Your SlideShare is downloading.
×

×

Saving this for later?
Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.

Text the download link to your phone

Standard text messaging rates apply

Like this presentation? Why not share!

679

Published on

No Downloads

Total Views

679

On Slideshare

0

From Embeds

0

Number of Embeds

0

Shares

0

Downloads

36

Comments

0

Likes

2

No embeds

No notes for slide

- 1. ORTHOGRAPHIC PROJECTIONS OF POINTS, LINES, PLANES, AND SOLIDS. 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. 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 HP X Y 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 IN Point A is Placed In 2ND QUADRANT different A quadrants and it’s Fv & Tv are brought in same plane for Observer to see HP clearly. Fv is visible as it is a view on VP. But as Tv is is a view on Hp, it is rotated downward 900, In clockwise direction.The In front part of Hp comes below xy line and the part behind Vp HP comes above. Observe and note the process. A POINT A IN RD 3 QUADRANT POINT A IN 1ST QUADRANT VP a’ VP a’ A a HP OBSERVER OBSERVER a a HP OBSERVER a’ a a’ VP OBSERVER VP A POINT A IN 4TH QUADRANT
- 5. Basic concepts for drawing projection of point 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. FV does not give any information about position of the point w.r.t. VP 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. TV does not give any information about position of the point w.r.t. HP
- 6. PROJECTIONS OF A POINT IN FIRST QUADRANT. POINT A ABOVE HP & INFRONT OF VP For Tv For Tv PICTORIAL PRESENTATION a’ A Y X POINT A IN HP & INFRONT OF VP POINT A ABOVE HP & IN VP a’ A Y For Fv For Fv Y a’ a X X a For Tv PICTORIAL PRESENTATION A a For Fv ORTHOGRAPHIC PRESENTATIONS OF ALL ABOVE CASES. Fv above xy, Tv below xy. Fv above xy, Tv on xy. VP VP Fv on xy, Tv below xy. VP a’ X a’ Y X a Y a’ X a a HP HP HP Y
- 7. PROJECTIONS OF A POINT IN SECOND QUADRANT. POINT A ABOVE HP & BEHIND VP For Tv For Fv A Y X
- 8. 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 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 .
- 9. Orthographic Pattern For Tv (Pictorial Presentation) Note: Fv is a vertical line Showing True Length & Tv is a point. a’ V.P 1. A Line perpendicular to Hp & // to Vp . A FV b’ Y Fo r B V.P. a’ Fv b’ X Y Fv TV a b Tv a b X H.P. Orthographic Pattern (Pictorial Presentation) 2. . V.P A Line // to Hp & // to Vp . F.V V.P. Note: Fv & Tv both are // to xy & both show T. L. For Tv b’ B a’ Fv b’ a’ A X Y Fo rF v b X a Y a V. T. H.P. Tv b
- 10. .P . V 3. b’ a’ b’ . F.V B F. V. A Line inclined to Hp and parallel to Vp V.P. Fv inclined to xy Tv parallel to xy. θ Y X θ (Pictorial presentation) A . T.V X θ a’ Y a b T.V. b a H.P. Orthographic Projections 4. A Line inclined to Vp and parallel to Hp .P. V a’ V.P. Tv inclined to xy Fv parallel to xy. a’ . b’ F .V A Ø B (Pictorial presentation) Ø T.V. b’ X Y a a Fv Ø Tv b H.P. b
- 11. For Tv For Tv 5. b’ F.V Y a’ For Fv A X a On removal of object i.e. Line AB Fv as a image on Vp. Tv as a image on Hp, β α Y a’ For Fv A X T.V. B . F.V B α b’ . V.P (Pictorial presentation) . . V.P A Line inclined to both Hp and Vp b a β T.V. b V.P. b’ FV a’ α X Y Orthographic Projections Fv is seen on Vp clearly. To see Tv clearly, HP is rotated 900 downwards, a Hence it comes below xy. β TV H.P. b 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)
- 12. 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). Orthographic Projections Means Fv & Tv of Line AB are shown below, with their apparent Inclinations α &β V.P. V.P. V.P. b’ b’ FV a’ α a’ Y Fv β θ TL a’ θ X Y a TV H.P. b’ b1’ b 1’ FV X a Note the procedure When True Length is known, How to locate FV & TV. (Component a’b2’ of TL is drawn which is further rotated to determine FV) β TV b1 Here TV (ab) is not // to XY line Hence it’s corresponding FV a’ b’ is not showing True Length & True Inclination with Hp. H.P. Y a b1 Ø β Tv b 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. b2 ’ X TV b α TL H.P. TL b b2 Here a’b1’ is component of TL ab1 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’b1‘) TV can be drawn.
- 13. 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). 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, b1’ b’ A: 20mm ↑HP,20 mm →VP θ: True inclination of the B: 55mm ↑HP,75 mm →VP line with HP = 26º a0 b0 = 50 mm To draw FV & TV, To find TL, θ ,Ø, HT & VT h X h’ a’ θ v’ α : Inclination of FV of the line with HP/XY =83 TL 20 v 55 b2’ α b0 a0 Y 50 20 a Φ β b1 TL Ø: True inclination of the line with VP = 41º β : Inclination of TV of the line with VP/XY h’v’a’b’ are always collinear. h v a b are always collinear. =8 3 75 b b2 List of questions of type I : 10.13,10.7,5,11,12, 10.28,18,23,26,10.18,10.23
- 14. 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 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. 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.
- 15. Line inclined to both HP & VP Type –II PROBLEM Given (i) T.L., θ and Ø, (ii) T.L., F.V., T.V. A 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. to draw projections, find α, β,H.T. and V.T. b1’ b’ Given, A: 25mm ↑HP,25 mm →VP TL=70mm, θ=30º, Ø=45º 70 a’ b2’ 30° 25 X To draw/find, FV, TV, HT & VT h v’ Y h’ v 25 b1 45° a 70 b b2 List of questions of type II : 10.8,10.11,10.12,10.14,10.17, 10.10,1,4,6,8,
- 16. 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. B2 b’ A2 Ø X v’ a’ h a A1 θ b B1 Let a’b’ and ab are FV & TV of a line. Now draw the problem by a’ and b’ For solvingperpendiculars attrapezoidal method draw perpendiculars at a and b Measure distance of a’from the XY line and a from the XY line mark it on on the perpendicular drawn and mark itthe perpendicular drawn from a’. from a. Similarly measure the distance of b’from b the XY line line and it on it on from the XYand mark mark the the perpendicular drawn from b. b’. Mark the points as A1and B1 respectively YMark the points as A2and B2 respectively and join them. This will give True length of and join them. This will give True length of the line. the line. Extend this line to intersect the top view. Extend this intersection gives front view. The point ofline to intersect the HT and the The point of intersection gives be and angle between TL and TV will VT θ. the angle between TL and FV will be 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.
- 17. 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
- 18. 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 line AB Given, TL=75mm,TV=65mm,FV=50mm A is in HP & 12mm→VP To find θ & Ø b1 ’ b’ 50 Ans. θ=30º 75 Ans. Ø=48º a’ X Y 12 30º a Hint: Draw ab1=65mm // to XY. Because when TV is // to XY, FV gives TL. b1 65 48º 75 65 b b2
- 19. 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, To find, True Length, θ,Ø, H.T. and V.T. A is in HP & 25mm ←V.P. B is in VP & 50 ↑ V.P. a0b0=75mm v’ 75 b’ b1’ h a b1 93 Ans. TL= a’b1’=93 mm 93 25 X 50 33º θ=33º b2 ’ a’ h’ b 15º v b2 Y Ø=15º
- 20. 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 AB A is 20mm ↑ HP & 25mm →V.P. B is 40mm ↑ & 65mm → V.P. To find θ & Ø b1 ’ b’ 65 a’ Hint2:Draw locus of b’ 40mm above XY & locus of b 65 mm below XY b2’ 40 TL=65mm Hint1:Mark a’ 20mm above H.P & a 25mm below XY 20 18º Ans. θ=18º X Y 25 Ans. Ø=38º 38º b1 65 a 65 b b2
- 21. 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 Given, To find, True Length, θ,Ø, H.T. and V.T. a0b0=50mm A is 20mm ↑ HP & 30mm →V.P. B is 10mm ↓ & 40mm ← V.P. b b2 91 v’ X v b2 ’ 50 h’ b1 ’ 30 b’ a 50º 10 20 h 40 a’ Y 20º Ans. θ=20º b1 Ans. Ø=50º
- 22. 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 Given, To find, True Length, θ,Ø, H.T. and V.T. A0B0=50mm A is 20mm ↑ HP & 30mm →V.P. B is 10mm ↓ & 40mm ← V.P. B1 A2 b TL 20 Ø 40 θ a’ h 91 v’ X 10 b’ TL 30 50 Y Ans. θ=20º TL=91 Ans. Ø=50º a A1 B2
- 23. 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. Q2 To find, True Length, θ,Ø, H.T. and V.T. 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. 55 q’ L. T. P1 X 45 L. T. 30 12 p Ø Y h v’ T.L.=102 mm θ= 34º θ p’ ø= 56º P2 Q1 q
- 24. 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. b’ Given, b1’ T.L.=90mm, θ=45º, β=60º A is in the H.P. & 12mm→V.P. To find/draw, 90 F.V.,T.V. & Ø Ans. Ø = 38º a’ 12 X a 45º 60º 38º b1 90 b b2 Y
- 25. 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. Given, TL=75mm, A: in H.P., B: in V.P., θ=30º, b1 ’ b’ To draw F.V. and T.V. To find H.T. and V.T. TL a’ X Ø=45º a b2’ θ ø TL h a b b’ v’ b1 ’ 75 75 75 X a’ h’ 30º v 45º b b2 b2 ’ 45º Y b2 Y
- 26. 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 Given, T.L.=90mm θ = 30º A is 12mm ↑H.P. &20mm→V.P. F.V.=65mm b1 ’ b’ To find/draw, 65 T.V. & Ø 90 a’ 12 Y 20 X 30° b1 44° a Ans: Ø = 44º 90 b b2
- 27. 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. 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 C c’ Assume any arbitrary length of AB & AC. Also assume any distances of A & B from HP & VP c2 ’ c1 ’ Ans. 112º 112° b’ 120° a’ Y X a b c2 120° c c1
- 28. 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. v’ Assume that the line is in the first quadrant. Given, a0b0= 70 mm, h’v= 110 mm, A: 10 mm ↑ H.P., α =60º, To find, β=30º b1’ 190 b’ T.L . True Length, Ans: HT is 63 mm in front of V.P. VT is 190 mm above the H.P. θ = 56º T.L.= 146 mm ø = 16º θ, Ø, H.T. and 10 X 56º b0 a0 h’ b2’ 30º T.L. a h 16º 70 110 Y v b2 b 63 V.T. a’ 60º b1
- 29. 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. v’ 32º b’ b1’ . T.L 55 mm, h’v= 80 mm, A: 10 mm ↑ H.P., b2’ a’ X 10 a0b0= 50 To find, True Length, θ, Ø b0 b a0 h’ 35 Given, a h v b2 T.L. 20º 50 80 b1 Y Ans. : T.L.= 65 mm θ = 32º ø = 20º
- 30. 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. b1 ’ b’ To find, True Length, θ, Ø a’ h’ b2’ 23º 30º Y 30 v v’ X . T.L 45 10 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. Ans. : T.L.= 66mm θ = 23º ø = 37º 100 h a 37º b1 T. L. b b2
- 31. 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. Given, p0q0= 90, h’v= 0 mm, P: 20 mm ↑ H.P., Q: 45 mm←V.P., p0h’=35 mm To find, True Length, θ, Ø q2 q 20 X Ans. : T.L.= 127 mm θ = 24º ø = 35º p0 q2’ .L. T h’ v h v’ T.L . 45 p’ Y q0 35º q1 p q’ 24º 35 90 q1’
- 32. 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, To find, Ø = 40º, A is 20mm↑HP, B is 50 mm ↑ HP, FV=65mm, TL, θ & HT ↑ HP b1 ’ b’ 85 50 a’ 21º v’ h 10 b3’ X 40º b1 a v h’ 20 VT is 10mm 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. Y Ans, TL = 85 mm, θ = 21º & HT is 17 mm behind VP b3
- 33. 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 Given, To find/draw, A0B0=65mm F.V., T.V., T.L., v’ A is 25mm ←V.P.& is ↓H.P. B is 12mm →V.P. & is above HP θ = 40º b’ b3 ’ b1 ’ Ans. TL= a’b1’=98 mm v’ b1 b3 h 20 h’ X a’ v 40º 12 25 a b 65 Y
- 34. 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. b’ b3’ 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. . T.L a’ a3 ’ v’ b3 10 h h’ 30º v Y b0 a0 Ans. TL= a3’b3’=98mm 75 VT is 13 mm above the H.P. a 60 20 X 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 b
- 35. 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 b1’ b’ L. T. 65 15 h’ a h Y 45º 38º 12 v 30º a’ X 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. b1 b3 ’ v’ Ans: TL=75mm θ = 38º HT is 12 mm in front of VP b b3
- 36. 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 b’ L T. h’ v 30ºa’ 15 X 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. a v’ b1 ’ . Y 45º 38º b1 h b3 ’ Ans: TL=81mm θ = 38º b b3
- 37. 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 100 20º a 5 a0 h’ Y a3 30 v h 20 10 v’ X b2 b 40 a’ a3 ’ 42º θ = 42º ø = 20º 10 0 HT is 5 mm behind the VP b’ b3 ’
- 38. 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.
- 39. 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. B1’ Given, To find, Ø = 40º, A is 20mm↑HP, B is 50 mm ↑ HP, FV=65mm, VT is 10mm ↑ HP 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. TL, θ & HT 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 b’ A1 ’ 65 40º X v 21º h h’ 20 v’ 10 50 a’ Y a Ans: A1’B1’=TL=85mm Ans:HT is 17 mm behind VP Ans:θ = 21º b
- 40. 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.. To draw/find, F.V. and T.V, TL of portion in second quadrant 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. b h b2 75 HBis the portion of the line in II quadrant as its FV and TV are above XY line b’ h’ v b1 ’ b1 Y 40 25 X 42 b2’ a β=45º v’ a’ θ=30º Ans: HT= 49 mm behind the V.P. TL of HB=42mm
- 41. 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. To draw/find, F.V. and T.V, TL of portion in second quadrant θ. ø α Given, = 30º, H.T.: 45 mm←VP, in the HP V.T.: 30 mm↑HP. In the VP P:10 mm ↓HP. p Q; in first quadrant. TL=150 mm HV is the portion of the line in II quadrant as its FV and TV are above XY line v1 q’ h q1’ v’ 45 75 30 10 X v1’ 150 h’ α=30º θ=24º v3’ ø=37º v Y v3 p’ q Ans: θ=24º TL of HV=75 mm ø=37º
- 42. 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 110 v1’ v’ q’ 130 h’ X 50 q1 ’ v Y θ=23º p’ q2 55 40 q h 130 ø=19º p v1
- 43. 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, TV =50 mm, C is 15mm ↓ HP & 50 mm → VP, D is 15 mm → VP FV & to find θ & Ø v’ d’ To mark HT & VT Hint 1: Cut an arc of 50 mm from c on locus of D to get the TV of the line Hint 2: Make TV (cd), // to XY so that FV will give TL 75 d1 ’ h’ X Y 15 v 50 c’ θ=48º h 50 d2 ’ d2 d Locus of D 75 c Ø=28º d1 Ans: θ=48º Ans: Ø=28º
- 44. 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, To draw, 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 FV & TV q’ p2 X p1 p1’ p m’ 50 30º 50 p’ 50 q2 ’ 20 p2’ q1 ’ 45º m q1 50 q q2 Y
- 45. 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. Given, To draw, 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 q’ 37.5 m’ X p1’ p1 20 .5 62 20 p q1’ .5 62 37.5 Y p’ m 50 q FV & TV 50 q1
- 46. 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. Given, TL =65mm, A: in the HP & 15mm →VP B: in third quadrant To draw, FV & TV VP θ = 30º, Ø=60º b b2 65 b 15 b2’ a’ Y 30º 15 X 60º a b’ 65 a’ X 30º 60º b1 b1’ Y a b” b’ a” HP
- 47. 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 4.8m b’ 3.6m 3.6m B b2 ’ m .2 4 A b a’ 4.2m X b2 .L. T 4.8m a Ans. T.L.= 7.32m Y
- 48. 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. 0.3m T.V. As the dimensions are in metres, we will use a scale 1cm = 0.5m or 1:50 0.3 b’ m T.L. m X m 1.2m Y 2.1m a b1 m 2.7 m 2.7 3.0m 1.8m A 1.2 1.8m 2.1 3.0m a’ B b1’ F.V . b Ans. T.L.= 4.65m

Be the first to comment