Source: https://www.scribd.com/presentation/61862269/Engineering-Graphics
Timestamp: 2019-04-22 00:51:26+00:00

Document:
OF POINTS, LINES, PLANES, AND SOLIDS.
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.
SAME SYSTEM OF NOTATIONS SHOULD BE FOLLOWED INCASE NUMBERS, LIKE 1, 2, 3 – ARE USED.
1ST Quad. THIS QUADRANT PATTERN.VP 2 n d Quad. . Y Observer X Y X HP 3rd Quad. 4th Quad. IT IS FURTHER USED TO UNDERSTAND ILLUSTRATION PROPERLLY. IF OBSERVED ALONG X-Y LINE ( IN RED ARROW DIRECTION) WILL EXACTLY APPEAR AS SHOWN ON RIGHT SIDE AND HENCE.
Observe and note the process. it is rotated downward 900. But as Tv is is a view on Hp.The In front part of Hp comes below xy line and the part behind Vp HP comes above. Fv is visible as it is a view on VP. In clockwise direction. VP a’ a VP a’ POINT A IN 1ST QUADRANT A HP OBSERVER OBSERVER a a HP OBSERVER OBSERVER A POINT A IN R D 3 QUADRANT a’ a a’ A POINT A IN 4TH QUADRANT VP VP .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.
Tv below xy. POINT A ABOVE HP & INFRONT OF VP For Tv PICTORIAL PRESENTATION POINT A ABOVE HP & IN VP For Tv POINT A IN HP & INFRONT OF VP a’ A Y X a’ a X A Y For F v PICTORIAL PRESENTATION For Tv For F v a’ X Y a A a For F v ORTHOGRAPHIC PRESENTATIONS OF ALL ABOVE CASES. Fv on xy. VP a’ X Y X VP a’ a Y X VP a’ a Y a HP HP HP . Tv below xy.PROJECTIONS OF A POINT IN FIRST QUADRANT. Tv on xy. Fv above xy. Fv above xy.
LINE INCLINED TO VP & PARALLEL TO HP. LINE INCLINED TO BOTH HP & VP.TO DRAW IT’S PROJECTIONS .MEANS FV & TV. POSITION OF IT’S ENDS WITH HP & VP IT’S INCLINATIONS WITH HP & VP WILL BE GIVEN.PROJECTIONS OF STRAIGHT LINES. STUDY ILLUSTRATIONS GIVEN ON NEXT PAGE SHOWING CLEARLY THE NATURE OF FV & TV OF LINES LISTED ABOVE AND NOTE RESULTS. SIMPLE CASES OF THE LINE 1. 2. 3. LINE INCLINED TO HP & PARALLEL TO VP. AIM:. 5. A VERTICAL LINE ( LINE PERPENDICULAR TO HP & // TO VP) LINE PARALLEL TO BOTH HP & VP. . 4. INFORMATION REGARDING A LINE means IT’S LENGTH.
For Tv A Line // to Hp & // to Vp X . Tv H. X V. Fo rF v V.P. Orthographic Pattern (Pictorial Presentation) 2. Fv a’ 1. a’ Fv b’ Y a b b a V.P. . L.V .P a’ F. V. T. .P.P. V.(Pictorial Presentation) a’ For Tv Orthographic Pattern Note: Fv is a vertical line Showing True Length & Tv is a point. b’ B A Y Fo rF v Note: Fv & Tv both are // to xy & both show T.P A FV b’ Y B TV a b A Line perpendicular to Hp & // to Vp b’ Y X X Tv a b H.
Fv inclined to xy Tv parallel to xy.
Tv inclined to xy Fv parallel to xy.
Orthographic Projections Fv is seen on Vp clearly.
Hence it comes below xy.
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).
Here TV (ab) is not // to XY line Hence it’s corresponding FV a’ b’ is not showing True Length & True Inclination with Hp.
True Length & True Inclination with Hp.
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.
2) Angle of TL with Hp Study and memorize it as a CIRCUIT DIAGRAM 3) Angle of TL with Vp – Ø And use in solving various problems.Distances of b & b’ from xy 10) Distance between End Projectors Fv TL a’ X a θ α LTV 1’ LFV Ø Y 1 θ & Ø &β α NOTE this Construct with a’ Construct with a β Tv b’ & b1’ on same locus. Distance between End Projectors.Distances of a & a’ from xy 9) Position of B. 5) Angle of TV with xy – b’ b1’ Important TEN parameters to be remembered with Notations used here onward 6) LTV (length of FV) – Component (a-1) 7) LFV (length of TV) – Component (a’-1’) 8) Position of A.P.1) The most important diagram showing graphical relations True Length ( TL) – a’ b1’ & a b among all important parameters of this topic. θ & Ø . 4) Angle of FV with xy – θ β α V. made horizontal & further extended to locate TL. b b1 Also Remember True Length is never rotated. TL H.P. b & b1 on same locus. It’s horizontal component is drawn & it is further rotated to locate view. Views are always rotated.
75mm on both lines. 3) Take 300 angle from a’ & 400 from a and mark TL I. 8) From b’ drop a projector down TL b’ b’1 θ a’ X a Ø LFV 1 Y TV TL b b1 .) 7) Extend it up to locus of a’ and rotating a’ as center locate b’ as shown. 5) Draw horizontal lines (Locus) from both points. Line AB is 75 mm long and it is 300 & 400 Inclined to Hp & Vp respectively. Draw projections. Line is in 1st quadrant. 4) Join both points with a’ and a resp. Join a’ b’ as Fv.e. Name those points b1’ and b1 respectively. 2) Locate a’ 12mm above xy line & a 10mm below xy line. 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. End A is 12mm above Hp and 10 mm in front of Vp.GROUP (A) PROBLEM 1) GENERAL CASES OF THE LINE INCLINED TO BOTH HP & VP ( based on 10 parameters). FV SOLUTION STEPS: 1) Draw xy line and one projector.
It is horizontal component of TL & is LFV.Draw a vertical line from b1 up to locus of a and name it 1. 6. 4. b’1 LOCUS OF FV 55 a’ 0 TL X Φ 45 0 a LFV 1 y TV b TL LOCUS OF b b1 . b’ Solution Steps:1.Draw one projector for a’ & a 3. Drop a projector from b’ on locus from point b1 and name intersecting point b. 2.Draw locus from b’ and from a’ with TL distance cut point b1‘ 10.PROBLEM 2: Line AB 75mm long makes 450 inclination with Vp while it’s Fv makes 550. End A is 10 mm above Hp and 15 mm in front of Vp.Locate a’ 10mm above x-y & Tv a 15 mm below xy. 8.Draw a line 450 inclined to xy from point a and cut TL 75 mm on it and name that point b1 Draw locus from point b1 5.Continue it to locus of a’ and rotate upward up to the line of Fv and name it b’. 7.This a’ b’ line is Fv. It will be true angle of line with HP. Line a b is Tv of line AB.Take 550 angle from a’ for Fv above xy line.Draw x-y line.Join a’ b1’ as TL and measure it’s angle at a’.If line is in 1st quadrant draw it’s projections and find it’s inclination with Hp. 9.
Draw Fv 500 to xy from a’ and mark b’ Cutting 55mm on it. 6.Similarly draw Tv 600 to xy from a & drawing projector from b’ Locate point b and join a b. 3.Then rotating views as shown.Locate a’ 10 mm above xy and a 15 mm below xy line. 5.Draw xy line and one projector. inclinations of line with Hp & Vp. b’ b’1 FV TL X a’ θ 500 y a Φ 600 b TL b1 .PROBLEM 3: Fv of line AB is 500 inclined to xy and measures 55 mm long while it’s Tv is 600 inclined to xy line.Draw locus from these points.find TL. 4. If end A is 10 mm above Hp and 15 mm in front of Vp. draw it’s projections. SOLUTION STEPS: 1. locate True Lengths ab1 & a’b1’ and their angles with Hp and Vp. 2.
PROBLEM 4 :Line AB is 75 mm long . Join a’ b’1 points.Draw locus from these points. End A is 10 mm above Hp and 15 mm in front of Vp. mark b’1 point on it. 10.Draw xy line and one projector.Cut 60mm distance on locus of a’ & mark 1’ on it as it is LTV. 6. With same steps below get b1 point and draw also locus from it. Now rotating one of the components I.Locate a’ 10 mm above xy and a 15 mm below xy line. Draw projections of line AB if end B is in first quadrant. Draw locus from b’1 8. 9.e. 2. 3. b’ b’1 SOLUTION STEPS: 1. a-1 locate b’ and join a’ with it to get Fv. 4. 7. 5.It’s Fv and Tv measure 50 mm & 60 mm long respectively.From 1’ draw a vertical line upward and from a’ taking TL ( 75mm ) in compass.Find angle with Hp and Vp. Locate tv similarly and measure & Φ Angles θ FV TL θ a’ LTV 1’ X a Y Φ LFV 1 TV b TL b1 .Similarly Similarly cut 50mm on locus of a and mark point 1 as it is LFV.
4. End D is 15 mm in front of Vp and it is above Hp.e. measures 50 mm. 7 Then draw one projector from d to meet this arc in d’ point & join c’ d’ 8. 6. d’ d’1 LOCUS OF d’ & d’1 SOLUTION STEPS: 1.From d1 draw a vertical line upward up to xy I.Draw xy line and one projector.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.Draw locus from these points. up to locus of c’ and draw an arc as shown. of a 75 mm long Line CD.Locate c’ on xy and c 50mm below xy line. End C is in Hp and 50 mm in front of Vp.V. Draw projections of CD and find angles with Hp and Vp. 3.Draw locus of d 15 mm below xy 5. 2.& join both with c.Measure Angles θ & Φ TL X c’ θ d TL FV d1 Y LOCUS OF d & d1 TV c Φ . Draw locus of d’ and cut 75 mm on it from c’ as TL 9.PROBLEM 5 :- T.
( IT IS CALLED H.:- It is a point on Vp.T.GROUP (B) PROBLEMS INVOLVING TRACES OF THE LINE...:H.P. THAT POINT IS CALLED TRACE OF THE LINE ON V. WHERE EVER TOUCHES V.) SIMILARLY.T. Hence it is called Fv of a point in Vp.P.P.( Here onward named as v It is a point on Hp. A LINE ITSELF OR IT’S EXTENSION. A LINE ITSELF OR IT’S EXTENSION.T.( IT IS CALLED V. WHERE EVER TOUCHES H. Hence it’s Tv comes on XY line.T. Hence it’s Fv comes on XY line. Hence it is called Tv of a point in Hp.P. TRACES OF THE LINE:THESE ARE THE POINTS OF INTERSECTIONS OF A LINE ( OR IT’S EXTENSION ) WITH RESPECTIVE REFFERENCE PLANES. THAT POINT IS CALLED TRACE OF THE LINE ON H.( Here onward named as ’h’ ) ) .) V.
This point is VT a’ 1. . Begin with TV. (WHEN PROJECTIONS ARE GIVEN.) Begin with FV. Extend TV up to XY line. Now extend Fv to meet this projector.) 1. (WHEN PROJECTIONS ARE GIVEN. 2. b These points are used to solve next three problems.VT’ always co-linear.h’. 4. Name this point v ( as it is a Tv of a point in Vp) Draw one projector from v. 4. 5.b’ STEPS TO LOCATE HT.v . 2. 3. 4. Points h’ & v always on x-y line. 3. HT & h’ always on one projector. Extend FV up to XY line. TV . FV .HT always co-linear. VT’ HT Observe & note :1. x v h’ FV y a TV 3. VT’ & v always on one projector. 2. Name this point h’ ( as it is a Fv of a point in Hp) Draw one projector from h’. Now extend Tv to meet this projector. This point is HT STEPS TO LOCATE VT.
touching xy as h’ and below it. End A is 15 mm above Hp and it’s VT is 10 mm below Hp. Line’s Tv makes 300 with XY line. Name extension of Fv.determine inclinations with Hp & Vp and locate HT. locate v on xy. Draw projector from vt. θ 450 x v y a ∅ b b1 .line. Draw locus of VT. From v take 300 angle downward as Tv and it’s inclination can begin with v.PROBLEM 6 :. Draw projections of line AB. one projector and 300 locate fv a’ 15 mm above xy. on extension of Tv. Now rotating views as usual TL and it’s inclinations can be found. Draw projector from b’ and locate b I. as fv-h’-vt’ lie on one st. VT.Tv point. 10 mm below xy & extending Fv to this locus locate VT. b’ b’1 a’ 15 SOLUTION STEPS:h’ Draw xy line. locate HT.e. 10 HT Take 450 angle from a’ and VT’ marking 60 mm on it locate point b’.Fv of line AB makes 450 angle with XY line and measures 60 mm.
& Locate v on xy above VT. It’s Fv is 450 inclined to xy while it’s HT & VT are 45mm and 30 mm below xy respectively. Draw locus 100 mm below xy for points b & b1 a Draw loci for VT and HT. Locate HT below h’ as shown. 30 mm & 45 mm below xy respectively. Now as usual rotating views find TL and it’s inclinations. 100 Φ TL TV b b1 LOCUS OF b & b1 . Draw projections and find TL with it’s inclinations with Hp & VP. Draw projector upward and locate b’ Make a b & a’b’ dark. Then join v – HT – and extend to get top view end b. one projector and locate a’ 10 mm above xy.PROBLEM 7 : One end of line AB is 10mm above Hp and other end is 100 mm in-front of Vp. b’ b’ 1 LOCUS OF b’ & b’1 FV a’ TL 450 θ X 45 10 30 v h’ Y HT VT’ SOLUTION STEPS:Draw xy line. Take 450 angle from a’ and extend that line backward to locate h’ and VT.
Now just like previous two problems. Draw projections. locate traces and find TL of line & inclinations with Hp and Vp. 2. 80 mm apart and locate HT & VT .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. 3.Locate h’ and v on xy as usual.Draw xy line and two projectors. VT b’ b’1 55 SOLUTION STEPS:1. End A is 10 mm above Hp. Extending certain lines complete Fv & Tv And as usual find TL and it’s inclinations. FV Locus of a’ 10 h’ 35 a HT 80 a’ TL v b b1 X θ 50 y TV TL Φ . 35 mm below xy and 55 mm above xy respectively on these projectors.PROBLEM 8 :. VT is 35 mm below Hp while it’s HT is 45 mm in front of Vp.
if v & VT’ are considered as first point . angles θ & Φ can be constructed with points VT’ & V respectively. a THIS CONCEPT IS USED TO SOLVE NEXT THREE PROBLEMS. then true inclinations of line with Hp & Vp i. & From point VT’ & h’ angles α & θ can be drawn. TL TV b b1 .Instead of considering a & a’ as projections of first point.e. b’ b1’ FV a’ X v TL Φ θ VT’ Y Then from point v & HT angles β & Φ can be drawn.
one projector Φ (450) and locate on it VT and V.Draw projections of the line and it’s HT. VT’ Where it intersects with locus of a’ name it a1’ as it is TL of that part.b’ Join it with VT’ and mark intersection point (with locus of a1’ ) and name it a’ Now as usual locate points a and b and h’ and HT. a1 TV 10 0m m b b1 . Draw it’s component on locus of VT’ & further rotate to get other end of Fv i. θ (300) Take 300 from VT and draw a line.PROBLEM 9 :Line AB 100 mm long is 300 and 450 inclined to Hp & Vp respectively. End A is 10 mm above Hp and it’s VT is 20 mm below Hp . HT From a1’ cut 100 mm (TL) on it and locate point b1’ Now from v take 450 and draw a line downwards a & Mark on it distance VT-a1’ I.e. 20 Draw locus of a’ 10 mm above xy.TL of extension & name it a1 Extend this line by 100 mm and mark point b1.e. FV b’ b 1’ 10 0 mm Locus of a & a1’ a’ a1’ Y SOLUTION STEPS:10 v h’ X Draw xy.
Also locate HT. 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 10 :A line AB is 75 mm long. So first take those angles from VT & v Properly. find inclinations with Hp & Vp. X v 600 450 HT h’ Φ θ a’ a1’ Y a a1 TV 75 mm b b1 . Draw projections. Line is in first quadrant. 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. b’ FV 75 b 1’ mm Locus of a & a1’ 15 20 VT’ SOLUTION STEPS:Similar to the previous only change is instead of line’s inclinations. views inclinations are given.
If line is 75mm long.The projectors drawn from VT & end A of line AB are 40mm apart. b b1 . VT of line is 20 mm below Hp. draw it’s projections. End A is 15mm above Hp and 25 mm in front of Vp.PROBLEM 11 :. find inclinations with HP & Vp b’ FV b 1’ a’ X 25 15 20 v VT’ θ Φ a a1’ m 5m 7 Y Draw two projectors for VT & end A Locate these points and then 40mm TV YES ! YOU CAN COMPLETE IT.
V.. B A a’ α X 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 AVP with VP = Inclination of TV with XY line a β b . B Line AB is in AVP as shown in above figure no 2.(Looking in arrow direction) Here one can clearly see that the Inclination of AIP with HP = Inclination of FV with XY line α β A A.I.P. b’ A. & PROFILE PLANE.GROUP (C) CASES OF THE LINES IN A.P.P. A.I . It’s TV (a b) is shown projected on Hp..V.P.
TV & FV both are vertical.V . 2. TV θ b” HT Y b HP Results:1.LINE IN A PROFILE PLANE ( MEANS IN A PLANE PERPENDICULAR TO BOTH HP & VP) For T. OBSERVE CAREFULLY ABOVE GIVEN ILLUSTRATION AND 2nd SOLVED PROBLEM. It’s Side View shows True Length ( TL) 3.V. Sum of it’s inclinations with HP & VP equals to 900 ( θ + Φ = 900 ) 4. ORTHOGRAPHIC PATTERN OF LINE IN PROFILE PLANE VP a’ A FV VT PP a” LSV a’ Φ b’ X B a b For b’ a F. It’s HT & VT arrive on same projector and can be easily located From Side View. . hence arrive on one single projector.
well then? Locus of b’ b b1 You sure can complete it as previous problems! Go ahead!! . Draw projections. makes 300 angle with Hp and lies in an Aux.Vertical Plane 450 inclined to Vp.PROBLEM 12 :. fine angle with Vp and Ht.Line AB 80 mm long. End A is 15 mm above Hp and VT is 10 mm below X-y line. b’ Locus of b’ b1’ Locus of a’ & a1’ a’ h’ 45 0 a1’ Y 15 X 10 v VT HT Φ θ a AVP 450 to VP Simply consider inclination of AVP as inclination of TV of our line.
of point B i. Find true angles with ref. means it is lying in a profile plane.planes and it’s traces.PROBLEM 13 :.A line AB.b’15 mm above xy as a X it is above Hp. 75mm long. (This is also VT of line.V. It will be HT.) From this point draw locus to left & get a’ HT Extend SV up to Hp. SOLUTION STEPS:- VT a’ (VT) a” Φ Side View ( True Length ) Front view After drawing xy line and one projector Locate top view of A I. 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.Draw the projections of the line when sum of it’s Inclinations with HP & Vp is 900. has one end A in Vp. Other end B is 15 mm above Hp and 50 mm in front of Vp. 50 mm below xy asit is 50 mm in front of Vp Draw side view structure of Vp and Hp top view and locate S. b’ Locate Fv of B i.e.and Tv of B i.e. b’’ From this point cut 75 mm distance on Vp and b Mark a’’ as A is in Vp.e.e point a on xy as It is in Vp. As it is a Tv Rotate it and bring it on projector of b. VP θ HP b” (HT) Y .
CHECK YOUR ANSWERS WITH THE SOLUTIONS GIVEN IN THE END. Now looking for views in given ARROW directions. ALL THE BEST !! . Indirectly information regarding Fv & Tv of some line or lines. YOU are supposed to draw projections & find answers. Here various problems along with actual pictures of those situations are given for you to understand those clearly. It’s relation with Ground ( HP ) And a Wall or some vertical object ( VP ) will be given. Off course you must visualize the situation properly.APPLICATIONS OF PRINCIPLES OF PROJECTIONS OF LINES IN SOLVING CASES OF DIFFERENT PRACTICAL SITUATIONS. 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. In these types of problems some situation in the field or some object will be described .
find distance between them If flower is 1. Consider suitable scale. whose P & Q are walls meeting at 900.. TV B Wall Q Wall P A FV . Orange B is 4M & 1.5 M and orange is 3.5 M above the ground. Drawing projection.5 M from walls P & Q respectively.PROBLEM 14:-Two objects. a flower (A) and an orange (B) are within a rectangular compound wall.5M from walls P & Q respectively. Flower A is 1M & 5.
00 m above ground and those are 1.2 m & 1. Then find real distance between them by drawing their projections.3M THICK FV .5 m and 3.PROBLEM 15 :.Two mangos on a tree A & B are 1. If the distance measured between them along the ground and parallel to wall is 2.5 m from a 0.3 m thick wall but on opposite sides of it. TV B A 0.6 m.
PROBLEM 16 :. 25mm. 45mm and 65mm long respectively.All equally inclined and the shortest is vertical.B & C are on ground and end O is 100mm above ground. ob & oc are three lines. is TV of three rods OA. TV O m 25m 65 mm C A FV 45 mm B . Draw their projections and find length of each along with their angles with ground.oa.This fig. OB and OC whose ends A.
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.PROBLEM 17:. N Do wn wa rd 5 1 G ra die nt 1 :5 12 M A B E C S . 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.
from a tower. on the ground.PROBLEM 18: A person observes two objects. 15 M high. At the angles of depression 300 & 450. O 300 450 A S N B W . Draw projections of situation and find distance of objects from observer and from tower also. Object A is is due North-West direction of observer and object B is due West direction. A & B.
5m and 7. TV C 15 M A 300 4.5 m above ground.5M . Determine by drawing their projections.The poles are 10 M apart.PROBLEM 19:-Guy ropes of two poles fixed at 4.5 M 450 B FV 10 M 7.Length of each rope and distance of poles from building. are attached to a corner of a building 15 M high. make 300 and 450 inclinations with ground respectively.
Determine graphically length and angle of each rod with flooring.PROBLEM 20:. 7 M FV .2 M and 0.A tank of 4 M height is to be strengthened by four stay rods from each corner by fixing their other ends to the flooring. TV 4M 1. as shown.7 M from two adjacent walls respectively.2 M 0. at a point 1.
TV Hook H 5M D A 2M C 1.A horizontal wooden platform 2 M long and 1.5 M wide is supported by four chains from it’s corners and chains are attached to a hook 5 M above the center of the platform. Draw projections of the objects and determine length of each chain along with it’s inclination with ground. 5 M FV B .PROBLEM 21:.
Ceiling TV Bulb Side w all Front wall H Switch L D Observ er .5m high.5m above the flooring. Draw the projections an determine real distance between the bulb and switch. A room is of size 6.3. A switch is placed in one of the corners of the room.5m L .5m D. An electric bulb hangs 1m below the center of ceiling.PROBLEM 22. 1.
PROBLEM 23:A PICTURE FRAME 2 M WIDE AND 1 M TALL IS RESTING ON HORIZONTAL WALL RAILING MAKES 350 INCLINATION WITH WALL.5 M ABOVE WALL RAILING. IT IS ATTAACHED TO A HOOK IN THE WALL BY TWO STRINGS. DETERMINE LENGTH OF EACH CHAIN AND TRUE ANGLE BETWEEN THEM TV 350 1.5 M 1M FV 2M Wall railing . THE HOOK IS 1.
Draw projections of CD and find angles with Hp and Vp. End C is 15 mm below Hp and 50 mm in front of Vp.Means. End D is 15 mm in front of Vp and it is above Hp.24 T.Means.V. REMEMBER: BELOW HP.Fv below xy BEHIND V p. measures 50 mm. d’ d’1 LOCUS OF d’ & d’1 X c’ θ d TL FV TL Y d1 LOCUS OF d & d1 TV c Φ . of a 75 mm long Line CD.Tv above xy.SOME CASES OF THE LINE IN DIFFERENT QUADRANTS. PROBLEM NO.
b’ FV b’1 LOCUS OF b’ & b’1 a Φ TV TL TL X a’ θ b b1 Y LOCUS OF b & b1 70 . Draw projections and find it’s inclinations with Ht. End B in Vp. Distance between projectors is 70mm. Vt.PROBLEM NO.25 End A of line AB is in Hp and 25 mm behind Vp.and 50mm above Hp.
a 35 Φ FV b’ TL b’1 LOCUS OF b’ & b’1 X 25 a’ θ =300 p p’ p’1 y TL LOCUS OF b & b1 TV b b1 . find inclination with Vp and traces. There is a point P on AB contained by both HP & VP. Line is 300 inclined to Hp.26 End A of a line AB is 25mm below Hp and 35mm behind Vp.PROBLEM NO. Draw projections.
VT.27 End A of a line AB is 25mm above Hp and end B is 55mm behind Vp. b Draw projections. find TL and angles and HT.PROBLEM NO. If both it’s HT & VT coincide on xy in a point. 35mm from projector of A and within two projectors. The distance between end projectors is 75mm. b1 TV 55 a’ 25 θ Vt Ht TL X Y FV TL b’ b’1 a Φ 35 75 .
What will be given in the problem? 1.V. & S.PROJECTIONS OF PLANES In this topic various plane figures are the objects.V. Description of the plane figure. T. In which manner it’s position with HP & VP will be described? 1.Inclination of it’s SURFACE with one of the reference planes will be give . . What is usually asked in the problem? To draw their projections means F.) Study the illustration showing surface & side inclination given on next page. 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. It’s position with HP and VP.V.
CASE OF A RECTANGLE – OBSERVE AND NOTE ALL STEPS.Reduced Shape VP d’ c’ a’ ’ b ORTHOGRAPHIC FV. For F For F .Line // to xy VP a’ b’ a b HP d’ c’ d c ORTHOGRAPHIC FV.Apparent Shape TV-Previous Shape VP d1’ a1’ b1’ d1 c1’ a1 b1 d1 c1 a1 c1 A HP B HP b1 C . v Fo r F.V. ORTHOGRAPHIC TV-True Shape FV. SURFACE INCLINED TO HP PICTORIAL PRESENTATION For Tv ONE SMALL SIDE INCLINED TO VP PICTORIAL PRESENTATION For T. SURFACE PARALLEL TO HP PICTORIAL PRESENTATION For T.V.Inclined to XY TV.V .V.
STEP 3. And If surface is assumed // to VP – It’s FV will show True Shape. on previous page illustration ).It’s TV will show True Shape. By making surface inclined to the resp plane & project it’s other view. (Ref. STEP 2.If in problem surface is inclined to HP – assume it // HP Or If surface is inclined to VP – assume it // to VP 2. 4. 3. While drawing this True Shape – keep one side/edge ( which is making inclination) perpendicular to xy line ( similar to pair no. ASSUMPTIONS FOR INITIAL POSITION: (Initial Position means assuming surface // to HP or VP) 1. 2nd pair B on previous page illustration ) Now Complete STEP 3. Hence begin with drawing TV or FV as True Shape. After this. (Ref. 3nd pair C on previous page illustration ) APPLY SAME STEPS TO SOLVE NEXT ELEVEN PROBLEMS . A Now Complete STEP 2.PROCEDURE OF SOLVING THE PROBLEM: IN THREE STEPS EACH PROBLEM CAN BE SOLVED:( As Shown In Previous Illustration ) STEP 1.consider side/edge inclination and draw 3rd ( final) Fv & Tv. Now consider surface inclination & draw 2nd Fv & Tv. By making side inclined to the resp plane & project it’s other view. Now if surface is assumed // to HP. Assume suitable conditions & draw Fv & Tv of initial position.
TV 4. Surface inclined to which plane? ------HP 2.Problem 1: Rectangle 30mm and 50mm sides is resting on HP on one small side which is 300 inclined to VP. draw rectangle below X-Y drawing one small side vertical. Surface // to Hp Read problem and answer following questions 1. So which view will show True shape? --. Draw it’s projections.while the surface of the plane makes 450 inclination with HP. Hence begin with TV. Which side will be vertical? ---One small side. Assumption for initial position? ------// to HP 3. Surface inclined to Hp d’c’ a’b’ X a b d c c’d’ a’ b’ 450 a1 b1 d1 b1 c’1 d’1 b’1 300 a’1 Y Side Inclined to Vp a1 c1 c1 d1 .
c’ c’1 a’1 side inclined to Hp c’1 b’ b’1 a 300 b’1 b 450 X a b c a1 c c1 b1 Y Surface // to Vp Surface inclined to Vp . Which side will be vertical? ------longest side. draw triangle above X-Y keeping longest side vertical.Surface inclined to which plane? ------VP 2. Assumption for initial position? ------// to VP 3.FV 4.Problem 2: A 300 – 600 set square of longest side 100 mm long.Draw it’s projections (Surface & Side inclinations directly given) a’ a’1 Read problem and answer following questions 1 . So which view will show True shape? --. is in VP and 300 inclined to HP while it’s surface is 450 inclined to VP. Hence begin with FV.
Surface inclined to which plane? ------VP 2. Draw it’s projections Read problem and answer following questions 1 .FV 4. First TWO steps are similar to previous problem. So which view will show True shape? --. draw triangle above X-Y keeping longest side vertical. Which side will be vertical? ------longest side. (Surface inclination directly given. One end of longest side is 10 mm and other end is 35 mm above HP.Problem 3: A 300 – 600 set square of longest side 100 mm long is in VP and it’s surface 450 inclined to VP. Hence begin with FV. Side inclination indirectly given) a’ a’1 c’ c’1 a’1 c’1 35 X b’ a b c b’1 a b 450 b’1 10 Y a1 c c1 b1 . Assumption for initial position? ------// to VP 3. Note the manner in which side inclination is given. 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.
taking one side vertical.TV 4.draw pentagon below X-Y line. d’ d’1 e’1 a’1 e1 a1 d b c b1 c1 e1 d1 d1 c1 c’e’ X b’ a’ c’1 b’1 Y a1 300 c’e’ e d’ b’ a’ 450 a b1 .HP 2.// to HP 3. Surface inclined to which plane? ------.Problem 4: A regular pentagon of 30 mm sides is resting on HP on one of it’s sides with it’s surface 450 inclined to HP. Assumption for initial position? -----. Read problem and answer following questions 1. Hence begin with TV. Which side will be vertical? -------. So which view will show True shape? --. Draw it’s projections when the side in HP makes 300 angle with VP SURFACE AND SIDE INCLINATIONS ARE DIRECTLY GIVEN.any side.
So which view will show True shape? --. Keep a’b’ on xy & d’ 30 mm above xy.HP 2. Assumption for initial position? -----. SURFACE INCLINATION INDIRECTLY GIVEN SIDE INCLINATION DIRECTLY GIVEN: Read problem and answer following questions 1.draw pentagon below X-Y line. b’ a’ d’ 30 a’ b’ e1 a1 e1 d1 b1 d1 c1 d’1 c’e’ e’1 a’1 a1 300 c’1 b’1 Y X c’e’ e d’ a d b c b1 c1 . Which side will be vertical? --------any side.// to HP 3. Draw projections when side in HP is 300 inclined to VP. Surface inclined to which plane? ------. Hence begin with TV. Hence redraw 1st Fv as a 2nd Fv making above arrangement.TV 4. taking one side vertical. ONLY CHANGE is the manner in which surface inclination is described: One side on Hp & it’s opposite corner 30 mm above Hp.Problem 5: A regular pentagon of 30 mm sides is resting on HP on one of it’s sides while it’s opposite vertex (corner) is 30 mm above HP.
Which diagonal horizontal? ---------. So which view will show True shape? --. b c1 c 2 .HP b1 b 2. 3. draw it’s projections.6 inclination of Tv of that diagonal is 4. Hence begin with TV. c’ d’ b’ b’1 a’1 30 a0 1 c’1 d’1 X a’ b’d’ d c’ a’ 450 d1 Y d1 a c1 c a Read problem and answer following questions b1 c1 1 1.e. is X-Y line.// to HP The difference in these two problems is in step 3 only.TV In problem no.e. a1 c1 is marked and final TV was completed.It could be drawn directly as shown in 3rd step. Assumption for initial position? -----.draw rhombus below While in no.Longer given. Draw it’s projections. c’ d’ b’ b’1 a’1 0 a1 30 c1 b1 b1 c’1 d’1 X a’ a b’d’ d c’ a’ a 1 450 Y d1 TL d1 c Note the difference in construction of 3rd step in both solutions.Problem 6: A rhombus of diagonals 40 mm and 70 mm long respectively has one end of it’s longer diagonal in HP while that diagonal is 350 inclined to HP. If the topview of the same diagonal makes 400 inclination with VP. Surface inclined to which plane? ------. Hence here angle of TL is taken. taking longer diagonal // to X-Y given.Study illustration carefully.7 angle of diagonal itself I. Is drawn and then LTV I. it’s TL.locus of c1 Problem 7: A rhombus of diagonals 40 mm and 70 mm long respectively having one end of it’s longer diagonal in HP while that diagonal is 350 inclined to HP and makes 400 inclination with VP.
While in no. is given. Draw it’s projections.8 inclination of Tv of that AC is given. Surface inclined to which plane? ------.e.Problem 8: A circle of 50 mm diameter is resting on Hp on end A of it’s diameter AC which is 300 inclined to Hp while it’s Tv is 450 inclined to Vp. So which view will show True shape? --. Hence here angle of TL is taken. a1 c1 is marked and final TV was completed. d’ b’ c’ a’ b’ d’ d c’ b’1 a’1 a 1 c’1 d’1 d 300 1 a’ d1 ca Note the difference in construction of 3rd step in both solutions. taking longer diagonal // to X-Y b b1 Problem 9: A circle of 50 mm diameter is resting on Hp on end A of it’s diameter AC which is 300 inclined to Hp while it makes 450 inclined to Vp.Draw it’s projections. The difference in these two problems is in step 3 only. it’s TL.HP 2.Study illustration carefully.e. a b 1 c1 b c TL 1 1 b1 . a’ X b’ d’ d c’ a’ d’ b’ 300 c’ b’1 450 c’1 d’1 Y a’1 a 1 d1 ca d 1 a 1 c1 b c 1 1 Read problem and answer following questions 1.locus of c1 Is drawn and then LTV I.TV 4.9 angle of AC itself i. Which diameter horizontal? ---------AC Hence begin with TV. Assumption for initial position? -----.It could be drawn directly as shown in 3rd step.// to HP 3. In problem no.draw rhombus below X-Y line.
So do the construction accordingly AND note the case carefully. Surface inclined to which plane? ------. Hence it’s both Tv & Fv must arrive on one single projector. AB // to X-Y The problem is similar to previous problem of circle – no. Like 9th problem True Length inclination of dia.draw CIRCLE below X-Y line. Which diameter horizontal? ---------AB Hence begin with TV.the the SUM of it’s inclinations with HP & VP is 900. Draw projections of circle. Means Line AB lies in a Profile Plane.HP 2.// to HP 3. Assumption for initial position? -----.Diameter AB. Read problem and answer following questions 1. X 300 600 Y SOLVE SEPARATELY ON DRAWING SHEET GIVING NAMES TO VARIOUS POINTS AS USUAL.Problem 10: End A of diameter AB of a circle is in HP A nd end B is in VP..9. AS THE CASE IS IMPORTANT TL .AB is definitely expected but if you carefully note .TV 4. So which view will show True shape? --. 50 mm long is 300 & 600 inclined to HP & VP respectively. taking DIA. But in the 3rd step there is one more change.
Keep a’b’ on xy & d’e’ 25 mm above xy. taking longer diagonal // to X-Y ONLY CHANGE is the manner in which surface inclination is described: One side on Hp & it’s opposite side 25 mm above Hp. Draw it’s projections.// to HP 3. Which diameter horizontal? ---------AC Hence begin with TV.Problem 11: A hexagonal lamina has its one side in HP and Its apposite parallel side is 25mm above Hp and In Vp.TV 4.HP 2. e’ d’ e’1 f’1 d’1 c1’ 25 f’ c’ X a’ b’ f c’ f’ d’e’ b’ a’ f1 e1 d1 c1 a’1 e1 f1 a1 b’1 d1 c1 b1 Y a b c e d a1 b1 As 3rd step redraw 2nd Tv keeping side DE on xy line. Assumption for initial position? -----.draw rhombus below X-Y line. Hence redraw 1st Fv as a 2nd Fv making above arrangement. . Take side of hexagon 30 mm long. Because it is in VP as said in problem. So which view will show True shape? --. Read problem and answer following questions 1. Surface inclined to which plane? ------.
It’s plane is 450 inclined to Vp. 4.Hence TV in this case will be always a LINE view. 3. (Here keep line joining point of contact & centroid of fig.It may remain parallel or inclined to Vp. Locate it’s centroid position And join it with point of suspension. g 450 c Similarly solve next problem of Semi-circle . a. 60 mm long altitude Is freely suspended from one corner of Base side.g c First draw a given triangle With given dimensions. Draw it’s projections. a’ a’1 C b’ H G H/3 g’ b’1 g’1 c’1 Y b c’ X A B b a. IMPORTANT POINTS 1.Assuming surface // to Vp.FREELY SUSPENDED CASES. draw true shape in suspended position as FV. AS shown in 1st FV.In this case the plane of the figure always remains perpendicular to Hp.Always begin with FV as a True Shape but in a suspended position. 2. vertical ) 5. Problem 12: An isosceles triangle of 40 mm long base side.
Draw its projections.In this case the plane of the figure always remains perpendicular to Hp.g d e g p.Hence TV in this case will be always a LINE view. (Here keep line joining point of contact & centroid of fig. vertical ) 5. d e . 2. AS shown in 1st FV. Locate it’s centroid position And join it with point of suspension.Always begin with FV as a True Shape but in a suspended position.It may remain parallel or inclined to Vp.414R a b c a p. 0. 4. 3. A 20 mm a’ p’ b’ g’ c’ d’ e’ Y b c P G CG X First draw a given semicircle With given diameter. draw true shape in suspended position as FV.Assuming surface // to Vp.IMPORTANT POINTS Problem 13 :A semicircle of 100 mm diameter is suspended from a point on its straight edge 30 mm from the midpoint of that edge so that the surface makes an angle of 450 with VP. 1.
BY USING AUXILIARY PLANE METHOD WHAT WILL BE THE PROBLEM? Description of final Fv & Tv will be given. Follow the below given steps: 1. 2.(By using x1y1 aux. It will be the required answer i. True Shape.) (It’s other view must be // to xy) 3.To determine true shape of plane figure when it’s projections are given. THEN AND THEN IT’S OTHER VIEW WILL SHOW TRUE SHAPE: NOW FINAL VIEWS ARE ALWAYS SOME SHAPE. Then among all lines of Fv & Tv select a line showing True Length (T. Study Next Four Cases .plane) THEN BY MAKING IT // TO X2-Y2 WE GET TRUE SHAPE. Project view on x1-y1 ( it must be a line view) 5. Draw the given Fv & Tv as per the given information in problem. Draw x2-y2 // to this line view & project new view on it. You are supposed to determine true shape of that plane figure.e. 4. Draw x1-y1 perpendicular to this line showing T.L. NOT LINE VIEWS: SO APPLYING ABOVE METHOD: WE FIRST CONVERT ONE VIEW IN INCLINED LINE VIEW . The facts you must know:If you carefully study and observe the solutions of all previous problems.L. You will find IF ONE VIEW IS A LINE VIEW & THAT TOO PARALLEL TO XY LINE.
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.
PROBLEM 16: Fv & Tv both are circles of 50 mm diameter. Determine true shape of an elliptical plate. ADOPT SAME PROCEDURE.
This figure is Tv of some plane whose Fv is A line 450 inclined to xy. AS ONE VIEW (FV) IS ALREADY A LINE VIEW. Determine it’s true shape. a1 IN THIS CASE ALSO TRUE LENGTH IS NOT AVAILABLE IN ANY VIEW. DISTANCES OF PREVIOUS TV a d REMEMBER!! b c .Problem 17 : Draw a regular pentagon of 30 mm sides with one side 300 inclined to xy. X X1 TR U b1 E SH AP E c1 a’ b’ e’ e1 d1 c’ 300 Y1 d’ 450 Y e ALWAYS FOR NEW FV TAKE DISTANCES OF PREVIOUS FV AND FOR NEW TV.. BUT ACTUALLY WE DONOT REQUIRE TL TO FIND IT’S TRUE SHAPE. SO JUST BY DRAWING X1Y1 // TO THIS VIEW WE CAN PROJECT VIEW ON IT AND GET TRUE SHAPE: STUDY THE ILLUSTRATION.
called Cylinder Prisms Cone Pyramids Triangular Square Pentagonal Hexagonal Triangular Square Pentagonal Hexagonal ( A solid having six square faces) Cube Tetrahedron ( A solid having Four triangular faces) . Group B Solids having base of some shape and just a point as a top. those are classified & arranged in to two major groups.To understand and remember various solids in this subject properly. SOLIDS Group A Solids having top and base of same shape apex.
Sections of solids( top & base not parallel) Frustum of cone & pyramids. ( top & base parallel to each other) . Square Prism Top Rectangular Face Longer Edge Corner of base Square Pyramid Apex Slant Edge Cylinder Cone Apex Base Base Edge of Base Triangular Base Face Base Edge of Base Corner of base Generators Imaginary lines generating curved surface of cylinder & cone.SOLIDS Dimensional parameters of different solids.
P On it’s base.P RESTING ON V.P On one generator.V. (Hp) Y X While observing Tv.V. x-y line represents Horizontal Plane.V.P On it’s base. T. (Axis inclined to Hp And // to Vp) LYING ON H. (Axis inclined to Hp And // to Vp) F.P On one point of base circle.P On one generator. LYING ON V. Axis inclined to Vp And // to Hp STANDING ON V. F. T. F. On one point of base circle.) RESTING ON H. X While observing Fv. Axis perpendicular to Vp Axis inclined to Vp And // to Hp And // to Hp .V.V. (Axis perpendicular to Hp And // to Vp. (Vp) Y T.STANDING ON H.V. x-y line represents Vertical Plane.
STEPS TO SOLVE PROBLEMS IN SOLIDS Problem is solved in three steps: STEP 1: ASSUME SOLID STANDING ON THE PLANE WITH WHICH IT IS MAKING INCLINATION. CONSIDERING REMAINING INCLINATION. GENERAL PATTERN ( THREE STEPS ) OF SOLUTION: GROUP B SOLID.IT’S FV WILL BE TRUE SHAPE OF IT’S BASE OR TOP. ASSUME IT STANDING ON VP) IF STANDING ON HP . STEP 3: IN LAST STEP.IT’S TV WILL BE TRUE SHAPE OF IT’S BASE OR TOP: IF STANDING ON VP . CONE AXIS INCLINED HP GROUP A SOLID. ( IF IT IS INCLINED TO HP. BEGIN WITH THIS VIEW: IT’S OTHER VIEW WILL BE A RECTANGLE ( IF SOLID IS CYLINDER OR ONE OF THE PRISMS): IT’S OTHER VIEW WILL BE A TRIANGLE ( IF SOLID IS CONE OR ONE OF THE PYRAMIDS): DRAW FV & TV OF THAT SOLID IN STANDING POSITION: STEP 2: CONSIDERING SOLID’S INCLINATION ( AXIS POSITION ) DRAW IT’S FV & TV. CYLINDER AXIS AXIS VERTICAL INCLINED HP GROUP B SOLID. CONE AXIS AXIS VERTICAL INCLINED HP GROUP A SOLID. CYLINDER AXIS INCLINED HP AXIS INCLINED VP AXIS INCLINED VP AXIS er TO VP AXIS INCLINED VP AXIS er TO VP AXIS INCLINED VP Three steps If solid is inclined to Hp Three steps If solid is inclined to Hp Three steps If solid is inclined to Vp Three steps If solid is inclined to Vp Study Next Twelve Problems and Practice them separately !! . DRAW IT’S FINAL FV & TV. ASSUME IT STANDING ON HP) ( IF IT IS INCLINED TO VP.
OF CUBE ( WITH SIDE VIEW) OF TRUE LENGTH INCLINATION WITH HP & VP. 9 PROBLEM NO. 8 PROBLEM NO. (AUXILIARY PLANE) CASE OF A FRUSTUM (AUXILIARY PLANE) . 2. 3. 12 GENERAL CASES OF SOLIDS INCLINED TO HP & VP CASES OF CUBE & TETRAHEDRON CASE CASE CASE OF FREELY SUSPENDED SOLID WITH SIDE VIEW. 4 PROBLEM NO. 5 & 6 PROBLEM NO.CATEGORIES OF ILLUSTRATED PROBLEMS! PROBLEM NO. 10 & 11 PROBLEM NO. CASES OF COMPOSITE SOLIDS.1. 7 PROBLEM NO.
Select farthest point to observer and draw all lines (remaining)from it. 40 mm base sides and axis 60 mm long. 2.P).Then construct remaining inclination with Vp ( Vp containing axis ic the center line of 2nd Tv. 5.Assume it standing on Hp.) o 1 1.Name all points as shown in illustration. as per the procedure. ( a triangle) 4.Draw square of 40mm sides with one side vertical Tv & taking 50 mm axis project Fv.P.Draw proper outline of new view DARK. Draw its projections. has a triangular face on the ground and the vertical plane containing the axis makes an angle of 450 with the VP. 2.Make visible lines dark and hidden dotted. as it is nearer to Vp) & project final Fv. 6. 4.Make it 450 to xy as shown take apex near to xy.dotted.e. 7.o’c’d’ face on xy. Take apex nearer to VP Solution Steps : Triangular face on Hp . 3. 1 (APEX AWAY FROM V. o’ a’b’ a’1 b’1 X a’b’ a b o c’ d’ d c d’1 a1 c’1 d1 a1 o’1 d o1 1 c1 b 1(APEX Y a1 c1 c’d’ o’ o1 d1 For dark and dotted lines b1 b c1 NEARER TO V.Draw 2nd Fv in lying position I. Select nearest point to observer and draw all lines starting from it-dark. A square pyramid. And project it’s Tv.Problem 1. .It’s Tv will show True Shape of base( square) 3. Decide direction of an observer. means it is lying on Hp: 1.
Name all points as shown in illustration. Solution Steps: Resting on Hp on one generator.Then construct remaining inclination with Vp ( generator o1e1 300 to xy as shown) & project final Fv.Draw 2nd Fv in lying position I. 2.o’e’ on xy.dotted.Make visible lines dark and hidden dotted.Draw 40mm dia.Draw proper outline of new vie DARK. as per the procedure. 2. 4. means lying on Hp: 1. Select nearest point to observer and draw all lines starting from it-dark. And project it’s Tv below xy. 7. 3.e. ( a triangle) 4. o’ a’ h’b ’ c’ g’ d’f’ h’1 g’1 o’ a’1 b’1 X a’ h’b’ h c’ g ’ g f’ d’ e’ f f’1 g1 f1 g1 f1 e e1 h1 a1 b1 c1 o1 e’1 c’ d’1 1 h1 a1 Y o1 e’ 30 o1 a b c d e1 d1 c1 b1 d1 . Circle as Tv & taking 50 mm axis project Fv.Assume it standing on Hp. 6. 5. Decide direction of an observer.Problem 2: A cone 40 mm diameter and 50 mm axis is resting on one generator on Hp which makes 300 inclination with Vp Draw it’s projections. Select farthest point to observer and draw all lines (remaining) from it.It’s Tv will show True Shape of base( circle ) 3. For dark and dotted lines 1.
Problem 3: Resting on Vp on one point of base. Solution Steps: 4’ 4’d’ 3’ c’ a’ d’ 4’ c’ d’ 3’ c’ 1’ 2’ b’ c1 1’ a’ 1’ 3’ a’ c X 2’ b’ a bd c 450 b’ 2’ 350 d1 a1 Y b1 3 4 2 a bd 1 24 3 1 24 3 1 .Assume it standing on Vp A cylinder 40 mm diameter and 50 mm will show axis is resting on one point of a base 2..e. 3. as per the procedure. Hp.Make visible lines dark and hidden dotted.Draw Circle as Fv & taking 50 mm axis 0 circle on Vp while it’s axis makes 45 ( a Rectangle) with Vp and Fv of the axis 350 with 4. 6. 7.Draw 2nd Tv making axis 450 to xy And project it’s Fv above xy.Then construct remaining inclination with Hp ( Fv of axis I.Name all points as shown in illustration.True Shape of base & top( circle ) project Tv.It’s Fv40mm dia. means inclined to Vp: 1. 5. Draw projections. center line of view to xy as shown) & project final Tv.
as per the procedure. 8. Draw it’s projections. ( a triangle) 5.xy.Name all points as shown in illustration. such that it’s one slant edge is vertical and a triangular face through it is perpendicular to Vp.It’s Tv will show True Shape of base( square) 3. 6.Solution Steps : Problem 4:A square pyramid 30 mm base side and 50 mm long axis is resting on it’s apex on Hp. as those will not be visible from top. 4.Draw 2nd Fv keeping o’a’ slant edge vertical & project it’s Tv 7.Draw a corner case square of 30 mm sides as Tv(as shown) Showing all slant edges dotted.e. a’ b’d’ c’ a’ b’d ’ c’ a’1 d’1 b’1 c’1 Y d1 c1 X d a o’ o’ o’1 d1 b1 c1 a1 1 o b1 b o c a1 o1 . 1. Then as usual project final Fv.taking 50 mm axis project Fv.Make visible lines dark and hidden dotted. 2.Then redrew 2nd Tv as final Tv keeping a1o1d1 triangular face perpendicular to Vp I.Assume it standing on Hp but as said on apex.( inverted ).
In final Tv draw same diagonal is perpendicular to Vp as said in problem. 6.Assuming standing on Hp.Draw 2nd Fv in which 1’-p’ line is vertical means c’-3’ diagonal must be horizontal.Problem 5: A cube of 50 mm long edges is so placed on Hp on one corner that a body diagonal is parallel to Hp and perpendicular to Vp Draw it’s projections..a square with all sides equally inclined to xy. Then as usual project final FV. a’ a’1 b’d ’ d’1 c’ a’ b’d’ p’ c’ 3’ d’1 c’1 p’ X d d1 d1 c1 3’ 1’ 1’ 1’ Y b1 a c a1 c1 a1 b b1 . begin with Tv.From 1’ drop a perpendicular on this and name it p’ 4. Solution Steps: 1.Project Fv and name all points of FV & TV. 2. .Draw a body-diagonal joining c’ with 3’( This can become // to xy) 3.Now as usual project Tv.
o’ is finalized by slant edge length) Then complete Fv. Solid of four faces. o’ TL o’ a’ a’1 c’1 o’1 90 0 a’ b’ c’ c b’ c’ b’1 450 Y c1 a1 o1 c1 o1 b1 a o b b1 a1 . Axis length generally not given. Like cube it is also described by One X dimension only. 50 mm. And like all previous problems solve completely. cut on axis line & mark o’ (as axis is not known. IMPORTANT: Tetrahedron is a special type of triangular pyramid in which base sides & slant edges are equal in length. In 2nd Fv make face o’b’c’ vertical as said in problem. Solution Steps As it is resting assume it standing on Hp.Problem 6:A tetrahedron of 50 mm long edges is resting on one edge on Hp while one triangular face containing this edge is vertical and 450 inclined to Vp. name those & axis line. Draw projections. an equilateral triangle as side case as shown: First project base points of Fv on xy.. From a’ with TL of edge. Begin with Tv .
CG H CG H/2 H/4 GROUP A SOLIDS ( Cylinder & Prisms) GROUP B SOLIDS ( Cone & Pyramids) . for different solids are shown below.FREELY SUSPENDED SOLIDS: Positions of CG. from base. on axis.
4.As 2nd Fv. is freely suspended from one corner of base so that a plane containing it’s axis remains parallel to Vp. then line joining point of contact & C.G.Project Fv & locate CG position on axis – ( ¼ H from base. redraw first keeping line g’d’ vertical.) X H/4 a’ b’ c’ e’ e d’ a o b c do 1 d1 b1 c1 . ( Here axis shows inclination with Hp. 1.a regular pentagon.corner case. assume solid standing on Hp initially.) So in all such cases. Solution Steps: In all suspended cases axis shows inclination with Hp.Hence assuming it standing on Hp. LINE o’ d’g’ VERTICAL d’ c’e’ e” a’b’ a” d” c” b” o” e1 a1 Y FOR SIDE VIEW H g’ g’ IMPORTANT: When a solid is freely suspended from a corner. drew Tv . Draw it’s three views.) and name g’ and Join it with corner d’ 3.As usual project corresponding Tv and then Side View looking from.Problem 7: A pentagonal pyramid 30 mm base sides & 60 mm long axis. remains vertical. 2.
Project it’s Tv drawing dark and dotted lines as per the procedure. from all points of Fv.With standard method construct Left-hand side view. draw horizontal lines.After this.Project corresponding Fv. above xy. For dark & dotted lines locate observer on left side of Fv as shown. ( Draw a 450 inclined Line in Tv region ( below xy).) Problem 8: A cube of 50 mm long edges is so placed on Hp on one corner that a body diagonal through this corner is perpendicular to Hp and parallel to Vp Draw it’s three views.& name all points as usual in both views. Name points of intersections and join properly. Project horizontally all points of Tv on this line and reflect vertically upward.Assuming it standing on Hp begin with Tv. a’’ a’ a’ b’d’ c’ d’’ b’’ ’ b’d c’ X d 1’ 1’ d1 c’’ 1’ Y a c a1 c1 b b . to meet these lines. 2. 5.Join a’1’ as body diagonal and draw 2nd Fv making it vertical (I’ on xy) 4.Solution Steps: 1. a square of corner case. 3.
See carefully the final Tv and inclination taken there. So the same construction done in those Problems is done here also. 40 mm base diameter and 60 mm long axis is resting on Hp on one point of base circle such that it’s axis makes 450 inclination with Hp and 400 inclination with Vp. Draw it’s projections. o’ o’1 a’1 h’1 g’1 f’1 o1 a’ 450 b’1 c’1 d’1 ’ h’b ’ c’g ’ d’f X a’ h’b’ c’ g’ g f’ d’ e’ g1 f e h1 a1 b1 c1 1 400 e’1 y e’ Axis True Length h f1 e1 d1 Axis Tv Length Axis Tv Length a b c d o1 f1 e1 d1 c1 b1 Locus of Center 1 1 g1 h1 a1 .Problem 9: A right circular cone. In previous all cases 2nd inclination was done by a parameter not showing TL. So assuming it standing on HP begin as usual.Like Tv of axis is inclined to Vp etc.7 & 9 from projections of planes topic. But here it is clearly said that the axis is 400 inclined to Vp. o’ This case resembles to problem no. Means here TL inclination is expected.
V. Draw x1y1 450 inclined to xy and project aux. Project Tv of both solids. It’s base side is 30 mm & axis is 60 mm long resting on Hp on one edge of base. Note the observer’s directions Shown by arrows and further steps carefully. 40 mm base side 60 mm axis is lying on Hp on one rectangular face with axis perpendicular to Vp. y Problem 10: A triangular prism.Draw FV & TV of both solids.V.Fv on it. Y 1 . Fv from x1y1 line. Mark the distances of first FV from first xy for the distances of aux. One square pyramid is leaning on it’s face F. centrally with axis // to vp.F. Aux.Project another FV X on an AVP 450 inclined to VP.X 1 Steps : Draw Fv of lying prism ( an equilateral Triangle) And Fv of a leaning pyramid. 450 45 0 to Vp ) (A VP T.V.
.Tv on it by using similar Steps like previous problem. o’ TL STEPS: Draw a regular hexagon as Tv of standing prism With one side // to xy and name the top points. Now join it’s alternate corners a-c-e and the triangle formed is base of a tetrahedron as said.The base of tetrahedron is a triangle formed by joining alternate corners of top of prism. is standing on Hp on it’s base with one base edge // to Vp. Project an auxiliary Tv on AIP 450 inclined to Hp. and complete Fv of tetrahedron.Draw projections of both solids. Locate center of this triangle & locate apex o Extending it’s axis line upward mark apex o’ By cutting TL of edge of tetrahedron equal to a-c. a’ Fv X f e Y 450 (A IP 45 0 b’ f’ c’ e’ d’ Y1 to Hp ) Aux. A tetrahedron is placed centrally on the top of it.Project it’s Fv – a rectangle and name it’s top. Draw an AIP ( x1y1) 450 inclined to xy And project Aux.Problem 11:A hexagonal prism of base side 30 mm longand axis 40 mm long.Tv e1 f1 d1 a1 b1 c1 o1 Tva b o d c X1 .
Problem 12: A frustum of regular hexagonal pyramid is standing on it’s larger base On Hp with one base side perpendicular to Vp.e. Base side is 50 mm long .Tv d1 e1 c1 b1 e Tv a 1 d 5 4 3 X1 a1 2 c b . top side is 30 mm long and 50 mm is height of frustum.1’ TL 5 4 1 3 2 X a’ b’ e’ c’ d’ Y Aux. Project it’s Aux. a’.Tv on an AIP parallel to one of the slant edges showing TL.Draw it’s Fv & Tv. Fv 1’ 2’5’ 3’4’ Y1 AIP // to slant edge Showing true length i.
DEVELOPMENT. 2. INTERSECTIONS. STUDY CAREFULLY THE ILLUSTRATIONS GIVEN ON NEXT SIX PAGES ! .ENGINEERING APPLICATIONS OF THE PRINCIPLES OF PROJECTIONS OF SOLIDES. 1. SECTIONS OF SOLIDS. 3.
I. ember:ter launching a section plane her in FV or TV.This section plane appears as a straight line in FV.e. A. (A) tion Plane perpendicular to Hp and inclined to Vp.) OTE:.P. An object ( here a solid ) is cut by some imaginary cutting plane to understand internal details of that object. A. his is a definition of an Aux.This section plane appears as a straight line in TV. utting actions means section planes are recommended. CT IN SE SECTIONING A SOLID. .) OTE:. Vertical Plane i.V.e. his is a definition of an Aux. far as possible the smaller part is sumed to be removed.P. OBSERVER ASSUME UPPER PART REMOVED E AN PL ON FV. tion ction Plane perpendicular to Vp and inclined to Hp. the part towards observer assumed to be removed. (B) ASSUME LOWER PART REMOVED OBSERVER SE CT ON IN PLA TV NE .The action of cutting is called SECTIONING a solid & The plane of cutting is called SECTION PLANE. Inclined Plane i.
. For TV Fo rT ru e Sh ap e SECTION PLANE TRUE SHAPE Of SECTION x Apparent Shape of section SECTION LINES (450 to XY) y SECTIONAL T.V.ILLUSTRATION SHOWING IMPORTANT TERMS IN SECTIONING.
Section Plane Ellipse Section PlaneTriangle Through Generators Through Apex Par abo la Section Plane Parallel to end generator. Ellipse Section Plane Hyperbola Parallel to Axis. Sq. Pyramid through all slant edges . Trapezium Cylinder through generators.Typical Section Planes & Typical Shapes Of Sections.
Aeroplanes and many more. Development is different drawing than PROJECTIONS. Pressure Vessels. their sections and frustums. ENGINEERING APLICATION: THERE ARE SO MANY PRODUCTS OR OBJECTS WHICH ARE DIFFICULT TO MANUFACTURE BY CONVENTIONAL MANUFACTURING PROCESSES. It is a shape showing AREA. LATERLAL SURFACE IS THE SURFACE EXCLUDING SOLID’S TOP & BASE. means it’s a 2-D plain drawing. Trays. THOSE ARE FABRICATED IN SHEET METAL INDUSTRY BY USING DEVELOPMENT TECHNIQUE. WHAT IS OUR OBJECTIVE IN THIS TOPIC ? To learn methods of development of surfaces of different solids. Study illustrations given on next page carefully. 4. Ships. Shovels. CUT OPEN IT FROM ONE SIDE AND UNFOLD THE SHEET COMPLETELY. . THEN THE SHAPE OF THAT UNFOLDED SHEET IS CALLED DEVELOPMENT OF LATERLAL SUEFACES OF THAT OBJECT OR SOLID. THERE IS A VAST RANGE OF SUCH OBJECTS. 2. Large Pipe sections. EXAMPLES:Boiler Shells & chimneys. 1. no edges can remain hidden Important points. Body & Parts of automotives. BECAUSE OF THEIR SHAPES AND SIZES.DEVELOPMENT OF SURFACES OF SOLIDS. Feeding Hoppers. And hence DOTTED LINES are never shown on development. Hence all dimensions of it must be TRUE dimensions. But before going ahead. As it is representing shape of an un-folded sheet. Boxes & Cartons. note following 3. MEANING:ASSUME OBJECT HOLLOW AND MADE-UP OF THIN SHEET.
(Lateral surface is the surface excluding top & base) Cylinder: H L A Rectangle D π D H= Height D= base diameter No. S = Edge of base H S S H= Height S = Edge of base Cube: Six Squares.Development of lateral surfaces of different solids. Tetrahedron: Four Equilateral Triangles All sides equal in length S S Cone: (Sector of circle) Pyramids: (No.of triangles) + .of Rectangles L θ R=Base circle radius. R 3600 θ L Prisms: = L= Slant edge. L=Slant height.
STUDY NEXT NINE PROBLEMS OF SECTIONS & DEVELOPMENT . + L= Slant edge of pyramid L1 = Slant edge of cut part.FRUSTUMS DEVELOPMENT OF FRUSTUM OF CONE DEVELOPMENT OF FRUSTUM OF SQUARE PYRAMID Base side Top side L L1 L L1 θ θ = R L 3600 R= Base circle radius of cone L= Slant height of cone L1 = Slant height of cut part.
Make remaining part of solid dark. Draw section lines in it. It is required true shape. Join those points in sequence and show Section lines in it. 30 mm base side & 50 mm axis is standing on Hp on it’s base whose one side is perpendicular to Vp.Problem 1: A pentagonal prism . Make existing parts dev. Mark distances of points of Sectioned part from Tv.e. Also draw true shape of section and Development of surface of remaining solid. in sequence as shown. lines. c b . through mid point of axis.dark. Join them in sequence in st. A. Side view.plane in Fv as described. on above projectors from x1y1 and join in sequence. sec. A E c’ d’ A d” e” a” e d B C D E A c” b” X1 a’ b’ e’ X Y DEVELOPMENT For Development: Draw development of entire solid. For True Shape: a Draw x1y1 // to sec. plane Draw projectors on it from cut points.Tv & sec. E RU T PE HA S C B D Y1 Solution Steps:for sectional views: Draw three views of standing prism. It is cut by a section plane 450 inclined to Hp. Locate sec. Name from cut-open edge I. Draw Fv. Mark the cut points on respective edges. Project points where edges are getting Cut on Tv & Sv as shown in illustration.
on above projectors from x1y1 and join in sequence. Draw section lines in it.Mark the cut points on respective edges. 50 mm base diameter and 70 mm axis is standing on it’s base on Hp.Draw projections. A. Project points where generators are getting Cut on Tv & Sv as shown in illustration. It cut by a section plane 450 inclined to Hp through base end of end generator. h g” h”f” a”e” b”d” c” Y F G f e For Development: Draw development of entire solid. Join them in sequence in curvature. O TI C SE N SE PL CT AN IO E N U TR E SH E AP OF Solution Steps:for sectional views: Draw three views of standing cone.e. Mark distances of points of Sectioned part from Tv. true shape of section and development of surfaces of remaining solid. Make remaining part of solid dark.Join those points in sequence and show Section lines in it.V . sectional views. A B DEVELOPMENT C D Y1 o’ SECTIONAL S. Make existing parts dev.dark. It is required true shape. Locate sec. in sequence as shown.Problem 2: A cone. plane Draw projectors on it from cut points.plane in Fv as described. Name from cut-open edge i. H A a b c d SECTIONAL T.V X1 X E a’ h’b’ c’ g’ f’ d’ e’ g For True Shape: Draw x1y1 // to sec.
Follow similar solution steps for Sec.Problem 3: A cone 40mm diameter and 50 mm axis is resting on one generator on Hp( lying on Hp) which is // to Vp. Draw it’s projections.It is cut by a horizontal section plane through it’s base center.True shape – Development as per previous problem! o’ a’ h’b ’ c ’g ’ d’ f ’ HORIZONTAL SECTION PLANE DEVELOPMENT A B C D o’ Y X a’ h’b’ c’ g’ g f’ d’ e’ e’ g1 f f1 e e1 d d1 c1 SECTIONAL T.views . development of the surface of the remaining part of cone. Draw sectional TV.V (SHOWING TRUE SHAPE OF SECTION) O h1 a1 b1 A o1 H G E F h a b O c ..
a’ b’ c’ f’ d’ e’ Note the steps to locate Points 1.6 1 8 c1 C D E F A B C 2 7 TR UE 3 SH AP EO 6 Y1 FS EC T IO N 4 5 DEVELOPMENT . there is c1 edge.8 7 Y d’ e’ f1 a1 e1 A.C. It is cut by a section plane normal to Hp and 300 inclined to Vp bisecting axis.V. Draw sec. a b c X1 d 4 . a’ b’ 5 c’ f’ 6 X f e 1 8 1 . Use similar steps for sec.B.views & true shape.. 6 in sec. 4 Problem 4: A hexagonal prism. then to 1st Fv and Then on 2nd Fv.V. 3 2 SECTIONAL F.2 3 . Views. Here it is Tv and in boundary.V. AS SECTION PLANE IS IN T. CUT OPEN FROM BOUNDRY EDGE C1 FOR DEVELOPMENT. always cut open object from From an edge in the boundary of the view in which sec.Fv: Those are transferred to 1st TV.plane appears as a line. 30 mm base side & 55 mm axis is lying on Hp on it’s rect. 2 .E.face with axis // to Vp. true shape & development.Hence it is opened from c and named C.F.7 b1 d1 5 .P300 inclined to Vp Through mid-point of axis.A. 5. NOTE: for development.D.
V. passing through mid-point of axis.hexagonal pyramid is shown in figure.UE 2 TR PE HA S 3 4 5 Y1 O’ Problem 5:A solid composed of a half-cone and half.v.Draw F. ( take radius of cone and each side of hexagon 30mm long and axis 70mm. 4 3 2 1 D 7 4 5 6 F G A E e SECTIONAL 7 TOP VIEW. 1 d .It is cut by a section plane 450 inclined to Hp. sectional T.) A ME NT DE VE LO P 1 6 7 Note: B Fv & TV 8f two solids sandwiched Section lines style in both: Development of half cone & half pyramid: C 4’ X1 2’ 6’ 1’ 7’ X d’e’ c’f’ f 6 5 4 3 2 c b a g’b’ g a’ Y O 3’ 5’ F.v...true shape of section and development of remaining part of the solid.
Semicircle being dev. then draw the projections of cone showing that curve. L=Slant height.of a cone it’s radius is slant height of cone. . DEVELOPMENT.. divide it in 8 Parts and inscribe in it a largest circle as shown.If the semicircle is development of a cone and inscribed circle is some VIEWS FROM GIVEN urve on it.e.( L ) Then using above formula find R of base of cone. Take o -1 distance from dev.mark on TL i. E 4 3 5 o’ R=Base circle radius.o’a’ on Fv & bring on o’b’ and name 1’ Similarly locate all points on Fv. Using this data draw Fv & Tv of cone and form 8 generators and name. 2.Name intersecting points 1. 3 etc.TO roblem 6: Draw a semicircle 0f 100 mm diameter and inscribe in it a largest DRAW PRINCIPAL ircle. Then project all on Tv on respective generators and join by smooth curve. R 3600 C θ L D F G 6 H = + L 1’ 7’ 6’ 2’ 2 B 1 3’ 5’ 4’ X a’ h’ b’ c’ g’ g d’f’ e’ Y θ O 7 A L A h 6 5 f a 7 1 o 4 e b 3 2 d c Solution Steps: Draw semicircle of given diameter.
If the semicircle is development of a cone and rhombus is some curve on it. TO DRAW PRINCIPAL VIEWS FROM GIVEN DEVELOPMENT. R 3600 θ L = + b 1 2 3 d c . Solution Steps: Similar to previous Problem: F 5 6 7 o’ D C 2 2’ 6’ 3’ 5’ E 4 3 G B 4’ 1 H X a’ h’ b’ 1’ 7’ c’ g’ g f’ d’ e’ Y A θ O L A h 7 6 5 f a 4 e R=Base circle radius.Problem 7:Draw a semicircle 0f 100 mm diameter and inscribe in it a largest rhombus. then draw the projections of cone showing that curve. L=Slant height.
Problem 8: A half cone of 50 mm base diameter, 70 mm axis, is standing on it’s half base on HP with it’s flat face parallel and nearer to VP.An inextensible string is wound round it’s surface from one point of base circle and brought back to the same point.If the string is of shortest length, find it and show it on the projections of the cone.
TO DRAW A CURVE ON PRINCIPAL VIEWS FROM DEVELOPMENT.
Concept: A string wound from a point up to the same Point, of shortest length Must appear st. line on it’s Development. Solution steps: Hence draw development, Name it as usual and join A to A This is shortest Length of that string. Further steps are as usual. On dev. Name the points of Intersections of this line with Different generators.Bring Those on Fv & Tv and join by smooth curves. Draw 4’ a’ part of string dotted As it is on back side of cone.
Draw Fv & Tv & dev.as usual On all form generators & name. Construction of curve Helix:: Show 8 generators on both views Divide axis also in same parts. Draw horizontal lines from those points on both end generators. 1’ is a point where first horizontal Line & gen. b’o’ intersect. 2’ is a point where second horiz. Line & gen. c’o’ intersect. In this way locate all points on Fv. Project all on Tv.Join in curvature. For Development: Then taking each points true Distance From resp.generator from apex, Mark on development & join.
WHEN ONE SOLID PENETRATES ANOTHER SOLID THEN THEIR SURFACES INTERSECT AND AT THE JUNCTION OF INTERSECTION A TYPICAL CURVE IS FORMED, WHICH REMAINS COMMON TO BOTH SOLIDS. THIS CURVE IS CALLED CURVE OF INTERSECTION AND IT IS A RESULT OF INTERPENETRATION OF SOLIDS.
PURPOSE OF DRAWING THESE CURVES:WHEN TWO OBJECTS ARE TO BE JOINED TOGATHER, MAXIMUM SURFACE CONTACT BETWEEN BOTH BECOMES A BASIC REQUIREMENT FOR STRONGEST & LEAK-PROOF JOINT.
Curves of Intersections being common to both Intersecting solids, show exact & maximum surface contact of both solids.
Lines of Intersections. Curves of Intersections.
Intersection of a Cylindrical main and Branch Pipe. A Feeding Hopper In industry. . An Industrial Dust collector. SHOWING CURVES OF INTERSECTIONS.SOME ACTUAL OBJECTS ARE SHOWN. Pump lid having shape of a hexagonal Prism and Hemi-sphere intersecting each other. Intersection of two cylinders. Two Cylindrical surfaces. A machine component having two intersecting cylindrical surfaces with the axis at acute angle to each other. BY WHITE ARROWS. Forged End of a Connecting Rod.
CYLINDER TO CONE COMMON SOLUTION STEPS One solid will be standing on HP Other will penetrate horizontally.PRISM TO SQ. REFFER ILLUSTRATIONS AND NOTE THE COMMON CONSTRUCTION FOR ALL 1. 2.V. .SQARE PRISM TO CONE ( from top ) 8. Similarly in case of penetration from top it is not required. Name views as per the illustrations. Draw three views of standing solid.) can be marked on respective generators in Fv and Tv. Accordingly those should be joined by curvature or straight lines.SQ. Note: Incase cone is penetrating solid Side view is not necessary. And other points from SV should be brought to Tv first and then projecting upward To Fv. Dark and dotted line’s decision should be taken by observing side view from it’s right side as shown by arrow.CYLINDER TO CYLINDER2.) The points which are on standard generators or edges of standing solid.CONE TO CYLINDER 4.SQ.PRISM ( SKEW POSITION) 7.PRISM TO SQ.V. ( in S.PRISM TO CYLINDER 3.TRIANGULAR PRISM TO CYLNDER 5.PRISM 6.SQ.FOLLOWING CASES ARE SOLVED. Beginning with side view draw three Views of penetrating solids also. On it’s S. mark number of points And name those(either letters or nos.
and 70mm axis is completely penetrated by another of 40 mm dia.and 70 mm axis horizontally Both axes intersect & bisect each other. Draw projections showing curves of intersections. 1’ 2’4’ 3’ 4” 1”3” CASE 1.Problem: A cylinder 50mm dia. CYLINDER STANDING & CYLINDER PENETRATING 2” a’ b ’h’ c’g’ d’f’ a” h” g” f” b” c” d” e” Y a’ X 4 1 3 2 .
and 70 mm axis. All faces of prism are equally inclined to Hp. by a square prism of 25 mm sides.PRISM PENETRATING 1’ 2’4’ 3’ 4” 1”3” 2” a’ a’ a” d” b’ d’ c’ X 4 b’ d’ c’ b” c” Y 1 3 2 . horizontally. Both axes CYLINDER STANDING Intersect & bisect each other. SQ.and 70mm axis is completely penetrated CASE 2.Problem: A cylinder 50mm dia. & Draw projections showing curves of intersections.
Problem: A cylinder of 80 mm diameter and 100 mm axis is completely penetrated by a cone of 80 mm diameter and 120 mm long axis horizontally. CYLINDER STANDING & CONE PENETRATING 2’ 4’ 3’ X 1 28 37 Y 46 5 . Draw projections showing curve of intersections. 7’ 6’ 8’ 1’ 5’ CASE 3.Both axes intersect & bisect each other.
Problem: A sq.PRISM STANDING Intersects & bisect each other. & Draw projections showing curves of intersections. SQ.and 70mm axis is completely penetrated CASE 4.and 70 mm axis. All faces of prisms are equally inclined to Vp. horizontally.PRISM PENETRATING 1’ 2’4’ 3’ 4” 1”3” 2” a’ b’ d’ c’ X 4 a’ b’ d’ c’ d” a” b” c” Y 1 3 2 . Both axes SQ.prism 30 mm base sides. by another square prism of 25 mm sides.
Problem: A cylinder 50mm dia.and 70mm axis is completely penetrated by a triangular prism of 45 mm sides.and 70 mm axis, horizontally. One flat face of prism is parallel to Vp and Contains axis of cylinder. Draw projections showing curves of intersections.
Problem: A sq.prism 30 mm base sides.and 70mm axis is completely penetrated by another square prism of 25 mm side s.and 70 mm axis, horizontally. Both axes Intersect & bisect each other.Two faces of penetrating prism are 300 inclined to Hp. Draw projections showing curves of intersections.
Problem: A cone70 mm base diameter and 90 mm axis is completely penetrated by a square prism from top with it’s axis // to cone’s axis and 5 mm away from it. a vertical plane containing both axes is parallel to Vp. Take all faces of sq.prism equally inclined to Vp. Base Side of prism is 0 mm and axis is 100 mm long. Draw projections showing curves of intersections.
2 3 7. Draw projections showing curves of intersection. The axis of the CONE STANDING cylinder is parallel to Hp and Vp and intersects axis of the cone at a point & 28 mm above the base. 64 c’g’ d’f’ e’ 8 7 6 5 g” g”h” 1 2 3 4 4 X 5 a”e” b”d” c” Y g h f a e b c d . CASE 8.Problem: A vertical cone. base diameter 75 mm and axis 100 mm long. CYLINDER PENETRATING o’ o” 1 2 3 5 a’ b’h’ 1 8. is completely penetrated by a cylinder of 45 mm diameter.
ALL THREE DIMENSIONAL AXES ARE MENTAINED AT EQUAL INCLINATIONS WITH EACH OTHER. HERE NO SPECIFIC RELATION AMONG H. NOW OBSERVE BELOW GIVEN DRAWINGS. L & D AXES IS MENTAINED. . SIZE & APPEARANCE OF AN OBJECT PRIOR TO IT’S PRODUCTION.( 1200) TYPICAL CONDITION. HENCE IT IS CALLED ISOMETRIC DRAWING L H H H PURPOSE OF ISOMETRIC DRAWING IS TO UNDERSTAND OVERALL SHAPE. L & D AXES. THEIR ACTUAL SIZES CAN BE MEASURED DIRECTLY FROM IT. HERE ONE CAN FIND EDUAL INCLINATION AMONG H. L & D AXES. SIMILAR OR EQUAL. IN THIS 3-D DRAWING OF AN OBJECT. EACH IS 1200 INCLINED WITH OTHER TWO. ALL THESE DRAWINGS MAY BE CALLED 3-DIMENSIONAL DRAWINGS. 3-D DRAWINGS CAN BE DRAWN IN NUMEROUS WAYS AS SHOWN BELOW. OR PHOTOGRAPHIC OR PICTORIAL DRAWINGS. ONE CAN NOTE SPECIFIC INCLINATION AMONG H.ISOMETRIC DRAWING IT IS A TYPE OF PICTORIAL PROJECTION IN WHICH ALL THREE DIMENSIONS OF AN OBJECT ARE SHOWN IN ONE VIEW AND IF REQUIRED. ISO MEANS SAME.
LINES AND PLANES: The three lines AL. while drawing isometric projection.) It forms a reducing scale which Is used to draw isometric drawings and is called Isometric scale. In practice.815 or 9 / 11 ( approx. This is conveniently done by constructing an isometric scale as described on next page.SOME IMPORTANT TERMS: ISOMETRIC AXES. ISOMETRIC SCALE: When one holds the object in such a way that all three dimensions are visible then in the process all dimensions become proportionally inclined to observer’s eye sight and hence appear apparent in lengths. The lines parallel to these axes are called Isometric Lines. meeting at point A and making 1200 angles with each other are termed Isometric Axes. it is necessary to convert true lengths into isometric lengths for measuring and marking the sizes. The planes representing the faces of of the cube as well as other planes parallel to these planes are called Isometric Planes. A H . This reduction is 0. AD and AH.
. From point A. with line AB draw 300 and 450 inclined lines AC & AD resp on AD.SCALE. Mark divisions of true length and from each division-point draw vertical lines upto AC line. 1 ISOM 2 THS NG LE 450 B Isometric scale [ Line AC ] required for Isometric Projection CONSTRUCTION OF ISOM. The divisions thus obtained on AC give lengths on isometric scale.TYPES OF ISOMETRIC DRAWINGS Drawn by using True scale ( True dimensions ) ISOMETRIC VIEW Drawn by using Isometric scale ( Reduced dimensions ) ISOMETRIC PROJECTION 4 D 3 H E RU T 1 LE S TH 2 NG C 4 3 H 0 0 A 300 .
or T. IF THE FIGURE IS TOP VIEW.V. Then first draw isom. H & L 1 AXES ARE REQUIRED. D L D L C D C A B B B TRIANGLE C B B H 3 A L D IF THE FIGURE IS FRONT VIEW. D & L AXES ARE REQUIRED.ISOMETRIC OF PLANE FIGURES AS THESE ALL ARE 2-D FIGURES WE REQUIRE ONLY TWO ISOMETRIC AXES. of that rectangle and then inscribe that shape 1 A E 4 D 1 4 D D C 3 L 1 A C B L 3 2 B C 3 2 B 2 .V. 2 B 3 A PENTAGON 1 3 L A 1 2 4 H E D E A 2 Shapes containing Inclined lines should be enclosed in a rectangle as shown. A 1 SHAPE H RECTANGLE D A Isometric view if the Shape is F.
STUDY IT. FOR CONSTRUCTION USE RHOMBUS METHOD SHOWN HERE. THEN USE H & L AXES FOR ISOMETRIC WHEN IT IS FRONT VIEW.STUDY Z ILLUSTRATIONS 2 DRAW ISOMETRIC VIEW OF A CIRCLE IF IT IS A TV OR FV. FIRST ENCLOSE IT IN A SQUARE. A 3 D 1 4 B A 3 4 D C 1 C 2 2 B . IT’S ISOMETRIC IS A RHOMBUS WITH D & L AXES FOR TOP VIEW.
STUDY Z ILLUSTRATIONS 25 R 3 DRAW ISOMETRIC VIEW OF THE FIGURE SHOWN WITH DIMENTIONS (ON RIGHT SIDE) CONSIDERING IT FIRST AS F.V. AND THEN T. IF FRONT VIEW 50 MM 100 MM IF TOP VIEW .V.
CIRCLE IF THE FIGURE IS For Isometric of Circle/Semicircle use Rhombus method. D & L of sides equal to Diameter of circle always. H & L AXES ARE REQUIRED. topic ENGG. IF THE FIGURE IS FRONT VIEW. For Isometric of Circle/Semicircle use Rhombus method. ( Ref. Construct it of sides equal to diameter of circle always.ISOMETRIC OF PLANE FIGURES SHAPE HEXAGON IF F. 4 AS THESE ALL ARE 2-D FIGURES WE REQUIRE ONLY TWO ISOMETRIC AXES.) AXES ARE SEMI CIRCLE REQUIRED. Previous two pages. Construct Rhombus TOP VIEW.V. ( Ref.) .V. IF T. CURVES.
4 4 E D E D D 1 A C B L 3 D L 1 A B C 3 2 2 .P.P. (Height is added from center of pentagon) ISOMETRIC VIEW OF BASE OF PENTAGONAL PYRAMID STANDING ON H.5 STUDY Z ILLUSTRATIONS ISOMETRIC VIEW OF PENTAGONAL PYRAMID STANDING ON H.
4 H 1 A E D L C 3 ISOMETRIC VIEW OF 2 B HEXAGONAL PRISM STANDING ON H. .P.P.6 STUDY Z ILLUSTRATIONS ISOMETRIC VIEW OF PENTAGONALL PRISM LYING ON H.
.P.P.7 STUDY Z ILLUSTRATIONS CYLINDER STANDING ON H. CYLINDER LYING ON H.
( with flat face // to H.P.P.) .P. ( ON IT’S SEMICIRCULAR BASE) HALF CYLINDER LYING ON H.8 STUDY Z ILLUSTRATIONS HALF CYLINDER STANDING ON H.
ON IT’S LARGER BASE.STUDY Z ILLUSTRATIONS ISOMETRIC VIEW OF A FRUSTOM OF SQUARE PYRAMID STANDING ON H. 9 60 FV X Y 40 20 TV .P.
STUDY ILLUSTRATION 10 PROJECTIONS OF FRUSTOM OF PENTAGONAL PYRAMID ARE GIVEN. THEN REDUCE THE TOP TO 20 MM SIDES AND JOIN WITH THE PROPER BASE CORNERS. ISOMETRIC VIEW OF FRUSTOM OF PENTAGONAL PYRAMID FV x 60 SOLUTION STEPS: FIRST DRAW ISOMETRIC OF IT’S BASE. 60 MM ABOVE THE BASE PENTAGON CENTER. 20 60 y 1 A E 4 THEN DRAWSAME SHAPE AS TOP. TV 40 20 B D 2 C 3 40 . DRAW IT’S ISOMETRIC VIEW.
STUDY Z ILLUSTRATIONS ISOMETRIC VIEW OF 11 A FRUSTOM OF CONE STANDING ON H. ON IT’S LARGER BASE.P. 60 FV X Y 40 20 TV .
DRAW ISOMETRIC VIEW OF THE PAIR. IS CENTRALLY PLACED ON THE TOP OF A CUBE OF 50 MM LONG EDGES. 12 50 30 50 .STUDY Z ILLUSTRATIONS PROBLEM: A SQUARE PYRAMID OF 30 MM BASE SIDES AND 50 MM LONG AXIS.
DRAW IT AS USUAL. c a p o b a p b o c SOLUTION HINTS. . SO DRAW TRIANGLE AS A TV. IT CAN NOT BE DRAWN DIRECTLY. SEPARATELY AND NAME VARIOUS POINTS AS SHOWN. THEN ADD HEIGHT FROM IT’S CENTER AND COMPLETE IT’S ISOMETRIC AS SHOWN.STUDY Z ILLUSTRATIONS 13 PROBLEM: A TRIANGULAR PYRAMID OF 30 MM BASE SIDES AND 50 MM LONG AXIS. AFTER THIS PLACE IT ON THE TOP OF CUBE AS SHOWN. BUT FOR PYRAMID AS IT’S BASE IS AN EQUILATERAL TRIANGLE. DRAW ISOMETRIC VIEW OF THE PAIR. TO DRAW ISOMETRIC OF A CUBE IS SIMPLE.SUPPORT OF IT’S TV IS REQUIRED. IS CENTRALLY PLACED ON THE TOP OF A CUBE OF 50 MM LONG EDGES.
DRAW ISOMETRIC VIEW. IT’S FV & TV ARE SHOWN. 14 FV 30 10 30 30 D 50 + 50 TV .STUDY Z ILLUSTRATIONS PROBLEM: A SQUARE PLATE IS PIERCED THROUGH CENTRALLY BY A CYLINDER WHICH COMES OUT EQUALLY FROM BOTH FACES OF PLATE.
IT’S FV & TV ARE SHOWN.STUDY Z ILLUSTRATIONS 15 PROBLEM: A CIRCULAR PLATE IS PIERCED THROUGH CENTRALLY BY A SQUARE PYRAMID WHICH COMES OUT EQUALLY FROM BOTH FACES OF PLATE. FV 30 10 30 40 SQUARE TV 60 D . DRAW ISOMETRIC VIEW.
Draw it’s isometric view.V. 40 FV 40 X Y TV 50 D 30 D . & T.V.STUDY Z ILLUSTRATIONS 10 20 16 F. of an object are given.
NAME IT P.STUDY ILLUSTRATIONS Z ISOMETRIC PROJECTIONS OF SPHERE & HEMISPHERE 17 R r Isom. WITH RADIUS ‘R’ DRAW CIRCLE. 3. Then around ‘C’ construct Rhombus of Sides equal to Isometric Diameter. . ‘C’ AS CENTER. 2. LENGTH ‘ r mm’ AND LOCATE CENTER OF SPHERE “C” 4. P C = Center of Sphere. LOCATE IT’S CENTER. For this use iso-scale. Draw lower semicircle only. Then construct ellipse in this Rhombus as usual And Complete Isometric-Projection of Hemi-sphere. P = Point of contact R = True Radius of Sphere r = Isometric Radius. FIRST DRAW ISOMETRIC OF SQUARE PLATE. TO DRAW ISOMETRIC PROJECTION OF A SPHERE 1. FROM PDRAW VERTICAL LINE UPWARD. Scale ct io n 450 300 r R r P r R R C r C Is oDi re r R r P TO DRAW ISOMETRIC PROJECTION OF A HEMISPHERE Adopt same procedure. THIS IS ISOMETRIC PROJECTION OF A SPHERE.
R r 450 300 . USE THIS SCALE FOR ALL DIMENSIONS IN THIS PROBLEM.PROBLEM: A HEMI-SPHERE IS CENTRALLY PLACED ON THE TOP OF A FRUSTOM OF CONE. DRAW ISOMETRIC PROJECTIONS OF THE ASSEMBLY. 18 STUDY ILLUSTRATIONS Z r 50 D R 30 D 50 r r P 50 D FIRST CONSTRUCT ISOMETRIC SCALE.
19 3’ 4’ 4 3 1’2’ X 1 a 1 d 4 Y 2 o 2 3 c b .DRAW ISOMETRIC VIEW OF SECTION OF PYRAMID.STUDY Z ILLUSTRATIONS A SQUARE PYRAMID OF 40 MM BASE SIDES AND 60 MM AXIS IS CUT BY AN INCLINED SECTION PLANE THROUGH THE MID POINT OF AXIS AS SHOWN.
20 STUDY Z ILLUSTRATIONS F.V.V. 50 X 20 O Y 25 O 25 20 . & T. of an object are given. Draw it’s isometric view.
Draw it’s isometric view.V.STUDY Z ILLUSTRATIONS 21 F.V. 35 FV 35 x O 10 y 10 20 30 40 70 O TV . & T. of an object are given.
Draw it’s isometric view. and S.V.V.of an object are given.V. & T. FV 30 30 10 30 10 30 SV x y TV ALL VIEWS IDENTICAL .STUDY Z ILLUSTRATIONS 22 F.
& T.STUDY Z ILLUSTRATIONS 24 F. ALL VIEWS IDENTICAL FV SV x y 10 40 60 40 60 TV . Draw it’s isometric view. and S.of an object are given.V.V.V.
V. & T. and S. ALL VIEWS IDENTICAL FV SV 25 x 10 y 40 60 40 60 TV .of an object are given.V.STUDY Z ILLUSTRATIONS F.V. Draw it’s isometric view.
V. & T.of an object are given. and S. Draw it’s isometric view.SIDE VIEW 20 20 x 20 30 20 O y 50 20 20 20 O TOP VIEW . STUDY Z ILLUSTRATIONS ORTHOGRAPHIC PROJECTIONS FRONT VIEW L.H.V.26 F.V.
V. Draw it’s isometric view.V.V.V. O 30 60 S. 30 SQUARE 40 20 50 10 20 O F.of an object are given. . and S.STUDY Z ILLUSTRATIONS 27 F.
& T. FV 40 O 10 30 D 45 10 50 O 80 TV . Draw it’s isometric view.STUDY Z ILLUSTRATIONS 28 F.V.V. of an object are given.
29 FV X 10 40 O 100 10 Y 25 TV 30 R 10 30 10 25 O 20 D . & T. Draw it’s isometric view.V. of an object are given.V.STUDY Z ILLUSTRATIONS F.
V. SLOT 10 35 50 X 10 O 20 D Y TV 60 D 30 D O . Draw it’s isometric view. of an object are given. & T. 30 30 FV RECT.STUDY Z ILLUSTRATIONS F.V.
. Draw it’s isometric view.STUDY Z ILLUSTRATIONS 31 F.V.V.V. 80 25 S. and S.of an object are given.V. 10 25 15 25 40 10 25 25 O O F.
STUDY Z ILLUSTRATIONS F.V. Draw it’s isometric view. & T.V. of an object are given. 32 450 30 FV 40 X O 30 D Y 40 TV O 40 15 .
Draw it’s isometric view. and S.STUDY Z ILLUSTRATIONS 33 F. HEX PART 20 30 20 40 20 50 O 20 15 O 100 30 60 .V.V.of an object are given.
V. 80 . & T.V. 20 40 34 F. of an object are given.STUDY Z ILLUSTRATIONS F. Draw it’s isometric view.V. 30 10 X O Y 10 30 10 30 O T.V.
of an object are given. Draw it’s isometric view. 10 X O Y FV LSV . 35 STUDY Z ILLUSTRATIONS 10 10 15 25 25 X O FV Y 50 10 LSV 36 NOTE THE SMALL CHZNGE IN 2ND FV & SV.V.F. and S. DRAW ISOMETRIC ACCORDINGLY.V.
V. LEFT S.V. 37 30 20 10 20 15 10 15 O 30 15 15 X O 50 Y F. and S.of an object are given.STUDY Z ILLUSTRATIONS F.V. .V. Draw it’s isometric view.
Draw it’s isometric view.V.STUDY Z ILLUSTRATIONS 38 F.of an object are given.V.V.V. 40 S. 30 10 60 O 30 O 40 F. . and S.
5.EXERCISES: PROJECTIONS OF STRAIGHT LINES 1. when it is lying in profile plane and inclined at 350 to HP. Draw the projections of line when sum of inclinations with HP and VP is 900. The line is inclined at 300 to VP and its VT is 10mm above HP. A line AB is in first quadrant. Its end A is in HP and 20mm in front of VP. Draw the projections of AB and determine its true length and HT and inclination with HP.Draw projections and find TL and inclination of line with Vp. The front view of the line measures 55mm. 6. while other end B is in first quadrant. End A is 10 mm above Hp. Draw the three views of line AB. 2. 4. 3. The point A is 35mm above HP and 25mm in front VP. Determine also its traces. & 15 mm in front of Vp.Distance between projectors is 50 mm. The distance between the end projectors is 75mm. The VT is 15mm above HP. A line AB is 75mm long and lies in an auxiliary inclined plane (AIP) which makes an angle of 450 with the HP. Line AB lies in an AVP 500 inclined to Vp while line is 300 inclined to Hp. Locate traces also. Draw the projections of the line AB and find its inclination with HP and VP. The projections through its VT and end A are 50mm apart. The end A is in VP and 20mm above HP. has its one end A in VP and other end B 15mm above HP and 50mm in front of VP. A line AB measures 100mm. A line AB 75 mm long. . 80mm long. Determine the true angles of inclination and show traces. Its ends A and B are 25mm and 65mm in front of VP respectively. Draw the projections of line and determine its HT and Inclinations with HP and VP.
A thin straight wire connects the point to the switch in one of the corners of the room and 2m above the floor.APPLICATIONS OF LINES Room . Peg A is 1. Find the distance between the ends of the two pegs. Distances being measured from the center line of the wall. Draw the projections of the and its length and slope angle with the floor.5m from a 0. Determine graphically the real distance between the top corner and its diagonally apposite bottom corners. Determine graphically the distance between the motors. Determine the real distance between the ranges.25m thick wall but on apposite sides of it. 10) Two fan motors hang from the ceiling of a hall 12m x 5m x 8m high at heights of 4m and 6m respectively. measured along ground and parallel to the wall is 3m. consider appropriate scale 9) Two pegs A and B are fixed in each of the two adjacent side walls of the rectangular room 3m x 4m sides. Also find the distance of each motor from the top corner joining end and front wall. 1. Peg B is 2m above the floor. 8) A room is of size 6m\5m\3. . An electric point hang in the center of ceiling and 1m below it.5m above the floor. 1m from other side wall and protruding 0.3m from the wall.5m and 2. The distance between the apples.2m from the longer side wall and is protruding 0. compound wall cases 7) A room measures 8m x 5m x4m high.2m from the wall. 11) Two mangos on a two tree are 2m and 3m above the ground level and 1. Also find the height of the roof if the shortest distance between peg A and and center of the ceiling is 5m.5m high.
NORTH. From C angles and elevations for A and B are 450 and 300 respectively. 13) A straight road going up hill from a point A due east to another point B is 4km long and has a slop of 250. CD and EF are lying along the corners of equilateral triangle lying on the ground of 100mm sides. Another straight road from B due 300 east of north to a point C is also 4 kms long but going downward and has slope of 150. 14) An electric transmission line laid along an uphill from the hydroelectric power station due west to a substation is 2km long and has a slop of 300. 8m and 12m respectively. PIPE LINES. 16) Two hill tops A and B are 90m and 60m above the ground level respectively. 15) Two wire ropes are attached to the top corner of a 15m high building. Draw their projections and find real distance between their top ends.. From B angle of elevation of A is 450. B and C. Draw the projections of the wire ropes. 20m above the ground. . The rope makes 300 angle of depression and is attached to the top of a 2m high pole.ROADS.POLES. The pole in the top view are 2m apart.EAST-SOUTH WEST. Determine the two distances between A. 12)Three vertical poles AB. Another line from the substation. SLOPE AND GRADIENT CASES. is 4km long and laid on the ground level. Determine the length and slope of the proposed telephone line joining the the power station and village. Their lengths are 5m. Find the length and slope of the straight road connecting A and C. The other end of one wire rope is attached to the top of the vertical pole 5m high and the rope makes an angle of depression of 450. running W 450 N to village. They are observed from the point C.
4. Draw its projections if a) the TV of same diameter is 450 inclined to Vp. 10.PROJECTIONS OF PLANES:1. Draw projections of a circle 40 mm diameter resting on Hp on a point A on the circumference with its surface 300 inclined to Hp and 450 to Vp. de = 25mm and angles abc 1200. Draw a rectangular abcd of side 50mm and 30mm with longer 350 with XY. A rhombus of 60mm and 40mm long diagonals is so placed on Hp that in TV it appears as a square of 40mm long diagonals. Side PQ rest on ground and makes 300 with Vp. Draw its FV. A 300-600 set-square of 40mm long shortest side in Hp appears is an isosceles triangle in its TV. 9. 7. cd =50mm. bc = 60mm.Determine it’s true shape. long respectively. The point A and B are 25 and 50mm above Hp respectively. 8. 6. A thin triangle PQR has sides PQ = 60mm.B and E are 15. A thin regular pentagon of 30mm sides has one side // to Hp and 300 inclined to Vp while its surface is 450 inclines to Hp. A top view of plane figure whose surface is perpendicular to Vp and 600 inclined to Hp is regular hexagon of 30mm sides with one side 300 inclined to xy. A circle of 50mm diameter has end A of diameter AB in Hp and AB diameter 300 inclined to Hp. An isosceles triangle having base 60mm long and altitude 80mm long appears as an equilateral triangle of 60mm sides with one side 300 inclined to XY in top view. Draw its projections.0 .Draw a pentagon abcde having side 500 to XY. 5. with the side ab =30mm. A figure is a TV of a plane whose ends A. Draw a suitable Fv and determine its true shape. representing TV of a quadrilateral plane ABCD. Draw its projections. Complete the projections and determine the true shape of the plane figure. Draw its projections. cde 1250. Point P is 30mm in front of Vp and R is 40mm above ground. and RP = 50mm. Draw projections of it and find its inclination with Hp. 25 and 35mm above Hp respectively. QR = 80mm. 2. 3. OR b) Diameter AB is in profile plane.
Draw another top view on an auxiliary plane inclined at 500 to the HP.s 8. Take edge of base 25mm and axis length as 125mm. Draw the projections (three views) when the edge of base apposite to the point of suspension makes an angle of 300 to VP. 2. A hexagonal prism. Draw an another front view on an AVP inclined at 300 to edge on which it is resting so that the base is visible. Draw the projections when the axis is parallel to Vp. 7.PROJECTIONS OF SOLIDS 1. . The prism is resting with one of its corners in VP and axis inclined at 300 to VP and parallel to HP. A cone base diameter 50mm and axis 70mm long. 3. Draw the projections of a square prism resting on an edge of base on HP. The axis is inclined at 400 to ground and at 300 to VP. above ground. Draw the projections. Draw the projections of a square prism of 25mm sides base and 50mm long axis. Draw its projections. A square pyramid of side 30mm and axis 60 mm long has one of its slant edges inclined at 450 to HP and a plane containing that slant edge and axis is inclined at 300 to VP. Draw the front view and the top view when the plane containing its axis is perpendicular to HP and makes an angle of 450 with VP. The axis makes an angle of 300 with VP and 450 with HP. base 40mm side and height 75mm rests on one edge on its base on the ground so that the highest point in the base is 25mm. has an edge of the base parallel to the HP and inclined at 450 to the VP. 6. 4. A pentagonal pyramid. is freely suspended from a point on the rim of its base. base 30mm sides and axis 75mm long. Its axis makes an angle of 600 with the HP. A right pentagonal prism is suspended from one of its corners of base. Draw the three views of a cone having base 50 mm diameter and axis 60mm long It is resting on a ground on a point of its base circle. Take base side 30mm and axis length 60mm. 5.
the sides of a bar being 450 to VP. A cube of 40mm long edges is resting on the ground with its vertical faces equally inclined to the VP. Draw the front and top views of the solids. Draw the front view and the top view of the solids. Project another top view on an AIP making 450 with the HP 10.A square bar of 30mm base side and 100mm long is pushed through the center of a cylindrical block of 30mm thickness and 70mm diameter.A cube of 50mm long edges is resting on the ground with its vertical faces equally inclined to VP. . A hexagonal pyramid . 11. Draw side view also. 9. Draw the front view.A circular block. A right circular cone base 25mm diameter and height 50mm is placed centrally on the top of the cube so that their axis are in a straight line. 12. and a face of the prism makes an angle of 300 with HP. which comes out equally on both sides of the block. 75mm diameter and 25mm thick is pierced centrally through its flat faces by a square prism of 35mm base sides and 125mm long axis. project another top view on an AIP making an angle of 450 with the HP.CASES OF COMPOSITE SOLIDS. Draw the projections of the solids when the combined axis is parallel to HP and inclined at 300 to VP. is placed centrally on the top of the cube so that their axes are in a straight line and two edges of its base are parallel to VP. top view and side view of the solid when the axis of the bar is inclined at 300 to HP and parallel to VP. so that the bar comes out equally through the block on either side. base 25mm side and axis 50mm long.
It is cut by a section plane perpendicular HP and parallel to VP. Draw the projections of the solid and show its development. Also find the inclination of the section plane with HP. Determine the inclination of cutting plane with HP. is resting on its edge on HP which is perpendicular toVP. 7) The pyramidal portion of a half pyramidal and half conical solid has a base ofthree sides. Draw the front view of the sectional top view and development of surface. true shape of section and development of surface of a cut pyramid containing apex. edge of base 30mm and axis 75mm. A plane parallel to VP cuts the solid at a distance of 10mm from the top view of the axis. 2) A hexagonal pyramid. Draw the development of the lower half of the cylinder. The length of axis is 80mm. The smallest generator on the cylinder is 20mm long after it is cut by a section plane. The plane cuts the axis at 10mm above the base. long has one of its generators in VP and parallel to HP. It is cut by section plane perpendicular to VP and inclined at 450 to HP. Also develop the lateral surface of the cut solid. Draw the sectional FV. each 30mm long. 5) A vertical cylinder cut by a section plane perpendicular to VP and inclined to HP in such a way that the true shape of a section is an ellipse with 50mm and 80mm as its minor and major axes. The solid rest on its base with the side of the pyramid base perpendicular to VP. A horizontal section plane passing through midpoint of vertical solid diagonal cuts the cube. The axis makes an angle of 300to HP.SECTION & DEVELOPMENT 1) A square pyramid of 30mm base sides and 50mm long axis is resting on its base in HP. Draw sectional front view and true shape of section. true shape of section and develop the lateral surface of the cone containing the apex. 4) A cube of 50mm long slid diagonal rest on ground on one of its corners so that the solid diagonal is vertical and an edge through that corner is parallel to VP. Edges of base is equally inclined to VP. the solid is cut by a section plane perpendicular to both HP and VP. Draw the projections and show the true shape of the section. 3) A cone of base diameter 60mm and axis 80mm. and passing through the mid point of the axis. It is cut by a section plane perpendicular to VP such that the true shape of section is regular hexagon.Draw the sectional top view and true shape of section. . Draw the projections showing the sectional view. 6) A cube of 75mm long edges has its vertical faces equally inclined to VP.
the HT of which makes an angle of 300 with the reference line passes through center of base. It is cut by a section plane perpendicular to VP in such a way that the true shape of a section is a parabola having base 40mm. Draw the top view. . Also show the path in front and top views 12) A cube of 65 mm long edges has its vertical face equally inclined to the VP. Draw three views of solid when it is resting on its cut face in HP. 9) A cone diameter of base 50mm and axis 60mm long is resting on its base on ground. Draw the projections of the cut pyramid.8) A hexagonal pyramid having edge to edge distance 40mm and height 60mm has its base in HP and an edge of base perpendicular to VP. so that the true shape of the section is a regular hexagon. Determine the inclination of the cutting plane with the HP and draw the sectional top view and true shape of the section. base 50mm side and axis 100mm long is lying on ground on one of its triangular faces with axis parallel to VP. Draw the development of the cut pyramid and show the shortest path of particle P. perpendicular to VP. It is cut by a section plane. It is cut by a section plane parallel to HP and passing through a point on the axis 25mm from the apex. It is cut by a section plane. perpendicular to VP and passing through a point on the axis 10mm from the base. A vertical section plane. true shape of section and development of remaining surface of cone removing its apex. resting the larger part of the pyramid. A particle P. 11) Hexagonal pyramid of 40mm base side and height 80mm is resting on its base on ground. initially at the mid point of edge of base. Also draw the lateral surface development of the pyramid. the apex being retained. 10) A hexagonal pyramid. starts moving over the surface and reaches the mid point of apposite edge of the top face. Draw three views showing section. sectional front view and the development of surface of the cut pyramid containing apex.
. whose P & Q are walls meeting at 900. a flower (A) and an orange (B) are within a rectangular compound wall. find distance between them If flower is 1.5M x ns w L (a T y Wall P 1.5M & 1 M from walls P & Q respectively.V.5M & 5. Flower A is 1.5 M above the ground.PROBLEM 14:-Two objects. Drawing projection. Consider suitable scale.5M a’ 1.5 M and orange is 3.5M Wall Q F. b’ er) b’1 3.5M a 3.6M 1M B Wall P A b Wall Q 5.5M from walls P & Q respectively. Orange B is 3. .
6 m. If the distance measured between them along the ground and parallel to wall is 2.00 m above ground and those are 1.3 m thick wall but on opposite sides of it.6m REAL DISTANCE BETWEEN MANGOS A & B IS = a’ b1 ’ .3m Wall thickness 1.5m (GL) X 0.5 m and 3.5 m from a 0.00 m a’ 1. Then find real distance between them by drawing their projections.2m a 2.2 m & 1. b’ b 1’ TV B 3.3M THICK 1.Two mangos on a tree A & B are 1.5m WALL A Y b FV 0.PROBLEM 15 :.
This fig.B & C are on ground and end O is 100mm above ground.PROBLEM 16 :oa.All equally inclined and the shortest is vertical. o’ Tv O C A TL2 100 TL1 Fv y B TL 3 x b1’ b’ a’ a 25 a1’ c’ c1’ o b 45 65 Answers: TL1 TL2 & TL3 c . Draw their projections and find length of each along with their angles with ground. OB and OC whose ends A. ob & oc are three lines. 45mm and 65mm long respectively. is TV of three rods OA. 25mm.
East.A pipe line from point A has a downward gradient 1:5 and it runs due South .PROBLEM 17:. 12m a’ 5 1 b’ TL ( an Do wn wa rd FV sw e Gra die nt 1 :5 5 1 r) N c’2 y EAST x W c’ N c’1 A 12 M B E a 450 b C 150 TV c DU E S TL ( answer) = a’ c’ SO UT H 2 = Inclination of pipe line BC -E AS T SOUTH . Pipe line from B runs 150 Due East of South and meets pipe line from A at point C. Another Point B is 12 M from A and due East of A and in same level of A. Draw projections and find length of pipe line from B and it’s inclination with ground.
A & B. Object A is is due North-West direction of observer and object B is due West direction. from a tower. At the angles of depression 300 & 450.PROBLEM 18: A person observes two objects. o’ 450 30 0 15M a’1 a’ a b’ N O 300 450 W b Answers: Distances of objects from observe o’a’1 & o’b’ From tower oa & ob o E N A S B W S . on the ground. Draw projections of situation and find distance of objects from observer and from tower also. 15 M high.
5M Answers: Length of Rope BC= b’c’2 Length of Rope AC= a’c’1 Distances of poles from building = ca & cb c . c1’ c’ c’2 TV C 300 b’ 15M 7.PROBLEM 19:-Guy ropes of two poles fixed at 4.5m and 7. make 300 and 450 inclinations with ground respectively. Determine by drawing their projections.5M a’ 4.5M 450 15 M A 12M 300 4.Length of each rope and distance of poles from building.5 M 450 a b B FV 10 M 7.5 m above ground. are attached to a corner of a building 15 M high.The poles are 10 M apart.
2 M 0.7 M b 1.2 M and 0.7 M from two adjacent walls respectively. as shown.A tank of 4 M height is to be strengthened by four stay rods from each corner by fixing their other ends to the flooring. M 1 . = A b’1 b’ a X Y 4M B 7 0. Determine graphically length and angle of each rod with flooring. FV a’ True Length TV Answers: Length of each rod = a’b’1 Angle with Hp. at a point 1.PROBLEM 20:.2 M FV TV .
A horizontal wooden platform 2 M long and 1.PROBLEM 21:. = 1. 5 M FV B .5 M h A b 2M 2M C a Answers: Length of each chain = a’d’1 Angle with Hp. Draw projections of the objects and determine length of each chain along with it’s inclination with ground.5 M wide is supported by four chains from it’s corners and chains are attached to a hook 5 M above the center of the platform. h’ TV TL 5M H Hook x d’1 a’d’ 5M b’c’ y (GL) d c D 1.
5m above the flooring.3. A room is of size 6.a’ b’1 . An electric bulb hangs 1m below the center of ceiling. Draw the projections an determine real distance between the bulb and switch.5m high. 6.5m 1m 3.5m L .5m D.5m 1. 1. A switch is placed in one of the corners of the room.PROBLEM 22.Bulb A-Switch Answer :.5 x a L Ceiling b’ b’1 TV Bulb Front wall H Side w all a’ y Switch D Observ er 5m b B.
DETERMINE LENGTH OF EACH CHAIN AND TRUE ANGLE BETWEEN THEM h’ (chains) TV a’b’ 1. THE HOOK IS 1.5 M A 1M 350 B 1M c’d’ X a1 ad (wall railing) D Y 2M FV C Wall railing (frame) h (chains) b1 bc Answers: Length of each chain= hb1 True angle between chains = .PROBLEM 23:A PICTURE FRAME 2 M WIDE AND 1 M TALL IS RESTING ON HORIZONTAL WALL RAILING MAKES 350 INCLINATION WITH WALL.5M 1.5 M ABOVE WALL RAILING. IT IS ATTAACHED TO A HOOK IN THE WALL BY TWO STRINGS.
Another World, Or, The Fourth Dimension 1905--Schofield, Alfred T.

References: V. 
 V. 
 V. 
 V. 
 V. 
 V.

 v. 
 V. 
 V. 
 V. 

V. 

V. 

V. 

V. 

V. 

V. 

V. 

V.