

NOTATIONS

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

IT’S FRONT VIEW a’ a’ b’

SAME SYSTEM OF NOTATIONS SHOULD BE FOLLOWED INCASE NUMBERS, LIKE 1, 2, 3 – ARE USED.

OBJECT POINT A LINE AB

IT’S TOP VIEW a a b

IT’S SIDE VIEW a” a” b”

X

Y

1ST Quad.2nd Quad.

3rd Quad. 4th Quad.

X Y

VP

HP

Observer

THIS QUADRANT PATTERN, IF OBSERVED ALONG X-Y LINE ( IN RED ARROW DIRECTION)WILL EXACTLY APPEAR AS SHOWN ON RIGHT SIDE AND HENCE,IT IS FURTHER USED TO UNDERSTAND ILLUSTRATION PROPERLLY.

HP

VPa’

a

A

POINT A IN1ST QUADRANT

OBSERVER

VP

HP

POINT A IN2ND QUADRANT

OBSERVER

a’

a

A

OBSERVER

a

a’

POINT A IN3RD QUADRANT

HP

VP

A

OBSERVER

a

a’POINT A IN4TH QUADRANT

HP

VP

A

Point A is Placed In different quadrants

and it’s Fv & Tv are brought in same plane for Observer to see

clearly. Fv is visible asit is a view on

VP. But as Tv is is a view on Hp,

it is rotateddownward 900, In clockwise direction.The In front part of

Hp comes below xy line and the part behind Vp comes above.

Observe and note the process.

FV & TV of a point always lie in the same vertical line

FV of a point ‘P’ is represented by p’. It shows position of the point with respect to HP only. While drawing FV, XY line represents the HP and the plane of paper represents the VP.

If the point P lies above HP, p’ lies above the XY line.If the point P lies in the HP, p’ lies on the XY line.If the point P lies below the HP, p’ lies below the XY line.

TV of a point ‘P’ is represented by p. It shows position of the point with respect to VP only. While drawing TV, XY line represents the VP and the plane of paper represents the HP.

If the point P lies in front of VP, p lies below the XY line.If the point P lies in the VP, p lies on the XY line.If the point P lies behind the VP, p lies above the XY line.

Basic concepts for drawing projection of point

FV does not give any information about position of the point w.r.t. VP

TV does not give any information about position of the point w.r.t. HP

A

a

a’A

a

a’

Aa

a’

X

Y

X

Y

X

YFor Fv

For Tv

For Fv

For Tv

For Tv

For Fv

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.

PICTORIAL PRESENTATION

PICTORIAL PRESENTATION

ORTHOGRAPHIC PRESENTATIONS OF ALL ABOVE CASES.

X Y

a

a’

VP

HP

X Y

a’

VP

HP

a X Y

a

VP

HP

a’

Fv above xy,Tv below xy.

Fv above xy,Tv on xy.

Fv on xy,Tv below xy.

PROJECTIONS OF A POINT IN SECOND QUADRANT.POINT A ABOVE HP

& BEHIND VP

X

Y

A

For Fv

For Tv

SIMPLE CASES OF THE LINE

1. A VERTICAL LINE ( LINE PERPENDICULAR TO HP & // TO VP)

2. LINE PARALLEL TO BOTH HP & VP.

3. LINE INCLINED TO HP & PARALLEL TO VP.

4. LINE INCLINED TO VP & PARALLEL TO HP.

5. LINE INCLINED TO BOTH HP & VP.

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

PROJECTIONS OF STRAIGHT LINES.INFORMATION REGARDING A LINE means

IT’S LENGTH, POSITION OF IT’S ENDS WITH HP & VP

IT’S INCLINATIONS WITH HP & VP WILL BE GIVEN. AIM:- TO DRAW IT’S PROJECTIONS - MEANS FV & TV.

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’ b’Fv

Tv

For Fv

For Tv

For Tv

For Fv

Note:Fv is a vertical line

Showing True Length&

Tv is a point.

Note:Fv & Tv both are

// to xy &

both show T. L.

1.

2.

A Line perpendicular

to Hp &

// to Vp

A Line // to Hp

& // to Vp

Orthographic Pattern

Orthographic Pattern

(Pictorial Presentation)

(Pictorial Presentation)

A Line inclined to Hp and

parallel to Vp

(Pictorial presentation) X

Y

V.P.

A

B

b’

a’

b

a

F.V.

T.V.

A Line inclined to Vp and

parallel to Hp

(Pictorial presentation) Ø

V.P.

a b

a’

b’

BAØ

F.V.

T.V.

X Y

H.P.

V.P.

F.V.

T.V.a b

a’

b’

X Y

H.P.

V.P.

Øa

b

a’ b’

Tv

Fv

Tv inclined to xyFv parallel to xy.

3.

4.

Fv inclined to xyTv parallel to xy.

Orthographic Projections

X

Y

V.P.

For Fva’

b’

a b

B

A

For Tv

F.V.

T.V.

X

Y

V.P.

a’

b’

a b

F.V.

T.V.

For Fv

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.

Note These Facts:-Both Fv & Tv are inclined to xy.

(No view is parallel to xy)Both Fv & Tv are reduced

lengths.(No view shows True Length)

Orthographic ProjectionsFv is seen on Vp clearly.

To see Tv clearly, HP is rotated 900 downwards,Hence it comes below xy.

On removal of objecti.e. Line AB

Fv as a image on Vp.Tv as a image on Hp,

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’

Here TV (ab) is not // to XY lineHence it’s corresponding FV

a’ b’ is not showing True Length &

True Inclination with Hp.

In this sketch, TV is rotated and made // to XY line.

Hence it’s corresponding FV a’ b1’ Is showing

True Length &

True Inclination with Hp.

Note the procedureWhen Fv & Tv known,

How to find True Length.(Views are rotated to determineTrue Length & it’s inclinations

with Hp & Vp).

Note the procedureWhen True Length is known,

How to locate FV & TV.(Component a’b2’ of TL is drawn

which is further rotatedto determine FV)

a

a’

b’

b

b1’

TL

b2

Ø

TL

Fv

Tv

Orthographic Projections Means Fv & Tv of Line AB

are shown below,with their apparent Inclinations

&

Here a’b1’ is component of TL ab1 gives length of FV.

Hence it is brought Up to Locus of a’ and further rotatedto get point b’. a’ b’ will be Fv.

Similarly drawing componentof other TL(a’b1‘) TV can be drawn.

b1

b2’

a’

X Y

a

20

2050

55

b’

75

b1

b1’

b2’

b2

θ α

Φ β

TL=83

TL=83

b

θ: True inclination of the line with HP = 26º

α : Inclination of FV of the line with HP/XY

Ø: True inclination of the line with VP = 41ºβ : Inclination of TV of the line with VP/XY

List of questions of type I : 10.13,10.7,5,11,12, 10.28,18,23,26,10.18,10.23

h’

h

v

v’

A: 20mm ↑HP,20 mm →VPB: 55mm ↑HP,75 mm →VP

End A of a line AB is 20mm above HP & 20mm in front of VP while its end B is 55mm above HP and 75mm in front of VP. The distance between end projectors of the line is 50mm. Draw projections of the line and find its true length and true inclination with the principal planes. Also mark its traces.

Given,

a0 b0 = 50 mm

a0 b0

To draw FV & TV,To find TL, θ ,Ø, HT & VT

Type-I:Given projections (FV & TV) of the line. To find True length & true inclination of the line with HP (θ) and with VP(Φ).

h’v’a’b’ are always collinear.

h v a b are always collinear.

Problems for finding TL, θ,Ø, HT & VT

Problem1: End A of a line AB is in the HP and 12 mm in front of VP, while its end B is 45mm above HP and 75mm in front of VP. The distance between their end projectors is 30 mm. Draw its projections and find its true length and true inclinations with the principle planes. Also mark its traces.Problem 2: End A of a line AB is 25mm below the HP and 20 mm behind VP, while its end B is 45 mm below HP and 65mm behind VP. The distance between their end projectors is 45 mm. Draw its projections and find its true length and true inclinations with the principle planes. Also mark its traces.

Problem 4: End A of a line AB is 25mm below the HP and 20 mm behind VP, while its end B is 40mm below HP and 65mm behind VP. The distance between their end projectors is 45 mm. Draw its projections and find its true length and true inclinations with the principle planes. Also mark its traces.

Problem 3: End A of a line AB is 70 mm above HP and 20 mm in front of VP, while its end B is 20mm above HP and 50mm in front of VP. The distance between their end projectors is 60 mm. Draw its projections and find its true length and true inclinations with the principle planes. Also mark its traces.

X Y

25

a’

25

70

b1’

a

b2

70

30°

45° b1

b

b’

b2’

h’

h

v

v’

List of questions of type II : 10.8,10.11,10.12,10.14,10.17, 10.10,1,4,6,8,

Line inclined to both HP & VP

Type –II Given (i) T.L., θ and Ø,

(ii) T.L., F.V., T.V. to draw projections, find α, β,H.T. and V.T.

PROBLEMA line AB, 70mm long, has its end A 25 mm above HP and 25mm in front of VP. It is inclined at 30° to HP and 45°to VP. Draw its projections and mark its traces.

Given,A: 25mm ↑HP,25 mm →VP

TL=70mm, θ=30º, Ø=45º

To draw/find,FV, TV, HT & VT

Trapezoidal Method

This method is used to determine true length, true inclination, θ and Ø when projections of the line (FV & TV) are already given.

Let a’b’ and ab are FV & TV of a line.

X Ya’

a

b’

b

For solving the problem by trapezoidal method draw perpendiculars at a and bMeasure distance of a’ from the XY line and mark it on the perpendicular drawn from a.Similarly measure the distance of b’ from the XY line and mark it on the perpendicular drawn from b.Mark the points as A1and B1 respectively and join them. This will give True length of the line.Extend this line to intersect the top view. The point of intersection gives HT and the angle between TL and TV will be θ.

A1

B1

θ

h

Now draw perpendiculars at a’ and b’

Measure distance of a from the XY line and mark it on the perpendicular drawn from a’.

Similarly measure the distance of b from the XY line and mark it on the perpendicular drawn from b’.

Mark the points as A2and B2 respectively and join them. This will give True length of the line.

A2

B2

v’

Extend this line to intersect the front view. The point of intersection gives VT and the angle between TL and FV will be Ø.

Ø

Note: If the distances of the points are measured on the opposite sides of XY line, then perpendiculars are drawn on opposite sides and, points are also marked on opposite sides.

Compulsory problems in projection of Straight Lines

Type I 10.13, 5, 10.17,10.18, 10.23

Type II 10.11,10.12,10.14, 6, 8,10.10

Type III 10.19

Q10.11 The top view of a 75mm long line AB measures 65mm,while its front view measures 50mm. Its one end A is in HP and12mm in front of VP. Draw the projections of AB and determine its inclination with HP and VP

X Ya’

12

ab1

Hint: Draw ab1=65mm // to XY. Because when TV is // to XY, FV gives TL.

75

b1’b’

50

65

65

b

75

b2

Given,

TL=75mm,TV=65mm,FV=50mm

A is in HP & 12mm→VP

To draw FV &TV of the line AB

To find θ & Ø

30º

48º

Ans. θ=30º

Ans. Ø=48º

Q 5. The end A of a line AB is in the HP and 25 mm behind the VP. The end B is in the VP and 50 mm above the HP. The distance between the end projectors is 75 mm. draw the projections of AB and determine its true length, traces and inclination with two planes.

Given,

A is in HP & 25mm ←V.P.

B is in VP & 50 ↑ V.P. a0b0=75mm

To find, True Length, θ,Ø, H.T. and V.T.

X Y

75

a’

50

b’

25

a

b

b1’

b1

93 Ans. TL= a’b1’=93 mm

33º

θ=33ºb2’

b2 15º

93

Ø=15º

h’

h

v

v’

Q10.12 A line AB, 65mm long has its end A 20mm above H.P. and 25mm in front of VP. The end B is 40mm above H.P. and 65mm in front of V.P. Draw the projections of AB and show its inclination with H.P. and V.P.

X Y

20

a’

25

a

4065

Given,

TL=65mm

A is 20mm ↑ HP & 25mm →V.P. B is 40mm ↑ & 65mm → V.P.

To draw FV &TV of the line AB

To find θ & Ø

Hint1:Mark a’ 20mm above H.P & a 25mm below XY

Hint2:Draw locus of b’ 40mm above XY & locus of b 65 mm below XY

65

b1’

b1

b

b’

b2’

b2

65

18º

38º

Ans. θ=18º

Ans. Ø=38º

Q10.13:The projectors of the ends of a line AB are 5cm apart. The end A is 2cm above the H.P and 3cm in front of V.P. The end B is1cm below H.P. and 4cm behind the V.P. Determine the true length and traces of AB, and its inclination with the two planes

X Y50

20

a’

30

a40

b

10

b’

b1

91

Ans. θ=20º20º

b2’

b2

50ºAns. Ø=50º

h’

h

v

v’

Given,

a0b0=50mm A is 20mm ↑ HP & 30mm →V.P. B is 10mm ↓ & 40mm ← V.P.


b1’

Q10.13:The projectors of the ends of a line AB are 5cm apart. The end A is 2cm above the H.P and 3cm in front of V.P. The end B is1cm below H.P. and 4cm behind the V.P. Determine the true length and traces of AB, and its inclination with the two planes

X Y50

20

a’

30

a40

b

10

b’

91

Ans. θ=20º

Ans. Ø=50º

h

v’

Given,

A0B0=50mm A is 20mm ↑ HP & 30mm →V.P. B is 10mm ↓ & 40mm ← V.P.


A1

B1

θ

TL

TL=91

A2

B2T

L

Ø

Q10.18: The ends of a line PQ are on the same projector. The end P is 30 mm below the HP and 12 mm behind the VP. The Q is 55 mm above the HP and 45mm in front of the VP. Determine the true length and traces of PQ and its inclination with two planes. Given, p0q0= 0mm P is 30mm ↓ HP & 12mm ←V.P. Q is 55mm ↑ & 45mm → V.P.


X Y

30

p’

12

p

55

q’

45

q

For solving the problem by trapezoidal method, first draw perpendiculars at p & q but on opposite sides as p’ & q’ are on the opposite sides of the xy line.

P1

Q1

h

θθ= 34º

T.L. T.L.=102 mm

P2

Q2

T.L.

v’

Ø

ø= 56º

Q10.14:A line AB, 90mm long, is inclined at 45º to the H.P. and its top view makes an angle of 60º with the V.P. The end A is in the H.P. and 12mm in front of V.P. Draw its front view and find its true inclination with the V.P.

YXa’

a

12

90

45º

b1’

b160º

b

b’ Given,

T.L.=90mm, θ=45º, β=60º A is in the H.P. & 12mm→V.P.

To find/draw,

F.V.,T.V. & Ø

90

b2

38º

Ans. Ø = 38º

Q 4:A line AB, 75mm long, is in the second quadrant with the end A in the H.P. and the end B in the V.P. The line is inclined at 30º to the H.P. and at 45º to the V.P. Draw the projections of AB and determine its traces.

X Y

Given,

A: in H.P., TL=75mm,

θ=30º,

Ø=45º

B: in V.P.,

To find H.T. and V.T.To draw F.V. and T.V.

a’30º

75

b1’

45º

75

a

75

45º b2

X Y

a’

a

b1’

θ

TL

ø

TL

b2

b2’

b’

b2’

b’

b

b

h’

h

v

v’

10.17:A line AB, 90mm long, is inclined at 30 to the HP. Its end A is 12mm above the HP and 20mm in front of the VP. Its FV measures 65mm. Draw the TV of AB and determine its inclination with the VP

X Y

12

a’

20

a

90

30°

b1’

65

b’

b1

b

90

b2

44°

Ans: Ø = 44º

Given,

T.L.=90mm θ = 30º A is 12mm ↑H.P. &20mm→V.P. F.V.=65mm

To find/draw,

T.V. & Ø

Q10.23:Two lines AB & AC make an angle of 120 between them in their FV & TV. AB is parallel to both the HP & VP. Determine the real angle between AB & AC.

X Y

c’

b’

ab

a’120°

c

120°

c1

c1’

c2

c2’

112°

C

Ans. 112º

Given,AB parallel to both HP and VP, a’c’ & ac inclined at 120 with a’b’ and ab respectively

To find,True angle between AB & AC

Assume any arbitrary length of AB & AC. Also assume any distances of A & B from HP & VP

Q10.28: The distance between the end projectors of a line AB is 70 mm and the projectors through traces are 110 mm apart. The end A of the line is 10 mm above the H.P. If the top view and the front view of the line make 30° and 60° with XY line respectively. Draw the projections of the line and determine (i) the traces, (ii) the angles with the H.P. and the V.P. and (iii) the true length of the line.Assume that the line is in the first quadrant.

X Y

Given, a0b0= 70 mm, h’v= 110 mm, A: 10 mm ↑ H.P.,

α =60º, β=30ºTo find, True Length, θ, Ø, H.T. and V.T.

a0

70

b010

a’ 60º

b’

h’

110

v30º

a

h

b

v’

Ans: HT is 63 mm in front of V.P.

63

VT is 190 mm above the H.P.

190

b1

b1’

56º

θ = 56ºT.L.= 146 mm

b2’

b2

16º

ø = 16º

T.L.

T.L.

Q18: The projectors drawn from H.T. and V.T. of a straight line AB are 80 mm apart while those drawn from its ends are 50 mm apart. The H.T. of the line 35 mm in front of V.P. ,the V.T. is 55 mm above the H.P and the end A is 10 mm above the H.P. Draw the projections of AB and determine its length and inclinations with the reference planes.

Given, a0b0= 50 mm, h’v= 80 mm, A: 10 mm ↑ H.P.,

To find, True Length, θ, Ø

X Yh’

80

v

35

h

55

v’

a’

a

a0

10

50

b’

b

b0

b1

b1’

T.L.

Ans. : T.L.= 65 mm

32º

θ = 32º

b2’

b2

T.L.

ø = 20º20º

Q23: The front view of a line makes an angle of 30º with the xy. The H.T. of the line is 45 mm in front of the V.P., while its V.T. is 30 mm below the H.P. One end of the line is 10 mm above the H.P. and the other end is 100 mm in front of the V.P.

Draw the projections of the line and determine (i) its true length, and (ii) its inclinations with the H.P. and the V.P.

X Y

Given, α =30º, H.T.: 45 mm→V.P., V.T. : 30 mm ↓H.P. A: 10 mm ↑ H.P., B: 100 mm→V.P.


10

a’

h’

45

h

30

v’

v

100

a

b’

b

b1’

b1

T.L.

Ans. : T.L.= 66mm

23º

θ = 23º

b2’

b2

T.L.

37º

ø = 37º

30º

Q26: The projectors of the ends of a line PQ are 90 mm apart. P is 20 mm above the H.P. while Q is 45 mm behind the V.P. The H.T. and the V.T. of the line coincide with each other on xy, between the two end projectors and 35 mm away from the projector of the end P. Draw the projections of PQ and determine its true length and inclinations with the two planes.

X Yp0

90


Given, p0q0= 90,

h’v= 0 mm, P: 20 mm ↑ H.P., Q: 45 mm←V.P., p0h’=35 mm

q0

20p’ 45

q

35

h’v h v’

q’p

q1

q1’

T.L.

Ans. : T.L.= 127 mm24º

θ = 24º

q2

T.L.

35º

ø = 35º

q2’

Q10.19 A line AB, inclined at 40º to the V.P. has its end 50mm and 20mm above the H.P. the length of its front view is 65mm and its V.T. is 10mm above the H.P. determine .the true length of AB its inclination with the H.P. and its H.T.

Given,

Ø = 40º, A is 20mm↑HP,

B is 50 mm ↑ HP, FV=65mm,

VT is 10mm ↑ HP

To find,

TL, θ & HT

X Y

50

b’

20

a’

10

v’

h’

b3’

v40º

b3b

ab1

b1’

85

21º h

Ans,

TL = 85 mm, θ = 21º & HT is 17 mm behind VP

Note: Here we do not know position of a and b in top view, but we know v’ and v. So for time being we consider the line as v’b’ instead of a’b’. Now if we make v’b’(FV) parallel to xy, we will get TL of VB in top view at angle Ø.

Q10.16:The end A of a line AB is 25 mm behind the V.P. and is below the H.P. The end B is 12 mm in front of the VP and is above the HP The distance between the projectors is 65mm. The line is inclined at 40 to the HP and its HT is 20 mm behind the VP. Draw the projections of the line and determine its true length and the VT

X Y

Given,

A0B0=65mm

A is 25mm ←V.P.& is ↓H.P. B is 12mm →V.P. & is above HP θ = 40º

To find/draw,

F.V., T.V., T.L., v’25

a

6512

b

20

h

h’

b3

40º

b3’b’

a’v

v’

b1

b1’

Ans. TL= a’b1’=98 mm

Q10. A line AB is in the first quadrant. Its end A and B are 20 mm and 60 mm in front of the V.P. respectively. The distance between the end projectors is 75 mm. The line is inclined at 30º to the H.P. and its H.T. is 10 mm above the xy. Draw the projections of AB and determine its true length and the V.T.Given, AB is in first quadrant, A is 20mm→V.P., B is 60 mm → V.P., a0b0= 75 mm, θ = 30º H.T. is 10mm↑ xy

To draw F.V. and T.V. to find,TL, θ & V.T.

X Y

75

a0 b0

20

a

b60

10

h

h’

b3

b3’

30º

Note: Here we do not know position of a’ and b’ in front view, but we know h’ and h. So for time being we consider the line as hb instead of ab. Now if we make hb(TV) parallel to xy, we will get TL of HB in front view at angle θ.

b’

a’Note: As b3’ and b’ lies on the same locus, similarly a3’ and a’ will also lie on the same locus. And a3’b3’ will be true length of AB

a3’

T.L.

Ans. TL= a3’b3’=98mmv

v’

VT is 13 mm above the H.P.

Q9. The front view of a line AB measures 65 mm and makes an angle of 45º with xy. A is in the H.P. and V.T. of the line is 15 mm below the H.P. The line is inclined at 30º to the V.P. draw the projections of AB and find its true length and inclination with the H.P. Also locate its H.T.Given, FV=65mm,

α = 45º, A is inHP, VT is 15 mm ↓HP Ø = 30º,

To find,

TL, θ & HT

X Y45º

a’

b’

15

v’

v

b3’

30º

b3b

ab1

b1’

T.L.

38ºh’

h 12

Ans: TL=75mm

HT is 12 mm in front of VP

θ = 38º

65


Q 33. The front view of a line AB measures 70 mm and makes an angle of 45º with xy. A is in the H.P. and V.T. of the line is 15 mm below the H.P. The line is inclined at 30º to the V.P. draw the projections of AB, and find its true length and inclination with the H.P. Also locate its H.T.Given, FV=70 mm,

α = 45º, A is in HP, VT is 15mm ↓HP Ø = 30º,

To find,

TL, θ & HT

X Y45º

b’

15

v’

a’v

b3’

30º

b3b

ab1

b1’

T.L.

38º

Ans: TL=81mm

θ = 38º

h’

h


Q34. A line AB measures 100 mm. The projectors through its V.T. and the end A are 40 mm apart. The point A is 30 mm below the H.P. and 20 mm behind the V.P. The V.T. is 10 mm above the H.P. Draw the projections of the line and determine its H.T. and inclinations with the H.P. and the V.P.Given, TL=100 mm, a0v= 40 mm, A: 30mm↓ H.P., 20mm ← V.P. VT is 10mm ↑HP

To draw FV & TV and find, HT,θ, & ø

X Yv

40

a0

10

v’

30

a’

20

a

a3

a3’

100

b3’b’

b

42º

θ = 42º

b2

100

ø = 20º

20º

h’

h5

HT is 5 mm behind the VP

Q9. The front view of a line AB measures 65 mm and makes an angle of 45º with xy. A is in the H.P. and V.T. of the line is 15 mm below the H.P. The line is inclined at 30º to the V.P. draw the projections of AB and find its true length and inclination with the H.P. Also locate its H.T.

Q10. A line AB is in the first quadrant. Its end A and B are 20 mm and 60 mm in front of the V.P. respectively. The distance between the end projectors is 75 mm. The line is inclined at 30º to the H.P. and its H.T. is 10 mm above the xy. Draw the projections of AB and determine its true length and the H.T.

Q33. The front view of a line AB measures 70 mm and makes an angle of 45º with xy. A is in the H.P. and V.T. of the line is 15 mm below the H.P. The line is inclined at 30º to the V.P. draw the projections of AB, and find its true length and inclination with the H.P. Also locate its H.T.

Q34. A line AB measures 100 mm. The projectors through its V.T. and the end A are 40 mm apart. The point A is 30 mm below the H.P. and 20 mm behind the V.P. The V.T. is 10 mm above the H.P. Draw the projections of the line and determine its H.T. and inclinations with the H.P. and the V.P.

Q10.9. A line PQ 75 mm long, has its end P in the V.P. and the end Q in the H.P. The line is inclined at 30 to the H.P. and at 60 to the V.P. Draw its projections.

Q10.19 A line AB, inclined at 40º to the V.P. has its end 50mm and 20mm above the H.P. the length of its front view is 65mm and its V.T. is 10mm above the H.P. determine .the true length of AB its inclination with the H.P. and its H.T.

Given,

Ø = 40º, A is 20mm↑HP, B is 50 mm ↑ HP, FV=65mm, VT is 10mm ↑ HP

To find,

TL, θ & HT

X Y

50

b’

20

a’

10

v’

h’v

65

40º

a

b

h21º

A1’

B1’

Ans: A1’B1’=TL=85mm

Ans:HT is 17 mm behind VP

Ans:θ = 21º

Step1: For solving the problem by trapezoidal method, draw a line at 40º(Ø) from VT’. Then draw perpendiculars from a’ and b’ on this line.

Step2: Then draw projectors from a’ and b’ and mark the distance of b’B1’ on the projector of b’ below XY. Similarly mark the distance a’A1’ on the projector of a’ below XY

Q10.25 The straight line AB is inclined at 30º to H.P., while its top view at 45º to XY line. The end A is 25 mm in front of the V.P. and it is below the H.P. The end B is 75 mm behind the V.P. and it is above the H.P. Draw the projections of the line when its V.T. is 40 mm below the H.P. Find the true length of the portion of the straight line which is in the second quadrant and locate its H.T..

Given,

θ=30º ,

β = 45º, A is 25mm →V.P. & ↓ H.P. B is 75mm ← V.P. & ↑ H.P V.T is 40 mm ↓ H.P.

To draw/find, F.V. and T.V, TL of portion in second quadrant H.T.

X Y

25

a β=45º

75

b

v

v’

40

b1

θ=30º

b1’b’

a’

h’

h

HBis the portion of the line in II quadrant as its FV and TV are above XY line

b2

b2’

42

Ans: HT= 49 mm behind the V.P.TL of HB=42mm

Q10.26 The front view of a line PQ makes an angle of 30º with XY line (ground line). The H.T. of the line is 45 mm behind the V.P. while its V.T. is 30 mm above H.P. The end P of the line is 10 mm below H.P. and the end Q is in the first quadrant. The line is 150 mm long. Draw the projections of the line and determine the true length of the portion of the line in second quadrant. Also find the angle of the line with H.P. and V.P.

Given, α = 30º, H.T.:

45 mm←VP, in the HP

V.T.: 30 mm↑HP. In the VP

P:10 mm ↓HP. Q; in first quadrant. TL=150 mm

To draw/find, F.V. and T.V, TL of portion in second quadrant θ. ø

X Yh’

45

h

10

p’

30

v’

vα=30º

p v1

v1’

150

q1’

q’

q

θ=24º

HV is the portion of the line in II quadrant as its FV and TV are above XY line

Ans: θ=24º

v3’

v3

75

TL of HV=75 mm

ø=37º

ø=37º

Q10.27 The end P of a line PQ 130 mm long, is 55 mm in front of the VP. The H.T. of the line is 40 mm in front of the V.P. and the V.T. is 50 mm above the HP. The distance between H.T. and V.T. is 110 mm. Draw the projections of the line PQ and determine its angles with the H.P. and the V.P.

Given, T.L.= 130 mm, P:55 mm →V.P., H.T.: 40 mm→V.P. , V.T.: 50 mm↑H.P., h’v = 110 mm

To draw/find, F.V. and T.V,θ. ø

X Y

110

h’

40

h

v

50

v’

55

p

p’

v1

v1’

130

q1’q’

qθ=23º

q2

ø=19º

130

Q6. The top view of a 75mm long line CD measures 50 mm. C is 50 mm in front of the VP & 15mm below the HP. D is 15 mm in front of the VP & is above the HP. Draw the FV of CD & find its inclinations with the HP and the VP. Show also its traces.

X Y

15

50Given,

TL = 75 mm, TV =50 mm, C is 15mm ↓ HP & 50 mm → VP, D is 15 mm → VP

To draw,

FV & to find θ & Ø

To mark HT & VT

c’

c

Locus of D

Hint 1: Cut an arc of 50 mm from c on locus of D to get the TV of the line

d

Hint 2: Make TV (cd), // to XY so that FV will give TL

d1

d1’d’

θ=48º Ans: θ=48º

d2

50

75

Ø=28º

Ans: Ø=28º75

h’

h

v

v’

d2’

Q10.10 A line PQ 100 mm long is inclined at 30º to the H.P. and at 45º to the V.P. Its mid point is in the V.P. and 20 mm above the H.P. Draw its projections, if its end P is in the third quadrant and Q is in the first quadrant.

Given,

TL = 100, θ = 30º,Ø=45º, Mid point M is 20mm↑HP & in the VP End P in third quadrant & End Q in first quadrant

To draw,

FV & TV

X Y

20

m’

m

30º

q1’

p1’

50

50

50

q2

p2

50 q2’

q’

p’

q

p

q1

p1

p2’

45º

Problem 3: The front view of a 125 mm long line PQ measures 75 mm while its top view measures 100 mm. Its end Q and the mid point M are in the first quadrant. M being 20 mm from both the planes. Draw the projections of line PQ.

X Y20

20

m’

m

q1’

p1’

p1q1

q’

p’

50 50

62.5

62.5

37.5

37.5

p

q

Given,

TL = 125mm, FV=75mm, TV=100mm, Mid point M is 20mm↑HP &20mm in front of the VP End Q and mid point M are in first quadrant

To draw,

FV & TV

Q8:A line AB 65 mm long has its end A in the H.P. & 15 mm in front of the V.P. The end B is in the third quadrant. The line is inclined at 30 to the H.P. and at 60 to the V.P. Draw its projections.

X Y

15

a

a’30º

b1’

b2

65

65

60º b1

b

b’

b2’ HP

VP

X Y

15

a”30º

60º

b”b’

a’

a

b

Given,TL =65mm, A: in the HP & 15mm →VPB: in third quadrant θ = 30º, Ø=60º

To draw,FV & TV

4.8m

4.2m

3.6m

A

B

X Y

4.8m

3.6m

a’

b’

4.2m

a

bb2’

b2

T.L.

Ans. T.L.= 7.32m

Q11:A room is 4.8 m X 4.2 m X 3.6 m high, determine graphically the distance between a top corner and the bottom corner diagonally opposite to it.

Taking scale 1cm=0.5m

PROBLEM 12 :- Two oranges on a tree are respectively 1.8 m and 3.0 m above ground, and those are 1.2 m & 2.1 m from a 0.3 m thick wall, but on the opposite sides of it. The distance between the oranges, measured along the ground and parallel to the wall is 2.7m. Determine the real distance between the oranges.

1.2m

1.8

m

0.3 m

2.7m

3.0

m

2.1m

F.V.

X Y1.2m

1.8

m

2.1m

3.0

m

0.3m

A

B

a’

b’T

.V.

a

2.7

m

b

b1

b1’

T.L.

Ans. T.L.= 4.65m

As the dimensions are in metres, we will use a scale 1cm = 0.5m or 1:50

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