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TO DRAW PROJECTIONS OF ANY OBJECT,
ONE MUST HAVE FOLLOWING INFORMATION
A) 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 PROJECTIONS
OF 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 NAMEING DIFFERENT 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”
HP
VP
a’
a
A
POINT A IN
1 ST^ QUADRANT
OBSERVER
VP
HP
POINT A IN
2 ND^ QUADRANT
OBSERVER
a’
a
A
OBSERVER
a
a’
POINT A IN 3 RD^ QUADRANT
HP
VP
A
OBSERVER
a
a’ POINT A IN 4 TH^ 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 as it is a view on VP. But as Tv is is a view on Hp, it is rotated downward 90^0 , 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.
If the point lies above HP, p’ lies above the XY line.
If the point lies in the HP, p’ lies on the XY line.
If the point 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.
If the point lies in front of VP, p lies below the XY line.
If the point lies in the VP, p lies on the XY line.
If the point lies behind the VP, p lies below the XY line.
Basic concepts for drawing projection of point
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 & TV OF 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’ Fv b’
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.
A Line
perpendicular
to Hp & // to Vp
A Line // to Hp & // to Vp
Orthographic Pattern
Orthographic Pattern
(Pictorial Presentation)
(Pictorial Presentation)
X
Y
V.P.
a’ For Fv
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)
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 Projections Fv is seen on Vp clearly.
To see Tv clearly, HP is
rotated 90^0 downwards,
Hence it comes below xy.
On removal of object i.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
b 1
b 1 ’
TL
X Y
H.P.
V.P.
a
b
FV
TV
a’
b’
Here TV (ab) is not // to XY line Hence 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’ b 1 ’ Is showing True Length & True Inclination with Hp.
Note the procedure When Fv & Tv known, How to find True Length. (Views are rotated to determine True Length & it’s inclinations with Hp & Vp).
Note the procedure When True Length is known, How to locate FV & TV. (Component a’b 2 ’ of TL is drawn which is further rotated to determine FV)
a
a’
b’
b
b 1 ’
TL
b 2
Ø
TL
Fv
Tv
Orthographic Projections Means Fv & Tv of Line AB are shown below, with their apparent Inclinations &
Here a’b 1 ’ is component of TL ab 1 gives length of FV. Hence it is brought Up to Locus of a’ and further rotated to get point b’. a’ b’ will be Fv. Similarly drawing component of other TL(a’b 1 ‘) TV can be drawn.
b 1
b 2 ’
a’
b’
a
b
X Y
b’ 1
b 1
Ø
GROUP (A) GENERAL CASES OF THE LINE INCLINED TO BOTH HP & VP
PROBLEM 1)^ ( based on 10^ parameters).
Line AB is 75 mm long and it is 30^0 & 400 Inclined to Hp & Vp respectively. End A is 12mm above Hp and 10 mm in front of Vp. Draw projections. Line is in 1st^ quadrant.
SOLUTION STEPS:
**1) Draw xy line and one projector.
- Locate a’ 12mm above xy line**
**& a 10mm below xy line.
- Take 30**^0 angle from a’ & 40^0 **from a and mark TL I.e. 75mm on both lines. Name those points b 1 ’ and b 1 respectively.
- Join both points with a’ and a resp.
- Draw horizontal lines (Locus) from both points.
- Draw horizontal component of TL a b 1 from point b 1 and name it 1. ( the length a-1 gives length of Fv as we have seen already.)
- Extend it up to locus of a’ and rotating a’ as center locate b’ as shown. Join a’ b’ as Fv.
- From b’ drop a projector down ward & get point b. Join a & b**
LFV
TL
TL
FV
TV
X y
a
a’
b 1
0
TL
b’^ b’ 1
LFV
FV TL
b
TV
PROBLEM 2:
Line AB 75mm long makes 45^0 inclination with Vp while it’s Fv makes 55^0.
End A is 10 mm above Hp and 15 mm in front of Vp.If line is in 1st^ quadrant
draw it’s projections and find it’s inclination with Hp.
LOCUS OF b
LOCUS OF b Solution Steps:- 1.Draw x-y line. 2.Draw one projector for a’ & a 3.Locate a’ 10mm above x-y & Tv a 15 mm below xy. 4.Draw a line 45^0 inclined to xy from point a and cut TL 75 mm on it and name that point b 1 Draw locus from point b 1 5.Take 55^0 angle from a’ for Fv above xy line. 6.Draw a vertical line from b 1 up to locus of a and name it 1****. It is horizontal component of TL & is LFV. 7.Continue it to locus of a’ and rotate upward up to the line of Fv and name it b’ .This a’ b’ line is Fv.
8. Drop a projector from b’ on locus from point b 1 and name intersecting point b****. Line a b is Tv of line AB. 9.Draw locus from b’ and from a’ with TL distance cut point b 1 ‘ 10.Join a’ b 1 ’ as TL and measure it’s angle at a’. It will be true angle of line with HP.
X Y
a’
a
b’ 1
LTV
TL
b 1
b’
b
LFV
TV
FV
TL
PROBLEM 4 :-
Line AB is 75 mm long .It’s Fv and Tv measure 50 mm & 60 mm long respectively. End A is 10 mm above Hp and 15 mm in front of Vp. Draw projections of line AB if end B is in first quadrant.Find angle with Hp and Vp.
SOLUTION STEPS:
1.Draw xy line and one projector. 2.Locate a’ 10 mm above xy and a 15 mm below xy line. 3.Draw locus from these points. 4.Cut 60mm distance on locus of a’ & mark 1’ on it as it is LTV. 5.Similarly Similarly cut 50mm on locus of a and mark point 1 as it is LFV. 6.From 1’ draw a vertical line upward and from a’ taking TL ( 75mm ) in compass, mark b’ 1 point on it. Join a’ b’ 1 points.
- Draw locus from b’ 1
- With same steps below get b 1 point
and draw also locus from it.
- Now rotating one of the components I.e. a-1 locate b’ and join a’ with it to get Fv.
- Locate tv similarly and measure
Angles ^ &
TRACES OF THE LINE:-
THESE ARE THE POINTS OF INTERSECTIONS OF A LINE ( OR IT’S EXTENSION )
WITH RESPECTIVE REFFERENCE PLANES.
A LINE ITSELF OR IT’S EXTENSION, WHERE EVER TOUCHES H.P.,
THAT POINT IS CALLED TRACE OF THE LINE ON H.P.( IT IS CALLED H.T.)
SIMILARLY, A LINE ITSELF OR IT’S EXTENSION, WHERE EVER TOUCHES V.P.,
THAT POINT IS CALLED TRACE OF THE LINE ON V.P.( IT IS CALLED V.T.)
V.T .:- It is a point on Vp.
Hence it is called Fv of a point in Vp.
Hence it’s Tv comes on XY line.( Here onward named as v )
H.T .:- It is a point on Hp.
Hence it is called Tv of a point in Hp.
Hence it’s Fv comes on XY line .( Here onward named as ’h’ )
GROUP (B) PROBLEMS INVOLVING TRACES OF THE LINE.
x (^) y
b’ b’ 1
a
v
VT’
a’
HT
b
h’
b 1
PROBLEM 6 :- Fv of line AB makes 45^0 angle with XY line and measures 60 mm. Line’s Tv makes 30^0 with XY line. End A is 15 mm above Hp and it’s VT is 10 mm below Hp. Draw projections of line AB,determine inclinations with Hp & Vp and locate HT, VT.
SOLUTION STEPS:-
Draw xy line, one projector and locate fv a’ 15 mm above xy. Take 45^0 angle from a’ and marking 60 mm on it locate point b’. Draw locus of VT, 10 mm below xy & extending Fv to this locus locate VT. as fv-h’-vt’ lie on one st.line. Draw projector from vt, locate v on xy. From v take 30^0 angle downward as Tv and it’s inclination can begin with v. Draw projector from b’ and locate b I.e.Tv point. Now rotating views as usual TL and it’s inclinations can be found. Name extension of Fv, touching xy as h’ and below it, on extension of Tv, locate HT.
a’
b’
FV
LOCUS OF b’ & b’ 1
X Y
VT’
v
HT
h’
LOCUS OF b & b 1
a
b
TV
b’ 1
TL
TL
b 1
PROBLEM 7 :
One end of line AB is 10mm above Hp and other end is 100 mm in-front of Vp. It’s Fv is 45^0 inclined to xy while it’s HT & VT are 45mm and 30 mm below xy respectively. Draw projections and find TL with it’s inclinations with Hp & VP.
SOLUTION STEPS:-
Draw xy line, one projector and locate a’ 10 mm above xy. Draw locus 100 mm below xy for points b & b 1
Draw loci for VT and HT, 30 mm & 45 mm below xy respectively. Take 45^0 angle from a’ and extend that line backward to locate h’ and VT, & Locate v on xy above VT. Locate HT below h’ as shown. Then join v – HT – and extend to get top view end b. Draw projector upward and locate b’ Make a b & a’b’ dark. Now as usual rotating views find TL and it’s inclinations.
