OF POINTS, LINES & PLANES

1 ORTHOGRAPHIC PROJECTIONS. OF points , LINES & PLANES . To draw projections of any object, one must have following information: A) OBJECT. {With its description, well defined}. B) OBSERVER. {Always observing perpendicular to resp. Ref. Plane}. C) LOCATION OF OBJECT. {Means its position with reference to & }. 2. NOTATIONS. Following notations should be followed while naming Different views in orthographic projections. OBJECT POINT A line AB. IT'S TOP VIEW a ab IT'S FRONT VIEW a a b . IT'S SIDE VIEW a a b . Same system of notations should be followed incase numbers, like 1, 2, 3 are used. TERMS ABOVE' & BELOW' WITH RESPECT TO AND TERMS INFRONT' & BEHIND' WITH RESPECT TO 3. VP. 2nd Quad. 1ST Quad. Y. Observer X Y HP. X. 3rd Quad. 4th Quad. This quadrant pattern, if observed along x-y line (in red arrow direction) will exactly appear as shown on right side and hence it is further used to understand illustration properly.

2 4. POINT A IN POINT A IN. 2 NDQUADRANT VP. A 1ST QUADRANT. VP. A a . Point A is a . placed In different a quadrants HP. OBSERVER. and it's FV & TV. are brought in HP. same plane for OBSERVER. Observer to see a clearly. FV is visible as it is a view on VP. But as TV is is a view on HP, it is rotated a downward 900, in clockwise direction. The HP. front part of HP OBSERVER. OBSERVER. HP comes below XY line and the part behind VP a comes above. a . A a . POINT A IN A POINT A IN. RD VP 4TH QUADRANT. 3 QUADRANT VP. 5. PROJECTIONS OF A POINT IN FIRST QUADRANT. POINT A ABOVE HP POINT A ABOVE HP POINT A IN HP. & INFRONT OF VP & IN VP & INFRONT OF VP. For TV. For TV. PICTORIAL PICTORIAL For TV. PRESENTATION A PRESENTATION. a a . A Y. Y. Y a . a a X a X X A. ORTHOGRAPHIC PRESENTATIONS. OF ALL ABOVE CASES. FV above XY, FV above XY, FV on XY, TV below XY.

3 TV on XY. TV below XY. VP VP VP. a a . X Y X Y X. a Y. a a a HP HP HP. 6. PROJECTIONS OF STRAIGHT LINES . INFORMATION REGARDING A line MEANS: It's length Position of it's ends with HP & VP. It's inclinations with HP & VP will be given. AIM:- To draw it's projections - means FV & TV. SIMPLE CASES OF THE line . 1. A vertical line ( line perpendicular to HP & parallel to VP). 2. line parallel to both HP & VP. 3. line inclined to HP & parallel to VP. 4. line inclined to VP & parallel to HP. 5. line inclined to both HP & VP. 7. For TV Orthographic Pattern (Pictorial Presentation) a . Note: a . FV is a vertical line A Showing True Length FV. 1. FV &. TV is a point. b . A line b . perpendicular Y. X Y. to HP B. & TV a (b). TV a (b). parallel to VP X. Orthographic Pattern (Pictorial Presentation) For TV Note: FV & TV both are 2. // to XY a FV b.

4 B &. A line B both show T. L. // to HP a . & A Y X Y. // to VP. b a b TV. X. a 8. FV inclined to XY 3. b . TV parallel to XY b . A line inclined to HP B. a . and Y. parallel to VP a X Y.. (Pictorial presentation) A b a b X. a Orthographic Projections TV inclined to XY 4. FV parallel to XY. b a FV b . A line inclined to VP. and a . parallel to HP . A B X Y. (Pictorial presentation) a .. a b TV. b 9. EXAMPLE PROBLEMS ON points . PROBLEM 1: A point A is 20 mm above HP and 30 mm in front of VP. Draw its projections Solution steps: 1) Draw reference line XY. 2) Mark a point a at a distance of 20 mm above XY. 3) Through this point draw a perpendicular line to XY and mark the top view a at a distance of 30 mm below XY. VP. Orthographic projection 30. A. a . a HP. 20 Y 20. X Y. a 30. X. a a 10. PROBLEM 2: A point D is 20 mm below HP and 30 mm in front of VP.

5 Draw its projections. Solution steps: 1) Draw reference line XY. 2) Mark a point d at a distance of 20 mm below XY. 3) Through this point draw a perpendicular line to XY and mark the top view d at a distance of 30 mm above XY. Y HP Orthographic projection d X Y. X 20. 20. 30 30 d . d D. d 11. PROBLEM 3: Draw the projections of the following points on the same ground line , keeping the distance between projectors equal to 25 mm. (i) Point A, 20 mm above HP, 25 mm behind VP;. (ii) Point B, 25 mm below HP, 20 mm behind VP;. (iii) Point C, 20 mm below HP, 30 mm in front of VP;. (iv) Point D, 20 mm above HP, 25 mm in front of VP;. (v) Point E, on HP, 25 mm behind VP;. (vi) Point F, on VP, 30 mm above HP;. Solution: f . e a b d a 30. 25. 25 20. 20 20. X e f Y. 20. 25 30 25. ` c . b d . c 12. Parallel to VP and inclined to HP. PROBLEM 4: Draw projections of a 80 mm long line PQ.

6 Its end P is 10 mm above HP and 10 mm in front of VP. The line is parallel to VP and inclined to HP at 30 . Solution steps: 1) Draw the plan and elevations of the end point P. 2) Draw plan PQ of the line at an angle of 30 to XY. 3) Draw the projector of Q. 4) From the elevation of end point P draw a line parallel to XY meeting projector of Q at Q . 5) P Q is the elevation and PQ is the plan of the line . Q . 30 . P . 10 10. X Y. P Q. 13. Parallel to VP and inclined to HP. PROBLEM 5: A straight line AB of 40 mm length has one of its ends A, at 10 mm from the HP and 15 mm from the VP. Draw the projections of the line if it is parallel to the VP and inclined at 30 to the HP. Assume the line to be located in each of the four quadrants by turns. (EXAMPLE). b . b . 30 . a . 10. a 30 . b X Y. 15. 15. a . 10. X Y. a ( Quadrant 1) b (Quadrant 2).

7 A b X Y. 15. 10. a . 15. X Y. 10. a b 30 . a . 30 . (Quadrant 4) b . (Quadrant 3) 14. b . Parallel to HP and inclined to VP. PROBLEM 6: A straight line AB of 40 mm length is parallel to the HP and inclined at 30 to the VP. Its end point A is 10 mm from the HP and 15 mm from the VP. Draw the projections of the line AB, assuming it to be located in all the four quadrants by turns. a b b 10. X Y 30 . a 15. b . 15. a a . 10. 30 . X Y. ( Quadrant 2). ( Quadrant 1). b b X Y. 10. a b . 15. 30 . a a 30 . 15. X Y b 10. a b ( Quadrant 4). ( Quadrant 3) 15. For TV For TV. 5. A line inclined to both b HP and VP b . (Pictorial presentation). B. B.. Y. Y. On removal of object a . a line AB. FV as a image on VP. TV as a image on HP, A. A . X. X a b a b b . FV. a . Orthographic Projections X Y. FV is seen on VP clearly. Note:- To see TV clearly, HP is rotated Both FV & TV are inclined to 900 downwards, XY.

8 A (No view is parallel to XY). Hence it comes below XY. Both FV & TV are reduced TV. lengths b (No view shows True Length). 16. Note the procedure 17. Note the procedure Orthographic Projections When FV & TV known, When True Length is known, Means FV & TV of line AB How to find True Length. How to locate FV & TV. are shown below, with their (Views are rotated to determine (Component a-1 of TL is drawn apparent inclinations & True Length & it's inclinations which is further rotated with HP & VP). to determine FV). b b b 1 b b 1 . FV FV. TL . a a a 1 . X Y. X Y X Y. 1. a b2 a a TV.. TV TV. b b b b1. In this sketch, TV is rotated Here a -1 is component Here TV (ab) is not // to XY. and made // to XY line . of TL ab1 gives length of FV. line Hence it's corresponding Hence it is brought upto Hence it's corresponding FV. FV, a' b1' is showing Locus of a' and further rotated a' b' is not showing True Length to get point b'.

9 A' b' will be FV. True Length &. & Similarly drawing component True Inclination with HP. True Inclination with HP. of other TL (a' b1 ) TV can be drawn. 1) True Length (TL) a' b1' & a b1. Diagram showing graphical relations 2) Angle of TL with HP - TEN important parameters among all important parameters of this topic. 3) Angle of TL with VP .. to be remembered 4) Angle of FV with XY with notations 5) Angle of TV with XY used here onward Distance between End Projectors. 6) LTV (length of FV) Component (a-1). b b1 . 7) LFV (length of TV) Component (a'-1'). 8) Position of A- Distances of a & a' from XY. 9) Position of B- Distances of b & b' from XY. 10) Distance between End Projectors 1 . a . LTV. NOTE. X Y & Construct with a'. a LFV 1 & Construct with a . b' & b1' on same locus.. b & b1 on same locus. Also remember b b1 True Length is never rotated.

10 It's horizontal component is drawn & it is further rotated to locate view. Views are always rotated, made horizontal & further extended to locate TL, & . 18. INCLINED TO HP & VP. PROBLEM 7: line AB is 75 mm long and it is 300 & 400 inclined to HP & VP respectively. End A is 12mm above HP and 10 mm in front of VP. Draw projections. line is in 1st quadrant. b b 1. Solution steps: 1) Draw XY line and one projector. 2) Locate a 12mm above XY line FV. TL. & a 10mm below XY line . 3) Take 300 angle from a & 400. from a and mark TL, , . 75mm on both LINES . Name a . those points b1 and b respectively. X Y. 4) Draw horizontal component of LFV. TL a b1 from point b1 and a 1. name it 1. (the length a-1. gives length of FV as we have seen already). 5) Extend it up to locus of a and rotating a' as center locate b TV TL. as shown. Join a b as FV.