Projection of Points and Lines, Exercises of Engineering Drawing and Graphics

Download Projection of Points and Lines and more Exercises Engineering Drawing and Graphics in PDF only on Docsity! 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 NEXT 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 ̎ b̎ A a a’ A a a’ A a 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. 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’ 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) X Y H.P. V.P. X Y  H.P. V.P. a b TV a’ b’ FV TV b2 b1’ 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’ b1’ 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-1 of TL is drawn which is further rotated to determine Fv) 1 a a’ b’ 1’ b  b1 ’   TL b1 Ø TL Fv Tv Orthographic Projections Means Fv & Tv of Line AB are shown below, with their apparent Inclinations  &  Here a -1 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.  The most important diagram showing graphical relations among all important parameters of this topic. Study and memorize it as a CIRCUIT DIAGRAM And use in solving various problems. True Length is never rotated. It’s horizontal component is drawn & it is further rotated to locate view. Views are always rotated, made horizontal & further extended to locate TL,  & Ø Also Remember Important TEN parameters to be remembered with Notations used here onward Ø    1) True Length ( TL) – a’ b1’ & a b 2) Angle of TL with Hp - 3) Angle of TL with Vp – 4) Angle of FV with xy – 5) Angle of TV with xy – 6) LFV (length of FV) – Component (a-1) 7) LTV (length of TV) – Component (a’-1’) 8) Position of A- Distances of a & a’ from xy 9) Position of B- Distances of b & b’ from xy 10) Distance between End Projectors X Y H.P. V.P. 1a b  b1 Ø TLTV a’ b’ 1’ b1 ’  TLFV  LTV LFV Distance between End Projectors. & Construct with a’ Ø & Construct with a b & b1 on same locus. b’ & b1’ on same locus. NOTE this a’ b’ a b X Y b’1 b1 Ø  GROUP (A) GENERAL CASES OF THE LINE INCLINED TO BOTH HP & VP ( based on 10 parameters).PROBLEM 1) Line AB is 75 mm long and it is 300 & 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. 2) Locate a’ 12mm above xy line & a 10mm below xy line. 3) Take 300 angle from a’ & 400 from a and mark TL I.e. 75mm on both lines. Name those points b1’ and b1 respectively. 4) Join both points with a’ and a resp. 5) Draw horizontal lines (Locus) from both points. 6) Draw horizontal component of TL a b1 from point b1 and name it 1. ( the length a-1 gives length of Fv as we have seen already.) 7) Extend it up to locus of a’ and rotating a’ as center locate b’ as shown. Join a’ b’ as Fv. 8) From b’ drop a projector down ward & get point b. Join a & b I.e. Tv. 1 LFV TL TL FV TV X Y a’ 1’ a b’1 LTV TL b1 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. 7. Draw locus from b’1 8. With same steps below get b1 point and draw also locus from it. 9. Now rotating one of the components I.e. a-1 locate b’ and join a’ with it to get Fv. 10. Locate tv similarly and measure Angles  & X Yc’ c LOCUS OF d & d1d d1 d’ d’1 TV FV TL TL   LOCUS OF d’ & d’1 PROBLEM 5 :- T.V. of a 75 mm long Line CD, measures 50 mm. End C is in Hp and 50 mm in front of Vp. End D is 15 mm in front of Vp and it is above Hp. Draw projections of CD and find angles with Hp and Vp. SOLUTION STEPS: 1.Draw xy line and one projector. 2.Locate c’ on xy and c 50mm below xy line. 3.Draw locus from these points. 4.Draw locus of d 15 mm below xy 5.Cut 50mm & 75 mm distances on locus of d from c and mark points d & d1 as these are Tv and line CD lengths resp.& join both with c. 6.From d1 draw a vertical line upward up to xy I.e. up to locus of c’ and draw an arc as shown. 7 Then draw one projector from d to meet this arc in d’ point & join c’ d’ 8. Draw locus of d’ and cut 75 mm on it from c’ as TL 9.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’ b1  300 450 PROBLEM 6 :- Fv of line AB makes 450 angle with XY line and measures 60 mm. Line’s Tv makes 300 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. 15 10 SOLUTION STEPS:- Draw xy line, one projector and locate fv a’ 15 mm above xy. Take 450 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 300 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 30 45 10 LOCUS OF b’ & b’1 X Y 450 VT’ v HT h’ LOCUS OF b & b1 100 a b TV b’1  TL  TL b1 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 450 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 & b1 Draw loci for VT and HT, 30 mm & 45 mm below xy respectively. Take 450 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. X y HT VT h’ a’ v b’ a b 80 50 b’1   TL TL FV TV b 1 10 35 55 Locus of a’ PROBLEM 8 :- Projectors drawn from HT and VT of a line AB are 80 mm apart and those drawn from it’s ends are 50 mm apart. End A is 10 mm above Hp, VT is 55 mm above HP while it’s HT is 35 mm in front of VP. Draw projections, locate traces and find TL of line & inclinations with HP and VP. SOLUTION STEPS:- 1.Draw xy line and two projectors, 80 mm apart and locate HT & VT , 35 mm below xy and 55 mm above xy respectively on these projectors. 2.Locate h’ and v on xy as usual. 3.Now just like previous two problems, Extending certain lines complete Fv & Tv And as usual find TL and it’s inclinations. PROBLEM 10 :- A line AB is 75 mm long. It’s Fv & Tv make 450 and 600 inclinations with X-Y line resp End A is 15 mm above Hp and VT is 20 mm below Xy line. Line is in first quadrant. Draw projections, find inclinations with Hp & Vp. Also locate HT. X Y VT’ v 15 20 Locus of a & a1’ a1’ 75 mm b1’ b1 a1 75 mm b’ a’ b a FV TV HT h’ 450 600   SOLUTION STEPS:- Similar to the previous only change is instead of line’s inclinations, views inclinations are given. So first take those angles from VT & v Properly, construct Fv & Tv of extension, then determine it’s TL( V-a1) and on it’s extension mark TL of line and proceed and complete it. PROBLEM 11 :- The projectors drawn from VT & end A of line AB are 40mm apart. End A is 15mm above Hp and 25 mm in front of Vp. VT of line is 20 mm below Hp. If line is 75mm long, draw it’s projections, find inclinations with HP & Vp X Y 40mm 15 20 25 v VT’ a’ a a1’ b1’b’ b TV FV 75mm b1   Draw two projectors for VT & end A Locate these points and then YES ! YOU CAN COMPLETE IT. X A.I.P. GROUP (C) CASES OF THE LINES IN A.V.P., A.I.P. & PROFILE PLANE.  a’ b’ Line AB is in AIP as shown in above figure no 1. It’s FV (a’b’) is shown projected on Vp.(Looking in arrow direction) Here one can clearly see that the Inclination of AIP with HP = Inclination of FV with XY line Line AB is in AVP as shown in above figure no 2.. It’s TV (a b) is shown projected on Hp.(Looking in arrow direction) Here one can clearly see that the Inclination of AVP with VP = Inclination of TV with XY line A.V.P.  A B  a b B A PROBLEM 13 :- A line AB, 75mm long, has one end A in Vp. Other end B is 15 mm above Hp and 50 mm in front of Vp.Draw the projections of the line when sum of it’s Inclinations with HP & Vp is 900, means it is lying in a profile plane. Find true angles with ref.planes and it’s traces. a b HT VT X Y a’ b’ Side View ( True Length ) a” b” (HT) (VT) HP V P Front view top view SOLUTION STEPS:- After drawing xy line and one projector Locate top view of A I.e point a on xy as It is in Vp, Locate Fv of B i.e.b’15 mm above xy as it is above Hp.and Tv of B i.e. b, 50 mm below xy asit is 50 mm in front of Vp Draw side view structure of Vp and Hp and locate S.V. of point B i.e. b’’ From this point cut 75 mm distance on Vp and Mark a’’ as A is in Vp. (This is also VT of line.) From this point draw locus to left & get a’ Extend SV up to Hp. It will be HT. As it is a Tv Rotate it and bring it on projector of b. Now as discussed earlier SV gives TL of line and at the same time on extension up to Hp & Vp gives inclinations with those panes.   APPLICATIONS OF PRINCIPLES OF PROJECTIONS OF LINES IN SOLVING CASES OF DIFFERENT PRACTICAL SITUATIONS. In these types of problems some situation in the field or some object will be described . It’s relation with Ground ( HP ) And a Wall or some vertical object ( VP ) will be given. Indirectly information regarding Fv & Tv of some line or lines, inclined to both reference Planes will be given and you are supposed to draw it’s projections and further to determine it’s true Length and it’s inclinations with ground. Here various problems along with actual pictures of those situations are given for you to understand those clearly. Now looking for views in given ARROW directions, YOU are supposed to draw projections & find answers, Off course you must visualize the situation properly. CHECK YOUR ANSWERS WITH THE SOLUTIONS GIVEN IN THE END. ALL THE BEST !! Wall P Wall Q A B PROBLEM 14:-Two objects, a flower (A) and an orange (B) are within a rectangular compound wall, whose P & Q are walls meeting at 900. Flower A is 1M & 5.5 M from walls P & Q respectively. Orange B is 4M & 1.5M from walls P & Q respectively. Drawing projection, find distance between them If flower is 1.5 M and orange is 3.5 M above the ground. Consider suitable scale.. TV FV PROBLEM 17:- A pipe line from point A has a downward gradient 1:5 and it runs due East-South. Another Point B is 12 M from A and due East of A and in same level of A. Pipe line from B runs 200 Due East of South and meets pipe line from A at point C. Draw projections and find length of pipe line from B and it’s inclination with ground. A B C Downward Gradient 1:5 1 5 12 M N E S N W S PROBLEM 18: A person observes two objects, A & B, on the ground, from a tower, 15 M high, At the angles of depression 300 & 450. Object A is is due North-West direction of observer and object B is due West direction. Draw projections of situation and find distance of objects from observer and from tower also. A B O 300 450 4.5 M 7.5M 300 450 10 M 15 M FV TV A B C PROBLEM 19:-Guy ropes of two poles fixed at 4.5m and 7.5 m above ground, are attached to a corner of a building 15 M high, make 300 and 450 inclinations with ground respectively.The poles are 10 M apart. Determine by drawing their projections,Length of each rope and distance of poles from building. PROBLEM 22. A room is of size 6.5m L ,5m D,3.5m high. An electric bulb hangs 1m below the center of ceiling. A switch is placed in one of the corners of the room, 1.5m above the flooring. Draw the projections an determine real distance between the bulb and switch. Switch Bulb Front wall Ceiling Side wall Observer TV L D H PROBLEM 23:- A PICTURE FRAME 2 M WIDE AND 1 M TALL IS RESTING ON HORIZONTAL WALL RAILING MAKES 350 INCLINATION WITH WALL. IT IS ATTAACHED TO A HOOK IN THE WALL BY TWO STRINGS. THE HOOK IS 1.5 M ABOVE WALL RAILING. DETERMINE LENGTH OF EACH CHAIN AND TRUE ANGLE BETWEEN THEM 350 1.5 M 1 M 2 M Wall railing FV TV PROBLEM 14:-Two objects, a flower (A) and an orange (B) are within a rectangular compound wall, whose P & Q are walls meeting at 900. Flower A is 1.5m & 1 m from walls P & Q respectively. Orange B is 3.5m & 5.5m from walls P & Q respectively. Drawing projection, find distance between them if flower is 1.5 m and orange is 3.5 m above the ground. Consider suitable scale.. a b a’ b’ b’1 x y 1.5M 3,5M 1M 1.5M 3.5M 5.5M Wall P Wall Q TL (answer) A B Wall Q Wall P F.V. PROBLEM 17:- A pipe line from point A has a downward gradient 1:5 and it runs due South - East. Another Point B is 12 m from A and due East of A and in same level of A. Pipe line from B runs 150 Due East of South and meets pipe line from A at point C. Draw projections and find length of pipe line from B and it’s inclination with ground. A B C Downward Gradient 1:5 1 5 12 m N E S 1 5 a b c x y 150 450 12m N EAST SOUTH W DUE SOUTH -EAST a’ b’ c’2c’ c’1 TL ( answer) TL ( answer) = a’ c’2 = Inclination of pipe line BC FV TV PROBLEM 18: A person observes two objects, A & B, on the ground, from a tower, 15 M high, At the angles of depression 300 & 450. Object A is is due North-West direction of observer and object B is due West direction. Draw projections of situation and find distance of objects from observer and from tower also. N W S A B O 300 450 W S E N o a b o’ a’1 b’a’ 300 450 15M Answers: Distances of objects from observe o’a’1 & o’b’ From tower oa & ob 7.5M10 M FV TV B 4.5 M 300 450 15 M A C PROBLEM 19:-Guy ropes of two poles fixed at 4.5m and 7.5 m above ground, are attached to a corner of a building 15 M high, make 300 and 450 inclinations with ground respectively.The poles are 10 M apart. Determine by drawing their projections,Length of each rope and distance of poles from building. c’ a b c a’ b’ c1’ c’2 12M 15M 4.5M 7.5M 300 450 Answers: Length of Rope BC= b’c’2 Length of Rope AC= a’c’1 Distances of poles from building = ca & cb PROBLEM 22. A room is of size 6.5m L ,5m D,3.5m high. An electric bulb hangs 1m below the center of ceiling. A switch is placed in one of the corners of the room, 1.5m above the flooring. Draw the projections an determine real distance between the bulb and switch. Switch Bulb Front wall Ceiling Side wall Observer TV L D H B- Bulb A-Switch Answer :- a’ b’1 a b x y a’ b’ b’1 6.5m 3.5m 5m 1m 1.5 PROBLEM 23:- A PICTURE FRAME 2 M WIDE AND 1 M TALL IS RESTING ON HORIZONTAL WALL RAILING MAKES 350 INCLINATION WITH WALL. IT IS ATTAACHED TO A HOOK IN THE WALL BY TWO STRINGS. THE HOOK IS 1.5 M ABOVE WALL RAILING. DETERMINE LENGTH OF EACH CHAIN AND TRUE ANGLE BETWEEN THEM 350 1.5 M 1 M 2 M Wall railing FV TV A B C D ad h bc a1 b1 a’b’ c’d’ (wall railing) (frame) (chains) Answers: Length of each chain= hb1 True angle between chains = (chains) X Y h’ 1.5M 1M X Y c’ c LOCUS OF d & d1d d1 d’ d’1 TV FV TL TL   LOCUS OF d’ & d’1 PROBLEM NO.24 T.V. of a 75 mm long Line CD, measures 50 mm. End C is 15 mm below Hp and 50 mm in front of Vp. End D is 15 mm in front of Vp and it is above Hp. Draw projections of CD and find angles with Hp and Vp. SOME CASES OF THE LINE IN DIFFERENT QUADRANTS. REMEMBER: BELOW HP- Means- Fv below xy BEHIND V p- Means- Tv above xy.