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Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
Edp  st line(new)
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Edp st line(new)

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  • 1. ORTHOGRAPHIC PROJECTIONS OF POINTS, LINES, PLANES, AND SOLIDS.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. 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.
  • 2. NOTATIONS FOLLOWING NOTATIONS SHOULD BE FOLLOWED WHILE NAMEING DIFFERENT VIEWS IN ORTHOGRAPHIC PROJECTIONS. OBJECT POINT A LINE AB IT’S TOP VIEW a ab IT’S FRONT VIEW a’ a’ b’ IT’S SIDE VIEW a” a” b”SAME SYSTEM OF NOTATIONS SHOULD BE FOLLOWED INCASE NUMBERS, LIKE 1, 2, 3 – ARE USED.
  • 3. VP 2 nd Quad. 1ST Quad. Y Observer X Y HP X 3rd Quad. 4th Quad.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.
  • 4. Point A is POINT A IN VP POINT A IN Placed In 2ND QUADRANT 1ST QUADRANT VP a’ different A A quadrants a’and it’s Fv & Tv a are brought insame plane for HP OBSERVERObserver to see clearly. HP OBSERVER Fv is visible as it is a view onVP. But as Tv is ais a view on Hp, it is rotateddownward 900, In clockwise direction.The In front part of aHp comes below xy line and thepart behind Vp HP comes above. HP OBSERVER OBSERVER Observe and note the a process. a’ A a’ POINT A IN A POINT A IN 3 QUADRANT RD VP 4TH QUADRANT VP
  • 5. Basic concepts for drawing projection of pointFV & TV of a point always lie in the same vertical lineFV of a point ‘P’ is represented by p’. It shows position of the pointwith 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 withrespect 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 above the XY line.
  • 6. PROJECTIONS OF A POINT IN FIRST QUADRANT. POINT A ABOVE HP POINT A ABOVE HP POINT A IN HP & INFRONT OF VP & IN VP & INFRONT OF VP For Tv For Tv PICTORIAL PICTORIAL For Tv PRESENTATION A PRESENTATION a’ a’ For A Fv Y Y Y For Fv a a’ aX a X X A For Fv ORTHOGRAPHIC PRESENTATIONS OF ALL ABOVE CASES. Fv above xy, Fv above xy, Fv on xy, Tv below xy. Tv on xy. Tv below xy. VP VP VP a’ a’ X Y X Y X a’ Y a a a HP HP HP
  • 7. 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.SIMPLE CASES OF THE LINE1. 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 PAGESHOWING CLEARLY THE NATURE OF FV & TVOF LINES LISTED ABOVE AND NOTE RESULTS .
  • 8. For Tv Orthographic Pattern (Pictorial Presentation) V.P. a’ Note: a’ Fv is a vertical line . A Fv V.P Showing True Length 1. FV & Tv is a point. b’ A Line b’perpendicular Y X Y Fo to Hp B r Fv & TV a b Tv a b // to Vp X H.P. Orthographic Pattern (Pictorial Presentation) For Tv Note: V.P. Fv & Tv both are 2. // to xy a’ Fv b’ . A Line V.P F.V . b’ & B both show T. L. // to Hp a’ & A Y X Y // to Vp Fo rF v b a b Tv X V. T. a H.P.
  • 9. Fv inclined to xy V.P. Tv parallel to xy. b’ 3. V.P . b’ . F.V A Line inclined to Hp B V. θ F. a’ and θ Y parallel to Vp a’ X Y θ(Pictorial presentation) A b a T.V. b . X T.V a H.P. Orthographic Projections Tv inclined to xy V.P. 4. Fv parallel to xy. V .P. a’ Fv b’ A Line inclined to Vp F .V . b’ and a’ parallel to Hp A Ø B X Y(Pictorial presentation) a Ø Ø Tv a b T.V. b H.P.
  • 10. For Tv For Tv 5. A Line inclined to both b’ Hp and Vp b’ . . (Pictorial presentation) V.P V.P B . B F.V . F.V α α Y Y On removal of object a’ For a’ For i.e. Line AB Fv Fv Fv as a image on Vp. A A Tv as a image on Hp, β β XX a T.V. b a T.V. b V.P. b’ FV a’ α X Y Orthographic Projections Note These Facts:- Fv is seen on Vp clearly. Both Fv & Tv are inclined to xy. To see Tv clearly, HP is a β (No view is parallel to xy) rotated 900 downwards, Both Fv & Tv are reduced Hence it comes below xy. TV lengths. (No view shows True Length) H.P. b
  • 11. Orthographic Projections Note the procedure Note the procedure Means Fv & Tv of Line AB When Fv & Tv known, When True Length is known, are shown below, How to find True Length. How to locate FV & TV. with their apparent Inclinations (Views are rotated to determine (Component a’b2’ of TL is drawn True Length & it’s inclinations which is further rotated α &β with Hp & Vp). to determine FV) V.P. V.P. V.P. b’ b’ b 1’ b’ b1’ FV FV Fv TL TL α b2 ’ a’ α a’ θ a’ θX Y X Y X Y b1 a β b1 a Ø a β TV β TL TV TV Tv H.P. b H.P. b H.P. b b2 Here TV (ab) is not // to XY line In this sketch, TV is rotated Here a’b1’ is component Hence it’s corresponding FV and made // to XY line. of TL ab1 gives length of FV. a’ b’ is not showing Hence it’s corresponding Hence it is brought Up to True Length & FV a’ b1’ Is showing Locus of a’ and further rotated True Inclination with Hp. True Length to get point b’. a’ b’ will be Fv. & Similarly drawing component True Inclination with Hp. of other TL(a’b1‘) TV can be drawn.
  • 12. Projection of straight line Line inclined to both HP & VPType-IGiven projections (FV & TV) of the line. To find True length & trueinclination of the line with HP (θ) and with VP(Φ). PROBLEM 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.
  • 13. b’ b1 ’ θ: True inclination of the line with HP = 24º TL 55 a’ b2’ α : Inclination of FV of θ α the line with HP/XY HT VT’ 20X Y v h’ 50 Ø: True inclination of 20 Φ β b1 the line with VP = 41º a β : Inclination of TV of 75 the line with VP/XY TL b b2
  • 14. Line inclined to both HP & VPType –IIGiven (i) T.L., θ and Ø, (ii) T.L., F.V., T.V.to draw projections, find α, β,H.T. and V.T. PROBLEM A line AB, 70mm long, has its end A 20 mm above HP and 20mm in front of VP. It is inclined at 30° to HP and 45°to VP. Draw its projections and mark its traces.
  • 15. b’ b1’ 70 a’ 30° HT b2’15 VT’ YX h’ v20 b1 a 45° 70 b2 b
  • 16. 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 To draw FV &TV of the lineGiven, Hint: Draw ab1=65mm // to XY. ABTL=75mm,TV=65mm,FV=50mm Because when TV is // to XY, FV To find θ & Ø gives TL.A is in HP & 12mm VP → b’ b1 ’ Ans. θ=31º 50 75 Ans. Ø=49º a’X Y 31º 12 65 b1 a 49º 75 65 b b2
  • 17. 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. Given, To draw FV &TV of the line Hint1:Mark a’ 20mm above AB H.P & a 25mm below XY TL=65mm To find θ & Ø A is 20mm ↑ HP & 25mm →V.P. b’ b1 ’ Hint2:Draw locus of b’ 40mm B is 40mm ↑ & 65mm → V.P. above XY & locus of b 65 mm below XY a’ 65 b2 ’ 40 18º 20X Y 25 38º Ans. θ=18º b1 a Ans. Ø=38º 65 65 b b2
  • 18. Q10.13:The projectors of the ends of a line AB are 5cm apart. The end A is2cm above the H.P and 3cm in front of V.P. The end B is1cm below H.P. and4cm behind the V.P. Determine the true length and traces of AB, and itsinclination with the two planes Given, To find, A0B0=50mm True Length, θ,Ø, H.T. and V.T. A is 20mm ↑ HP & 30mm →V.P. B is 10mm ↓ & 40mm ← V.P. b b2 a’ 40 HT b2 ’ 20 VT’ 91X v h’ Y 50 10 b’ 20º 30 Ans. θ=20º a 50º b1 Ans. Ø=50º
  • 19. Q10.14:A line AB, 90mm long, is inclined at 45 to the H.P. and its top view makes an angle of 60 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. b’ Given, b1’ T.L.=90mm, θ=45º, β=60º A is in the H.P. & 12mm→V.P. To find/draw, F.V.,T.V. & Ø 90Ans. Ø = 38º a’ X Y 45º b1 12 60º 38º a 90 b b2
  • 20. Q10.16:The end A of a line AB is 25 mm behind the V.P. and is belowthe H.P. The end B is 12 mm in front of the VP and is above the HP Thedistance between the projectors is 65mm. The line is inclined at 40 tothe HP and its HT is 20 mm behind the VP. Draw the projections of theline and determine its true length and the VTGiven, To find/draw,A0B0=65mm F.V., T.V., T.L., VT’ A is 25mm ←V.P.& is ↓H.P. B is 12mm →V.P. & is above HP θ = 40º b1 ’ b’ b2 ’ Ans. TL= a’b2’=123 mm VT’ a b2 b1 HT 25 20X h’ v Y 12 40º a’ b 65
  • 21. 10.17:A line AB, 90mm long, is inclined at 30 to the HP. Its end A is 12mm above the HP and20mm in front of the VP. Its FV measures 65mm. Draw the TV of AB and determine itsinclination with the VP b’ b1 ’ 90 a’ 65 30° 12 X Y 20 44° b1 a Ans: Ø = 44º 90 b b2
  • 22. Q10.23:Two lines AB & AC make an angle of 120 between them in their FV & TV. AB isparallel to both the HP & VP. Determine the real angle between AB & AC. C c2 ’ c1 ’ c’ 112° Ans. 112º b’ 120° a’ X Y b a c2 c1 120° c
  • 23. 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 isin the third quadrant. The line is inclined at 30 to the H.P. and at 60 to the V.P. Draw itsprojections. VP b b2 b 65 15 a’ b2’ X Y a”X Y a’ 30º HP 30º 60º 15 60º 65 b1 a a b’ b” b’ b1’
  • 24. Q10.19 A line AB, inclined at 40º to the V.P. has its end 50mm and 20mm above theH.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, To find, Ø = 40º, A is 20mm↑HP, TL, θ & HT B is 50 mm ↑ HP, FV=65mm, VT is 10mm ↑ HP b1 ’ b’ 85 a’ 50 b2 ’ 21º HT VT’ 20 10X Y 40º v h’ b1 Ans, a TL = 85 mm, θ = 21º & HT is 17 mm behind VP b2
  • 25. 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 is65mm 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. B1’ Step1: For solving the problem byGiven, To find, trapezoidal method, draw a line at 40º(Ø)Ø = 40º, A is 20mm↑HP, from VT’. Then draw perpendiculars from B is 50 mm ↑ HP, TL, θ & HT a’ and b’ on this line.FV=65mm, VT is 10mm ↑ HP Step2: Then draw projectors from a’ and b’ and mark the distance of b’B1’ on the projector of b’ below XY. Similarly mark b’ the distance a’A1’ on the projector of a’ below XY 65 A1 ’ a’ 50 40º VT’ 21º HT 20X 10 v h’ Y a Ans: A1’B1’=TL=85mm Ans:HT is 17 mm behind VP Ans:θ = 21º b
  • 26. 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. Given, To draw, TL = 75 mm, FV =50 mm, FV & to find θ & Ø C is 15mm ↓ HP & 50 mm → VP, VT’ To mark HT & VT D is 15 mm → VP Hint 1: Cut an arc of 50 mm d’ d1 ’ from c on locus of D to get the TV of the line Hint 2: Make TV (cd), // to XY 75 so that FV will give TL h’X Y v 15 d d2 c’ θ=48º Ans: θ=48º Locus of D 50 Ans: Ø=28º 50 75 HT c d1 Ø=28º
  • 27. Q10.10 A line PQ 100 mm long is inclined at 30º to the H.P. and at 45º to the V.P. Itsmid point is in the V.P. and 20 mm above the H.P. Draw its projections, if its end P is inthe third quadrant and Q is in the first quadrant. Given, To draw, TL = 100, θ = 30º, Mid point M is 20mm↑HP & in the VP FV & TV End P in third quadrant & End Q in first quadrant q’ q1 ’ p2 p p2’ 50 m’ 50 q2 ’ 30º 20 p1 50 45º X q1 Y m p1’ p’ 50 q q2
  • 28. Problem 3: The front view of a 125 mm long line PQ measures 75 mm while its top viewmeasures 100 mm. Its end Q and the mid point M are in the first quadrant. M being 20 mmfrom both the planes. Draw the projections of line PQ.
  • 29. For Tv For Tv 5. A Line inclined to both b’ Hp and Vp b’ . . (Pictorial presentation) V.P V.P B . B F.V . F.V α α Y Y On removal of object a’ For a’ For i.e. Line AB Fv Fv Fv as a image on Vp. A A Tv as a image on Hp, β β XX a T.V. b a T.V. b V.P. B b’ FV a’ α A X Y Orthographic Projections Note These Facts:- Fv is seen on Vp clearly. Both Fv & Tv are inclined to xy. To see Tv clearly, HP is a β (No view is parallel to xy) rotated 900 downwards, Both Fv & Tv are reduced Hence it comes below xy. TV lengths. (No view shows True Length) H.P. b

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