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# Edp projection of planes

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### Edp projection of planes

1. 1. PROJECTIONS OF PLANES In this topic various plane figures are the objects. What is usually asked in the problem? To draw their projections means F.V, T.V. & S.V. What will be given in the problem? 1. Description of the plane figure. 2. It’s position with HP and VP. In which manner it’s position with HP & VP will be described? Upto 16th sheet1.Inclination of it’s SURFACE with one of the reference planes will be given .2. Inclination of one of it’s EDGES with other reference plane will be given (Hence this will be a case of an object inclined to both reference Planes.) Study the illustration showing surface & side inclination given on next page.
2. 2. CASE OF A RECTANGLE – OBSERVE AND NOTE ALL STEPS.SURFACE PARALLEL TO HP SURFACE INCLINED TO HP ONE SMALL SIDE INCLINED TO VP PICTORIAL PRESENTATION PICTORIAL PRESENTATION PICTORIAL PRESENTATION For T.V. For Tv For T.V. For Fo For Fv r F F.V .V. . ORTHOGRAPHIC ORTHOGRAPHIC ORTHOGRAPHIC TV-True Shape FV- Inclined to XY FV- Apparent Shape FV- Line // to xy TV- Reduced Shape TV-Previous Shape d’ VP VP VP c’ d1’ c1’ a’ d’ a1’ b1’ b’ c’ a’ ’ b d1 a d a1 d1 c1 b c b1 c1 a1 HP A HP B HP C b1
3. 3. PROCEDURE OF SOLVING THE PROBLEM:IN THREE STEPS EACH PROBLEM CAN BE SOLVED :( As Shown In Previous Illustration )STEP 1. Assume suitable conditions & draw Fv & Tv of initial position.STEP 2. Now consider surface inclination & draw 2nd Fv & Tv.STEP 3. After this,consider side/edge inclination and draw 3rd ( final) Fv & Tv.ASSUMPTIONS FOR INITIAL POSITION:(Initial Position means assuming surface // to HP or VP)1.If in problem surface is inclined to HP – assume it // HP Or If surface is inclined to VP – assume it // to VP2. Now if surface is assumed // to HP- It’s TV will show True Shape. And If surface is assumed // to VP – It’s FV will show True Shape.3. Hence begin with drawing TV or FV as True Shape.4. While drawing this True Shape – keep one side/edge ( which is making inclination) perpendicular to xy line ( similar to pair no. on previous page illustration ). ANow Complete STEP 2. By making surface inclined to the resp plane & project it’s other view. (Ref. 2nd pair B on previous page illustration )Now Complete STEP 3. By making side inclined to the resp plane & project it’s other view. (Ref. 3nd pair C on previous page illustration ) APPLY SAME STEPS TO SOLVE NEXT ELEVEN PROBLEMS
4. 4. Q12.4: A regular pentagon of 25mm side has one side on the ground. Its plane is inclined at45º to the HP and perpendicular to the VP. Draw its projections and show its traces Hint: As the plane is inclined to HP, it should be kept parallel to HP with one edge perpendicular to VP c’ d’ ’ b a’ b’ e’ d’ c’ e’ ’ a 45º X Y b b1 a a1 c c1 25 e e1 d d1
5. 5. Q.12.5:Draw the projections of a circle of 5 cm diameter having its plane vertical and inclinedat 30º to the V.P. Its centre is 3cm above the H.P. and 2cm in front of the V.P. Show also itstraces 50 Ø 4’ 41’ 3’ 5’ 31’ 51 ’ 2’ 6’ 61’ 21’ 1’ 7’ 11’ 71’ 12’ 8’ 121’ 81’ 30 91 ’ 11’ 9’ 111’ X 10’ Y 101’ 1 20 2, 12 3, 30º 11 4, 1 2, 3, 4, 5, 6, 7 8 10 5, 12 11 10 9 9 6, 8 7
6. 6. Problem 5 : draw a regular hexagon of 40 mm sides, with its two sides vertical. Draw a circleof 40 mm diameter in its centre. The figure represents a hexagonal plate with a hole in it andhaving its surfacre parallel to the VP. Draw its projections when the surface is vertical abdinclined at 30º to the VP. a a’ a1’ 11’ 1’ f1’ 121’ 21’ b1’ f’ 12’ 2’ b’ 11’ 3’ 111’ 31’ 10’ 4’ 101’ 41’ 5’ 91’ 51’ 9’ e1’ c1’ e’ 8’ 6’ c’ 81’ 61’ 7’ 71’ d’ 30º d 1’ X Y e f a d b c 10 e f 9 a d 11 8 10 9 8 1 2 3 4 12 1 11 12 7 6 5 7 2 6 3 4 b c 5
7. 7. Problem 1 : Draw an equilateral triangle of 75 mm sides and inscribe a circle in it. Draw theprojections of the figure, when its plane is vertical and inclined at 30º to the VP and one of thesides of the triangle is inclined at 45º to the HP. a 1’ a’ 11’ 1’ 121’ 21’ 12’ 2’ c 1’ 11’ 3’ c’ 111’ 31’ 10’ 4’ 101’ 41’ 9’ 5’ 91’ 51’ 8’ 6’ 81’ 61’ 75 7’ 71’ b’ 30º b1’ 45º X Y a a b c 10 b 9 10 9 8 1 2 3 4 11 8 6 5 12 1 11 12 7 7 2 6 3 4 c 5
8. 8. Q12.7: Draw the projections of a regular hexagon of 25mm sides, having one of itsside in the H.P. and inclined at 60 to the V.P. and its surface making an angle of 45ºwith the H.P. Side on the H.P. making 60° with the VP. Plane inclined to HP at 45°and ┴ to VP Plane parallel to HP e’ e1’ d’ d1’ f’ f1’ c’ c1’ a’ b’ c’ f’ d’e’ b’ 45º a1 ’ X a’ Y f f1 60º b1 ’ f1 a e a1 e1 e1 a1 d1 b d b1 d1 c c1 b1 c1
9. 9. Q12.6: A square ABCD of 50 mm side has its corner A in the H.P., its diagonal AC inclined at30º to the H.P. and the diagonal BD inclined at 45º to the V.P. and parallel to the H.P. Draw itsprojections. Keep AC parallel to the H.P. Incline AC at 30º to the H.P. & BD perpendicular to V.P. i.e. incline the edge view Incline BD at 45º to the V.P. (considering inclination of (FV) at 30º to the HP AC as inclination of the plane) c’ c1’ b’ d’ b1’ d1 ’ b’ a’ d’ c’ a’ 30º X Y 45º 45º a1’ b1 b b1 c1 a c a1 c1 a1 d1 50 d d1
10. 10. Q: Draw a rhombus of 100 mm and 60 mm long diagonals with longer diagonal horizontal. The figureis the top view of a square having 100 mm long diagonals. Draw its front view. c’ c1 b’d’ d1 b1 a’ b’ d’ c’ a’ X Y 60 a1 b b1 a1 d1 b1 60 a c a1 c1 100 c1 d d1 100 100
11. 11. Q4: Draw projections of a rhombus having diagonals 125 mm and 50 mm long, the smallerdiagonal of which is parallel to both the principal planes, while the other is inclined at 30º tothe H.P. Keep AC parallel to the H.P. & Incline AC at 30º to the H.P. Make BD parallel to XY BD perpendicular to V.P. (considering inclination of AC as inclination of the plane and inclination of BD as inclination c’ of edge) c1’ b’ d’ d1’ b1 ’ b’ d’ c’ a’ a’ 30º X Y a1’ 125 b b1 b1 c1 a1 50 a c c1 a1 d d1 d1
12. 12. Q 2:A regular hexagon of 40mm side has a corner in the HP. Its surface inclined at45° tothe HP and the top view of the diagonal through the corner which is in the HP makes anangle of 60° with the VP. Draw its projections. Top view of the diagonal Plane inclined to HP making 60° with the VP. Plane parallel to HP at 45°and ┴ to VP d’ d1’ c’ ’ c1 ’ e e1’ b’ b1’ f’ f 1’ b’ c’ Y a’ a’ f’ e’ d’ 45° a1’ X 60° f1 f1 e1 a1 f e e1 b1 a d d1 a1 d1 c1 b c b1 c1
13. 13. Q7:A semicircular plate of 80mm diameter has its straight edge in the VP and inclined at 45to HP.The surface of the plate makes an angle of 30 with the VP. Draw its projections. Plane inclined at 30º to the Plane in the V.P. with V.P. and straight edge in the St.edge in V.P. and straight edge ┴ to H.P H.P. inclined at 45º to the H.P. 11 11’ ’ 1’ 2’ 21’ 21 ’ 3’ 31 ’ 31 ’ 4’ Ø 80 41 ’ 71 ’ 41 ’ 5’ 51 ’ 51 61 ’ ’ 6’ 71’ 61’ 7’ 45º 71 11 X 30º Y 1 2 3 4 1 7 21 7 6 5 2 61 6 3 51 31 41 5 4
14. 14. Problem 12.8 : Draw the projections of a circle of 50 mm diameter resting on the HP on point Aon the circumference. Its plane inclined at 45º to the HP and (a) The top view of the diameter ABmaking 30º angle with the VP (b) The the diameter AB making 30º angle with the VP 6, 7 5, 8 9 4, 10 3, 2, 11 12X 1 45º Y 1 2, 3, 4, 5, 6, 7 12 11 10 9 8 4 3 5 2 6 1 7 11 12 8 11 9 10
15. 15. Q12.10: A thin rectangular plate of sides 60 mm X 30 mm has its shorter side in the V.P. andinclined at 30º to the H.P. Project its top view if its front view is a square of 30 mm long sides A rectangle can be seen as a F.V. (square) is drawn first Incline a1’b1’ at 30º to the square in the F.V. only when its H.P. surface is inclined to VP. So for the first view keep the plane // to VP & shorter edge ┴ c1’ to HP 60 b’ c’ b1 ’ c1’ d1’ b1’ 30 a1’ a’ d’ a1 ’ d1’ b1 a1 30º X Y c a a d b b 60 c c1 d1 d
16. 16. Q12.11: A circular plate of negligible thickness and 50 mm diameter appears as an ellipse inthe front view, having its major axis 50 mm long and minor axis 30 mm long. Draw its topview when the major axis of the ellipse is horizontal.A circle can be seen as aellipse in the F.V. only when its Incline the T.V. till the Incline the F.V. till thesurface is inclined to VP. So distance between the end major axis becomesfor the first view keep the projectors is 30 mm horizontalplane // to VP. 50 Ø 4’ 41’ 41 ’ 3’ 5’ 31’ 51 ’ 51’ 31’ 2’ 6’ 21’ 61 ’ 21 ’ 61 ’ 11 ’ 71’ 1’ 7’ 71 ’ 11’ 12’ 8’ 121’ 81’ 121’ 81 ’ 111’ 91’ 11’ 9’ 111’ 101’ Y 91 ’ X 10’ 11 ’ 101’ 121’ 21’ 30 1 2, 31’ 12 111’ 3, 11 1 2, 3, 4, 5, 6, 7 8 101’ 41 ’ 12 11 10 9 4, 10 91’ 51’ 5, 9 81’ 61 ’ 6, 8 7
17. 17. Problem 9 : A plate having shape of an isosceles triangle has base 50 mm long andaltitude 70 mm. It is so placed that in the front view it is seen as an equilateral triangle of50 mm sides an done side inclined at 45º to xy. Draw its top view c1’ a’ a1’ a1’ c’ c1’ 50 b’ b1’ b1’ 45º a.b 70 c a.b a1 b1 c c1
18. 18. Problem 1: Read problem and answer following questionsRectangle 30mm and 50mm 1. Surface inclined to which plane? ------- HPsides is resting on HP on one 2. Assumption for initial position? ------// to HPsmall side which is 300 inclined 3. So which view will show True shape? --- TVto VP,while the surface of the 4. Which side will be vertical? ---One small side.plane makes 450 inclination with Hence begin with TV, draw rectangle below X-YHP. Draw it’s projections. drawing one small side vertical. Surface // to Hp Surface inclined to Hp d’c’ c’1 d’1 c’d’ a’b’ a’ b’ 450 b’1 a’1 YX 300 a a1 d1 a1 d Side Inclined to Vp b1 b c b1 c1 d1 c1
19. 19. X1 4’ 3’ 5’ 4” 41 ’ 4” 2’ 31’ 51 ’ 6’ 5”3” 5”3” 6”2” 21’ 61 ’ 6”2” 1’ 7’ 7”1” 11 ’ 71 ’ 7”1” 12’ 121’ 8’ 81’ 8”12” 8”12” 11’ 41 9’ 60º 9”11” 111’ 31 91 ’ 51 9”11” 101’ 10’ 10” YX 21 61 11 71 121 81 111 91 101 Y1
20. 20. X1 4’ 3’ 5’ 4” 4” 2’ 6’ 5”3” 5”3” 6”2” 6”2” 1’ 7’ 7”1” 7”1” 41 ’ 31’ 51 ’ 12’ 8’ 21 ’ 8”12” 8”12” 61’ 41 11’ 9’ 60º 9”11” 31 9”11” 51 10’X 21 61 11’ 10” 71 ’ Y 11 71 121’ 81 ’ 121 81 111 91 111’ 91 ’ 101 Y1 101’
21. 21. The top view of a plate, the surface of which is inclined at 60º to the HP is a circle of 60 mmdiameter. Draw its three views. X1 7’ 7” 6’ 8’ 6” 8” 5’ 9’ 5” 9” 4’ 10’ 4” 10” 3’ 11’ 3” 11” 2’ 12’ 600 2” 12”X 1’ Y 1” 4 3 5 2 6 1 7 12 8 11 9 10 60 Y1
22. 22. Problem 12.9: Read problem and answer following questionsA 300 – 600 set square of longest side 1 .Surface inclined to which plane? ------- VP100 mm long, is in VP and 300 inclined 2. Assumption for initial position? ------// to VPto HP while it’s surface is 450 inclined 3. So which view will show True shape? --- FVto VP.Draw it’s projections 4. Which side will be vertical? ------longest side.(Surface & Side inclinations directlygiven) Hence begin with FV, draw triangle above X-Y keeping longest side vertical. a’ a’1 c’ c’1 side inclined to Hp c’1 a’1 b’1 b’1 b’ 300 X a b 450 a1 b1 Y a c b c1 c Surface // to Vp Surface inclined to Vp
23. 23. Problem 3: Read problem and answer following questionsA 300 – 600 set square of longest side 1 .Surface inclined to which plane? ------- VP100 mm long is in VP and it’s surface 2. Assumption for initial position? ------// to VP450 inclined to VP. One end of longest 3. So which view will show True shape? --- FVside is 10 mm and other end is 35 mm 4. Which side will be vertical? ------longest side.above HP. Draw it’s projections Hence begin with FV, draw triangle above X-Y(Surface inclination directly given. keeping longest side vertical.Side inclination indirectly given) First TWO steps are similar to previous problem. Note the manner in which side inclination is given. a’ a’1 End A 35 mm above Hp & End B is 10 mm above Hp. So redraw 2nd Fv as final Fv placing these ends as said. c’ c’1 c’1 a’1 35 b’1 b’1 b’ X 10 Y a a1 b 450 b1 a c b c1 c
24. 24. Problem 4: Read problem and answer following questionsA regular pentagon of 30 mm sides is 1. Surface inclined to which plane? ------- HPresting on HP on one of it’s sides with it’s 2. Assumption for initial position? ------ // to HPsurface 450 inclined to HP. 3. So which view will show True shape? --- TVDraw it’s projections when the side in HP 4. Which side will be vertical? -------- any side.makes 300 angle with VP Hence begin with TV,draw pentagon below SURFACE AND SIDE INCLINATIONS X-Y line, taking one side vertical. ARE DIRECTLY GIVEN. d’ d’1 c’e’ e’1 c’1 b’ a’ X b’ a’ c’e’ d’ 450 b’1 Y a’1 a1 e e1 300 e1 a a1 b1 d1 d d1 c1 b b1 c c1
25. 25. Problem 5: Read problem and answer following questions A regular pentagon of 30 mm sides is resting 1. Surface inclined to which plane? ------- HP on HP on one of it’s sides while it’s opposite 2. Assumption for initial position? ------ // to HP vertex (corner) is 30 mm above HP. 3. So which view will show True shape? --- TV Draw projections when side in HP is 300 4. Which side will be vertical? --------any side. inclined to VP. Hence begin with TV,draw pentagon below SURFACE INCLINATION INDIRECTLY GIVEN X-Y line, taking one side vertical. SIDE INCLINATION DIRECTLY GIVEN: ONLY CHANGE is the manner in which surface inclination is described: One side on Hp & it’s opposite corner 30 mm above Hp. d’ d’1Hence redraw 1st Fv as a 2nd Fv making above arrangement. Keep a’b’ on xy & d’ 30 mm above xy. c’e’ c’1 30 e’1 X b’ a’ c’e’ d’ a’ b’ a’1 b’1 Y 300 e1 a1 e e1 a a1 b1 d d1 d1 c1 b b1 c c1
26. 26. c’ c’1 d’ b’1 Problem 8: A circle of 50 mm diameter is a’ b’ d’ c’ b’ resting on Hp on end A of it’s diameter AC a’ 300 a’1 d’1 Y X which is 300 inclined to Hp while it’s Tv 450 a d d1 d is 450 inclined to Vp.Draw it’s projections. 1 1 a ca c1 1 b cRead problem and answer following questions 1 11. Surface inclined to which plane? ------- HP b b12. Assumption for initial position? ------ // to HP3. So which view will show True shape? --- TV The difference in these two problems is in step 3 only.4. Which diameter horizontal? ---------- AC In problem no.8 inclination of Tv of that AC is Hence begin with TV,draw rhombus below given,It could be drawn directly as shown in 3 rd step. X-Y line, taking longer diagonal // to X-Y While in no.9 angle of AC itself i.e. it’s TL, is given. Hence here angle of TL is taken,locus of c 1 Is drawn and then LTV I.e. a1 c1 is marked and Problem 9: A circle of 50 mm diameter is final TV was completed.Study illustration carefully. resting on Hp on end A of it’s diameter AC which is 300 inclined to Hp while it makes c’ c’1 d’ b’1 450 inclined to Vp. Draw it’s projections. a’ b’ d’ c’ b’ a’ a’1 d’1 d d1 d a 300 1 1 Note the difference in TL a ca c1 construction of 3rd step 1 c b in both solutions. 1 1 b b1
27. 27. Read problem and answer following questionsProblem 10: End A of diameter AB of a circle is in HP 1. Surface inclined to which plane? ------- HP A nd end B is in VP.Diameter AB, 50 mm long is 2. Assumption for initial position? ------ // to HP 300 & 600 inclined to HP & VP respectively. 3. So which view will show True shape? --- TV Draw projections of circle. 4. Which diameter horizontal? ---------- AB Hence begin with TV,draw CIRCLE below X-Y line, taking DIA. AB // to X-YThe problem is similar to previous problem of circle – no.9.But in the 3rd step there is one more change.Like 9th problem True Length inclination of dia.AB is definitely expectedbut if you carefully note - the the SUM of it’s inclinations with HP & VP is 900.Means Line AB lies in a Profile Plane.Hence it’s both Tv & Fv must arrive on one single projector.So do the construction accordingly AND note the case carefully.. 300X Y 600 SOLVE SEPARATELY ON DRAWING SHEET TL GIVING NAMES TO VARIOUS POINTS AS USUAL, AS THE CASE IS IMPORTANT
28. 28. Problem 11: Read problem and answer following questions A hexagonal lamina has its one side in HP and 1. Surface inclined to which plane? ------- HP Its apposite parallel side is 25mm above Hp and 2. Assumption for initial position? ------ // to HP In Vp. Draw it’s projections. Take side of hexagon 30 mm long. 3. So which view will show True shape? --- TV 4. Which diameter horizontal? ---------- ACONLY CHANGE is the manner in which surface inclination Hence begin with TV,draw rhombus belowis described: X-Y line, taking longer diagonal // to X-YOne side on Hp & it’s opposite side 25 mm above Hp.Hence redraw 1st Fv as a 2nd Fv making above arrangement.Keep a’b’ on xy & d’e’ 25 mm above xy. e’ e’1 d’1 d’ f’ 25 c’ f’1 c1’ X a’ b’ c’ f’ d’e’ b ’ a’1 b’1 Y a’ e1 d1 f f1 f1 c1 a e a1 e1 a1 b1 As 3rd step b d b1 d1 redraw 2nd Tv keeping c1 side DE on xy line. c Because it is in VP as said in problem.
29. 29. FREELY SUSPENDED CASES . IMPORTANT POINTS 1.In this case the plane of the figure always remains perpendicular to Hp.Problem 12: 2.It may remain parallel or inclined to Vp.An isosceles triangle of 40 mm long 3.Hence TV in this case will be always a LINE view.base side, 60 mm long altitude Is 4.Assuming surface // to Vp, draw true shape in suspended position as FV.freely suspended from one corner of (Here keep line joining point of contact & centroid of fig. vertical )Base side.It’s plane is 450 inclined to 5.Always begin with FV as a True Shape but in a suspended position. Vp. Draw it’s projections. AS shown in 1st FV. a’1 a’ C b’1 b’ g’ g’1 H G c’ c’1 H/3 X Y A B b a, b a,g c g 450 First draw a given triangle With given dimensions, Locate it’s centroid position c And Similarly solve next problemjoin it with point of suspension. of Semi-circle
30. 30. IMPORTANT POINTSProblem 13 1.In this case the plane of the figure always remains perpendicular to Hp.:A semicircle of 100 mm diameter 2.It may remain parallel or inclined to Vp. is suspended from a point on its 3.Hence TV in this case will be always a LINE view. straight edge 30 mm from the midpoint 4.Assuming surface // to Vp, draw true shape in suspended position as FV.of that edge so that the surface makes (Here keep line joining point of contact & centroid of fig. vertical ) an angle of 450 with VP. 5.Always begin with FV as a True Shape but in a suspended position.Draw its projections. AS shown in 1st FV. A a’ 20 mm p’ P G b’ CG g’ c’ e’ d’ X Y 0.414R b c First draw a given semicircle a With given diameter, b c a p,g d e p, Locate it’s centroid position g And djoin it with point of suspension. e
31. 31. To determine true shape of plane figure when it’s projections are given. BY USING AUXILIARY PLANE METHOD WHAT WILL BE THE PROBLEM? Description of final Fv & Tv will be given. You are supposed to determine true shape of that plane figure. Follow the below given steps: 1. Draw the given Fv & Tv as per the given information in problem. 2. Then among all lines of Fv & Tv select a line showing True Length (T.L.) (It’s other view must be // to xy) 3. Draw x1-y1 perpendicular to this line showing T.L. 4. Project view on x1-y1 ( it must be a line view) 5. Draw x2-y2 // to this line view & project new view on it. It will be the required answer i.e. True Shape. The facts you must know:- If you carefully study and observe the solutions of all previous problems, You will find IF ONE VIEW IS A LINE VIEW & THAT TOO PARALLEL TO XY LINE, THEN AND THEN IT’S OTHER VIEW WILL SHOW TRUE SHAPE: NOW FINAL VIEWS ARE ALWAYS SOME SHAPE, NOT LINE VIEWS: SO APPLYING ABOVE METHOD: Study NextWE FIRST CONVERT ONE VIEW IN INCLINED LINE VIEW .(By using x1y1 aux.plane) Four Cases THEN BY MAKING IT // TO X2-Y2 WE GET TRUE SHAPE.
32. 32. Problem 14 Tv is a triangle abc. Ab is 50 mm long, angle cab is 300 and angle cba is 650. a’b’c’ is a Fv. a’ is 25 mm, b’ is 40 mm and c’ is 10 mm above Hp respectively. Draw projections of that figure and find it’s true shape.s per the procedure- irst draw Fv & Tv as per the data.n Tv line ab is // to xy hence it’s other view a’b’ is TL. So draw x 1y1 perpendicular to it. roject view on x1y1.a) First draw projectors from a’b’ & c’ on x1y1.b) from xy take distances of a,b & c( Tv) mark on these projectors from x 1y1. Name points a1b1 & c1.c) This line view is an Aux.Tv. Draw x2y2 // to this line view and project Aux. Fv on it. for that from x1y1 take distances of a’b’ & c’ and mark from x 2y= on new projectors.Name points a’1 b’1 & c’1 and join them. This will be the required true shape. Y1 a1b1 Y2 900 b’ b’1 15 TL a’ 15 C1 10 C’ X1 X X2 a’1 Y c c’1 TRUE SHAPE ALWAYS FOR NEW FV TAKE DISTANCES OF PREVIOUS FV 300 650 AND FOR NEW TV, DISTANCES a b OF PREVIOUS TV 50 mm REMEMBER!!
33. 33. Problem 15: Fv & Tv of a triangular plate are shown. Determine it’s true shape.USE SAME PROCEDURE STEPS 50OF PREVIOUS PROBLEM: 25BUT THERE IS ONE DIFFICULTY: c’ 15NO LINE IS // TO XY IN ANY VIEW. a’ 1’MEANS NO TL IS AVAILABLE. 20IN SUCH CASES DRAW ONE LINE b’// TO XY IN ANY VIEW & IT’S OTHER 10 X YVIEW CAN BE CONSIDERED AS TL 15 x1FOR THE PURPOSE. a c TLHERE a’ 1’ line in Fv is drawn // to xy. 40 90 0HENCE it’s Tv a-1 becomes TL. 1 c’1 a’1 y2 bTHEN FOLLOW SAME STEPS ANDDETERMINE TRUE SHAPE. c1 b’1(STUDY THE ILLUSTRATION) y1 x2 ALWAYS FOR NEW FV TAKE b1 DISTANCES OF PREVIOUS FV AND FOR NEW TV, DISTANCES TRUE d1 OF PREVIOUS TV SHAP E REMEMBER!!
34. 34. PROBLEM 16: Fv & Tv both are circles of 50 mm diameter. Determine true shape of an elliptical plate.ADOPT SAME PROCEDURE.a c is considered as line // to xy.Then a’c’ becomes TL for the purpose.Using steps properly true shape can be 50D y1Easily determined. b’ b1 y2Study the illustration. TL ac1 1 a’ c’ b’1 c’1 d’ d X1 1 X d Y X2 ALWAYS, FOR NEW FV a’1 TAKE DISTANCES OF d’1 PREVIOUS FV AND a c TRUE SHAPE FOR NEW TV, DISTANCES OF PREVIOUS TV REMEMBER!! 50 D. b
35. 35. Problem 17 : Draw a regular pentagon of 30 mm sides with one side 300 inclined to xy. This figure is Tv of some plane whose Fv is A line 450 inclined to xy. TR U E Determine it’s true shape. b1 SH AP a1 E c1IN THIS CASE ALSO TRUE LENGTHIS NOT AVAILABLE IN ANY VIEW. X1BUT ACTUALLY WE DONOT REQUIRETL TO FIND IT’S TRUE SHAPE, AS ONE a’ e1 d1VIEW (FV) IS ALREADY A LINE VIEW.SO JUST BY DRAWING X1Y1 // TO THIS b’VIEW WE CAN PROJECT VIEW ON IT e’AND GET TRUE SHAPE: c’ Y1 d’STUDY THE ILLUSTRATION.. 450 X 300 Y e d ALWAYS FOR NEW FV a TAKE DISTANCES OF PREVIOUS FV AND FOR NEW TV, DISTANCES OF PREVIOUS TV c REMEMBER!! b