Unit 1 Projection of straight lines I.pdf
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1) The document discusses orthographic projections of points, lines, and planes. It provides notation for different views, such as front view (FV) and top view (TV). 2) Key concepts covered include locating an object relative to horizontal and vertical planes using quadrants, and drawing the FV and TV based on the object's location. Examples are given for points located in different quadrants. 3) Projections of straight lines are also discussed, including lines parallel or inclined to the planes. True length, reduced length, and inclinations of views are important parameters. 4) Several example problems are provided to demonstrate how to draw orthographic projections of points and lines in different configurations. Steps are
Unit 1 Projection of straight lines I.pdf
- 1. ORTHOGRAPHIC PROJECTIONS OF POINTS, LINES & PLANES
- 2. 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 H.P. & V.P.} 2
- 3. NOTATIONS Following notations should be followed while naming Different views in orthographic projections. ITâS FRONT VIEW a´ a´ b´ Same system of notations should be followed incase numbers, like 1, 2, 3 â are used. OBJECT POINT A LINE AB ITâS TOP VIEW a a b ITâS SIDE VIEW a´´ a´´ b´´ TERMS âABOVEâ & âBELOWâ WITH RESPECT TO H.P. AND TERMS âINFRONTâ & âBEHINDâ WITH RESPECT TO V.P. 3
- 4. X Y 1ST Quad. 2nd Quad. 3rd Quad. 4th Quad. X Y VP HP Observer 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. 4
- 5. HP VP a´ a A POINT A IN 1ST QUADRANT OBSERVER VP HP POINT A IN 2ND QUADRANT OBSERVER a´ a A OBSERVER a a´ POINT A IN 3RD QUADRANT HP VP A OBSERVER a a´ POINT A IN 4TH QUADRANT HP VP A Point A is placed In different quadrants and itâs FV & TV are brought in same plane for Observer to see clearly. FV is visible as it is a view on VP. But as TV is is a view on HP, it is rotated downward 900, in clockwise direction. The front part of HP comes below XY line and the part behind VP comes above. 5
- 6. A a a´ A a a´ A a a´ X Y X Y X Y For TV For TV For TV POINT A ABOVE HP & INFRONT OF VP POINT A IN HP & INFRONT OF VP POINT A ABOVE HP & IN VP PROJECTIONS OF A POINT IN FIRST QUADRANT PICTORIAL PRESENTATION PICTORIAL PRESENTATION ORTHOGRAPHIC PRESENTATIONS OF ALL ABOVE CASES. X Y a a´ VP HP X Y a´ VP HP a X Y a VP HP a´ FV above XY, TV below XY. FV above XY, TV on XY. FV on XY, TV below XY. 6
- 7. 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. PROJECTIONS OF STRAIGHT LINES INFORMATION REGARDING A LINE MEANS: ⢠Itâs length ⢠Position of itâs ends with HP & VP ⢠Itâs inclinations with HP & VP will be given. AIM:- To draw itâs projections - means FV & TV. 7
- 8. X Y X Y b´ a´ b a a (b) a´ b´ B A TV FV A B X Y H.P. V.P. a´ b´ a (b) FV TV X Y H.P. V.P. a b a´ b´ FV TV For TV For TV Note: FV is a vertical line Showing True Length & TV is a point. Note: FV & TV both are // to XY & both show T. L. 1. 2. A line perpendicular to HP & parallel to VP A line // to HP & // to VP Orthographic Pattern Orthographic Pattern (Pictorial Presentation) (Pictorial Presentation) 8
- 9. A line inclined to HP and parallel to VP (Pictorial presentation) X Y A B b´ a´ b a ďą ďą A line inclined to VP and parallel to HP (Pictorial presentation) Ă a b a´ b´ B A Ă X Y H.P. V.P. T.V. a b a´ b´ ďą X Y H.P. V.P. Ă a b a´ b´ TV FV TV inclined to XY FV parallel to XY 3. 4. FV inclined to XY TV parallel to XY Orthographic Projections 9
- 10. a´ X Y 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. a 30 20 HP a VP A a´ X Y a 20 30 10 Orthographic projection
- 11. PROBLEM 2: A point D is 20 mm below HP and 30 mm in front of VP. 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. d 20 30 D d´ HP X Y X Y 30 20 d d´ 11 Orthographic projection
- 12. 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; X Y 25 20 a a´ 12 20 25 c b c´ b´ 30 20 20 25 ` d d´ e e´ 25 30 f f´ Solution:
- 13. PROBLEM 4: Draw projections of a 80 mm long line PQ. 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. 30° X Y P Q P´ Q´ 10 10 13 Parallel to VP and inclined to HP
- 14. 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) 30° X Y a b a´ b´ 15 10 14 Parallel to VP and inclined to HP 30° X Y a b a´ 15 10 b´ 15 10 X Y 30° a a´ b b´ 10 15 X Y a´ a 30° b b´ ( Quadrant 1) (Quadrant 2) (Quadrant 3) (Quadrant 4)
- 15. 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. 15 Parallel to HP and inclined to VP 30° X Y a b a´ 15 10 b´ 15 10 X Y 30° a a´ b b´ 10 15 X Y a´ a 30° b b´ 30° X Y a b a´ 15 10 b´ ( Quadrant 1) ( Quadrant 4) ( Quadrant 3) ( Quadrant 2)
- 16. X Y a´ b´ a b B A ďĄ ď˘ For TV T.V. X Y a´ b´ a b ďĄ ď˘ T.V. For TV B A X Y ďĄ ď˘ H.P. V.P. a b FV TV a´ b´ A Line inclined to both HP and VP (Pictorial presentation) 5. Note:- Both FV & TV are inclined to XY. (No view is parallel to XY) Both FV & TV are reduced lengths (No view shows True Length) Orthographic Projections FV is seen on VP clearly. To see TV clearly, HP is rotated 900 downwards, Hence it comes below XY. On removal of object i.e. Line AB FV as a image on VP. TV as a image on HP, 16
- 17. X Y H.P. V.P. X Y ď˘ H.P. V.P. a b TV a´ b´ FV TV b2 b1´ TL X Y ďĄ ď˘ H.P. V.P. a b FV TV a´ b´ Here TV (ab) is not // to XY line Hence itâs corresponding FV aâ bâ is not showing True Length & True Inclination with HP. In this sketch, TV is rotated and made // to XY line. Hence itâs corresponding FV, aâ b1âis showing True Length & True Inclination with HP. Note the procedure When FV & TV known, How to find True Length. (Views are rotated to determine True Length & itâs inclinations with HP & VP). Note the procedure When True Length is known, How to locate FV & TV. (Component a-1 of TL is drawn which is further rotated to determine FV) 1 a a´ b´ 1´ b ď˘ b1´ ďą ďĄ b1 Ă Orthographic Projections Means FV & TV of Line AB are shown below, with their apparent inclinations ďĄ & ď˘ Here a -1 is component of TL ab1 gives length of FV. Hence it is brought upto Locus of aâ and further rotated to get point bâ. aâ bâ will be FV. Similarly drawing component of other TL (aâ b1â) TV can be drawn. ďą 17
- 18. Diagram showing graphical relations among all important parameters of this topic. True Length is never rotated. Itâs horizontal component is drawn & it is further rotated to locate view. Views are always rotated, made horizontal & further extended to locate TL, ďą & Ă Also remember TEN important parameters to be remembered with notations used here onward Ă ďĄ ď˘ ďą 1) True Length (TL) â aâ b1â & a b1 2) Angle of TL with HP - 3) Angle of TL with VP â 4) Angle of FV with XY â 5) Angle of TV with XY â 6) LTV (length of FV) â Component (a-1) 7) LFV (length of TV) â Component (aâ-1â) 8) Position of A- Distances of a & aâ from XY 9) Position of B- Distances of b & bâ from XY 10) Distance between End Projectors X Y H.P. V.P. 1 a b ď˘ b1 Ă LFV a´ b´ 1´ b1´ ďĄ ďą LTV Distance between End Projectors. ďĄ ďą & Construct with aâ Ă ď˘ & Construct with a b & b1 on same locus. bâ & b1â on same locus. NOTE 18
- 19. a´ b´ a b X Y b´1 b1 Ă ďą 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. Solution steps: 1) Draw XY line and one projector. 2) Locate a´ 12mm above XY line & a 10mm below XY line. 3) Take 300 angle from a´ & 400 from a and mark TL, i.e., 75mm on both lines. Name those points b1´ and b respectively. 4) 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) 5) Extend it up to locus of a and rotating aâ as center locate b´ as shown. Join a´ b´ as FV. 6) From b´ drop a projector downward & get point b. Join a & b, i.e., TV. LFV TL TL FV TV 19 1 INCLINED TO HP & VP
- 20. X Y a a´ b1 1 b´ 1 b´ LFV 550 b PROBLEM 8: Line AB 75mm long makes 450 inclination with VP while itâs FV makes 550. End A is 10 mm above HP and 15 mm in front of VP. If line is in 1st quadrant draw itâs projections and find itâs inclination with HP. LOCUS OF b1 LOCUS OF b1´ Solution Steps:- 1.Draw xy line. 2.Draw one projector for aâ & a 3.Locate a´ 10mm above XY & a 15 mm below XY. 4.Draw a line 450 inclined to XY from point a and cut TL 75 mm on it and name that point b1. 5.Draw locus from point b1. 6.Take 550 angle from a´ for FV above XY line. 7.Draw a vertical line from b1 up to locus of a and name it 1. It is horizontal component of TL & is LFV. 8.Continue it to locus of a´ and rotate upward up to the line of FV and name it b´. This a´ b´ line is FV. 9. Drop a projector from b´ on locus from point b1 and name intersecting point b. Line ab is TV of line ab. 10.Draw locus from b´ and with TL distance cut point b1´ 11.Join a´ b1´ as TL and measure itâs angle Îą at a´. It will be true angle of line with HP. 20 Îą FINDING INCLINATION WITH HP
- 21. X aâ Y a bâ 500 b 600 b1 bâ1 PROBLEM 9: FV of line AB is 500 inclined to XY and measures 55 mm long while itâs TV is 600 inclined to XY line. If end A is 10 mm above HP and 15 mm in front of VP, draw itâs projections, find TL, inclinations of line with HP & VP. Solution steps: 1.Draw XY line and one projector. 2.Locate aâ 10 mm above XY and a 15 mm below XY line. 3.Draw locus from these points. 4.Draw FV 500 from aâ and mark bâ cutting 55mm on it. 5.Similarly draw TV 600 from a & drawing projector from bâ locate point b and join a b. 6.Then rotating views as shown, locate True Lengths ab1 & aâb1â and their angles with HP and VP. 21 FINDING TL AND INCLINATIONS
- 22. X Y aâ 1â a bâ1 LTV b1 1 bâ b LFV ď ďą PROBLEM 10:- Line AB is 75 mm long. Itâs FV and TV measure 50 mm & 60 mm long respectively. An end is 10 mm above HP and 15 mm in front of VP. Draw projections of line AB if end B is in first quadrant. Find angle with HP and VP. 22 SOLUTION STEPS: 1.Draw XY line and one projector. 2.Locate aâ 10 mm above XY and a 15 mm below XY line. 3.Draw locus from these points. 4.Cut 60mm distance on locus of aâ & mark 1â on it as it is LTV. 5.Similarly cut 50mm on locus of a and mark point 1 as it is LFV. 6.From 1â draw a vertical line upward and from aâ taking TL (75mm ) in compass, mark bâ1 point on it. Join aâ bâ1 points. 7. Draw locus from bâ1 8. With same steps below get b1 point and draw also locus from it. 9. Now rotating one of the components i.e., a-1 locate bâ and join aâ with it to get FV. 10. Locate TV similarly and measure angles ďą and ď FINDING ANGLE WITH HP & VP
- 23. X Y câ c LOCUS OF d d d1 dâ1 ďą ď LOCUS OF dâ PROBLEM 11:- TV of a 75 mm long line CD, measures 50 mm. End C is in HP and 50 mm in front of VP. End D is 15 mm in front of VP and it is above HP. Draw projections of CD and find angles with HP and VP. 23 SOLUTION STEPS: 1.Draw XY line and one projector. 2.Locate câ on XY and c 50mm below XY line. 3.Draw locus from these points. 4.Draw locus of d 15 mm below XY. 5.Cut 50mm & 75 mm distances on locus of d from c and mark points d & d1 as these are TV and TL. Join both with c. 6.From d1 draw a vertical line upward up to XY i.e., up to locus of câ and draw an arc as shown. 7 Then draw one projector from d to meet this arc in dâ point & join câ dâ 8. Draw locus of dâ and cut 75 mm on it from câ as TL 9.Measure anglesďą and ď FINDING ANGLE WITH HP & VP dâ
- 24. X Y PROBLEM 9:- Two straight lines PQ and QR make an angle of 120° between them in front and top views. PQ is 60 mm long and is parallel to and 15 mm from both H.P. and V.P. Determine the true angle between PQ and QR, if point R is 50 mm above H.P. (EXAMPLE) SOLUTION STEPS: 1. Draw a reference line xy. Mark point p´ at 15 mm above xy and point p at 15 mm below xy. 2. Draw 60 mm long lines p´q´ and pq, parallel to xy. 3. Draw a line from point q´, inclined at 120° to xy such that it meets the horizontal line at 50 mm above xy at point r´. Join q´r´ and p´r´. 4. Draw a line from point q, inclined at 120° to xy such that it meets the projector from r´ at a point r. Join qr and pr. 5. As lines pq and p´q´ are parallel to xy, they represent the true length of side PQ. Here PQ = 60 mm. 6. Draw an arc with centre p and radius pr to meet the horizontal line from p at point r1. Project point r1 to meet horizontal lines from point r´ at point r1 â. Join p´r1 â to represent the TL of the line PR. Here, PR = p´ r1 ´= 94 mm. 7. Draw an arc with centre q and radius qr, to meet the horizontal line at r2. Project point r2 to meet horizontal lines form point r´ at point r´2. Join q´ r2 ´ to represent the TL of line QR. Here, QR = q´ r2 ´ = 53mm. 8. Draw actual triangle PQR taking true lengths, i.e., 60 mm, 94 mm and 53 mm. Measure the inclined angle PQR as the actual angle between sides PQ and QR. Here, it is 112°. Q R r´ r1 â r´2 r2 r1 p pâ q qâ r 15 15 60 24 P 50 FINDING TRUE ANGLE
- 25. TRACES OF THE LINE:- These are the points of intersections of a line ( or itâs extension ) with respect to reference planes. A line itself or its extension, where ever touches H.P., that point is called TRACE OF THE LINE ON H.P. (It is called H.T.) Similarly, a line itself or itâs extension, where ever touches V.P., that point is called TRACE OF THE LINE ON V.P. (it is called V.T.) V.T.:- It is a point on VP. Hence it is called FV of a point in VP. Hence itâs TV comes on XY line.( Here onward denoted as âvâ) H.T.:- It is a point on HP. Hence it is called TV of a point in HP. Hence itâs FV comes on XY line.( Here onward denoted as âhâ ) PROBLEMS INVOLVING TRACES OF THE LINE 25